Preparación para Cálculo: Gráficas, Modelos y Funciones

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CHAPTER P
Preparation for Calculus
Section P.1
Graphs and Models.................................................................................2
Section P.2
Linear Models and Rates of Change....................................................11
Section P.3
Functions and Their Graphs.................................................................22
Section P.4
Fitting Models to Data..........................................................................34
Review Exercises ..........................................................................................................37
Problem Solving ...........................................................................................................43
INSTRUCTOR USE ONLY
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NOT FOR SALE
C H A P T E R P
Preparation for Calculus
Section P.1 Graphs and Models
1. y
32 x 3
4 x2
7. y
x-intercept: (2, 0)
x
3
2
0
2
3
y
5
0
4
0
5
y-intercept: (0, 3)
Matches graph (b).
y
9 x2
2. y
6
x-intercepts: 3, 0 , 3, 0
(0, 4)
y-intercept: (0, 3)
(− 2, 0)
−6
Matches graph (d).
3. y
(2, 0)
x
−4
4
6
−2
(− 3, − 5)
3 x2
(3, − 5)
−4
−6
3, 0 , 3, 0
x-intercepts:
8. y
y-intercept: (0, 3)
Matches graph (a).
x x
3
4. y
2
2
x3
x
0
1
2
3
4
5
6
y
9
4
1
0
1
4
9
x-intercepts: 0, 0 , 1, 0 , 1, 0
y
y-intercept: (0, 0)
10
Matches graph (c).
8
(0, 9)
(6, 9)
6
4
1x 2
2
5. y
2
(5, 4)
(1, 4)
(2, 1)
(4, 1)
x
x
4
2
0
2
4
y
0
1
2
3
4
−6 −4 −2
9. y
−2
4
2
6
(3, 0)
x 2
y
6
x
5
4
3
2
1
0
1
y
3
2
1
0
1
2
3
(4, 4)
4
(2, 3)
(0, 2)
(−2, 1)
y
x
−4
−2
2
(−4, 0)
4
6
−2
4
(− 5, 3)
6. y
5 2x
(− 4, 2) 2
(− 1, 1)
(− 3, 1)
x
1
0
1
2
5
2
3
4
y
7
5
3
1
0
1
3
−6
−4
(1, 3)
(0, 2)
x
(− 2, 0)
2
−2
y
8
(− 1, 7)
(0, 5)
4
2
−6 − 4 −2
−2
(1, 3)
(2, 1)
(3, −1)
x
( , −3)
3))
( ( (4,
INSTRUCTOR
ST
USE ONLY
−
−44
2
5,0
2
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.1
x 1
10. y
14. y
x
3
2
1
0
1
2
3
y
2
1
0
1
0
1
2
1
x 2
6
4
3
2
1
0
2
y
14
12
1
Undef.
1
1
2
1
4
y
4
5
4
3
2
3
(3, 2)
2
(− 2, 1)
(− 1, 1)
(2, 1)
(0, 12 )
(2, 14 )
x
x
−3 −2
1
−1
(− 1, 0)
3
x
y
(−3, 2)
Graph
Graphs and Models
2
3
(1, 0)
(0, − 1)
−2
−1
(− 6, − 14 )
(− 4, − 12 )
1 2 3
−2
−3
−4
−5
(−3, − 1)
x 6
11. y
x
0
1
4
9
16
y
6
5
4
3
2
15. y
5 x
5
(−4.00, 3)
(2, 1.73)
y
−6
6
2
−3
x
−4
4
−2
8
12
(9, − 3)
16
(16, −2)
−4
(4, −4)
(1, − 5)
−6
(0, −6)
−8
2, y
2, 1.73
(b)
x, 3
4, 3
16. y
x 2
12. y
(a)
y
3
52
3 | 1.73
5 4
x5 5 x
6
x
2
1
0
2
7
14
y
0
1
2
2
3
4
(−0.5, 2.47)
−9
9
(1, −4)
y
−6
5
4
(14, 4)
3
(7, 3)
(− 1, 1)
2
(2, 2)
(0, 2)
x
(− 2, 0)
13. y
5
10
15
0.5, y
0.5, 2.47
(b)
x, 4
1.65, 4 and x, 4
17. y
2x 5
y-intercept: y
20
x-intercept: 0
3
x
x
3
2
1
0
1
2
3
y
1
32
3
Undef.
3
3
2
1
y
(1, 3)
3
(2, 32 (
2
(3, 1)
(−3, −1)
(a)
1
x
−3 −2 −1
−1
−2
1
2
3
18. y
20 5
1, 4
5; 0, 5
2x 5
5
2x
x
5;
2
5, 0
2
4x2 3
2
3
y-intercept: y
40
x-intercept: 0
4 x2 3
3
3; 0, 3
4 x2
None. y cannot equal 0.
(− 2, − 32 (
INSTRUCTOR
S
USE ONLY
(−1,
(−1,
1, −3)
− 3))
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4
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Chapter P
Preparation
paration for Calculus
Calc
x2 x 2
19. y
y-intercept: y
02 0 2
y
2; 0, 2
x-intercepts: 0
x2 x 2
20. y
2
x
2, 1; 2, 0 , 1, 0
ª¬3 0 1º¼
x 2 3x
x-intercepts: 0
3x 1
x 4x
y
0; 0, 0
x-intercepts: 0
x3 4 x
x
0, r 2; 0, 0 , r 2, 0
2
0 16 0
x-intercepts: 0
x 16 x 2
2
0
y
0; 0, 0
x
26. y
4 x 4 x
x
x
0, 4, 4; 0, 0 , 4, 0 , 4, 0
x2 1
y
y-intercept: 02 y 02 4 y
0; 0, 0
0
y-intercept: y
0
x-intercept: x 2 0 x 2 4 0
y-intercept: y
x 1
2
0, 3; 0, 0 , 3, 0
25. x 2 y x 2 4 y
x x2 x 2
x 16 x
3x 1
x
0
2
x x3
0
3
2
0; 0, 0
3
0 40
2x 0 1
1; 1, 0
4 x2
x2 1
3x 2
1
x2
1
3
x
r
x-intercept: 0
2 x
5x 1
0
2
x
x
4;
4, 0
0, 2
Note: x
x2 1
3
3
3 § 3 ·
;¨
, 0 ¸¸
3 ¨© 3
¹
x
2;
02 1
x2 1
2x
2 x
5x 1
y -intercept: y
2x 0
x2 1
2 0
50 1
0; 0, 0
1; 0, 1
y
1; 0, 1
x
20 y-intercept: y
02 1
x 1
0
x2 1
x-intercept:
x-intercept: 0
23. y
02 3 0
y
x 2 x 1
y-intercept: y
22. y
2
3x 1
y-intercept: y
0
2
21. y
x 2 3x
24. y
3 3 is an extraneous solution.
27. Symmetric with respect to the y-axis because
x
y
28. y
2
6
x 2 6.
x2 x
No symmetry with respect to either axis or the origin.
29. Symmetric with respect to the x-axis because
y
2
y2
x3 8 x.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.1
30. Symmetric with respect to the origin because
3
y
x
y
x x
y
x x.
40. y
x
3
1
2 0
3
0
2x 1 Ÿ 2x
3
3
1, y -intercept
1Ÿ x
23 , x-intercept
y
Intercepts: 0, 1 , 32 , 0
31. Symmetric with respect to the origin because
x y
xy
4.
2
Symmetry: none
(0, 1)
(− 32 , 0)
x
32. Symmetric with respect to the x-axis because
x y
2
−1
10.
xy 2
5
2x 1
3
y
3
Graphs
Graph and Models
1
2
−1
−2
4
33. y
x 3
No symmetry with respect to either axis or the origin.
34. Symmetric with respect to the origin because
x y 4 x
xy 2
0
4 x2
x
y
9 0
0
9 x Ÿ x2
y
36. y
r 3, x-intercepts
y
9 x
2
(0, 9)
9 x2
4
1
2
(− 3, 0)
−6 −4 −2
2x2 x
(3, 0)
x
2
−2
4
6
x 2x 1
2
x2 1
is symmetric with respect to the y-axis
x
because y
37. y
9 Ÿ x
6
42. y
x
9, y -intercept
2
Symmetry: y-axis
x
.
x2 1
y
2
10
x
2
9 x2
Intercepts: 0, 9 , 3, 0 , 3, 0
0.
35. Symmetric with respect to the origin because
y
41. y
x
2
2
x2
.
x 1
2
1
y
020 1
0
x 2x 1 Ÿ x
0, y -intercept
0, 12 , x-intercepts
y
Intercepts: 0, 0 , 12 , 0
x3 x is symmetric with respect to the y-axis
5
Symmetry: none
4
3
x
because y
3
x
x3 x
x3 x .
2
(− 12 , 0)
38. y x
1
(0, 0)
x
3 is symmetric with respect to the x-axis
−3
−2
−1
1
2
3
because
y x
3
43. y
x3 2
y x
3.
y
03 2
0
x 2 Ÿ x3
39. y
2 3x
y
230
0
2 3 x Ÿ 3x
Intercepts: 0, 2 ,
2, y -intercept
3
2 Ÿ x
3 2, x-intercept
Intercepts: 3 2, 0 , 0, 2
2, y -intercept
2 Ÿ x
2
, x-intercept
3
y
Symmetry: none
5
4
y
2, 0
3
3
(0, 2)
Symmetry: none
2
1
−1
(− 3 2, 0)
(0, 2)
−3 −2
( (
2
,0
3
1
x
−1
1
2
3
x
2
3
−1
INSTRUCTOR USE ONLY
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6
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Chapter P
Preparation
paration for Calculus
Calc
44. y
x3 4 x
y
03 4 0
y3
47. x
0, y -intercept
x3 4 x
0
x x2 4
0
x x 2 x 2
0
0 Ÿ y
x
0, x-intercept
x
y
3
x
4 x
3
Ÿ x
y3
Symmetry: origin
y
Intercepts: 0, 0 , 2, 0 , 2, 0
y
0, y -intercept
Intercept: (0, 0)
0, r 2, x-intercepts
x
y3
4
x3 4 x
3
x3 4 x
2
(0, 0)
Symmetry: origin
y
−1
(0, 0)
(2, 0)
1
3
48. x
x
−3
4
y2 4
y2 4
0
y 2 y 2
0
x5
x
3
−4
−2
45. y
2
−3
(−2, 0)
−1
1
−2
3
−3
x
−4 −3 −2 − 1
y
0 0 5
x
x 5
r 2, y -intercepts
x
02 4
4, x-intercept
Intercepts: 0, 2 , 0, 2 , 4, 0
0, y -intercept
0 Ÿ x
y
0, 5, x-intercepts
Intercepts: 0, 0 , 5, 0
x
y
2
4
y2 4
y
y
Symmetry: x-axis
3
Symmetry: none
3
2
(− 5, 0)
(0, 0)
−4 −3 −2 −1
(0, 2)
x
1
(− 4, 0)
2
−5
−2
x
−1
1
−3
(0, −2)
−4
−3
46. y
25 x 2
y
25 0
49. y
2
25
5, y -intercept
8
x
25 x 2
0
y
25 x 2
8
Ÿ Undefined Ÿ no y -intercept
0
0
5 x 5 x
0
8
x
0 Ÿ No solution Ÿ no x-intercept
r 5, x-intercept
Intercepts: none
x
Intercepts: 0, 5 , 5, 0 , 5, 0
25 x
y
2
25 x 2
y
8
Ÿ y
x
8
x
Symmetry: origin
y
Symmetry: y-axis
8
y
6
7
6
(−5, 0)
4
3
2
1
4
(0, 5)
2
x
−2
2
4
6
8
(5, 0)
INSTRUCTOR
NSTR
N
STR
USE
S ONLY
x
−44 −3
− 3 −2
−2 −
−11
1 2 3 4 5
−2
−2
−3
−3
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NOT FOR SALE
Section P.1
53. y 2 x
10
x2 1
50. y
y
10
02 1
y
10
x2 1
10, y -intercept
0 Ÿ No solution Ÿ no x-intercepts
r
y
Intercept: (0, 10)
10
2
x
10
x2 1
1
10
7
9
2
x 9
y
r
x9
y
r
x 9
0
x 9
0
12
y
Graphs and Models
Graph
r
r 3, y -intercepts
9
9, x-intercept
x
(0, 10)
09
Intercepts: 0, 3 , 0, 3 , 9, 0
Symmetry: y-axis
2
y
x
9 Ÿ y2 x
9
2
− 6 − 4 −2
x
2
4
Symmetry: x-axis
6
y
51. y
6 x
y
6 0
6
4
2
(0, 3)
−2
(0, − 3)
(−9, 0)
6, y -intercept
x
−10
6 x
0
6
x
x
r 6, x-intercepts
−6
6 x
54. x 2 4 y 2
6 x
Symmetry: y-axis
4 Ÿ y
r
y
y
8
(0, 6)
6
2
4
2
6
r
x
2
4
x
r 2, x-intercepts
(6, 0)
− 4 −2
−2
4 02
2
4 x2
2
2
x
−8
r
x2 4 0
4
(− 6, 0)
2
−4
Intercepts: 0, 6 , 6, 0 , 6, 0
y
−6 −4 −2
8
4
2
r1, y -intercepts
4
Intercepts: 2, 0 , 2, 0 , 0, 1 , 0, 1
−4
−6
−8
x
52. y
6 x
y
60
2
4 y
2
4 Ÿ x2 4 y 2
4
Symmetry: origin and both axes
y
6, y -intercept
6
3
6 x
0
6 x
0
2
(0, 1)
x, x-intercept
6
(2, 0)
(−2, 0)
Intercepts: (0, 6), (6, 0)
−3
−1
−2
1
x
3
(0, −1)
−3
Symmetry: none
y
8
(0, 6)
4
2
(6, 0)
x
2
4
6
8
INSTRUCTOR USE ONLY
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8
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Chapter P
55. x 3 y 2
Preparation
paration for Calculus
Calc
x y
8 Ÿ y
8 x
4x y
7 Ÿ y
4x 7
6 x
3
8 x
4x 7
15
5x
60
3
3
x
57.
6
6 x
3y2
r
y
r
y
2
x 30
r
2, y -intercepts
The corresponding y-value is y
Point of intersection: (3, 5)
6
x
6, x-intercept
2 , 0, Intercepts: 6, 0 , 0,
2
x 3 y
6 Ÿ x 3y2
2
58. 3x 2 y
4 Ÿ y
4x 2 y
10 Ÿ y
3x 4
2
3x 4
4 x 10
2
4 x 10
6
Symmetry: x-axis
y
4
3
( 0,
2
2)
1
7x
14
x
2
(6, 0)
x
−1
1
−2
2
3
6
7
1.
Point of intersection: 2, 1
−3
−4
56. 3x 4 y 2
8
59. x 2 y
6 Ÿ y
6 x2
x y
4 Ÿ y
4 x
6 x
2
4 x
0
x2 x 2
0
x 2 x 1
x
2, 1
3x 8
4 y2
r
y
3x 4 0
3x 4
2
4 x 10
2
The corresponding y-value is y
( 0, − 2 )
y
5.
3x 2
4
3
r 2
0 2
4
Ÿ no solution Ÿ no y -intercepts
r
2
The corresponding y-values are y
y
3x
8
Points of intersection: 2, 2 , 1, 5
x
8
, x-intercept
3
Intercept:
8, 0
3
3x 4 y
2
8 Ÿ 3x 4 y 2
Symmetry: x-axis
y
6
8
5 for x
x
3 y2 Ÿ y2
y
x 1
3 x
x 1
3 x
x2 2x 1
0
x2 x 2
x
1 or x
3 x
2
x1 x 2
2
4
2
−2
−2
−4
The corresponding y-values are y
( 83, 0)
x
2
6
8
10
and y
2 and
1 .
8
60.
2 for x
1 for x
2 for x
1
2.
Points of intersection: 1, 2 , 2, 1
−6
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.1
61. x 2 y 2
5 Ÿ y2
5 x2
x y
1Ÿ y
x 1
64. y
y
x 1
5 x
2
x 2x 1
0
2x2 2 x 4
x
1 or x
y = x 4 − 2x 2 + 1
1 for x
(0, 1)
2 x 1 x 2
−3
2
2 for x
25 Ÿ y 2
25 x 2
3 x y
15 Ÿ y
3 x 15
25 x 2
3x 15
25 x 2
9 x 2 90 x 225
0
10 x 2 90 x 200
0
x 2 9 x 20
0
x 5 x 4
y = 1 − x2
Points of intersection: 1, 0 , 0, 1 , 1, 0
Analytically, 1 x 2
0
x4 x2
0
x2 x 1 x 1
x
1, 0, 1.
x2 4 x
y
4
y=
5
3 for x
4
x+6
(3,
y=
−x 2 − 4x
3)
(−2, 2)
−7
5 .
0 for x
x4 2x2 1
x6
65. y
The corresponding y-values are y
2
−2
Points of intersection: 4, 3 , 5, 0
−4
−2
2
4 or x
Points of intersection: 2, 2 , 3,
x3 2 x 2 x 1
x6
Analytically,
x6
x2 3x 1
4
3
(1, 0)
2.
62. x 2 y 2
and y
(−1, 0)
1
Points of intersection: 1, 2 , 2, 1
y
1 x2
2
The corresponding y-values are y
63. y
x4 2 x2 1
2
x
9
2
5 x2
and y
Graphs and Models
Graph
y = x3 − 2x2 + x − 1
(2, 1)
x2 4x
x2 4 x
x2 5x 6
0
x3 x 2
0
6
(0, −1)
x
3 | 3, 1.732
3, 2.
(−1, −5)
66. y
−8
y = − x2 + 3x − 1
y
2x 3 6
6 x
Points of intersection: 1, 5 , 0, 1 , 2, 1
Analytically, x 2 x x 1
3
2
x 3x 1
x x 2x
0
x x 2 x 1
0
3
2
x
7
2
y=6−x
(1, 5)
(3, 3)
−4
8
−1
1, 0, 2.
y = −⏐2x − 3⏐+ 6
Points of intersection: (3, 3), (1, 5)
Analytically, 2 x 3 6
2x 3
2x 3
x
x or 2 x 3
x
x
1.
3 or
6 x
x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
10
NOT FOR SALE
Chapter P
Preparation
paration for Calculus
Calc
67. (a) Using a graphing utility, you obtain
y
0.005t 2 0.27t 2.7.
(b)
71. y
(a) 1, 4 :
16
0
30
0
(c) For 2020, t
y
40.
0.005 40
2, 1 :
1
k 2
(c)
0, 0 :
0
k 0
(d)
1, 1 : 1
(a) 1, 1 :
68. (a) Using a graphing utility, you obtain
y
0.24t 2 12.6t 40.
(b)
2, 4 :
330
(c)
0, 0 :
The model is a good fit for the data.
y
(d)
3, 3 :
30.
0.24 30
3
k 1
4
8k Ÿ k
18
Ÿ k can be any real number.
3
k Ÿ k
1
12
4k 1
1
4k
k
1
4
4
2
4k 2
16
8k
k
2
0
2
4k 0
k can be any real number.
20
30
(c) For 2020, t
Ÿ k
3
4kx
The GDP in 2020 will be $21.5 trillion.
5
3
0.27 40 2.7
21.5
(b)
k1
4
(b)
72. y 2
2
kx3
2
12.6 30 40
3
2
4k 3
9
12k
k
9
12
3
4
554
The number of cellular phone subscribers in 2020
will be 554 million.
C
69.
2.04 x 5600
R
3.29 x
5600
3.29 x 2.04 x
5600
1.25 x
x
5600
1.25
4480
10,770
0.37
x2
x 4 x 3 x 8 has intercepts at
4, x
3, and x
8.
y
x 32 x 4 x 52 has intercepts at
x
32 , x
4, and x
5.
2
75. (a) If (x, y) is on the graph, then so is x, y by y-axis
symmetry. Because x, y is on the graph, then so
is x, y by x-axis symmetry. So, the graph is
symmetric with respect to the origin. The converse is
not true. For example, y
x3 has origin symmetry
but is not symmetric with respect to either the x-axis
or the y-axis.
400
0
y
x
74. Answers may vary. Sample answer:
To break even, 4480 units must be sold.
70. y
73. Answers may vary. Sample answer:
(b) Assume that the graph has x-axis and origin
symmetry. If (x, y) is on the graph, so is x, y by
100
0
If the diameter is doubled, the resistance is changed by
approximately a factor of 14. For instance,
y 20 | 26.555 and y 40 | 6.36125.
x-axis symmetry. Because x, y is on the graph,
then so is x, y
x, y by origin
symmetry. Therefore, the graph is symmetric with
respect to the y-axis. The argument is similar for
y-axis and origin symmetry.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.2
11
x3 x :
76. (a) Intercepts for y
y
03 0
0 ; 0, 0
x-intercepts: 0
x3 x
x x2 1
y -intercept:
Linear Models and Rat
Rate
Rates of Change
x x 1 x 1;
0, 0 , 1, 0 1, 0
Intercepts for y
x 2 2:
y -intercept:
y
0 2
x-intercepts: 0
x 2
2 ; 0, 2
2
None. y cannot equal 0.
(b) Symmetry with respect to the origin for y
y
x
3
x
x3 x.
Symmetry with respect to the y-axis for y
x
y
2
2
x 2 2 because
x 2 2.
x3 x
(c)
x3 x because
x2 2
x3 x 2 x 2
0
x 2 x x 1
0
2
2 Ÿ y
x
6
Point of intersection : (2, 6)
Note: The polynomial x 2 x 1 has no real roots.
77. False. x-axis symmetry means that if 4, 5 is on the
graph, then 4, 5 is also on the graph. For example,
4, 5 is not on the graph of x
4, 5 is on the graph.
y 2 29, whereas
§ b r
79. True. The x-intercepts are ¨
¨
©
b 2 4ac ·
, 0 ¸.
¸
2a
¹
§ b
·
80. True. The x-intercept is ¨ , 0 ¸.
2
a
©
¹
f 4 .
78. True. f 4
Section P.2 Linear Models and Rates of Change
1. m
2
2. m
0
3. m
1
4. m
12
5. m
2 4
53
6. m
7 1
2 1
6
3
2
y
(− 2, 7)
7
6
5
3
2
6
2
(1, 1)
1
3
− 4 −3 −2 −1
x
1
3
4
y
3
2
(5, 2)
1
x
−1
1
2
3
5
6
7
−2
−3
−4
(3, −4)
−5
INSTRUCTOR USE ONLY
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12
NOT FOR SALE
Chapter P
Preparation
paration for Calculus
Calc
16
44
7. m
5
, undefined.
0
11.
y
m = −2
m is undefined.
m=−3
2
The line is vertical.
m=1
8
y
6
(3, 4)
4
7
2
(4, 6)
6
5
−6 −4
4
x
2
−2
4
8 10
3
2
m = −3
x
−2 −1
1
2
y
12.
(4, 1)
1
3
5
6
1
m=3
(−2, 5)
8. m
5 5
53
6
m=0
4
0
2
0
m=3
x
−6
The line is horizontal.
−2
2
4
−2
y
1
13. Because the slope is 0, the line is horizontal and its
equation is y
2. Therefore, three additional points are
x
−1
−1
1
2
3
4
5
6
−2
0, 2 , 1, 2 , 5, 2 .
−3
−4
14. Because the slope is undefined, the line is vertical and its
equation is x
4. Therefore, three additional points
(3, −5) (5, −5)
−6
are 4, 0 , 4, 1 , 4, 2 .
9. m
2 1
3 6
1 § 3·
¨ ¸
2 © 4¹
1
2
1
4
15. The equation of this line is
2
y
y 7
3 x 1
y
3x 10.
Therefore, three additional points are (0, 10), (2, 4), and
(3, 1).
3
2
−3
(− 12 , 23 )
(− 34 , 16 )
−2
1
16. The equation of this line is
x
2
3
−1
y 2
2x 2
y
2 x 2.
−2
−3
Therefore, three additional points are 3, 4 , 1, 0 ,
and (0, 2).
10. m
§ 3· § 1·
¨ ¸ ¨ ¸
© 4¹ © 4¹
§7· §5·
¨ ¸¨ ¸
©8¹ © 4¹
1
3
8
8
3
17.
y
3x 3
4
4y
3x 12
0
3x 4 y 12
y
y
5
3
4
(0, 3)
2
1
2
( 78 , 34 )
1
x
−2
−1
1
−1
( 54 , − 14 )
x
−4 −3 −2 −1
1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.2
18. The slope is undefined so the line is vertical.
x
5
x 5
0
Linear Models and Rat
Rate
Rates of Change
21. y 2
3x3
y 2
3x 9
y
3x 11
0
3x y 11
y
1
13
y
x
−4 −3 −2 −1
(−5, −2)
1
−1
3
2
−2
1
−3
x
−2 −1
−1
−4
1
2
3
5
6
(3, −2)
−2
−5
4
−3
−4
−5
19.
y
2x
3
3y
2x
y 4
22.
2x 3y
0
53 x 2
5 y 20
y
3 x 6
3x 5 y 14
0
4
y
3
5
2
4
(− 2, 4)
(0, 0)
x
1
2
3
4
2
−1
1
x
y
4
y 4
0
20.
−3
y
−2
−1
1
23. (a) Slope
'y
'x
2
1
3
(b)
5
x
10 ft
(0, 4)
3
30 ft
2
By the Pythagorean Theorem,
1
−3
−2
x
−1
1
2
y
Population (in millions)
24. (a)
(b) The slopes are:
310
(9, 307)
305
300
(5, 295.8)
295
290
(8, 304.4)
(7, 301.6)
(6, 298.6)
(4, 293)
t
4
5
6
7
8
9
Year (4 ↔ 2004)
x2
302 102
x
10 10 | 31.623 feet.
295.8 293.0
5 4
298.6 295.8
65
301.6 298.6
7 6
304.4 301.6
87
307.0 304.4
98
1000
2.8
2.8
3.0
2.8
2.6
The population increased least rapidly from 2008 to 2009.
(c) Average rate of change from 2004 to 2009:
307.0 293.0
9 4
14
5
2.8 million per yr
(d)
For 2020, t
20 and y | 16 2.8 293.0
ª¬Equivalently, y | 11 2.8 307.0
337.8 million.
337.8.º¼
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
14
NOT FOR SALE
Chapter P
Preparation
paration for Calculus
Calc
4x 3
25. y
33. y
2 x 1
4 and the y-intercept is 0, 3 .
The slope is m
y
3
26. x y
1
x 1
y
The slope is m
1
1 and the y-intercept is (0, 1).
x
−2
27. x 5 y
−1
1
−1
20
15 x 4
y
Therefore, the slope is m
15 and the y-intercept is
34. y
1x 1
3
y
(0, 4).
2
28. 6 x 5 y
15
1
6x 3
5
y
x
−3 − 2 − 1
3
(0, −1)
Therefore, the slope is m
6 and the y-intercept is
5
−2
−3
0, 3 .
29. x
−4
4
The line is vertical. Therefore, the slope is undefined and
there is no y-intercept.
35. y 2
3 x 1
2
y
3x 1
2
2
y
1
30. y
2
4
The line is horizontal. Therefore, the slope is m
the y-intercept is 0, 1 .
0 and
3
2
1
x
−4 −3 −2
3
31. y
1
2
3
4
−2
−3
y
−4
2
1
x
−3 −2 −1
1
2
3
4
5
36. y 1
3x 4
y
3 x 13
−2
−4
y
−5
−6
16
12
32. x
4
y
x
−16 −12 − 8
3
4
−4
8
−8
2
1
x
1
2
3
5
37. 2 x y 3
−1
y
−2
0
2x 3
y
1
x
−2
−1
2
3
−1
−2
−3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.2
38. x 2 y 6
0
4
2
−8
−6
x
6
x6
5
, undefined
0
0
y
x
−2
15
The line is horizontal.
y
− 10
83
66
43. m
12 x 3
y
Linear Models and Rat
Rate
Rates of Change
(6, 8)
8
−4
6
−6
4
(6, 3)
2
80
2
40
2x0
m
39.
y 0
y
0
x
−2
y
2
4
8
−2
(4, 8)
8
6
2x
4
2x y
2
2 2
31
44. m
(0, 0)
−4
y
2
y 2
0
x
−2
2
4
6
0
2
0
y
40. m
7 2
1 2
y 2
41. m
9
3
−3
y
3x 4
0
3x y 4
8
x5
3
8
40
x
3
3
0
4
4
y 2
1 x 1
y 2
x 1
x y 3
4
3
4
(1, −2)
(3, − 2)
6
−4
45. m
y
62
3 1
2
(− 2, − 2)
2
−4
x
−6 −4
8
3
8 x 3 y 40
1
−1
4
3x 2
y
x
−1
(1, 7)
6
y 2
y 0
42. m
8
3 x 2
80
25
1
y
3
9
8
7
6
5
4
3
2
1
−1
(2, 8)
7 3
2 4
1
0
2
y
(5, 0)
1 2 3 4
x
6 7 8 9
0
−2
y
7
46. m
6
(−3, 6)
5
3
0
(1, 2)
2
1
x
−4 −3 −2 −1
11
2
y
4
( 12 , 72 )
3
3
y 4
1
11
4
1
2
1
2
3
11
x0
2
11
3
x 2
4
22 x 4 y 3
§ 3· § 1·
¨ ¸ ¨ ¸
© 4¹ © 4¹
§7· §5·
¨ ¸¨ ¸
©8¹ © 4¹
1
3
8
2
x
−4 −3 −2 −1
1
2
3
4
8
3
y
1
y 4
5·
8 §
¨x ¸
3©
4¹
12 y 3
32 x 40
32 x 12 y 37
( 0, 34 )
1
3
2
1
0
( 78 , 34 )
x
−2
−1
1
−1
( 54 , − 14 )
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
16
Chapter P
x
3
x 3
0
47.
Preparation
paration for Calculus
Calc
52.
y
2
1
(3, 0)
1
2
x
4
x
y
a
a
3 4
a
a
1
a
a
1
1
1
1Ÿ x y
x y 1
−1
−2
53.
48. m
b
a
b
xb
a
y
b
x y
a
x
y
a
b
x
y
2a
a
2
9
2a
a
9 4
2a
5
b
1
1
1
2a
5
2
a
1
x
2 52
y
y
1
5
2
x
2y
5
5
x 2y
(0, b)
1
5
x 2y 5
(a, 0)
54.
x
y
49.
2
3
3x 2 y 6
50.
x
y
2
2
3
y
3x
2
2
3x y
3x y 2
x
y
51.
a
a
1
2
a
a
3
a
a
23
1
a
0
2
a
x
4
3
2
a
4
3
y
43
3x 3 y 4
1
1
a
x y
0
1
2
2
3
1
1
0
x
y
a
a
x
1
0
1
4
3
0
55. The given line is vertical.
1
1
(a) x
7, or x 7
0
(b) y
2, or y 2
0
56. The given line is horizontal.
3 Ÿ x y
3
(a) y
0
x y 3
0
(b) x
1, or x 1
0
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.2
57. x y
2
61. 5 x 3 y
y
x 2
m
1
(a)
y 5
1x 2
y 5
x 2
x y 3
(b)
Linear Models and Rate
Rat
Rates of Change
(a)
y 5
1 x 2
y 5
x 2
x y 7
0
y
5x
3
m
5
3
y 78
5 x 3
3
4
24 y 21
0
40 x 30
40 x 24 y 9
0
y 78
53 x 34
40 y 35
24 x 18
(b)
0
24 x 40 y 53
58. x y
0
7
y
x 7
m
1
y 2
1 x 3
y 2
x 3
(a)
x y 1
62. 3x 4 y
0
(b) y 2
1x 3
y 2
x 3
59. 4 x 2 y
(a)
4y
3 x 7
y
34 x 74
m
34
y 5
34 x 4
y 5
34 x 3
3 x 12
3x 4 y 8
3
2 x 32
y
7
4 y 20
x y 5
0
0
(b) y 5
4 x 4
3
m
(a) y 1
2
2 x 2
y 5
4 x 16
3
3
3 y 15
4 x 16
y 1
2x 4
0
0
(b)
2x y 3
y 1
12 x 2
2y 2
x 2
x 2y 4
60. 7 x 4 y
4y
y
m
(a)
V
1850 when t
V
250 t 2 1850
2.
250t 1350
8
64. The slope is 4.50.
7x 8
7
x 2
4
7
4
V
156 when t
V
4.5 t 2 156
2.
4.5t 147
1
y 2
5·
7§
¨x ¸
4©
6¹
1
2
24 y 12
7
35
x 4
24
42 x 35
42 x 24 y 23
4 x 3 y 31
63. The slope is 250.
0
y (b)
17
65. The slope is 1600.
V
17,200 when t
V
1600 t 2 17,200
2.
1600t 20,400
66. The slope is 5600.
0
V
245,000 when t
1
y 2
4§
5·
¨x ¸
7©
6¹
V
5600 t 2 245,000
42 y 21
24 x 20
24 x 42 y 41
2.
5600t 256,200
0
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
18
NOT FOR SALE
Chapter P
Preparation
paration for Calculus
Calc
67. m1
10
2 1
1
m2
2 0
2 1
71. Equations of altitudes:
a b
y
x a
c
x
b
a b
y
x a
c
Solving simultaneously, the point of intersection is
§ a 2 b2 ·
¨ b,
¸.
c
©
¹
2
3
m1 z m2
The points are not collinear.
6 4
70
11 4
m2
5 0
m1 z m2
10
7
7
5
68. m1
y
(b, c)
The points are not collinear.
(a, 0)
69. Equations of perpendicular bisectors:
c
y 2
a b§
a b·
¨x ¸
2 ¹
c ©
c
2
a b§
b a·
¨x ¸
2 ¹
c ©
y x
(− a, 0)
§b c·
72. The slope of the line segment from ¨ , ¸ to
© 3 3¹
Setting the right-hand sides of the two equations equal
and solving for x yields x
0.
§ a2 b2 ·
¨ b,
¸ is:
c
©
¹
Letting x
0 in either equation gives the point of
intersection:
m1
§ a 2 b2 c 2 ·
¨ 0,
¸.
2c
©
¹
3a 2 3b 2 c 2
§ a 2 b2 c 2 ·
¨ 0,
¸ is:
2c
©
¹
(b, c)
)
m2
( a +2 b , 2c )
ª a 2 b 2 c 2 2c º c 3
¬
¼
0 b3
3a 2 3b 2 3c 2 2c 2
x
(a, 0)
(−a, 0)
3a 2 3b 2 c 2
2bc
§b c·
The slope of the line segment from ¨ , ¸ to
© 3 3¹
y
(
3c
2b 3
This point lies on the third perpendicular bisector,
x
0.
b − a, c
2
2
ª a 2 b 2 cº c 3
¬
¼
b b3
b 3
m1
70. Equations of medians:
y
y
y
y
6c
3a 2 3b 2 c 2
2bc
m2
Therefore, the points are collinear.
c
x
b
c
x a
3a b
c
xa
3a b
( b −2 a , 2c )
(b, c)
( a +2 b , 2c )
x
(−a, 0)
(0, 0) (a, 0)
§b c·
Solving simultaneously, the point of intersection is ¨ , ¸.
© 3 3¹
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section P.2
73. ax by
0 and
(b) The line is parallel to the y-axis if b
a z 0.
0 and
(c) Answers will vary. Sample answer: a
b 8.
5 x 8 y
(b) Lines a and b have negative slopes.
(c) Lines c and e appear parallel.
Lines d and f appear parallel.
(d) Lines b and f appear perpendicular.
5 and
Lines b and d appear perpendicular.
75. Find the equation of the line through the points (0, 32)
and (100, 212).
4
1 5x 4
8
y
5x 1
8
2
m
F 32
(d) The slope must be 52 .
Answers will vary. Sample answer: a
b
2.
5x 2 y
F
5 and
C
1 5 x 4
2
5 and b
2
52 x 2
4
5x 6 y
8
76. C
For x
W1
0.07 s 2000
New job offer: W2
0.05s 2300
77. (a) Current job:
1 5 F 160
9
5F 9C 160
0
72q, C | 22.2q.
For F
3.
5 x 3y
2
180
9
100
5
9 C 0
5
9 C 32
5
or
4
y
(b)
19
74. (a) Lines c, d, e and f have positive slopes.
4
(a) The line is parallel to the x-axis if a
b z 0.
(e) a
Linear Models and Rate
Rat
Rates of Change
0.51x 200
137, C
0.51 137 200
$269.87.
3500
(15,000, 3050)
0
1500
20,000
Using a graphing utility, the point of intersection is (15,000, 3050).
Analytically, W1
W2
0.07 s 2000
0.05s 2300
0.02 s
300
s
So, W1
W2
15,000
0.07 15,000 2000
3050.
When sales exceed $15,000, the current job pays more.
(c) No, if you can sell $20,000 worth of goods, then W1 ! W2 .
(Note: W1
3400 and W2
78. (a) Depreciation per year:
875
5
3300 when s
20,000.)
1000
$175
875 175 x
y
where 0 d x d 5.
0
6
0
875 175 2
(b) y
(c)
200
875 175 x
175 x
675
$525
x | 3.86 years
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
20
Chapter P
NOT FOR SALE
Preparation
paration for Calculus
Calc
79. (a) Two points are (50, 780) and (47, 825).
The slope is
825 780
45
m
15.
47 50
3
p 780
82. The tangent line is perpendicular to the line joining the
point 4, 3 and the center of the circle, (1, 1).
y
15 x 50
4
15 x 750 780
p
15 x 1530
or
−6
−2
1
1530 p
15
x
(b)
2
(1, 1)
x
2
−2
4
(4, −3)
−6
50
Slope of the line joining (1, 1) and 4, 3 is
13
1 4
0
1600
Tangent line:
0
If p
855, then x
(c) If p
795, then x
80. (a) y
18.91 3.97 x
x
quiz score, y
(b)
4
.
3
45 units.
1
1530 795
15
y 3
49 units
y
0
3
x4
4
3
x6
4
3 x 4 y 24
test score
83. x y 2
100
1 2 1 1 2
0 Ÿ d
12 12
5
2
0
20
4 2 3 3 10
0
(c) If x
18.91 3.97 17
17, y
86.4.
(d) The slope shows the average increase in exam score
for each unit increase in quiz score.
(e) The points would shift vertically upward 4 units.
The new regression line would have a y-intercept
4 greater than before: y
22.91 3.97 x.
84. 4 x 3 y 10
81. The tangent line is perpendicular to the line joining the
point (5, 12) and the center (0, 0).
15
12 12
2
4
2
2 2.
1 is 1, 1 . The
distance from the point 1, 1 to 3x 4 y 10
8
4
(0, 0) 8
7
5
42 32
1 0 11 5
86. A point on the line 3x 4 y
(5, 12)
−8 −4
0 Ÿ d
85. A point on the line x y 1 is (0, 1). The distance
from the point (0, 1) to x y 5 0 is
d
y
5 2
2
x
d
16
3 1 4 1 10
3 4
2
−8
2
3 4 10
5
0 is
9
.
5
−16
Slope of the line joining (5, 12) and (0, 0) is
12
.
5
The equation of the tangent line is
5
y 12
x 5
12
5
169
y
x 12
12
5 x 12 y 169 0.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.2
87. If A
If B
0, then Ax C
Ax1 By1 C
A2 B 2
.
C A. The distance to x1 , y1 is
0 is the vertical line x
Ax1 C
A
§ C ·
x1 ¨
¸
© A ¹
Ax1 By1 C
A2 B 2
.
(Note that A and B cannot both be zero.) The slope of the line Ax By C
The equation of the line through x1 , y1 perpendicular to Ax By C
Ay Ay1
B
x x1
A
Bx Bx1
Bx1 Ay1
Bx Ay
y y1
21
C B. The distance to x1 , y1 is
0 is the horizontal line y
By1 C
B
§ C ·
y1 ¨
¸
© B ¹
d
d
0, then By C
Linear Models and Rate
Rat
Rates of Change
0 is A B.
0 is:
The point of intersection of these two lines is:
Ÿ A2 x ABy
Bx1 Ay1 Ÿ B 2 x ABy
Ax By
AC
C
Bx Ay
1
B x1 ABy1
2
A B x
AC B x1 ABy1
By adding equations 1 and 2
x
AC B x1 ABy1
A2 B 2
2
2
2
2
2
Ax By
C
Ÿ
ABx B 2 y
Bx1 Ay1 Ÿ ABx A2 y
Bx Ay
A B y
2
2
y
BC
ABx1 A2 y1
3
4
BC ABx1 A y1 By adding equations 3 and 4
2
BC ABx1 A2 y1
A2 B 2
§ AC B 2 x1 ABy1 BC ABx1 A2 y1 ·
,
¨
¸ point of intersection
A2 B 2
A2 B 2
©
¹
The distance between x1 , y1 and this point gives you the distance between x1 , y1 and the line Ax By C
2
d
ª AC B 2 x1 ABy1
º
ª BC ABx1 A2 y1
º
x1 » «
y1 »
«
2
2
2
2
A
B
A
B
¬
¼
¬
¼
2
ª AC ABy1 A2 x1 º
ª BC ABx1 B 2 y1 º
«
» «
»
2
2
A B
A2 B 2
¬
¼
¬
¼
2
ª A C By1 Ax1 º
ª B C Ax1 By1 º
«
» «
»
2
2
A B
A2 B 2
¬
¼
¬
¼
88. y
d
mx 4 Ÿ mx 1 y 4
m3 1 1 4
A B
m 1
2
2
The distance is 0 when m
2
2
A2 B 2 C Ax1 By1
A B
2
2
2
2
Ax1 By1 C
A2 B 2
0
Ax1 By1 C
2
2
0.
2
3m 3
m2 1
1. In this case, the line y
x 4 contains the point (3, 1).
8
−9
(−1, 0)
9
INSTRUCTOR USE ONLY
−4
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© Cengage Learning. All Rights Reserved.
22
NOT FOR SALE
Chapter P
Preparation
paration for Calculus
Calc
89. For simplicity, let the vertices of the rhombus be (0, 0),
(a, 0), (b, c), and a b, c , as shown in the figure.
c
and
a b
The slopes of the diagonals are then m1
c
. Because the sides of the rhombus are
ba
m2
equal, a
y2* y1*
x2* x1*
y2 y1
x2 x1
y
b c , and you have
2
2
2
c2
b2 a 2
c
c
˜
a b ba
m1m2
91. Consider the figure below in which the four points are
collinear. Because the triangles are similar, the result
immediately follows.
c2
c 2
(x 2 , y2 )
1.
(x *2 , y*2 )
(x1, y1 )
(x *1, y*1 )
Therefore, the diagonals are perpendicular.
y
x
(b, c)
(a + b, c)
1. Let L3 be a line with
1 m2 , then m1m2
92. If m1
slope m3 that is perpendicular to L1. Then m1m3
x
(0, 0)
(a , 0)
1.
m3 Ÿ L 2 and L3 are parallel. Therefore,
So, m2
L 2 and L1 are also perpendicular.
90. For simplicity, let the vertices of the quadrilateral be
(0, 0), (a, 0), (b, c), and (d, e), as shown in the figure.
The midpoints of the sides are
§a · §a b c · §b d c e·
§d e·
, ¸, ¨
,
¨ , 0 ¸, ¨
¸, and ¨ , ¸.
2¹ © 2
2 ¹
©2 ¹ © 2
© 2 2¹
The slope of the opposite sides are equal:
c
c e e
0
2
2
2
a b a
b d
d
2
2
2
2
e
c
ce
0
2
2
2
a
d
a b b d
2
2
2
2
c
b
93. True.
ax by
c1 Ÿ y
bx ay
c2 Ÿ y
m2
a
c
x 1 Ÿ m1
b
b
b
c2
Ÿ m2
x
a
a
a
b
b
a
1
m1
1 m1 is negative.
94. False; if m1 is positive, then m2
95. True. The slope must be positive.
e
a d
96. True. The general form Ax By C
horizontal and vertical lines.
0 includes both
Therefore, the figure is a parallelogram.
y
(d, e)
( b +2 d , c 2+ e )
(b, c)
( d2 , 2e )
(a +2 b , 2c )
x
(0, 0)
( a2 , 0) (a, 0)
Section P.3 Functions and Their Graphs
1. (a) f 0
70 4
(b) f 3
7 3 4
(c) f b
7b 4
(d) f x 1
4
2. (a) f 4
4 5
1
1
25
(b) f 11
11 5
16
4
7b 4
(c) f 4
45
7 x 1 4
7 x 11
(d) f x ' x
9
3
x 'x 5
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
5 02
3. (a) g 0
(b) g
5
5
5 2
2
5
(c) g 2
5
(d) g t 1
2
55
54
5 t 1
2
Section P.3
Functions and T
Their Graphs
5. (a) f 0
cos 2 0
§ S·
(b) f ¨ ¸
© 4¹
§ § S ··
cos¨ 2¨ ¸ ¸
© © 4 ¹¹
§S ·
(c) f ¨ ¸
©3¹
§ § S ··
cos¨ 2¨ ¸ ¸
© © 3 ¹¹
(d) f S
cos 2 S
6. (a) f S
sin S
0
1
5 t 2 2t 1
4 2t t 2
42 4 4
4. (a) g 4
g 32
(b)
3
2
2
0
3 4
2
9 5
4
2
c c4
c 4c
2
(c) g c
(d) g t 4
3
t 4
2
2
7.
8.
9.
f x 'x f x
45
8
2
t 44
t 4 t
t 3 8t 2 16t
x 'x
3
x3
'x
'x
f x f 1
3x 1 3 1
3x 1
x 1
x 1
x 1
f x f 2
1
x 2
f x f 1
x 1
11. f x
4x2
x2 5
Domain: f, f
Range: >5, f
x3
Domain: f, f
Range: f, f
14. h x
4 x2
Domain: f, f
Range: f, 4@
§ 2S ·
(c) f ¨ ¸
© 3 ¹
§ 2S ·
sin ¨ ¸
© 3 ¹
3
2
§ S·
(d) f ¨ ¸
© 6¹
§ S·
sin ¨ ¸
© 6¹
x3
1
2
0
3
1
§ 5S ·
sin ¨ ¸
© 4 ¹
'x
2S
3
2
2
1
2
2
3 x 2 3x'x 'x , 'x z 0
3, x z 1
x 1 1
x3 x 0
x 1
Range: >0, f
13. f x
cos
0
x 2
Domain: f, f
12. g x
2
1
§ S·
cos¨ ¸
© 2¹
§ 5S ·
(b) f ¨ ¸
© 4 ¹
'x
1 x 1
1
˜
x 2 x 1 1
10.
x 3 3 x 2 'x 3 x 2 ' x
cos 0
23
x 1
x 1
x x 1 x 1
x 1
2 x
x 2
1
x 11
x 1
x 11 x 1
, x z 2
x x 1, x z 1
15. g x
6x
Domain: 6 x t 0
x t 0 Ÿ >0, f
Range: >0, f
16. h x
x 3
Domain: x 3 t 0 Ÿ >3, f
Range: f, 0@
17. f x
16 x 2
16 x 2 t 0 Ÿ x 2 d 16
Domain: > 4, 4@
Range: >0, 4@
Note: y
16 x 2 is a semicircle of radius 4.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
24
Chapter P
NOT FOR SALE
Preparation
paration for Calculus
Calc
x 3
18. f x
1
sin x 1 2
26. h x
Domain: f, f
Range: >0, f
19. f t
sec
St
z
4
1
z 0
2
1
sin x z
2
sin x St
4
2n 1 S
Ÿ t z 4n 2
2
Domain: all t z 4n 2, n an integer
Range: f, 1@ ‰ >1, f
20. h t
x 3 z 0
nS , n an integer
Domain: all x z 3
Domain: f, 3 ‰ 3, f
Range: f, f
21. f x
3
x
1
x2 4
28. g x
Domain: all x z 0 Ÿ f, 0 ‰ 0, f
Domain: all x z r 2
x 2
x 4
Domain: all x z 4
22. f x
Domain: f, 2 ‰ 2, 2 ‰ 2, f
Range: all y z 1
29. f x
[Note: You can see that the range is all y z 1 by
graphing f.]
x x2 4 z 0
x 2 x 2 z 0
Range: f, 0 ‰ 0, f
23. f x
1
x3
27. f x
x3 z 0
cot t
Domain: all t
S
5S
2nS ,
2nS , n integer
6
6
Domain: all x z
1 x
x t 0 and 1 x t 0
­2 x 1, x 0
®
¯2 x 2, x t 0
(a) f 1
2 1 1
(b) f 0
20 2
2
(c) f 2
22 2
6
(d) f t 2 1
x t 0 and x d 1
Domain: 0 d x d 1 Ÿ >0, 1@
1
2 t2 1 2
2t 2 4
(Note: t 2 1 t 0 for all t.)
Domain: f, f
24. f x
x 3x 2
2
Range: f, 1 ‰ >2, f
x 3x 2 t 0
2
x 2 x 1 t 0
30. f x
Domain: x t 2 or x d 1
2
°­ x 2, x d 1
® 2
°̄2 x 2, x ! 1
Domain: f, 1@ ‰ >2, f
(a) f 2
2
2
2
2
1 cos x
(b) f 0
02 2
2
(c) f 1
12 2
3
25. g x
1 cos x z 0
(d) f s 2 2
cos x z 1
Domain: all x z 2nS , n an integer
2 s2 2
6
2
2
2 s 4 8s 2 10
(Note: s 2 2 ! 1 for all s.)
Domain: f, f
INSTRUCTOR USE ONLY
Range: >2, f
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.3
°­ x 1, x 1
®
°̄ x 1, x t 1
31. f x
(a) f 3
3 1
y
4
5
4
1 1
0
(c) f 3
3 1
2
(d) f b 1
2
1
b 1 1
b
2
x
2
−3
Domain: f, f
1
2
3
Range: f, f
­° x 4, x d 5
®
2
°̄ x 5 , x ! 5
(a) f 3
3 4
(b) f 0
0 4
2
(c) f 5
5 4
3
(d) f 10
10 5
1
2
y
9 x2
37. f x
5
Domain: >3, 3@
1
4
Range: >0, 3@
2
1
x
−4 −3 −2 −1
1
2
3
4
−2
−3
25
Domain: >4, f
x
38. f x
4 x2
Domain: >2, 2@
Range: >0, f
4 x
33. f x
−1
−1
Domain: f, f
Range: f, 0@ ‰ >1, f
32. f x
25
1 x3 3
4
36. f x
(b) f 1
2
Functions and T
Their Graphs
Range: ª¬2, 2 2 º¼ | >2, 2.83@
y
8
Domain: f, f
y-intercept: 0, 2
6
Range: f, f
x-intercept: 4
y
2
4
x
−4
2, 0
−2
2
3
4
(0, 2)
(− 2, 0(
4
x
34. g x
x
y
−4 −3 −2
2
3
4
−2
6
Domain: f, 0 ‰ 0, f
1
−1
−3
4
−4
2
Range: f, 0 ‰ 0, f
x
2
4
6
39. g t
3 sin S t
y
3
2
x 6
35. h x
1
y
t
1
3
3
Domain:
x 6 t 0
x t 6 Ÿ >6, f
Range: >0, f
2
1
x
3
6
9
12
Domain: f, f
Range: >3, 3@
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
26
Chapter P
40. h T
5 cos
NOT FOR SALE
Preparation
paration for Calculus
Calc
T
2
49. y 2
x2 1 Ÿ y
x2 1
y is not a function of x because there are two values of y
for some x.
Domain: f, f
Range: >5, 5@
50. x 2 y x 2 4 y
y
x2
x2 4
0 Ÿ y
y is a function of x because there is one value of y for
each x.
5
4
3
2
1
−2π
51. The transformation is a horizontal shift two units to the
right.
θ
2π
Shifted function: y
−5
20
1
mi min during the first 4
40
2
minutes. The student is stationary for the next 2 minutes.
62
Finally, the student travels
1 mi min during
10 6
the final 4 minutes.
41. The student travels
42.
r
x 2
52. The transformation is a vertical shift 4 units upward.
Shifted function: y
sin x 4
53. The transformation is a horizontal shift 2 units to the
right and a vertical shift 1 unit downward.
Shifted function: y
x 2
2
1
54. The transformation is a horizontal shift 1 unit to the left
and a vertical shift 2 units upward.
d
27
Shifted function: y
18
55. y
9
x 1
3
2
f x 5 is a horizontal shift 5 units to the left.
Matches d.
t1
t2
t
t3
56. y
43. x y
0 Ÿ y
2
r
x
y is not a function of x. Some vertical lines intersect the
graph twice.
x2 4 y
44.
0 Ÿ y
x2 4
57. y
58. y
45. y is a function of x. Vertical lines intersect the graph at
most once.
59. y
46. x y
2
4
y
r
f x 4 is a horizontal shift 4 units to the right,
followed by a reflection in the x-axis. Matches a.
f x 6 2 is a horizontal shift to the left 6
units, and a vertical shift upward 2 units. Matches e.
60. y
4 x2
f x 2 is a reflection in the y-axis, a
reflection in the x-axis, and a vertical shift downward 2
units. Matches c.
y is a function of x. Vertical lines intersect the graph at
most once.
2
f x 5 is a vertical shift 5 units downward.
Matches b.
f x 1 3 is a horizontal shift to the right 1 unit,
and a vertical shift upward 3 units. Matches g.
y is not a function of x. Some vertical lines intersect the
graph twice.
47. x 2 y 2
16 Ÿ y
r 16 x 2
y is not a function of x because there are two values of y
for some x.
48. x 2 y
16 Ÿ y
16 x 2
y is a function of x because there is one value of y for
each x.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.3
61. (a) The graph is shifted 3 units to the left.
Functions and T
Their Graphs
27
(f) The graph is stretched vertically by a factor of 14 .
y
y
4
4
2
x
−6
−4
−2
2
x
4
−4
−2
−2
2
4
6
−4
−6
−6
(b) The graph is shifted 1 unit to the right.
(g) The graph is a reflection in the x-axis.
y
y
4
2
2
x
−2
2
4
6
x
8
−4
−2
−2
2
4
6
−2
−4
−4
−6
(h) The graph is a reflection about the origin.
y
(c) The graph is shifted 2 units upward.
6
y
4
6
4
x
−6
2
2
4
−2
x
−4
−4
−2
2
4
−4
6
−2
(d) The graph is shifted 4 units downward.
y
x
−4
−2
2
4
6
−2
−4
−6
−8
(e) The graph is stretched vertically by a factor of 3.
y
x
−4
−2
4
6
−2
−4
−6
−8
−10
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
28
Chapter P
62.
(a) g x
NOT FOR SALE
Preparation
paration for Calculus
Calc
f x 4
g 6
f 2
g 0
f 4
(e)
1
3
g x
2f x
g 2
2f 2
g 4
2 f 4
2
6
The graph is stretched vertically by a factor of 2.
The graph is shifted 4 units to the right.
y
y
(2, 2)
2
4
1
3
x
2
−5 −4 −3 −2 −1
(6, 1)
1
1
2
3
x
−1
1
2
3
5
6
−3
7
−4
−2
−5
(0, − 3)
−4
(− 4, − 6)
f x 2
g x
(b)
g 0
f 2
g 6
f 4
(f)
1
3
The graph is shifted 2 units to the left.
g x
1 f
2
x
g 2
1 f
2
2
g 4
1 f
2
4
1
2
32
The graph is stretched vertically by a factor of 12 .
y
y
4
3
2
2
(2, 12 )
1
(0, 1)
x
5
x
−7 −6 −5 −4 −3
−1
(− 6, − 3)
1
4
3
1
−1
−2
( − 4, − 32 ) −− 23
−3
−4
−4
2
3
−5
−6
g x
f x 4
g 2
f 2 4
g 4
f 4 4
(c)
(g)
5
1
The graph is shifted 4 units upward.
g x
f x
g 2
f 2
g 4
f 4
1
3
The graph is a reflection in the y-axis.
y
y
6
(2, 5)
5
3
4
(−2, 1) 2
1
x
2
− 3 − 2 −1
−1
1
(− 4, 1)
1
2
−3
3
f x 1
g 2
f 2 1
g 4
f 4 1
(h)
0
4
The graph is shifted 1 unit downward.
f x
g 2
f 2
1
g 4
f 4
3
The graph is a reflection in the x-axis.
y
y
5
(2, 0)
x
4
3
2
1
5
g x
2
5
4
−5
g x
1
3
(4, − 3)
−4
−2
(d)
2
−2
x
−5 −4 −3 − 2 −1
1
2
3
(−4, 3)
4
3
2
−3
(− 4, −4)
−4
1
−5
− 5 − 4 −3 −2 −1
−1
−6
−2
x
3
(2, − 1)
−3
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section P.3
63. f x
3x 4, g x
29
4
(a) f x g x
3x 4 4
3x
(b) f x g x
3x 4 4
3x 8
(c) f x ˜ g x
3x 4 4
(d) f x g x
3x 4
4
64. f x
Functions and T
Their Graphs
12 x 16
3
x 1
4
x 2 5 x 4, g x
x 1
(a) f x g x
x2 5x 4 x 1
x2 6 x 5
(b) f x g x
x2 5x 4 x 1
x2 4 x 3
(c) f x ˜ g x
x2 5x 4 x 1
x3 5 x 2 4 x x 2 5 x 4
x3 6 x 2 9 x 4
f 0
0
(b) g f 1
g1
0
(c) g f 0
g 0
1
(d) f g 4
f 15
(e) f g x
f x2 1
(f) g f x
g
x2 1
x
2
1
g D f x
x 1, x t 0
0
§ § 1 ··
(b) f ¨ g ¨ ¸ ¸
© © 2 ¹¹
§S ·
f¨ ¸
©2¹
§S ·
sin ¨ ¸
©2¹
1
f Sx
(f) g f x
g sin x
x
g f x
g x2
2
x, x t 0
x2
x
Domain: f, f
68. f x
x 2 1, g x
f D g x
g D f
f g x
cos x
f cos x
cos 2 x 1
Domain: f, f
0
g D f x
§ § S ··
g ¨ sin ¨ ¸ ¸
© © 4 ¹¹
(e) f g x
x
for x t 0.
sin 2S
§ 2·
g ¨¨
¸¸
© 2 ¹
f g x
No. Their domains are different. f D g
Sx
f 2S
§ § S ··
(d) g ¨ f ¨ ¸ ¸
© © 4 ¹¹
x
Domain: >0, f
15
g 0
x2 , g x
f
(a) f g 2
(c) g f 0
x 4, x z 1
f D g x
x
sin x, g x
x 1
67. f x
65. (a) f g 1
66. f x
x 4 x 1
x2 5x 4
x 1
(d) f x g x
g x2 1
cos x 2 1
Domain: f, f
§ 2·
S ¨¨
¸¸
© 2 ¹
S
2
2
No, f D g z g D f .
sin S x
S sin x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
30
NOT FOR SALE
Chapter P
Preparation
paration for Calculus
Calc
3
, g x
x
69. f x
f D g x
x2 1
70.
2
§ 3·
¨ ¸ 1
© x¹
9 x
x2
9
1
x2
negative.
+
−2
No, f D g z g D f .
72.
f D g 3
+
+ − − +
−1 − 1
2
0
+
+
1
+
x
2
Domain: f, 12 º¼, 0, f
f 1
f g 3
4
2
(b) g f 2
g1
(c) g f 5
g 5 , which is undefined
(d)
f D g 3
f g 3
f 2
(e)
g D f 1
g f 1
g 4
(f)
f g 1
ADr t
1 2x
x
intervals where 1 2x and x are both positive, or both
2
Domain: all x z 0 Ÿ f, 0 ‰ 0, f
71. (a)
1
2
x
You can find the domain of g D f by determining the
g f x
§ 3·
g¨ ¸
© x¹
§1·
g¨ ¸
© x¹
g D f x
Domain: all x z r1 Ÿ f, 1 ‰ 1, 1 ‰ 1, f
g D f x
1
x 2
x 2
f
Domain: 2, f
3
x2 1
f x2 1
f g x
f D g x
3
2
f 4 , which is undefined
Art
A 0.6t
S 0.6t
2
0.36S t 2
A D r t represents the area of the circle at time t.
73. F x
Let h x
2x 2
x 2 and f x
2 x, g x
Then, f D g D h x
f g 2x
x.
f 2x 2
2x 2
2x 2
F x.
[Other answers possible]
74. F x
Let f x
4 sin 1 x
4 x, g x
f D g Dh x
sin x and h x
f g1 x
1 x. Then,
f sin 1 x
4 sin 1 x
F x.
[Other answers possible]
75. (a) If f is even, then
3, 4
2
is on the graph.
(b) If f is odd, then
3, 4
2
is on the graph.
76. (a) If f is even, then 4, 9 is on the graph.
(b) If f is odd, then 4, 9 is on the graph.
77. f is even because the graph is symmetric about the y-axis. g is neither even nor odd. h is odd because the graph is symmetric
about the origin.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.3
78. (a) If f is even, then the graph is symmetric about the
y-axis.
f x
2
x
4 x
2
x2 4 x2
f x
f is even.
6
f
31
x2 4 x2
79. f x
y
Functions and T
Their Graphs
4
x2 4 x2
0
x2 2 x 2 x
0
f x
2
x
−6 −4 −2
−2
2
4
6
0, 2, 2
Zeros: x
−4
−6
3
80. f x
(b) If f is odd, then the graph is symmetric about the
origin.
f x
x
3
3 x
f x
f is odd.
y
3
f x
6
f
x
0 Ÿ x
x
0 is the zero.
4
2
81. f x
x
−6 −4 −2
−2
2
4
6
x cos x
f x
−4
x cos x
x cos x
f x
f is odd.
−6
f x
x cos x
Zeros: x
0
S
0,
nS , where n is an integer
2
sin 2 x
82. f x
f x
sin 2 x
sin x sin x
0 Ÿ sin x
0
sin x sin x
sin 2 x
f is even.
sin 2 x
nS , where n is an integer
Zeros: x
4 6
2 0
83. Slope
10
2
y 4
5 x 2
y 4
5 x 10
y
5 x 6
5
For the line segment, you must restrict the domain.
5 x 6, 2 d x d 0
f x
y
6
(−2, 4)
81
7
53
2
7
y 1
x3
2
7
21
y 1
x
2
2
7
19
y
x
2
2
For the line segment, you must restrict the domain.
7
19
f x
x , 3 d x d 5
2
2
84. Slope
y
4
2
(5, 8)
8
x
−6 −4 −2
2
4
6
6
4
−4
−6
(0, − 6)
2
(3, 1)
x
−2
2
4
6
8
−2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter P
Preparation
paration for Calculus
Calc
85. x y 2
0
y2
x
y
x
f x
x, x d 0
89. Answers will vary. Sample answer: In general, as the
price decreases, the store will sell more.
y
Number of sneakers sold
32
y
3
2
1
x
x
−5 −4 −3 −2 −1
Price (in dollars)
1
−2
90. Answers will vary. Sample answer: As time goes on, the
value of the car will decrease
−3
y
86. x 2 y 2
y2
36 x 2
y
Value
36
36 x 2 , 6 d x d 6
y
t
8
4
2
x
−4 −2
−2
2
4
c x2
y
91.
−4
c x2
y2
x2 y 2
87. Answers will vary. Sample answer: Speed begins and
ends at 0. The speed might be constant in the middle:
Speed (in miles per hour)
y
c, a circle.
For the domain to be >5, 5@, c
25.
92. For the domain to be the set of all real numbers, you
must require that x 2 3cx 6 z 0. So, the
discriminant must be less than zero:
3c
2
46 0
9c 2 24
c 2 83
x
Time (in hours)
88. Answers will vary. Sample answer: Height begins a few
feet above 0, and ends at 0.
23
93. (a) T 4
y
(b) If H t
8
3
c 6 c 23
8
3
6
16q, T 15 | 23q
T t 1 , then the changes in temperature
Height
will occur 1 hour later.
(c) If H t
T t 1, then the overall temperature
would be 1 degree lower.
x
Distance
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.3
94. (a) For each time t, there corresponds a depth d.
97. f x
(b) Domain: 0 d t d 5
x x 2
If 0 d x 2, then f x
d
2 x 2.
x x2
x x2
If x t 2, then f x
30
33
x x2
If x 0, then f x
Range: 0 d d d 30
(c)
Functions and T
Their Graphs
2.
2 x 2.
25
So,
20
15
10
f x
5
t
1
2
3
4
5
6
98. p1 x
(d) d 4 | 18. At time 4 seconds, the depth is
Average number of
acres per farm
x3 x 1 has one zero. p2 x
x3 x has
three zeros. Every cubic polynomial has at least one
zero. Given p x
Ax3 Bx 2 Cx D, you have
approximately 18 cm.
y
95. (a)
­2 x 2, x d 0
°
0 x 2.
®2,
°2 x 2,
x
t 2
¯
500
p o f as x o f and p o f as x o f if
400
A ! 0. Furthermore, p o f as x o f and
300
p o f as x o f if A 0. Because the graph has
200
no breaks, the graph must cross the x-axis at least one time.
100
x
99. f x
10 20 30 40 50 60
Year (0 ↔ 1960)
2 n 1
" a3 x
3
a1 x
ª¬a2 n 1 x 2 n 1 " a3 x3 a1xº¼
(b) A 25 | 445 Answers will vary.
96. (a)
a2 n 1 x
f x
Odd
25
100
0
0
2
§ x ·
(b) H ¨ ¸
© 1.6 ¹
§ x ·
§ x ·
0.002¨ ¸ 0.005¨ ¸ 0.029
© 1.6 ¹
© 1.6 ¹
0.00078125 x 2 0.003125 x 0.029
100. f x
a2 n x
2n
a2 n 2 x
2n 2
" a2 x
2
a0
a2 n x 2 n a2 n 2 x 2 n 2 " a2 x 2 a0
f x
Even
101. Let F x
f x g x where f and g are even. Then F x
So, F x is even. Let F x
F x
f x g x
f x g x
f x g x
F x.
f x g x where f and g are odd. Then
ª
¬ f x ºª
¼¬ g x º¼
f x g x
F x.
So, F x is even.
102. Let F x
F x
f x g x where f is even and g is odd. Then
f x g x
f x ª
¬ g x º¼
f x g x
F x .
So, F x is odd.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
34
Chapter P
NOT FOR SALE
Preparation
paration for Calculus
Calc
y 2
03
103. By equating slopes,
02
x 3
6
x 3
6
2
x 3
y 2
y
106. True
107. True. The function is even.
2x
,
x 3
2
§ 2x ·
x2 ¨
¸ .
© x 3¹
x2 y2
L
x 24 2 x
104. (a) V
3x
2
9 x 2 and
3x 2 . So, 3 f x z f 3x .
3f x
109. False. The constant function f x
0 has symmetry
with respect to the x-axis.
110. True. If the domain is ^a`, then the range is ^ f a `.
2
Domain: 0 x 12
(b)
x 2 then, f 3x
108. False. If f x
111. First consider the portion of R in the first quadrant:
x t 0, 0 d y d 1 and x y d 1; shown below.
1100
y
The area of this region is
3.
1 12
2
2
−1
12
−100
1
Maximum volume occurs at x
4. So, the
dimensions of the box would be 4 u 16 u 16 cm.
(c)
x
length and width
1
24 2 1
volume
1ª¬24 2 1 º¼
2
2 ª¬24 2 2 º¼
24 2 3
3ª¬24 2 3 º¼
2
4
24 2 4
4 ª¬24 2 4 º¼
2
5
24 2 5
5ª¬24 2 5 º¼
2
6
24 2 6
6 ª¬24 2 6 º¼
2
3
x 2 , then f 3
f 3
(1, 0) 2
−1
By symmetry, you obtain the entire region R:
y
800
The area of R is 4 32
2
(− 2, 1)
(2, 1)
6.
972
1024
980
x
−2
1
2
(2, −1)
(−2, − 1)
−2
864
c be constant polynomial.
112. Let g x
The dimensions of the box that yield a maximum
volume appear to be 4 u 16 u 16 cm.
105. False. If f x
(2, 1)
x
−1 (0, 0)
484
2
24 2 2
2
(0, 1)
9, but
Then f g x
f c and g f x
c. Because this is true for all real numbers c,
So, f c
f is the identity function: f x
3 z 3.
c.
x.
Section P.4 Fitting Models to Data
1. (a) and (b)
2. (a)
y
y
15
1000
14
13
900
12
11
800
10
700
9
8
600
7
x
x
900
1050
1200
7
1350
Yes, the data appear to be approximately linear.
The data can be modeled by equation
y
0.6 x 150. (Answers will vary).
8
9 10 11 12 13 14 15
The data do not appear to be linear.
(b) Quiz scores are dependent on several variables such
as study time, class attendance, and so on. These
variables may
y change
g from one qquiz to the next.
INSTRUCTOR USE ONLY
(c) When x
1075, y
0.6 1075 150
795.
795
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section P.4
3. (a) d
(b)
Fitting Models
M
Mo
to Data
35
7. (a) Using graphing utility,
S
180.89 x 2 205.79 x 272.
0.066 F
10
(b)
d = 0.066F
0
25,000
110
0
14
0
0
The model fits the data well.
55, then d | 0.066 55
(c) If F
3.63 cm.
(d)
9.7t 0.4
4. (a) s
(b)
(c) When x
(e)
−1
5
The model fits the data well.
2.5, s
24.65 meters second.
5. (a) Using a graphing utility, y
23,860
| 4.37
5460
When the height is doubled, the breaking strength
increases approximately by a factor of 4.
−5
0.122 x 2.07
The correlation coefficient is r | 0.87.
(b)
2370
| 4.06
584
The breaking strength is approximately 4 times
greater.
45
(c) If t
2, S | 583.98 pounds.
8. (a) Using a graphing utility
t
0.0013s 2 0.005s 1.48.
(b)
15
60
25
95
0
0
500
0
(c) Greater per capita energy consumption by a country
tends to correspond to greater per capita gross
national income. The three countries that most
differ from the linear model are Canada, Japan, and
Italy.
(d) Using a graphing utility, the new model
is y
0.142 x 1.66.
The correlation coefficient is r | 0.97.
(c) According to the model, the times required to attain
speeds of less than 20 miles per hour are all about
the same. Furthermore, it takes 1.48 seconds to reach
0 miles per hour, which does not make sense.
(d) Adding 0, 0 to the data produces
0.0009 s 2 0.053s 0.10.
t
(e) Yes. Now the car starts at rest.
9. (a) y
(b)
1.806 x3 14.58 x 2 16.4 x 10
300
6. (a) Trigonometric function
(b) Quadratic function
(c) No relationship
0
(d) Linear function
7
0
4.5, y | 214 horsepower.
(c) If x
10. (a) T
(b)
2.9856 u 104 p 3 0.0641 p 2 5.282 p 143.1
350
110
0
150
(c) For T
300q F , p | 68.29 lb in.2 .
(d) The model is based on data up to 100 pounds per
quare inch
square
inch.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
36
Chapter P
11. (a) y1
NOT FOR SALE
Preparation
paration for Calculus
Calc
0.0172t 3 0.305t 2 0.87t 7.3
y2
0.038t 2 0.45t 3.5
y3
0.0063t 3 0.072t 2 0.02t 1.8
(b)
20
y1 + y2 + y3
y1
y2
y3
0
11
0
y1 y2 y3
0.0109t 3 0.195t 2 0.40t 12.6
For 2014, t
14. So,
y1 y2 y3
0.0109 14
3
0.195 14
2
0.40 14 12.6
| 15.31 cents/mile
12. (a) N1
N2
(b)
1.89t 46.8
Linear model
0.0485t 2.015t 27.00t 42.3
3
2
Cubic model
100
N1
N2
0
20
40
(c) The cubic model is the better model.
(d) N 3
0.414t 2 11.00t 4.4
Quadratic model
100
0
20
40
The model does not fit the data well.
(e) For 2014, t
24 and
N1 | 92.16 million
N 2 | 115.524 million
The linear model seems too high. The cubic model is better.
(f) Answers will vary.
13. (a) Yes, y is a function of t. At each time t, there is one
and only one displacement y.
(c) One model is y
(d)
0.35 sin 4S t 2.
4
(b) The amplitude is approximately
2.35 1.65 2
(0.125, 2.35)
0.35.
(0.375, 1.65)
The period is approximately
2 0.375 0.125
0.5.
0
0.9
0
The model appears to fit the data.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
Review Exercises for
fo
f Chapter P
56.37 25.47 sin 0.5080t 2.07
14. (a) S t
(b)
37
(d) The average is the constant term in each model.
83.70qF for Miami and 56.37qF for Syracuse.
100
(e) The period for Miami is 2S 0.4912 | 12.8. The
period for Syracuse is 2S 0.5080 | 12.4. In both
M(t)
cases the period is approximately 12, or one year.
0
(f ) Syracuse has greater variability because
25.47 ! 7.46.
13
0
The model is a good fit.
(c)
15. Answers will vary.
100
16. Answers will vary.
S(t)
0
13
0
The model is a good fit.
17. Yes, A1 d A2 . To see this, consider the two triangles of areas A1 and A2 :
T2
T1
a1
γ1
β1
α1
b2
β2
c1
For i
γ2
a2
b1
α2
c2
1, 2, the angles satisfy D i E i J i
S . At least one of D1 d D 2 , E1 d E 2 , J 1 d J 2 must hold.
Assume D1 d D 2 . Because D 2 d S 2 (acute triangle), and the sine function increases on >0, S 2@, you have
A1
1 b c sin D d 1 b c sin D
1
1
2 1 1
2 2 2
d 12 b2c2 sin D 2
A2
Review Exercises for Chapter P
1. y
5x 8
x
0: y
50 8
y
0: 0
5x 8 Ÿ x
2. y
8 Ÿ 0, 8 , y-intercept
8
, 0 , x-intercept
5
x 2 8 x 12
2
8 0 12
x
0: y
y
0: x 2 8 x 12
0
x 3
x4
x
0: y
03
0 4
y
0: 0
x 3
Ÿ x
x 4
x 3
x 4
x
0: y
03
y
0: x 3
12 Ÿ 0, 12 , y -intercept
x 6 x 2
3. y
4. y
8
Ÿ
5
0 Ÿ x
2, 6 Ÿ 2, 0 , 6, 0 , x -intercepts
3
§ ·
Ÿ ¨ 0, ¸, y-intercept
4
© 4¹
3 Ÿ 3, 0 , x-intercept
0 4
x 4
3 4
0 Ÿ x
3 2
6 Ÿ 0, 6 , y -intercept
3, 4 Ÿ 3, 0 , 4, 0 , x-intercepts
x 2 4 x does not have symmetry with respect to either axis or the origin.
origin
INSTRUCTOR USE ONLY
5. y
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© Cengage Learning. All Rights Reserved.
38
Chapter P
Preparation
paration for Calculus
Calc
6. Symmetric with respect to y-axis because
4
2
y
x
x
y
x 4 x 2 3.
3
7. Symmetric with respect to both axes and the origin because:
x2 5
y
x 5
2
y2
y
2
2
y
2
x2 5
x 5
2
8. Symmetric with respect to the origin because:
x y
2
xy
2.
y
y
2
2
x
2
5
x 5
2
03 4 0
y-intercept: y
0
0, 0
1
x 3
2
9. y
x3 4 x
11. y
x3 4 x
0
x x 4
0
x x 2 x 2
0
x-intercepts:
2
y-intercept: y
1
0 3
2
3
x
(0, 3)
1
x-intercept: x 3
2
1
x
2
x
0, 2, 2
0, 0 , 2, 0 , 2, 0
0
Symmetric with respect to the origin because
x
3
3
4 x
x3 4 x
x3 4 x .
y
6
4
6, 0
3
Symmetry: none
1
(−2, 0)
y
−4 −3
(0, 0) (2, 0)
−1
6
1
3
x
4
−2
4
−3
(0, 3)
−4
2
(6, 0)
−2
2
4
x
6
y x 9
2
−4
10. y
9 x
y2
12.
−2
y-intercept: y 2
x2 4
0
y-intercept: y
2
4
x-intercept: 02
x2 4
0
2 x 2 x
0
x
x
4
x 4.
r3
9 x Ÿ x
9
Symmetric with respect to the x-axis because
r2
Symmetric with respect to the y-axis because
2
9 Ÿ y
9, 0
y
2, 0 , 2, 0
2
9 0
0, 3 , 0, 3
4
(0, 4)
x-intercepts:
0
2
x 9
y2 x 9
0.
y
5
4
(0, 3)
(9, 0)
2
1
x
y
−1
−2
5
(0, 4)
−4
−5
3
1 2 3 4 5 6 7
9
(0, −3)
2
1
(−2, 0)
−3
(2, 0)
−1
−11
x
INSTRUCTOR
NST
S
USE ONLY
1
3
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises for
fo
f Chapter P
2 4 x
13. y
2 4 0
y-intercept: y
2 4
16. 2 x 4 y
9 Ÿ y
6x 4 y
7 Ÿ y
2x 9
4
2x 9
6x 7
4
6x 7
2x 9
4
6x 7
4
4
0, 4
x-intercept: 2 4 x
0
4 x
0
4 x
0
8x
16
x
4
x
2
4, 0
For x
Symmetry: none
5
(0, 4)
2
5 Ÿ y
x 5
x y
1Ÿ y
x 1
x 5
x2 1
2
1
(4, 0)
x
1
2
3
4
5
x4 4
14. y
5
4
4
x y
17.
3
−1
62 7
2, y
§ 5·
Point of intersection: ¨ 2, ¸
© 4¹
y
−1
0 4 4
y-intercept: y
4 4
4 4
0
0, 0
2
0
x2 x 6
0
x 3 x 2
x
3 or x
For x
3, y
2, y
2
35
8.
2 5
x-intercepts: x 4 4
0
For x
x 4
4
Points of intersection: 3, 8 , 2, 3
x 4
x
4 or x 4
8
x
4
0
1 Ÿ y2
1 x2
x y
1Ÿ y
x 1
Symmetry: none
y
x 1
1 x
2
x2 2x 1
0
2x2 2x
0
2x x 1
x
0 or x
4
−2
−2
2
(8, 0)
4
6
8
x
10
−4
−6
0, y
01
For x
1, y
1 1
1 5 x 1
3
x y
5 Ÿ y
x5
1 5 x 1
3
x 5
5
3 x 15
4
For x
8x
2
x
2, y
1.
0.
Points of intersection: 0, 1 , 1, 0
1 Ÿ y
16
1
For x
15. 5 x 3 y
5 x 1
2
1 x2
6
(0, 0)
3.
18. x 2 y 2
0, 0 , 8, 0
2
39
y
19.
( 5, 52 )
3
2
1
x5
2 5
3.
( 32 , 1 )
1
Point of intersection is: 2, 3
Slope
2
3
x
4
§5·
¨ ¸ 1
© 2¹
§ 3·
5¨ ¸
© 2¹
5
3
2
7
2
3
7
INSTRUCTOR USE ONLY
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40
NOT FOR SALE
Chapter P
Preparation
paration for Calculus
Calc
20. The line is horizontal and has slope 0.
25. y
y
(−7, 8)
(−1, 8)
6
y
Slope: 0
7
y -intercept: 0, 6
5
4
6
3
4
2
1
2
−8
−6
−4
x
−4 −3 −2 −1
x
−2
1
2
3
4
2
−2
26. x
y
3
21. y 5
7
x 3
4
Slope: undefined
y 5
7
x 21
4
4
Line is vertical.
4 y 20
7 x 21
3
2
1
x
− 5 −4
− 2 −1
−1
7 x 4 y 41
0
−2
−3
y
2
27. y
x
−8 −6 −4 −2
2
4
−4
6
8
(3, −5)
−6
−8
(
0, − 41
4
−10
1
4x 2
y
Slope: 4
4
y -intercept: 0, 2
2
3
(
1
x
−4 −3 −2 −1
1
2
3
4
2
4
6
2
3
4
6
8
−2
−3
22. Because m is undefined the line is vertical.
x
8 or x 8
28. 3 x 2 y
0
2y
y
3x 12
3
x 6
2
3
Slope : 2
y -intercept: 0, 6
4
2
(−8, 1)
x
−4
y
6
y
6
−6
12
−2
2
−2
4
2
x
− 4 −2
−2
8
−4
−4
23.
y 0
23 x y
23 x 2
2x 3y 6
3
y 0
2
1
−4 −3
x
−1
1
2
3
y
4y x
−3
−4
24. Because m
y 4
y
30.
0, the line is horizontal.
0 x 5
4 or y 4
m
3
(− 3, 0)
0
29.
y
y
m
y 5
8
0
20
1
80
4
1
x 0
4
1
x
4
0
5 y 25
6
5 y 2 x 15
(5, 4)
2
−2
3
2
1
x
−4
1 5
10 5
2
x 5
5
2 x 10
0
−1
1
−2
−3
−4
6
15
2
5
y
8
6
4
x
−4
y
4
2
4
6
−2
2
x
−2
−2
2
4
−4
INSTRUCTOR USE ONLY
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NOT FOR SALE
Review Exercises fo
ffor Chapter P
y 5
7 x 3
16
34. (a) C
16 y 80
7 x 21
(b) R
7 x 16 y 101
(c)
31. (a)
0
(b) 5 x 3 y
5 x 3
3
3 y 15
5 x 15
0
30t
22.75t 36,500
36,500
t | 5034.48 hours to break even
5x 4
35. f x
5 x 3 y 30
(a) f 0
50 4
4
8
(b) f 5
55 4
29
(c) f 3
5 3 4
(c) 3x 4 y
4y
3x 8
y
3
x 2
4
(d) f t 1
4
Perpendicular line has slope .
3
3 y 15
4 x 3 y 27
0 or y
4
x 9
3
(d) Slope is undefined so the line is vertical.
5t 1 4
5t 9
3
(a) f 3
3
(b) f 2
23 2 2
(c) f 1
1
(d) f c 1
3
2 3
27 6
8 4
2 1
3
c 1
21
4
1 2
1
2c 1
x
3
c 3c 3c 1 2c 2
x 3
0
c3 3c 2 c 1
3
2
x 2
3
2 x 4
y 4
32. (a)
3 y 12
2 x 3 y 16
37. f x
2
4x2
f x 'x f x
4 x 'x
'x
0
y 4
1x 2
y
x 2
0
x y 2
(c) m
4 1
26
4 x 2 x'x 'x
2
8 x'x 4 'x
'x
8 x 4'x,
3
4
3
x2
4
3 x 6
38. f x
2x 6
f 1
21 6
0
f x f 1
(d) Because the line is horizontal the slope is 0.
x 1
0
4x2
2
4 x2
'x
4 y 16
y 4
2
'x
4 x 8 x'x 4 'x
3x 4 y 22
4
4 x2
2
y 4
y
2
'x
(b) x y
0 has slope 1. Slope of the perpendicular
line is 1.
33. The slope is 850.
V
850t 12,500.
V 3
11
x3 2 x
36. f x
4
x 3
3
4 x 12
y 5
22.75t 36,500
30t
7.25t
3 has slope 53 .
y 5
9.25t 13.50t 36,500
41
850 3 12,500
$9950
2
'x z 0
4
2x 6 4
x 1
2x 6 4
x 1
2x 2
x 1
2 x 1
x 1
2, x z 1
INSTRUCTOR USE ONLY
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42
NOT FOR SALE
Chapter P
x2 3
f x
39.
Preparation
paration for Calculus
Calc
x3 3x 2
47. f x
Domain: f, f
6
Range: >3, f
(0, 0)
−6
6
6 x
40. g x
(2, − 4)
Domain: 6 x t 0
6
6 t x
(a) The graph of g is obtained from f by a vertical shift
down 1 unit, followed by a reflection in the x-axis:
f, 6@
Range: >0, f
x3 3x 2 1
(b) The graph of g is obtained from f by a vertical shift
upwards of 1 and a horizontal shift of 2 to the right.
x 1
f x
41.
ª¬ f x 1º¼
g x
Domain: f, f
f x 2 1
g x
x2
Range: f, 0@
48. (a) Odd powers: f x
42.
2
x 1
Domain: all x z 1; f, 1 ‰ 1, f
h x
6
y
r
−3
3
The graphs of f, g, and h all rise to the right and fall
to the left. As the degree increases, the graph rises
and falls more steeply. All three graphs pass through
the points (0, 0), (1, 1), and 1, 1 and are
3
1
x
2
−1
4
8 10 12 14
symmetric with respect to the origin.
−2
−3
h
6
Function of x because
there is one value for
y for each x.
f
4
−3
The graphs of f, g, and h all rise to the left and to the
right. As the degree increases, the graph rises more
steeply. All three graphs pass through the points
0, 0 , 1, 1 , and 1, 1 and are symmetric with
1
x
−3
−2
−1
1
2
3
y
x 2
x 2
3
0
2
respect to the y-axis.
4
3
2
1
x
−2 −1
1
3
4
5
6
−2
All of the graphs, even and odd, pass through the
origin. As the powers increase, the graphs become
flatter in the interval 1 x 1.
(b) y
−3
−4
x 7 will look like h x
even more steeply. y
9 y2
Not a function of x
since there are two
values of y for some x.
x6
5
3
46. x
x4 , h x
g
4
y
0
y is a function of x
because there is one
value of y for each x.
x2 , g x
Even powers: f x
−4
45. y
x5
−2
2
44. x 2 y
1
f
4
Not a function because
there are two values of
y for some x.
x3 , h x
2
h
y
x6
3x2
g
2
Range: all y z 0; f, 0 ‰ 0, f
43. x y 2
x, g x
3
h x
y
x5 , but rise and fall
x8 will look like
x 6 , but rise even more steeply.
4
2
1
x
− 12 − 9 − 6 − 3
−1
3
6
12
−2
INSTRUCTOR
UC
C
USE ONLY
−4
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffo
for Chapter P
49. (a) f x
x2 x 6
43
52. (a) Using a graphing utility, you obtain
2
100
0.043x 2 4.19 x 56.2.
y
The leading coefficient is
positive and the degree is
even so the graph will rise
to the left and to the right. − 4
(b)
50
10
− 25
(b) g x
x3 x 6
10
2
The leading coefficient is
positive and the degree is
odd so the graph will rise
to the right and fall to
the left.
(c) h x
x x 6
3
80
0
300
(c) For x
y
−2
10
− 100
(d) For x
y
200
50. (a) 3 (cubic), negative leading coefficient
(b) 4 (quartic), positive leading coefficient
(c) 2 (quadratic), negative leading coefficient
(d) 5, positive leading coefficient
(b)
4.19 26 56.2
34 :
0.043 34
2
4.19 34 56.2
53. (a) Yes, y is a function of t. At each time t, there is one
and only one displacement y.
(b) The amplitude is approximately
0.25 0.25 2
0.25. The period is
approximately 1.1.
(c) One model is y
(d)
1.204 x 64.2667
51. (a) y
2
| $36.6 thousand
10
− 800
0.043 26
| $23.7 thousand
3
The leading coefficient is
−4
positive and the degree is
even so the graph will rise
to the left and to the right.
26 :
1
1
§ 2S ·
cos¨ t ¸ |
cos 5.7t
4
4
© 1.1 ¹
0.5
(1.1, 0.25)
70
0
2.2
(0.5, −0.25)
−0.5
0
The model appears to fit the data.
33
0
(c) The data point (27, 44) is probably an error. Without
this point, the new model is
y
1.4344 x 66.4387.
Problem Solving for Chapter P
x2 6 x y 2 8 y
1. (a)
x 6 x 9 y 8 y 16
2
2
x3
2
y 4
2
0
(c) Slope of line from (6, 0) to (3, 4) is
9 16
25
Center: (3, 4); Radius: 5
4
(b) Slope of line from (0, 0) to (3, 4) is .
3
3
Slope of tangent line is . So,
4
3
3
y 0
x 0 Ÿ y
x, Tangent line
4
4
Slope of tangent line is
y 0
3
(d) x
4
3
x
2
x
40
36
4
.
3
3
. So,
4
3
x 6 Ÿ y
4
3
9
x 4
2
9
2
3
3
9
x , Tangent line
4
2
9·
§
Intersection: ¨ 3, ¸
4¹
©
INSTRUCTOR USE ONLY
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44
NOT FOR SALE
Chapter P
2. Let y
Preparation
paration for Calculus
Calc
mx 1 be a tangent line to the circle from the
point (0, 1). Because the center of the circle is at 0, 1
­1, x t 0
®
¯0, x 0
(c) H x
and the radius is 1 you have the following.
x y 1
2
x mx 1 1
2
2
2
m 1 x 4mx 3
2
2
y
1
4
1
2
0
−4 −3 −2 −1
−1
3
1
x
2
3
4
−2
Setting the discriminant b 2 4ac equal to zero,
−3
−4
16m 4 m 1 3
0
16m 2 12m 2
12
4m 2
12
m
r
2
1
2
­1, x d 0
®
¯0, x ! 0
(d) H x
y
4
3x 1 and y
Tangent lines: y
3. H x
3
3
3 x 1.
2
x
­1, x t 0
®
¯0, x 0
−4 −3 −2 −1
−1
2
3
4
−2
−3
−4
y
­1
° , x t 0
®2
°0, x 0
¯
4
3
(e)
2
1
1
H x
2
x
−4 −3 −2 −1
−1
1
1
2
3
4
y
−2
−3
4
−4
3
2
­1, x t 0
®
¯2, x 0
(a) H x 2
1
x
−4 −3 −2 −1
−1
1
2
3
4
−2
−3
y
−4
4
3
­1, x t 2
®
¯2, x 2
(f ) H x 2 2
2
1
x
−4 −3 −2 −1
−1
1
2
3
y
4
4
−3
3
−4
­1, x t 2
®
¯0, x 2
(b) H x 2
1
x
−4 −3 −2 −1
−1
1
2
3
4
−2
−3
y
−4
4
3
2
1
x
−4 −3 −2 −1
−1
1
2
3
4
−2
−3
−4
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffo
for Chapter P
4. (a) f x 1
45
f x
(f )
y
y
4
4
2
−3
x
−1
1
−4
3
x
−2
−2
−2
−4
−4
(b) f x 1
(g) f
2
4
2
4
x
y
y
4
4
2
x
−4
4
−4
−2
x
−2
−2
−4
−4
(c) 2 f x
5. (a) x 2 y
y
4
Ax
−4
2
4
−2
xy
§ 100 x ·
x¨
¸
2 ¹
©
x2
50 x
2
Domain: 0 x 100 or 0, 100
x
−2
100 x
2
100 Ÿ y
(b)
1600
−4
(d) f x
0
y
4
Maximum of 1250 m 2 at x
2
−4
110
0
(c) A x
x
−2
2
50 m, y
25 m.
12 x 2 100 x
12 x 2 100 x 2500 1250
4
−2
12 x 50
−4
A 50
(e) f x
x
2
1250
1250 m 2 is the maximum.
50 m, y
25 m
y
4
2
−4
x
−2
2
4
−2
−4
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter P
46
Preparation
paration for Calculus
Calc
6. (a) 4 y 3 x
Ax
300 3 x
4
300 Ÿ y
§ 300 3 x ·
x¨
¸
2
©
¹
x 2y
8. Let d be the distance from the starting point to the beach.
Average speed
3 x 2 300 x
2
Domain: 0 x 100
(b)
y
4000
3500
3000
2500
2000
distance
time
2d
d
d
120 60
2
1
1
120 60
80 km h
1500
1000
9. (a) Slope
500
x
25
50
75
100
94
32
5. Slope of tangent line is less
4 1
2 1
3. Slope of tangent line is greater
than 5.
Maximum of 3750 ft 2 at x
50 ft, y
37.5 ft.
(b) Slope
32 x 2 100 x
(c) A x
than 3.
32 x 2 100 x 2500 3750
32 x 50
A 50
2
4.41 4
2.1 2
less than 4.1.
(c) Slope
3750
3750 square feet is the maximum area,
where x
50 ft and y
(d) Slope
37.5 ft.
4.1. Slope of tangent line is
f 2 h f 2
2 h 2
2
7. The length of the trip in the water is
length of the trip over land is
4 x2
2
total time is T
2 h 4
h
22 x 2 , and the
2
1 3 x . So, the
1 3 x
4h h 2
h
4 h, h z 0
2
4
hours.
(e) Letting h get closer and closer to 0, the slope
approaches 4. So, the slope at (2, 4) is 4.
10.
y
4
3
2
(4, 2)
1
x
1
2
3
4
5
−1
(a) Slope
32
94
1
1
. Slope of tangent line is greater than .
5
5
(b) Slope
2 1
4 1
1
1
. Slope of tangent line is less than .
3
3
(c) Slope
2.1 2
4.41 4
(d) Slope
f 4 h f 4
4 h 4
(e)
4 h 2
h
10
10
. Slope of tangent line is greater than .
41
41
4 h 2
h
4 h 2
˜
h
4 h 2
4 h 2
As h gets closer to 0, the slope gets closer to
4 h 4
h
4 h 2
1
, h z 0
4 h 2
1
1
. The slope is at the point (4, 2).
4
4
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffo
for Chapter P
11. f x
47
1
1 x
y
(a) Domain: all x z 1 or f, 1 ‰ 1, f
Range: all y z 0 or f, 0 ‰ 0, f
§ 1 ·
f¨
¸
©1 x ¹
(b) f f x
1
§ 1 ·
1¨
¸
©1 x ¹
1
1 x 1
1 x
1 x
x
x 1
x
Domain: all x z 0, 1 or f, 0 ‰ 0, 1 ‰ 1, f
§ x 1·
f¨
¸
© x ¹
(c) f f f x
1
§ x 1·
1¨
¸
© x ¹
1
1
x
x
Domain: all x z 0, 1 or f, 0 ‰ 0, 1 ‰ 1, f
(d) The graph is not a line. It has holes at (0, 0) and (1, 1).
y
2
1
x
−2
1
2
−2
y
12. Using the definition of absolute value, you can rewrite the equation.
y y
4
x x
3
2
y ! 0
­2 y,
®
¯0,
­2 x, x ! 0
.
®
x d 0
¯0,
y d 0
1
x
−4 −3 − 2 − 1
For x ! 0 and y ! 0, you have 2 y
I
x2
2I
x 3
x.
3
4
−3
−4
x x is as follows.
I
x2 y 2
(b)
2
2
−2
2x Ÿ y
For any x d 0, y is any y d 0. So, the graph of y y
13. (a)
1
2I
x 3
2
y
y2
8
x2 6x 9
2x2
x3
y2
2 x2 y 2
6
x 6x 9
0
x 6x 9 y
2x 2 y
2
2
2
6 r
x
36 36
2
3 r 18
| 1.2426, 7.2426
x
0
1
2
2
2
x y 6x 9
2
2
x 3
2
y2
Circle of radius
2
2
0
18
−8
− 4 −2
−2
x
2
4
−6
18 and center 3, 0 .
3
INSTRUCTOR USE ONLY
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48
Chapter P
14. (a)
Preparation
paration for Calculus
Calc
I
x y2
kI
2
x 4
2
x4
y2
2
y2
k x2 y 2
k 1 x2 8x k 1 y2
If k
16
2 is a vertical line. Assume k z 1.
1, then x
8x
y2
k 1
8x
16
x2 y2
2
k 1
k 1
16
k 1
16
16
2
k 1
k 1
x2 2
4 ·
§
2
¨x ¸ y
k 1¹
©
3, x 2
(b) If k
2
y2
16k
k 1
2
, Circle
12
y
6
4
2
−6
−4
x
−2
2
4
−2
−4
16k
4
o 0.
o 0 and
2
k 1
k 1
(c) As k becomes very large,
The center of the circle gets closer to (0, 0), and its radius approaches 0.
d1d 2
1
y2º
¼
1
x 1 º y4
¼
1
y 2 ª¬2 x 2 2º¼ y 4
1
x4 2x2 1 2x2 y 2 2 y 2 y 4
1
x4 2x2 y2 y 4 2x2 2 y 2
0
15.
ªx 1
¬
x 1
2
x 1
2
y 2 ºª x 1
¼¬
2
y2 ª x 1
¬
x2 1
2
2
2
2
y
2
(− 2 , 0)
x y
2
Let y
0. Then x 4
So, 0, 0 ,
2x2 Ÿ x
2, 0 and 0 or x 2
2
2
2x y
2
1
( 2 , 0)
2
x
−2
2.
2
−1
(0, 0)
−2
2, 0 are on the curve.
INSTRUCTOR USE ONLY
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NOT FOR SALE
C H A P T E R 1
Limits and Their Properties
Section 1.1
A Preview of Calculus..........................................................................50
Section 1.2
Finding Limits Graphically and Numerically .....................................51
Section 1.3
Evaluating Limits Analytically ............................................................62
Section 1.4
Continuity and One-Sided Limits........................................................75
Section 1.5
Infinite Limits .......................................................................................86
Review Exercises ..........................................................................................................94
Problem Solving .........................................................................................................101
INSTRUCTOR USE ONLY
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NOT FOR SALE
C H A P T E R 1
Limits and Their Properties
Section 1.1 A Preview of Calculus
1. Precalculus: 20 ft/sec 15 sec
300 ft
(a)
2. Calculus required: Velocity is not constant.
Distance | 20 ft/sec 15 sec
6x x2
7. f ( x)
y
10
300 ft
8
P
6
3. Calculus required: Slope of the tangent line at x
the rate of change, and equals about 0.16.
4. Precalculus: rate of change
slope
1 bh
2
5. (a) Precalculus: Area
(b) Calculus required: Area
1 5
2
2 is
x
0.08
4
−2
2
4
10 sq. units
(b) slope
m
bh
| 2 2.5
5 sq. units
6. f ( x)
(a)
x
8
6 x x2 8
x 2 4 x
x2
x2
4 x,x z 2
4 3
For x
3, m
For x
2.5, m
4 2.5
1.5
3
2
For x
1.5, m
4 1.5
2.5
5
2
1
y
P(4, 2)
2
(c) At P 2, 8 , the slope is 2. You can improve your
approximation by considering values of x close to 2.
8. Answers will vary. Sample answer:
x
1
(b) slope
2
m
3
4
5
The instantaneous rate of change of an automobile’s
position is the velocity of the automobile, and can be
determined by the speedometer.
x 2
x 4
x 2
x 2
x 2
1
,x z 4
x 2
x
1: m
x
3: m
x
5: m
1
1
3
1 2
1
| 0.2679
3 2
1
| 0.2361
5 2
(c) At P 4, 2 the slope is
1
4 2
1
4
0.25.
You can improve your approximation of the slope at
x
4 by considering x-values very close to 4.
INSTRUCTOR USE ONLY
50
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.2
Finding Limits Graphically and Numerically
51
9. (a) Area | 5 52 53 54 | 10.417
5
5
5
5
Area | 12 5 1.5
52 2.5
53 3.5
54 4.5
| 9.145
(b) You could improve the approximation by using more rectangles.
51
2
15
1
2
10. (a) D1
(b) D2
5
2
1
2
16 16 | 5.66
2
5
53
2
1
5
54
3
2
1
5
1
4
2
| 2.693 1.302 1.083 1.031 | 6.11
(c) Increase the number of line segments.
Section 1.2 Finding Limits Graphically and Numerically
1.
x
3.9
3.99
3.999
4.001
4.01
4.1
f (x)
0.2041
0.2004
0.2000
0.2000
0.1996
0.1961
lim
x 4
| 0.2000
3x 4
x
2.9
2.99
2.999
3
3.001
3.01
3.1
f (x)
0.1695
0.1669
0.1667
?
0.1666
0.1664
0.1639
x o 4 x2
2.
x 3
| 0.1667
9
lim
x o3 x2
3.
5.
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
0.5132
0.5013
0.5001
?
0.4999
0.4988
0.4881
lim
x 1 1
1·
§
| 0.5000 ¨ Actual limit is .¸
x
2¹
©
x
2.9
2.99
2.999
3.001
3.01
3.1
f (x)
–0.0641
–0.0627
–0.0625
–0.0625
–0.0623
–0.0610
ª1 x 1 º¼ 1 4
lim ¬
| –0.0625
x o3
x 3
1 ·
§
¨ Actual limit is .¸
16 ¹
©
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.9983
0.99998
1.0000
1.0000
0.99998
0.9983
lim
sin x
| 1.0000
x
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.0500
0.0050
0.0005
–0.0005
–0.0050
–0.0500
xo0
6.
1·
§
¨ Actual limit is .¸
6¹
©
x
xo0
4.
1·
§
¨ Actual limit is .¸
5¹
©
cos x 1
| 0.0000
x
Actual limit is 1. Make sure you use radian mode.
INSTRUCTOR USE ONLY
lim
xo0
Actual limit is 0. Make sure yyou use radian mode.
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NOT FOR SALE
Chapter 1
52
7.
x
0.9
0.99
0.999
1.001
1.01
1.1
f (x)
0.2564
0.2506
0.2501
0.2499
0.2494
0.2439
lim
x 2
| 0.2500
x 6
x o1 x 2
8.
9.
Limits
its and Their Properties
x
–4.1
–4.01
–4.001
–4
–3.999
–3.99
–3.9
f (x)
1.1111
1.0101
1.0010
?
0.9990
0.9901
0.9091
x o 4 x2
lim
x 4
| 1.0000
9 x 20
x
0.9
0.99
0.999
1.001
1.01
1.1
f (x)
0.7340
0.6733
0.6673
0.6660
0.6600
0.6015
x4 1
| 0.6666
x o1 x 6 1
x
–3.1
–3.01
–3.001
–3
–2.999
–2.99
–2.9
f (x)
27.91
27.0901
27.0090
?
26.9910
26.9101
26.11
x 3 27
| 27.0000
x o 3 x 3
Actual limit is 27.
x
–6.1
–6.01
–6.001
–6
–5.999
–5.99
–5.9
f (x)
–0.1248
–0.1250
–0.1250
?
–0.1250
–0.1250
–0.1252
lim
10 x 4
| 0.1250
x 6
lim
11.
x o 6
12.
1.9
1.99
1.999
2
2.001
2.01
2.1
f (x)
0.1149
0.115
0.1111
?
0.1111
0.1107
0.1075
lim
x ( x 1) 2 3
| 0.1111
x 2
1·
§
¨ Actual limit is .¸
9¹
©
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
1.9867
1.9999
2.0000
2.0000
1.9999
1.9867
lim
sin 2 x
| 2.0000
x
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.4950
0.5000
0.5000
0.5000
0.5000
0.4950
xo0
14.
1·
§
¨ Actual limit is .¸
8¹
©
x
xo2
13.
Actual limit is 1.
2·
§
¨ Actual limit is .¸
3¹
©
lim
10.
1·
§
¨ Actual limit is .¸
4¹
©
lim
tan x
x o 0 tan 2 x
| 0.5000
Actual limit is 2. Make sure you use radian mode.
1·
§
¨ Actual limit is .¸
2¹
©
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 1.2
15. lim 4 x
xo0
xo0
left, f x approaches 12 , whereas as x approaches 0
1
from the right, f x approaches 4.
17. lim f x
xo2
18. lim f x
x o1
19. lim
53
(d) lim f x does not exist. As x approaches 0 from the
1
x o3
16. lim sec x
Finding Limits Graphically and Numerically
x 2
xo2
x 2
lim 4 x
2
(e) f 2 does not exist. The hollow circle at
lim x 2 3
4
2, 12 indicates that f 2 is not defined.
xo2
x o1
(f ) lim f x exists. As x approaches 2, f x approaches
xo2
1
: lim f
2 xo2
does not exist.
(g) f 4 exists. The black dot at 4, 2 indicates that
For values of x to the left of 2,
x 2
x 2
1, whereas
for values of x to the right of 2,
x 2
x 2
1.
2
20. lim
x o5 x 5
1
.
2
x
f 4
2.
(h) lim f x does not exist. As x approaches 4, the
xo4
values of f x do not approach a specific number.
does not exist because the function increases
25.
y
6
and decreases without bound as x approaches 5.
5
4
21. lim cos 1 x does not exist because the function
3
xo0
2
oscillates between –1 and 1 as x approaches 0.
1
x
− 2 −1
−1
22. lim tan x does not exist because the function increases
S
2
S
2
from
2
3
4
5
lim f x exists for all values of c z 4.
from the left and
decreases without bound as x approaches
1
−2
x oS 2
without bound as x approaches
f
xoc
26.
y
2
the right.
1
23. (a) f 1 exists. The black dot at (1, 2) indicates that
f 1
π
2
−π
2
2.
π
x
−1
(b) lim f x does not exist. As x approaches 1 from the
x o1
left, f (x) approaches 3.5, whereas as x approaches 1
from the right, f (x) approaches 1.
lim f x exists for all values of c z S .
xoc
27. One possible answer is
y
(c) f 4 does not exist. The hollow circle at
6
5
4, 2 indicates that f is not defined at 4.
4
f
(d) lim f x exists. As x approaches 4, f x approaches
xo4
2
2: lim f x
xo4
1
2.
x
−2 − 1
−1
1
2
3
4
5
24. (a) f 2 does not exist. The vertical dotted line
indicates that f is not defined at –2.
28. One possible answer is
y
4
(b) lim f x does not exist. As x approaches –2, the
x o 2
3
values of f x do not approach a specific number.
2
1
(c) f 0 exists. The black dot at 0, 4 indicates that
f 0
x
4.
−3
−2
−1
1
2
−1
INSTRUCTOR USE ONLY
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54
Chapter 1
NOT FOR SALE
Limits
its and Their Properties
29. You need f x 3
x 2
x 1 3
30. You need f x 1
x 1 3
x 2 0.4. So, take G
0.4. If 0 x 2 0.4, then
f x 3 0.4, as desired.
1
1
x 1
2 x
0.01. Let G
x 1
1
1
. If 0 x 2 , then
101
101
1
1
1
1
x 2 Ÿ1
x 11
101
101
101
101
100
102
Ÿ
x 1
101
101
100
Ÿ x 1 !
101
and you have
f x 1
1
1
x 1
2 x
1 101
100 101
x 1
1
100
31. You need to find such that 0 x 1 G implies
f x 1
1
1 0.1. That is,
x
1
0.1 1 0.1
x
1
1 0.1
1 0.1 x
9
1
11
x
10
10
10
10
!
x !
9
11
10
10
1 ! x 1 !
1
9
11
1
1
! x 1 ! .
9
11
So take G
f x 3
x2 1 3
x 2 4 0.2. That is,
0.2 x 2 4 0.2
4 0.2 x2
3.8 x
2
4.2
3.8 x
So take G
4 0.2
4.2
4.2 2
4.2 2 | 0.0494.
Then 0 x 2 G implies
1
1
x 1
11
11
1
1
x 1 .
11
9
4.2 2 x 2 4.2 2
3.8 2 x 2 4.2 2.
Using the first series of equivalent inequalities, you
obtain
f x 3
33. lim 3 x 2
xo2
x 2 4 0.2.
3(2) 2
8
L
3 x 2 8 0.01
Using the first series of equivalent inequalities, you
obtain
f x 1
32. You need to find such that 0 x 2 G implies
3.8 2 x 2 1
. Then 0 x 1 G implies
11
1
1 0.1.
x
0.01.
3 x 6 0.01
3 x 2 0.01
0 x 2 0.01
| 0.0033
3
So, if 0 x 2 G
G
0.01
, you have
3
3 x 2 0.01
3 x 6 0.01
3 x 2 8 0.01
f x L 0.01.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.2
x·
§
34. lim ¨ 6 ¸
x o 6©
3¹
6
6
3
36. lim x 2 6
L
4
x 4 x 4 0.01
x 4 G
0.01
, you have
9
0.01
0.01
x 4 9
x 4
0.03, you have
( x 4)( x 4) 0.01
x
0.01
3
x 2 16 0.01
x·
§
¨ 6 ¸ 4 0.01
3¹
©
x 2 6 22 0.01
f x L 0.01.
f x L 0.01.
35. lim x 2 3
xo2
22 3
1
37. lim x 2
6
Given H ! 0:
x 2 6 H
x 2 4 0.01
x 4 H
x 2 x 2 0.01
So, let G
x 2 x 2 0.01
G
H . So, if 0 x 4 G
H , you have
x4 H
0.01
x 2 x 2
x 2 6 H
If you assume 1 x 3, then G | 0.01 5
0.002.
So, if 0 x 2 G | 0.002, you have
x 2 x 2 0.01
4 2
xo4
L
x 2 3 1 0.01
x 2 0.002
0.01
| 0.00111.
9
So, if 0 x 4 G |
13 ( x 6) 0.01
2
0.01
x 4
If you assume 3 x 5, then G
x 6 0.03
So, if 0 x 6 G
L
x 2 16 0.01
0.01
0 x 6 0.03
22
55
x 2 6 22 0.01
x
0.01
3
13 ( x 6)
42 6
xo4
x·
§
¨ 6 ¸ 4 0.01
3¹
©
2
Finding Limits Graphically and Numerically
1
1
0.01 0.01
5
x 2
f x L H.
38. lim 4 x 5
4( 2) 5
x o 2
3
Given H ! 0:
4 x 5 ( 3) H
4x 8 H
x 2 4 0.01
4 x 2 H
x 2 3 1 0.01
x 2 f x L 0.01.
So, let G
H
4
H
4
G
.
So, if 0 x 2 G
x 2 H
4
, you have
H
4
4x 8 H
(4 x 5) ( 3) H
f x L H.
INSTRUCTOR USE ONLY
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56
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
39. lim 12 x 1
1 4
2
x o 4
1
42. lim 1
3
xo2
Given H ! 0: 1 1 H
Given H ! 0:
1 1
x
2
0 H
3 H
1 2
x
2
1
2
1
So, any G ! 0 will work.
H
So, for any G ! 0, you have
x 4 H
1 1 H
x 4 2H
f x L H.
So, let G
2H .
So, if 0 x 4 G
2H , you have
1
x 1
2
xo0
0
Given H ! 0:
x 4 2H
1
x 2
2
43. lim 3 x
3
H
x 0 H
3
x H3
3 H
So, let G
f x L H.
40.
3
3
4
1
13
4
3
13
4
H
3
x 94
4
H
3
4
3
x 0 H
x 3 H
So, let G
44. lim
xo4
4
2
x 3 H
H
x 2 H
x 2
4 H , you have
3
x 3 43 H
3
x 94
4
x
Given H ! 0:
4H .
3
So, if 0 x 3 G
3
x 1
4
x H
f x L H.
x 3 43 H
3
4
H 3 , you have
x H3
Given H ! 0:
3
x 1
4
G
H 3.
So, for 0 x 0 G
lim 34 x 1
x o3
x H
x 2 H
x 2
x 4 H
x 2
Assuming 1 x 9, you can choose G
3H . Then,
0 x 4 G
x 2
3H Ÿ x 4 H
Ÿ
13
H
4
x 2 H.
f x L H.
41. lim 3
xo6
3
Given H ! 0:
33 H
0 H
So, any G ! 0 will work.
So, for any G ! 0, you have
33 H
f x L H.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.2
45. lim x 5
5 5
x o 5
10
48. lim x 2 4 x
10
x o 4
x 5 10 H
Given H ! 0:
Finding Limits Graphically and Numerically
x 5 0
x2 4 x 0 H
x 5 H
x( x 4) H
x 5 H
x 4 H.
So for x 5 G
H , you have
because x 5 0
x 4 f x L H.
33
x o3
Given H ! 0:
H
, you have
5
1
H
x
f x L H.
49. lim f x
x oS
So, for 0 x 3 G
H , you have
x 3 0 H
50. lim f x
4
lim x
S
x oS
x 5 3
x 4
1
6
51. f x
f x L H.
lim f x
xo4
2
lim 4
x oS
x oS
x 3 H
x o1
.
x 4x 0 H
x 3 0 H
12 1
5
5
2
H.
47. lim x 2 1
H
H
x( x 4) H
0
x 3 H
So, let G
x
So for 0 x ( 4) G
x 5 10 H
46. lim x 3
H
If you assume 5 x 3, then G
x5 H
x 5 10 H
0
Given H ! 0:
x 5 10 H
So, let G
( 4) 2 4( 4)
57
0.5
Given H ! 0:
x2 1 2 H
−6
x 1 H
6
2
− 0.1667
The domain is >5, 4 ‰ 4, f . The graphing utility
x 1 x 1 H
x 1 H
§ 1·
does not show the hole at ¨ 4, ¸.
© 6¹
x 1
If you assume 0 x 2, then G
So for 0 x 1 G
x 1 H 3.
H
, you have
3
1
1
H H
3
x 1
x 3
x 4x 3
1
lim f x
x o3
2
52. f x
2
4
x 1 H
2
x2 1 2 H
f x 2 H.
−3
5
−4
The domain is all x z 1, 3. The graphing utility does not
§ 1·
show the hole at ¨ 3, ¸.
© 2¹
INSTRUCTOR USE ONLY
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58
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
x 9
x 3
53. f x
lim f x
x 3
x2 9
1
lim f x
x o3
6
54. f x
6
x o9
10
3
−9
0
3
10
0
−3
The domain is all x t 0 except x
9. The graphing
utility does not show the hole at 9, 6 .
The domain is all x z r3. The graphing utility does not
§ 1·
show the hole at ¨ 3, ¸.
© 6¹
9.99 0.79 ª¬ª
¬ t 1 º¼º¼
55. C t
(a)
16
0
6
8
(b)
t
3
3.3
3.4
3.5
3.6
3.7
4
C
11.57
12.36
12.36
12.36
12.36
12.36
12.36
lim C t
12.36
t o 3.5
(c)
t
2
2.5
2.9
3
3.1
3.5
4
C
10.78
11.57
11.57
11.57
12.36
12.36
12.36
The lim C t does not exist because the values of C approach different values as t approaches 3 from both sides.
t o3
5.79 0.99 ª¬ª
¬ t 1 º¼º¼
56. C t
(a)
12
0
6
4
(b)
t
3
3.3
3.4
3.5
3.6
3.7
4
C
7.77
8.76
8.76
8.76
8.76
8.76
8.76
lim C t
8.76
t o 3.5
(c)
t
2
2.5
2.9
3
3.1
3.5
4
C
6.78
7.77
7.77
7.77
8.76
8.76
8.76
The limit lim C t does not exist because the values of C approach different values as t approaches 3 from both sides.
t o3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.2
57. lim f x
25 means that the values of f approach 25 as
x o8
x gets closer and closer to 8.
58. In the definition of lim f x , f must be defined on both
xoc
sides of c, but does not have to be defined at c itself. The
value of f at c has no bearing on the limit as x approaches c.
59. (i) The values of f approach different numbers as x
approaches c from different sides of c:
Finding Limits Graphically and Numerically
4 3
2.48
Sr , V
3
4 3
(a) 2.48
Sr
3
1.86
r3
62. V
S
r | 0.8397 in.
(b)
y
4
2.45 d
V
2.45 d
4 3
S r d 2.51
3
0.5849 d
3
2
x
1
2
3
d 2.51
r 3 d 0.5992
0.8363 d r d 0.8431
2.51 2.48 0.03, G | 0.003
(c) For H
1
−4 −3 −2 −1
−1
59
4
−3
1 x
63. f x
−4
(ii) The values of f increase without bound as x
approaches c:
1x
1x
lim 1 x
e | 2.71828
xo0
y
y
7
6
5
4
3
3
(0, 2.7183)
2
2
1
1
x
x
−3 −2 −1
−1
2
3
4
−3 −2 −1
−1
5
1
2
3
4
5
−2
(iii) The values of f oscillate between two fixed numbers
as x approaches c:
y
x
f (x)
x
f (x)
–0.1
2.867972
0.1
2.593742
–0.01
2.731999
0.01
2.704814
–0.001
2.719642
0.001
2.716942
–0.0001
2.718418
0.0001
2.718146
–0.00001
2.718295
0.00001
2.718268
–0.000001
2.718283
0.000001
2.718280
4
3
x
−4 −3 −2
2
3
4
−3
−4
60. (a) No. The fact that f 2
4 has no bearing on the
existence of the limit of f x as x approaches 2.
(b) No. The fact that lim f x
xo2
4 has no bearing on
the value of f at 2.
61. (a) C
2S r
r
C
2S
6
2S
3
S
| 0.9549 cm
5.5
| 0.87535 cm
2S
6.5
When C
| 1.03451 cm
6.5: r
2S
So 0.87535 r 1.03451.
(b) When C
(c)
lim
2S r
5.5: r
6; H
0.5; G | 0.0796
INSTRUCTOR
S
USE ONLY
x o3 S
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© Cengage Learning. All Rights Reserved.
60
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
x 1 x 1
64. f x
69. False. Let
x
x
–1
–0.5
–0.1
0
0.1
0.5
1.0
f(x)
2
2
2
Undef.
2
2
2
lim f x
­x 4, x z 2
®
2
x
¯0,
f 2
0
lim x 4
lim f x
2
xo0
f x
xo2
Note that for
1 x 1, x z 0, f x
x 1 x 1
x
2 z 0
xo2
70. False. Let
2.
­x 4, x z 2
®
x
2
¯0,
f x
y
lim x 4
lim f x
3
xo2
71. f x
1
2 and f 2
xo2
0 z 2
x
x
−2
−1
1
x
lim
2
x o 0.25
−1
0.5 is true.
As x approaches 0.25
65.
0.002
x approaches 12
f x
(1.999, 0.001)
1 from either side,
4
0.5.
(2.001, 0.001)
72. f x
1.998
2.002
lim
0
xo0
Using the zoom and trace feature, G
2 G, 2 G
1.999, 2.001 .
x2 4
Note:
x 2
0.001. So
x
x
x is not defined on an open interval
f x
containing 0 because the domain of f is x t 0.
x 2 for x z 2.
73. Using a graphing utility, you see that
sin x
1
x
sin 2 x
lim
2, etc.
xo0
x
lim
66. (a) lim f x exists for all c z 3.
xo0
xoc
(b) lim f x exists for all c z 2, 0.
xoc
So, lim
67. False. The existence or nonexistence of f x at
x
c has no bearing on the existence of the limit
of f x as x o c.
xo0
tan x
x
tan 2 x
lim
xo0
x
lim
xo0
L1 and lim f x
xoc
n.
1
xo0
So, lim
xoc
sin nx
x
74. Using a graphing utility, you see that
68. True
75. If lim f x
0 is false.
2,
tan nx
x
etc.
n.
L2 , then for every H ! 0, there exists G1 ! 0 and G 2 ! 0 such that
x c G1 Ÿ f x L1 H and x c G 2 Ÿ f x L2 H . Let equal the smaller of G1 and G 2 . Then for
x c G , you have L1 L2
L1 f x f x L2 d L1 f x f x L2 H H . Therefore,
L1 L2 2H . Since H ! 0 is arbitrary, it follows that L1
L2 .
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.2
mx b, m z 0. Let H ! 0 be given. Take
76. f x
H
G
m
.
H
If 0 x c G
m
, then
m x c H
mx mc H
Finding Limits Graphically and Numerically
79. The radius OP has a length equal to the altitude z of the
h
h
triangle plus . So, z 1 .
2
2
Area triangle
h·
1 §
b¨1 ¸
2 ©
2¹
Area rectangle
bh
Because these are equal,
mx b mc b H
which shows that lim mx b
1 §
h·
b¨1 ¸
2 ©
2¹
bh
h
2
2h
5
h
2
1
h
2
.
5
1
mc b.
xoc
0 means that for every H ! 0 there
77. lim ª¬ f x Lº¼
xoc
61
exists G ! 0 such that if
0 x c G,
P
then
f x L 0 H.
h
O
This means the same as f x L H when
b
0 x c G.
So, lim f x
xoc
78. (a)
L.
3 x 1 3x 1 x 2 0.01
1
9 x2 1 x2 100
1
4
2
9x x 100
1
2
10 x 1 90 x 2 1
100
So, 3x 1 3x 1 x 2 0.01 ! 0 if
10 x 1 0 and 90 x 1 0.
2
Let a, b
2
1
§
,
¨
90
©
1 ·
¸.
90 ¹
For all x z 0 in a, b , the graph is positive.
You can verify this with a graphing utility.
(b) You are given lim g x
xoc
80. Consider a cross section of the cone, where EF is a
3, BC
2.
diagonal of the inscribed cube. AD
Let x be the length of a side of the cube.
Then EF
x 2.
By similar triangles,
A
EF
AG
BC
AD
x 2
3 x
2
3
Solving for x,
3 2x
6 2x
3 2 2x
E
B
G
D
F
C
6
x
L ! 0. Let
6
3 2 2
9 2 6
| 0.96.
7
1
L. There exists G ! 0 such that
2
0 x c G implies that
H
L
. That is,
2
L
L
g x L 2
2
L
3L
g x
2
2
g x L H
For x in the interval c G , c G , x z c, you
have g x !
L
! 0, as desired.
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Chapter 1
62
NOT FOR SALE
Limits
its and Their Properties
Section 1.3 Evaluating Limits Analytically
1.
8. lim 2 x 3
6
2( 4) 3
x o 4
−4
8
9. lim x 2 3 x
3
10. lim x3 1
2
x o 3
−6
(a) lim h x
0
(b) lim h x
5
xo4
x o 1
2.
xo2
2
3
8 3
3 3
1
11. lim 2 x 2 4 x 1
99
8 1
2 3
x o3
2
12. lim 2 x3 6 x 5
0
−5
x 1
13. lim
x o3
x 3
12
14. lim 3 12 x 3
x 9
(a) lim g x
2.4
(b) lim g x
4
xo4
xo0
31
61 5
1
2
3 12(2) 3
xo2
3
24 3
15. lim x 3
2
4 3
16. lim 3 x 2
4
3(0) 2
x o 4
3.
7
7
265
10
g x
3
21
x o1
0
4 3 1
18 12 1
10
5
3
2
27
3
1
4
−
xo0
17. lim
f x
xo0
lim f x | 0.524
x oS 3
§
¨
©
x o 5 x 3
5
5 3
19. lim
x
4
1
12 4
1
5
20. lim
3x 5
x 1
3(1) 5
11
3 5
2
21. lim
3x
x 2
37
7 2
21
3
7
22. lim
x 6
x 2
3 6
3 2
9
5
3
5
x o1 x 2
S·
¸
6¹
x o1
4.
10
−5
xo7
10
− 10
x o3
t t 4
f t
5
18. lim
0
(a) lim f t
t o4
23. (a) lim f x
x o1
0
(b) lim g x
xo4
5
(b) lim f t
t o 1
51
43
xo2
23
x o 3
( 3) 4
81
24. (a) lim f x
3 7
4
(b) lim g x
xo4
(c) lim g f x
42
4
64
g 4
8
8
2
4
g f 1
x o 3
6. lim x 4
16
5
2
(c) lim g f x
x o1
5. lim x3
( 2) 4
1
2
x cos x
(a) lim f x
(b)
1
xo2 x
−4
4
64
16
g 4
16
INSTRUCTOR USE ONLY
7. lim 2 x 1
xo0
20 1
1
x o 3
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.3
4 1
25. (a) lim f x
x o1
3
30. lim sin
(b) lim g x
31
2
(c) lim g f x
g3
2
x o3
x o1
xo2
xo4
3
(b) lim g x
x o 21
21 6
3
g 21
3
(c) lim g f x
xo4
lim sin x
sin
x oS 2
S
28. lim tan x
tan S
Sx
cos
x oS
29. lim cos
x o1
3
0
S
1
2
3
5 lim g x
32. lim cos 3x
cos 3S
1
xoc g
lim f x lim g x
xoc
lim f x
lim sin x
sin
5S
6
1
2
lim cos x
cos
5S
3
1
2
§S x ·
35. lim tan ¨ ¸
x o3
© 4 ¹
tan
3S
4
1
§S x ·
36. lim sec¨ ¸
xo7
© 6 ¹
sec
7S
6
2 3
3
lim g x
xoc
38. (a) lim ª¬4 f x º¼
xoc
xoc g
39. (a) lim ¬ª f x ¼º
lim g x
xoc
3
xoc
(b) lim
xoc
f x
(d) lim ª¬ f x º¼
xoc
ª lim f x º
¬« x o c
¼»
2
34
5
6
3 lim f x
xoc
ª lim f x º
¬« x o c
¼»
§ 3·
2¨ ¸
© 4¹
3
4
11
4
3
2
8
3
3
4
lim f x
32
2
xoc
3
4
xoc
(c) lim ª¬3 f x º¼
xoc
x o 5S 3
8
ª lim f x ºª lim g x º
«¬x o c
»«
»¼
¼¬ x o c
xoc
x
3 2
lim f x lim g x
xoc
lim f x
f x
4(2)
xoc
(b) lim ª¬ f x g x º¼
xoc
(d) lim
3 2
xoc
4 lim f x
(c) lim ª¬ f x g x º¼
xoc
x o 5S 6
3
2
xoc
x
1
10
ª lim f x º ª lim g x º
¬« x o c
¼» ¬«x o c
¼»
(c) lim ª¬ f x g x º¼
xoc
f x
52
xoc
(b) lim ª¬ f x g x º¼
xoc
(d) lim
0
2
sec 0
33.
63
1
2
37. (a) lim ª¬5 g x º¼
xoc
2
x oS
21
S 2
sin
31. lim sec 2 x
34.
27.
Sx
xo0
2 42 3 4 1
26. (a) lim f x
Evaluating Limits Analytically
64
40. (a) lim 3 f x
2
(b) lim
xoc
34
3
xoc
lim f x
f x
xoc
18
lim 18
xoc
12
32
4
32
8
(c) lim ª¬ f x º¼
xoc
2
(d) lim ¬ª f x ¼º
xoc
23
3
lim f x
xoc
27
18
ª lim f x º
¬«x o c
¼»
27
3
2
2
ª lim f x º
¬«x o c
¼»
3
27
2
23
27
729
23
9
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
64
Chapter 1
Limits
its and Their Properties
x 2 3x
x
agree except at x
41. f x
lim f x
x( x 3)
and g x
x
0.
lim ( x 3)
lim g x
xo0
xo0
xo0
x 3
x3 8
and g x
x 2
2.
45. f x
at x
03
3
lim f x
lim x 2 2 x 4
lim g x
xo2
x 2 2 x 4 agree except
xo2
xo2
22 2(2) 4
5
12
12
−5
4
−1
−9
9
0
x4 5x2
x2
agree except at x
42. f x
lim f x
x 2 ( x 2 5)
and g x
x2
0.
lim ( x 2 5)
lim g x
xo0
xo0
xo0
x2 5
02 5
5
x3 1
and g x
x 1
46. f x
x
x 2 x 1 agree except at
1.
lim f x
lim x 2 x 1
lim g x
x o 1
x o 1
x o 1
( 1) 2 (1) 1
2
−6
3
6
7
−6
−4
x 1
( x 1)( x 1)
and
x 1
x 1
x 1 agree except at x
1.
4
−1
2
43. f x
g x
lim f x
lim x 1
lim g x
x o 1
x o 1
x o 1
47. lim
x
x
48. lim
2x
4x
x o 0 x2
1 1
2
x o 0 x2
3
−3
3x 2 5 x 2
( x 2)(3 x 1)
and
x 2
x 2
3 x 1 agree except at x
2.
lim f x
lim 3 x 1
lim g x
x o 2
x 4
16
x o 4 x2
x o 2
x o 2
3( 2) 1
7
lim
xo4
lim
5 x
25
x o5 x2
lim
x o5
lim
1
x2 x 6
x o 3
x2 9
5
1
2
1
4 4
1
8
x 5
1
lim
x o 3
lim
1
55
1
10
x 3 x 2
x 3 x 3
x 2
x o 3 x 3
−3
2
x 5 x 5
3
−4
lim
1
x 4
x 4 x 4
x o5 x 5
51. lim
1
0 1
xo0 x 4
2
4
xo4 x 4
50. lim
1
lim
xo0 x 1
2x
lim
2
0 4
49. lim
g x
x
xo0 x x 4
4
−4
44. f x
lim
xo0 x x 1
3 2
3 3
5
6
5
6
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.3
x2 2 x 8
x o 2 x2 x 2
52. lim
lim
xo2
lim
xo4
x 5 3
x 4
x 2 x 1
x 4
2 4
21
54. lim
x o3
x 1 2
x 3
xo0
x 5 x
x 5 9
x 1 2
˜
x 3
x 1 2
x 1 2
lim
1
x 1 2
1
4 2
5
2 x x
2
2 x x
lim
1
1
x
4
4
58. lim
xo0
x
59. lim
'x o 0
60. lim
'x o 0
61. lim
lim
lim
2 x 2
2 x 2 x 2 x
xo0
1
x 5 lim
xo0 3 5
1
5 x2
2
1
2 1
x3
1
(3)3
x
x 3 x
lim
lim
1
4(4)
lim
xo0 3 2 x 2'x 2 x
'x
lim
2 'x
lim 2
'x o 0 'x
x 2 2 x'x 'x
2
x2
lim
'x
'x 2 x 'x
'x o 0
2 x 'x 1 x 2 2 x 1
'x
x 2 2 x'x 'x
lim
'x o 0
'x o 0
'x o 0
'x
1
2 2
2
4
lim 2 x 'x
2x
2
1
9
2
'x o 0
lim 2 x 'x 2
3
2 5
1
16
'x
x 'x
5
5
10
4 x 4
4 x 4
'x o 0
'x o 0
62. lim
1
2
2
1
2 x lim
lim
3 x3x
'x o 0
'x
x 'x
3 3 x
x
1
lim
xo0 4 x 4
'x
2
˜
5
5
xo0
2 x 'x 2 x
x 'x
5
xo0
2 x 2
lim
1
6
1
4
x 5 x 5 2
1
9 3
x 3
x 3 ª¬ x 1 2º¼
lim
x o3
˜
5
x 5 xo0
xo0
xo4
x 5 5
xo0
1
1
x
3
3
57. lim
xo0
x
x 5 3
x 5 x
lim
xo0
1
x 5 3
lim
lim
x o3
lim
xo0
x 5 3
x 5 3
x 4
xo0 x
56. lim
2
xo4
x o3
55. lim
6
3
x 5 3
˜
x 4
lim
xo4
lim
65
x 2 x 4
xo2 x 1
53. lim
Evaluating Limits Analytically
x3
lim
x3 3 x 2 'x 3x 'x
'x
3
2
2 x 2'x 1 x 2 2 x 1
'x
2x 2
x3
'x
'x o 0
'x 3 x 3 x'x 'x
2
lim
2
'x o 0
2
lim 3x 2 3 x'x 'x
2
3x 2
INSTRUCTOR USE ONLY
'x o 0
'x
'x o 0
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66
NOT FOR SALE
Chapter 1
63. lim
xo0
64. lim
ª§ sin x ·§ 1 ·º
lim Ǭ
¸¨ ¸»
¬ x ¹© 5 ¹¼
sin x
5x
sin x 1 cos x
lim
cos T tan T
lim
T
T o0
S 1
I oS
71.
3 0
0
72.
ª sin x 1 cos x º
˜
»
x
x
¼
x o 0«
¬
x2
xo0
70. lim I sec I
1
5
ª § 1 cos x ·º
lim «3¨
¸»
xo0
x
«¬ ©
¹»¼
x
66. lim
§1·
1¨ ¸
©5¹
xo0 ©
3 1 cos x
xo0
65. lim
Limits
its and Their Properties
lim
cos x
lim sin x
x o S 2 cot x
lim
S
1
x oS 2
1 tan x
cos x sin x
lim
x o S 4 sin x cos x
x o S 4 sin x cos x cos 2 x
sin x cos x
lim
1 0
0
x o S 4 cos x sin x cos x
sin T
1
x o S 4 cos x
T
T o0
1
lim
lim sec x
x oS 4
2
67. lim
xo0
sin x
x
lim
ª sin x
º
sin x»
x
¼
1 sin 0
x o 0«
¬
tan 2 x
xo0
x
sin 2 x
x o 0 x cos 2 x
68. lim
lim
1 0
2
0
73. lim
ª sin x sin x º
˜
lim «
»
xo0
cos 2 x ¼
¬ x
t o0
0
74. lim
sin 3t
2t
§ sin 3t ·§ 3 ·
lim¨
¸¨ ¸
3t ¹© 2 ¹
1 cos h
ho0
2
h
lim
75. f x
xo0
ª1 cos h
º
1 cos h »
h
¼
§1·
21¨ ¸1
© 3¹
h o 0«
¬
0 0
x 2 x
3
2
ª § sin 2 x ·§ 1 ·§ 3x ·º
lim «2¨
¸»
¸¨ ¸¨
¬ © 2 x ¹© 3 ¹© sin 3 x ¹¼
sin 2 x
x o 0 sin 3 x
69. lim
§ 3·
1¨ ¸
© 2¹
t o 0©
2
3
0
2
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
0.358
0.354
0.354
?
0.354
0.353
0.349
lim
x 2 x
2
It appears that the limit is 0.354.
2
−3
3
−2
The graph has a hole at x
Analytically, lim
xo0
x 2 x
0.
2
xo0
lim
xo0 x
˜
x 2 x 2 2
xo0
x 2 2
x 2 lim
2
2
1
x 2 1
2
2 2
2
| 0.354.
4
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 1.3
76. f x
Evaluating Limits Analytically
67
4 x
x 16
x
15.9
15.99
15.999
16
16.001
16.01
16.1
f (x )
–0.1252
–0.125
–0.125
?
–0.125
–0.125
–0.1248
4
x
It appears that the limit is –0.125.
1
0
20
−1
The graph has a hole at x
16.
4 x
x o16 x 16
Analytically, lim
77. f x
lim
x o16
x 4
1
x 4
lim
x 4
x o16
1
.
8
1
1
2 x
2
x
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x )
–0.263
–0.251
–0.250
?
–0.250
–0.249
–0.238
It appears that the limit is –0.250.
3
−5
1
−2
The graph has a hole at x
0.
1
1
2
2
x
Analytically, lim
xo0
x
78. f x
lim
2 2 x
xo0
22 x
˜
x
1
x
xo0 2 2 x
lim
˜
1
x
lim
1
xo0 2 2 x
1
.
4
x 5 32
x 2
x
1.9
1.99
1.999
1.9999
2.0
2.0001
2.001
2.01
2.1
f (x )
72.39
79.20
79.92
79.99
?
80.01
80.08
80.80
88.41
It appears that the limit is 80.
100
−4
3
−25
The graph has a hole at x
x 5 32
xo2 x 2
Analytically, lim
2.
lim
x 2 x 4 2 x3 4 x 2 8 x 16
x 2
xo2
lim x 4 2 x3 4 x 2 8 x 16
xo2
80.
INSTRUCTOR USE ONLY
(Hint:
Hint Use long division to factor x5 32. )
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68
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
79. f t
sin 3t
t
t
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (t )
2.96
2.9996
3
?
3
2.9996
2.96
It appears that the limit is 3.
4
− 2
2
−1
The graph has a hole at t
Analytically, lim
t o0
80. f x
sin 3t
t
0.
§ sin 3t ·
lim 3¨
¸
© 3t ¹
t o0
31
3.
cos x 1
2x2
x
–1
–0.1
–0.01
0.01
0.1
1
f (x )
–0.2298
–0.2498
–0.25
–0.25
–0.2498
–0.2298
It appears that the limit is –0.25.
1
−
−1
The graph has a hole at x
Analytically,
0.
cos x 1 cos x 1
˜
2x
cos x 1
cos 2 x 1
2 x 2 cos x 1
sin 2 x
2 x cos x 1
2
1
sin 2 x
˜
2
x
2 cos x 1
ª sin 2 x
º
1
lim « 2 ˜
»
2 cos x 1 »¼
¬ x
x o 0«
§ 1 ·
1¨ ¸
© 4¹
1
4
0.25
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.3
81. f x
sin x 2
x
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x )
–0.099998
–0.01
–0.001
?
0.001
0.01
0.099998
Evaluating Limits Analytically
69
It appears that the limit is 0.
1
− 2
2
−1
The graph has a hole at x
sin x
xo0
x
0.
§ sin x 2 ·
lim x¨
¸
xo0
© x ¹
2
Analytically, lim
01
0.
82. f x
sin x
3
x
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x )
0.215
0.0464
0.01
?
0.01
0.0464
0.215
§ sin x ·
lim 3 x 2 ¨
¸
© x ¹
0 1
0.
It appears that the limit is 0.
2
−3
3
−2
The graph has a hole at x
sin x
Analytically, lim 3
xo0
x
83. lim
'x o 0
84. lim
'x o 0
f x 'x f x
'x
f x 'x f x
'x
0.
xo0
lim
lim
lim
85. lim
'x o 0
'x
lim
> 6( x 'x) 3@ > 6 x 3@
'x
6'x
'x
'x o 0
f x 'x f x
'x
lim ( 6)
2
lim
'x o 0
lim
3 x 3'x 2 3x 2
'x
lim
'x o 0
'x
'x 2 x 'x 4
x 2 2 x'x 'x 2 4 x 4'x x 2 4 x
'x o 0
'x
2x 4
'x o 0
x
lim
'x o 0
x 'x x
x 'x 3
lim
lim 2 x 'x 4
'x
'x o 0 'x
3'x
6 x 6'x 3 6 x 3
'x
4 x 'x x 2 4 x
x 'x 'x
lim
'x o 0 'x
6
'x o 0
x 'x
'x o 0
86. lim
'x o 0
'x o 0
lim
lim
'x
'x o 0
'x o 0
f x 'x f x
3 x 'x 2 3 x 2
'x o 0
x 'x 'x
lim
x
'x o 0
x
1
x 'x ˜
x 'x x 'x 1
x
2 x
x
x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
70
Chapter 1
87. lim
Limits
its and Their Properties
1
1
x
x
x
'
3
3
lim
'x o 0
'x
x 3 x 'x 3
1
˜
lim
'x o 0 x 'x 3 x 3
'x
f x 'x f x
'x
'x o 0
lim
'x
x 'x 3 x 3 'x
lim
1
x 'x 3 x 3
'x o 0
'x o 0
88. lim
2
1
1
2
( x 'x) 2
x
lim
'x o 0
'x
f x 'x f x
'x
'x o 0
1
x 3
lim
x 2 x 'x
'x o 0 x 2
lim
2
x 'x 'x
x 2 ª¬ x 2 2 x'x ('x) 2 º¼
x 2 ( x 'x) 2 'x
'x o 0
lim
2
2 x'x 'x
2
'x o 0 x 2 ( x 'x ) 2 'x
lim
2 x 'x
'x o 0 x 2 ( x 'x ) 2
2 x
x4
89. lim 4 x 2 d lim f x d lim 4 x 2
xo0
xo0
xo0
2
x3
93. f x
x sin
4 d lim f x d 4
xo0
Therefore, lim f x
xo0
0.5
4.
− 0.5
0.5
90. lim ¬ªb x a ¼º d lim f x d lim ¬ªb x a ¼º
xoa
xoa
xoa
b d lim f x d b
xoa
Therefore, lim f x
xoa
91. f x
1
x
b.
− 0.5
1·
§
lim ¨ x sin ¸
x¹
94. h x
x sin x
0
x o 0©
x cos
1
x
6
0.5
−2
2
− 0.5
−6
lim x sin x
0
− 0.5
0
xo0
92. f x
1·
§
lim ¨ x cos ¸
x¹
x o 0©
0.5
95. (a) Two functions f and g agree at all but one point
g x for all x in the
(on an open interval) if f x
x cos x
interval except for x
6
c, where c is in the interval.
x 1
( x 1)( x 1)
and
x 1
x 1
x 1 agree at all points except x
2
(b) f x
− 2
2
g x
−6
1.
(Other answers possible.)
INSTRUCTOR
NS
USE ONLY
lim x cos x
xo0
0
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.3
L as x o c, then lim f x exists and equals L
xoc
f x
xoc g
71
97. If a function f is squeezed between two functions h and
g, h x d f x d g x , and h and g have the same limit
96. An indeterminant form is obtained when evaluating a
limit using direct substitution produces a meaningless
fractional expression such as 0 0. That is,
lim
Evaluating Limits Analytically
x
for which lim f x
lim g x
xoc
xoc
0
98. (a) Use the dividing out technique because the numerator and denominator have a common factor.
x2 x 2
x o 2
x 2
( x 2)( x 1)
x 2
lim ( x 1)
2 1
lim
lim
x o 2
x o 2
3
(b) Use the rationalizing technique because the numerator involves a radical expression.
x 4 2
x
lim
xo0
lim
xo0
lim
xo0 x
lim
xo0
99. f x
x, g x
sin x, h x
x 4 2
x
x 4 2
x 4 2
( x 4) 4
x 4 2
1
x 4 2
1
4 2
sin x
x
1
4
101. s t
3
lim
f
g
t o2
h
−5
16t 2 500
s2 st
2t
lim
16 2
2
500 16t 2 500
t o2
2t
lim
436 16t 500
2t
5
2
t o2
−3
When the x-values are "close to" 0 the magnitude of f is
approximately equal to the magnitude of g. So,
g
f | 1 when x is "close to" 0.
lim
t o2
lim
16 t 2 4
2 t
16 t 2 t 2
2t
lim 16 t 2
t o2
100. f x
x, g x
sin 2 x, h x
sin 2 x
x
t o2
64 ft/sec
The paint can is falling at about 64 feet/second.
2
g
−3
3
h
f
−2
When the x-values are "close to" 0 the magnitude of g is
"smaller" than the magnitude of f and the magnitude of g
is approaching zero "faster" than the magnitude of f. So,
g
f | 0 when x is "close to" 0.
INSTRUCTOR USE ONLY
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72
NOT FOR SALE
Chapter 1
102. s t
Limits
its and Their Properties
16t 2 500
500
16
0 when t
§5 5 ·
s¨¨
¸ st
2 ¸¹
©
lim
§5 5 ·
5 5
t o ¨¨
¸¸
t
© 2 ¹
2
5 5
sec. The velocity at time a
2
5 5
is
2
0 16t 2 500
lim
§5 5 ·
t o ¨¨
¸¸
© 2 ¹
5 5
t
2
125 ·
§
16¨ t 2 ¸
4 ¹
©
lim
§5 5 ·
5 5
t o ¨¨
¸¸
t
© 2 ¹
2
§
5 5 ·§
5 5·
16¨¨ t ¸¨ t ¸
¸¨
2
2 ¸¹
©
¹©
lim
§5 5 ·
5 5
t o ¨¨
¸¸
t
© 2 ¹
2
ª
§
5 5 ·º
lim «16¨¨ t ¸»
5 5«
2 ¸¹¼»
to
©
¬
80 5 ft/sec
2
| 178.9 ft/sec.
The velocity of the paint can when it hits the ground is about 178.9 ft/sec.
103. s t
lim
t o3
4.9t 2 200
s3 st
3t
lim
4.9 3
2
200 4.9t 2 200
3t
t o3
lim
t o3
lim
4.9 t 2 9
3t
4.9 t 3 t 3
3t
t o3
lim ª
¬ 4.9 t 3 º¼
t o3
29.4 m/sec
The object is falling about 29.4 m/sec.
104. 4.9t 2 200
lim
t oa
sa st
a t
0 when t
lim
t oa
lim
200
4.9
20 5
sec. The velocity at time a
7
20 5
is
7
0 ¬ª4.9t 2 200¼º
a t
4.9 t a t a
a t
t oa
ª
§
20 5 ·º
lim «4.9¨¨ t ¸»
20 5 «
7 ¸¹»¼
to
©
¬
28 5 m/sec
7
| 62.6 m/sec.
The velocity of the object when it hits the ground is about 62.6 m/sec.
105. Let f x
1 x and g x
1/ x. lim f x and lim g x do not exist. However,
x o0
ª 1 § 1 ·º
lim ª f x g x º¼
lim « ¨ ¸»
x o 0¬
xo0 x
© x ¹¼
¬
and therefore does not exist.
x o0
lim >0@
xo0
0
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 1.3
106. Suppose, on the contrary, that lim g x exists. Then,
xoc
f x d f x d f x
which is a contradiction. So, lim g x does not exist.
lim ª f x º¼ d lim f x d lim f x
x oc¬
xoc
x c G . Because f x b
xoc
Therefore, lim f x
0 H for
b b
lim x
lim xx
xoc
(b) Given lim f x
For every H ! 0, there exists G ! 0 such that
f x L H whenever 0 x c G . Since
n 1
f x L d f x L H for
ª lim xº ª lim x
«¬x o c »¼ «¬x o c
n 1 º
c ª« lim xx
¬x o c
»¼
c ª« lim xº» ª« lim x n 2 º»
¬x o c ¼ ¬ x o c
¼
n2 º
c c lim xx n 3
xoc
xoc
c .
­ 4, if x t 0
®
¯4, if x 0
f x
f x L H b whenever 0 x c G . So,
whenever 0 x c G , we have
b f x L H or
bf x bL H
which implies that lim ¬ªbf x ¼º
110. Given lim f x
xoc
lim f x
lim 4
xo0
4.
xo0
lim f x does not exist because for
xo0
x 0, f x
4 and for x t 0, f x
4.
114. The graphing utility was set in degree mode, instead of
radian mode.
115. The limit does not exist because the function approaches
1 from the right side of 0 and approaches 1 from the
left side of 0.
bL.
xoc
L.
113. Let
n
L, there exists G ! 0 such that
lim f x
x c G , then lim f x
»¼
109. If b
0, the property is true because both sides are
equal to 0. If b z 0, let H ! 0 be given. Because
xoc
L:
xoc
xoc
"
0.
xoc
x n , n is a positive integer, then
108. Given f x
xoc
0 d lim f x d 0
a G ! 0 such that f x b H whenever
n
xoc
b, show that for every H ! 0 there exists
every H ! 0, any value of G ! 0 will work.
73
0.
xoc
xoc
xoc
107. Given f x
0, then lim ª¬ f x º¼
If lim f x
112. (a)
because lim f x exists, so would lim ª¬ f x g x º¼ ,
xoc
Evaluating Limits Analytically
0:
2
For every H ! 0, there exists G ! 0 such that
f x 0 H whenever 0 x c G .
Now f x 0
−3
f x 0 H for
f x
x c G . Therefore, lim f x
−2
0.
xoc
116. False. lim
x oS
M f x d f x g x d M f x
111.
lim M f x
d lim f x g x d lim M f x
xoc
3
xoc
xoc
118. False. Let
0 d lim f x g x d 0
f x
xoc
Therefore, lim f x g x
xoc
0.
0
0
S
117. True.
M 0 d lim f x g x d M 0
xoc
sin x
x
­x x z 1
,
®
¯3 x 1
Then lim f x
x o1
c
1.
1 but f 1 z 1.
INSTRUCTOR USE ONLY
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74
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
119. False. The limit does not exist because f x approaches
3 from the left side of 2 and approaches 0 from the right
side of 2.
sec x 1
x2
123. f x
(a) The domain of f is all x z 0, S /2 nS .
4
(b)
−3
2
− 3
2
6
3
2
−2
−2
120. False. Let f x
1 x 2 and g x
2
The domain is not obvious. The hole at x
apparent.
1
(c) lim f x
xo0
2
2
x .
Then f x g x for all x z 0. But
lim f x
xo0
lim g x
0.
xo0
(d)
121. lim
x o0
1 cos x
x
lim
x o0
1 cos x 1 cos x
˜
x
1 cos x
1 cos x
2
lim
x o 0 x 1 cos x
sec x 1
x2
sin x
x o 0 x 1 cos x
sin x
sin x
lim
˜
x o0 x
1 cos x
So, lim
x o0
0
lim
1 § sin 2 x ·
1
¨ 2 ¸
© x ¹ sec x 1
x o 0 cos 2 x
§1·
11 ¨ ¸
© 2¹
sin x º
ª sin x º ª
»
«lim
» «lim
¬x o 0 x ¼ ¬x o 01 cos x ¼
1 0
sec x 1
x2
124. (a) lim
xo0
1 cos x
x2
sec 2 x 1
x 2 sec x 1
1 § sin 2 x ·
1
¨
¸
cos 2 x © x 2 ¹ sec x 1
tan 2 x
x sec x 1
2
2
lim
sec x 1 sec x 1
˜
x2
sec x 1
lim
xo0
1
.
2
1 cos x 1 cos x
˜
1 cos x
x2
122. f x
­0, if x is rational
®
¯1, if x is irrational
1 cos 2 x
x o 0 x 1 cos x
g x
­0, if x is rational
®
¯x, if x is irrational
sin 2 x
1
˜
x o 0 x2
1 cos x
lim
§·
1¨ ¸
© 2¹
lim f x does not exist.
No matter how “close to” 0 x is, there are still an infinite
number of rational and irrational numbers so that
lim f x does not exist.
xo0
lim g x
2
lim
xo0
xo0
0 is not
0
when x is "close to" 0, both parts of the function are
“close to” 0.
1
2
(b) From part (a),
1 cos x
1
|
Ÿ 1 cos x
2
x2
1
| x 2 Ÿ cos x
2
1
| 1 x 2 for x
2
| 0.
(c) cos 0.1 | 1 1
2
0.1
2
0.995
(d) cos 0.1 | 0.9950, which agrees with part (c).
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.4
Continuity and One-Sided
OneOne
Limits
75
Section 1.4 Continuity and One-Sided Limits
lim f x
3
6. (a)
(b) lim f x
3
(b)
(c) lim f x
3
(c) lim f x does not exist.
1. (a)
x o 4
x o 4
xo4
lim f x
0
lim f x
2
x o 1
x o 1
x o 1
The function is continuous at x
on f, f .
1
7. lim
lim f x
2
lim f x
2
(c) lim f x
2
2. (a)
(b)
x o 2
x o 2
x o 2
x o 8 x 8
lim f x
0
(b) lim f x
0
3. (a)
x o 3
x o 3
(c) lim f x
x o3
(b)
lim f x
x o 3
lim f x
x o 3
(c) lim f x
x o 3
2
x 2
9. lim
x 5
2
25
x o 5 x
2.
1
88
1
16
2
2 2
1
2
x 5
lim
x o 5 ( x 5)( x 5)
1
lim
4 x
16
10. lim
0
( x 4)
lim
1
4 4
3.
3
x
lim
11.
x o 3
3
x2 9
x
x 9
2
3
3 because
12. lim
x o 4
3
(b) lim f x
3
x o 2
x 2
x 4
lim
x o 4
x o 4
lim
x o 4
(c) lim f x does not exist
xo2
13. lim
The function is NOT continuous at x
2.
lim
14.
1
1
'
x
x
x
15. lim
'x
'x o 0
lim
'x o 0
x x 'x
x x 'x
x
x o 0 x
˜
1
'x
lim
'x
'x o 0 x x 'x
lim
lim
x o 0
x
x
x 10
x o10 x 10
˜
1
8
decreases without bound as x o 3.
lim
lim f x
1
x o 4 x 4
does not exist because
x 2
˜
x 4
x o 3
x o 2
lim
x o 4 ( x 4)( x 4)
2
x o 4 x
The function is NOT continuous at x
f 3
4 z lim f x .
5. (a)
1
10
x o 5 x 5
The function is NOT continuous at x
4. (a)
8. lim
x o 2
The function is continuous at x
1.
The function is NOT continuous at x
4 and is continuous
lim
x 2
x 2
x 4
x 4
x 2
1
x 2
1
4 2
1
4
1
x 10
x o10 x 10
1
1
'x
1
'x o 0 x x 'x
1
x x 0
1
x2
INSTRUCTOR USE ONLY
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76
Chapter 1
16.
lim
NOT FOR SALE
Limits
its and Their Properties
x 'x
2
x 'x x 2 x
'x
'x o 0
lim
x 'x x 2 x
'x
'x o 0
lim
2
x 2 2 x 'x 'x
2 x 'x 'x
2
'x
'x
lim 2 x 'x 1
'x o 0
'x o 0
2x 0 1
17. lim f x
lim
x o 3
x o 3
x 2
2
5
2
x o 3
9 12 6
x o 3
lim x 2 4 x 2
lim f x
x o 3
1
x2 4
27. f x
lim x 2 4 x 6
18. lim f x
2x 1
9 12 2
x o 3
2 and
has discontinuities at x
x
2 because f 2 and f 2 are not defined.
3
1
x2 1
x 1
28. f x
Since these one-sided limits disagree, lim f x
x o3
does not exist.
has a discontinuity at x
1 because f 1 is not
defined.
x o1
lim x 1
x o1
2
lim f x
lim x3 1
2
19. lim f x
x o1
lim f x
x o1
29. f x
2
has discontinuities at each integer k because
lim f x z lim f x .
2
x o1
axb x
x o k
20. lim f x
x o1
lim 1 x
x o1
0
30. f x
21. lim cot x does not exist because
x oS
lim cot x and lim cot x do not exist.
x oS 22.
x oS lim sec x does not exist because
x oS 2
lim sec x and
xo S 2 23. lim 5a xb 7
x o 4
axb
x o 2
53 7
32. f t
8
22 2
lim 2 a xb
c x f·
§
26. lim¨1 dd gg¸
x o1
e 2 h¹
©
x
1 1
9 t 2 is continuous on >3, 3@.
3
lim f x . f is continuous on >1, 4@.
3
x o 0
6
has a nonremovable discontinuity at
x
0 because lim f x does not exist.
xo0
5
36. f x
2 4
1.
2
and
x o 3
x o1
34. g 2 is not defined. g is continuous on >1, 2 .
35. f x
2 3
2 z lim f x
49 x 2 is continuous on >7, 7@.
33. lim f x
x o 0
x o3
x o 3
1 because f 1
lim sec x do not exist.
xo S 2 25. lim 2 a xb does not exist because
lim 2 a xb
x 1
­ x,
°
2,
x 1 has a discontinuity at
®
°2 x 1, x ! 1
¯
31. g x
3 for 3 d x 4
24. lim 2 x a xb
x
x ok
6.
x
4
has a nonremovable discontinuity at
x 6
6 because lim f x does not exist.
xo6
37. f x
x 2 9 is continuous for all real x.
38. f x
x 2 4 x 4 is continuous for all real x.
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.4
1
4 x2
39. f x
1
2 x 2 x
has nonremovable
r2 because lim f x and
discontinuities at x
xo2
lim f x do not exist.
Continuity and One-Sided
OneOne
Limits
x 2
x x 6
48. f x
has a nonremovable discontinuity at x
3 because
lim f x does not exist, and has a removable
1
is continuous for all real x.
x2 1
lim f x
3x cos x is continuous for all real x.
42. f x
cos
Sx
2
is continuous for all real x.
x 7
has a nonremovable discontinuity at x
lim f x does not exist.
x
is not continuous at x
0, 1.
x2 x
x
1
0 is
Because 2
for x z 0, x
x x
x 1
a removable discontinuity, whereas x 1 is a
nonremovable discontinuity.
x
has nonremovable discontinuities at
x 4
2 because lim f x and lim f x
2 and x
2
xo2
x o 2
do not exist.
45. f x
46. f x
x 5
( x 5)( x 5)
has a nonremovable discontinuity at x
lim f x does not exist, and has a removable
x o 5
discontinuity at x
lim f x
x o5
47. f x
1
x o5 x 5
x 2
x 2 3 x 10
1
.
10
has a nonremovable discontinuity at x
lim f x does not exist.
x 2
x 2 x 5
lim
­ x, x d 1
® 2
¯x , x ! 1
51. f x
has a possible discontinuity at x
2.
f 1
lim f x
2.
x o1
lim f x
lim x 2
x o1
x o1
f 1
lim f x
1
1
.
7
1½
°
¾ lim f x
1° x o1
¿
1, therefore, f is continuous for
­2 x 3, x 1
® 2
x t1
¯x ,
f 1
12
lim f x
x o1
lim f x
3.
1
x o1
x o1
2 because
x o 2 x 5
lim x
x o1
has a possible discontinuity at x
1.
1.
1
f is continuous at x
all real x.
x o5
lim f x
5 because
x o5
52. f x
has a nonremovable discontinuity at x
5 because
lim f x does not exist, and has a removable
x o 2
x 5
5 because
lim
discontinuity at x
x 5
50. f x
3.
5 because
7 because
x o 7
1.
x
is continuous for all real x.
x2 1
x 5
x 2 25
1
.
5
x 7
43. f x
x
1
lim
x o 2 x 3
x o 2
49. f x
44. f x
2 because
discontinuity at x
41. f x
x 2
( x 3)( x 2)
2
x o3
x o 2
40. f x
77
f 1
1.
1
lim 2 x 3
x o1
lim x 2
x o1
1
1½
°
f x
¾ lim
° x o1
¿
1
lim f x
x o1
f is continuous at x
all real x.
1, therefore, f is continuous for
INSTRUCTOR USE ONLY
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78
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
­x
° 1, x d 2
®2
°3 x, x ! 2
¯
53. f x
has a possible discontinuity at x
1.
2.
2
1
2
f 2
2
½
2°
°
f x does not exist.
¾ xlim
o2
1°
°¿
§x
·
lim ¨ 1¸
¹
lim 3 x
lim f x
x o 2
x o 2 © 2
lim f x
x o 2
2.
x o 2
Therefore, f has a nonremovable discontinuity at x
x d 2
­2 x,
® 2
x
x
x ! 2
4
1,
¯
54. f x
has a possible discontinuity at x
1.
2.
2 2
f 2
2.
4
lim 2 x
lim f x
x o 2
x o 2
4
lim x 2 4 x 1
lim f x
x o 2
x o 2
½
° lim f x does not exist.
¾
3° x o 2
¿
Therefore, f has a nonremovable discontinuity at x
55. f x
2.
­ Sx
°tan ,
4
®
° x,
¯
2.
x 1
­ Sx
°csc ,
6
®
°2,
¯
56. f x
x t1
­ Sx
°tan , 1 x 1
4
®
° x,
x d 1 or x t 1
¯
has possible discontinuities at x
1.
2.
3.
f 1
1
1.
1
lim f x
1
lim f x
1
f 1
lim f x
f 1
lim f x
x o 1
x o1
f is continuous at x
all real x.
x 3 ! 2
­ Sx
°csc , 1 d x d 5
6
®
°
x 1 or x ! 5
¯2,
1, x
f 1
x 3 d 2
x o1
x o1
r1, therefore, f is continuous for
has possible discontinuities at x
1, x
5.
S
f 5
csc
1.
f 1
2.
lim f x
2
3.
f 1
lim f x
csc
x o1
6
2
x o1
f is continuous at x 1 and x
continuous for all real x.
57. f x
5S
6
lim f x
2
xo5
f 5
2
lim f x
x o5
5, therefore, f is
csc 2 x has nonremovable discontinuities at
integer multiples of S 2.
Sx
has nonremovable discontinuities at each
2
2k 1, k is an integer.
58. f x
tan
59. f x
ax 8b has nonremovable discontinuities at
each integer k.
60. f x
5 a xb has nonremovable discontinuities at
INSTRUCTOR USE ONLY
each
ach integer k.
k.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.4
61. f 1
63. f 2
3
Find a so that lim ax 4
3
a1 4
3
a
7.
x o1
62. f 1
x o 2
3
a1 5
3
Let a
65. Find a and b such that lim ax b
a b
x o 1
a b
2
3a b
2
4a
4
1
b
2 1
2 and lim ax b
x2 a2
xoa x a
lim x a
2a
8 Ÿ a
4.
Find a such 2a
4
x o 0
a
x o 0
4.
x o 3
3a b
2.
1
sin x 2
72. f g x
lim
xoa
2.
x d 1
­ 2,
°
® x 1, 1 x 3
°
x t 3
¯2,
f x
a
8
22
2.
a
xoa
lim
x o 0
lim g x
x o1
8 Ÿ a
4 sin x
x
lim a 2 x
64. lim g x
x o 0
Find a so that lim ax 5
79
8
Find a so that lim ax 2
3
66. lim g x
Continuity and One
OneOne-Sided Limits
Continuous for all real x
73. y
axb x
Nonremovable discontinuity at each integer
67. f g x
x 1
2
0.5
Continuous for all real x
68. f g x
5 x3 1
−3
3
5 x3 1
− 1.5
Continuous for all real x
69. f g x
1
x 5 6
2
1
x 1
Nonremovable discontinuities at x
1
x 2 x 15
74. h x
2
r1
1
( x 5)( x 3)
2
5 and x
Nonremovable discontinuities at x
3
2
70. f g x
1
x 1
−8
Nonremovable discontinuity at x
all x ! 1
71. f g x
tan
−2
x
2
Not continuous at x
7
1; continuous for
2
°­x 3 x, x ! 4
®
°̄2 x 5, x d 4
75. g x
r S , r 3S , r 5S , ... Continuous on
the open intervals ..., ( 3S , S ), ( S , S ), (S , 3S ),...
Nonremovable discontinuity at x
4
10
−2
8
−2
INSTRUCTOR USE ONLY
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80
NOT FOR SALE
Chapter 1
76. f x
f 0
Limits
its and Their Properties
­ cos x 1
, x 0
°
x
®
°5 x,
x t 0
¯
50
3
0
lim f x
cos x 1
lim
x o 0
x o 0
lim f x
x
lim 5 x
x o 0
xo0
x
77. f x
0
x o 0
Therefore, lim f x
−7
0
0
2
−3
f 0 and f is continuous on the entire real line.
0 was the only possible discontinuity.
x
x2 x 2
­2 x 4, x z 3
®
x
3
¯1,
84. f x
Continuous on f, f
lim 2 x 4
Since lim f x
x o3
78. f x
x 1
x
3
85. f x
3
−4
sec
−2
The graph appears to be continuous on the interval
>4, 4@. Because f 0 is not defined, you know that
Sx
f has a discontinuity at x
0. This discontinuity is
removable so it does not show up on the graph.
4
Continuous on:
!, 6, 2 , 2, 2 , 2, 6 , 6, 10 , !
82. f x
4
x 3
x
Continuous on >3, f
81. f x
sin x
x
x
Continuous on >0, f
80. f x
cos
2 z 1,
f is continuous on (f, 3) and (3, f).
Continuous on 0, f
79. f x
x o3
86. f x
x3 8
x 2
1
x
14
Continuous on (f, 0) and (0, f)
83. f x
­ x2 1
, x z1
°
®x 1
°2,
x 1
¯
Since lim f x
x o1
−4
lim
x2 1
x o1 x 1
lim
lim ( x 1)
2,
x o1
f is continuous on (f, f).
4
0
x o1
( x 1)( x 1)
x 1
The graph appears to be continuous on the interval
>4, 4@. Because f 2 is not defined, you know that
f has a discontinuity at x
2. This discontinuity is
removable so it does not show up on the graph.
87. f x
1 x 4 x 3 4 is continuous on the interval
12
>1, 2@. f 1
37
and
12
f 2
83 . By the Intermediate
Value Theorem, there exists a number c in >1, 2@ such
that f c
0.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
x3 5 x 3 is continuous on the interval >0, 1@.
88. f x
3 and f 1
f 0
3. By the Intermediate Value
Theorem, there exists a number c in >0, 1@ such that
f c
0.
x 2 2 cos x is continuous on >0, S @.
89. f x
3 and f S
f 0
S 2 1 | 8.87 ! 0. By the
Intermediate Value Theorem, f c
0 for at least one
value of c between 0 and S .
90. f x
f 1
f 4
5
§S x ·
tan ¨ ¸ is continuous on the interval >1, 4@.
x
© 10 ¹
§S ·
5 tan ¨ ¸ | 4.7 and
© 10 ¹
5
§ 2S ·
tan ¨ ¸ | 1.8. By the Intermediate
4
© 5 ¹
Section 1.4
Continuity and One Sided Limits
94. h T
tanT 3T 4 is continuous on >0, 1@.
91. f x
T | 0.91. Using the root feature, you obtain
T | 0.9071.
f 0
x2 x 1
95. f x
f is continuous on >0, 5@.
1 and f 5
f 0
1 11 29
x2 x 1
11
x x 12
0
x 4 x 3
0
4 or x
3
2
x
1 and f 1
4 is not in the interval.
3x
So, f 3
By the Intermediate Value Theorem, f c
0 for at
11.
x2 6x 8
96. f x
1
29
The Intermediate Value Theorem applies.
c
f x is continuous on >0, 1@.
f is continuous on >0, 3@.
least one value of c between 0 and 1. Using a graphing
utility to zoom in on the graph of f x , you find that
f 0
x | 0.68. Using the root feature, you find that
x | 0.6823.
The Intermediate Value Theorem applies.
92. f x
2
f ( x) is continuous on >0, 1@.
f (0)
1 and f (1)
x2 6x 8
0
x2 x 4
0
x
c
2
By the Intermediate Value Theorem, f c
0 for at
1
8 and f 3
1 0 8
x x 3x 1
4
0 for at
least one value of c between 0 and 1. Using a graphing
utility to zoom in on the graph of h T , you find that
0.
x3 x 1
tan(1) 1 | 0.557.
By the Intermediate Value Theorem, h c
Value Theorem, there exists a number c in >1, 4@ such
that f c
4 and h 1
h0
81
2 x
So, f 2
2 or x
4
4 is not in the interval.
0.
least one value of c between 0 and 1. Using a graphing
utility to zoom in on the graph of f x , you find that
x | 0.37. Using the root feature, you find that
x | 0.3733.
93. g t
2 cos t 3t
g is continuous on >0, 1@.
g 0
2 ! 0 and g 1 | 1.9 0.
By the Intermediate Value Theorem, g c
0 for at
least one value of c between 0 and 1. Using a graphing
utility to zoom in on the graph of g t , you find that
t | 0.56. Using the root feature, you find that
t | 0.5636.
5636
INSTRUCTOR USE ONLY
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82
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
x3 x 2 x 2
97. f x
100. Answers will vary. Sample answer:
y
f is continuous on >0, 3@.
2 and f 3
f 0
5
4
3
2
1
19
2 4 19
The Intermediate Value Theorem applies.
x3 x 2 x 2
4
x3 x 2 x 6
0
x 2 x x 3
0
x
2
2
So, f 2
1
−2
−3
x o 3
2
x o 3
continuous if g x
4.
20
3
6
x x
6x 6
x 5x 6
0
x 2 x3
0
2
x
c
3 x
2 or x
r1.
102. A discontinuity at c is removable if the function f can be
made continuous at c by appropriately defining (or
redefining) f c . Otherwise, the discontinuity is
nonremovable.
x 4
(a) f x
x 4
sin x 4
(b) f x
x 4
The Intermediate Value Theorem applies.
x2 x
x 1
x and
x 1. Then f and g are continuous for all real
x, but f g is not continuous at x
ª5 º
f is continuous on « , 4». The nonremovable
¬2 ¼
discontinuity, x 1, lies outside the interval.
2
0. For example, let f x
2
g x
35
§5·
f¨ ¸
and f 4
2
6
© ¹
35
20
6 6
3
3 because
101. If f and g are continuous for all real x, then so is f g
(Theorem 1.11, part 2). However, f g might not be
x2 x
x 1
98. f x
3 4 5 6 7
The function is not continuous at x
lim f x
1z 0
lim f x .
x 2 x 3 has no real solution.
c
x
−2 −1
(c) f x
x
x t 4
­1,
°
°0,
®
°1,
°0,
¯
4 x 4
4
x
x 4
4 is nonremovable, x
3
4 is removable
y
2 is not in the interval.
4
3
So, f 3
6.
2
1
99. (a) The limit does not exist at x
c.
(b) The function is not defined at x
c.
(c) The limit exists at x
c, but it is not equal to the
value of the function at x
c.
(d) The limit does not exist at x
c.
x
−6 −4 −2
2
−1
4
6
−2
103. True
1.
2.
3.
f c
L is defined.
lim f x
L exists.
xoc
f c
lim f x
xoc
All of the conditions for continuity are met.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.4
g x , x z c, then
104. True. If f x
lim f x
Continuity and One Sided Limits
§ c t 2 fg
·
25¨ 2dd
g t¸
© e 2 h
¹
110. N t
lim g x (if they exist) and at least one of
xoc
xoc
these limits then does not equal the corresponding
c.
function value at x
105. False. A rational function can be written as
P x Q x where P and Q are polynomials of degree m
and n, respectively. It can have, at most, n
discontinuities.
83
t
0
1
1.8
2
3
3.8
N t
50
25
5
50
25
5
Discontinuous at every positive even integer. The
company replenishes its inventory every two months.
N
x o1
107. The functions agree for integer values of x:
3 a xb
g x
3 a xb
f x
3 x
3 x
3 a xb
g x 1
2 a xb.
3 0
30
20
10
t
2
3, g 12
3 1
4.
108. lim f t | 28
4
6
8
10 12
Time (in months)
111. Let s t be the position function for the run up to the
campsite. s 0
s 20
For example,
f 12
40
3 x½°
¾ for x an integer
°¿
However, for non-integer values of x, the functions
differ by 1.
f x
Number of units
50
106. False. f 1 is not defined and lim f x does not exist.
0 t
0 corresponds to 8:00 A.M.,
k (distance to campsite)). Let r t be the
position function for the run back down the mountain:
r 0
k , r 10
0. Let f t
st rt.
When t
0 (8:00 A.M.),
f 0
s0 r0
0 k 0.
t o 4
lim f t | 56
s 10 r 10 ! 0.
t o 4
When t
At the end of day 3, the amount of chlorine in the pool
has decreased to about 28 oz. At the beginning of day 4,
more chlorine was added, and the amount is now about
56 oz.
Because f 0 0 and f 10 ! 0, then there must be a
0 t d 10
­0.40,
°
0.40
0.05
9
,
t
t
! 10, t not an integer
a b
®
°0.40 0.05 t 10 , t ! 10, t an integer
¯
109. C t
10 (8:00 A.M.), f 10
value t in the interval >0, 10@ such that f t
f t
0, then s t r t
st
r t . Therefore, at some time t, where
0. If
0, which gives us
0 d t d 10, the position functions for the run up and the
run down are equal.
4 3
S r be the volume of a sphere with radius r.
3
500S
V is continuous on >5, 8@. V 5
| 523.6 and
3
2048S
V 8
| 2144.7. Because
3
523.6 1500 2144.7, the Intermediate Value
Theorem guarantees that there is at least one value r
between 5 and 8 such that V r
1500. (In fact,
112. Let V
C
0.7
0.6
0.5
0.4
0.3
0.2
0.1
t
2
4
6
8 10 12 14
There is a nonremovable discontinuity at each integer
greater than or equal to 10.
r | 7.1012.)
Note: You could also express C as
Ct
0 t d 10
­°0.40,
®
°̄0.40 0.05a10 t b, t ! 10
INSTRUCTOR USE ONLY
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84
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
113. Suppose there exists x1 in >a, b@ such that
f x1 ! 0 and there exists x2 in >a, b@ such that
­0, 0 d x b
®
¯b, b x d 2b
118. (a) f x
f x2 0. Then by the Intermediate Value Theorem,
y
f x must equal zero for some value of x in
2b
> x1, x2 @ or > x2 , x1@ if x2 x1 . So, f would have a zero in
>a, b@, which is a contradiction. Therefore, f x ! 0 for
all x in >a, b@ or f x 0 for all x in >a, b@.
b
x
b
114. Let c be any real number. Then lim f x does not exist
NOT continuous at x
xoc
because there are both rational and irrational numbers
arbitrarily close to c. Therefore, f is not continuous at c.
0, then f 0
continuous at x
0 and lim f x
xo0
y
If x z 0, then lim f t
0 for x rational, whereas
tox
lim f t
2b
kx z 0 for x irrational. So, f is not
lim kt
tox
0. So, f is
0.
tox
continuous for all x z 0.
(a)
b
­1, if x 0
°
0
®0, if x
°1,
if x ! 0
¯
116. sgn x
x
b
119. f x
(b) lim sgn x
1
x o 0
2b
Continuous on >0, 2b@.
1
lim sgn x
x o 0
b.
­x
0 d x d b
°° 2 ,
®
°b x , b x d 2b
°̄
2
(b) g x
115. If x
2b
­°1 x 2 , x d c
®
x, x ! c
°̄
f is continuous for x c and for x ! c. At x
(c) limsgn x does not exist.
need 1 c
xo0
y
c
4
3
2
1 r
c, you
c. Solving c c 1, you obtain
2
1 4
2
1 r
2
5
.
2
120. Let y be a real number. If y
0, then x
y ! 0, then let 0 x0 S 2 such that
1
x
−4 −3 −2 −1
1
2
3
4
−2
M
−3
tan x0 ! y this is possible since the tangent
function increases without bound on >0, S 2 . By the
−4
Intermediate Value Theorem, f x
117. (a)
0. If
S
tan x is
continuous on >0, x0 @ and 0 y M , which implies
60
that there exists x between 0 and x0 such that
tan x
y. The argument is similar if y 0.
50
40
30
20
10
t
5
10
15 20
25 30
(b) There appears to be a limiting speed and a possible
cause is air resistance.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 1.4
Continuity and One Sided Limits
85
x c2 c
,c ! 0
x
121. f x
Domain: x c 2 t 0 Ÿ x t c 2 and x z 0, ª¬c 2 , 0 ‰ 0, f
lim
x o0
x c2 c
x
Define f 0
122. 1.
2.
x o0
x c2 c
˜
x
x c2 c
x c c
2
1 2c to make f continuous at x
xo0
x c2 c2
x ª x c cº
¬
¼
1
1
2c
x c c
2
xa xb
f c exists.
'x o 0
lim
xo0
2
123. h x
lim f c 'x
lim f x
xoc
lim
0.
f c is defined.
[Let x
3.
lim
15
c 'x. As x o c, 'x o 0]
lim f x
f c.
xoc
−3
Therefore, f is continuous at x
3
c.
−3
h has nonremovable discontinuities at
x
f 2 x f1 x . Because f1 and f 2 are continuous on >a, b@, so is f.
124. (a) Define f x
f 2 a f1 a ! 0 and f b
f a
r1, r 2, r 3, !.
f 2 b f1 b 0
By the Intermediate Value Theorem, there exists c in >a, b@ such that f c
f 2 c f1 c
f c
(b) Let f1 x
0 Ÿ f1 c
0.
f2 c
cos x, continuous on >0, S 2@, f1 0 f 2 0 and f1 S 2 ! f 2 S 2 .
x and f 2 x
So by part (a), there exists c in >0, S 2@ such that c
cos c .
Using a graphing utility, c | 0.739.
125. The statement is true.
If y t 0 and y d 1, then y y 1 d 0 d x 2 , as
desired. So assume y ! 1. There are now two cases.
Case l: If x d y 12 , then 2 x 1 d 2 y and
y y 1
y y 1 2y
d x 1
2
2y
2
126. P 1
P 02 1
P0
P2
P 12 1
P1
P5
P 22 1
P2
2
2
1
1
1
2
1
5
Continuing this pattern, you see that P x
x for
infinitely many values of x. So, the finite degree
x for all x.
polynomial must be constant: P x
x2 2x 1 2 y
d x2 2 y 2 y
x2
Case 2: If x t y 12
x 2 t y 12
2
y 2 y 14
! y2 y
y y 1
In both cases, y y 1 d x 2 .
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 1
86
Limits
imits and Their Properties
Section 1.5 Infinite Limits
1.
lim 2
x
x 4
f
lim 2
x
x2 4
f
x o 2
x o 2
2
1
x 4
6. f x
As x approaches 4 from the left, x 4 is a small
negative number. So,
f.
lim f x
1
2. lim
x o 2 x 2
1
lim
x o 2 x 2
3.
4.
f
f
4
Sx
lim tan
4
x o 2
Sx
lim sec
Sx
x o 2
x o 2
5. f x
x 4
2
2
As x approaches 4 from the left or right, x 4 is a
small positive number. So,
f
lim f x
lim f x
x o 4
4
1
x 4
f
x o 4
f.
1
8. f x
x 4
2
2
As x approaches 4 from the left or right, x 4 is a
As x approaches 4 from the left, x 4 is a small
negative number. So,
small positive number. So,
f
x o 4
1
7. f x
f
4
lim f x
f.
lim f x
x o 4
x o 2
lim sec
As x approaches 4 from the right, x 4 is a small
positive number. So,
f
Sx
lim tan
x o 4
lim f x
lim f x
x o 4
x o 4
f.
As x approaches 4 from the right, x 4 is a small
positive number. So,
f
lim f x
x o 4
9. f x
1
x2 9
x
–3.5
–3.1
–3.01
–3.001
2.999
–2.99
–2.9
–2.5
f x
0.308
1.639
16.64
166.6
166.7
16.69
1.695
0.364
lim f x
f
lim f x
f
x o 3
x o 3
2
−6
6
−2
10. f x
x
x2 9
x
–3.5
–3.1
–3.01
f x
1.077
5.082
50.08
lim f x
f
lim f x
f
x o 3
x o 3
–3.001
2.999
–2.99
–2.9
–2.5
500.1
499.9
49.92
4.915
0.9091
2
−6
6
INSTRUCTOR USE ONLY
−2
−2
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NOT FOR SALE
Section 1.5
11. f x
x2
x 9
x
–3.5
–3.1
–3.01
–3.001
2.999
–2.99
–2.9
–2.5
f x
3.769
15.75
150.8
1501
1499
149.3
14.25
2.273
lim f x
f
lim f x
f
IIn
Infinite Limits
87
2
x o 3
x o 3
4
−6
6
−4
12. f x
Sx
cot
3
x
–3.5
–3.1
–3.01
–3.001
2.999
–2.99
–2.9
–2.5
f x
1.7321
9.514
95.49
954.9
954.9
95.49
9.514
1.7321
lim f x
f
lim f x
f
x o 3
x o 3
4
−6
6
−4
1
x2
13. f ( x)
lim
1
x o 0 x
f
2
t 1
t2 1
17. g (t )
Therefore, x
lim
1
x o 0 x
No vertical asymptotes because the denominator
is never zero.
2
0 is a vertical asymptote.
3s 4
3s 4
2
s 16
( s 4)( s 4)
3s 4
3s 4
f and lim 2
lim 2
s o 4 s 16
s o 4 s 16
18. h( s )
14. f ( x)
lim
x o 3
lim
x o 3
2
( x 3)3
2
x 3
3
f
3
f
2
x 3
Therefore, x
Therefore, s
lim
19. f ( x)
x2
lim 2
x o 2 x 4
x2
f and lim 2
x o 2 x 4
lim
Therefore, x
2 is a vertical asymptote.
x2
lim 2
x o 2 x 4
x2
f and lim 2
x o 2 x 4
Therefore, x
2 is a vertical asymptote.
16. f ( x)
x2
2
x 4
3s 4
2
16
f and lim
f
lim
2
x o1 x
g ( x)
lim g ( x)
xo2
f
3
( x 2)( x 1)
f and lim
2
x o 2 x
3
x 2
f
2 is a vertical asymptote.
3
x 2
Therefore, x
20.
No vertical asymptotes because the denominator
is never zero.
3
x 2
Therefore, x
f
3x
x2 9
3
x2 x 2
2
x o 2 x
3s 4
2
16
s o 4 s
4 is a vertical asymptote.
Therefore, s
x2
( x 2)( x 2)
15. f ( x)
4 is a vertical asymptote.
s o 4 s
3 is a vertical asymptote.
f
f and lim
2
x o1 x
3
x 2
f
1 is a vertical asymptote.
x3 8
( x 2)( x 2 2 x 4)
x 2
x 2
x 2 2 x 4, x z 2
4 4 4 12
There are no vertical asymptotes. The graph has a
ole at x
hole
2.
INSTRUCTOR USE ONLY
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88
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
4 x2 x 6
21. f x
x x3 2 x 2 9 x 18
x x 2 x2 9
f and lim f ( x)
x o 0
Therefore, x
t o 2
t o2
f
x o 3
Therefore, x
3 is a vertical asymptote.
Let n be any integer.
4
3( 3 3)
2
9
Therefore, the graph has vertical asymptotes at x
Therefore, the graph has holes at x
2 and x
3.
lim
x o 2 n 1
2
x 1 x 1 x 3
f and lim h( x)
1 is a vertical asymptote.
x o1
Therefore, x
lim h( x)
x o 3
f
x o 1
f and lim h( x)
lim h( x)
x o1
f or f
f ( x)
t
sin t
0 for t
27. s t
sin t
f
Therefore, the graph has a hole at x
nS , where n is an integer.
f or f (for n z 0)
lim s (t )
1 is a vertical asymptote.
3 3
( 3 1)( 3 1)
sinS x
cosS x
2n 1
0 for x
, where n is an integer.
2
Therefore, the graph has vertical asymptotes at
2n 1
x
.
2
x 3
, x z 3
( x 1)( x 1)
Therefore, x
t o nS
Therefore, the graph has vertical asymptotes at
t
nS , for n z 0.
3
4
3.
lim s t
1
t o0
Therefore, the graph has a hole at t
x 2 x 15
x3 5 x 2 x 5
x 5 x 3
2
23. f x
53
52 1
T
cos T
x 3
,x z 5
x2 1
lim f ( x)
tan T
28. g T
x 5 x2 1
x o5
n.
tanS x
26. f x
cosS x
x2 9
x3 3x 2 x 3
x 3 x 3
lim h( x)
f or f
lim f ( x )
xon
x o 3
x o 1
2.
1
sinS x
cscS x
25. f x
2 and lim f ( x)
22. h x
1
16
Therefore, the graph has a hole at t
4
2(2 3)
lim f ( x)
xo2
f
2 is a vertical asymptote.
2
(2 2)(22 4)
lim h(t )
f and lim f ( x)
lim f ( x)
t o 2
Therefore, t
f
0 is a vertical asymptote.
x o 3
f and lim h(t )
lim h(t )
4
, x z 3, 2
x x 3
x o 0
t 2 t 2 t2 4
t
,t z 2
t 2 t2 4
4 x3 x2
lim f ( x)
tt 2
t 2 2t
t 4 16
24. h t
0 for T
lim
T o S nS
g (T )
0.
sin T
T cos T
S
2
nS , where n is an integer.
f or f
2
15
26
Therefore, the graph has vertical asymptotes at
T
There are no vertical asymptotes. The graph has a hole
at x
5.
S
2
nS .
lim g (T )
T o0
1
Therefore,, the ggraph
p has a hole at T
INSTRUCTOR USE ONLY
0.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.5
29. lim
x2 1
lim x 1
x o 1 x 1
2
x o 1
Removable discontinuity at x
36. lim
2
x o 2 x
1
37.
2
−3
x2
4
4
4 4
lim
x2 x 6
x o 3
x 3
x 3 x 2
3
1
lim
x o 3 x 2
−5
30.
38.
x2 2x 8
x 1
x o 1
f
x2 2 x 8
x 1
f
lim
lim
x o 1
1
8
1·
§
39. lim ¨1 ¸
x¹
x o 0 ©
f
1·
§
40. lim ¨ 6 3 ¸
x ¹
x o 0 ©
f
41.
−8
31.
§
lim ¨ x 2 x o 4 ©
x2 1
x o 1 x 1
f
Sx·
§x
42. lim ¨ cot
¸
2 ¹
x o 3 © 3
x2 1
x o 1 x 1
f
43. lim
2
1
lim
44.
8
xo S 2
2
x
45. lim
x oS x 2
cot x
x o 0
46. lim
x o 1
x o 0
47.
lim
xo 1 2 1
48.
2
−3
3
−2
lim
1
34. lim
x o1
0
lim ¬ª x 2 tan xº¼
49. f x
0
f
2
Sx
f
f
x2 x 1
x3 1
lim
x
lim
x o 1 2 cos
lim x 2 tan S x
lim f x
1
x 1
x sin x
xo 1 2 f
x o 1 x 1
f
x sec S x
x2 x 1
x 1 x2 x 1
1
x o1 x 1
x o1
33.
5
8
1
x 1
Removable discontinuity at
x
3x 1
f
lim
x o S csc x
−8
32. lim
3x 1 2 x 1
2x 3 2x 1
f
cos x
3
sin x 1
1
5
f
x o 0 sin x
Vertical asymptote at x
−3
xo 1 2 2 ·
¸
x 4¹
lim
lim
lim
lim
4
− 10
6x2 x 1
2
x o 1 2 4 x 4 x 3
lim
xo 1 2 2x 3
Vertical asymptote at x
89
1
2
x 3
lim
x o 3
IIn
Infinite Limits
f
3
−4
5
−3
x
35. lim
x o 2 x 2
f
INSTRUCTOR USE ONLY
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90
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
x 1 x2 x 1
x3 1
x2 x 1
x2 x 1
lim f x
lim x 1
0
53. A limit in which f x increases or decreases without
50. f x
x o1
bound as x approaches c is called an infinite limit. f is
not a number. Rather, the symbol
x o1
f
lim f x
xoc
4
says how the limit fails to exist.
−8
54. The line x
c is a vertical asymptote if the graph of f
approaches r f as x approaches c.
8
−4
55. One answer is
1
x 2 25
lim f x
f
51. f x
x 3
x 6 x 2
f x
x o 5
1
has no vertical
x2 1
56. No. For example, f x
0.3
x 3
.
x 2 4 x 12
asymptote.
−8
8
y
57.
3
− 0.3
52. f x
sec
lim f x
x o 4
2
Sx
1
x
8
f
−2
−1
1
3
−1
−2
6
m0
58. m
−9
lim m
v o c
−6
59. (a)
1 v2 c2
9
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.0411
0.0067
0.0017
0
0
0
lim
x sin x
x
0
0.5
x o 0
−1.5
lim
v o c
m0
1 v2 c2
f
1.5
−0.25
(b)
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.0823
0.0333
0.0167
0.0017
0
0
lim
x sin x
x2
0
0.25
x o 0
− 1.5
1.5
−0.25
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.5
(c)
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.1646
0.1663
0.1666
0.1667
0.1667
0.1667
IIn
Infinite Limits
91
0.25
− 1.5
1.5
− 0.25
x o 0
lim
x sin x
x3
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.3292
0.8317
1.6658
16.67
166.7
1667.0
(d)
0.1667 1 6
1.5
−1.5
1.5
− 1.5
lim
x o 0
x sin x
x4
or n ! 3, lim
x o 0
60. lim P
V o 0
f
x sin x
xn
f.
f
62. (a) Average speed
As the volume of the gas decreases, the pressure
increases.
27
61. (a) r
625 49
2 15
(b) r
(c)
625 225
lim
x o 25
2x
625 x
2
50
7
ft sec
12
50
50 y 50 x
3
ft sec
2
f
Total distance
Total time
2d
d x d y
2 xy
y x
2 xy
50 x
2 xy 50 y
50 x
2 y x 25
25 x
x 25
y
Domain: x ! 25
(b)
(c)
x
30
40
50
60
y
150
66.667
50
42.857
lim
x o 25
25 x
x 25
f
As x gets close to 25 mi/h, y becomes larger and
larger.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
92
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
1
1
bh r 2T
2
2
63. (a) A
1
1
2
10 10 tan T 10 T
2
2
50 tan T 50 T
§ S·
Domain: ¨ 0, ¸
© 2¹
(b)
T
0.3
0.6
0.9
1.2
1.5
f T
0.47
4.21
18.0
68.6
630.1
100
0
1.5
0
f
lim A
(c)
T o S 2
64. (a) Because the circumference of the motor is half that of the saw arbor, the saw makes 1700 2
850 revolutions per minute.
(b) The direction of rotation is reversed.
§ §S
··
(c) 2 20 cot I 2 10 cot I : straight sections. The angle subtended in each circle is 2S ¨ 2¨ I ¸ ¸
2
¹¹
© ©
So, the length of the belt around the pulleys is 20 S 2I 10 S 2I
S 2I .
30 S 2I .
60 cot I 30 S 2I
Total length
§ S·
Domain: ¨ 0, ¸
© 2¹
(d)
(e)
I
0.3
0.6
0.9
1.2
1.5
L
306.2
217.9
195.9
189.6
188.5
450
2
0
0
60S | 188.5
lim L
(f )
Io S 2 (All the belts are around pulleys.)
(g)
lim L
I o 0
f
65. False. For instance, let
f x
x 1
or
x 1
g x
x
.
x2 1
2
66. True
67. False. The graphs of
y
tan x, y
cot x, y
sec x and y
csc x have
vertical asymptotes.
68. False. Let
f x
­1
° , x z 0
®x
°3, x
0.
¯
The graph of f has a vertical asymptote at x
0, but
INSTRUCTOR USE ONLY
f 0
3.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 1.5
1
and g x
x2
69. Let f x
1
x o 0 x2
1
, and c
x4
f and lim g x
70. Given lim f x
xoc
93
0.
1·
§1
f, but lim ¨ 2 4 ¸
x o 0© x
x ¹
1
x o 0 x4
f and lim
lim
In
Infinite
I
Limits
§ x2 1·
lim ¨
¸
4
xo0
© x ¹
f z 0.
L:
xoc
(1) Difference:
g x . Then lim h x
Let h x
L, and lim ª¬ f x g x º¼
xoc
lim ª f x h x º¼
f, by the Sum Property.
xoc ¬
xoc
(2) Product:
If L ! 0, then for H
L 2 ! 0 there exists G1 ! 0 such that g x L L 2 whenever 0 x c G1.
So, L 2 g x 3L 2. Because lim f x
xoc
f then for M ! 0, there exists G 2 ! 0 such that
f x ! M 2 L whenever x c G 2 . Let G be the smaller of G1 and G 2 . Then for 0 x c G ,
you have f x g x ! M 2 L L 2
f. The proof is similar for L 0.
M . Therefore lim f x g x
xoc
(3) Quotient: Let H ! 0 be given.
There exists G1 ! 0 such that f x ! 3L 2H whenever 0 x c G1 and there exists G 2 ! 0 such that
g x L L 2 whenever 0 x c G 2 . This inequality gives us L 2 g x 3L 2. Let G be the
smaller of G1 and G 2 . Then for 0 x c G , you have
g x
f x
3L 2
3L 2H
Therefore, lim
g x
xoc f
71. Given lim f x
xoc
lim
g x
xoc f
H.
0.
x
f, let g x
1. Then
0 by Theorem 1.15.
x
72. Given lim
x oc f
1
x
1
xoc f x
Then, lim
0. Suppose lim f x exists and equals L.
xoc
lim 1
xoc
lim f x
xoc
1
is defined for all x ! 3.
x 3
Let M ! 0 be given. You need G ! 0 such that
1
f x
! M whenever 3 x 3 G .
x 3
73. f x
Equivalently, x 3 1
L
0.
1
whenever
M
x 3 G , x ! 3.
1
. Then for x ! 3 and
M
1
1
x 3 G,
!
M and so f x ! M .
8
x 3
So take G
This is not possible. So, lim f x does not exist.
xoc
1
1
is defined for all x 5. Let N 0 be given. You need G ! 0 such that f x
N whenever
x 5
x 5
1
1
1
whenever
whenever x 5 G , x 5. Equivalently,
5 G x 5. Equivalently, x 5 !
N
x 5
N
74. f x
1
. Note that G ! 0 because N 0. For x 5 G and
N
1
1
N.
N , and
x 5
x 5
x 5 G , x 5. So take G
x 5,
1
1
!
G
x 5
INSTRUCTOR USE ONLY
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94
Chapter 1
NOT FOR SALE
Limits
its and Their Properties
Review Exercises for Chapter 1
1. Calculus required. Using a graphing utility, you can estimate the length to be 8.3. Or, the length is slightly longer
than the distance between the two points, approximately 8.25.
11
−9
9
−1
9 1
2. Precalculus. L
2
31
2
| 8.25
x 3
x 2 7 x 12
3. f x
x
2.9
2.99
2.999
3
3.001
3.01
3.1
f (x)
–0.9091
–0.9901
–0.9990
?
–1.0010
–1.0101
–1.1111
lim f x | 1.0000 (Actual limit is 1.)
x o3
6
−6
12
−6
x 4 2
x
4. f x
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
0.2516
0.2502
0.2500
?
0.2500
0.2498
0.2485
6. g x
2 x
x 3
lim f x | 0.2500
xo0
Actual limit is 14 .
0.5
−5
5
0
4x x2
x
5. h x
x4 x
x
(a) lim h x
40
(b) lim h x
4 1
xo0
x o 1
4 x, x z 0
(a) lim g x does not exist
4
x o3
5
(b) lim g x
xo0
2 0
0 3
0
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 1
7. lim x 4
1 4
x o1
95
5
Let H ! 0 be given. Choose G
H . Then for 0 x 1 G
H , you have
x 1 H
x 4 5 H
f x L H.
x
8. lim
x o9
9
3
Let H ! 0 be given. You need
x 3 H Ÿ
x 3 H
x 3
x 3 Ÿ x 9 H
Assuming 4 x 16, you can choose G
So, for 0 x 9 G
x 9 5H x 3.
5H .
5H , you have
x 3H
x 3 H
f x L H.
9. lim 1 x 2
xo2
1 22
3
Let H ! 0 be given. You need
1 x 2 3 H Ÿ x 2 4
x 2 x 2 H Ÿ x 2 Assuming 1 x 3, you can choose G
So, for 0 x 2 G
x 2 H
5
H
5
H
5
1
H
x 2
.
, you have
H
x 2
x 2 x 2 H
x2 4 H
4 x2 H
1 x 2 3 H
f x L H.
10. lim 9
x o5
9. Let H ! 0 be given. G can be any positive
number. So, for 0 x 5 G , you have
99 H
13. lim t 2
4 2
14. lim 3 x 3
3
t o4
x o 5
f x L H.
15. lim x 2
2
11. lim x 2
x o 6
( 6) 2
12. lim 5 x 3
xo0
36
16. lim ( x 4)3
xo7
50 3
( 5) 3
6 2
xo6
6
2
(7 4)3
2.45
3
8
2
16
33
27
3
INSTRUCTOR USE ONLY
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96
Chapter 1
Limits
its and Their Properties
4
17. lim
4
4 1
xo4 x 1
x
xo2 x 1
18. lim
4
3
2
2 1
2
xo4
2
41
2
t 2
19. lim 2
t o 2 t 4
21. lim
1
t o 2 t 2
lim
t 2 16
t o4 t 4
2
5
x 3 1
˜
x 4
lim
xo4
x 4
22. lim
lim
xo0
t o4
1
x 3 1
1
2
lim
4 x 2
˜
x
lim
1
4 x 2
4 x 2
4 x 2
1
4
lim
1 x 1
x x 1
4 x 2
x
xo0
8
xo0
ª1 x 1 º¼ 1
23. lim ¬
x o0
x
x o0
1
lim
x o0 x 1
1
24. lim
1 s 1
s o0
s
ª1
lim «
s o 0«
¬
1 s 1
s
˜
1
1
ª1 1 s º¼ 1
lim ¬
s o0 ª
s 1 1 s 1º
¬
¼
25. lim
xo0
26.
§ x ·§ 1 cos x ·
lim ¨
¸¨
¸
x
¹
© sin x ¹©
1 cos x
sin x
xo0
4x
4S 4
lim
x o S 4 tan x
27. lim
1
lim
'x
lim
cos S 'x 1
'x o 0
1 s ª1
¬
'x
1 s 1º
¼
1
2
0
sin S 6 cos 'x cos S 6 sin 'x 1 2
'x
'x o 0
1
'x o 0 2
28. lim
1
lim
S
sin ª¬ S 6 'xº¼ 1 2
'x o 0
1
1 s 1º
»
1 s 1»
¼
s o0
1 0
x 3 1
lim
xo4
t o4
x 3 1
x 3 1
x 3 1
lim
xo4
1
4
(t 4)(t 4)
t 4
lim(t 4)
4 4
20. lim
x 3 1
x 4
lim
'x o 0
˜
cos 'x 1
'x
lim
'x o 0
3 sin 'x
˜
2
'x
0
3
1
2
3
2
cos S cos 'x sin S sin 'x 1
'x
ª cos 'x 1 º
sin 'x º
ª
lim «
sin S
» 'lim
x o 0«
'
'x »¼
x
¬
¬
¼
'x o 0
0 0 1
29. lim ª¬ f x g x º¼
xoc
ª lim f x ºª lim g x º
»«x o c
¬«x o c
¼¬
¼»
( 6) 12
f ( x)
30. lim
x o c g ( x)
6
lim g ( x )
1
2
xoc
31. lim ª¬ f x 2 g x º¼
lim f ( x) 2 lim g ( x)
xoc
3
lim f ( x)
xoc
0
12
32. lim ª¬ f x º¼
xoc
2
xoc
xoc
6 2 12
5
ª lim f ( x )º
¬«x o c
¼»
6
2
2
36
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 1
33. f x
97
2x 9 3
x
1
−1
1
0
The limit appears to be
1
.
3
x
–0.01
–0.001
0
0.001
0.01
f (x)
0.3335
0.3333
?
0.3333
0.331
lim f x | 0.3333
xo0
lim
xo0
34. f x
2x 9 3
˜
x
2x 9 3
2x 9 3
lim
(2 x 9) 9
x o0 xª
º
¬ 2 x 9 3¼
2
lim
xo0
2x 9 3
2
1
3
9 3
¬ª1 x 4 º¼ 1 4
x
35. f x
x 3 125
x 5
100
3
−8
1
−6
−4
0
−3
The limit appears to be 1
16
The limit appears to be 75.
x
–0.01
–0.001
0
0.001
0.01
x
–5.01
–5.001
–5
–4.999
–4.99
f (x)
–0.0627
–0.0625
?
–0.0625
–0.0623
f (x)
75.15
75.015
?
74.985
74.85
lim f x | 0.0625
xo0
1
1
4
4
x
lim
xo0
x
lim f x | 75.000
1
16
x o 5
4 ( x 4)
x o 0 ( x 4)4( x )
lim
x3 125
x o 5 x 5
lim
1
x o 0 ( x 4)4
lim
x 5 x 2 5 x 25
lim
x o 5
x 5
lim x 2 5 x 25
x o 5
5
2
5 5 25
75
1
16
INSTRUCTOR USE ONLY
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98
Chapter 1
NOT FOR SALE
Limits
its and Their Properties
cos x 1
x
36. f x
1
37. v
− 10
lim
−1
The limit appears to be 0.
lim
t o4
–0.01
–0.001
0
0.001
0.01
f (x)
0.005
0.0005
0
–0.0005
–0.005
lim
4.9 t 16
4t
4.9 t 4 t 4
t o4
4t
lim ª
¬ 4.9 t 4 º¼
t o4
lim f x | 0.000
xo0
cos x 1
lim
xo0
x
4 t
2
ª
¬ 4.9 16 250º¼ ª¬4.9t 250º¼
lim
t o4
4 t
10
x
s4 st
t o4
The object is falling at about 39.2 m/sec.
cos x 1 cos x 1
lim
˜
xo0
cos x 1
x
lim
39.2 m/sec
cos 2 x 1
x o 0 x (cos x 1)
lim
sin 2 x
x o 0 x (cos x 1)
§ sin x ·§ sin x ·
lim ¨
¸¨
¸
x ¹© cos x 1 ¹
xo0 ©
§0·
1¨ ¸
© 2¹
0
38. 4.9t 2 250
When a
lim
50
, the velocity is
7
sa st
t oa
50
sec
7
0 Ÿ t
a t
ª4.9a 2 250¼º ¬ª4.9t 2 250¼º
lim ¬
t oa
a t
4.9 t 2 a 2
lim
t oa
a t
4.9 t a t a
lim
t oa
a t
lim ª
4.9
t a º¼
¬
t oa
4.9 2a
§
¨a
©
50 ·
¸
7¹
70 m/sec.
The velocity of the object when it hits the ground is about 70 m/sec.
39. lim
1
x o 3 x 3
40. lim
x o 6 x
x 6
2
36
1
33
lim
1
6
41. lim
x o 4
x 2
x 4
lim
x o 4
x 6
lim
x 2
x 2
˜
x 4
x 2
x 4
x o 4 ( x 4)
x o 6 ( x 6)( x 6)
1
x o 6 x 6
1
12
lim
lim
x o 4
x 2
1
x 2
1
4
x 3
x 3
1
INSTRUCTOR USE ONLY
42. lim
x o 3
x 3
lim
x o 3
x 3
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 1
43. lim f x
0
xo2
11
44. lim g x
x o1
2
11
t o1
t o1
1 1 1
2
and lim h t
t o1
1.
21 1
x o 2
x o1
0 because
x 3
x 2 3 x 18
x 3
( x 3)( x 6)
3
48. lim a x 1b does not exist. There is a break in the graph
1
, x z 3
x 6
xo4
at x
x( x 1)
and has a removable discontinuity at x
1
1.
lim f ( x)
lim
xo0
x o 0 ( x 1)( x 1)
54. f ( x)
47. lim 2a xb 1
1
,x z 0
( x 1)( x 1)
2
x o 1
2
2
s o 2
x
3
has nonremovable discontinuities at x
r1
because lim f ( x) and lim f ( x) do not exist,
45. lim h t does not exist because lim h t
46. lim f s
x
x x
53. f ( x)
4.
49. f ( x)
x 2 4 is continuous for all real x.
has a nonremovable discontinuity at x
6
because lim f ( x) does not exist, and has a
50. f ( x)
x x 20 is continuous for all real x.
removable discontinuity at x
xo6
2
lim f ( x )
4
51. f ( x)
has a nonremovable discontinuity at
x 5
x
5 because lim f x does not exist.
1
x2 9
lim
55. f 2
x o 2
5
r3
2c
1
because lim f ( x ) and lim f ( x) do not exist.
c
x o 3
56. lim x 1
2
lim x 1
4
x o1
x o 3
Find b and c so that lim x 2 bx c
2 and lim x 2 bx c
x o1
Consequently you get
1 b c
Solving simultaneously,
b
57. f x
x o 3
2
and 9 3b c
3 and
4x 1 x 2
4 x2 7 x 2
x 2
x 2
Continuous on (f, 2) ‰ ( 2, f). There is a
58. f ( x)
x 4
Continuous on >4, f
1
2
4.
4.
4.
60. f x
Continuous on f, f
59. f ( x)
c
5.
ax 3b
lim a x 3b k 3 where k is an integer.
x o k
3 x 2 7
removable discontinuity at x
1
.
9
5
c2 6
x o3
3 because
Find c so that lim cx 6
1
( x 3)( x 3)
has nonremovable discontinuities at x
1
x o 3 x 6
x o 3
x o5
52. f ( x)
99
2.
lim a x 3b
x o k
k 2 where k is an integer.
Nonremovable discontinuity at each integer k
Continuous on k , k 1 for all integers k
3x 2 x 1
3x 2 x 2
x 1
x 1
lim f x
lim 3 x 2
5
61. f x
x o1
x o1
Removable discontinuity at x
1
Continuous on f, 1 ‰ 1, f
INSTRUCTOR USE ONLY
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100
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
­5 x, x d 2
®
¯2 x 3, x ! 2
62. f x
lim 5 x
3
x o 2
lim 2 x 3
x o 2
2
x o 3 x
1
2
63. f is continuous on >1, 2@. f 1
1 0 and
13 ! 0. Therefore by the Intermediate Value
64. C x
Therefore, x
3 is a vertical asymptote.
Therefore, x
12.80 2.50 ª¬a xb 1º¼ ,
x ! 0
lim
5
x 4
x 2
lim f x
4
(b) lim f x
4
(a)
x o 2
x o 2
Therefore, x
lim
(b) lim f x
(c) lim f x
lim
lim
3
lim
x2 2 x 1
x 1
x o1
74.
0 is a vertical asymptote.
5
4
f
lim
4
f
2
x o 1 x
4
x o 1 x
lim
x 1
1
x o 1
1·
§
77. lim ¨ x 3 ¸
x ¹
x o 0 ©
f
3
x o 1 x
5
x o 2 ( x 2)
x
lim
x o 1 2 2x 1
x 1
1
76.
5
( x 2) 4
x o 2 ( x 2)
n.
f
lim
75.
f
Therefore, x
f
8 is a vertical asymptote.
73. lim
f
3
2x 1
64
2
x o 8 x
Therefore, the graph has vertical asymptotes at x
3
x
x o 0 x
68. f x
xon
0
x o1
x o 0 x
f and lim
0
x o 0
67. f x
8 is a vertical asymptote.
f or f
lim f x
(a) Domain: f, 0@ ‰ >1, f
f
1
sin S x
0 for x
n, where n is an integer.
sin S x
x 1x
2x 1
64
2
x o 8 x
csc S x
72. f x
xo2
f
2x 1
( x 8)( x 8)
f and lim
2x 1
64
Therefore, x
x2
6 is a vertical asymptote.
2
x o 8 x
(c) lim f x does not exist.
66. f x
x o 6 36 2
x o 8 x
ª x 2º
x 2«
»
«¬ x 2 »¼
f
x2
6x
f and lim
2x 1
64
lim
6x
6 is a vertical asymptote.
2x 1
x 2 64
71. g x
6x
( x 6)( x 6)
x o 6 36 x2
Therefore, x
f
2
f and lim
2
6x
x o 6 36 10
65. f x
6x
x ! 0
2
x3
x o 3 x 9
6x
36 x 2
12.80 2.50 ª¬a xb 1º¼ ,
0
f
f and lim
lim
C has a nonremovable
discontinuity at each
integer 1, 2, 3,!.
x3
9
x3
x o 3 x 9
x o 6 36 x
25
2
x o 3 x
2
70. f x
0.
f and lim
3 is a vertical asymptote.
Theorem, there is at least one value c in 1, 2 such that
2c 3 3
x3
9
Therefore, x
lim
Continuous on f, 2 ‰ 2, f
x3
( x 3)( x 3)
2
lim
Nonremovable discontinuity at x
f 2
x3
x 9
69. f x
78. lim
x o 2 3
1
x 4
2
lim
1
x 1
lim
1
x2 1 x 1
1
3
1
4
f
INSTRUCTOR USE ONLY
Therefore, x
2 is a vertical asymptote.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffor Chapter 1
sin 4 x
x o 0 5 x
ª 4 § sin 4 x ·º
lim « ¨
¸»
x o 0 ¬ 5 © 4 x ¹¼
sec x
x
x o 0
f
csc 2 x
x
xo0
1
x o 0 x sin 2 x
79. lim
80. lim
81. lim
4
5
82. lim
cos 2 x
x
83. C
80,000 p
, 0 d p 100
100 p
x o 0
f
(a) C 15 | $14,117.65
f
lim
(b) C 50
$80.000
(c) C 90
$720,000
(d)
lim
80,000 p
p
p o100 100 f
tan 2x
x
84. f x
(a)
101
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f(x)
2.0271
2.0003
2.0000
2.0000
2.0003
2.0271
lim
xo0
tan 2 x
x
2
­ tan 2 x
, x z 0
°
.
® x
°2,
x
0
¯
(b) Yes, define f x
Now f x is continuous at x
0.
Problem Solving for Chapter 1
1. (a) Perimeter 'PAO
x2 y 1
2
x2 x2 1
Perimeter 'PBO
(b) r x
x2 x2 1
x 1
(c)
2
2
x2 y2 1
x2 x4 1
x 1
2
y2 x2 y2 1
x 1
2
x4 x2 x4 1
2
x2 x4 1
x4 x2 x4 1
x
4
2
1
0.1
0.01
Perimeter 'PAO
33.02
9.08
3.41
2.10
2.01
Perimeter 'PBO
33.77
9.60
3.41
2.00
2.00
r x
0.98
0.95
1
1.05
1.005
lim r x
x o 0
1 01
1 01
2
2
1
INSTRUCTOR USE ONLY
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102
Chapter 1
NOT FOR SALE
Limits
its and Their Properties
x
2
y
2
1
bh
2
1
bh
2
1
1 x
2
1
1 y
2
(b) a x
Area 'PBO
Area 'PAO
x2 2
x 2
x
4
2
1
0.1
0.01
Area 'PAO
2
1
12
1 20
1 200
Area 'PBO
8
2
12
1 200
1 20,000
a x
4
2
1
1 10
1 100
lim x
0
2. (a) Area 'PAO
Area 'PBO
(c)
lim a x
x o 0
x o 0
x2
2
x
S 3. So,
3. (a) There are 6 triangles, each with a central angle of 60q
Sº
ª1
6« 1 sin »
3¼
¬2
ª1 º
6« bh»
¬2 ¼
Area hexagon
h = sin θ
h = sin 60°
1
1
θ
60°
Error
3 3
| 2.598.
2
Area (Circle) Area (Hexagon)
S 3 3
| 0.5435
2
(b) There are n triangles, each with central angle of T
(c)
n sin 2S n
2S º
ª1
n « 1 sin »
n¼
¬2
2S n. So,
An
ª1 º
n « bh»
¬2 ¼
n
6
12
24
48
96
An
2.598
3
3.106
3.133
3.139
2
.
(d) As n gets larger and larger, 2S n approaches 0. Letting x
which approaches 1 S
4. (a) Slope
(b) Slope
40
30
4
3
mx
x,
25 x 2 4
x 3
sin 2S n
2n
2S n
(d) lim mx
x o3
3
Tangent line: y 4
4
x, y
sin 2S n
S
sin x
S
x
S.
y
(c) Let Q
2S n, An
25 x 2
3
x 3
4
3
25
x 4
4
lim
x o3
lim
x o3
lim
x o3
lim
x o3
25 x 2 4
˜
x 3
25 x 2 4
25 x 2 4
25 x 2 16
x 3
25 x 2 4
3 x 3 x
x 3
25 x 2 4
3 x
25 x 4
2
6
4 4
3
4
This is the slope of the tangent line at P.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffor Chapter 1
5. (a) Slope
12
5
(d) lim mx
lim
x o5
xo5
x 5 12 xo5
169 x 2 12
x 5
a bx x
6.
Letting a
x 5
12 169 x 2
xo0
b
3 Setting
10
12 12
5
12
3
˜
a bx a bx 3
3
a bx 3
x
a bx 3
lim
b
3 bx 3
3 simplifies the numerator.
3 bx x
So, lim
169 x 2
This is the same slope as part (b).
a bx x
3
169 x 2
x 5 12 xo5
mx
169 x 2
x 2 25
lim
lim
169 x 2
144 169 x 2
lim
x, 169 x 2
x, y
169 x 2 12 ˜
x 5
12 xo5
5
(b) Slope of tangent line is .
12
5
y 12
x 5
12
5
169
y
x Tangent line
12
12
(c) Q
12 103
3
bx
lim
xo0 x
3 bx 3, you obtain b
3
xo0
3
6. So, a
7. (a) 3 x1 3 t 0
3 and b
6.
(d) lim f x
x o1
x1 3 t 3
x t 27
(b)
x o1
lim
x o1
Domain: x t 27, x z 1 or >27, 1 ‰ 1, f
lim
0.5
3 x1 3 2
˜
x 1
lim
x o1
x o1
12
3 x1 3 2
3 x1 3 4
x 1
3 x1 3 2
x1 3 1
x1 3 1 x 2 3 x1 3 1
3 x1 3 2
1
lim
−30
3 x1 3 2
x
23
x
13
1
3 x1 3 2
−0.1
(c)
3 27
lim f x
13
1
111 2 2
2
1
12
27 1
x o 27 2
28
1
14
| 0.0714
8. lim f x
x o 0
lim f x
x o 0
lim a 2 2
x o 0
ax
lim
x o 0 tan x
Thus,
a2 2
a
a a 2
0
a 2 a 1
0
2
a2 2
tan x
§
a¨ because lim
x
o
0
x
©
9. (a) lim f x
xo2
·
1¸
¹
3: g1 , g 4
(b) f continuous at 2: g1
(c)
lim f x
x o 2
3: g1 , g3 , g 4
INSTRUCTOR USE ONLY
a
1, 2
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104
NOT FOR SALE
Chapter 1
Limits
its and Their Properties
y
10.
(a) f 14
a4b
4
f 3
c1f
de 3 gh
0
f 1
a1b
1
3
2
1
x
−1
1
−1
1
lim f x
0
lim f x
f
lim f x
f
x o1
x o1
x o 0
−2
x o 0
y
11.
(a)
f 1
a1b a1b
f 0
0
4
3
(c) f is continuous for
all real numbers except
x
0, r1, r 12 , r 13 , !
(b) lim f x
1 1
0
(b) lim f x
1
lim f x
1
lim f x
1
x o1
x o1
2
1
2
0 1
1
f 2.7
3 2
1
f
1
x
− 4 − 3 − 2 −1
1
2
3
4
−2
x o1 2
(c) f is continuous for
all real numbers except
x
0, r1, r 2, r 3, !
−3
−4
192,000
v0 2 48
r
v2
12. (a)
192,000
r
v v0 48
r
192,000
v 2 v0 2 48
2
Let v0
v
48
v v0 2.17
r
1920
v 2 v0 2 2.17
vo0
r
10,600
2
v v0 2 6.99
vo0
Let v0
0
(iii) lim Pa , b x
0
(iv) lim Pa , b x
1
x o a
x o b
x o b
(c) Pa , b is continuous for all positive real numbers
except x
1920
2.17 v0 2
2.17 mi sec
lim r
(ii) lim Pa , b x
2
Let v0
(c)
1
x o a
1920
v0 2 2.17
r
2
b
(b) (i) lim Pa , b x
4 3 mi sec.
1920
r
lim r
x
a
192,000
48 v0 2
lim r
(b)
2
2
1
vo0
2
y
13. (a)
a, b.
(d) The area under the graph of U, and above the x-axis,
is 1.
| 1.47 mi/sec .
14. Let a z 0 and let H ! 0 be given. There exists
G1 ! 0 such that if 0 x 0 G1 then
f x L H . Let G
10,600
6.99 v0 2
0 x 0 G
x 6.99 | 2.64 mi sec.
Because this is smaller than the escape velocity for
Earth, the mass is less.
G1 a . Then for
G1 a , you have
G1
a
ax G1
f ax L H .
As a counterexample, let
­1, x z 0
.
a
0 and f x
®
0
¯2, x
Then lim f x
1
xo0
L, but
INSTRUCTOR USE ONLY
lim
im f ax
xo0
lim f 0
xo0
lim 2
xo0
2.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
C H A P T E R 2
Differentiation
Section 2.1
The Derivative and the Tangent Line Problem.................................106
Section 2.2
Basic Differentiation Rules and Rates of Change.............................122
Section 2.3
Product and Quotient Rules and Higher-Order Derivatives.............135
Section 2.4
The Chain Rule...................................................................................150
Section 2.5
Implicit Differentiation.......................................................................163
Section 2.6
Related Rates ......................................................................................176
Review Exercises ........................................................................................................188
Problem Solving .........................................................................................................195
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
C H A P T E R
Differentiation
2
Section 2.1 The Derivative and the Tangent Line Problem
1. At x1 , y1 , slope
0.
At x2 , y2 , slope
5
.
2
2. At x1 , y1 , slope
2.
3
8. Slope at 3, 4
'x
lim ( 6 'x )
'x o 0
6
(4, 5)
f (4) − f (1) = 3
9. Slope at 0, 0
f 0 't f 0
lim
(1, 2)
lim
x
(c) y
2
3
4
f 4 f 1
4 1
3
x 1 2
3
1x 1 2
5
lim 3 't
10. Slope at 1, 5
So,
2
4(1 't ) 5
't
3 5 x is a line. Slope
5
6. g x
3
x 1 is a line. Slope
2
3
2
lim
lim
't o 0
lim
9 5
4 4 'x 'x
'x o 0
lim 4 'x
7
fc x
lim
f x 'x f x
'x
'x o 0
7 7
lim
'x o 0
'x
lim 0
0
gc x
'x
'x o 0
f x
12. g x
'x
2
11.
'x o 0
g 2 'x g 2
2 'x
4 4('t ) 5
6('t ) ('t ) 2
lim
't o 0
't
lim (6 't ) 6
This slope is steeper than the slope of the line
f 4 f 1
fc1.
through 1, 2 and 4, 5 . So,
4 1
5. f x
2
't
't o 0
1
5 4.75
0.25
1
f 4 f 3
!
.
43
'x o 0
1 2 't 't
lim
(b) The slope of the tangent line at 1, 2 equals f c 1 .
7. Slope at 2, 5
1 't
lim
5 2
3
|
43
f 4 f 1
4 1
't
't o 0
f 4 f 1
4 1
f 4 f 3
h 1 't h 1
lim
't o 0
x 1
4. (a)
3
't o 0
x 1 f 1
0
't
't o 0
6
2
3 't 't
1
1
't
't o 0
f(1) = 2
2
6
'x o 0
f (4) = 5
3
4
2
6 'x 'x
lim
2
'x
'x o 0
4
( 4)
5 9 6 'x 'x
lim
y
5
2
'x
'x o 0
f (4) − f(1)
y=
(x − 1) + f(1) = x + 1
4−1
3. (a), (b)
'x
5 3 'x
lim
52 .
At x2 , y2 , slope
f 3 'x f 3
lim
'x o 0
2
4
3
lim
'x o 0
lim
g x 'x g x
'x
3 3
'x o 0
0
lim
'x o 0 'x
'x
0
'x
4
INSTRUCTOR USE ONLY
'x o 0
106
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ection 2.1
13.
10 x
f x
fc x
15. h s
f x 'x f x
lim
'x o 0
lim
The Derivative and the Tangent L
Line Problem
'x
10 x 'x 10 x
hc s
'x
10 x 10'x 10 x
lim
'x o 0
'x
10'x
lim
'x o 0
'x
10
lim 10
'x o 0
'x o 0
14.
7x 3
f x
fc x
lim
'x o 0
lim
'x
7 x 'x 3 7 x 3
16.
'x
7 x 7 'x 3 7 x 3
lim
'x o 0
'x
7 'x
lim
'x o 0 'x
lim 7
7
'x o 0
f x
fc x
lim
2
s
3
h s 's h s
's
2
2 ·
§
3 s 's ¨ 3 s ¸
3
3 ¹
©
lim
's o 0
's
2
2
2
3 s 's 3 s
3
3
3
lim
's o 0
's
2
's
2
lim 3
's o 0 's
3
's o 0
f x 'x f x
'x o 0
17.
3
f x
fc x
107
5
lim
2
x
3
f x 'x f x
'x
'x o 0
2
2 ·
§
x 'x ¨ 5 x ¸
3
3 ¹
©
lim
'x o 0
'x
2
2
2
5 x 'x 5 x
3
3
3
lim
'x o 0
'x
2
'x
lim 3
'x o 0
'x
2
§ 2·
lim ¨ ¸
'x o 0 © 3 ¹
3
5
x2 x 3
lim
f x 'x f x
'x
'x o 0
lim
x 'x
2
x 'x 3 x 2 x 3
'x
'x o 0
x 2 x 'x 'x
2
lim
2
'x
'x o 0
lim
x 'x 3 x 2 x 3
2 x 'x 'x
2
'x
'x
lim 2 x 'x 1
'x o 0
2x 1
'x o 0
18.
f x
fc x
x2 5
lim
f x 'x f x
'x
'x o 0
lim
x 'x
2
5 x2 5
'x
'x o 0
x 2 x 'x 'x
2
lim
5 x2 5
'x
'x o 0
lim
2
2 x 'x 'x
'x
lim 2 x 'x
2
'x o 0
INSTRUCTOR USE ONLY
'x o 0
2x
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108
Chapter 2
19.
f x
fc x
NOT FOR SALE
Differentiation
ferentiation
x3 12 x
f x 'x f x
lim
'x
'x o 0
ª x 'x 3 12 x 'x º ª x3 12 xº
¼
¼ ¬
lim ¬
'x o 0
'x
lim
2
x3 3 x 2 'x 3 x 'x
'x
3 x 'x 3x 'x
2
2
'x
lim 3 x 3 x 'x 'x
2
2
fc x
12 'x
12
'x o 0
f x
3
'x
'x o 0
20.
12 x 12 'x x3 12 x
'x
'x o 0
lim
3
3 x 2 12
x3 x 2
lim
f x 'x f x
'x
'x o 0
ª x 'x 3 x 'x 2 º ª x3 x 2 º
¼
¼ ¬
lim ¬
'x o 0
'x
lim
2
x3 3x 2 'x 3 x 'x
'x
3 x 'x 3 x 'x
2
2
'x
lim 3x 3x 'x 'x
2
2
'x o 0
f x
fc x
f x 'x f x
'x
lim
'x o 0 'x x 'x 1 x 1
x3 x 2
2
3x2 2 x
1
x 'x 1 x 1
1
x2
lim
2
f x 'x f x
'x
'x o 0
1
lim
x 'x
2
x x 'x
2
lim
'x o 0
lim
'x o 0
lim
1
x2
'x
'x o 0
'x o 0
1
x 1
f x
fc x
'x
1
1
'
1
x
x
x
1
lim
'x o 0
'x
x 1 x 'x 1
lim
'x o 0 'x x 'x 1 x 1
'x o 0
2 x ' x 'x
22.
'x o 0
lim
3
2 x 'x
1
x 1
lim
2
'x
'x o 0
21.
x 2 2 x 'x ' x
'x
'x o 0
lim
3
2
2
'x x 'x x 2
2 x 'x 'x
2
2
'x x 'x x 2 x 'x
2
x 'x x 2
2 x
x4
2
3
x
INSTRUCTOR USE ONLY
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copied or duplicated, or posted to a publicly accessible website,
website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ection 2.1
23.
The Derivative and the Tangent L
Line Problem
109
x 4
f x
fc x
f x 'x f x
lim
'x o 0
'x
lim
x 'x 4 'x
'x o 0
x 4
§
˜ ¨¨
©
x 'x 4 x 'x 4 x 4·
¸
x 4 ¸¹
x 'x 4 x 4
lim
'x o 0 'x ª
º
¬ x 'x 4 x 4 ¼
1
1
x 'x 4 x 4
x 4 lim
'x o 0
24.
x 4
2
1
x 4
4
x
f x
fc x
f x 'x f x
lim
'x
4
x 'x
'x
'x o 0
lim
'x o 0
4
lim
'x o 0
4
x
x 4 x 'x §
˜ ¨¨
'x x x 'x
©
x x x 'x ·
¸
x 'x ¸¹
4 x 4 x 'x
lim
'x o 0 'x
x
x 'x
x x 'x
4
lim
'x o 0
x 'x
x
x 4
x
x x
x
x
x 'x
2
x
x2 3
25. (a) f ( x)
fc x
lim
(b)
f x 'x f x
'x
'x o 0
(− 1, 4)
ª x 'x 2 3º x 2 3
¼
lim ¬
'x o 0
'x
lim
x 2 2 x'x 'x
2
3 x2 3
'x
'x o 0
lim
8
2 x'x 'x
'x
lim 2 x 'x
−3
3
−1
(c) Graphing utility confirms
dy
dx
2 at 1, 4 .
2
'x o 0
'x o 0
2x
At 1, 4 , the slope of the tangent line is
m
2 1
2.
The equation of the tangent line is
y 4
2 x 1
y 4
2x 2
y
2x 2
INSTRUCTOR USE ONLY
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110
Chapter 2
NOT FOR SALE
Differentiation
ferentiation
x2 2 x 1
26. (a) f ( x)
fc x
f x 'x f x
lim
'x
'x o 0
ª x 'x 2 2 x 'x 1º ª x 2 2 x 1º
¼
¼ ¬
lim ¬
'x o 0
'x
ª x 2 2 x'x 'x 2 2 x 2'x 1º ª x 2 2 x 1º
¼
¬
¼ ¬
lim
'x o 0
'x
2 x'x 'x
lim
2
2'x
'x
lim 2 x 'x 2
'x o 0
2x 2
'x o 0
21 2
At 1, 2 , the slope of the tangent line is m
4.
The equation of the tangent line is
y 2
4 x 1
y 2
4x 4
y
4 x 2.
(b)
8
(1, 2)
− 10
8
−4
(c) Graphing utility confirms
27. (a)
dy
dx
4 at 1, 2 .
x3
f x
fc x
lim
f x 'x f x
'x
'x o 0
lim
x 'x
x3
'x
'x o 0
lim
3
3x 'x 3 x 'x
2
3
2
'x
3
'x
'x o 0
lim 3 x 3 x 'x 'x
2
2
'x o 0
At 2, 8 , the slope of the tangent is m
y 8
12 x 2
y 8
12 x 24
y
12 x 16.
(b)
'x
'x
'x o 0
lim
2
x3 3 x 2 'x 3x 'x
3x 2
32
2
12. The equation of the tangent line is
10
(2, 8)
−5
5
−4
(c) Graphing utility confirms
dy
dx
12 at 2, 8 .
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ection 2.1
28. (a)
The Derivative and the Tangent L
Line Problem
111
x3 1
f x
fc x
lim
f x 'x f x
'x
'x o 0
ª x 'x 3 1º x3 1
¼
lim ¬
'x o 0
'x
lim
x3 3 x 2 'x 3x 'x
2
'x
1 x3 1
'x
'x o 0
lim ª3 x 2 3x 'x 'x º
¼
2
3x 2
'x o 0 ¬
At 1, 0 , the slope of the tangent line is m
y 0
3x 1
y
3x 3.
(b)
3
3 1
2
3. The equation of the tangent line is
6
(−1, 0)
−9
9
−6
(c) Graphing utility confirms
29. (a)
f x
dy
dx
3 at 1, 0 .
x
fc x
lim
f x 'x f x
'x
'x o 0
lim
'x o 0
x 'x 'x
lim
x 'x x 'x ˜
x
x
x 'x x
lim
'x o 0 'x
'x o 0
x
x 'x 1
x 'x x
1
x
2
x
At 1, 1 , the slope of the tangent line is m
1
2 1
1
.
2
The equation of the tangent line is
y 1
y 1
y
(b)
1
x 1
2
1
1
x 2
2
1
1
x .
2
2
3
(1, 1)
−1
5
−1
(c) Graphing utility confirms
dy
dx
1
at 1, 1 .
2
INSTRUCTOR USE ONLY
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112
Chapter 2
30. (a)
NOT FOR SALE
Differentiation
ferentiation
x 1
f x
fc x
lim
(b)
4
(5, 2)
f x 'x f x
−2
'x o 0
'x
lim
x 'x 1 'x
'x o 0
x 1
§
˜ ¨¨
©
x 'x 1 x 'x 1 x 1·
¸
x 1 ¸¹
10
−4
x 'x 1 x 1
lim
'x o 0 'x
x 'x 1 1
x 'x 1 lim
'x o 0
x 1
x 1
(c) Graphing utility confirms
2
1
x 1
dy
dx
1
2 51
At 5, 2 , the slope of the tangent line is m
1
at 5, 2 .
4
1
.
4
The equation of the tangent line is
1
y 2
x 5
4
1
5
y 2
x 4
4
1
3
y
x .
4
4
31. (a)
f x
fc x
x lim
4
x
f x 'x f x
(b)
− 12
'x
'x o 0
4
4·
§
¨x ¸
x 'x ©
x¹
lim
'x o 0
'x
x x 'x x 'x 4 x x 2 x 'x 4 x 'x
lim
'x o 0
x 'x x 'x
x3 2 x 2 'x x 'x
2
− 10
(c) Graphing utility confirms
dy
3
at 4, 5 .
dx
4
x3 x 2 'x 4 'x
x 'x x 'x
'x o 0
x 'x x 'x
2
lim
12
(− 4, − 5)
x 'x lim
6
2
4 'x
x 'x x 'x
'x o 0
x x 'x 4
2
lim
'x o 0
x 4
x2
2
x x 'x
1
4
x2
At 4, 5 , the slope of the tangent line is m
1
4
4
2
3
.
4
The equation of the tangent line is
3
y 5
x 4
4
3
y 5
x 3
4
3
y
x 2.
4
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ection 2.1
32. (a)
6
x 2
f x 'x f x
x f x
fc x
lim
6
6
x 'x 2
x 2
lim
'x o 0
lim
113
34. Using the limit definition of derivative, f c x
4 x.
Because the slope of the given line is –4, you have
'x
'x o 0
The Derivative and the Tangent L
Line Problem
4x
4
x
1.
At the point 1, 2 the tangent line is parallel to
4x y 3
'x
6 x 12 6 x 'x 2
y 2
4 x 1
y
4 x 2.
'x x 'x 2 x 2
'x o 0
6 x 12 6 x 6'x 12
lim
'x o 0 'x x 'x 2 x 2
35. From Exercise 27 we know that f c x
6'x
lim
x 2
3x 2
3
x
r1.
6
2
Therefore, at the points 1, 1 and 1, 1 the tangent
At 0, 3 , the slope of the tangent line is
64
m
32 .
lines are parallel to 3x y 1
These lines have equations
0.
y 1
3 x 1 and y 1
3x 1
3x 2
3 x 2.
The equation of the tangent line is
y 3
y 3
y
y
3
x 0
2
3
x
2
3
x 3.
2
(b)
3x 2
x
8
(c) Graphing utility confirms
dy
dx
3
at 0, 3 .
2
Because the slope of the given line is 2, you have
2
1Ÿ x
0. These lines have
y 3
3 x 1 and y 1
3x 1
3x
3x 4.
2 x.
0. The equation of this line is
y
37. Using the limit definition of derivative,
1
fc x
.
2x x
1
Because the slope of the given line is , you have
2
1
1
2
2x x
x
At the point 1, 1 the tangent line is parallel to
2x y 1
r1.
lines are parallel to 3x y 4
equations
y
33. Using the limit definition of derivative, f c x
1
3
2
Therefore, at the points 1, 3 and 1, 1 the tangent
−6
x
3x 2 .
Because the slope of the given line is 3, you have
(0, 3)
2x
y
36. Using the limit definition of derivative, f c x
6
− 10
3x 2 .
Because the slope of the given line is 3, you have
'x o 0 'x x 'x 2
x 2
0. The equation of this line is
1.
Therefore, at the point 1, 1 the tangent line is parallel to
y 1
2 x 1
x 2y 6
y
2 x 1.
y 1
y 1
y
0. The equation of this line is
1
x 1
2
1
1
x 2
2
1
3
x .
2
2
INSTRUCTOR USE ONLY
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114
Chapter 2
Differentiation
ferentiation
38. Using the limit definition of derivative,
1
fc x
.
32
2 x 1
1
Because the slope of the given line is , you have
2
1
1
32
2
2 x 1
y
3
2
f′
1
x
1
2
3
x 1
1
x 1Ÿ x
0. The equation of the tangent line is
6
43. The slope of the graph of f is negative for x 0 and
positive for x ! 0. The slope is undefined at x
0.
y
1
x 2
2
1
x 2.
2
y
5
2.
At the point 2, 1 , the tangent line is parallel to
y 1
4
32
1
x 2y 7
42. The slope of the graph of f is –1 for x 4, 1 for
4.
x ! 4, and undefined at x
2
1
f′
x
−2 −1
1
2
3
4
39. The slope of the graph of f is 1 for all x-values.
y
−2
4
44. The slope is positive for 2 x 0 and negative for
0 x 2. The slope is undefined at x
r 2, and 0 at
3
2
f′
x
x
−3 −2 −1
−1
1
0.
3
2
y
−2
f′
2
1
40. The slope of the graph of f is 0 for all x-values.
x
−2
y
−1
1
2
−1
2
−2
1
f′
−2
−1
1
x
45. Answers will vary.
Sample answer: y
2
−1
−2
x
y
4
41. The slope of the graph of f is negative for
4.
x 4, positive for x ! 4, and 0 at x
3
2
1
x
y
−4 −3 −2 −1
−1
4
f′
−6
4
−3
−4
x
−4
3
−2
2
−6 −4 −2
−2
2
2
4
6
46. Answers will vary.
Sample answer: y
−8
x
y
4
3
2
1
x
−4 −3 −2
1
2
3
4
−2
−3
INSTRUCTOR USE ONLY
−4
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ection 2.1
47. g 4
5 because the tangent line passes through 4, 5 .
50
47
gc 4
48. h 1
The Derivative and the Tangent L
Line Problem
55. Let x0 , y0 be a point of tangency on the graph of f.
By the limit definition for the derivative,
fc x
4 2 x. The slope of the line through 2, 5 and
5
3
4 because the tangent line passes through
x0 , y0 equals the derivative of f at x0 :
1, 4 .
64
3 1
hc 1
2
4
1
2
5 3 x and c
49. f x
x and c
2
51. f x
x 2 and c
6
52. f x
2
x and c
2 x0 4 2 x0
2
8 8 x0 2 x0 2
0
x0 2 4 x0 3
0
x0 1 x0 3 Ÿ x0
1, 3
3, 3 , and the corresponding slopes are 2 and –2. The
equations of the tangent lines are:
y 5
2 x 2
y 5
2 x 2
y
2x 1
y
2 x 9
3, f x f
y
7
6
(2, 5)
5
4
y
3
2
(3, 3)
(1, 3)
1
2
1
−3 −2 −1
5 y0
Therefore, the points of tangency are 1, 3 and
3 x 2
f x
4 2 x0
5 4 x0 x0
9
2 and f c x
53. f 0
5 y0
2 x0
1
50. f x
3
115
x
−2
1
2
3
6
x
2
−1
3
56. Let x0 , y0 be a point of tangency on the graph of f. By
−2
−3
54. f 0
the limit definition for the derivative, f c x
f
4, f c 0
0; f c x 0 for x 0, f c x ! 0
for x ! 0
Answers will vary: Sample answer: f x
slope of the line through 1, 3 and x0 , y0 equals the
derivative of f at x0 :
x2 4
y
f
12
10
8
6
2 x. The
3 y0
1 x0
2 x0
3 y0
1 x0 2 x0
3 x0
2 x0 2 x0 2
2
x0 2 2 x0 3
0
x0 3 x0 1
0 Ÿ x0
3, 1
4
Therefore, the points of tangency are 3, 9 and
2
x
−6 −4 −2
2
4
6
1, 1 , and the corresponding slopes are 6 and –2. The
equations of the tangent lines are:
y 3
6 x 1
y 3
2 x 1
y
6x 9
y
2 x 1
y
10
(3, 9)
8
6
4
(−1, 1)
−8 −6 −4 −2
−2
x
2
4
6
(1, −3)
INSTRUCTOR USE
S ONLY
−4
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© Cengage Learning. All Rights Reserved.
116
NOT FOR SALE
Chapter 2
Differentiation
ferentiation
x2
f x
57. (a)
fc x
lim
f x 'x f x
2
x 'x
lim
3
x2
x3 3 x 2 'x 3 x 'x
'x o 0
'x
x 2 2 x 'x 'x
2
x2
'x
lim
'x 2 x 'x
'x
lim 2 x 'x
'x 3 x 2 3x 'x 'x
lim 3 x 3 x 'x 'x
2
'x o 0
2x
'x o 0
At x
1, f c 1
2 and the tangent line is
y 1
2 x 1
or
At x
0, f c 0
0 and the tangent line is y
At x
1, f c 1
2 and the tangent line is
y
2 x 1.
3
2
3x 2
1, g c 1
3 and the tangent line is
y 1
3x 1
or
At x
0, g c 0
0 and the tangent line is y
At x
1, g c 1
3 and the tangent line is
y 1
x3
2
At x
0.
2 x 1.
'x
'x
'x o 0
'x o 0
y
2
lim
'x o 0
lim
g x 'x g x
'x
lim
'x
'x o 0
lim
'x o 0
x 'x x3
'x o 0
'x
'x
'x o 0
lim
(b) g c x
3x 1
or
y
y
3 x 2.
0.
3 x 2.
2
2
−3
−3
3
3
−2
For this function, the slopes of the tangent lines are
sometimes the same.
−3
For this function, the slopes of the tangent lines are
always distinct for different values of x.
58. (a) g c 0
3
(b) g c 3
0
(c) Because g c 1
83 , g is decreasing (falling) at x
1.
(d) Because g c 4
7
, g is increasing (rising) at x
3
4.
(e) Because g c 4 and g c 6 are both positive, g 6 is greater than g 4 , and g 6 g 4 ! 0.
(f) No, it is not possible. All you can say is that g is decreasing (falling) at x
2.
1 2
x
2
59. f x
(a)
6
−6
6
−2
fc 0
0, f c 1 2
1 2, f c 1
(b) By symmetry: f c 1 2
1, f c 2
1 2, f c 1
2
1, f c 2
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ection 2.1
The Derivative and the Tangent L
Line Problem
117
y
(c)
4
f′
3
2
1
x
−4 −3 −2
1
2
3
4
−2
−3
−4
f x 'x f x
lim
'x o 0
'x
(d) f c x
1
1
2
x 'x x 2
2
2
lim
'x o 0
'x
1 2
1
2
x 2 x 'x 'x x 2
2
2
lim
'x o 0
'x
'x ·
§
lim ¨ x ¸
2 ¹
'x o 0©
x
1 3
x
3
60. f x
(a)
6
−9
9
−6
fc 0
0, f c 1 2
1 4, f c 1
(b) By symmetry: f c 1 2
(c)
1, f c 2
1 4, f c 1
4, f c 3
1, f c 2
9
4, f c 3
9
y
f′
5
4
3
2
1
x
− 3 − 2 −1
(d) f c x
1
−1
lim
2
3
f x 'x f x
'x
1
1
3
x 'x x3
3
3
lim
'x o 0
'x
1 3
1
2
3
x 3 x 2 'x 3 x 'x 'x x3
3
3
lim
'x o 0
'x
1
2º
ª
lim x 2 x 'x 'x »
x2
'x o 0 «
3
¬
¼
'x o 0
f x 0.01 f x
61. g x
f x 0.01 f x
62. g x
0.01
ª2 x 0.01 x 0.01 2 2 x x 2 º100
¬
¼
2 2 x 0.01
0.01
3 x 0.01 3 x 100
8
3
f
g
g
f
−2
−1
4
8
−1
The graph of g x is approximately the graph of
−1
The graph of g x is approximately the graph of
fc x
3
.
INSTRUCTOR USE
S ONLY
fc x
2 2 x.
2
x
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118
NOT FOR SALE
Chapter 2
63. f 2
Differentiation
ferentiation
242
2.1 4 2.1
4, f 2.1
3.99 4
fc 2 |
2.1 2
ª¬Exact: f c 2
0.1
1 3
2
2, f 2.1
2.31525
4
2.31525 2
fc 2 |
3.1525 ª¬Exact: f c 2
2.1 2
3.99
0º¼
gc 0
64. f 2
65. f x
fc 3
x 2 5, c
lim
lim
x o3
lim
x,c
lim
x 3
70. f x
x 3
x 3 x 3
fc 4
1
x3 2 x 2 1, c
lim
71. f x
fc 6
72. g x
lim x 2
x o2
f x f 6
x 6
x 3
4
g c 3
2
lim
x o 3
lim
x o6
1
x 6
.
6.
g x g 3
x 3
13
x 2
13
3
,c
x 3
0
x o 3
x 3
f x f 2
lim
x o 3
1
x3
23
.
Does not exist.
x 6 x 20
3.
Therefore g x is not differentiable at x
x 2
x 2 x 2 2 x 10
73. h x
x 7 ,c
x 2
lim x 2 2 x 10
13
lim
3
xo2
6
23
x 2
xo2
xo2
,c
x 2x 1 1
x 3 6 x, c
lim
lim
xo6
23
3
16
lim
2
x x 2
lim
x o2
x 2
lim
x 6
x 6
0
xo6
x6
2
2
xo2
3
4x
lim xo4
Does not exist.
x o2
lim
3 x 4
Therefore f x is not differentiable at x
lim
fc 2
x 4
3 3
x
4
f x f 2
x 2
3
68. f x
f x f 4
lim
xo4
lim
x 1
x o2
4
x o 4 4x x 4
x o1
f c 2
0.
x 4
12 3 x
lim
x o 4 4x x 4
x2 x 0
lim
x o1
x 1
x x 1
lim
x o1 x 1
lim x 1
67. f x
o f.
1
o f.
x
x
3
,c
x
lim
g x g1
x o1
. Does not exist.
xo4
x o3
lim
x
x
As x o 0 ,
x 3
lim x 3
6
gc 1
1
x
x
Therefore g x is not differentiable at x
x2 5 9 5
x 2 x, c
xo0
x
3º¼
x
lim
x 0
As x o 0 ,
x o3
66. g x
0
g x g 0
xo0
3
f x f 3
x o3
69. g x
18
hc 7
7
lim
h x h 7
x 7
lim
x7 0
x 7
x o7
x o7
lim
x7
x o7 x 7
.
Does not exist.
INSTRUCTOR USE ONLY
Therefor h x iss not differentiable at x
Therefore
77.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ection 2.1
x 6 ,c
74. f x
119
84. f is differentiable for all x z 1.
6
f is not continuous at x
f x f 6
xo6
x 6
x 6 0
lim
xo6
x 6
fc 6
The Derivative and the Tangent L
Line Problem
lim
1.
3
lim
x 6
xo6 x 6
.
−4
5
Does not exist.
Therefore f x is not differentiable at x
75. f x is differentiable everywhere except at
x
r3. (Sharp turns in the graph)
4. (Sharp turn in the graph)
r2. (Discontinuities)
lim
f x f 1
x 1
1 the tangent line is vertical.)
f x f 1
x 1
x 1 0
80. f x is differentiable everywhere except at
lim
x o1
x 1 0
x 1
f x f 1
x 1
x o1
lim
1 x2 0
x 1
lim
1 x2
˜
x 1
x o1
x o1
x
x 5 is differentiable everywhere except at
5. There is a sharp corner at x
1 x2
f.
1 x2
(Vertical tangent)
−1
The limit from the right does not exist since f is
undefined for x ! 1. Therefore, f is not differentiable at
x 1.
11
−1
4x
82. f x
is differentiable everywhere except at
x 3
x
3. f is not defined at x
3. (Vertical asymptote)
87. f x
lim
x o1
−8
­° x 1 3 , x d 1
®
2
°̄ x 1 , x ! 1
The derivative from the left is
15
f x f 1
x 1
3
0
x 1
2
0.
The derivative from the right is
0.
lim
x o1
f x f 1
x 1
lim
x 1
2
0
x 1
lim x 1
0.
x o1
x o1
5
−6
x o1
x 1
x o1
x 2 5 is differentiable for all x z 0. There is a
sharp corner at x
lim
lim x 1
12
−6
83. f x
1 x2
1 x
lim x o1
5.
7
1.
The derivative from the left does not exist because
0. (Discontinuity)
81. f x
1.
x 1
1 x2
lim
x
lim
x o1
The one-sided limits are not equal. Therefore, f is not
differentiable at x 1.
86. f x
79. f x is differentiable on the interval 1, f . (At
x
lim
x o1
x o1
78. f x is differentiable everywhere except at
x
x 1
The derivative from the right is
77. f x is differentiable everywhere except at
x
85. f x
The derivative from the left is
3. (Discontinuity)
76. f x is differentiable everywhere except at
x
−3
6.
6
The one-sided limits are equal. Therefore, f is
differentiable at x 1. f c 1
0
INSTRUCTOR USE ONLY
−3
3
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120
NOT FOR SALE
Chapter 2
Differentiation
ferentiation
­ x, x d 1
® 2
¯x , x ! 1
88. f x
90. Note that f is continuous at x
lim
x o1
f x f 1
x 1
lim
x 1
lim 1
x o1 x 1
x o1
1.
The derivative from the left is
The derivative from the right is
f x f 1
x 1
lim
x o1
x 1
x 1
2
lim
x o1
­1
° x 1, x 2
®2
° 2x ,
x t 2
¯
f x
The derivative from the left is
lim x 1
x o1
f x f 2
lim
x2
x o 2
2.
The one-sided limits are not equal. Therefore, f is not
differentiable at x 1.
89. Note that f is continuous at x
lim
x o 2
f x f 2
x 2
The derivative from the left is
f x f 2
x2
xo2
lim
lim
x o 2
x2 1 5
x2
lim x 2
4.
x o 2
lim
x o 2
The derivative from the right is
f x f 2
lim
lim
4x 3 5
lim
x 2
x 2
x o 2
lim
x o 2
lim
x o 2
x o 2
§1
·
¨ x 1¸ 2
2
©
¹
lim
x2
x o 2
1
x 2
1
lim 2
.
2
x2
xo2
The derivative from the right is
2.
2
°­x 1, x d 2
®
°̄4 x 3, x ! 2
f x
2.
x o 2
lim 4
x o 2
4.
The one-sided limits are equal. Therefore, f is
differentiable at x
2. f c 2
4
91. (a) The distance from 3, 1 to the line mx y 4
0 is d
2x 2
˜
x2
2x 2
2x 2
2x 4
x2
2x 2
2x2
x2
2
2x 2
2x 2
1
.
2
The one-sided limits are equal. Therefore, f is
1·
§
differentiable at x
2. ¨ f c 2
¸
2¹
©
Ax1 By1 C
m 3 11 4
3m 3
A B
m 1
m2 1
2
2
2
y
3
2
1
x
1
(b)
2
3
4
5
−4
4
−1
The function d is not differentiable at m
1. This corresponds to the line y
x 4, which passes through
the point 3, 1 .
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section
ection 2.1
x 2 and f c x
92. (a) f x
The Derivative and the Tangent L
Line Problem
121
2x
y
5
4
f
3
2
1
x
−4 −3 −2 −1
f'
1
2
3
4
−3
x3 and g c x
(b) g x
3x 2
y
3
g′
2
g
1
x
−2
−1
1
2
−1
(c) The derivative is a polynomial of degree 1 less than the original function. If h x
nx n 1.
x 4 , then
(d) If f x
fc x
x n , then hc x
lim
'x o 0
f x 'x f x
'x
4
x 'x x 4
'x o 0
'x
lim
lim
x 4 4 x3 'x 6 x 2 'x
'x o 0
lim
4 x 'x
'x
'x 4 x3 6 x 2 'x 4 x 'x
2
3
'x
'x
x 4 , then f c x
4
x4
3
'x
'x o 0
So, if f x
2
lim 4 x3 6 x 2 'x 4 x 'x
'x o 0
2
'x
3
4 x3.
4 x3 which is consistent with the conjecture. However, this is not a proof because you
must verify the conjecture for all integer values of n, n t 2.
93. False. The slope is lim
'x o 0
94. False. y
f 2 'x f 2
'x
x 2 is continuous at x
.
2, but is not differentiable at x
2. (Sharp turn in the graph)
95. False. If the derivative from the left of a point does not equal the derivative from the right of a point, then the derivative does
not exist at that point. For example, if f x
x , then the derivative from the left at x
0 is –1 and the derivative from the
right at x
0 is 1. At x
0, the derivative does not exist.
96. True—see Theorem 2.1.
INSTRUCTOR USE ONLY
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122
NOT FOR SALE
Chapter 2
Differentiation
ferentiation
°­x sin 1 x , x z 0
®
x
0
°̄0,
97. f x
Using the Squeeze Theorem, you have x d x sin 1 x d x , x z 0. So, lim x sin 1 x
0
xo0
f is continuous at x
f x f 0
lim
x 0
xo0
f 0 and
0. Using the alternative form of the derivative, you have
lim
x sin 1 x 0
x 0
xo0
1·
§
lim ¨ sin ¸.
x¹
x o 0©
Because this limit does not exist ( sin 1 x oscillates between –1 and 1), the function is not differentiable at x
2
°­x sin 1 x , x z 0
®
x
0
°̄0,
g x
Using the Squeeze Theorem again, you have x 2 d x 2 sin 1 x d x 2 , x z 0. So, lim x 2 sin 1 x
xo0
and g is continuous at x
lim
0.
g x g 0
xo0
x 0
lim
x sin 1 x 0
x 0
Therefore, g is differentiable at x
98.
g 0
0. Using the alternative form of the derivative again, you have
2
xo0
0
lim x sin
xo0
0, g c 0
1
x
0.
0.
3
−3
3
−1
As you zoom in, the graph of y1
x 2 1 appears to be locally the graph of a horizontal line, whereas the graph
x 1 always has a sharp corner at 0, 1 . y2 is not differentiable at 0, 1 .
of y2
Section 2.2 Basic Differentiation Rules and Rates of Change
1. (a)
(b)
2. (a)
(b)
y
x1 2
yc
1 1 2
x
2
yc 1
1
2
y
x3
yc
3x 2
yc 1
3
y
x 1 2
yc
12 x 3 2
yc 1
12
y
x 1
yc
x 2
yc 1
1
3. y
12
yc
0
4.
f x
9
fc x
0
5. y
x7
yc
7 x6
6. y
x12
yc
7. y
yc
8. y
yc
12 x11
1
x5
x 5
5 x 6
3
x7
3 x 7
3 7 x 8
5
x6
21
x8
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
ion 2.2
9. y
yc
10. y
yc
11.
5
18. y
yc
1
5x4 5
1 3 4
x
4
1
4 x3 4
f x
x 11
fc x
1
6x 3
21. y
x 2 12 cos x
yc
2 x 12 sin x
fc t
4t 3
yc
t 2 3t 1
23. y
2 x 12 x 2
7 sin x
22. y
2t 3
gc x
cos T sin T
S sin t
2t 2 3t 6
x 2 4 x3
2
gc t
f t
15. g x
S
S cos t
6
yc
sin T cos T
20. g t
gc x
14. y
6 x 2 12 x
2
yc
yc
cos x
1
3 sin x
x
1
2 3 cos x
x
5
24. y
3
2x
16. y
4 x 3x3
yc
4 9x2
17. s t
sc t
yc
2 cos x
5
3 x 4 2 sin x
8
Rewrite
Differentiate
15
2 sin x
8x4
Simplify
5
2x2
y
5 2
x
2
yc
5 x 3
yc
5
x3
26. y
3
2x4
y
3 4
x
2
yc
6 x 5
yc
6
x5
3
y
6 3
x
125
yc
18 4
x
125
yc
18
125 x 4
y
S
2
x 2
yc
2S 3
x
9
yc
2S
9 x3
1
2 x3 2
6
5x
S
3x
9
29. y
x
x
y
x 1 2
yc
1
x 3 2
2
yc
30. y
4
x 3
y
4 x3
yc
12 x 2
yc
12 x 2
31. f x
3t 2 10t 3
25. y
28. y
5 3
x 2 cos x
8
t 3 5t 2 3t 8
Function
27. y
123
2 x3 6 x 2 1
S
19. y
x1 4
x
12. g x
13.
x1 5
x
1 4 5
x
5
4
Basic Differentiation
Differen tiation Rules and Rat
Rate
Rates of Change
8
x2
8 x 2 , 2, 2
fc x
16 x 3
fc 2
2
16
x3
32.
f t
2
fc t
4t 2
fc 4
1
4
4
t
2 4t 1 , 4, 1
4
t2
INSTRUCTOR USE ONLY
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124
Chapter 2
33.
f x
12 75 x3 , 0, 12
fc x
21 x 2
5
fc 0
0
34.
y
8x
yc 1
8
y
36.
44.
2 x 4 3, 1, 1
yc
35.
Differentiation
ferentiation
45.
2
4 x 1 , 0, 1
16 x 2 8 x 1
39.
40.
yc
32 x 8
yc 0
32 0 8
f x
2 x 4 , 2, 8
42.
43.
fc 2
8 16
f T
4 sin T T , 0, 0
fcT
4 cos T 1
fc 0
41 1
8
f x
x3 3x 2 4
x2
fc x
1
8x 5x2
yc
3x 2 1
gc t
2 sin t
gc S
0
f x
x 2 5 3 x 2
fc x
2 x 6 x 3
f x
x3 2 x 3x 3
fc x
3x 2 9 x
50.
2x 2
t2 4
t3
3
x2
8x fc x
8 6 x 3
6
x3
9
3x 2 4
x
4
52.
4 x3 3x 2
x
8x 3
x2 8x 9
x 63 x
fc x
1 1 2
x
2 x 2 3
2
f t
t 2 3 t1 3 4
fc t
2 1 3 1 2 3
t
t
3
3
f x
6
x 5 cos x
6 x1 2 5 cos x
fc x
3 x 1 2 5 sin x
3
5 sin x
x
x
2
x2 3
2
1
23
3t1 3
3t
y
x 4 3x2 2
yc
4 x3 6 x
At 1, 0 : yc
8 x 3x 2
1
2
3
fc x
53. (a)
x1 2 6 x1 3
2
3 cos x
2 x 1 3 3 cos x
x
2
2
x 4 3 3 sin x
4 3 3 sin x
3
3x
f x
2
41
Tangent line:
6
x3
4 x 2 3x
2 x 4 3x3
f x
12
2t 4
t
8
5
x2
x3 x
8 x3 9 x 2
t 2 4t 3
2t 12t 4
f x
51.
4 x 2 2 5 x 1
8x x 2 2 x 2 3x
yc
3
x 3 4 x 2
4 x3 2 x 5
x
x x2 1
2
x3
x3 8
x3
8
x3
47. y
49.
2 cos t 5, S , 7
fc x
2
48. y
gt
f x
2 2 x 3
2
4 x 16
gc t
fc x
hc x
8
fc x
41. g t
2 x x 2
46. h x
2 x 16 x 32
38.
2x4 x
x3
3
2
37.
f x
3
61
2
y 0
2 x 1
y
2x 2
2x y 2
(b)
0
3
−2
2
(1, 0)
INSTRUCTOR USE ONLY
−1
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NOT FOR SALE
Section
ion 2.2
54. (a)
Basic Differentiation
Differen tiation Rules and Rat
Rate
Rates of Change
y
x3 3x
57. y
yc
3x 2 3
yc
At 2, 2 : yc
32
2
Tangent line:
(b)
x4 2 x2 3
4 x3 4 x
4x x2 1
3
9
y 2
9 x 2
y
9 x 16
9 x y 16
4x x 1 x 1
yc
0 Ÿ x
0, r1
Horizontal tangents: 0, 3 , 1, 2 , 1, 2
0
3
x3 x
58. y
(2, 2)
−4
125
yc
5
3 x 2 1 ! 0 for all x.
Therefore, there are no horizontal tangents.
−3
1
x2
59. y
55. (a)
2
f x
4
x3
3
x 7 4
2
fc x
At 1, 2 : f c 1
2 x 3 4
3
2
yc
3
x 1
2
3
7
x 2
2
0
3x 2 y 7
5
(1, 2)
0 Ÿ x
2x
At x
0, y
0
1.
Horizontal tangent: 0, 9
x sin x, 0 d x 2S
61. y
yc
−2
2
cannot equal zero.
x3
x2 9
60. y
y
(b)
Therefore, there are no horizontal tangents.
y 2
Tangent line:
2 x 3
yc
3
2 x7 4
x 2
1 cos x
0
cos x
1 Ÿ x
At x
S: y
S
S
Horizontal tangent: S , S
7
−1
56. (a)
y
x 2 x 2 3x
yc
3x 2 2 x 6
2
x3 x 2 6 x
At 1, 4 : yc
31
Tangent line:
y 4
1 x 1
y
x 3
21 6
x y 3
(b)
0
1
3 x 2 cos x, 0 d x 2S
62. y
yc
3 2 sin x
0
sin x
3
Ÿ x
2
S
At x
At x
10
−5
5
(1, − 4)
− 10
S
3
2 x
So, x
2S
3
: y
3S 3
3
2S
: y
3
2 3S 3
3
3
§S
Horizontal tangents: ¨¨ ,
©3
63. k x 2
or
3S 3 · § 2S 2 3S 3 ·
¸¸, ¨¨ ,
¸¸
3
3
¹ © 3
¹
6 x 1 Equate functions.
6
3 and k 9
Equate derivatives.
18 1 Ÿ k
8.
INSTRUCTOR USE ONLY
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126
Differentiation
ferentiation
2 x 3 Equate functions.
64. kx 2
65.
NOT FOR SALE
Chapter 2
2kx
2
So, k
k
x
k
2
x
3
x 3 Equate functions.
4
3
Equate derivatives.
4
So, k
3 2
x and
4
3 2
x
4
x
66. k
1
§ 1·
, and ¨ ¸ x 2
x
© x¹
2
2x
1
x
2x 3 Ÿ x
3 Ÿ k
1
.
3
69. The graph of a function f such that f c ! 0 for all x and
the rate of change of the function is decreasing
(i.e., f cc 0 ) would, in general, look like the graph
below.
3
x 3
4
3 Ÿ x
x
2 Ÿ k
3.
x 4 Equate functions.
x
70. (a) The slope appears to be steepest between A and B.
(b) The average rate of change between A and B is
greater than the instantaneous rate of change at B.
Equate derivatives.
(c)
So, k
2
x and
x
x
x 4 Ÿ 2x
2
2x 3 Ÿ x
y
3
3
x 3Ÿ x
4
4
3
Ÿ x
2
k
2
Equate derivatives.
x4Ÿ x
4Ÿk
y
f
4.
B C
A
67.
kx 3
x 1 Equate equations.
3kx 2
1
So, k
1
and
3x 2
x
x 1
3
,k
2
f x 6 Ÿ gc x
fc x
72. g x
2 f x Ÿ gc x
2fc x
73. g x
5 f x Ÿ g c x
5 f c x
74. g x
3 f x 1 Ÿ gc x
3fc x
4
.
27
kx 4
4 x 1 Equate equations.
4kx3
4
So, k
1
and
x3
x
71. g x
x 1
x
§1· 4
¨ 3 ¸x
©x ¹
x
E
Equate derivatives.
§ 1 · 3
¨ 2 ¸x
© 3x ¹
1
x
3
68.
D
Equate derivatives.
4x 1
4x 1
1
and k
3
27.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ion 2.2
y
75.
Basic Differentiation
Differen tiation Rules and Rat
Rate
Rates of Change
76.
127
y
3
f′
2
1
f
1
f
x
−2
x
−3 −2 −1
1
2
3
−1
1
3
4
f′
−2
−3
−4
If f is linear then its derivative is a constant function.
f x
ax b
fc x
a
If f is quadratic, then its derivative is a linear function.
77. Let x1 , y1 and x2 , y2 be the points of tangency on y
f x
ax 2 bx c
fc x
2ax b
x 2 6 x 5, respectively.
x 2 and y
The derivatives of these functions are:
yc
2x Ÿ m
2 x1 and yc
m
2 x1
2 x2 6
x1
x2 3
x2 2 6 x2 5 x12
y2 y1
x2 x1
m
2 x2 6
x2 2 6 x2 5:
x12 and y2
Because y1
2 x 6 Ÿ m
x2 x1
x2 2 6 x2 5 x2 3
2 x2 6
2
2 x2 6
x2 x2 3
x2 2 6 x2 5 x2 2 6 x2 9
2 x2 6 2 x2 3
2 x2 12 x2 14
4 x2 2 18 x2 18
2
2 x2 2 6 x2 4
0
2 x2 2 x2 1
0
x2
x2
1 Ÿ y2
0, x1
2 and y1
1 or 2
4
So, the tangent line through 1, 0 and 2, 4 is
y 0
§ 4 0·
¨
¸ x 1 Ÿ y
© 2 1¹
4 x 4.
So, the tangent line through 2, 3 and 1, 1 is
y 1
5
5
(2, 4)
4
3
3
2
2
1
1
−1
2 x 1.
y
y
4
§ 3 1·
¨
¸ x 1 Ÿ y
© 2 1¹
(1, 0) 2
(2, 3)
(1, 1)
x
x
3
−1
2
3
−2
x2
2 Ÿ y2
3, x1
1 and y1
1
INSTRUCTOR USE ONLY
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128
Chapter 2
Differentiation
ferentiation
78. m1 is the slope of the line tangent to y
x Ÿ yc
y
1 Ÿ m1
1
Ÿ x2
x
79.
1Ÿ x
At x
r1, m2
f x
3x sin x 2
fc x
3 cos x
1
Ÿ yc
x
1 and y
The points of intersection of y
x
x. m2 is the slope of the line tangent to y
x and y
1
Ÿ m2
x2
1
.
x2
1 x are
r1.
1. Because m2
1 m1 , these tangent lines are perpendicular at the points of intersection.
82.
Because cos x d 1, f c x z 0 for all x and f does not
f x
x5 3x3 5 x
fc x
5x 9x 5
4
fc x
Because 5 x 9 x t 0, f c x t 5. So, f does not
2
have a tangent line with a slope of 3.
81.
fc x
1 1 2
x
2
1
0 y
4 x
x
0 y
5 x
10 2 x
x2 y
10 2 x
4 x
2
xy
4 x
2
x
4 x
2x
x
4, y
1
2
x
§2·
x2 ¨ ¸
© x¹
2 x
10 2 x
x , 4, 0
f x
2
2
x2
2
4
2
, 5, 0
x
2
2
x
f x
have a horizontal tangent line.
80.
1 x. Because
4x
10
x
5
,y
2
4
5
§5 4·
The point ¨ , ¸ is on the graph of f. The slope of the
©2 5¹
8
§5·
tangent line is f c¨ ¸
.
25
© 2¹
x
y Tangent line:
4
5
25 y 20
2
8 x 25 y 40
The point 4, 2 is on the graph of f.
Tangent line: y 2
4y 8
0
0 2
x 4
4 4
x 4
8§
5·
¨x ¸
25 ©
2¹
8 x 20
0
83. f c 1 appears to be close to 1.
fc1
1
x 4y 4
3.64
0.77
3.33
1.24
84. f c 4 appears to be close to 1.
fc 4
1
16
−10
19
−1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section
ion 2.2
Basic Differentiation
Differen tiation Rules and Rat
Rate
Rates of Change
129
85. (a) One possible secant is between 3.9, 7.7019 and 4, 8 :
y 8
8 7.7019
x4
4 3.9
2.981 x 4
S x
2.981x 3.924
y 8
y
20
(4, 8)
−2
12
−2
3 12
3
2
x Ÿ fc 4
2
2
3 x 4 8 3x 4
(b) f c x
T x
3
The slope (and equation) of the secant line approaches that of the tangent line at
4, 8 as you choose points closer and closer to 4, 8 .
(c) As you move further away from 4, 8 , the accuracy of the approximation T gets worse.
20
f
T
−2
12
−2
(d)
'x
–3
–2
–1
–0.5
–0.1
0
0.1
0.5
1
2
3
f 4 'x
1
2.828
5.196
6.548
7.702
8
8.302
9.546
11.180
14.697
18.520
T 4 'x
–1
2
5
6.5
7.7
8
8.3
9.5
11
14
17
86. (a) Nearby point: 1.0073138, 1.0221024
1.0221024 1
x 1
1.0073138 1
3.022 x 1 1
Secant line: y 1
y
2
(1, 1)
−3
3
(Answers will vary.)
(b) f c x
−2
3x 2
3x 1 1
T x
3x 2
(c) The accuracy worsens as you move away from 1, 1 .
2
(1, 1)
−3
3
f
T
−2
(d)
'x
–3
–2
–1
–0.5
–0.1
0
0.1
0.5
1
2
3
f x
–8
–1
0
0.125
0.729
1
1.331
3.375
8
27
64
T x
–8
–5
–2
–0.5
0.7
1
1.3
2.5
4
7
10
The accuracy decreases more rapidly than in Exercise 85 because y
87. False. Let f x
fc x
gc x
x and g x
x 1. Then
x, but f x z g x .
x3 is less "linear" than y
88. True. If f x
g x c, then
fc x
gc x .
gc x 0
x3 2 .
INSTRUCTOR USE ONLY
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130
NOT FOR SALE
Chapter 2
Differentiation
ferentiation
89. False. If y
S 2 , then dy dx
90. True. If y
xS
3 f x , then g c x
92. False. If f x
1
xn
nx
fc t
0, 0 Ÿ f c 0
3fc x .
x , then
fc 2
f S 6 f 0
12 0
3
S 6 0
S 6 0
S
2 1
fc t
2t
f 3.1 f 3
2.61 2
0.1
3.1 3
fc x
1: v 1
32 ft/sec
When t
2: v 2
64 ft/sec
§
(e) v¨¨
©
Average rate of change:
95. f x
When t
0
1362
Ÿt
16
t2
6.2
1362 ·
¸
4 ¸¹
§
32¨¨
©
98.
16t 2 22t 220
vt
32t 22
v3
118 ft/sec
st
16t 2 22t 220
112 height after falling 108 ft
1, 1 Ÿ f c 1
1
16t 2 22t 108
0
1·
§
¨ 2, ¸ Ÿ f c 2
2¹
©
1
4
2 t 2 8t 27
0
t
Average rate of change:
1 2 1
2 1
2 1
1362 ·
¸
4 ¸¹
st
Instantaneous rate of change:
f 2 f 1
1362
| 9.226 sec
4
8 1362 | 295.242 ft/sec
6.1
>1, 2@
48 ft/sec
32t
6
At 3.1, 2.61 : f c 3.1
1
,
x
1
x2
sc t
(d) 16t 2 1362
Instantaneous rate of change:
At 3, 2 : f c 3
1298 1346
2 1
(c) v t
>3, 3.1@
t 2 7,
s 2 s1
4
(These are the same because f is a line of slope 4.)
94. f t
32t
vt
(b)
13 9
1
| 0.955
16t 2 1362
97. (a) s t
4.
Instantaneous rate of change is the constant 4. Average
rate of change:
f 2 f 1
3
| 0.866
2
Average rate of change:
n
.
x n 1
4. So, f c 1
1
§S 1 ·
§S ·
¨ , ¸ Ÿ f c¨ ¸
© 6 2¹
©6¹
n
>1, 2@
4t 5,
93. f t
cos x
Instantaneous rate of change:
91. True. If g x
fc x
fc x
1 S.
n 1
ª Sº
sin x, «0, »
¬ 6¼
96. f x
1 S ˜ x, then
1S 1
dy dx
0. ( S 2 is a constant.)
v2
1
2
2
32 2 22
86 ft/sec
99.
st
4.9t 2 v0t s0
4.9t 2 120t
vt
9.8t 120
v5
9.8 5 120
71 m/sec
v 10
9.8 10 120
22 m/sec
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
ion 2.2
4.9t 2 v0t s0
4.9t 2 s0
4.9t 2
s0
0 when t
2
4.9 5.6
vt
s
5.6.
10
| 153.7 m
1t Ÿ v t
2
101. From 0, 0 to 4, 2 , s t
1 mi/min.
2
30 mi/h for 0 t 4
1 60
2
8
(10, 6)
6
4
(4, 2)
2
(6, 2)
t
(0, 0)
0 for 4 t 6. Finally, from
Similarly, v t
131
104. This graph corresponds with Exercise 101.
Distance (in miles)
100. s t
Basic Differentiation Rules and Rat
Rate
Rates of Change
2
4
6
8
10
Time (in minutes)
6, 2 to 10, 6 ,
t 4 Ÿ vt
st
1 mi/min.
s3 ,
105. V
60 mi/h.
dV
ds
3s 2
Velocity (in mi/h)
v
When s
60
6 cm,
dV
ds
108 cm3 per cm change in s.
50
40
s2 ,
106. A
30
20
dA
ds
2s
10
When s
t
2
4
6
8
10
6 m,
Time (in minutes)
Rv
5
t Ÿ vt
6
5
mi/min.
6
Bv
0 for 6 t 8.
Similarly, v t
(c) T v
(d)
Finally, from 8, 5 to 10, 6 ,
1t 1 Ÿ v t
2
st
0.417v 0.02.
(b) Using a graphing utility,
50 mi/h for 0 t 6
5
60
6
vt
0.0056v 2 0.001v 0.04.
Rv Bv
0.0056v 2 0.418v 0.02
80
T
1 mi/min
2
12 m 2 per m change in s.
107. (a) Using a graphing utility,
(The velocity has been converted to miles per hour.)
102. From 0, 0 to 6, 5 , s t
dA
ds
30 mi h.
B
R
v
0
Velocity (in mph)
50
120
0
40
(e)
30
20
10
t
2
4
6
8
10
dT
dv
0.0112v 0.418
For v
40, T c 40 | 0.866
For v
80, T c 80 | 1.314
For v
100, T c 100 | 1.538
Time (in minutes)
(The velocity has been converted to miles per hour.)
40 mi/h
2 mi/min
3
v
6 min
0 mi/h
0 mi/min
0 mi/min 2 min
v
60 mi/h
4 mi
0 mi
1 mi/min
1 mi/min 2 min
2 mi
(f ) For increasing speeds, the total stopping distance
increases.
s
Distance (in miles)
103. v
2 mi/min
3
10
8
(10, 6)
6
(6, 4)
4
(8, 4)
2
(0, 0)
t
2
4
6
8
10
Time (in minutes)
INSTRUCTOR USE ONLY
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132
Chapter 2
Differentiation
ferentiation
108.
C
gallons of fuel used cost per gallon
dC
dx
§ 15,000 ·
¨
¸ 3.48
© x ¹
52,200
x2
x
10
15
20
25
30
35
40
C
5220
3480
2610
2088
1740
1491.4
1305
dC dx
–522
–232
–130.5
–83.52
–58
–42.61
–32.63
52,200
x
The driver who gets 15 miles per gallon would benefit more. The rate of change at x
at x
35.
1
at 2 c and sc t
2
109. s t
Average velocity:
15 is larger in absolute value than that
at
s t0 't s t0 't
t0 't t0 't
ª 1 2 a t0 't 2 cº ª 1 2 a t0 't 2 c º
¬
¼ ¬
¼
2't
1 2 a t0 2 2t0 't 't
2
1 2 a t0 2 2t0't 't
2
2 't
2at0 't
2 't
1,008,000
6.3Q
Q
1,008,000
6.3
Q
C
110.
dC
dQ
C 351 C 350 | 5083.095 5085 | $1.91
When Q
111. y
dC
| $1.93.
dQ
350,
at0
sc t 0
instantaneous velocity at t
1
,x ! 0
x
1
2
x
112. y
yc
At a, b , the equation of the tangent line is
y 1
a
1
x a
a2
or
Because the parabola passes through 0, 1 and 1, 0 ,
1, 0 : 0
a1
x
2
.
a2
a
§ 2·
The y-intercept is ¨ 0, ¸.
© a¹
you have:
a0
y
The x-intercept is 2a, 0 .
ax 2 bx c
0, 1 : 1
t0
2
b0 c Ÿ c
1
2
b1 1 Ÿ b
a 1
The area of the triangle is A
1
bh
2
1
§2·
2a ¨ ¸
2
©a¹
2.
y
ax 2 a 1 x 1.
So, y
From the tangent line y
x 1, you know that the
2
( )
(a, b) = a, a1
derivative is 1 at the point 1, 0 .
1
yc
2ax a 1
1
2a 1 a 1
1
a 1
a
2
b
a 1
Therefore, y
x
1
2
3
3
2 x 2 3 x 1.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
ion 2.2
113. y
x3 9 x
yc
3x 2 9
Basic Differentiation
Differen tiation Rules and Rat
Rate
Rates of Change
133
Tangent lines through 1, 9 :
y 9
3x 2 9 x 1
x3 9 x 9
3x3 3x 2 9 x 9
0
2 x3 3x 2
x
0 or x
x2 2x 3
3
2
The points of tangency are 0, 0 and
3
, 81
.
2
8
At 0, 0 , the slope is yc 0
3
, 81
,
2
8
the slope is yc 32
9. At
94 .
Tangent Lines:
y 0
9 x 0 and
y 81
8
94 x 23
y
94 x 27
4
9 x
y
9x y
9 x 4 y 27
0
114. y
x2
yc
2x
0
(a) Tangent lines through 0, a :
y a
2x x 0
x a
2 x2
a
x2
a
x
2
r
a , a , the slope is yc
The points of tangency are r a , a . At
At a , a , the slope is yc a
a
2 a .
2 a .
Tangent lines: y a
2 a x a and y a
2 a x y
2 ax a
y
2 ax a
a
Restriction: a must be negative.
(b) Tangent lines through a, 0 :
y 0
2x x a
2
2 x 2 2ax
0
x 2 2ax
x
x x 2a
The points of tangency are 0, 0 and 2a, 4a 2 . At 0, 0 , the slope is yc 0
Tangent lines: y 0
y
0 x 0 and y 4a 2
4a x 2 a
y
4ax 4a 2
0
0. At 2a, 4a 2 , the slope is yc 2a
4a.
Restriction: None, a can be any real number.
INSTRUCTOR USE ONLY
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134
NOT FOR SALE
Chapter 2
Differentiation
ferentiation
3
x d 2
°­ax ,
® 2
x
b
x ! 2
,
°̄
115. f x
f must be continuous at x
lim f x
lim ax
x o 2
3
2 to be differentiable at x
8a
x o 2
lim x 2 b
lim f x
x o 2
x o 2
fc x
2
°­3ax , x 2
®
x ! 2
°̄2 x,
½
8a
°
¾
4 b ° 8a 4
¿
For f to be differentiable at x
3a 2
2
4
a
1
3
b
8a 4
b
2, the left derivative must equal the right derivative.
43
119. You are given f : R o R satisfying
x 0
­cos x,
®
¯ax b, x t 0
1Ÿ b
f 0
b
fc x
­sin x, x 0
®
x ! 0
¯a,
So, a
0.
cos 0
Answer: a
117. f1 x
4 b
22
12a
116. f x
2.
0, b
1
for all real numbers x and
n
all positive integers n. You claim that
f x
mx b, m, b  R.
For this case,
fc x
1
sin x is differentiable for all x z nS , n an
integer.
Note first that f c x 1
and f c x
You can verify this by graphing f1 and f 2 and observing
the locations of the sharp turns.
2fc x
118. Let f x
fc x
f x 'x f x
'x
cos x cos 'x sin x sin 'x cos x
lim
'x o 0
'x
cos x cos 'x 1
§ sin 'x ·
lim sin x¨
lim
¸
'x o 0
'x o 0
'x
© 'x ¹
'x o 0
0 sin x 1
sin x
m.
f x 2 f x 1
,
1
f x 1 f x . From * you have
f x 2 f x
¬ª f x 2 f x 1 º¼ ª¬ f x 1 f x º¼
fc x 1 fc x .
cos x.
lim
ª¬m x n bº¼ >mx b@
n
m
Furthermore, these are the only solutions:
sin x is differentiable for all x z 0.
f2 x
f x n f x
* fc x
Thus, f c x
fc x 1.
Let g x
f x 1 f x.
f 1 f 0.
Let m
g 0
Let b
f 0 . Then
gc x
fc x 1 fc x
g x
constant
fc x
f x 1 f x
Ÿ f x
mx b.
g 0
0
m
g x
m
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 2.3
Product and Quotie
Quotient Rules and Higher-Orde
Higher-Order Derivatives
135
Section 2.3 Product and Quotient Rules and Higher-Order Derivatives
1. g x
gc x
x2 3 x2 4x
7.
f x
x2 3 2x 4 x2 4x 2 x
2 x 4 x 6 x 12 2 x 8 x
3
2
3
2
x
x2 1
x2 1 1 x 2x
fc x
2 2 x3 6 x 2 3x 6
2. y
yc
8. g t
3x 4 x3 5
gc t
3x 4 3x 2 x3 5 3
hc t
2t 5 6t 3t 2 1 2
2t 5
2t 5
2t 5
1 1 2
t
2
9. h x
1
1
2t 3 2 1 2 t 3 2
2t
2
5 32
1
t 12
2
2t
1 5t 2
2t1 2
4. g s
gc s
hc x
1 5t 2
2 t
s s2 8
x1 2
x 1
x
x 1
3
1
x3 1 x 1 2 x1 2 3 x 2
2
2
x3 1
x3 1 6 x3
1 5 x3
1 1 2
s
2
10. f x
fc x
5s 2 8
2 s
x2
2 x 1
x 1 2 x x 2 x 1 2
2
x 1
2
2
4 x3 2 2 x x3 2
x 1
f x
x3 cos x
fc x
x3 sin x cos x 3 x 2
3x3 2 2 x
3 x 2 cos x x3 sin x
2
x 1
x 2 3 cos x x sin x
x3
x 2
2
x 1
gc x
2
x x3 1
2
1 32
s 4s 1 2
2
5 32
4
s 12
s
2
6. g x
2
2 x1 2 x3 1
2s 3 2 5.
2
2
3
s1 2 s 2 8
s1 2 2 s s 2 8
2
6t 2 30t 2
t1 2 1 t 2
t1 2 2t 1 t 2
2
3t 2 1
2t 5
12 x 3 12 x 2 15
t 1 t2
x2 1
12t 2 30t 6t 2 2
9 x 3 12 x 2 3 x3 15
3. h t
2
x2 1
4 x3 12 x 2 6 x 12
1 x2
2
x sin x
§ 1 ·
x cos x sin x¨
¸
©2 x ¹
1
sin x
x cos x 2 x
11. g x
gc x
2
2
2
sin x
x2
x 2 cos x sin x 2 x
x
2
2
x cos x 2 sin x
x3
INSTRUCTOR USE ONLY
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136
Chapter 2
12.
f t
fc t
NOT FOR SALE
Differentiation
ferentiation
cos t
t3
t 3 sin t cos t 3t 2
t
13.
3
2
t sin t 3 cos t
t4
f x
x3 4 x 3x 2 2 x 5
fc x
x3 4 x 6 x 2 3x 2 2 x 5 3 x 2 4
6 x 4 24 x 2 2 x3 8 x 9 x 4 6 x3 15 x 2 12 x 2 8 x 20
15 x 4 8 x3 21x 2 16 x 20
fc 0
14.
20
y
x 2 3x 2 x3 1
yc
x 2 3 x 2 3 x 2 x3 1 2 x 3
16.
f x
fc x
3x 4 9 x3 6 x 2 2 x 4 3x3 2 x 3
5 x 4 12 x3 6 x 2 2 x 3
15.
yc 2
52
f x
x2 4
x 3
4
12 2
62
3
2
22 3
x 4
9
x 3
fc 3
2
x 3
17.
2
x 6x 4
2
x 3
2
16 4
13
2
2
8
x 4
2x2 6x x2 4
fc1
x 4
x 4 x 4
x 3 2x x2 4 1
fc x
x 4
x 4
x 4 1 x 4 1
2
1
4
18.
2
8
3 4
8
49
2
f x
x cos x
fc x
x sin x cos x 1
§S ·
f c¨ ¸
©4¹
S§ 2·
2
¨¨
¸
2
4 © 2 ¸¹
f x
fc x
§S ·
f c¨ ¸
©6¹
cos x x sin x
2
4S
8
sin x
x
x cos x sin x 1
x2
x cos x sin x
x2
S 6
3 2 12
S 2 36
3 3S 18
S2
3
3S 6
S2
Function
Rewrite
Differentiate
Simplify
19. y
x 2 3x
7
y
1 2
3
x x
7
7
yc
2
3
x 7
7
yc
2x 3
7
20. y
5x2 3
4
y
5 2
3
x 4
4
yc
10
x
4
yc
5x
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.3
Function
Product and Quotie
Quotient Rules and Higher-Orde
Higher-Order Derivatives
Rewrite
Differentiate
Simplify
21. y
6
7 x2
y
6 2
x
7
yc
12 3
x
7
yc
12
7 x3
22. y
10
3x3
y
10 3
x
3
yc
30 4
x
3
yc
10
x4
23. y
4x3 2
x
y
4 x1 2 , x ! 0
yc
2 x 1 2
yc
2
,x ! 0
x
24. y
2x
x1 3
y
2x2 3
yc
4 1 3
x
3
yc
4
3 x1 3
26.
f x
25.
f x
fc x
4 3x x 2
x2 1
x 2 1 3 2 x 4 3 x x 2 2 x
x2 1
2
fc x
3 x 2 3 2 x3 2 x 8 x 6 x 2 2 x3
x2 1
x2 4
3x 1
2
x 1
2
5 x 2 20 x 20
x2 4
x2 1
2
x2 4
2
2
2
x2 4 2x 5 x2 5x 6 2x
2
3 x2 2x 1
x 1
x2 5x 6
x2 4
2 x3 5 x 2 8 x 20 2 x3 10 x 2 12 x
3x 2 6 x 3
x2 1
137
2
5 x2 4 x 4
x 2
3
2
x 1
2
x 2
5 x 2
,x z 1
2
x 2
2
x 2
2
x 2
5
2
2
2
, x z 2, 2
Alternate solution:
f x
x2 5x 6
x2 4
x 3 x 2
x 2 x 2
x 3
, x z 2
x 2
fc x
x 2 1 x 3 1
x 2
2
5
x 2
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
138
Chapter 2
Differentiation
ferentiation
27.
f x
4 ·
§
x¨1 ¸
x 3¹
©
fc x
1
x 4x
x 3
30.
2
x 2 6 x 9 12
x 3
2
x2 6x 3
x 3
ª x 1º
x4 «
»
¬ x 1¼
f x
2 º
ª
x 4 «1 x 1»¼
¬
fc x
ª x 1 x 1 º ª x 1º
3
» «
x4 «
2
» 4x
«¬
»¼ ¬ x 1¼
x 1
hc s
f x
fc x
x
x 3
3 x1 3
2
s 6 4s 3 4
6s 5 12 s 2
x2 3
32. h x
3x 1
3 x1 2 x 1 2
x
3 1 2
1
3x 1
x
x 3 2
2
2
2 x3 2
x
56
s3 2
31. h s
ª 2 x2 x 2 º
»
2 x3 «
2
«¬ x 1
»¼
3
5 1 6
x
x 2 3
6
5
1
23
16
6x
x
fc x
ª 2 º ª x2 1 º
» «
» 4 x3
x4 «
2
2
¬« x 1 ¼» ¬« x 1 ¼»
29.
x1 3 x1 2 3
Alternate solution:
2
f x
28.
x 3
x
§1
·
§1
·
x1 3 ¨ x 1 2 ¸ x1 2 3 ¨ x 2 3 ¸
©2
¹
©3
¹
5 1 6
x 2 3
x
6
5
1
23
6 x1 6
x
fc x
x 3 4 4x 1
x 3
3
f x
hc x
6s 2 s 3 2
3
x 6 9 x 4 27 x 2 27
6 x5 36 x3 54 x
6 x x4 6x2 9
6 x x2 3
2
Alternate solution:
f x
fc x
34. g x
35.
3x 1
x
3x 1
x1 2
33.
§1·
x1 2 3 3 x 1 ¨ ¸ x 1 2
© 2¹
x
1 1 2
3x 1
x
2
x
3x 1
2 x3 2
1 ·
§2
x2 ¨ ¸
1¹
x
x
©
2 1x
f x
2x 1
x 2 3x
x 2 3x 2 2 x 1 2 x 3
fc x
x 2 3x
2
2x2 6x 4x2 8x 3
x 2 3x
2x2 2 x 3
x 2 3x
2
2
2x2 2x 3
x2 x 3
2
x2
x 1
2x 2 x2 2 x 1 x2 2 x
x 1 2 x x2 1
x2 2 x 2
gc x
2
f x
2 x3 5 x x 3 x 2
fc x
6 x 2 5 x 3 x 2 2 x3 5 x 1 x 2 2 x3 5 x x 3 1
x 1
2x 1
x x 3
x 3
2
x 1
2
x 1
2
6 x 2 5 x 2 x 6 2 x3 5 x x 2 2 x3 5 x x 3
6 x 4 5 x 2 6 x3 5 x 36 x 2 30 2 x 4 4 x3 5 x 2 10 x 2 x 4 5 x 2 6 x3 15 x
10 x 4 8 x3 21x 2 10 x 30
INSTRUCTOR USE ONLY
Note: You could simplify first: f x
2 x3 5 x x 2 x 6
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.3
36.
Product and Quotie
Quotient Rules and Higher-Orde
Higher-Order Derivatives
f x
x3 x x 2 2 x 2 x 1
fc x
3 x 2 1 x 2 2 x 2 x 1 x 3 x 2 x x 2 x 1 x3 x x 2 2 2 x 1
139
3x 4 5 x 2 2 x 2 x 1 2 x 4 2 x 2 x 2 x 1 x5 x3 2 x 2 x 1
3x6 5 x 4 2 x 2 3x5 5 x3 2 x 3x 4 5 x 2 2
2 x 6 2 x 4 2 x5 2 x3 2 x 4 2 x 2
2 x 6 2 x 4 4 x 2 x5 x3 2 x
7 x6 6 x5 4 x3 9 x 4 x 2
37.
f x
fc x
x2 c2
x2 c2
x2 c2 2 x x2 c2 2 x
f x
4 xc 2
40.
46. h x
hc x
c2 x2
c2 x2
c x
2 x c x
2
2
c x
2
39.
2
x c2
2
2
fc x
gc t
2
x2 c2
38.
45. g t
2
2x
47. y
4 xc 2
2
c2 x2
yc
f t
t 2 sin t
fc t
t 2 cos t 2t sin t
f T
T 1 cos T
fcT
T 1 sin T cos T 1
41. f t
fc t
t t cos t 2 sin t
42.
f x
fc x
43.
48. y
t sin t cos t
t2
x3
f x
x tan x
fc x
1 sec 2 x
x cot x
1 csc 2 x
yc
49. y
x 3 cos x sin x 3x 2
yc
1 3 4
t
6 csc t cot t
4
1
6 csc t cot t
4t 3 4
1
12 sec x
x
x 1 12 sec x
x 2 12 sec x tan x
3 1 sin x
3 3 sin x
2 cos x
2 cos x
3 cos x 2 cos x 3 3 sin x 2 sin x
2 cos x
2
sec x
x
x sec x tan x sec x
x2
sec x x tan x 1
x2
sin x
x3
44. y
t1 4 6 csc t
6 cos 2 x 6 sin x 6 sin 2 x
4 cos 2 x
3
1 tan x sec x tan 2 x
2
3
sec x tan x sec x
2
cos T T 1 sin T
cos t
t
t sin t cos t
t2
t 6 csc t
1
12 sec x tan x
x2
2
2
4
2
x cos x 3 sin x
x4
yc
csc x sin x
csc x cot x cos x
cos x
cos x
sin 2 x
cos x csc 2 x 1
tan 2 x
cos x cot 2 x
50. y
cot 2 x
yc
x sin x cos x
x cos x sin x sin x
x cos x
INSTRUCTOR USE ONLY
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NOT FOR SALE
140
Chapter 2
51.
f x
x 2 tan x
fc x
x 2 sec 2 x 2 x tan x
52.
Differentiation
ferentiation
56.
f x
§ x2 x 3 · 2
¨
¸ x x 1
2
© x 1 ¹
fc x
2
x x sec 2 x 2 tan x
f x
sin x cos x
fc x
sin x sin x cos x cos x
cos 2 x
x5 2 x3 2 x 2 2
x2 1
2
Form of answer may vary.
2 x sin x x 2 cos x
53. y
yc
2 x cos x 2 sin x x 2 sin x 2 x cos x
T
57. g T
1 sin T
4 x cos x 2 x 2 sin x
1 sin T T cos T
gc T
54. h T
hc T
5T sec T tan T 5 sec T T sec 2 T tan T
§ x 1·
¨
¸ 2x 5
© x 2¹
55. g x
Form of answer may vary.
58.
ª x 2 1 x 1 1 º
§ x 1·
»
¨
¸ 2 2x 5 «
2
© x 2¹
«¬
»¼
x2
gc x
f T
sin T
1 cos T
fcT
1
cos T 1
cos T 1
1 cos T
2
(Form of answer may vary.)
2x2 8x 1
x2
2
1 sin T
5T sec T T tan T
2
Form of answer may vary.
y
59.
1 csc x
1 csc x
1 csc x csc x cot x 1 csc x csc x cot x
yc
60.
61.
§S ·
yc¨ ¸
©6¹
2 2
f x
tan x cot x
fc x
0
fc1
0
ht
hc t
hc S
62.
1 csc x
3
1 2
2
2 csc x cot x
1 csc x
2
4 3
2
1
63. (a)
f x
x3 4 x 1 x 2 ,
fc x
x 4 x 1 1 x 2 3x 4
x3 4 x 1 3x3 6 x 2 4 x 8
4 x3 6 x 2 8 x 9
sec t
t
t sec t tan t sec t 1
fc1
sec t t tan t 1
t2
sec S S tan S 1
1
S2
S2
sin x sin x cos x
fc x
sin x cos x sin x sin x cos x cos x
3; Slope at 1, 4
Tangent line: y 4
t2
f x
(b)
S
2
cos
S
2
3x 1
3
−1
3
−6
(c) Graphing utility confirms
sin 2 x cos 2 x
sin
3 x 1 Ÿ y
(1, − 4)
sin x cos x sin 2 x sin x cos x cos 2 x
§S ·
f c¨ ¸
©4¹
1, 4
3
dy
dx
3 at 1, 4 .
1
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.3
64. (a)
f x
x 2 x 2 4 , 1, 5
fc x
x 2 2 x x2 4 1
Product and Quotie
Quotient Rules and Higher-Orde
Higher-Order Derivatives
2x 4x x 4
2
2
3x 2 4 x 4
fc1
3; Slope at 1, 5
Tangent line:
y 5
3x 1 Ÿ y
(b)
§S ·
¨ , 1¸
©4 ¹
f x
tan x,
fc x
sec2 x
§S ·
f c¨ ¸
©4¹
2;
67. (a)
§S ·
Slope at ¨ , 1¸
©4 ¹
Tangent line:
3x 8
y 1
S·
§
2¨ x ¸
4¹
©
y 1
2x 4x 2 y S 2
3
−3
141
S
2
0
6
(b)
(1, − 5)
4
( (
π
,1
4
−
− 15
(c) Graphing utility confirms
3 at 1, 5 .
fc x
x 4
f c 5
5 4
4
2
x 4
4
4;
2
Tangent line: y 5
−4
(c) Graphing utility confirms
x
5, 5
,
x 4
x 4 1 x1
f x
65. (a)
dy
dx
§S ·
¨ , 2¸
©3 ¹
sec x tan x
f x
68. (a)
sec x,
2
fc x
Slope at 5, 5
4x 5 Ÿ y
§S ·
f c¨ ¸
©3¹
4 x 25
§S ·
Slope at ¨ , 2 ¸
©3 ¹
2 3;
Tangent line:
8
(b)
S·
§
2 3¨ x ¸
3¹
©
y 2
(− 5, 5)
−8
§S ·
2 at ¨ , 1¸.
©4 ¹
dy
dx
6 3 x 3 y 6 2 3S
1
(b)
0
6
−6
(c) Graphing utility confirms
66. (a)
dy
dx
fc x
x 3
6
1
fc 4
6;
(c) Graphing utility confirms
6
x 3
69.
f x
fc x
8
;
x2 4
fc 2
x 4
2
16 2
4 4
10
y 1
−8
(c) Graphing utility confirms
dy
dx
y
6 at 4, 7 .
2, 1
x2 4 0 8 2 x
6 x 31
(4, 7)
§S ·
2 3 at ¨ , 2 ¸.
©3 ¹
dy
dx
2
Slope at 4, 7
8
−5
−2
2
Tangent line:
y 7
6 x 4 Ÿ y
(b)
−
x 3
,
4, 7
x 3
x 3 1 x 3 1
f x
( π3 , 2(
4 at 5, 5 .
2y x 4
2
2
16 x
x 4
2
2
1
2
1
x 2
2
1
x 2
2
0
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
142
70.
Chapter 2
NOT FOR SALE
Differentiation
ferentiation
2
x2 9
54 3
f c 3
9 9
y 3
2
2y x 6
f x
256 16 4
x2
x 1
2
x2 1 2 x x2 2x
x2 1
x 1
fc x
75.
f x
fc x
0 when x
2
x 2 16
x2
x 1
x 1 2x x2 1
0 when x
76.
f x
fc x
fc x
0 when x
24 4 x 2
x2 6
2
x 2
25
2
16
x 25
25
0
2 x 2 2 x 3
0 or x
2
2.
Horizontal tangents are at 0, 0 and 2, 4 .
2
4
5
fc x
x 1
x4
x2 7
x2 7 1 x 4 2x
x2 7
x2 7
2
25
2x 1
x2
x x 2
2
2
x2 7 2x2 8x
24 16
102
f x
2
x 1
x 1
2
fc x
x2 6 4 4 x 2 x
25 y 2 x 16
0.
x2 2x
256 16 x 2
4x § 4 ·
; ¨ 2, ¸
x 6 © 5¹
y
2
Horizontal tangent is at 0, 0 .
12
x 2
25
12
16
x 25
25
0
x2 6
2
2x
2
2
y 73.
2
12
25
202
25 y 12 x 16
fc 2
fc x
8·
§
¨ 2, ¸
5¹
©
x 2 16
y
fc x
x2 9
x 2 16 16 16 x 2 x
8
y 5
f x
2
f x
1
2
2
16 x
;
2
x 16
fc x
f c 2
54 x
1
x 3
2
1
x 3
2
0
y
72.
74.
x 2 9 0 27 2 x
fc x
71.
3·
§
¨ 3, ¸
2¹
©
27
;
x 9
f x
2
fc x
2
x2 8x 7
x 7
2
0 for x
2
1, 7; f 1
x 7 x 1
x2 7
1
, f 7
2
2
1
14
§ 1·
§ 1·
f has horizontal tangents at ¨1, ¸ and ¨ 7, ¸ .
2
©
¹
© 14 ¹
2 x 1 x 2
2 x 1
x3
1, and f 1
1.
Horizontal tangent at 1, 1 .
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.3
77.
x 1
x 1
x 1 x 1
f x
fc x
x 1
2y x
6 Ÿ y
2
1
2
x 1
2
x 1
2
x 1
2y + x = 7
x 1
x 1
2
graph of f.
y
y = −x + 4
r2
y = − 4x + 1
y
f(x) =
x+1
x−1
(3, 2)
f(x) =
x
x−1
0, f 3
2
1
1
x 2
2
1
7
x 2
2
(2, 2)
x
−4
−2
2
4
(12 , −1)
−2
5 x x 1
1
1 x
x 1
4x 5
x 1 x 1
1
x 1
4x 5 x 1
x 1
4 x 10 x 4
0
x 2 2x 1
0 Ÿ x
2
x
2
1
2
x, x x 1 be a point of tangency on the
Let x, y
4
(−1, 0)
4
2
2
6
−4
−6
fc x
143
6
6
−2
2
(−1, 5)
y 2
2
1
1
x 3; Slope: 2
2
1
x 1 Ÿ y
2
1
x 3 Ÿ y
2
y 0
x
x 1
x 1 x
78. f x
x 1
1, 3; f 1
x
−6 − 4 −2
2
Product and Quotie
Quotient Rules and Higher-Orde
Higher-Order Derivatives
2y + x = −1
§1·
f¨ ¸
1, f 2
© 2¹
Two tangent lines:
1
,2
2
§1·
2; f c¨ ¸
© 2¹
4, f c 2
y 1
1·
§
4¨ x ¸ Ÿ y
2¹
©
4 x 1
y 2
1 x 2
x 4
79. f c x
gc x
g x
Ÿ y
x 2 3 3x 1
x 2
2
6
2
x 2
x 2 5 5x 4 1
x 2
5x 4
x 2
2
1
6
x 2
3x
2x 4
x 2
x 2
2
f x 2
f and g differ by a constant.
80. f c x
gc x
g x
x cos x 3 sin x 3x 1
x2
x cos x 2 sin x 2 x 1
x2
sin x 2 x sin x 3x 5 x
x
x
x cos x sin x
x2
x cos x sin x
x2
f x 5
f and g differ by a constant.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
144
Chapter 2
81. (a) pc x
Differentiation
ferentiation
f c x g x f x gc x
pc 1
82. (a) pc x
§ 1·
1 4 6¨ ¸
© 2¹
f c 1 g 1 f 1 gc 1
g x
32
83. Area
Ac t
84. V
4 2 4 1
C
x ·
§ 200
100¨ 2 ¸, 1 d x
x 30 ¹
© x
dC
dx
§ 400
·
30
¸
100¨ 3 2
¨ x
x 30 ¸¹
©
4
12
16
2
3
4
At
6t 5
t
9t1 2 5 1 2
t
2
18t 5
cm 2 /sec
2 t
S r 2h
§1
©2
S t 2¨
1§ 3 1 2
1 2 ·
¨ t t ¸S
2© 2
¹
Vc t
85.
2
g x
qc 7
1
3
4
g x f c x f x gc x
(b) qc x
2
3 1 7 0
qc 4
1
8 10
2
pc 4
1
g x f c x f x gc x
(b) qc x
f c x g x f x gc x
6t 3 2 5t1 2
·
t¸
¹
1 32
t 2t1 2 S
2
3t 2
S in.3 /sec
4t1 2
dC
$38.13 thousand 100 components
dx
dC
(b) When x 15:
$10.37 thousand 100 components
dx
dC
(c) When x
20:
$3.80 thousand 100 components
dx
As the order size increases, the cost per item decreases.
(a) When x
86.
10:
Pt
4t º
ª
500 «1 t 2 »¼
50
¬
Pc t
ª 50 t 2 4 4t 2t º
»
500 «
2
«
»
2
50 t
¬
¼
ª
º
200 4t 2 »
500 «
« 50 t 2 2 »
¬
¼
ª
º
50 t 2 »
2000 «
« 50 t 2 2 »
¬
¼
Pc 2 | 31.55 bacteria/h
87. (a)
sec x
d
>sec x@
dx
(b)
csc x
d
>csc x@
dx
(c)
cot x
d
>cot x@
dx
1
cos x
d ª 1 º
«
»
dx ¬ cos x ¼
cos x 0 1 sin x
cos x
2
sin x
cos x cos x
1
sin x
˜
cos x cos x
sec x tan x
cos x
sin x sin x
1
cos x
˜
sin x sin x
csc x cot x
1
sin x
d ª 1 º
«
»
dx ¬ sin x ¼
sin x 0 1 cos x
sin x
2
cos x
sin x
d ª cos x º
«
»
dx ¬ sin x ¼
sin x sin x cos x cos x
2
sin x cos 2 x
sin 2 x
1
sin 2 x
csc 2 x
INSTRUCTOR USE ONLY
sin x
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.3
88.
f x
sec x
g x
csc x, >0, 2S
fc x
gc x
sec x tan x
csc x cot x Ÿ
csc x cot x
sec x tan x
1
sin x
˜
cos x cos x
1 Ÿ
1
cos x
˜
sin x sin x
112.4t 1332
89. (a) h t
91.
2.9t 282
pt
400
3000
92.
2
2
10
10
0
0
112.4t 1332
2.9t 282
(c) A
93.
10
2
10
94.
0
A represents the average health care expenses per
person (in thousands of dollars).
(d) Ac t |
3407.5
t 98.53
|
2
27,834
8.41t 2 1635.6t 79,524
Ac t represents the rate of change of the average
health care expenses per person per year t.
sin T
r h
h
(b)
sin 3 x
cos3 x
f x
x 4 2 x3 3x 2 x
fc x
4 x3 6 x 2 6 x 1
f cc x
12 x 2 12 x 6
f x
4 x5 2 x3 5 x 2
fc x
20 x 4 6 x 2 10 x
f cc x
80 x3 12 x 10
f x
4 x3 2
fc x
6 x1 2
f cc x
3 x 1 2
f x
x 2 3 x 3
fc x
2x 9x4
f cc x
2 36 x 5
1 Ÿ tan 3 x
1 Ÿ tan x
1
p(t)
h(t)
90. (a)
1 Ÿ
145
3S 7S
,
4 4
x
(b)
Product and Quotie
Quotient Rules and Higher-Orde
Higher-Order Derivatives
hc T
hc 30q
r
r h
r csc T
r csc T r
95.
f x
fc x
f cc x
3
x
2
x
x 1
x 1 1 x1
x 1
2
36
x5
1
x 1
2
2
x 1
3
r csc T 1
r csc T ˜ cot T
§S ·
hc¨ ¸
©6¹
3960 2 ˜
3
7920 3 mi/rad
INSTRUCTOR USE ONLY
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146
Chapter 2
96.
f x
NOT FOR SALE
Differentiation
ferentiation
x 2 3x
x 4
103.
x 4 2 x 3 x 2 3x 1
fc x
x 4
x 4
f cc x
2
2 g c x hc x
fc 2
2 g c 2 hc 2
2 2 4
2
x 4
2g x h x
2
2 x 5 x 12 x 3x
2
f x
fc x
2
x 8 x 12
x 2 8 x 16
2
0
104.
2 x 8 x 2 8 x 12 2 x 8
x 4
4
f x
4 h x
fc x
hc x
fc 2
hc 2
x 4 ª¬ x 4 2 x 8 2 x 2 8 x 12 º¼
x 4
4
105.
x 4 2 x 8 2 x 2 8 x 12
x 4
2 x 2 16 x 32 2 x 2 16 x 24
x 4
g x
f x
h x
h x g c x g x hc x
fc x
3
2
ª¬h x º¼
h 2 g c 2 g 2 hc 2
fc 2
3
2
ª¬h 2 º¼
1 2 3 4
56
x 4
3
1
2
10
f x
x sin x
fc x
x cos x sin x
f x
g xh x
f cc x
x sin x cos x cos x
fc x
g x hc x h x g c x
x sin x 2 cos x
fc 2
g 2 hc 2 h 2 g c 2
97.
106.
3 4 1 2
f x
sec x
fc x
sec x tan x
f cc x
sec x sec 2 x tan x sec x tan x
98.
14
sec x sec 2 x tan 2 x
99.
4
fc x
x2
f cc x
2x
f cc x
2 2 x 1
f ccc x
2 x 2
107. The graph of a differentiable function f such that
f 2
0, f c 0 for f x 2, and f c ! 0 for
2 x f would, in general, look like the graph
below.
y
4
3
100.
2
2
x2
1
x
1
101.
f ccc x
2
f 4 x
1
2 x 1 2
2
102. f 4 x
x
2x 1
3
2
4
One such function is f x
1
x
2
x 2 .
108. The graph of a differentiable function f such that
f ! 0 and f c 0 for all real numbers x would, in
general, look like the graph below.
y
f
5
x
2
f 6 x
0
f
x
INSTRUCTOR USE
S ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.3
y
109.
It appears that f is cubic, so
f c would be quadratic and
f cc would be linear.
f′
2
f
1
Product and Quotie
Quotient Rules and Higher-Orde
Higher-Order Derivatives
113.
147
y
f′
f″
1
x
−2
−1
1
2
−1
π
2
2π
f ′′
f′
x
−2
−3
−4
f″
y
114.
y
110.
f′
3
It appears that f is quadratic
so f c would be linear and
f cc would be constant.
f
f ′′
2
1
π
2
x
−2
−1
2
−1
3
x
π
4
−2
36 t 2 , 0 d t d 6
115. v t
y
111.
4
3
2
1
f′
1 2 3 4 5
f″
−3
−4
−5
vc t
v3
27 m/sec
a3
6 m/sec2
116. v t
at
112.
2t
The speed of the object is decreasing.
x
−3 −2 −1
at
100t
2t 15
2t 15 100 100t 2
vc t
2t 15
y
f ′′
3
f′
1500
(a) a 5
¬ª2 5 15º¼
2
1
(b) a 10
x
−4
−3
−1
−1
(c) a 20
117. s t
8.25t 2 66t
vt
sc t
16.50t 66
at
vc t
16.50
1500
2t 15
2
2.4 ft/sec 2
2
1500
ª¬2 10 15º¼
2
2
| 1.2 ft/sec 2
2
| 0.5 ft/sec 2
1500
ª¬2 20 15º¼
t(sec)
0
1
2
3
4
s(t) (ft)
0
57.75
99
123.75
132
vt
sc t ft/sec
66
49.5
33
16.5
0
at
vc t ft/sec 2
–16.5
–16.5
–16.5
–16.5
–16.5
Average velocity on:
57.75 0
57.75
>0, 1@ is
10
99 57.75
41.25
>1, 2@ is
2 1
123.75 99
24.75
>2, 3@ is
3 2
132 123.75
8.25
>3, 4@ is
43
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
148
NOT FOR SALE
Chapter 2
118. (a)
Differentiation
ferentiation
s position function
y
v velocity function
16
12
a acceleration function
s
8
v
4
t
−1
1
4
5
6
7
a
(b) The speed of the particle is the absolute value of its velocity. So, the particle’s speed is slowing
down on the intervals 0, 4 3 and 8 3, 4 and it speeds up on the intervals 4 3, 8 3 and 4, 6 .
16
t= 8
3
12
8
v
speed
4
t
−4
1
3
5
6
7
−8
−12
119.
−16
t= 4
f x
xn
(a)
n!
f n x
n n 1 " 3 ˜ 2 ˜ 1 read "n factorial"
Note: n!
f x
120.
nn 1 n 2 " 2 1
f n x
121. f x
t=4
3
1
x
1
n
n n 1 n 2" 2 1
x n 1
n
1 n!
x n 1
g xh x
fc x
g x hc x h x g c x
f cc x
g x hcc x g c x hc x h x g cc x hc x g c x
g x hcc x 2 g c x hc x h x g cc x
f ccc x
g x hccc x g c x hcc x 2 g c x hcc x 2 g cc x hc x h x g ccc x hc x g cc x
g x hccc x 3 g c x hcc x 3 g cc x hc x g ccc x h x
f 4 x
g x h 4 x g c x hccc x 3 g c x hccc x 3 g cc x hcc x 3 g cc x hcc x 3 g ccc x hc x
g ccc x hc x g 4 x h x
g x h 4 x 4 g c x hccc x 6 g cc x hcc x 4 g ccc x hc x g 4 x h x
(b) f n x
g x hn x nn 1 n 2 " 2 1
1ª¬ n 1 n 2 " 2 1 º¼
g c x h n 1 x nn 1 n 2 " 2 1
3 2 1 ª¬ n 3 n 4 " 2 1 º¼
nn 1 n 2 " 2 1
2 1 ª¬ n 2 n 3 " 2 1 º¼
g cc x h n 2 x
g ccc x h n 3 x "
nn 1 n 2 " 2 1
g n 1 x hc x g n x h x
ª¬ n 1 n 2 " 2 1 º¼ 1
n!
n!
x g c x h n 1 x g cc x h n 2 x "
1! n 1 !
2! n 2 !
g x hn
Note: n!
n!
g n 1 x hc x g n x h x
n 1 !1!
n n 1 "3 ˜ 2 ˜ 1 (read "n factorial")
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 2.3
122.
ª¬ xf x º¼c
xf c x f x
ª¬ xf x º¼cc
xf cc x f c x f c x
ª¬ xf x º¼ccc
xf ccc x f cc x 2 f cc x
In general, ª¬ xf x º¼
123.
n
xf cc x 2 f c x
xf ccc x 3 f cc x
fc x
x n cos x nx n 1 sin x
When n
1: f c x
x cos x sin x
When n
2: f c x
x 2 cos x 2 sin x
When n
3: f c x
x3 cos x 3 x 2 sin x
4: f c x
x 4 cos x 4 x3 sin x
128.
2 cos x
2 sin x
2 sin x 2 sin x 3
y
3 cos x sin x
yc
3 sin x cos x
ycc
3 cos x sin x
ycc y
cos x
xn
x n cos x
129. False. If y
fc x
x n sin x nx n 1 cos x
dy
dx
dny
dx n
1: f c x
2: f c x
x sin x 2 cos x
x3
When n
3: f c x
x sin x 3 cos x
x4
When n
4: f c x
When n
1
, yc
x
hc c
2 x3 6 x 10
yc
6x2 6
ycc
12 x
yccc
12
yccc xycc 2 yc
0 when n ! 4.
132. True
133. True
134. True. If v t
135.
2 2
f c gc c g c f c c
0
y
f x gc x g x f c x .
f c 0 g c 0
x sin x 4 cos x
x5
1
2
, ycc
x2
x3
ª2º
ª 1º
x3 ycc 2 x 2 yc
x3 « 3 » 2 x 2 « 2 »
x
¬ ¼
¬ x ¼
125. y
f x g x , then
131. True
x sin x n cos x
.
x n 1
For general n, f c x
0
130. True. y is a fourth-degree polynomial.
x sin x n cos x
x n 1
x sin x cos x
x2
When n
3
3 cos x sin x 3 cos x sin x
f x
x n 1 x sin x n cos x
126.
yc
ycc
ycc y
x n cos x nx n 1 sin x.
For general n, f c x
2 sin x 3
y
127.
x n sin x
149
xf n x nf n 1 x .
f x
When n
124.
Product and Quotient
Quotie Rules and Higher-Orde
Higher-Order Derivatives
0.
2
°­x , x t 0
® 2
°̄ x , x 0
f x
x x
fc x
­2 x, x ! 0
®
¯ 2 x, x 0
f cc x
­2, x ! 0
®
¯ 2, x 0
0
vc t
c then a t
2 x
f cc 0 does not exist because the left and right
–12 x 12 x 2 6 x 2 6
24 x 2
derivatives do not agree at x
0.
INSTRUCTOR USE ONLY
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150
NOT FOR SALE
Chapter 2
136. (a)
Differentiation
ferentiation
fg c f cg c
fg cc f cg c f cg c f ccg
fg cc f ccg
(b)
fg cc
fg c f cg c
fg cc f cg c f cg c f ccg
True
fg cc 2 f cg c f ccg
z fg cc f ccg
137.
d
ª f x g x h x º¼
dx ¬
False
d
ª f x g x h x º¼
dx ¬
d
ª f x g x º¼ h x f x g x hc x
dx ¬
ª¬ f x g c x g x f c x º¼ h x f x g x hc x
f c x g x h x f x g c x h x f x g x hc x
Section 2.4 The Chain Rule
y
f g x
u
g x
y
f u
1. y
5x 8
u
5x 8
y
u4
2. y
1
x 1
u
x 1
y
u 1 2
3. y
x3 7
u
x3 7
y
u
4. y
3 tan S x 2
u
S x2
y
3 tan u
5. y
csc3 x
u
csc x
y
u3
6. y
sin
5x
2
u
5x
2
y
sin u
7. y
4x 1
yc
3 4x 1
2
5 2 x3
4
8. y
yc
gc x
10.
11.
3
12 4 x 1
4
3
3
3 x
60 x x 2
3
3 4 9x
4
3
12 4 9 x
3
3
3
13. y
108 4 9 x
3
14.
f x
fc x
23
fc t
2
1 3
9t 2
9
3
fc t
2
5t
1
1 2
5t
1
2
3
6
9t 2
15. y
yc
12
1
2 5t
12
4 3x 2
1 2
1
4 3x2
6 x
2
gc x
60 x 2 x
2
9
9t 2
5t
2
yc
f t
f t
4 3x 2
12. g x
54 2 x
2
9. g x
4
3
6x2 1
6x2 1
2 3
1 2
6x 1
12 x
3
x2 4x 2
4 3x 2
13
4x
6x2 1
4x
23
x2 4x 2
1 2
1 2
x 4x 2
2x 4
2
2 4 9 x2
3x
2 9 x2
3
6x2 1
2
12
x2
x2 4x 2
14
3 4
§1·
2 x
2¨ ¸ 9 x 2
4
© ¹
x
x
9 x2
34
9 x2
3
INSTRUCTOR
U
USE
E ONLY
4
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.4
16.
3
f x
yc
12 x 5
1
2 3
12 x 5
12
3
fc x
17. y
12 x 5
x 2
4
12 x 5
2
1 x 2
x 2
4 5t t
2
2
t 3
fc t
2 t 3
yc
3
2
yc
1
8x 5
x 1 x2
1 x2
2
12
1 2
12
ª1
º
x« 1 x2
2 x » 1 x 2
1
2
¬
¼
1 2
x2 1 x2
2
1
t 3
5
12
t 2
1 2
1 x2
12
ª x 2 1 x 2 º
¬
¼
1 2x2
26. y
3
yc
4
1 2
x 16 x 2
2
12
1 2
1 2§ 1
·
2 x ¸ x 16 x 2
x ¨ 16 x 2
2 ©2
¹
x3
x 3x 2 32
x 16 x 2
2 16 x 2
2 16 x 2
5
27. y
yc
32
x
x
x2 1
x 1
2
12
x2 1
12
1 2
§1·
1 x¨ ¸ x 2 1
2x
© 2¹
ª x2 1 1 2 º 2
¬«
¼»
x2 1
12
x2 x2 1
1 2
x2 1
3
x2 1
1
t2 2
1 2
x 1
2
28. y
32
ª¬ x 2 1 x 2 º¼
x2 1
1
t
yc
32
x2 1
x
x 4
x 4
12
x4 4 2x4
x4 4
4
x2 x 2
fc x
3
4
x 2 ª4 x 2 1 º x 2 2 x
¬
¼
1 2
1 4
x 4
4 x3
2
x4 4
1 x
3
f x
3
4
4
t2 2
1 2
1
t 2
3 2
1 2
2t
t 2
2
t
2
t2 2
23.
2
t 2 5t 4
3 t 2
3x 5
2
x 1 x2
3
2
gc t
2
25. y
1
1 2
3x 5
3x 5
1
3 2
3
3x 5
2
3
3
2 2x 5 1
2 x 5 ª¬6 x 2 x 5 º¼
5 2t
4
12 t 2
2
1 x2
3
2 3x 5
22. g t
x 3 2x 5
2
f t
21. y
fc x
2t 5
4 5t t 2
t 2
2
4 5t t
5 2t
yc
x 2x 5
1
1
sc t
23
151
3
f x
2x 5
18. s t
20. y
24.
1
1
4 5t t 2
19.
13
The
Th Chain Rule
32
4 x4
x4 4
32
4 x4
x4 4
3
3
2 x x 2 ª¬2 x x 2 º¼
2x x 2
3
3x 2
INSTRUCTOR USE ONLY
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152
Chapter 2
NOT FOR SALE
Differentiation
ferentiation
§ x 5·
¨ 2
¸
© x 2¹
29. g x
2
31.
§ 2
·
§ x 5 ·¨ x 2 x 5 2 x ¸
2¨ 2
¸
2
© x 2 ¹¨¨
¸¸
x2 2
©
¹
gc x
§ 1 2v ·
¨
¸
©1 v¹
fc v
2
§ 1 2v · §¨ 1 v 2 1 2v ·¸
3¨
¸
2
¸
© 1 v ¹ ©¨
1 v
¹
1 v
3
2 x 5 x 2 10 x 2
§ t2 ·
¨ 3
¸
©t 2¹
30. h t
2t 4t t
33.
2
4
3
2t 4 t
3
t3 2
3
t3 2
3
f x
x2 3
5
fc x
2 x2 3
x
5
2x 3
6 3x 2 2
gc x
3 2 x2 1
4
2x 1
x2 3
5
10 x 2 x 2 3
4
xº
¼»
4 2
4 x2 1
3
2x
24 x x 2 1
x 1
x 1
20 x x 2 3
3
2 x2 1
36. y
2
1 3x 2 4 x3 2
x x2 1
2
4
3x 2 9 x 2
4
9
2 x2 3
5
20 x 2 x 2 3
4
2x
4 3
2 x2 1
2
6 x 2 18 x 4
2
9
34. g x
35. y
2
2x 3
x 5 x2 3
2 ª10 x x 2 3
¬«
yc
3 3x 2 2
2
2 ·
§ 3
§ t 2 ·¨ t 2 2t t 3t ¸
2¨ 3
¸
2
¸¸
© t 2 ¹¨¨
t3 2
©
¹
hc t
3
2
2
§ 3x 2 2 · § 2 x 3 6 x 3x 2 2 ·
¸
3¨
¸ ¨
2
¸
2x 3
© 2 x 3 ¹ ©¨
¹
gc x
2
4
§ 3x 2 2 ·
¨
¸
© 2x 3 ¹
32. g x
3
x2 2
2
9 1 2v
2 x 5 2 10 x x 2
x2 2
3
f v
yc
2
The zero of yc corresponds to the point on the graph of y
where the tangent line is horizontal.
4 2
2x
x 1
1
2x x 1
32
yc has no zeros.
7
2
y
y
−1
y′
5
y′
−6
6
−1
−2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 2.4
x 1
x
37. y
yc
y
x 1 x
−5
x 1
1
x 1
2
2
1
y
sin 2 x
yc
2 cos 2 x
2
The slope of sin ax at the origin is a.
42. (a)
y
sin 3 x
g
yc
3 cos 3 x
g′
yc 0
−2
10
3
3 cycles in >0, 2S @
−2
dy
dx
cos x
2 cycles in >0, 2S @
6
y
yc
yc 0
1
x 1
g c has no zeros.
39.
sin x
153
1 cycle in >0, 2S @
(b)
x 1
gc x
4
−2
yc has no zeros.
y
yc 0
y′
2x x 1
38. g x
41. (a)
4
Th
The Chain Rule
cos S x 1
x
S x sin S x cos S x 1
x2
S x sin S x cos S x 1
x2
(b)
y
yc
yc 0
§ x·
sin ¨ ¸
© 2¹
§1·
§ x·
¨ ¸ cos¨ ¸
© 2¹
© 2¹
1
2
Half cycle in >0, 2S @
The zeros of yc correspond to the points on the graph of
y where the tangent lines are horizontal.
The slope of sin ax at the origin is a.
3
y
y
cos 4 x
dy
dx
4 sin 4 x
y
sin S x
dy
dx
S cos S x
45. g x
5 tan 3 x
43.
−5
5
y′
−3
40.
y
dy
dx
44.
1
x
1
1
2 x tan sec 2
x
x
x 2 tan
The zeros of yc correspond to the points on the graph of
y where the tangent lines are horizontal.
6
46. h x
hc x
y
−4
gc x
5
y′
−6
47. y
yc
15 sec 2 3 x
sec x 2
2 x sec x 2 tan x 2
sin S x
2
sin S 2 x 2
cos S 2 x 2 ª¬2S 2 xº¼
2S 2 x cos S x
2
2
48. y
cos 1 2 x
yc
sin 1 2 x
2S 2 x cos S 2 x 2
cos 1 2 x
2
2
2 1 2 x 2
4 1 2 x sin 1 2 x
2
INSTRUCTOR USE ONLY
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Chapter 2
154
49. h x
hc x
NOT FOR SALE
Differentiation
56. g T
sin 2 x cos 2 x
sin 2 x 2 sin 2 x cos 2 x 2 cos 2 x
cos 2 8T
cos 8T
2
gc T
2 cos 8T sin 8T 8
f T
1 sin 2 2T
4
fcT
2 14 sin 2T cos 2T 2
16 cos 8T sin 8T
2 cos 2 x 2 sin 2 x
2
2
57.
2 cos 4 x
1 sin 4 x
2
Alternate solution: h x
hc x
50. g T
gc T
1
cos 4 x 4
2
51.
f x
fc x
58. h t
cot x
sin x
59.
cos x
sin 2 x
fc t
6 sec S t 1 sec S t 1 tan S t 1 S
6S sec 2 S t 1 tan S t 1
cos v
csc v
1 cos x
sin 3 x
2
60.
cos v ˜ sin v
cos v cos v sin v sin v
2
61.
cos 2v
8 sec x ˜ sec x tan x
5 cos 2 S t
5 cos S t
3x 5 cos S x
dy
dx
3 5 sin S 2 x 2 2S 2 x
y
dy
dx
8 sec 2 x tan x
62. y
2
yc
10 cos S t sin S t S
63. y
2
tan 2 5T
fcT
2 tan 5T sec 2 5T 5
3x 5 cos S 2 x 2
3 10S 2 x sin S x
2
1
1
2
x sin 4 x 2
sin 2 x
4
4
1 1/2 1
1
2
2
cos 4 x 8 x
2 x cos 2 x
x
2
4
2 x
x sin x1/3 sin x
1/3
2/3
§1
· 1
cos x1/3 ¨ x 2/3 ¸ sin x
cos x
3
3
©
¹
yc
10 tan 5T sec 2 5T
sin tan 2 x
cos tan 2 x sec 2 2 x 2
2 cos tan 2 x sec2 2 x
cos sin tan S x
64. y
yc
65.
tan 5T
cos3 S t 1
1 ª cos x1/3
cos x º
« 2/3 »
2/3
3« x
sin x ¼»
¬
5S sin 2S t
f T
2
y
10S sin S t cos S t
55.
6S sin S t 1
2
4 sec x
gc t
3sec 2 S t 1
sin 4 x
2
54. g t
4 cot S t 2 csc 2 S t 2 S
f t
sin x sin x cos x 2 sin x cos x
cos v sin v
yc
2 cot 2 S t 2
2
2
53. y
1 sin 4T
2
4S cot S t 2 csc 2 S t 2
tan 2 12T º
¼
sin x 2 cos x
sin 3 x
gc v
hc t
sec 12T sec 2 12T 12 tan 12T sec 12T tan 12T 12
2
52. g v
sin 2T cos 2T
2 cos 4 x
sec 12T tan 12T
1 sec 1 T ªsec 2 1 T
2
2 ¬
2
1 sin 2T 2
4
sin sin tan S x ˜
y
yc
yc 1
x2 8x
1
sin tan S x
2
x2 8x
12
1 81
2
5
9
cos tan S x sec 2 S x S
S sin sin tan S x cos tan S x sec 2 S x
2 sin tan S x
, 1, 3
2x 4
1/ 2
1 2
x 8x
2x 8
2
1 4
1/ 2
2 x 8x
2
x 4
12
x2 8x
5
3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.4
66.
y
yc
1/5
3x3 4 x
,
71.
2, 2
4/5
1 3
3x 4 x
9 x2 4
5
9x 4
5 3x 4 x
yc 2
67.
68.
69.
fc x
5 x 2
5x 2
3
60
100
2
3x
x 2 3x
1
2
x 2 3x
2
fc 4
2
yc
1·
§
, ¨ 2, ¸
2¹
©
0
1
x
x3 2
3
2x 3
2
x2 2
73. (a) f x
§ 1·
, ¨ 4, ¸
© 16 ¹
3
3t 2
,
0, 2
t 1
t 1 3 3t 2 1
t 1
t 1
fc x
2x
2 x2 7
0
6
6
−2
74. (a)
2
f x
fc x
12
1
1 2
x x2 5
x x 5 , 2, 2
3
3
1/ 2
1/ 2
1 ª1 2
º 1
2 x » x2 5
x x 5
3 «¬ 2
3
¼
x2
2
x 5
4
1
3
33
3
2
2
fc 2
x 4
, 9, 1
2x 5
2x 5 1 x 4 2
2x 5
2x 5
§S 2 ·
, ¨ , ¸
©2 S¹
1
sin x
2 x
2 cos x
4, 5
8
5
2
1
3
x2 5
13
9
Tangent line:
13
y 2
x 2 Ÿ 13 x 9 y 8
9
2
2x 5 2x 8
fc 9
,
fc 4
−6
5
1/ 2
1/ 2
1 2
2x 7
4x
2
5
f x
12
(4, 5)
t 1
70.
2x2 7
fc x
3t 3 3t 2
fc 0
1
1 2
cos x
sin x
2
(b)
5
32
x 1 cos x
Tangent line:
8
y 5
x 4 Ÿ 8x 5 y 7
5
2 2 x 3
x2 3x
cos x
yc S /2 is undefined.
15 x 2
3
5
1
2 x 3 x
fc t
3 sec 4 x sec 4 x tan 4 x 4
yc 0
3
fc x
f t
yc
0, 25
2
72. y
f x
f x
26 sec3 4 x,
4/5
1
2
5
x3 2
f c 2
y
155
12 sec3 4 x tan 4 x
2
3
The
Th Chain Rule
(b)
0
6
2
−9
9
13
2x 5
2
−6
13
18 5
2
1
13
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 2
156
Differentiation
2
y
4 x3 3 ,
yc
2 4 x3 3 12 x 2
75. (a)
yc 1
1, 1
24 x 2 4 x3 3
yc
24
Tangent line:
y 1 24 x 1 Ÿ 24 x y 23
(b)
§S ·
yc¨ ¸
©4¹
0
§ 3S ·
3 sin ¨ ¸
© 4 ¹
3 2
2
2
2
3 2 §
S·
¨x ¸
2 ©
4¹
y
3 2
3 2S
x 2
8
(− 1, 1)
1
(b)
−2
2/3
f x
9 x2
fc x
1/3
2
9 x2
2 x
3
76. (a)
§S
2·
cos 3 x, ¨¨ , ¸¸
4
2
©
¹
3 sin 3 x
Tangent line: y 14
−2
y
78. (a)
4
fc1
38
1/3
π
2
4 x
3 9 x2
−2
1/3
2
3
f x
§S ·
tan 2 x, ¨ , 1¸
©4 ¹
fc x
2 tan x sec 2 x
§S ·
f c¨ ¸
©4¹
21 2
79. (a)
Tangent line:
2
y 4
x 1 Ÿ 2 x 3 y 14
3
(b)
( π4 , − 22 (
−π
2
, 1, 4
2
2
2
0
4
Tangent line:
6
(1, 4)
−2
y 1
5
(b)
S·
§
4¨ x ¸ Ÿ 4 x y 1 S
4¹
©
0
4
−1
77. (a)
f x
sin 2 x,
fc x
2 cos 2 x
fc S
S, 0
−
( (
−4
2
Tangent line:
2 x S Ÿ 2 x y 2S
y
80. (a) y
§S ·
2 tan 3 x, ¨ , 2 ¸
©4 ¹
yc
6 tan 2 x ˜ sec 2 x
0
2
(b)
0
π
,1
4
(π , 0)
§S ·
yc¨ ¸
©4¹
2
61 2
12
Tangent line:
−2
y 2
(b)
S·
§
12¨ x ¸ Ÿ 12 x y 2 3S
4¹
©
0
3
( π4 , 2(
−π
2
π
2
−1
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 2.4
81.
25 x 2
f x
25 x 2
fc x
1
25 x 2 2 x
2
fc 3
3
4
12
,
3, 4
25 x 2
fc x
fc1
x 2 x2
2 x2
2
x
Tangent line:
0
1 2
,
157
1, 1
for x ! 0
32
2 x2
2
Tangent line: y 1
3
x 3 Ÿ 3 x 4 y 25
4
y 4
x
f x
82.
The
Th Chain Rule
2 x 1 Ÿ 2x y 1
0
3
8
(1, 1)
(3, 4)
−2
−9
9
2
−1
−4
83.
0 x 2S
f x
2 cos x sin 2 x,
fc x
2 sin x 2 cos 2 x
2 sin x 2 4 sin 2 x
2 sin x sin x 1
0
sin x 1 2 sin x 1
0
2
1 Ÿ x
sin x
0
3S
2
S 5S
1
,
Ÿ x
2
6 6
S 3S 5S
,
,
6 2 6
sin x
Horizontal tangents at x
§ S 3 3 · § 3S ·
§ 5S 3 3 ·
Horizontal tangent at the points ¨¨ ,
¸¸, ¨ , 0 ¸, and ¨¨ , ¸
2 ¸¹
©6 2 ¹ © 2 ¹
© 6
f x
84.
fc x
x
2x 1
2x 1
86.
12
x 2x 1
1 2
2x 1
2x 1 x
2x 1
6 x3 4
fc x
18 x3 4
f cc x
54 x 2 2 x3 4 3 x 2 108 x x3 4
2x 1
32
2
2
32
0 Ÿ x
f x
1
x 6
fc x
x 6
2
f cc x
2x 6
3
1
Horizontal tangent at 1, 1
85.
54 x 2 x3 4
3x2
432 x x3 4 x3 1
87.
x 1
2
108 x x3 4 ª¬3x3 x3 4º¼
32
x 1
2x 1
3
f x
x 6
1
2
x 6
3
4
f x
5 2 7x
fc x
20 2 7 x
3
7
f cc x
420 2 7 x
2
140 2 7 x
7
3
2940 2 7 x
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
158
88.
90.
NOT FOR SALE
Chapter 2
f x
Differentiation
8
x 2
8x 2
2
2
f x
sin x 2
fc x
2 x cos x 2
f cc x
2 x ª¬2 x sin x 2 º¼ 2 cos x 2
89.
3
fc x
16 x 2
f cc x
48 x 2
f x
sec 2 S x
fc x
2 sec S x S sec S x tan S x
48
4
x 2
2 cos x 2 2 x 2 sin x 2
4
2S sec 2 S x tan S x
f cc x
2S sec 2 S x sec 2 S x S 2S tanS x 2S sec 2 S x tan S x
2S 2 sec 4 S x 4S 2 sec 2 S x tan 2 S x
2S 2 sec 2 S x sec 2 S x 2 tan 2 S x
2S 2 sec 2 S x 3 sec 2 S x 2
91.
92.
h x
1 3x 1 3 ,
9
hc x
1 3 3x 1 2 3
9
hcc x
2 3x 1 3
hcc 1
24
f x
fc x
f cc x
f cc 0
93.
95.
1, 64
9
f′
3x 1
3
2
2
1
18 x 6
x
−2
2
f
1
1 2
§ 1·
, ¨ 0, ¸
x 4
x 4
© 2¹
1
3 2
x 4
2
3
3
5 2
x 4
52
4
4 x 4
3
−2
−3
The zeros of f c correspond to the points where the graph
of f has horizontal tangents.
96.
y
f
3
f′
f
2
3
128
1
−3
−2
−1
f′
f x
cos x 2 ,
fc x
sin x 2 2 x
f cc x
2 x cos x 2 2 x 2 sin x 2
0, 1
x
2
3
−2
f is decreasing on f, 1 so f c must be negative there.
f is increasing on 1, f so f c must be positive there.
y
97.
0
1
−1
−3
2 x sin x 2
4 x 2 cos x 2 2 sin x 2
f cc 0
y
3
94.
§S
¨ ,
©6
f
2
·
3¸
¹
g t
tan 2t ,
gc t
2
2 sec 2t
g cc t
4 sec 2t ˜ sec 2t tan 2t 2
x
−3
−1
2
f′
2
8 sec 2t tan 2t
§S ·
g cc¨ ¸
©6¹
32 3
The zeros of f c correspond to the points where the graph
of f has horizontal tangents.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.4
98.
y
99. g x
gc x
4
3
2
159
f 3x
f c 3x 3 Ÿ g c x
3 f c 3x
f
100. g x
x
−1
−2
The
Th Chain Rule
f x2
4
f′
gc x
−3
f c x2 2 x Ÿ gc x
2 xf c x 2
−4
The zeros of f c correspond to the points where the graph
of f has horizontal tangents.
101. (a) g x
f x 2 Ÿ gc x
fc x
(b) h x
2 f x Ÿ hc x
2fc x
(c) r x
f 3 x Ÿ r c x
f c 3x 3
So, you need to know f c 3 x .
rc 0
3 f c 0
3
r c 1
3 f c 3
3 4
(d) s x
(b)
x
2
1
0
1
2
3
fc x
4
2
3
13
1
2
4
13
1
gc x
4
2
3
13
1
2
4
12
hc x
8
4
3
23
2
4
8
fc x 2
rc x
12
1
1
2
f x 2 Ÿ sc x
So, you need to know f c x 2 .
102. (a)
3 f c 3 x
sc 2
fc 0
f x
g xh x
fc x
g x hc x g c x h x
fc 5
3 2 6 3
f x
g h x
fc x
g c h x hc x
fc 5
g c 3 2
sc x
13
–4
13 , etc.
24
2 g c 3
Not possible, you need g c 3 to find f c 5 .
(c)
f x
fc x
fc 5
(d)
g x
h x
h x g c x g x hc x
ª¬h x º¼
2
3 6 3 2
3
2
12
9
4
3
3
f x
ª¬ g x º¼
fc x
3ª¬ g x º¼ g c x
fc 5
3 3
2
2
6
162
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
Chapter 2
160
Differentiation
4, g c 1
12 , f c 4
103. (a) h x
f g x ,g1
hc x
f c g x gc x
hc 1
f c g 1 gc 1
f c 4 gc 1
(b) s x
g f x , f 5
6, f c 5
sc x
gc f x
sc 5
gc f 5 f c 5
1
1 12
1
2
1, g c 6 does not exist.
fc x
g c 6 1
sc 5 does not exist because g is not differentiable at 6.
104. (a) h x
hc x
f c g x gc x
hc 3
f c g 3 gc 3
(b) s x
105. (a)
(a) Amplitude: A
fc 5 1
gc f x f c x
sc 9
gc f 9 f c 9
F
132,400 331 v
Fc
1 132,400 331 v
gc 8 2
1 2
y
2
2
1
132,400
331 v
2
v
(a)
2
1
5
St º
ª S
1.75« sin »
5¼
¬ 5
0.35S sin
St
5
100
132,400
331 v
2
0
30, F c | 1.016.
13
0
yc
>
1 12 sin 12t
3
100
@ 14 >12 cos 12t@
4 sin 12t 3 cos 12t
S 8, y
0
0.25 ft and v
The model is a good fit.
0.2 cos 8t
0.2>8 sin 8t @
When t
3, dT dt
13
0
4 ft/sec.
(c)
The maximum angular displacement is T
1 d cos 8t d 1).
dT
dt
5
St
1 cos 12t 1 sin 12t
3
4
When t
107. T
1.75 cos
S
109. (a) Using a graphing utility, you obtain a model similar to
T (t ) 56.1 27.6 sin 0.48t 1.86 .
(b)
106. y
2S
10
1
1 132,400 331 v
When v
yc
(b) v
1
30, F c | 1.461.
132,400 331 v
Fc
y
3.5
1.75
2
1.75 cos Z t
Period: 10 Ÿ Z
sc x
(b) F
1
2
g f x
When v
A cos Z t
108. y
f g x
20
0.2 (because
0
13
1.6 sin 8t
−20
1.6 sin 24 | 1.4489 rad/sec.
T c(t ) | 13.25 cos 0.48t 1.86
(d) The temperature changes most rapidly around spring
(March–May), and fall (Oct–Nov).
110. (a) According to the graph C c 4 ! C c 1 .
(b) Answers will vary.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.4
ª
º
3
»
400 «1 2
«
»
2
t
2
¬
¼
N
111.
Nc t
3
2400 t 2 2
t2 2
0 bacteria/day
(b) N c 1
4800 1
(c) N c 2
4800 2
9 2
3
rc 1
f c g 1 gc 1
3
Also, g c 1
16 2
V 0
10,000
V
10,000
t 1
10,000 t 1
(a)
5000
43 2
5000
8
g x
sin 2 x cos 2 x
gc x
2 sin x cos x 2 cos x sin x
f x
gc x
117. (a) If f x
5000
t 1
d
ª f x º¼
dx ¬
f c x 1
32
f c x
(b) If f x
f x , then
f ccc x
E 3 cos E x
f c x
f 4
E 4 sin E x
So, f c is odd.
fc x p
E 2 sin E x E 2 sin E x
0
u
118.
k
1 E 2 k sin E x
d
ªu º
dx ¬ ¼
uuc
E 2 k 1 cos E x
u2
f x for all x, then
f c x , which shows that f c is
periodic as well.
f 2 x , then g c x
(b) Yes, if g x
2 sec 2 x tan x, which
So, f c x is even.
625 dollars/year
E 2 sin E x
114. (a) Yes, if f x p
2 sec 2 x tan x and
fc x .
f cc x
1
fc x .
d
ª f x º¼
dx ¬
fc x
d
ª f x º¼
dx ¬
f c x 1
f 2 k 1 x
0
f x , then
E cos E x
k 1
0
are the same.
1 2
fc x
f 2k x
1 Ÿ gc x
2 tan x ˜ sec 2 x
1
and
2
5
.
8
5
. So, sc 4
4
g x 1
sin E x
(b) f cc x E 2 f x
(c)
1§ 5 ·
¨ ¸
2© 4 ¹
fc 4
sec 2 x
5000
| 1767.77 dollars/year
23 2
(c) V c 3
64
62
5 § 5·
, g c¨ ¸
2 © 2¹
(b) tan 2 x 1
k
3 2
§ 1·
10,000¨ ¸ t 1
© 2¹
Vc 1
113. f x
gc f 4 f c 4
Equivalently,
fc x
2 sec x ˜ sec x tan x
k
01
dV
dt
(b)
sc 4
5
.
4
0.
Taking derivatives of both sides, g c x
k
t 1
V
gc f x f c x
116. (a)
(f) The rate of change of the population is decreasing as
t o f.
112. (a)
(b) sc x
Note that f 4
19,200
| 3.3 bacteria/day
5832
3
0. So, r c 1
161
50
62
4 and f c 4
Note that g 1
14,400
| 10.8 bacteria/day
1331
3
4800 4
(e) N c 4
f c g x gc x
9600
| 44.4 bacteria/day
216
4800 3
(d) N c 3
115. (a) r c x
4800
| 177.8 bacteria/day
27
3
4 2
2
4800t
2t
(a) N c 0
1 2
400 1200 t 2 2
The
Th Chain Rule
2 f c 2x .
119.
d
ª f x º¼
dx ¬
fc x
fc x .
u2
1 2 1 2
d ª 2º
2uuc
u
u
¼
2
dx ¬
u
uc , u z 0
u
g x
3x 5
gc x
§ 3x 5 ·
3¨
,
¨ 3x 5 ¸¸
©
¹
x z
5
3
INSTRUCTOR USE ONLY
Because f c is periodic, so is g c.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
120.
Differentiation
f x
x2 9
fc x
§ x2 9 ·
¸, x z r 3
2 x¨ 2
¨ x 9 ¸
©
¹
121. h x
f x
tan x
f S 4
1
fc x
sec 2 x
fc S 4
2
f cc x
2
2 sec x tan x
f cc S 4
4
P1 x
2 x S 4 1
P2 x
1
2
4 x S 4 2 x S 4 1
2
123. (a)
x cos x
x
x sin x cos x,
x
hc x
122.
NOT FOR SALE
Chapter 2
162
x z 0
2 x S 4
(b)
f x
sin x
fc x
§ sin x ·
cos x¨
, x z kS
¨ sin x ¸¸
©
¹
2
2 x S 4 1
5
f
P2
P1
p
2
0
−1
(c) P2 is a better approximation than P1.
(d) The accuracy worsens as you move away from
x S 4.
124. (a)
f x
sec x
f S 6
fc x
sec x tan x
fc S 6
f cc x
sec x sec 2 x tan x sec x tan x
f cc S 6
2
3
2
3
10 3
9
sec3 x sec x tan 2 x
P1 x
2
2
x S 6 3
3
P2 x
S·
S·
1 § 10 ·§
2§
2
˜¨
¸¨ x ¸ ¨ x ¸ 2 © 3 3 ¹©
6¹
3©
6¹
3
2
2
S·
S·
2§
2
§ 5 ·§
¨
¸¨ x ¸ ¨ x ¸ 6¹
3©
6¹
3
© 3 3 ¹©
(b)
3
f
P2
−1.5
1.5
P1
−1
(c) P2 is a better approximation than P1.
(d) The accuracy worsens as you move away from x
125. False. If y
126. False. If f x
1 x
12
S 6.
1 1 x 1 2 1 .
2
, then yc
sin 2 2 x, then f c x
2 sin 2 x 2 cos 2 x .
127. True
128. True
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.5
f x
a1 sin x a2 sin 2 x " an sin nx
fc x
a1 cos x 2a2 cos 2 x " nan cos nx
fc 0
a1 2a2 " nan
129.
fc 0
a1 2a2 " nan
130.
ª
º
d « Pn x
»
dx « x k 1 n 1 »
¬
¼
xk 1
lim
x 0
xo0
˜
sin x
x
n
Pnc x Pn x n 1 x k 1 kx k 1
lim
xo0
f x
d1
sin x
x k 1 Pnc x n 1 kx k 1Pn x
2n 2
xk 1
n2
1 º
Ÿ
dx ¬ x 1»¼
Pn 1 x
xk 1
Pn 1 1
n 1 kPn 1
n« k
n2 d ª d n ª
1 ºº
« n« k
»
dx ¬ dx ¬ x 1»¼¼
kx k 1
dª 1 º
dx «¬ x k 1»¼
x 1
k
x k 1 Pnc x n 1 kx k 1Pn x
P1 x
2
x 1
k
2
Ÿ P1 1
You now use mathematical induction to verify that Pn 1
n
n 1 k Pn 1
Pn 1 1
f x
sin x
163
n 1 d n ª
xk 1
1,
f x f 0
xk 1
Pn x
For n
n 1
lim
xo0
Implicit Differentiation
Dif
D
n 1 k k n!
k
n 1
k . Also, P0 1
1.
n
k n! for n t 0. Assume true for n. Then
n 1 !.
Section 2.5 Implicit Differentiation
1.
x2 y 2
9
2 x 2 yyc
4.
25
2 x 2 yyc
0
yc
x
y
0
2 y x yc
y 3x 2
yc
y 3x 2
2y x
x2 y y 2 x
2
x yc 2 xy y 2 yxyc
0
6.
2
x 2 xy yc
16
1 1 2
1
x
y 1 2 yc
2
2
0
yc
x 1 2
y 1 2
y
x
2 x3 3 y 3
64
6 x 9 y yc
0
2
9 y yc
6x2
yc
6x2
9 y2
2
2
2
x1 2 y1 2
2
3x xyc y 2 yyc
x
y
x2 y 2
3.
7
2
0
yc
2.
x3 xy y 2
5.
yc
y 2 2 xy
y y 2x
x x 2y
x3 y 3 y x
0
2 3
3 x y yc 3 x y yc 1
0
3 x y 1 yc
1 3x 2 y 3
yc
1 3x 2 y 3
3x3 y 2 1
7.
3 2
3 2
2x2
3 y2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 2
164
8.
Differentiation
x2 y 1
xy
1
1 2
xy
xyc y
2
x
y
yc 2 xy
2 xy
2 xy x 2 yc
§ x
·
x 2 ¸ yc
¨¨
¸
© 2 xy
¹
y
cos x 4 sin 2 y yc
0
cos x
4 sin 2 y
yc
2 xy x yc
2 xy yc
sin S x cos S y
12.
xy
2
S cos S x S sin S y yc
0
x
4 xy
xy y
x 2x
2
yc
2
xy
12
3x 2 3 x 2 yc 6 xy 4 xyyc 2 y 2
0
6 xy 3 x 2 2 y 2
yc
6 xy 3 x 2 2 y 2
4 xy 3 x 2
4 cos x sin y
1
4>sin x sin y cos x cos y yc@
0
cos x tan y 1
x sec 2 y
yc
4 xy 3 x yc
2
x sec 2 y yc 1 tan y 1
cos x
x3 3 x 2 y 2 xy 2
cot y
x y
csc y yc
1 yc
14.
2
yc
15.
cos x cos y yc
sin x sin y
yc
sin x sin y
cos x cos y
cos S x
sin S y
x 1 tan y
13. sin x
xy
2
0
xy
2
2
2 sin S x cos S y ª¬S cos S x S sin S y ycº¼
y
2
x
yc
10.
1
2
2 xy 9.
sin x 2 cos 2 y
11.
1
1 csc 2 y
y
sin xy
yc
> xyc y@ cos xy
yc x cos xy yc
yc
tan x tan y
1
cot 2 y
tan 2 y
y cos xy
y cos xy
1 x cos xy
1
y
16. x
sec
1
yc
y2
sec 1 y tan 1 y
1
1
yc
sec tan
y2
y
y
17. (a) x 2 y 2
§1· §1·
y 2 cos¨ ¸ cot ¨ ¸
© y¹ © y¹
64
y
(b)
y2
64 x 2
y
r
12
64 x 2
y1 =
64 − x 2
4
12
4
x
− 12
−4
− 12
y2 = −
64 − x 2
(c) Explicitly:
dy
dx
r
1 2
1
64 x 2
2 x
2
(d) Implicitly: 2 x 2 yyc
0
yc
Bx
64 x
x
2
r
64 x
2
x
y
x
y
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.5
18. (a) 25 x 2 36 y 2
36 y
19. (a) 16 y 2 x 2
300
300 25 x
2
25 12 x
2
2
y
6
4
x 2 16
y2
x2
1
16
6
12 − x 2
6
x 2 16
4
y1 = 1
x 2 + 16
4
4
2
2
x
−6 −4 −2
2
−2
−4
−6
4
x
−6
6
y2 = − 5
−4
y2 = −
−6
1
4
x 2 + 16
(c) Explicitly:
1 2
5§1·
r ¨ ¸ 12 x 2
2x
6© 2¹
6 12 x 2
25 x
36 y
4
(d) Implicitly: 50 x 72 y ˜ yc
50 x
72 y
x2 y 2 4x 6 y 9
x2 4 x 4 y 2 6 y 9
x 2
y 3
2
y 3
2
y 3
y
x
16 y
(d) Implicitly: 16 y 2 x 2
16
32 yyc 2 x
0
32 yyc
2x
yc
2x
32 y
x
16 y
x 2
y 3
0
yc
2
1 2
1 2
2 x
x 16
2
4
rx
rx
2
r
4
4y
x 16
r
dy
dx
5x
B
20. (a)
6
−2
12 − x 2
6
(c) Explicitly:
dy
dx
x 2 16
16
y
(b)
y1 = 5
r
y
y
(b)
165
16
2
16 y
25
12 x 2
36
5
r
12 x 2
6
y2
Implicit Differentiation
Dif
D
25 x
36 y
0
9 4 9
4
4 x 2
r
2
4 x 2
3 r
2
4 x 2
2
y
(b)
1
y1 = − 3 +
4 − (x − 2)2
1
4
x
−1
−1
2
3
5
6
−2
−3
−4
−5
−6
y2 = −3 −
4 − (x − 2)2
(c) Explicitly:
dy
dx
2 1 2
1
r ª4 x 2 º
¼
2¬
B
x 2
4 x 2
x 2
y 3
ª
¬ 2 x 2 º¼
(d) Implicitly:
2 x 2 yyc 4 6 yc
0
2 yyc 6 yc
2x 4
yc 2 y 6
2 x 2
2
yc
2 x 2
2 y 3
INSTRUCTOR USE ONLY
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NOT FOR SALE
166
Chapter 2
Differentiation
21.
xy
6
xyc y 1
0
3 x 3 y yc
6 xyc 6 y
xyc
y
3 y 2 6 x yc
6 y 3x 2
yc
y
x
yc
6 y 3x 2
3y2 6x
2
At 6, 1 : yc
y3 x2
4
3 y yc 2 x
0
22.
2
1
6
27.
18 12
27 12
6
15
tan x y
x
1 yc sec x y
1
2
yc
22
At 2, 2 : yc
6 xy 1
2
At 2, 3 : yc
2x
3y2
yc
x3 y 3
26.
1 sec 2 x y
sec 2 x y
1
3
3 22
2
5
tan 2 x y
tan 2 x y 1
23.
x 2 49
x 2 49
y2
sin 2 x y
x 2 49 2 x x 2 49 2 x
2 yyc
x 2 49
2
At 0, 0 : yc
196 x
2 yyc
98 x
yc
y x 2 49
0
x cos y
1
x> yc sin y@ cos y
0
28.
2
x 49
2
3
x3 y 3
x3 3x 2 y 3xy 2 y 3
x3 y 3
x y
3x 2 y 3 xy 2
0
x 2 y xy 2
0
x yc 2 xy 2 xyyc y
0
2
2
x 2 xy yc
2
yc
At 1, 1 : yc
25.
cos y
x sin y
yc
2
At 7, 0 : yc is undefined.
24.
y 2 xy
1
cot y
x
cot y
x
§ S·
At ¨ 2, ¸ : yc
© 3¹
29.
2
x2
x 1
2
1
2 3
x2 4 y
8
x 2 4 yc y 2 x
0
y y 2x
yc
x x 2y
2 xy
x2 4
2 x ª¬8 x 2 4 º¼
1
23
5
x2 4
16 x
2 1 3
2
x
y 1 3 yc
3
3
0
x2 4
x
23
y
yc
At 8, 1 : yc
1
2
x 1 3
y 1 3
3
y
x
At 2, 1 : yc
32
64
2
1
2
§
¨ Or, you could just solve for y: y
©
8 ·
¸
x2 4 ¹
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section 2.5
4 x y2
x3
4 x 2 yyc y 2 1
3x 2
30.
3x 2 y 2
2y 4 x
yc
At 2, 2 : yc
x y
31.
2
2
4 x yc 8 xy
3
8 xy 4 x3 4 xy 2
4 yc x 2 y y 3 x 2
4 2 xy x3 xy 2
x y 6 xy
0
3x 3 y yc 6 xyc 6 y
0
32.
2
2
yc 3 y 2 6 x
6 y 3x 2
yc
6 y 3x 2
3 y2 6x
16 3 16 9
§ 4 8·
At ¨ , ¸ : yc
© 3 3¹
33.
2
y 3
2 y 3 yc
2
y 3
At 6, 1 : yc
2
13
7 x 2 6 3xy 13 y 2 16
0,
14 x 6 3 xyc 6 3 y 26 yyc
0
At
37.
2 y x2
y2 2x
3 x y
3x 4
0,
x 2 2 yyc 2 xy 2 18 x 8 yyc
0
y 3
37,
2 x 2 2 y 3 yc
6
1
2 1 3
2
x
y 1 3 yc
3
3
0
x 2
At 8, 1 : yc
y 3
6
Tangent line: y 4
6 x 4
y
6 x 28
3
6
3
8
x 3
6
3
8, 1
x 1 3
y 1 3
yc
x 2
yc
5,
0
y 3 yc
At 4, 4 : yc
4, 4
x2 3 y2 3
1
2 3
3
x 4
6
y
38.
18 x 2 xy 2
2x2 y 8 y
2 16 2 3 16 3
24
48 3
x 7
4, 2 3
18 4 2 4 12
At 4, 2 3 : yc
1 x 6
3
3
x2 y 2 9x2 4 y2
yc
4
5
8 3
8
Tangent line: y 1
6, 1
2
3, 1
6 3 y 14 x
26 y 6 3x
6 3 14 3
26 6 3 3
3, 1 : yc
1
y
x 2
x 2
Tangent line: y 2 3
Tangent line: y 1
34.
y
4
yc
2
32
40
64 9 8 3
4 x 5,
1 x 1
yc
0
3
Tangent line: y 1
36.
2 xy x3 xy 2
x2 y y3 x2
yc
3
1
2
4 x 2 yyc 4 y 3 yc 4 x 2 yc
At 1, 1 : yc
y
x
4x y
4 x 4 x yyc 4 xy 4 y yc
2
0
167
1, 1
2
4 x 2 yc y 8 x
2
xyc y
At 1, 1 : yc
2 x 2 y 2 2 x 2 yyc
3
1,
yc
2
2
xy
35.
Implicit Differentiation
Dif
D
13
§ y·
¨ ¸
© x¹
1
2
Tangent line: y 1
y
1
x 8
2
1
x 5
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 2
168
Differentiation
2
3 x2 y2
39.
6x y
2
100 x 2 y 2 ,
2 x 2 yyc
2
4, 2
42. (a)
100 2 x 2 yyc
At 4, 2 :
6 16 4 8 4 yc
100 8 4 yc
960 480 yc
800 400 yc
880 yc
x2
y2
6
8
x
y
yc
3
4
y
yc
4
yc
160
yc
2
11
y 2
Tangent line:
0
y x y
2
2x ,
y 2 x2 y 4
2 x2
2 yycx 2 2 xy 2 4 y 3 yc
4x
40.
2
(b)
1, 1
4
6 yc
2
yc
1
3
y
2 x 4
y
1x 2
3
3
x2
y2
2
8
yyc
x 4
yc
At 1, 2 : yc
(b)
1,
1, 2
43.
0
1
2
yc
2 x 4
Because
yc sec y
sec 2 y
y
y0 y y02
2
b2
b
x
yc
2 x 1
y y0
2 x 2 yyc
2
a2
b
xb 2
ya 2
x0 x
x2
02
2
a
a
44.
0 Ÿ yc
b 2 x
a2 y
b x0
x x0 , Tangent line at x0 , y0
a 2 y0
1
sec 2 y
1, you have
x0 x
yy
20
a2
b
1.
cos y
x
sin y ˜ yc
1
yc
sin 2 y cos 2 y
2
sin y
1, you have
y0 y
xx
02
b2
a
1.
S
2
1Ÿ y
y 2 x 4,
S
2
1 x2
1
1 x2
sin y
x0 x x02
2
a2
a
cos 2 y, 1 tan 2 y
2
x02
y2
02
2
a
b
0 Ÿ yc
x0b 2
x x0 , Tangent line at x0 , y0
y0 a 2
x02
y2
02
2
a
b
tan y
4x
y
1Ÿ
2x
2 yyc
2
2
a
b
2 y
3x
1
y
1Ÿ x
6
8
2
4
Tangent line.
2
1Ÿ
Note: From part (a),
Tangent line: y 2
x2
y2
2
2
a
b
x
y
2
2
a
b
Because
1 x 1
3
41. (a)
2
yy0
y2
02
2
b
b
Tangent line: y 1
2
2 x 3
y y0
At 1, 1 :
2 yc 2 4 yc
43
3 2
Tangent line: y 2
2
2
x
3
4x
3y
30
2
11
x 11
y
2
0
At 3, 2 : yc
2 x 4
11
11y 2 x 30
1, 3, 2
yc
1
,
sin y
0 y S
1
1 cos 2 y
1 cos 2 y
1
1 x2
1 x2
, 1 x 1
Note: From part (a),
1x
2 y
1
1Ÿ y
2
8
4
Tangent line.
1
x 1Ÿ y
2
2 x 4,
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 2.5
45.
x2 y2
2 x 2 yyc
4
Implicit Differentiation
Dif
D
x2 y 4 x
46.
0
x 2 yc 2 xy 4
x
y
yc
yc
ycc
y 1 xyc
x 2 ycc 2 xyc 2 xyc 2 y
y2
ª 4 2 xy º
x 2 ycc 4 x «
» 2y
2
¬ x
¼
4
cc
x y 4 x 4 2 xy 2 x 2 y
y x x y
y2
y x2
y3
4
3
y
x 4 ycc 16 x 8 x 2 y 2 x 2 y
2
x 4 ycc
ycc
47.
x2 y 2
2 x 2 yyc
yc
x yyc
1 yycc yc
2
169
5
0
4 2 xy
x2
0
0
0
0
6 x 2 y 16 x
6 xy 16
x3
36
0
x
y
0
0
2
§ x·
1 yycc ¨ ¸
© y¹
2
y y 3 ycc
ycc
48.
xy 1
xyc y
2x y2
xyc 2 yyc
2 y
x 2 y yc
2 y
yc
2 y
x 2y
xycc yc yc
0
x2
y 2 x2
y3
36
y3
2 2 yyc
2 yycc 2 yc
xycc 2 yycc
2 yc
2
2 yc
x 2 y ycc
2 yc
2
2 yc
2
2
ycc
§ 2 y ·
§ 2 y ·
2¨
¸ 2¨
¸
© x 2y ¹
© x 2y ¹
2 2 y ª¬ 2 y x 2 y º¼
x 2y
3
22 y 2 x y
x 2y
3
2 4 2 x 2 y 2 y xy y 2
x 2y
2 y 2 xy 2 x 4
2y x
3
3
2 5
2y x
10
3
x 2y
3
INSTRUCTOR USE ONLY
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NOT FOR SALE
170
Chapter 2
Differentiation
49.
y2
x3
2 yyc
3x 2
yc
3x 2
2y
ycc
2 x 3 yc 3 y 2
4 x2
3 x 2 xy
˜
2 y xy
3 y x3
˜
2x y2
3y
4 x2
4x2
y3
4x
3 y yc
4
2
3 y 2 ycc 6 y yc
2
yycc 2 yc
2
yc
ycc
51.
5
1 1 2
1
x
y 1 2 y c
2
2
0
yc
2
5
5
5
§
5·
At ¨¨ 2,
¸ : yc
5 ¸¹
©
2
2 § 4 ·
¨
¸
y © 3y2 ¹
1 4 4
ª 2 5 º
«
» 4 12
« 5 »
¬
¼
2
y Tangent line:
5
5
10 5 y 10
x 10 5 y 8
y
Tangent line:
2 y x2 1
32
9 y5
4
2 3
4x
3
8
5 3
4 4x
9
x At 9, 4 : yc
1 2 x x2
13
4x
2
−1
y
x2 1
(2, )
0
ycc
x2 1
−1
2 yc
x2 1 2 x2 2 x
2
1
0
ycc
Note: y
3x
4y
4
3y2
yc
x2 1 1 x 1 2 x
2 yyc
3y
2x
yc
2 x ª¬3 ˜ 3 y 2 x ¼º 6 y
50.
x 1
x2 1
y2
52.
x2 y2
25
2 x 2 yyc
0
yc
x
y
53.
32
9 4x
53
32
9 y5
1
10 5
1
x 2
10 5
x 2
0
9
At 4, 3 :
(9, 4)
y
−1
14
−1
x
Tangent line:
4
y 3
x 4 Ÿ 4 x 3 y 25
3
3
x 4 Ÿ 3x 4 y
4
Normal line: y 3
2
3
y 4
2 x 3 y 30
0
0
0
6
2
x9
3
(4, 3)
−9
9
−6
At 3, 4 :
Tangent line:
y 4
3
x 3 Ÿ 3x 4 y 25
4
0
4
x 3 Ÿ 4x 3y
3
Normal line: y 4
0
6
(−3, 4)
−9
9
INSTRUCTOR USE ONLY
−6
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
Section 2.5
54.
x2 y 2
36
2 x 2 yyc
0
yc
56.
x
y
6
Normal line: y
0
4x
2 yyc
4
yc
2
y
1 at 1, 2
y 2
1 x 1 , y
3 x. The centers of the
circles must be on the normal line and at a distance of 4
units from 1, 2 . Therefore,
8
x 1
(6, 0)
−12
2
ª¬ 3 x 2º¼
2
2 x 1
2
12
16
16
1 r 2 2.
x
−8
Centers of the circles: 1 2 2, 2 2 2 and
5
11 , slope is
11
At 5,
y Tangent line:
5x 1 2 2, 2 2 2
5
x 5
11
11
11 y 11
5 x 25
11 y 36
0
y Normal line:
11
x 5
5
11
5 y 5 11
5y 11x
Equations: x 1 2 2
2
x 1 2 2
2
y 2 2 2
2
0
50 x 32 yyc 200 160 yc
0
16
4:
25 16 16 y 2 200 4 160 y 400
12
16
200 50 x
160 32 y
yc
0
(5, 11)
y y 10
0
0Ÿ y
0,10
Horizontal tangents: 4, 0 , 4, 10
−8
Vertical tangents occur when y
x2 y2
r2
2 x 2 yyc
0
yc
x
y
y
x
slope of normal line
25 x 2 400 200 x 800 400
25 x x 8
slope of tangent line
5:
0
0 Ÿ x
0, 8
Vertical tangents: 0, 5 , 8, 5
y
Let x0 , y0 be a point on the circle. If x0
y0
x x0
x0
(− 4, 10)
0, then the
tangent line is horizontal, the normal line is vertical and,
hence, passes through the origin. If x0 z 0, then the
equation of the normal line is
y
2
Horizontal tangents occur when x
−12
y y0
y 2 2 2
57. 25 x 2 16 y 2 200 x 160 y 400
11x 5 11
8
55.
171
Equation of normal line at 1, 2 is
At 6, 0 ; slope is undefined.
Tangent line: x
y2
Implicit Differentiation
Dif
D
10
6
(− 8, 5)
(0, 5)
4
(− 4, 0)
x
−10 − 8 − 6 − 4
−2
2
y0
x
x0
which passes through the origin.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 2
172
Differentiation
58. 4 x 2 y 2 8 x 4 y 4
0
8 x 2 yyc 8 4 yc
0
y
(1, 0)
8 8x
2y 4
yc
Horizontal tangents occur when x
−1
4 4x
y 2
2
y2 8 1 4 y 4
3
4
(2, − 2)
−3
1:
0
y2 4 y
2
(0, − 2)
−4
41
x
1
−1
(1, − 4)
−5
y y 4
0 Ÿ y
0, 4
Horizontal tangents: 1, 0 , 1, 4
2:
Vertical tangents occur when y
4 x 2 2
2
8 x 4 2 4
0
4 x2 8x
4x x 2
0 Ÿ x
0, 2
Vertical tangents: 0, 2 , 2, 2
59. Find the points of intersection by letting y 2
2x 4x
2
x 3 x 1
6 and
4 x in the equation 2 x 2 y 2
0
The curves intersect at 1, r 2 .
2x 2 + y 2 = 6 4
Ellipse:
Parabola:
4 x 2 yyc
0
2 yyc
4
yc
yc
2
y
yc
1
yc
1
2x
y
6.
y 2 = 4x
(1, 2)
−6
6
(1, − 2)
−4
At 1, 2 , the slopes are:
yc
1
At 1, 2 , the slopes are:
yc
1
Tangents are perpendicular.
60. Find the points of intersection by letting y 2
2 x 3x
2
3
5 and 3x 2 x 5
Intersect when x
3
2
0
2
1.
2x 3y
2
5:
2 yyc
3x 2
4 x 6 yyc
0
yc
3x 2
2y
yc
2
x:
2x2 + 3y2 = 5
−2
3
y
5.
(1, 1)
Points of intersection: 1, r1
2
x3 in the equation 2 x 2 3 y 2
4
(1, − 1)
−2
y 2= x 3
2x
3y
At 1, 1 , the slopes are:
yc
3
2
yc
yc
2
3
2
3
At 1, 1 , the slopes are:
yc
3
2
INSTRUCTOR USE ONLY
Tangents are perpendicular.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
Section 2.5
x and x
61. y
sin y
y
x:
x
sin y:
yc
1
1
yccos y
yc
sec y
C2
y
Kx
2 x 2 yyc
0
yc
K
yc
yc
1
K=1
−3
62. Rewriting each equation and differentiating:
x 3 y 29
3
y
1§ 3
·
¨ 29 ¸
3© x
¹
yc
1
2
x
3
yc
2
x
x (3y − 29) = 3 15
x 3 = 3y − 3
−15
−2
65. Answers will vary. Sample answer: In the explicit form
of a function, the variable is explicitly written as a
function of x. In an implicit equation, the function is only
implied by an equation. An example of an implicit
5. In explicit form it would be
function is x 2 xy
x+y=0
y
3
C=2
(0, 0)
x
1
3
C=1
−3
−2
6
3 y 1
K = −1
3
x = sin y
x3
1. The curves
2
2
4
−4
x Kx K
are orthogonal.
Tangents are perpendicular.
−6
x
y
slopes is x y K
1
173
At the point of intersection x, y , the product of the
At 0, 0 , the slopes are:
yc
x2 y2
64.
Point of intersection: 0, 0
Implicit Differentiation
Dif
D
5 x2
y
x.
66. Answers will vary. Sample answer: Given an implicit
equation, first differentiate both sides with respect to x.
Collect all terms involving yc on the left, and all other
terms to the right. Factor out yc on the left side. Finally,
divide both sides by the left-hand factor that does not
contain yc.
67.
12
−3
For each value of x, the derivatives are negative
reciprocals of each other. So, the tangent lines are
orthogonal at both points of intersection.
63.
xy
C
x2 y2
K
xyc y
0
2 x 2 yyc
0
yc
x
y
y
x
yc
68. (a) The slope is greater at x
3
K = −1
−2
−3
3.
(b) The graph has vertical tangent lines at about
2, 3 and 2, 3 .
(c) The graph has a horizontal tangent line at about
0, 6 .
2
C=1
1994
Use starting point B.
C=4
−3
B
A
00
1. The curves are orthogonal.
2
00
18
At any point of intersection x, y the product of the
slopes is y x x y
18
1671
3
K=2
−2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 2
174
69. (a)
Differentiation
4 4 x2 y 2
x4
16 x x
2
4
y2
4x2 1 4
x
4
y
r
4y
2
(b)
4x2 3Ÿ9
y
36
x 16 x 36
4
2
x2
10
− 10
− 10
1 4
x
4
1 4
x
4
4 x2 16 r
1 y3
1
3
1
3
x 8 x2
23
.
1ª
3¬
7 7 x 8 7 23º.
¼
7 7 , and the line is
2
7 . So, there are four values of x:
7
7 3
1
3
7, yc
7, yc
1r
7 7 , and the line is
7 3
1
3
1ª
3¬
7 7 x 23 8 7 º.
¼
7 7 , and the line is
7 7 x 1
1
For x
y4
1
3
7 7 x 1
For x
28
8 x x 3 Ÿ yc
7, yc
7, yc
1
1
3
y2
7, 1 7 7 x 1
For x
8r
8r 2 7
28
To find the slope, 2 yyc
1
3
y2
− 10
256 144
2
7, 1 y1
10
y4
y1
y3
1 1 − 10
0
8r
For x
10
16 x 2 x 4
Note that x 2
7, 1 10
1
ª
3¬
7 3
1
3
7 7 x 23 8 7 º.
¼
7 7 , and the line is
7 7 x 1
1
ª
3¬
7 3
7 7 x 8 7 23 º.
¼
(c) Equating y3 and y4 :
1
3
7 7 x 1
7 3
7 7 x 1
7x If x
7
7 7 7x 7 7 7
8 7
, then y
7
1
3
7 7 x 1
7 7 x 1
7x 16 7
14 x
x
8 7
7
7 3
7
7 7 7x 7 7 7
§8 7 ·
, 5 ¸¸.
5 and the lines intersect at ¨¨
© 7
¹
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 2.5
70.
1
2
x
x y
1
dy
y dx
0
dy
dx
2
Implicit Differentiation
Dif
D
175
c
y
x
Tangent line at x0 , y0 : y y0
x-intercept: x0 x0
y-intercept: 0, y0 y0
x0
x x0
y0 , 0
x0
y0
y0 y0 x0
Sum of intercepts:
x0 71.
x0
x0 2
y0
y
x p q ; p, q integers and q ! 0
yq
xp
qy q 1 yc
x0
y0 y0
x0 72.
px p 1
p x p 1
˜
q y q 1
yc
p x p 1 y
˜
q
yq
p x p 1 p q
˜ p x
q
x
xn , n
So, if y
2
x2 y2
nx n 1.
c
2
c
100, slope
2 x 2 yyc
0
yc
§ 16 ·
x2 ¨ x2 ¸
©9 ¹
25 2
x
9
x
p p q 1
x
q
p q, then yc
y0
x
y
3
4
3
Ÿ y
4
4
x
3
100
100
r6
Points: 6, 8 and 6, 8
73.
x2
y2
4
9
2x
2 yyc
4
9
1,
4, 0
0
yc
9 x
4y
9 x
4y
y 0
x 4
9 x x 4
4 y2
But, 9 x 2 4 y 2
36 Ÿ 4 y 2
3
§
Points on ellipse: ¨1, r
2
©
§ 3
At ¨1,
© 2
·
3 ¸: yc
¹
9 x
4y
3
§
At ¨1, 2
©
·
3 ¸: yc
¹
3
2
Tangent lines: y
y
36 9 x 2 . So, 9 x 2 36 x
4 y2
36 9 x 2 Ÿ x
1.
·
3¸
¹
9
4 ª¬ 3 2
3 º¼
3
2
3
x 4
2
3
x 2 3
2
3
x 4
2
3
x 2 3
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 2
176
Differentiation
y2
74. x
1
2 yyc
yc
1
,
2y
x2
y2
32
8
2x
2 yyc
32
8
75. (a)
slope of tangent line
1
At 4, 2 : yc
Consider the slope of the normal line joining x0 , 0 and
y 2 , y on the parabola.
x, y
y 2 x0
1
2
1
2
1 , then
4
(a) If x0
1 , then
2
(c) If x0
2 x 4
y
2x 6
1
2
4
−6
1 1
4
2
y2
y2
0 Ÿ y
2
1
2
1, then y
6
14 , which is
−4
impossible. So, the only normal line is the x-axis
y
0.
(b) If x0
y 2
(b)
x0 y2
4
42
Slope of normal line is 2.
y 0
y 2 x0
2 y
x
4y
0 Ÿ yc
2x 6
x2
32
8
(c)
0. Same as part (a).
2
1
x 2 4 4 x 2 24 x 36
32
17 x 2 96 x 112
0
17 x 28 x 4
0 Ÿ x
x and there are three
normal lines.
§1 1 ·
The x-axis, the line joining x0 , 0 and ¨ ,
¸,
2¹
©2
4,
28
17
§ 28 46 ·
Second point: ¨ , ¸
© 17 17 ¹
1 ·
§1
and the line joining x0 , 0 and ¨ , ¸
2
2¹
©
If two normals are perpendicular, then their slopes are –1
and 1. So,
2 y
1
y 0
Ÿ y
y 2 x0
1
2
1
x0
4
and
12
1 4 x0
1 Ÿ
1
Ÿ x0
2
The perpendicular normal lines are y
x y
3
.
4
x 3
and
4
3
.
4
Section 2.6 Related Rates
1.
y
dy
dt
dx
dt
x
§ 1 · dx
¨
¸
© 2 x ¹ dt
dy
2 x
dt
(a) When x
dy
dt
4 and dx dt
1
3
2 4
3
4
3:
(b) When x
dx
dt
25 and dy dt
2 25 2
2:
20
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.6
2.
y
3x 2 5 x
dy
dt
dx
dt
dx
dt
dy
1
6 x 5 dt
5.
dy
dt
3 and
ª¬6 3 5º¼ 2
(b) When x
dx
dt
2 and
dy
dt
§ x · dy
¨ ¸
© y ¹ dt
1/2
10
8
(b) When x
1, y
2x
dx
dy
2y
dt
dt
0
dx
dt
§ y · dy
¨ ¸
© x ¹ dt
(b) When x
dx
dt
3, y
3
8
4
4, y
3
2
4
1:
41 2
2
2
12 x
6
1 x2
6:
(a) When x
2:
dy
dt
12 2
ª1 2 2 º
¬
¼
(b) When x
4, and dx dt
8 cm/sec
dx
1
,
6
1 x 2 dt
dx
2x
˜
2
dt
1 x2
dy
dt
§ x · dx
¨ ¸
© y ¹ dt
0 cm/sec
2
10:
5
8
dy
dt
(a) When x
40 2
1 x
3
2
25
8 cm/sec
0:
2x
4, and dy dt
1
6
4
y
dy
dt
1/2, and dx dt
x2 y 2
dy
dt
dy
dt
6.
8, y
4 1 2
(c) When x
0
(a) When x
4.
dy
dt
4:
4
dx
dt
1:
(b) When x
4
7
§ y · dx
¨ ¸
© x ¹ dt
dx
dt
dy
dt
26
dy
dx
x
y
dt
dt
dy
dt
dy
dt
2:
1
4
62 5
xy
3.
dx
dt
177
2x2 1
dx
2
dt
dy
dx
4x
dt
dt
(a) When x
6x 5
(a) When x
y
R
Related Rates
24
in./sec
25
2
0:
12 0
(c) When x
2:
dy
dt
12 2
1 2
0 in./sec
2
1 0
2
2
24
in./sec
25
8:
6
3, and dy dt
2:
3
2
INSTRUCTOR USE ONLY
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7.
y
dy
dt
dx
3
dt
dx
sec 2 x ˜
sec 2 x 3
dt
(b) When x
(c) When x
dy
dt
dy
dt
T
12b
2
s
cos
T
2
3 sec x
:
3
32
2
(b)
2
2
6 ft/sec
(b) When x
S
4
(c) When x
dy
dt
dr
dt
dA
dt
s
b
§1·
4¨ ¸
© 2¹
2 cm/sec
§ 2·
4¨¨
¸¸
© 2 ¹
13.
§ 3·
4¨¨
¸¸
© 2 ¹
3
4S r 2
dr
dt
(a) When r
:
§S ·
4 sin ¨ ¸
©3¹
4 3
Sr
3
V
dr
dt
dV
dt
2 2 cm/sec
dV
dt
2 3 cm/sec
9,
4S 9
When r
dV
dt
dx
dt
2
3
972S in.3 /min.
2
15,552S in.3 /min.
36,
4S 36
3
(b) If dr dt is constant, dV dt is proportional to r 2 .
14.
4 3 dV
Sr ,
3
dt
2 dr
4S r
dt
1 § dV ·
¨
¸
4S r 2 © dt ¹
V
dV
dt
dr
dt
10. Answers will vary. See page 149.
A
3 dt
s2
.
8
4 sin x
No, the rate dy dt is a multiple of dx dt.
11.
s 2 § 1 ·§ 1 ·
¨ ¸¨ ¸
2 © 2 ¹© 2 ¹
,
h
:
3
S dA
s
9. Yes, y changes at a constant rate.
a˜
6 dt
3s 2
.
8
θ
:
S
1
rad/min.
2
dT
dA
(c) If s and
is constant,
is proportional to cos T .
dt
dt
§S ·
4 sin ¨ ¸
©4¹
dy
dt
,
3 ft/sec
§S ·
4 sin ¨ ¸
©6¹
dy
dt
s2
sin T
2
s 2 § 3 ·§ 1 ·
¨
¸¨ ¸
2 ¨© 2 ¸¹© 2 ¹
S dA
When T
When T
dx
4
dt
dx
sin x ˜
sin x 4
dt
6
T
s2
dT
dT
cos T
where
2
dt
dt
dA
dt
cos x,
S
2 s sin
T
T·
s2 §
¨ 2 sin cos ¸
2©
2
2¹
0:
3 sec 2 0
(a) When x
dy
dt
2
12 ft/sec
S
3
Ÿ b
2
T
h
Ÿ h
s cos
2
s
1
1§
T ·§
T·
bh
¨ 2s sin ¸¨ s cos ¸
2
2©
2 ¹©
2¹
A
:
4
§ S·
3 sec 2 ¨ ¸
© 4¹
dy
dt
12. (a) sin
S
§ S·
3 sec2 ¨ ¸
© 3¹
dy
dt
y
Differentiation
tan x,
(a) When x
8.
NOT FOR SALE
Chapter 2
178
Sr2
4
(a) When r
dr
2S r
dt
dr
dt
dA
dt
2S 8 4
dA
dt
2S 32 4
(a) When r
8,
(b) When r
32,
64S cm 2 /min.
256S cm 2 /min.
1
800
4S r 2
30,
1
4S 30
(b) When r
dr
dt
800
2
800
2
cm/min.
9S
800
1
cm/min.
188S
60,
1
INSTRUCTOR USE ONLY
4S 60
2
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
x3
V
15.
dx
dt
dV
dt
17.
6
3x 2
dx
dt
2,
dV
dt
2
32
6
(b) When x
10,
dV
dt
2
179
1 §9 2·
S ¨ h ¸h
3 ©4 ¹
>because 2r
3h@
1 2
Sr h
3
3 10
72 cm3/sec.
dV
dt
10
dV
dt
9S 2 dh
dh
Ÿ
h
4
dt
dt
When h
dh
dt
1800 cm3/sec.
6
4 dV dt
9S h 2
15,
4 10
9S 15
8
ft/min.
405S
2
6x2
s
dx
dt
ds
dt
6
h
dx
12 x
dt
(a) When x
ds
dt
12 2 6
ds
dt
144 cm 2 /sec.
10,
720 cm 2 /sec.
12 10 6
1 2
Sr h
3
V
dV
dt
dV
dt
r
2,
(b) When x
18.
R
Related Rates
3S 3
h
4
(a) When x
16.
V
Section 2.6
25S 3
h
3 144
1 25 3
h
S
3 144
r
§
¨ By similar triangles,
5
©
h
Ÿ r
12
5 ·
h.¸
12 ¹
5
10
r
25S 2 dh
dh
h
Ÿ
144
dt
dt
When h
19.
8,
dh
dt
12
§ 144 · dV
¨
2¸
© 25S h ¹ dt
144
10
25S 64
h
9
ft/min.
10S
12
1
6
3
1
(a) Total volume of pool
1
2 12 6 1 6 12
2
1
1 6 6
2
Volume of 1 m of water
% pool filled
18
100%
144
18 m3
b=6
6h, you have
V
1
bh 6
2
18h 2
dV
dt
36h
dh
dt
1
dh
Ÿ
4
dt
1
144h
h=1
(see similar triangle diagram)
12.5%
3 6h h
2
12
(b) Because for 0 d h d 2, b
3bh
144 m3
1
144 1
1
m/min.
144
INSTRUCTOR USE ONLY
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Chapter 2
180
1
bh 12
2
6bh
dV
dt
dh
dh
Ÿ
dt
dt
20. V
(a)
NOT FOR SALE
Differentiation
12h
When h
6h 2
h
1 dV
12h dt
dV
dt
1 and
since b
dh
dt
2,
1
2
12 1
1
ft/min.
6
12 ft
3 ft
h ft
3 ft
(b) If
dh
dt
3
in./min
8
x2 y 2
21.
2x
dx
dy
2y
dt
dt
dy
dt
1
ft/min and h
32
0
x dx
˜
y dt
2 x
y
dy
dt
2 7
dy
dt
2 15
576
24,
When x
15, y
400
20,
When x
24, y
dy
dt
2 24
7,
24
20
7
25
y
2.
7
ft/sec.
12
x
3
ft/sec.
2
48
ft/sec.
7
1
xy
2
1 § dy
dx ·
y ¸
¨x
dt ¹
2 © dt
A
dA
dt
From part (a) you have x
1ª § 7 ·
º
7¨ ¸ 24 2 »
2 «¬ © 12 ¹
¼
dA
dt
tan T
sec2T
7, y
24,
dx
dt
2, and
dy
dt
7
. So,
12
527 2
ft /sec.
24
x
y
θ
25
y
dT
dt
1 dx
x dy
˜
2 ˜
y dt
y
dt
dT
dt
ª 1 dx
x dy º
2 ˜ »
cos T « ˜
y
dt ¼
¬ y dt
Using x
dT
dt
dx
dt
because
7, y
(c)
3 3
ft /min.
4
252
(a) When x
(b)
§1·
12 2 ¨ ¸
© 32 ¹
dV
dt
2 ft, then
x
2
7, y
24,
dx
dt
2,
dy
dt
2
7 § 7 ·º
§ 24 · ª 1
¸»
2 ¨ ¸ «
2¨
© 25 ¹ «¬ 24
24 © 12 ¹»¼
7
and cos T
12
24
, you have
25
1
rad/sec.
12
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 2.6
x2 y 2
22.
23. When y
181
25
dx
dy
2y
2x
dt
dt
dx
dt
When x
R
Related Rates
0
y dy
˜
x dt
0.15 y
x
dx
dt
2.5, y
18.75,
6, x
122 62
dy
§
¨ because
dt
©
·
0.15 ¸
¹
18.75
0.15 | 0.26 m/sec.
2.5
6 3, and s
x 2 12 y
5
y
2
x
108 36
12.
s
12 − y
( x, y )
x
y
12
2
x 2 12 y
2x
dx
dy
2 12 y 1
dt
dt
dx
dy
x
y 12
dt
dt
Also, x 2 y 2
2x
s2
122.
dx
dy
2y
dt
dt
So, x
ds
dt
ds
s
dt
2s
0 Ÿ
x dx
y dt
dy
dt
§ x dx ·
dx
y 12 ¨
¸
dt
© y dt ¹
12 x º
dx ª
«x x »
dt ¬
y ¼
ds
dx
Ÿ
dt
dt
s
6 3 3
˜
6
15
x dx
y dt
dy
dt
s
ds
.
dt
sy ds
˜
12 x dt
12 6
12 6 3
0.2
1
5 3
3
m/sec (horizontal)
15
1
m/sec (vertical)
5
24. Let L be the length of the rope.
144 x 2
L2
(a)
2L
dL
dt
dx
dt
dx
dt
L dL
˜
x dt
2x
4L
x
dL
§
¨ since
dt
©
·
4 ft/sec ¸
¹
4 ft/sec
13 ft
12 ft
When L
13:
x
L 144
dx
dt
169 144
2
4 13
5
52
5
5
10.4 ft/sec
Speed of the boat increases as it approaches the dock.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 2
182
(b) If
dx
dt
Differentiation
4, and L
dL
dt
x dx
L dt
dL
dt
dL
lim
L o12 dt
20
ft/sec
13
5
4
13
L2 144
4
L
x dx
L dt
lim
L o12
4
L
L2 144
0
x2 y 2
s2
25. (a)
13:
dx
dt
dy
dt
ds
2s
dt
450
600
dx
dy
2y
2x
dt
dt
x dx dt y dy dt
ds
dt
s2
902 x 2
x
20
dx
dt
ds
2s
dt
25
27.
2x
When x
ds
dt
s
y
dx
ds
Ÿ
dt
dt
x dx
˜
s dt
902 202
20, s
10 85,
50
| 5.42 ft/sec.
85
20
25
10 85
2nd
300
200
y
s
20 ft
x
3rd
100
x
−100
100
1st
s
x
200
300
90 ft
Home
When x
1
h
2
x2 y2
s2
dx
0
dt
2s
2x
dx
dt
When s
dx
dt
375 and
750 mi/h.
375
375
750
(b) t
300, s
225 450 300 600
ds
dt
26.
225 and y
30 min
dy
§
¨ because
dt
©
ds
dt
·
0¸
¹
s ds
x dt
10, x
10
240
5 3
28. s 2
902 x 2
x
90 20
dx
dt
ds
dt
100 25
480
3
75
25
x dx
˜
s dt
When x
ds
dt
5 3,
70
902 702
70, s
70
25
10 130
2nd
10 130,
175
| 15.35 ft/sec.
130
20 ft
160 3 | 277.13 mi/h.
x
3rd
1st
s
y
90 ft
x
Home
5 mi
s
x
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 2.6
29. (a)
15
6
y
Ÿ 15 y 15 x
y x
y
5
x
3
dx
dt
dy
dt
1
St
sin , x 2 y 2
2
6
31. x t
6y
2S
S 6
(a) Period:
5
5
5
3
25
ft/sec
3
1
,
4
y
§1·
1¨ ¸
© 4¹
dx
dt
St
1§S ·
¨ ¸ cos
2© 6 ¹
6
x
y
dy
dx
dt
dt
dt
10
ft/sec
3
x y
2
2x
y
y x
20
6
30. (a)
25
5
3
20 y 20 x
6y
14 y
20 x
y
10
x
7
dx
dt
dy
dt
So,
5
10 dx
7 dt
50
ft/sec
7
2
dx
dy
2y
dt
dt
dy
dt
Speed
10
5
7
§1·
12 ¨ ¸
©2¹
(c) When x
6
d y x
1
2
3
m.
2
§
3·
Lowest point: ¨¨ 0,
¸
2 ¸¹
©
15
(b)
183
12 seconds
1
,y
2
(b) When x
5 dx
˜
3 dt
R
Related Rates
2
15
and t
4
S
1:
cos
St
dy
dt
x dx
y dt
12
6
1
0 Ÿ
14
S
§S ·
cos¨ ¸
˜
15 4 12
©6¹
S § 1 · 3
¨ ¸
15 © 12 ¹ 2
S
24
5S
120
5S
m/sec
120
1
5
5S
.
120
20
6
x
y
(b)
d y x
dt
dy
dx
dt
dt
50
5
7
50 35
7
7
15
ft/sec
7
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 2
184
Differentiation
3
sin S t , x 2 y 2
5
32. x t
2S
(a) Period:
2 seconds
S
(b) When x
1
§ 3·
1¨ ¸
©5¹
3
,y
5
2
4
m.
5
§ 4·
Lowest point: ¨ 0, ¸
© 5¹
(c) When x
3
,y
10
§1·
1¨ ¸
© 4¹
dx
dt
3
S cos S t
5
x2 y 2
2x
dx
dy
2y
dt
dt
So,
dy
dt
15
3
and
4
10
3
sin S t Ÿ sin S t
5
1
Ÿ t
2
0 Ÿ
x dx
y dt
dy
dt
3 10 3
§S ·
˜ S cos¨ ¸
5
15 4
©6¹
9 5S
.
125
9 5S
| 0.5058 m/sec
125
33. Because the evaporation rate is proportional to the
k 4S r 2 . However, because
surface area, dV dt
V
4 3 S r 3 , you have
dV
dt
4S r 2
4S r 2
dr
Ÿ k
dt
1
R
35.
dR1
dt
dR2
dt
1 dR
˜
R 2 dt
dr
.
dt
Therefore, k 4S r 2
dr
.
dt
dx
dy
negative Ÿ
positive
dt
dt
When R1
34. (i) (a)
(b)
dy
dx
positive Ÿ
negative
dt
dt
dR
dt
(ii) (a)
dx
dy
negative Ÿ
negative
dt
dt
(b)
1
:
6
1
9S
25 5
Speed
2
dy
dx
positive Ÿ
positive
dt
dt
R
1
1
R1
R2
1
1.5
1
1
dR
dR2
˜ 1 2 ˜
2
R1
dt
R2
dt
50 and R2
75:
30
ª 1
º
1
2
30 «
1 1.5 »
2
2
«¬ 50
»¼
75
pV 1.3
36.
dV
dp
V 1.3
1.3 pV 0.3
dt
dt
dV
dp ·
§
V 0.3 ¨1.3 p
V ¸
dt
dt ¹
©
1.3 p
dV
dt
0.6 ohm/sec
k
0
0
V
dp
dt
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 2.6
37.
rg tan T
v2
32r tan T
v2 ,
32r sec 2T
r is a constant.
dT
dt
dv
dt
dy
dt
d
T
sec2T ˜
dt
dT
dt
dT
dt
v
dv
cos 2T .
16r
dt
L
θ
x
dT
dt
4
2
sin T
10
x
1 ft/sec
§ dT ·
cos T ¨ ¸
© dt ¹
dT
dt
10 dx
˜
x2
dt
10 1
25 5 21
2
120 sin 2 75q | 111.96 rad/h | 1.87 rad/min.
x
50
tan T
30 2S
S rad/sec
§ dT ·
50 sec 2T ¨ ¸
© dt ¹
θ
50 ft
25 21
x
(a) When T
30q,
dx
dt
200S
ft/sec.
3
(b) When T
60q,
dx
dt
200S ft/sec.
(c) When T
70q,
dx
| 427.43S ft/sec.
dt
10
θ
60S rad/min
1 § dx ·
¨ ¸
50 © dt ¹
dx
dt
2 21
| 0.017 rad/sec
525
x
3
rad/min.
2
75q,
dT
dt
§ dT ·
sec2T ¨ ¸
© dt ¹
10 dx
sec T
x 2 dt
25
10
1
2
252
25 102
90 rad/h
Police
dx
dt
§ 3·
120¨ ¸
© 4¹
dT
dt
1
rad/sec.
25
1
rad/min.
2
30 rad/h
60q,
(c) When T
2
. So,
2
, and cos T
41.
39.
120
4
dT
dt
x
1 § 2·
¨
¸ 4
50 ¨© 2 ¸¹
dT
dt
30q,
(b) When T
y
S
120 sin 2T
y=5
(a) When T
50, T
x 2 § 5 · dx
¨ ¸
L2 © x 2 ¹ dt
§1·
sin 2T ¨ ¸ 600
© 5¹
y
When y
5 dx
˜
x 2 dt
§ 52 ·§ 1 · dx
¨ 2 ¸¨ ¸
© L ¹© 5 ¹ dt
1 dy
50 dt
1
dy
cos 2T ˜
50
dt
50
5
§ 5 · dx
cos 2T ¨ 2 ¸
© x ¹ dt
4 m/sec
θ
185
600 mi/h
y
50
tan T
38.
dx
dt
dT
2
sec T
dt
dv
dt
dT
16r
sec 2T
v
dt
2v
dT
Likewise,
dt
y
,y
x
tan T
40.
R
Related Rates
INSTRUCTOR USE ONLY
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42.
NOT FOR SALE
Chapter 2
186
Differentiation
dT
dt
10 rev/sec 2S rad/rev
(a)
cos T
x
y
43. sin 18q
x
30
1 dx
30 dt
dT
dt
dx
dt
sin T
20S rad/sec
x dy
1 dx
˜
˜
2
y
dt
y dt
x dy
sin 18q 275 | 84.9797 mi/hr
˜
y dt
0
dx
dt
dT
dt
30 sin T 20S
30 sin T
y
x
18°
600S sin T
44. tan T
P
x
Ÿ x
50
x
x
(b)
2000
45. (a) dy dt
4
0
(b) y changes slowly when x | 0 or x | L. y changes
more rapidly when x is near the middle of the
interval.
dx dt
600S sin T is greatest when
sin T
1Ÿ T
S
2
dx dt is least when T
(d) For T
or 90q n ˜ 180q .
nS
or n ˜ 180q .
600S
1
2
300S cm/sec.
600S sin 60q
600S
46. x 2 y 2
nS
30q,
dx
600S sin 30q
dt
For T
60q,
dx
dt
3
2
300 3S cm/sec.
25; acceleration of the top of the ladder
First derivative: 2 x
dx
dy
2y
dt
dt
dx
dy
x
y
dt
dt
Second derivative: x
d2y
dt 2
0
0
d 2x
dx dx
d2y
dy dy
˜
y 2 ˜
2
dt
dt dt
dt
dt dt
d2y
dt 2
When x
50 sec 2 T
as fast as x changes.
− 2000
(c)
dT
dt
dT
2
2 50 sec T
dt
1
S
S
dT
cos 2 T , d T d
25
4
4
dt
3 dx dt means that y changes three times
dx
dt
30
θ
50 tan T
7, y
d2y
dt 2
24,
dy
dt
7
dx
, and
12
dt
0
2
2
§ 1 · ª d 2 x § dx ·
§ dy · º
¨ ¸ « x 2 ¨ ¸ ¨ ¸ »
© dt ¹
© dt ¹ »¼
© y ¹«¬ dt
2 (see Exercise 25). Because
2
1ª
2
§ 7· º
«7 0 2 ¨ ¸ »
24 «¬
© 12 ¹ »¼
1ª
49 º
4 24 ¬«
144 ¼»
d 2x
dx
is constant,
dt 2
dt
0.
1 ª 625 º
| 0.1808 ft/sec2
24 ¬« 144
44 ¼»
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 2.6
dL
dt
dL
L
dt
dx
dt
dx
x
dt
First derivative: 2 L
2x
Second derivative: L
d 2L
dL dL
˜
dt 2
dt dt
x
d 2x
dt 2
13, x
5,
dx
dt
d 2x
dx dx
˜
dt 2
dt dt
§ 1 · ª d 2 L § dL ·
§ dx · º
¨ ¸ «L 2 ¨ ¸ ¨ ¸ »
x
dt
dt
© ¹ ¬«
© ¹
© dt ¹ ¼»
2
d 2x
dt 2
When L
187
d 2x
dt 2
144 x 2 ; acceleration of the boat
47. L2
R
Related Rates
10.4, and
1ª
2
2
13 0 4 10.4 º
¼
5¬
1
1
>16 108.16@
>92.16@
5
5
dL
dt
2
4 (see Exercise 28). Because
d 2L
dL
is constant,
dt 2
dt
0.
18.432 ft/sec2
48. (a) Using a graphing utility,
(b)
ms
1.24449 s 3 72.7661 s 2 1416.428 s 9215.21.
dm
dt
3.73347 s 2 145.5322 s 1416.428
If
ds
dt
0.75 and t
7, then s
4.9t 2 20
49. y t
dy
dt
y1
9.8t
yc 1
9.8
4.9 20
19.7 and
dm
| 1.23 million/year.
dt
y
15.1
20
y
12
(0, 0)
By similar triangles:
When y
15.1:
20 x 240
20
At t
dx
dt
dx
dt
1,
dx
dt
ds
dt
20
x
20 x 240
x
x
y
x 12
xy
20 x 240
x 15.1
20 15.1 x
240
x
240
4.9
xy
dy
dx
y
dt
dt
x
dy
20 y dt
x
240 4.9
9.8 | 97.96 m/sec.
20 15.1
INSTRUCTOR USE ONLY
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188
NOT FOR SALE
Chapter 2
Differentiation
Review Exercises for Chapter 2
1.
f x
fc x
12
2.
f x 'x f x
lim
fc x
'x
'x o 0
5x 4
f x
12 12
lim
'x o 0
'x
0
lim
0
'x o 0 'x
3.
f x
fc x
f x 'x f x
lim
'x
'x o 0
ª5 x 'x 4º¼ 5 x 4
lim ¬
'x o 0
'x
5 x 5'x 4 5 x 4
lim
'x o 0
'x
5'x
lim
5
'x o 0 'x
x2 4x 5
f x 'x f x
lim
'x
'x o 0
ª x 'x 2 4 x 'x 5º ª x 2 4 x 5º
¼
¬
¼ ¬
lim
'x o 0
'x
x 2 2 x 'x 'x
lim
2
4 x 4 'x 5 x 2 4 x 5
'x
'x o 0
lim
2 x 'x 'x
f x
fc x
5. g x
gc 2
4 'x
'x
'x o 0
4.
2
lim 2 x 'x 4
'x o 0
2x 4
6
x
f x 'x f x
'x
6
6
6 x 6 x 6 'x
x
'
x
x
lim
lim
0
'x o 0
'
o
x
'x
'x x 'x x
lim
'x o 0
2 x 2 3 x, c
lim
xo2
2
lim
6 'x
lim
'x o 0 'x x 'x x
'x o 0
1
,c
x 4
6. f x
g x g 2
x 2
fc 3
2 x 3x 2
lim
xo2
lim
xo2
xo2
3
f x f 3
x 3
1
1
4
7
x
lim
xo3
x 3
7 x 4
lim
xo3 x 3 x 4 7
x 2
x 2 2x 1
x 2
lim 2 x 1
22 1
6
x2
xo3
2
lim
6
x 'x x
5
lim
xo3
1
x 47
1
49
7. f is differentiable for all x z 3.
8. f is differentiable for all x z 1.
9. y
25
yc
0
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Review Exercises ffor Chapter 2
10.
11.
f t
4t 4
fc t
16t 3
f x
x3 11x 2
fc x
3 x 2 22 x
12. g s
g' s
22.
3s 2s
5
23.
4
15s 4 8s 3
14.
x 33 x
6
1 2
3x
f x
x1 2 x 1 2
fc x
1 1 2
1
x
x 3 2
2
2
15. g t
gc t
16. h x
hc x
x
2 2
t
3
4 3
t
3
fcT
4 5 cos T
18. g D
4 cos D 6
19.
gc D
4 sin D
f T
3 cos T fcT
20. g D
gc D
fc1
6 4
f x
2 x 4 8, 0, 8
fc x
8 x3
fc 0
0
3 sin T 2
3
1
3
x
x2
fc 0
3 sin 0 2
F
25.
26.
100
T
(a) When T
4, F c 4
50 vibrations/sec/lb.
(b) When T
9, Fc 9
33 13 vibrations/sec/lb.
S
6l 2
dS
dl
12l
(a) When l
3,
dS
dl
12 3
36 in.2 /in.
(b) When l
5,
dS
dl
12 5
60 in.2 /in.
16t 2 v0t s0 ; s0
27. s t
(a) s t
sc t
5 sin D
2D
3
5 cos D
2
3
f x
fc x
27 3 x 4
fc 3
81
4
3
27 x 3 , 3, 1
81
x4
600, v0
30
16t 2 30t 600
vt
32t 30
(b) Average velocity
s3 s1
31
366 554
2
94 ft/sec
(c) v 1
32 1 30
62 ft/sec
v3
32 3 30
126 ft/sec
(d) s t
21.
2
200 T
Fc t
x 1
2 x3 2
sin T
4
cos T
3 sin T 4
27
x3
2
3 cos T 2T , 0, 3
8
8 4
x
5x4
5
32
32
x 5
5
5
5x
4T 5 sin T
6x 4
fcT
4
3t 3
f T
17.
fc x
f T
2 3
hc x
3x 2 4 x, 1, 1
6 x1 2 3 x1 3
24.
13. h x
f x
189
0
16t 2 30t 600
Using a graphing utility or the Quadratic Formula,
t | 5.258 seconds.
(e) When
t | 5.258, v t | 32 5.258 30 | 198.3 ft/sec.
1
INSTRUCTOR USE ONLY
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NOT FOR SALE
190
Chapter 2
Differentiation
28.
st
16t 2 s0
s 9.2
16 9.2
s0
1354.24
2
x4
cos x
35. y
s0
0
cos x 4 x3 x 4 sin x
yc
cos 2 x
The building is approximately 1354 feet high (or 415 m).
29.
f x
5x2 8 x2 4x 6
fc x
5 x 2 8 2 x 4 x 2 4 x 6 10 x
4 x3 cos x x 4 sin x
cos 2 x
sin x
x4
36. y
10 x3 16 x 20 x 2 32 10 x3 40 x 2 60 x
x 4 cos x sin x 4 x3
yc
20 x 3 60 x 2 44 x 32
x
4 5 x 3 15 x 2 11x 8
30. g x
gc x
yc
2 x3 5 x 3 3x 4 6 x 2 5
6 x3 15 x 18 x3 24 x 2 15 x 20
31. h x
hc x
32.
1
2
x
sin x yc
2t 5 cos t
fc t
2t 5 sin t cos t 10t 4
fc x
x cos x sin x
yc
x sin x cos x cos x
40. g x
3 x sin x x 2 cos x
gc x
f t
f x
2 x 2 sec 2 x 2 x tan x
f x
x 2 x 2 5 , 1, 6
fc x
x 2 2x x2 5 1
41.
x2 1 2x 1 x2 x 1 2 x
x2 1
2
x2 1
2 x2 4x x2 5
f c 1
f x
fc x
3 45
4x 1
y
4 x 10
2
2x 7
x2 4
fc x
fc 0
2
2 x 2 14 x 8
x2 4
2
x 4 2 x 6 x2 6 x 1 1
3 x 2 4 x 25
2
2 x 2 8 4 x 2 14 x
x2 4
x 4 x 2 6 x 1 , 0, 4
2 x 2 2 x 24 x 2 6 x 1
x2 4 2 2 x 7 2 x
x2 4
3x 2 4 x 5
4
Tangent line: y 6
42. f x
34.
3 x cos x 3 sin x x 2 sin x 2 x cos x
5 x cos x 3 x 2 sin x
x2 x 1
x2 1
x2 1
x sin x
x cos x
2t 5 sin t 10t 4 cos t
33.
2 x x 2 tan x
39. y
x1 2 sin x
x sin x
3 x 2 sec x tan x 6 x sec x
38. y
24 x3 24 x 2 30 x 20
x cos x 4 sin x
x5
2
3 x 2 sec x
37. y
2 x3 5 x 3x 4
4
0 0 25
Tangent line: y 4
25 x 0
y
25 x 4
2 x2 7 x 4
x2 4
25
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Review Exercises ffor Chapter 2
43.
f x
fc x
§1·
f c¨ ¸
© 2¹
x 1 §1
·
, ¨ , 3¸
x 1 ©2
¹
x 1 x 1
x 1
2
14
2
fc x
10 sin t 15 cos t
hcc t
x 1
10 cos t 15 sin t
51. v t
20 t 2 , 0 d t d 6
2
1·
§
8¨ x ¸
2¹
©
8 x 1
y
f x
10 cos t 15 sin t
hc t
2
8
Tangent line: y 3
44.
ht
50.
at
vc t
v3
20 3
11 m/sec
a3
2 3
6 m/sec 2
90t
4t 10
4t 10 90 90t 4
at
cos x sin x cos x sin x
cos x
4t 10
2
§S ·
f c¨ ¸
©2¹
2
1
(a) v 1
2
45.
46.
47.
48.
gt
8t 3 5t 12
gc t
24t 2 5
g cc t
48t
h x
2
6x
a1
S·
§
2¨ x ¸
2¹
©
2 x 1 S
y
(b) v 5
a5
7x
hcc x
36 x 4 14
f x
14 x
4 7x 3
15 x5 2
54. y
x2 6
fc x
75 3 2
x
2
yc
3 x2 6
f cc x
225 1 2
x
4
55. y
f x
20
5
1
x2 4
fc x
4x
4 5
f cc x
16 9 5
x
5
3
3
7
28 7 x 3
2x
6 x x2 6
3
2
2
x
15
20 x
yc
16
5 x9 5
56.
49.
0.36 ft/sec2
4
yc
x
18 ft/sec
50
225
252
7x 3
225
4
1 ft/sec 2
90 10
53. y
36
14
x4
15 ft/sec
30
225
152
2
12 x
2
2t 5
90 5
(c) v 10
hc x
2
90
| 6.43 ft/sec
14
225
| 4.59 ft/sec 2
49
a 10
3
225
4t 10
2
Tangent line: y 1
2
900
2 sin x
cos x
2t
2
52. v t
cos x § S ·
, ¨ , 1¸
cos x © 2 ¹
191
f T
3 tan T
fcT
3 sec2 T
f cc T
6 sec T sec T tan T
6 sec 2 T tan T
f x
fc x
57. y
yc
x2 4
1 x 2 4
2
5x 1
2x
x 4
2
5x 1
2
2 5x 1
2x
1
1
3
5
2
2
10
5x 1
3
5 cos 9 x 1
5 sin 9 x 1 9
45 sin 9 x 1
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 2
192
58. y
yc
Differentiation
1 cos 2 x 2 cos 2 x
2 sin 2 x 4 cos x sin x
0
yc
7
60. y
yc
12
23
f c 2
x
sin 2 x
2
4
1
1
cos 2 x 2
2
4
2
sin x
sec x sec x
7
5
23
fc 3
x 6x 1
4
x 5 6x 1
5
6 6x 1 1
4
6x 1
6x 1
4
30 x 6 x 1
6x 1
4
36 x 1
52
s3 5
fc s
s2 1
52
3s 2 s 3 5
fc x
s2 1
s s2 1
ª3s s 2 1 5 s 3 5 º
¬
¼
s s 1
32
8s 3s 25
32
f c 1
2
2x
8 1
ª 1
¬
1º
¼
2
8x
x 1
8
4
2
2
2
2
3x 1
, 4, 1
4x 3
4 x 3 3 3x 1 4
4x 3
2
12 x 9 12 x 4
2s
2
4x 3
3
13
4x 3
fc 4
3x
2
13
16 3
2
1
13
x2 1
3 x2 1
12
x2 1
hc x
5
2
32
1 2
1 2
x 1
2x
2
x2 1
3x
3 x 2 1 3x 2
64. h x
4 x2 1
fc x
s2 1
f x
fc x
68. f x
f s
4 x 2 1 , 1, 2
2
5
23
1
4
x 1
67. f x
5
2
63.
3x 1
1
2
34
sec6 x sec x tan x sec 4 x sec x tan x
30 x 6 x 1
62.
2x
2
5
sec5 x tan 3 x
yc
2
2 3
1 2
2x
x 1
3
fc x
sec5 x tan x sec 2 x 1
61. y
2 1 x3
x 2 1, 3, 2
3
66. f x
1
1 cos 2 x
2
3 x 2
1 2
1
1 x3
3 x 2
2
fc x
2>2 sin x cos x@ 4 sin x cos x
59. y
1 x3 , 2, 3
65. f x
§ x 5·
¨ 2
¸
© x 3¹
69.
x2 1
32
2
70.
§ 2
·
§ x 5 ·¨ x 3 1 x 5 2 x ¸
2¨ 2
¸¨
2
¸¸
© x 3 ¹¨
x2 3
©
¹
2 x 5 x 2 10 x 3
x 3
2
3
1
§S 1 ·
csc 2 x, ¨ , ¸
2
© 4 2¹
yc
csc 2 x cot 2 x
§S ·
yc¨ ¸
©4¹
3
32
y
y
§S ·
csc 3 x cot 3 x, ¨ , 1¸
©6 ¹
yc
3 csc 3x cot 3x 3 csc 2 3x
§S ·
yc¨ ¸
©6¹
71.
0
03
3
3
y
8x 5
yc
3 8x 5
ycc
24 2 8 x 5 8
2
8
24 8 x 5
2
384 8 x 5
INSTRUCTOR USE ONLY
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Review Exercises ffor Chapter 2
72.
y
1
1
! 5x 1
5x 1
yc
1 5 x 1
ycc
2
5 5x 1
5
5 2 5x 1
3
1
1
cos 8t sin 8t
4
4
1
1
sin 8t 8 cos 8t 8
4
4
2 sin 8t 2 cos 8t
y
76.
2
yc
50
5
5x 1
3
S
At time t
73.
f x
cot x
fc x
csc 2 x
f cc x
2 csc x csc x ˜ cot x
4
§S ·
y¨ ¸
©4¹
vt
y
sin 2 x
yc
2 sin x cos x
ycc
2 cos 2 x
sin 2 x
T
700
2
t 4t 10
700 t 2 4t 10
t 2 4t 10
(a) When t
Tc
Tc
§S ·
yc¨ ¸
©4¹
ª § S ·º
ª § S ·º
2 sin «8 ¨ ¸» 2cos «8 ¨ ¸»
¬ © 4 ¹¼
¬ © 4 ¹¼
64
2 x 2 yyc
0
1
2 yyc
2x
yc
2 ft/sec
x
y
2
x 2 4 xy y 3
6
2 x 4 xyc 4 y 3 y yc
0
78.
1,
2
1400 1 2
1 4 10
(b) When t
1
ft.
4
x2 y2
1400 t 2
Tc
1
1
4
2 0 2 1
77.
75. T
,
ª § S ·º 1
ª § S ·º
1
cos «8 ¨ ¸» sin «8 ¨ ¸»
4
4
4
¬ © ¹¼
¬ © 4 ¹¼
2 csc 2 x cot x
74.
| 18.667 deg/h.
2
4 x 3 y yc
2x 4 y
yc
2x 4 y
3 y2 4x
x3 y xy 3
4
x y c 3 x 2 y x3 y 2 yc y 3
0
2
3,
1400 3 2
9 12 10
| 7.284 deg/h.
2
79.
3
(c) When t
Tc
Tc
80.
5,
1400 5 2
25 20 10
(d) When t
193
| 3.240 deg/h.
2
3
x yc 3xy 2 yc
y3 3x 2 y
yc x3 3 xy 2
y3 3x 2 y
yc
y3 3x 2 y
x3 3xy 2
10,
1400 10 2
100 40 10
2
xy
x 4y
y
x
yc 2 y
2 x
1 4 yc
| 0.747 deg/h.
yc
xyc y
2
xy 8
x 8 xy yc
2
xy y
yc
2
xy y
2 x 4y y
x 8 xy
x 8 x 4y
y y 2 3x 2
x x2 3 y 2
xy yc
2x 9 y
9 x 32 y
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 2
194
Differentiation
x sin y
81.
y cos x
cos x y
x
1 yc sin x y
1
82.
x cos y yc sin y
y sin x yc cos x
yc x cos y cos x
y sin x sin y
yc sin x y
yc
y sin x sin y
cos x x cos y
yc
83. x 2 y 2
1 sin x y
sin x y
csc x 1 1
0
x
y
4
At 3, 1 , yc
3
(3, 1)
Tangent line:
y 1
3 x 3 Ÿ 3x y 10
Normal line:
y 1
1
x 3 Ÿ x 3y
3
84. x 2 y 2
20
2 x 2 yyc
−6
0
−4
3
2
y 4
Tangent line:
y
2 y 3x 10
y 4
Normal line:
A
86. Surface area
0
At 6, 4 , yc
6
0
x
y
yc
10
2 x 2 yyc
yc
1 sin x y
y
3 y 2 x 24
dx
dt
8
dA
dt
12 x
tanT
87.
3
x 6
2
3
x 5
2
0
dx
dt
dT
dt
2 § dT ·
sec T ¨ ¸
© dt ¹
dx
dt
2
x 6
3
2
x 8
3
0
6x2 , x
12 6.5 8
length of edge
624 cm 2 /sec
x
3 2S rad/min
dx
dt
tan 2T 1 6S
1
,
2
§1
·
6S ¨ 1¸
©4
¹
6S x 2 1
When x
dx
dt
2
15S
km/min
2
450S km/h.
P1
P2
1
θ
f
0
3
0
85.
x
y
dy
dt
dy
dt
x
2 units/sec
1
2
dx
dx
Ÿ
dt
x dt
2
x
dy
dt
(a) When x
1 dx
,
2 dt
(b) When x
1,
dx
dt
4 units/sec.
(c) When x
4,
dx
dt
8 units/sec.
4
x
2 2 units/sec.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Problem Solving ffor Chapter 2
88.
st
60 4.9t 2
sc t
9.8t
s
35
4.9t 2
25
195
60 4.9t 2
5
4.9
t
st
tan 30
1
3
xt
3s t
dx
dt
3
xt
s (t)
ds
dt
3 9.8
30°
5
| 38.34 m/sec
4.9
x(t )
Problem Solving for Chapter 2
2
r 2 , Circle
x2
y, Parabola
1. (a) x 2 y r
Substituting:
3
2
r2 y
y 2 2ry r 2
r2 y
y 2 2ry y
0
y y 2r 1
0
y r
−3
3
−1
Because you want only one solution, let 1 2r
1·
§
x 2 and x 2 ¨ y ¸
2¹
©
1
. Graph y
2
0 Ÿ r
2
1
.
4
(b) Let x, y be a point of tangency:
2
x2 y b
y
x 2 Ÿ yc
1 Ÿ 2 x 2 y b yc
0 Ÿ yc
x
, Circle
b y
2 x, Parabola
Equating:
3
x
b y
2x
2b y
−3
1
1
Ÿ b
2
b y
Also, x 2 y b
y y b
2
2
y 3
1
2
−1
x 2 imply:
1 and y
ª
1 ·º
§
1 Ÿ y « y ¨ y ¸»
2 ¹¼
©
¬
2
1Ÿ y 1
4
1Ÿ y
3
and b
4
5
4
§ 5·
Center: ¨ 0, ¸
© 4¹
Graph y
5·
§
x 2 and x 2 ¨ y ¸
4¹
©
2
1.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 2
196
Differentiation
2. Let a, a 2 and b, b 2 2b 5 be the points of tangency. For y
yc
2 x 2. So, 2a
2b 2 Ÿ a b
a b 2b 5
2
2
1b
a b
2
2 x and for y
x 2 2 x 5,
1 b. Furthermore, the slope of the common tangent line is
1, or a
b 2b 5
1b b
x 2 , yc
2
2b 2
1 2b b 2 b 2 2b 5
2b 2
1 2b
Ÿ 2b 2 4b 6
4b 2 6b 2
Ÿ
Ÿ 2b 2 2b 4
0
Ÿ b b2
2
0
Ÿ b 2 b 1
For b
1b
2, a
2: y 1
2 x
1, a
For b
4: y 4
0
2, 1
b
1 and the points of tangency are 1, 1 and 2, 5 . The tangent line has slope
1 Ÿ y
1b
2 x 1
2 and the points of tangency are 2, 4 and 1, 8 . The tangent line has slope
4x 2 Ÿ y
4x 4
y
10
8
6
4
−8 −6 −4 −2
x
2 4 6 8 10
−4
−6
3. (a) f x
P1 x
a0 a1 x
1
P1 0
a0 Ÿ a0
1
fc 0
0
P1c 0
a1 Ÿ a1
0
P1 x
1
cos x
P2 x
a0 a1 x a2 x 2
1
P2 0
a0 Ÿ a0
1
a1 Ÿ a1
0
2 a2 Ÿ a 2
f 0
(b) f x
f 0
fc 0
(c)
cos x
0
P2c 0
f cc 0
1
P2cc 0
P2 x
1 12 x 2
x
1.0
0.1
0.001
0
0.001
0.1
1.0
cos x
0.5403
0.9950
|1
1
|1
0.9950
0.5403
P2 x
0.5
0.9950
|1
1
|1
0.9950
0.5
P2 x is a good approximation of f x
12
cos x when x is near 0.
sin x
P3 x
a0 a1 x a2 x 2 a3 x3
f 0
0
P3 0
a0 Ÿ a0
0
fc 0
1
P3c 0
a1 Ÿ a1
1
f cc 0
0
P3cc 0
2 a2 Ÿ a 2
0
f ccc 0
1
P3ccc 0
6a3 Ÿ a3
16
P3 x
x 16 x 3
(d) f x
INSTRUCTOR USE ONLY
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NOT FOR SALE
Problem Solving ffor Chapter 2
x 2 , yc
4. (a) y
2 x, Slope
(d) Let a, a 2 , a z 0, be a point on the parabola y
4 at 2, 4
x2.
Tangent line: y 4
4x 2
Tangent line at a, a 2 is y
2a x a a 2 .
y
4x 4
Normal line at a, a 2 is y
1 2a x a a 2 .
(b) Slope of normal line: 1
4
To find points of intersection, solve:
x2
1
x 2
4
1
9
x y
4
2
1
9
x y
x2
4
2
Ÿ 4 x 2 x 18
Normal line: y 4
1
x
2a
1
1
x2 x 2a
16a 2
x2 1 ·
§
¨x ¸
4
a¹
©
0
Ÿ 4x 9 x 2
0
9
2, 4
x
§ 9 81 ·
Second intersection point: ¨ , ¸
© 4 16 ¹
(c) Tangent line: y
Normal line: x
pc x
x 1
4a
1 ·
§
r¨ a ¸
4
a¹
©
1
a Ÿ x
4a
x 1
4a
1 ·
§
¨ a ¸ Ÿ x
4a ¹
©
a
(Point of tangency)
a 1
2a
2a 2 1
2a
2a 2 1
.
2a
At 1, 3 :
1
Equation 1
A
= 14
Equation 2
3 A 2B C
B C D =
Adding Equations 1 and 3: 2 B 2 D
Adding Equations 2 and 4: 6 A 2C
Subtracting Equations 2 and 4: 4 B
1 2 2 B
2
4 and D
you obtain 4 A
8 Ÿ A
Equation 3
= 2
Equation 4
4
12
16
5. Subtracting 2 A 2C
2. Finally, C
1 4 2A
2
a b cos cx
4 and 6 A 2C
0. So, p x
12,
2 x3 4 x 2 5.
From Equation 3, b
bc sin cx
At 0, 1 : a b
B C D = 3
2
Subtracting Equations 1 and 3: 2 A 2C
fc x
1
4a
3 Ax 2 2 Bx C.
3A 2B C
6. f x
2
The normal line intersects a second time at x
At 1, 1 :
So, B
1 ·
§
¨a ¸
4
a¹
©
x 0
0
2
1
x a a2
2a
1
a2 2
1
1
a2 2 16a 2
Ax3 Bx 2 Cx D
5. Let p x
A
197
1
Equation 1
§S 3·
§ cS ·
At ¨ , ¸: a b cos¨ ¸
© 4 2¹
© 4 ¹
3
2
Equation 2
§ cS ·
bc sin ¨ ¸
© 4 ¹
1
Equation 3
From Equation 1, a
1 b. Equation 2 becomes
§ cS ·
1 b b cos¨ ¸
© 4 ¹
3
cS
Ÿ b b cos
2
4
1
1
§ cS ·
cos¨ ¸
c sin cS 4
c sin cS 4
© 4 ¹
§ cS ·
1 cos¨ ¸
© 4 ¹
1
2
1
§ cS ·
c sin ¨ ¸
2
© 4 ¹
Graphing the equation
g c
1
.
2
1
. So:
c sin cS 4
1
§ cS ·
§ cS ·
c sin ¨ ¸ cos¨ ¸ 1,
2
4
© ¹
© 4¹
you see that many values of c will work. One answer:
1
,a
2
3
Ÿ f x
2
3
1
cos 2 x
co
2
2
INSTRUCTOR USE ONLY
c
2,, b
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 2
198
Differentiation
x4
a2 x2 a2 y 2
a2 y2
a2 x2 x4
7. (a)
r
y
y2
y2
a x x
.
a
(b)
b
4
30
2
x3 a x
Graph y1
a2 x2 x4
and
a
2 2
x3 a x
y2
a2 x2 x4
a
Graph: y1
x 3 a x ; a, b ! 0
8. (a) b 2 y 2
b
x3 a x
b
and
.
(b) a determines the x-intercept on the right: a, 0 .
2
b affects the height.
a = 12
−3
(c) Differentiating implicitly:
3
2b 2 yyc
a=2
a=1
−2
yc
r a, 0 are the x-intercepts, along with 0, 0 .
(c) Differentiating implicitly:
3x 2 a x x3
3ax 2 4 x3
2b 2 y
Ÿ 3ax 2
4 x3
4 x3
2a 2 x 2a 2 yyc
3a
4x
yc
2a 2 x 4 x 3
2a 2 y
x
3a
4
x a2 2 x2
b2 y 2
3a ·
§ 3a · §
¨ ¸ ¨a ¸
4
4¹
© ¹ ©
y2
27 a 4
Ÿ y
256b 2
a2 y
2 2
§a ·
¨ ¸
©2¹
4
a
4
a2 y2
y2
y
3ax 2 4 x3
0
3
0 Ÿ 2x2
a2 Ÿ x
§a ·
a2 ¨ ¸ a2 y2
©2¹
2
ra
2
27 a 3 § 1 ·
¨ a¸
64 © 4 ¹
r
3 3a 2
16b
§ 3a 3 3a 2 · § 3a 3 3a 2 ·
Two points: ¨¨ ,
¸, ¨ ,
¸
16b ¸¹ ¨© 4
16b ¸¹
© 4
4
a
a2 y2
2
a4
4
a2
4
a
r
2
a· § a a·
§ a a· § a
, ¸, ¨
, ¸, ¨ , ¸,
Four points: ¨
2
2¹ ©
2
2
2 2¹
©
¹ ©
a·
§ a
, ¸
¨
2¹
2
©
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffor Chapter 2
y
9. (a)
Line determined by 0, 30 and 90, 6 :
y 30
30 6
x 0
0 90
When x
100: y
(0, 30)
30
199
(90, 6)
(100, 3)
x
90
100
24
x
90
4
100 30
15
4
x Ÿ y
15
4
x 30
15
10
! 3
3
Not drawn to scale
As you can see from the figure, the shadow determined by the man extends beyond the shadow determined by the child.
y
(b)
30
Line determined by 0, 30 and 60, 6 :
(0, 30)
y 30
30 6
x0
0 60
When x
70: y
(60, 6)
(70, 3)
x
60
70
2
xŸ y
5
2
70 30
5
2
x 30
5
2 3
Not drawn to scale
As you can see from the figure, the shadow determined by the child extends beyond the shadow determined by the man.
(c) Need 0, 30 , d , 6 , d 10, 3 collinear.
30 6
0d
63
24
Ÿ
d d 10
d
3
Ÿd
10
80 feet
(d) Let y be the distance from the base of the street light to the tip of the shadow. You know that dx /dt
For x ! 80, the shadow is determined by the man.
y
30
y x
Ÿ y
6
5
dy
x and
4
dt
5 dx
4 dt
5.
25
4
For x 80, the shadow is determined by the child.
y
30
y x 10
Ÿ y
3
10
100
dy
and
x 9
9
dt
10 dx
9 dt
50
9
Therefore:
dy
dt
­ 25
°° 4 , x ! 80
®
° 50 , 0 x 80
°̄ 9
dy / dt is not continuous at x
80.
ALTERNATE SOLUTION for parts (a) and (b):
(a) As before, the line determined by the man’s shadow is
ym
4
x 30
15
The line determined by the child’s shadow is obtained by finding the line through 0, 30 and 100, 3 :
y 30
30 3
x 0 Ÿ yc
0 100
By setting ym
Man: ym
Child: yc
yc
27
x 30
100
0, you can determine how far the shadows extend:
4
1
x
30 Ÿ x 112.5 112
15
2
27
1
x
0 Ÿ
30 Ÿ x 111.11 111
100
9
0 Ÿ
The man’s shadow is 112
1
1
111
2
9
7
1 ft beyond the child’s shadow.
18
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
200
NOT FOR SALE
Chapter 2
Differentiation
(b) As before, the line determined by the man’s shadow is
2
x 30
5
ym
For the child’s shadow,
y 30
30 3
x 0 Ÿ yc
0 70
Man: ym
0 Ÿ
So the child’s shadow is 77
1
dy
dt
1 2 3 dx
8
3
dt
7
75
9
77
7
9
7
ft beyond the man’s shadow.
9
2
1 2 3 dx
x
3
dt
x1 3 Ÿ
dx
dt
27
x 30
70
2
x
30 Ÿ x
75
5
27
700
x
0 Ÿ
30 Ÿ x
70
9
Child: yc
10. (a) y
12 cm/sec
dD
dt
x2 y 2 Ÿ
(b) D
1 2
dy ·
§ dx
x y2 ¨ 2x
2y ¸
2
dt ¹
© dt
x dx / dt y dy / dt
8 12 2 1
98
68
49
cm/sec
17
8 1 2 12
16
68
64 4
y
dT
Ÿ sec 2T ˜
x
dt
(c) tan T
x2 y 2
x dy / dt y dx / dt
x2
68
2
θ
8
From the triangle, sec T
68 8. So
dT
dt
64 68 64
4
rad/sec.
17
27
t 27 ft/sec
5
11. (a) v t
at
27
ft/sec 2
5
(b) v t
27
t 27
5
27 5
10
S5
2
0 Ÿ 27
t
5
27 Ÿ t
27 5 6
73.5 feet
5seconds
(c) The acceleration due to gravity on Earth is greater in magnitude than that on the moon.
12. E c x
lim
'x o 0
But, E c 0
E x 'x E x
'x
lim
'x o 0
For example: E x
E 'x E 0
'x
lim
'x o 0
lim
E x E 'x E x
'x
'x o 0
E 'x 1
'x
§ E 'x 1 ·
lim E x ¨
¸
'x
©
¹
'x o 0
1. So, E c x
E x Ec 0
E x lim
'x o 0
E 'x 1
'x
E x exists for all x.
ex.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffor Chapter 2
13. Lc x
lim
'x o 0
Also, Lc 0
lim
lim
'x o 0
L 'x L 0
'x
'x o 0
So, Lc x
14. (a)
L x 'x L x
'x
L x L 'x L x
'x
. But, L 0
lim
'x o 0
0 because L 0
201
L 'x
'x
L0 0
L0 L0 Ÿ L0
0.
Lc 0 for all x. The graph of L is a line through the origin of slope Lc 0 .
z (degrees)
0.1
0.01
0.0001
sin z
z
0.0174524
0.0174533
0.0174533
sin z
| 0.0174533
z
S
sin z
In fact, lim
.
z o0 z
180
(b) lim
z o0
(c)
d
sin z
dz
lim
sin z 'z sin z
'z
sin z ˜ cos 'z sin 'z ˜ cos z sin z
lim
'z o 0
'z
'z o 0
ª
ª
§ cos 'z 1 ·º
§ sin 'z ·º
lim «sin z ¨
¸» 'lim
«cos z ¨ 'z ¸»
z o0
'
z
©
¹
©
¹¼
¬
¼
¬
'z o 0
§ S ·
sin z 0 cos z ¨
¸
© 180 ¹
(d) S 90
§ S
·
sin ¨
90 ¸
© 180 ¹
sin
S
2
S
180
cos z
1
§ S
·
1
cos¨
180 ¸
© 180
¹
d
d
S
S z
c ˜ cos cz
C z
sin cz
dz
dz
180
(e) The formulas for the derivatives are more complicated in degrees.
C 180
ac t
15. j t
(a) j t is the rate of change of acceleration.
(b)
st
8.25t 2 66t
vt
16.5t 66
at
16.5
ac t
jt
0
The acceleration is constant, so j t
0.
(c) a is position.
b is acceleration.
c is jerk.
d is velocity.
INSTRUCTOR USE ONLY
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NOT FOR SALE
C H A P T E R 3
Applications of Differentiation
Section 3.1
Extrema on an Interval .......................................................................203
Section 3.2
Rolle’s Theorem and the Mean Value Theorem...............................211
Section 3.3
Increasing and Decreasing Functions
and the First Derivative Test ..............................................................220
Section 3.4
Concavity and the Second Derivative Test .......................................245
Section 3.5
Limits at Infinity .................................................................................264
Section 3.6
A Summary of Curve Sketching........................................................280
Section 3.7
Optimization Problems.......................................................................298
Section 3.8
Newton’s Method ...............................................................................315
Section 3.9
Differentials ........................................................................................324
Review Exercises ........................................................................................................330
Problem Solving .........................................................................................................348
INSTRUCTOR USE ONLY
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NOT FOR SALE
C H A P T E R 3
Applications of Differentiation
Section 3.1 Extrema on an Interval
1.
x2
x 4
f x
3.
x 4
2
x 4
2
0
f x
cos
Sx
fc x
S
sin
fc 0
0
fc 2
0
f x
4
x 2
x
x 4x
fc x
1 8 x 3
1
fc 2
0
2
8x
2
fc 0
4.
5.
x
x2 4 2x x2 2 x
fc x
2.
8. Critical number: x
2
2
9. Critical numbers: x
1, 3: absolute maxima (and relative maxima)
x
2: absolute minimum (and relative minimum)
x
2: neither
2
x
5: absolute maximum (and relative maximum)
11.
3 x
fc x
3 x ª 12 x 1
¬
2
8
x3
1 2
º ¼
x 1 3
1 2
ª¬ x 2 x 1 º¼
32 x 1
1 2
3x 2
f x
x 2
fc x
1 3
2
x 2
3
x3 3x 2
fc x
3x 2 6 x
12. g x
x4 8x2
gc x
4 x3 16 x
Critical numbers: x
32 x 1
0
f x
Critical numbers: x
x 1
f c 23
13. g t
gc t
23
14.
f x
6. Using the limit definition of the derivative,
lim
x o 0
f x f 0
x 0
f x f 0
x 0
lim
x o 0
lim
x o 0
4 x
4
x
4 x
4
x 0
1
1
f c 0 does not exist, because the one-sided derivatives
are not equal.
7. Critical number: x
x
t
fc x
hc x
0, 2
4x x2 4
0, 2, 2
1 2
12
ª1
º
1 » 4 t
t« 4 t
2
¬
¼
1
1 2
ª
4t
¬ t 2 4 t º¼
2
8 3t
2 4t
8
3
4x
x2 1
x2 1 4 4x 2 x
x2 1
Critical numbers: x
15. h x
3x x 2
4 t, t 3
Critical number: t
f c 2 is undefined.
lim
2, 5
Sx
f x
x o 0
1, 2, 3
x
10. Critical numbers: x
2
0
0: neither
x2 1
2
r1
sin 2 x cos x, 0 x 2S
2 sin x cos x sin x
Critical numbers in 0, 2S : x
2
4 1 x2
2
sin x 2 cos x 1
S
3
, S,
5S
3
2: absolute maximum (and relative maximum)
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203
204
Chapter 3
Applications
lications of Differentiation
16.
f T
2 sec T tan T , 0 T 2S
fcT
f x
2 x3 6 x, >0, 3@
2 sec T tan T sec 2 T
fc x
6x2 6
sec T 2 tan T sec T
Critical number: x
22.
ª § sin T ·
1 º
sec T «2¨
»
¸ cos
cos
T
T¼
¹
¬ ©
Critical number: 1, 4 Minimum
Right endpoint: 3, 36 Maximum
7S 11S
,
6 6
Critical numbers in 0, 2S : T
23.
>1, 2@
f x
3 x,
fc x
1 Ÿ no critical numbers
fc x
3 x 2 3 2 x,
fc x
2 x 1 3 2
>1, 1@
21 3 x
3
x
Critical number: 0, 0 Minimum
Right endpoint: 2, 1 Minimum
f x
f x
Left endpoint: 1, 5 Maximum
Left endpoint: 1, 4 Maximum
18.
1 not in interval.
1 x
Left endpoint: 0, 0
sec 2 T 2 sin T 1
17.
6 x2 1
Right endpoint: 1, 1
3
x 2, >0, 4@
4
3
Ÿ no critical numbers
4
24. g x
gc x
3
x1 3 , > 8, 8@
x
1
3x2 3
Critical number: x
Left endpoint: 0, 2 Minimum
0
Left endpoint: 8, 2 Minimum
Right endpoint: 4, 5 Maximum
Critical number: 0, 0
19. g x
gc x
2 x 8 x, >0, 6@
2
4x 8
Right endpoint: 8, 2 Maximum
4x 2
Critical number: x
25. g t
2
Left endpoint: 0, 0
gc t
Critical number: 2, 8 Minimum
>1, 1@
t2 3
2
Right endpoint: 6, 24 Maximum
1·
§
Left endpoint: ¨ 1, ¸ Maximum
4¹
©
5 x 2 , >3, 1@
Critical number: 0, 0 Minimum
2 x
§ 1·
Right endpoint: ¨1, ¸ Maximum
© 4¹
20. h x
hc x
Critical number: x
0
Left endpoint: 3, 4 Minimum
26.
f x
Critical number: 0, 5 Maximum
fc x
Right endpoint: 1, 4
21.
t2
,
t 3
6t
2
f x
x3 32 x 2 , >1, 2@
fc x
3x 2 3x
3x x 1
fc x
2x
,
x2 1
>2, 2@
x2 1 2 2x 2x
x2 1
2
2 2x2
x2 1
2
2 1 x2
x2 1
2
Left endpoint: 1, 52 Minimum
4·
§
Left endpoint: ¨ 2, ¸
5¹
©
Right endpoint: 2, 2 Maximum
Critical number: 1, 1 Minimum
Critical number: 0, 0
Critical number: 1, 1 Maximum
Critical number: 1, 12
§ 4·
endpoint: ¨ 2, ¸
Right endpoint
© 5¹
INSTRUCTOR USE ONLY
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Section 3.1
1
s 2
1
27. h s
hc s
s 2
s 2
1
205
a2 xb, >2, 2@
32. h x
, >0, 1@
Extrema oon an Interval
From the graph you see that the maximum value of h is 4
at x
2, and the minimum value is 0 for 1 x d 2.
2
y
4
1·
§
Left endpoint: ¨ 0, ¸ Maximum
2¹
©
3
Right endpoint: 1, 1 Minimum
1
t
, >1, 6@
t 3
t 3 1 t1
28. h t
−2
x
−1
3
1
2
f x
ª 5S 11S º
sin x, « ,
»
¬6 6 ¼
No critical numbers
fc x
cos x
1·
§
Left endpoint: ¨ 1, ¸ Minimum
2¹
©
Critical number: x
hc t
t 3
2
t 3
33.
2
§ 2·
Right endpoint: ¨ 6, ¸ Maximum
© 3¹
29. y
3 t 3,
For x 3, y
and yc
§ 5S 1 ·
Left endpoint: ¨ , ¸ Maximum
© 6 2¹
>1, 5@
3 t 3
§ 3S
·
Critical number: ¨ , 1¸ Minimum
© 2
¹
t
§ 11S 1 ·
Right endpoint: ¨
, ¸
2¹
© 6
1 z 0 on >1, 3
For x ! 3, y
3 t 3
3S
2
6t
and yc
1 z 0 on 3, 5@
So, x
3 is the only critical number.
34. g x
sec x,
gc x
Left endpoint: 1, 1 Minimum
ª S Sº
« 6 , 3 »
¬
¼
sec x tan x
§ S 2 · § S
·
Left endpoint: ¨ ,
¸ | ¨ , 1.1547 ¸
3¹ © 6
¹
© 6
Right endpoint: 5, 1
Critical number: 3, 3 Maximum
§S ·
Right endpoint: ¨ , 2 ¸ Maximum
©3 ¹
x 4 , >7, 1@
Critical number: 0, 1 Minimum
30. g x
g is the absolute value function shifted 4 units to the
4.
left. So, the critical number is x
Left endpoint: 7, 3
yc
3 sin x
>0, 2S @
S
Left endpoint: 0, 3 Maximum
Right endpoint: 1, 5 Maximum
Critical number: S , 3 Minimum
axb, >2, 2@
From the graph of f, you see
that the maximum value of f is
2 for x
2, and the minimum
value is –2 for 2 d x 1.
3 cos x,
Critical number in 0, 2S : x
Critical number: 4, 0 Minimum
31. f x
35. y
Right endpoint: 2S , 3 Maximum
y
2
1
36. y
x
−2
−1
1
2
yc
−2
§S x ·
tan ¨ ¸,
© 8 ¹
S
8
>0, 2@
§S x ·
sec 2 ¨ ¸ z 0
© 8 ¹
Left endpoint: 0, 0 Minimum
INSTRUCTOR USE ONLY
Right endpoint: 2, 1 Maximum
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Chapter 3
206
37. f x
NOT FOR SALE
Applications
lications of Differentiation
2x 3
2
,
2 x
42. f x
(a) Minimum: 0, 3
>0, 2
Left endpoint: 0, 1 Minimum
Maximum: 2, 1
3
(b) Minimum: 0, 3
(c) Maximum: 2, 1
38. f x
(0, 1)
−1
(d) No extrema
5
−1
5 x
x 4 2 x3 x 1,
43. f x
(a) Minimum: 4, 1
>1, 3@
32
Maximum: 1, 4
(b) Maximum: 1, 4
−1
(c) Minimum: 4, 1
(d) No extrema
39. f x
fc x
x 2x
2
Maximum: 1, 3
2 x 1 2x2 2 x 1
0
Right endpoint: 3, 31 Maximum
§1 r 3 3 ·
Critical points: ¨¨
, ¸¸ Minima
4¹
© 2
(b) Maximum: 3, 3
(c) Minimum: 1, 1
(d) Minimum: 1, 1
x
x cos ,
2
44. f x
>0, 2S @
3
4 x2
(1.729, 1.964)
(a) Minima: 2, 0 and 2, 0
Maximum: 0, 2
2
0
(b) Minimum: 2, 0
(c) Maximum: 0, 2
(d) Maximum: 1,
4 x3 6 x 2 1
1 1r 3
| 0.5, 0.366, 1.366
,
2
2
x
(a) Minimum: 1, 1
40. f x
3
−4
3
0
fc x
1
2
x
1
x
sin
2
2
Left endpoint: 0, 1 Minimum
Graphing utility: 1.729, 1.964 Maximum
41. f x
3
,
x 1
1, 4@
Right endpoint: 4, 1 Minimum
8
0
4
0
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.1
45. (a)
Extrema on
o an Interval
207
5
(1, 4.7)
0
Minimum: (0.4398, 1.0613)
1
(0.4398, −1.0613)
−2
f x
3.2 x5 5 x3 3.5 x,
fc x
16 x 15 x 3.5
(b)
16 x 15 x 3.5
4
2
4
2
15 r
15
>0, 1@
0
x2
2
4 16 3.5
2 16
15 r 449
32
15 449
| 0.4398
32
x
Left endpoint: 0, 0
Critical point: 0.4398, 1.0613 Minimum
Right endpoint: 1, 4.7 Maximum
46. (a)
3
§ 8·
Maximum: ¨ 2, ¸
© 3¹
(2, 83 )
0
3
0
f x
4
x 3 x,
3
fc x
º
4ª § 1·
1 2
12
1 3 x 1 »
x¨ ¸ 3 x
«
3¬ © 2¹
¼
(b)
>0, 3@
4
1 2 § 1 ·
3 x
¨ ¸ ª
¬ x 2 3 x º¼
3
© 2¹
2 6 3x
62x
22x
3 3x
3 3 x
3 x
Left endpoint: 0, 0 Minimum
§ 8·
Critical point: ¨ 2, ¸ Maximum
© 3¹
Right endpoint: 3, 0 Minimum
1 x3
47. f x
3 2
x 1 x3
2
3 2
3 4
x 4 x 1 x3
4
5 2
3
x 6 20 x3 8 1 x3
8
f cc x
f ccc x
Setting f ccc
x
>0, 2@
,
1 2
fc x
x3
12
20 r
0, you have x 6 20 x3 8
0.
400 4 1 8
2
3
10 r
108
3 1
In the interval >0, 2@, choose x
f cc 3 10 108
3
10 r
108
3 1 | 0.732.
| 1.47 is the maximum value.
INSTRUCTOR
NSTR
USE ONLY
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208
Chapter 3
48.
f x
NOT FOR SALE
Applications
lications of Differentiation
ª1 º
« 2 , 3»
¬
¼
1
,
x 1
2 x
2
fc x
x 1
2
C: neither
2
x2 1
D: relative minimum
E: relative maximum
F: relative minimum
3
G: neither
24 x 24 x3
f ccc x
x2 1
Setting f ccc
5
0, you have x
23
x 1
fc x
2 x 1 1 3
3
,
f ccc x
f 4 x
56
x 1
81
f 5 x
13 3
560
x 1
243
f
f
x
1
2
3
4
5
6
−3
55. (a) Yes
(b) No
56. (a) No
(b) Yes
57. (a) No
(b) Yes
5
58. (a) No
240 x 3 x 4 10 x 2 3
x 1
(b) Yes
6
24 is the maximum value.
1
on the interval 0, 1
x
59. P
VI RI 2
12 I 0.5 I 2 , 0 d I d 15
P
0 when I
0.
P
67.5 when I
Pc
12 I
When I
15.
0
Critical number: I
There is no maximum or minimum value.
y
f
−2
51. Answers will vary. Sample answer:
y
y
−2 −1
x 1
0
6
10 3
2
4
5
2
4
2
f 5 x
4
3
24 5 x 4 10 x 2 1
x
3
5
24 x 24 x3
x2 1
1
−3
>1, 1@
2
f ccc x
x
−2 −1
4
1
,
x 1
f x
1
−2
56
is the maximum value.
81
0
f
2
54.
7 3
8
x 1
27
4
3
4 3
92 x 1
50.
4
0, r1.
>0, 2@
f x
f cc x
f
y
53.
4
1
is the maximum value.
2
f cc 1
4
B: relative maximum
2 1 3x 2
f cc x
49.
52. A: absolute minimum
12 amps
12 amps, P
72, the maximum output.
No, a 20-amp fuse would not increase the power output.
P is decreasing for I ! 12.
2
1
x
1
2
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NOT FOR SALE
Section 3.1
60. x
v 2 sin 2T S
3S
,
dT d
32
4
4
61.
dT
is constant.
dt
dx dT
by the Chain Rule
dT dt
dx
dt
v 2 cos 2T dT
16
dt
Extrema on
o an Interval
S
6hs dS
dT
3s 2
2
3s 2 §
¨
2 ¨©
In the interval >S 4, 3S 4@, T
S 4, 3S 4 indicate
csc T
3cot T
minimums for dx dt and T
S 2 indicates a maximum
sec T
3
for dx dt. This implies that the sprinkler waters longest
T
S 4 and 3S 4. So, the lawn farthest from
the sprinkler gets the most water.
3 cos T · S
S
¸¸, d T d
sin T
6
2
¹
3csc T cot T csc 2 T
3s 2
csc T 2
when T
209
3cot T csc T
0
arcsec 3 | 0.9553 radians
§S ·
S¨ ¸
©6¹
6hs 3s 2
2
3
§S ·
S¨ ¸
©6¹
6hs 3s 2
2
3
S arcsec 3
6hs 3s 2
2
2
S is minimum when T
arcsec 3 | 0.9553 radian.
62. (a) Because the grade at A is 9%, A 500, 45
The grade at B is 6%, B 500, 30 .
y
A
B
9%
6%
−500
(b)
x
500
y
ax 2 bx c
yc
2ax b
At A: 2a 500 b
0.09
At B : 2a 500 b
0.06
Solving these two equations, you obtain
a
3
40,000
b
and
3
.
200
Using the points A 500, 45 and B 500, 30 , you obtain
45
3
2
§ 3 ·
500 ¨ ¸ 500 C
40,000
© 200 ¹
30
3
2
§ 3 ·
500 ¨ ¸ 500 C.
40,000
© 200 ¹
In both cases, C
(c)
75
. So, y
4
18.75
3
3
75
x2 x 40,000
200
4
x
–500
–400
–300
–200
–100
0
100
200
300
400
500
d
0
0.75
3
6.75
12
18.75
12
6.75
3
0.75
0
For 500 d x d 0, d
ax 2 bx c 0.09 x .
For 0 d x d 500, d
ax 2 bx c 0.06 x .
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210
NOT FOR SALE
Chapter 3
3
3
x 20,000
200
(d) yc
x
Applications
lications of Differentiation
3
20,000
˜
200
3
0
100
The lowest point on the highway is 100, 18 , is not directly over the origin.
63. True. See Exercise 25.
69. First do an example: Let a
64. True. This is stated in the Extreme Value Theorem.
4 and f x
Then R is the square 0 d x d 4, 0 d y d 4.
Its area and perimeter are both k
65. True
2
66. False. Let f x
g x
x
0 is a critical number of f.
x .x
f x k
67. If f has a maximum value at
x
c, then f c t f x for all x in I. So,
f c d f x for all x in I. So, f has a minimum
68.
Claim that all real numbers a ! 2 work. On the one
hand, if a ! 2 is given, then let f x
2a a 2 .
2a ½
­
® x, y : 0 d x d a, 0 d y d
¾
a 2¿
¯
R
k is a critical number of g.
value at x
has k
2a 2
:
a 2
Area
§ 2a ·
a¨
¸
© a 2¹
c.
f x
ax3 bx 2 cx d ,
fc x
3ax 2 2bx c
2b r
4b 2 12ac
6a
b r
b 2 3ac
3a
Zero critical numbers: b 2 3ac.
Example: a
b
c
1, d
2 a a 2 2 2a
a 2
x3 x 2 x
has no critical numbers.
One critical number: b
Example: a
2
1, b
c
critical number, x
0.
2a 2
.
a 2
To see that a must be greater than 2, consider
^ x, y : 0 d x d a, 0 d y d f x `.
R
0 f x
f attains its maximum value on >0, a@ at some point
P x0 , y0 , as indicated in the figure.
3ac.
d
y
0 f x
3
x has one
P ( x0 , y0 )
y0
f
Two critical numbers: b 2 ! 3ac.
Example:
a
c 1, b
2a 2
a 2
§ 2a ·
2a 2¨
¸
© a 2¹
Perimeter
a z 0
The quadratic polynomial can have zero, one, or two
zeros.
x
16.
Then the rectangle
2
xk
4.
R
2, d
two critical numbers: x
x3 2 x 2 x has
0 f x
1
1, .
3
O
A
x0
x
a
Draw segments OP and PA. The region R is bounded
by the rectangle 0 d x d a, 0 d y d y0 , so
area R
k d ay0 . Furthermore, from the figure,
y0 OP and y0 PA. So,
k
Perimeter R ! OP PA ! 2 y0 . Combining,
2 y0 k d ay0 Ÿ a ! 2.
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NOT FOR SALE
Section
ction 3.2
Rolle’s Theorem and the Mean Va
Value
V
Theorem
211
Section 3.2 Rolle's Theorem and the Mean Value Theorem
1. f x
1
x
f 1
1. But, f is not continuous on >1, 1@.
f 1
2. Rolle's Theorem does not apply to f x
>S , 3S @ because f is not continuous at x
f 1
fc x
f 1
3
33
0, 3 . Rolle's Theorem applies.
fc x
2 x 3
23
x x 2
0.
4 16 5
7
f 6
36 48 5
7
2, 6 . Rolle's Theorem applies.
x 2 x 1
2
f 2
f is continuous on >2, 6@ and differentiable on
x
2x 1
0 at x
6. f x
x2 6x
x x 6
fc x
2x 8
2x
x-intercepts: 0, 0 , 6, 0
0 at x
fc x
8. f x
4
x 1 x 2 x 3 , >1, 3@
f 1
11 1 2 13
0
f 3
31 3 2 33
0
1, 3 . Rolle's Theorem applies.
x 4
1
1 2
12
x 4
x x 4
2
1 2 § x
·
x 4
¨ x 4¸
2
©
¹
1 2
§3
·
¨ x 4¸ x 4
©2
¹
3 x
8 Ÿ x
f is continuous on >1, 3@. f is differentiable on
3.
x-intercepts: 4, 0 , 0, 0
fc x
0
c-value: 4
1.
2
11. f x
x
3
2
x 2 8 x 5, >2, 6@
10. f x
fc x
7. f x
3 Ÿ x
3
13
2x 6
0
2 x
x-intercepts: 1, 0 , 2, 0
fc x
0
c-value: 32
f is not differentiable at x
5. f x
30
2
2S .
1
2 x
2
f is continuous on >0, 3@ and differentiable on
1.
2 x2 3
4. f x
f 3
0
cot x 2 over
3. Rolle's Theorem does not apply to
f x
1 x 1 over >0, 2@ because f is not
differentiable at x
f 0
>0, 3@
x 2 3 x,
9. f x
0 at x
f x
x3 6 x 2 11x 6
fc x
3 x 2 12 x 11
x
c-values:
8
3
0
6r 3
3
6 3 6 3
,
3
3
x 1
x-intercepts: 1, 0 , 0, 0
1
1 2
12
3x 1
x 1
2
1 2 § x
·
3 x 1 ¨ x 1 ¸
©2
¹
fc x
3 x
fc x
3 x 1
1 2 § 3
·
¨ x 1¸
©2
¹
0 at x
2
3
INSTRUCTOR USE ONLY
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212
NOT FOR SALE
Chapter 3
12. f x
Applications
lications of Differentiation
x 4 x 2 , > 2, 4@
2
f 2
2 4 2 2
f 4
4 4 4 2
2
13.
2
0
0
f x
x 2 3 1, >8, 8@
f 8
8
f 8
8
23
23
1
3
1
3
f is continuous on > 2, 4@. f is differentiable on
f is continuous on >8, 8@. f is not differentiable on
2, 4 . Rolle's Theorem applies.
8, 8 because f c 0 does not exist. Rolle's Theorem
does not apply.
f x
x 4 x2 4 x 4
x3 12 x 16
fc x
3 x 2 12
0
14. f x
3x 2
12
f 0
x2
4 Ÿ x
f 1
f 3
0
0, 6 because f c 3 does not exist. Rolle's Theorem
does not apply.
c-value: 2
15. f x
f 6
f is continuous on >0, 6@. f is not differentiable on
r2
2 is not in the interval.
Note: x
3 x 3 , >0, 6@
x2 2x 3
, >1, 3@
x 2
1
2
2 1 3
0
1 2
32 2 3 3
0
3 2
f is continuous on >1, 3@.
(Note: The discontinuity x
fc x
2, is not in the interval.) f is differentiable on 1, 3 . Rolle's Theorem applies.
x 2 2x 2 x2 2 x 3 1
x 2
2
x2 4x 1
x 2
2
x
0
0
4 r 2 5
2
(Note: x
c-value: 2 16.
f x
f 1
f 1
2 5
5 is not in the interval.)
5
x2 1
, >1, 1@
x
1
2 r
2
1
0
1
12 1
1
17. f x
sin x, >0, 2S @
f 0
sin 0
f 2S
sin 2S
0
0
f is continuous on >0, 2S @. f is differentiable on
0, 2S . Rolle's Theorem applies.
0
f is not continuous on >1, 1@ because f 0 does not exist.
fc x
cos x
c-values:
S 3S
Rolle's Theorem does not apply.
2
,
0 Ÿ x
S 3S
2
,
2
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 3.2
cos x, >0, 2S @
18. f x
f 0
cos 0
f 2S
cos 2S
Rolle’s Theorem and the Mean Va
Value
V
Theorem
f is not continuous on
>S , 2S @ because f 3S 2
1
0, 2S . Rolle's Theorem applies.
sin x
x 1, >1, 1@
f 1
f 1
1, 1 because f c 0 does not exist. Rolle's Theorem
ª Sº
sin 3 x, «0, »
¬ 3¼
f 0
sin 3 ˜ 0
§S ·
f¨ ¸
©3¹
S·
§
sin ¨ 3 ˜ ¸
3¹
©
does not apply.
1
0
0
−1
1
ª Sº
f is continuous on «0, ». f is differentiable on
¬ 3¼
§ S·
¨ 0, ¸. Rolle's Theorem applies.
© 3¹
fc x
3 cos 3x
0
3x
S
Ÿ x
2
x x 1 3 , >0, 1@
f 0
S
6
f 1
0, 1 . (Note: f is not differentiable at x
f S
cos 2S
f S
cos 2S
1
1
3
1
f is continuous on >S , S @ and differentiable on
S , S . Rolle's Theorem applies.
fc x
x2
x2
2 sin 2 x
2 sin 2 x
0
sin 2 x
0
x
c-value:
S
S
S , , 0, , S
2
2
x
S
25. f x
S
c-values: , 0,
2
2
f 14
f 14
tan x, >0, S @
f 0
tan 0
0
f S
tan S
0
0. )
Rolle's Theorem applies.
fc x
6
0
f is continuous on >0, 1@. f is differentiable on
cos 2 x, >S , S @
20. f x
21. f x
−1
24. f x
S
c-value:
0
f is continuous on >1, 1@. f is not differentiable on
c-value: S
19. f x
f x
23.
S
0 Ÿ x
sec 3S 2 does not exist.
Rolle's Theorem does not apply.
f is continuous on >0, 2S @. f is differentiable on
fc x
sec x, >S , 2S @
22. f x
1
213
1
1
0
3 3 x2
1
3 3 x2
1
3
1
27
1
27
1
3
9
0
3
| 0.1925
9
1
−1
x tan S x, ª¬ 14 , 14 º¼
14 1
1 1
4
3
4
43
Rolle's Theorem does not apply
0.75
f is not continuous on >0, S @ because f S 2 does not
−0.25
0.25
exist. Rolle's Theorem does not apply.
− 0.75
INSTRUCTOR USE ONLY
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214
Chapter 3
26.
f x
NOT FOR SALE
Applications
lications of Differentiation
Sx
x
sin
, >1, 0@
2
6
f 0
0
f 1
y
29.
Tangent line
(c2, f (c2))
f is continuous on >1, 0@. f is differentiable on
(a, f (a))
t line
1, 0 . Rolle's Theorem applies.
Sx
1 S
Sx
cos
2
6
6
3
6
S
x
fc x
cos
6
S
arccos
3
f
Secan
(b, f (b))
(c1, f(c1))
a
x
b
Tangent line
0
y
30.
ªValue needed in 1, 0 .º¼
S ¬
| 0.5756 radian
f
x
c-value: –0.5756
a
0.02
b
31. f is not continuous on the interval >0, 6@. ( f is not
continuous at x
−1
2.)
0
32. f is not differentiable at x
smooth at x
2.
−0.01
16t 2 48t 6
27. f t
(a) f 1
f 2
fc t
38
32t 48
t
f has a discontinuity at x
f is not differentiable at x
3
sec
2
C 6
(b)
Cc x
2 x2 6x 9
0
x
6 r
x 3
x y 3
0
(b) f c x
2 x
1
Ÿ
(c) f c
§1·
f¨ ¸
© 2¹
1
5
4
108
Tangent line:
3r3 3
2
y x
19
4
4 y 19
4 x 4 y 21
In the interval
3, 6 : c
x 1
y
0
4
6 r 6 3
4
1
y 4
Secant line:
3
x 6x 9
2
1 4
2 1
(a) Slope
§ 1
·
3
¸
10¨ 2 2¸
¨ x
x 3 ¹
©
1
x2
3.
x2 5
35. f x
25
3
(a) C 3
3.
x 3 , >0, 6@
34. f x
0
x ·
§1
10¨ ¸
x
x
3¹
©
28. C x
1
, >0, 6@
x 3
33. f x
f c t must be 0 at some time in 1, 2 .
(b) v
2. The graph of f is not
33 3
| 4.098 | 410 components
2
(d)
c
1
2
19
4
1·
§
¨ x ¸
2¹
©
4 x 2
0
7
Secant
Tangent
f
−6
6
−1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 3.2
x 2 x 12
36. f x
differentiable on 0, 2 .
1
y 0
Secant line:
2x 1
(c) f c
f 1
c
y 12
3
c
2 1 3
x
3
fc x
on 2, 1 .
x
f 1 f 2
1 4
3
1 2
2x
1
x
1
2
1
0, 6 .
6x
2
12
x
r2 3
x
72
f 2 f 7
2 7
2 3.
1 1
2
r
3x 2 2
3
3x 2
1
x
r
03
9
1
2 2 x
2 2 x
3
2 x
3
2
9
4
2 x
differentiable on 1, 1 .
3 3
1 2. The
2 x is continuous on >7, 2@ and
44. f x
fc x
f 1 f 1
0. The Mean
2 x 1 is not differentiable at x
x3 2 x is continuous on >1, 1@ and
fc x
8
27
Mean Value Theorem does not apply.
72
In the interval 0, 6 : c
39. f x
3
differentiable on 7, 2 .
432 0
60
fc x
§ 2·
¨ ¸
© 3¹
x 1
is not continuous at x
x
Value Theorem does not apply.
2 x3 is continuous on >0, 6@ and differentiable on
6 0
1
42. f x
43. f x
f 6 f 0
2
8
27
c
1
2
2
3
1
10
x 2 is continuous on >2, 1@ and differentiable
38. f x
x
x 2 3 is continuous on >0, 1@ and differentiable on
f 1 f 0
Tangent
−15
2
0, 1 .
15
Secant
c
x3
2
41. f x
5
fc x
0
0
f
37. f x
4 x3 2
1
x 1
x y 13
−15
4 x3 8
0
12
Tangent line:
(d)
fc x
x
00
2
20
0
Ÿ
1
f 2 f 0
x 4
x y 4
(b) f c x
215
x 4 8 x is continuous on >0, 2@ and
40. f x
6 0
2 4
(a) Slope
Rolle’s Theorem and the Mean Va
Value
V
Theorem
x
3
c
1
3
1
3
1
4
1
4
1
3
3
3
INSTRUCTOR USE ONLY
c
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216
NOT FOR SALE
Chapter 3
Applications
lications of Differentiation
sin x is continuous on >0, S @ and differentiable
45. f x
on 0, S .
(a) – (c)
f S f 0
00
S 0
S
fc x
cos x
tangent
0
− 2
tangent
− 2
(b) Secant line:
f S f S
slope
S S
2
y S
cos x tan x is not continuous at
S 2. The Mean Value Theorem does not apply.
x
y
1 2 cos x
1
cos x
0
x
c
f
Tangent
− 0.5
§S ·
f¨ ¸
©2¹
2
Secant
S
S
2
(b) Secant line:
slope
y y
(c) f c x
f 2 f 1 2
2 3 1
2 1 2
52
2
3
Tangent lines:
2
2
x 2
3
3
2
x 1
3
1
x 1
2
x 1
2
x
2
3
(a) – (c)
In the interval >1 2, 2@: c
1 1 6 2
9
y
(c) f c x
·
2§
6
1¸¸
¨x 3 ¨©
2
¹
y 1
6
3
2
6
2
x 3
3
3
1
2x 5 2 6
3
y
S·
§
1¨ x ¸
2¹
©
x 2
(b) Secant line:
6
2
6
§ S
·
y ¨ 2¸
© 2
¹
y
1
6
Tangent line: y 1 S·
§
1¨ x ¸
2¹
©
x 2
1
y 1
2
1
6
§S
·
y ¨ 2¸
2
©
¹
y
Secant
f 9 f 1
slope
6 2º 1
¼
2
Tangent
f
6
2
6 2
ª1 ¬
2 1 r
2
3
3
2
1 r
S
x , >1, 9@
49. f x
3
2
r
2
2
§ S·
f ¨ ¸
© 2¹
−1
1
x
1
(a) – (c)
S S
2S
1x S
(c) f c x
x ª 1 º
, ,2
x 1 «¬ 2 »¼
47. f x
f c
secant 2
f
S 2
S
46. f x
2
0
x
c
x 2 sin x, >S , S @
48. f x
f 4
1
4
1
x 1
4
1
3
x 4
4
1
2
x
x
1
4
c
4
2
Tangent line: y 2
y
31
8
9 1
1
x 4
4
1
x 1
4
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 3.2
x 4 2 x 3 x 2 , >0, 6@
50. f x
Rolle’s Theorem and the Mean Va
Value
V
Theorem
f b and f c c
54. f a
0 where c is in the interval
a, b .
(a) – (c) 1000
(a)
Secant
Tangent
6
−100
f 6 f 0
900 0
6
6 0
150 x 0
y 0
y
g a
gb
gc x
f c x Ÿ gc c
f a k
0
Critical number of g: c
150
g x
f x k
g a k
gb k
gc x
fc x k
(b)
150 x
(c) f c x
f x k
Interval: >a, b@
(b) Secant line:
slope
g x
f
0
4 x3 6 x 2 2 x
150
gc c k
fc c
f a
0
Using a graphing utility, there is one solution in
0, 6 , x
c | 3.8721 and f c | 123.6721
Interval: >a k , b k @
Tangent line: y 123.6721
Critical number of g : c k
150 x 3.8721
150 x 457.143
y
(c)
4.9t 2 300
51. s t
s3 s0
(a) vavg
30
255.9 300
3
14.7 m/sec
(b) s t is continuous on >0, 3@ and differentiable on
0, 3 . Therefore, the Mean Value Theorem applies.
sc t
vt
(a)
9.8t
14.7
9.8
t
52. S t
1.5 sec
Sc t
55. f x
200 ª¬5 9 14 º¼ 200 ª¬5 9 2 º¼
12
1
2t
2
2t
t
§
·
9
¸
200¨
2
¨ 2t ¸
©
¹
1
28
f kx
§a·
g¨ ¸
©k¹
gc x
§b·
g¨ ¸
©k¹
kf c kx
§c·
g c¨ ¸
©k¹
kf c c
f a
0
Critical number of g :
9 ·
§
200¨ 5 ¸
2 t¹
©
S 12 S 0
12 0
g x
ªa b º
Interval: « , »
¬k k ¼
14.7 m/sec
450
7
(b)
217
450
7
c
k
x
0
­0,
®
¯1 x, 0 x d 1
No, this does not contradict Rolle's Theorem. f is not
continuous on >0, 1@.
56. No. If such a function existed, then the Mean Value
Theorem would say that there exists c  2, 2 such
that
fc c
f 2 f 2
2 2
6 2
4
2.
But, f c x 1 for all x.
2 7
2 7 2 | 3.2915 months
S c t is equal to the average value in April.
53. No. Let f x
x 2 on >1, 2@.
fc x
2x
fc 0
0 and zero is in the interval 1, 2 but
INSTRUCTOR USE ONLY
f 1 z f 2 .
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© Cengage Learning. All Rights Reserved.
218
Chapter 3
NOT FOR SALE
Applications
lications of Differentiation
62. (a) f is continuous on >10, 4@ and changes sign,
57. Let S t be the position function of the plane. If
t
0 corresponds to 2 P.M., S 0
0, S 5.5
f 8 ! 0, f 3 0 . By the Intermediate Value
2500
and the Mean Value Theorem says that there exists a
time t0 , 0 t0 5.5, such that
Theorem, there exists at least one value of x in
>10, 4@ satisfying f x 0.
2500 0
| 454.54.
5.5 0
Applying the Intermediate Value Theorem to the
velocity function on the intervals >0, t0 @ and >t0 , 5.5@,
S c t0
(b) There exist real numbers a and b such that
10 a b 4 and f a
f b
2.
v t0
Therefore, by Rolle’s Theorem there exists at least
one number c in 10, 4 such that f c c
0.
you see that there are at least two times during the
flight when the speed was 400 miles per hour.
0 400 454.54
This is called a critical number.
y
(c)
8
58. Let T t be the temperature of the object. Then
1500q and T 5
T 0
4
390q. The average
x
−8
temperature over the interval >0, 5@ is
−8
63. f is continuous on >5, 5@ and does not satisfy the
222qF/h.
conditions of the Mean Value Theorem. Ÿ f is not
differentiable on 5, 5 . Example: f x
59. Let S t be the difference in the positions of the 2
S1 t S2 t . Because
S 0
0, there must exist a time
At this time, v1 t0
8
6
(− 5, 5)
t0  0, 2.25 such that S c t0
v t0
x
y
bicyclists, S t
S 2.25
4
−4
390 1500
222q F/h.
50
By the Mean Value Theorem, there exist a time t0 ,
0 t0 5, such that T c t0
−4
0.
f(x) = ⏐x⏐
(5, 5)
4
2
v2 t0 .
x
−4
60. Let t
0 correspond to 9:13 A.M. By the Mean Value
a t0
85 35
1 30
2
4
−2
1 such that
Theorem, there exists t0 in 0, 30
vc t0
−2
64. f is not continuous on >5, 5@.
1500 mi/h 2 .
­1 x, x z 0
®
x
0
¯0,
Example: f x
§S x ·
3 cos 2 ¨ ¸, f c x
© 2 ¹
61. f x
§ S x ·§
§ S x · ·§ S ·
6 cos¨ ¸¨ sin ¨ ¸ ¸¨ ¸
© 2 ¹©
© 2 ¹ ¹© 2 ¹
y
§S x · §S x ·
3S cos¨ ¸ sin ¨ ¸
© 2 ¹ © 2 ¹
f (x) = 1x
4
2
(5, 15)
7
(a)
f′
x
f
2
(− 5, − 15)
−2
4
2
−5
−7
(b) f and f c are both continuous on the entire real line.
(c) Because f 1
f 1
0, Rolle's Theorem
applies on >1, 1@. Because f 1
0 and f 2
3,
Rolle's Theorem does not apply on >1, 2@.
(d) lim f c x
0
lim f c x
0
x o 3
INSTRUCTOR USE ONLY
x o 3
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 3.2
Rolle’s Theorem and the Mean Va
Value
V
Theorem
x5 x3 x 1
69. f c x
0
f is differentiable for all x.
f x
c
f 2
5
65. f x
f 1
2 and f 0
1, so the Intermediate Value
Theorem implies that f has at least one zero c in
>1, 0@, f c 0.
Suppose f had 2 zeros, f c1
f c2
So, f x
70. f c x
0. Then
Rolle's Theorem would guarantee the existence of a
number a such that
fc a
f c2 f c1
But, f c x
0.
66. f x
4x c
f 0
1Ÿ c
71. f c x
1 and f 1
8, so the Intermediate Value
Theorem implies that f has at least one zero c in
>0, 1@, f c 0.
Suppose f had 2 zeros, f c1
f c2
0. Then
Rolle's Theorem would guarantee the existence of a
number a such that
f c2 f c1
But f c x
x2 c
f 1
0 Ÿ 0
0.
0 has
6x 1
f x
3x 2 x c
f 2
7 Ÿ 7
3 22 2 c
3
3 x x 3.
2
73. False. f x
1 x has a discontinuity at x
0.
74. False. f must also be continuous and differentiable on
each interval. Let
3 x 1 sin x
f is differentiable for all x.
f S
1
x 2 1.
So, f x
So, f x
10 x 4 7 ! 0 for all x. So, f x
1 c Ÿ c
10 c Ÿ c
exactly one real solution.
67. f x
2x
f x
72. f c x
1
4 x 1.
So, f x
2 x5 7 x 1
fc a
4
f x
f is differentiable for all x.
f 0
5.
5 x 4 3x 2 1 ! 0 for all x. So, f has
exactly one real solution.
219
3S 1 0 and f 0
f x
1 ! 0, so the
x3 4 x
.
x2 1
Intermediate Value Theorem implies that f has at least
one zero c in >S , 0@, f c
0.
75. True. A polynomial is continuous and differentiable
everywhere.
Suppose f had 2 zeros, f c1
76. True
f c2
0. Then
Rolle's Theorem would guarantee the existence of a
number a such that
fc a
f c2 f c1
But f c x
x1 and x2 . Then by Rolle's Theorem, because
0.
3 cos x ! 0 for all x. So, f x
0 has
exactly one real solution.
68. f x
f 0
2S 2 1
Suppose f had 2 zeros, f c1
2S 1 ! 0. By the
f c2
0. Then
Rolle's Theorem would guarantee the existence of a
number a such that
But, f c x
f c2 f c1
p x2
pc c
0. But pc x
2n 1 x 2 n a z 0, because
roots.
Intermediate Value Theorem, f has at least one zero.
fc a
0, there exists c in x1 , x2 such that
p x1
n ! 0, a ! 0. Therefore, p x cannot have two real
2 x 2 cos x
3, f S
x 2 n 1 ax b has two real roots
77. Suppose that p x
78. Suppose f x is not constant on a, b . Then there exists
x1 and x2 in a, b such that f x1 z f x2 . Then by the
Mean Value Theorem, there exists c in a, b such that
fc c
0.
2 sin x t 1 for all x. So, f has exactly
f x2 f x1
x2 x1
z 0.
This contradicts the fact that f c x
0 for all x in
INSTRUCTOR USE ONLY
one real solution.
aa,, b .
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219
220
NOT FOR SALE
Chapter 3
Ax 2 Bx C , then
79. If p x
pc x
Applications
lications of Differentiation
fc x
f b f a
2 Ax B
1 cos x differentiable on
2
82. f x
b a
12 sin x
12 d f c x d 12 Ÿ f c x 1
Ab Bb C Aa Ba C
2
f, f .
for all real numbers.
2
So, from Exercise 62, f has, at most, one fixed point.
x | 0.4502
b a
A b2 a 2 B b a
83. Let f x
b a
b a ª¬ A b a Bº¼
all real numbers. By the Mean Value Theorem, for any
interval >a, b@, there exists c in a, b such that
b a
A b a B.
f b f a
A b a and x
So, 2Ax
b a 2 which is the
midpoint of >a, b@.
80. (a) f x
x2 , g x
x3 x 2 3x 2
f 1
g 1
1, f 2
Let h x
h 1
g 2
4
0. So, by Rolle's Theorem these
exists c  1, 2 such that
hc c
f c c gc c
0.
x3 3 x 2, hc x
3x 2 3
(b) Let h x
0 Ÿ x
c
f x g x . Then h a
f c c gc c
cos b cos a
sin c b a
hb
84. Let f x
b a cos c
b a cos c
sin a sin b d a b .
85. Let 0 a b. f x
x satisfies the hypotheses of
the Mean Value Theorem on >a, b@. Hence, there exists c
0.
the Mean Value Theorem, there exists c such that
c2 c1
fc c
b a
sin b sin a
in a, b such that
fc c
81. Suppose f x has two fixed points c1 and c2 . Then, by
f c2 f c1
sin x. f is continuous and differentiable for
f b f a
0 by
c, the tangent line to f is parallel to the
So, at x
tangent line to g.
fc c
sin c b a
sin b sin a
1
Rolle's Theorem, there exists c in a, b such that
hc c
sin c
all real numbers. By the Mean Value Theorem, for any
interval >a, b@, there exists c in a, b such that
So, at x
c, the tangent line to f is parallel to the
tangent line to g.
h x
fc c
b a
cos b cos a
b a
cos b cos a
cos b cos a d b a since sin c d 1.
f x g x . Then,
h2
cos x. f is continuous and differentiable for
c2 c1
c2 c1
So,
1
2 c
b a
f b f a
b a
b a
b a
.
b a
1
b a
.
2 c
2 a
1.
This contradicts the fact that f c x 1 for all x.
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test
1. (a) Increasing: 0, 6 and 8, 9 . Largest: 0, 6
(b) Decreasing: 6, 8 and 9, 10 . Largest: 6, 8
2. (a) Increasing: 4, 5 , 6, 7 . Largest: 4, 5 , 6, 7
(b) Decreasing: 3, 1 , 1, 4 , 5, 6 . Largest: 3, 1
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.3
3. f x
Increasing and Decreasing Functions and the First De
Derivative Test
x2 6x 8
4. y
x 1
221
2
From the graph, f is decreasing on f, 3 and
From the graph, f is increasing on f, 1 and
increasing on 3, f .
decreasing on 1, f .
Analytically, f c x
2 x 6.
Analytically, yc
2 x 1 .
Critical number: x
3
Critical number: x
1
Test intervals:
f x 3
3 x f
Test intervals:
f x 1
1 x f
Sign of f c x :
fc 0
fc ! 0
Sign of yc:
yc ! 0
yc 0
Conclusion:
Decreasing
Increasing
Conclusion:
Increasing
Decreasing
5. y
x3
3x
4
From the graph, y is increasing on f, 2 and 2, f , and decreasing on 2, 2 .
3x 2
3
4
Analytically, yc
Critical numbers: x
3 2
x 4
4
3
x 2 x 2
4
r2
Test intervals:
f x 2
2 x 2
2 x f
Sign of yc:
yc ! 0
yc 0
yc ! 0
Conclusion:
Increasing
Decreasing
Increasing
6. f x
x4 2 x2
From the graph, f is decreasing on f, 1 and 0, 1 , and increasing on 1, 0 and 1, f .
Analytically, f c x
4 x3 4 x
Critical numbers: x
0, r1.
Test intervals:
f x 1
1 x 0
0 x 1
1 x f
Sign of f c:
fc 0
fc ! 0
fc 0
fc ! 0
Conclusion:
Decreasing
Increasing
Decreasing
Increasing
7. f x
4x x 1 x 1 .
1
x 1
2
From the graph, f is increasing on f, 1 and decreasing on 1, f .
Analytically, f c x
2
x 1
3
.
No critical numbers. Discontinuity: x
1
Test intervals:
f x 1
1 x f
Sign of f c x :
fc ! 0
fc 0
Conclusion:
Increasing
Decreasing
INSTRUCTOR USE ONLY
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222
Chapter 3
8. y
NOT FOR SALE
Applications
lications of Differentiation
x2
2x 1
From the graph, y is increasing on f, 0 and 1, f , and decreasing on 0, 1 2 and 1 2, 1 .
2x 1 2x x2 2
Analytically, yc
2x 1
Critical numbers: x
2x x 1
2 x2 2 x
2
2x 1
2
2x 1
2
0, 1
Discontinuity: x
12
Test intervals:
f x 0
0 x 12
12 x 1
1 x f
Sign of yc:
yc ! 0
yc 0
yc 0
yc ! 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
9. g x
gc x
x2 2x 8
2x 2
Critical number: x
1
Test intervals:
f x 1
1 x f
Sign of g c x :
gc 0
gc ! 0
Conclusion:
Decreasing
Increasing
Increasing on: 1, f
Decreasing on: f, 1
10. h x
12 x x 3
hc x
12 3 x 2
3 4 x2
32 x 2 x
r2
Critical numbers: x
Test intervals:
f x 2
2 x 2
2 x f
Sign of hc x :
hc 0
hc ! 0
hc 0
Conclusion:
Decreasing
Increasing
Decreasing
Increasing on: 2, 2
Decreasing on: f, 2 , 2, f
11. y
yc
Domain: >4, 4@
x 16 x 2
2 x 2 8
2
16 x
16 x 2
2
Critical numbers: x
x 2 2 x 2 2
r2 2
Test intervals:
4 x 2 2
2 2 x 2 2
2 2 x 4
Sign of yc:
yc 0
yc ! 0
yc 0
Conclusion:
Decreasing
Increasing
Decreasing
Increasing on: 2 2, 2 2
INSTRUCTOR USE ONLY
Decreasing on: 4, 2 2 , 2 2, 4
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NOT FOR SALE
Section 3.3
12. y
yc
x Increasing and Decreasing Functions and the First De
Derivative Test
223
9
x
19
x2
x 3 x 3
x2 9
x2
x2
r3
Critical numbers: x
Discontinuity: x
0
Test intervals:
f x 3
3 x 0
0 x 3
3 x f
Sign of yc:
yc ! 0
yc 0
yc 0
yc ! 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing on: f, 3 , 3, f
Decreasing on: 3, 0 , 0, 3
13. f x
fc x
0 x 2S
sin x 1,
cos x
S 3S
Critical numbers: x
2
,
2
S
S
2
2
x 3S
2
3S
x 2S
2
Test intervals:
0 x Sign of f c x :
fc ! 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Increasing
§ S · § 3S
·
Increasing on: ¨ 0, ¸, ¨ , 2S ¸
2
2
©
¹ ©
¹
§ S 3S ·
Decreasing on: ¨ ,
¸
©2 2 ¹
14. h x
hc x
x
cos ,
0 x 2S
2
x
1
sin
2
2
Critical numbers: none
Test interval:
0 x 2S
Sign of hc x :
hc 0
Conclusion:
Decreasing
Decreasing on 0 x 2S
INSTRUCTOR USE ONLY
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224
NOT FOR SALE
Chapter 3
Applications
lications of Differentiation
x 2 cos x, 0 x 2S
15. y
yc
1 2 sin x
yc
0: sin x
1
2
7S 11S
,
6 6
Critical numbers: x
7S
11S
x 6
6
11S
x 2S
6
yc ! 0
yc 0
yc ! 0
Increasing
Decreasing
Increasing
Test intervals:
0 x Sign of yc :
Conclusion:
7S
6
§ 7S · § 11S
·
Increasing on: ¨ 0,
, 2S ¸
¸, ¨
6
6
©
¹ ©
¹
§ 7S 11S ·
Decreasing on: ¨ ,
¸
© 6 6 ¹
sin 2 x sin x, 0 x 2S
16. f x
fc x
2 sin x cos x cos x
2 sin x 1
0 Ÿ sin x
1
Ÿ x
2
7S 11S
,
6 6
S 3S
0 Ÿ x
cos x
cos x 2 sin x 1
2
,
2
Critical numbers:
S 7S 3S 11S
Test intervals:
0 x Sign of f c x :
fc ! 0
Conclusion:
Increasing
2
,
,
6
2
,
6
S
S
2
2
7S
3S
x 6
2
3S
11S
x 2
6
11S
x 2S
6
fc 0
fc ! 0
fc 0
fc ! 0
Decreasing
Increasing
Decreasing
Increasing
x 7S
6
§ S · § 7S 3S · § 11S
·
Increasing on: ¨ 0, ¸, ¨ ,
, 2S ¸
¸, ¨
© 2¹ © 6 2 ¹ © 6
¹
§ S 7S · § 3S 11S ·
Decreasing on: ¨ ,
¸, ¨ ,
¸
©2 6 ¹ © 2 6 ¹
17. (a)
f x
x2 4 x
fc x
2x 4
Critical number: x
(b)
2
Test intervals:
f x 2
2 x f
Sign of f c:
fc 0
fc ! 0
Conclusion:
Decreasing
Increasing
Decreasing on: f, 2
Increasing on: 2, f
(c) Relative minimum: 2, 4
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.3
18. (a)
f x
x 2 6 x 10
fc x
2x 6
225
3
Critical number: x
(b)
Increasing and Decreasing Functions and the First De
Derivative Test
Test intervals:
f x 3
3 x f
Sign of f c:
fc 0
fc ! 0
Conclusion:
Decreasing
Increasing
Decreasing on: f, 3
Increasing on: 3, f
(c) Relative minimum: 3, 1
19. (a)
(b)
f x
2 x 2 4 x 3
fc x
4 x 4
0
Critical number: x
1
Test intervals:
f x 1
1 x f
Sign of f c x :
fc ! 0
fc 0
Conclusion:
Increasing
Decreasing
Increasing on: f, 1
Decreasing on: 1, f
(c) Relative maximum: 1, 5
20. (a)
f x
3x 2 4 x 2
fc x
6x 4
Critical number: x
(b)
0
23
Test intervals:
f x 23
23 x f
Sign of f c x :
fc ! 0
fc 0
Conclusion:
Increasing
Decreasing
Increasing on: f, 23
Decreasing on: 23 , f
(c) Relative maximum: 23 , 23
INSTRUCTOR USE ONLY
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226
Chapter 3
21. (a)
(b)
NOT FOR SALE
Applications
lications of Differentiation
f x
2 x3 3 x 2 12 x
fc x
6 x 2 6 x 12
6 x 2 x 1
0
Critical numbers: x
2, 1
Test intervals:
f x 2
2 x 1
1 x f
Sign of f c x :
fc ! 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: f, 2 , 1, f
Decreasing on: 2, 1
(c) Relative maximum: 2, 20
Relative minimum: 1, 7
22. (a)
(b)
f x
x3 6 x 2 15
fc x
3 x 2 12 x
3x x 4
Critical numbers: x
0, 4
Test intervals:
f x 0
0 x 4
4 x f
Sign of f c x :
fc ! 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: f, 0 , 4, f
Decreasing on: 0, 4
(c) Relative maximum: 0, 15
Relative minimum: 4, 17
23. (a)
2
f x
x 1
x 3
fc x
3x 2 2 x 5
x 1 3x 5
1, 53
Critical numbers: x
(b)
x3 x 2 5 x 3
Test intervals:
f x 53
5 3 x 1
1 x f
Sign of f c:
fc ! 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: f, 53 and 1, f
Decreasing on: 53 , 1
(c) Relative maximum: 53 , 256
27
Relative minimum: 1, 0
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.3
24. (a)
(b)
2
f x
x 2
fc x
3x x 2
Increasing and Decreasing Functions and the First De
Derivative Test
227
x 1
Critical numbers: x
2, 0
Test intervals:
f x 2
2 x 0
0 x f
Sign of f c x :
fc ! 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: f, 2 , 0, f
Decreasing on: 2, 0
(c) Relative maximum: 2, 0
Relative minimum: 0, 4
25. (a)
f x
x5 5 x
5
fc x
x4 1
Critical numbers: x
(b)
1, 1
Test intervals:
f x 1
1 x 1
1 x f
Sign of f c x :
fc ! 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: f, 1 , 1, f
Decreasing on: 1, 1
4·
§
(c) Relative maximum: ¨ 1, ¸
5
©
¹
4·
§
Relative minimum: ¨1, ¸
5¹
©
26. (a)
f x
x 4 32 x 4
fc x
4 x3 32
4 x3 8
Critical number: x
(b)
27. (a)
2
f x 2
2 x f
Sign of f c x :
fc 0
fc ! 0
Conclusion:
Decreasing
Increasing
Decreasing on: f, 2
x1 3 1
fc x
1 2 3
x
3
1
3x 2 3
Critical number: x
Test intervals:
Increasing on: 2, f
f x
(b)
0
Test intervals:
f x 0
0 x f
Sign of f c x :
fc ! 0
fc ! 0
Conclusion:
Increasing
Increasing
Increasing on: f, f
(c) No relative extrema
(c) Relative minimum: 2, 44
INSTRUCTOR USE ONLY
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228
Chapter 3
28. (a)
NOT FOR SALE
Applications
lications of Differentiation
f x
x2 3 4
fc x
2 1 3
x
3
31. (a)
2
3x1 3
Critical number: x
(b)
0
f x
5 x 5
fc x
­ 1, x 5
®
¯1, x ! 5
x 5
x 5
Critical number: x
5
Test intervals:
f x 0
0 x f
Test intervals:
f x 5
5 x f
Sign of f c x :
fc 0
fc ! 0
Sign of f c x :
fc ! 0
fc 0
Conclusion:
Decreasing
Increasing
Conclusion:
Increasing
Decreasing
(b)
Increasing on: 0, f
Increasing on: f, 5
Decreasing on: f, 0
Decreasing on: 5, f
(c) Relative minimum: 0, 4
(c) Relative maximum: 5, 5
29. (a)
23
f x
x 2
fc x
2
1 3
x 2
3
32. (a)
13
2
Critical number: x
(b)
2
3x 2
f x
x 3 1
fc x
x 3
x 3
­ 1, x ! 3
®
¯1, x 3
3
Critical number: x
(b)
Test intervals:
f x 3
3 x f
fc ! 0
Sign of f c x :
fc 0
fc ! 0
Increasing
Conclusion:
Decreasing
Increasing
Test intervals:
f x 2
2 x f
Sign of f c:
fc 0
Conclusion:
Decreasing
Decreasing on: f, 2
Increasing on: 3, f
Increasing on: 2, f
Decreasing on: f, 3
(c) Relative minimum: 2, 0
(c) Relative minimum: 3, 1
30. (a)
13
f x
x 3
fc x
1
2 3
x 3
3
Critical number: x
(b)
1
3x 3
23
3
Test intervals:
f x 3
3 x f
Sign of f c:
fc ! 0
fc ! 0
Conclusion:
Increasing
Increasing
Increasing on: f, f
(c) No relative extrema
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.3
33. (a)
f x
fc x
1
x
1
2 2
x
Increasing and Decreasing Functions and the First De
Derivative Test
229
2x 2x2 1
x2
2
2
r
Critical numbers: x
Discontinuity: x
0
Test intervals:
f x Sign of f c:
fc ! 0
fc 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
(b)
2
2
§
2·
Increasing on: ¨¨ f, ¸¸
2
©
¹
and
0 x 2
2
2
x f
2
§ 2 ·
¨¨ 2 , f ¸¸
©
¹
and
§
2 ·
, 0 ¸¸
Decreasing on: ¨¨ 2
©
¹
2
x 0
2
§
2·
¨¨ 0,
¸¸
2
©
¹
§
·
2
, 2 2 ¸¸
(c) Relative maximum: ¨¨ 2
©
¹
§ 2
·
Relative minimum: ¨¨
, 2 2 ¸¸
© 2
¹
34. (a)
f x
fc x
x
x 5
x 5 x
x 5
5
2
x 5
2
No critical numbers
(b)
Discontinuity: x
5
Test intervals:
f x 5
5 x f
Sign of f c x :
fc 0
fc 0
Conclusion:
Decreasing
Decreasing
Decreasing on: f, 5 , 5, f
(c) No relative extrema
35. (a)
f x
fc x
x2
x 9
2
x2 9 2 x x2 2x
x 9
2
2
Critical number: x
0
Discontinuities: x
3, 3
18 x
x 9
2
2
INSTRUCTOR USE ONLY
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230
Chapter 3
(b)
NOT FOR SALE
Applications
lications of Differentiation
Test intervals:
f x 3
3 x 0
0 x 3
3 x f
Sign of f c x :
fc ! 0
fc ! 0
fc 0
fc 0
Conclusion:
Increasing
Increasing
Decreasing
Decreasing
Increasing on: f, 3 , 3, 0
Decreasing on: 0, 3 , 3, f
(c) Relative maximum: 0, 0
36. (a)
f x
fc x
x2 2x 1
x 1
x 1 2x 2 x2 2 x 1 1
x 1
x 1
x 3 x 1
2
x 1
2
3, 1
Critical numbers: x
(b)
x2 2x 3
2
Discontinuity: x
1
Test intervals:
f x 3
3 x 1
1 x 1
1 x f
Sign of f c x :
fc ! 0
fc 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing on: f, 3 , 1, f
Decreasing on: 3, 1 , 1, 1
(c) Relative maximum: 3, 8
Relative minimum: 1, 0
37. (a)
(b)
f x
2
°­4 x , x d 0
®
x ! 0
°̄2 x,
fc x
­2 x, x 0
®
x ! 0
¯2,
Critical number: x
0
Test intervals:
f x 0
0 x f
Sign of f c:
fc ! 0
fc 0
Conclusion:
Increasing
Decreasing
Increasing on: f, 0
Decreasing on: 0, f
(c) Relative maximum: 0, 4
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.3
38. (a)
f x
­2 x 1, x d 1
® 2
¯x 2, x ! 1
fc x
x 1
­2,
®
¯2 x, x ! 1
231
1, 0
Critical numbers: x
(b)
Increasing and Decreasing Functions and the First De
Derivative Test
Test intervals:
f x 1
1 x 0
0 x f
Sign of f c:
fc ! 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: f, 1
and
0, f
Decreasing on: 1, 0
(c) Relative maximum: 1, 1
Relative minimum: 0, 2
39. (a)
(b)
f x
­3 x 1, x d 1
®
2
¯5 x , x ! 1
fc x
x 1
­3,
®
¯2 x, x ! 1
Critical number: x
1
Test intervals:
f x 1
1 x f
Sign of f c:
fc ! 0
fc 0
Conclusion:
Increasing
Decreasing
Increasing on: f, 1
Decreasing on: 1, f
(c) Relative maximum: 1, 4
40. (a)
(b)
f x
­° x3 1,
x d 0
® 2
°̄ x 2 x, x ! 0
fc x
2
­
x 0
° 3x ,
®
°̄2 x 2, x ! 0
Critical numbers: x
0, 1
Test intervals:
f x 0
0 x 1
1 x f
Sign of f c:
fc 0
fc ! 0
fc 0
Conclusion:
Decreasing
Increasing
Decreasing
Increasing on: 0, 1
Decreasing on: f, 0
and
1, f
(c) Relative maximum: 1, 1
INSTRUCTOR USE ONLY
Note: 0, 1 is not a relative minimum
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232
Chapter 3
f x
41. (a)
fc x
NOT FOR SALE
Applications
lications of Differentiation
x
cos x, 0 x 2S
2
1
sin x
0
2
Critical numbers: x
S 5S
Test intervals:
0 x Sign of f c( x):
fc ! 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Increasing
6
,
6
S
S
6
6
5S
6
x 5S
x 2S
6
§ S · § 5S
·
Increasing on: ¨ 0, ¸, ¨ , 2S ¸
© 6¹ © 6
¹
§ S 5S ·
Decreasing on: ¨ ,
¸
©6 6 ¹
§S S 6 3 ·
(b) Relative maximum: ¨¨ ,
¸¸
12
©6
¹
§ 5S 5S 6 3 ·
Relative minimum: ¨¨ ,
¸¸
12
© 6
¹
5
(c)
2
0
0
42. (a)
f x
sin x cos x 5
fc x
cos 2 x
1
sin 2 x 5, 0 x 2S
2
Critical numbers:
S 3S 5S 7S
Test intervals:
0 x Sign of f c:
fc ! 0
Conclusion:
Increasing
4
,
4
,
4
,
4
S
S
3S
5S
x 4
4
5S
7S
x 4
4
7S
x 2S
4
4
4
fc 0
fc ! 0
fc 0
fc ! 0
Decreasing
Increasing
Decreasing
Increasing
x 3S
4
§ S · § 3S 5S · § 7S
·
Increasing on: ¨ 0, ¸, ¨ ,
¸, ¨ , 2S ¸
4
4
4
4
©
¹ ©
¹ ©
¹
§ S 3S · § 5S 7S ·
Decreasing on: ¨ ,
¸, ¨ ,
¸
©4 4 ¹ © 4 4 ¹
§ S 11 · § 5S 11 ·
(b) Relative maxima: ¨ , ¸, ¨ , ¸
©4 2¹ © 4 2¹
(c)
7
§ 3S 9 · § 7S 9 ·
Relative minima: ¨ , ¸, ¨ , ¸
© 4 2¹ © 4 2¹
2
0
3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.3
Increasing and Decreasing Functions and the First De
Derivative Test
f x
sin x cos x,
0 x 2S
fc x
cos x sin x
0 Ÿ sin x
43. (a)
cos x
Critical numbers: x
S 5S
Test intervals:
0 x Sign of f c( x):
fc ! 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Increasing
4
,
233
4
S
S
4
4
x 5S
4
5S
x 2S
4
§ S · § 5S
·
Increasing on: ¨ 0, ¸, ¨ , 2S ¸
© 4¹ © 4
¹
§ S 5S ·
Decreasing on: ¨ ,
¸
©4 4 ¹
§S
(b) Relative maximum: ¨ ,
©4
·
2¸
¹
§ 5S
Relative minimum: ¨ , © 4
(c)
·
2¸
¹
3
2
0
−3
f x
x 2 sin x,
fc x
1 2 cos x
44. (a)
0 x 2S
0 Ÿ cos x
Critical numbers:
2S 4S
,
3 3
Test intervals:
0 x Sign of f c( x):
Conclusion:
2S
3
1
2
2S
4S
x 3
3
4S
x 2S
3
fc ! 0
fc 0
fc ! 0
Increasing
Decreasing
Increasing
§ 2S · § 4S
·
Increasing on: ¨ 0,
¸, ¨ , 2S ¸
© 3 ¹ © 3
¹
§ 2S 4S ·
Decreasing on: ¨ ,
¸
© 3 3 ¹
§ 2S 2S
(b) Relative maximum: ¨ ,
© 3 3
· § 2S
·
3 ¸ | ¨ , 3.826 ¸
¹ © 3
¹
§ 4S 4S
Relative minimum: ¨ ,
© 3 3
· § 4S
·
3 ¸ | ¨ , 2.457 ¸
3
¹ ©
¹
(c)
7
0
2
INSTRUCTOR USE ONLY
−2
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234
Chapter 3
Applications
lications of Differentiation
0 x 2S
f x
cos 2 2 x ,
fc x
4 cos 2 x sin 2 x
45. (a)
0 Ÿ cos 2 x
0 or sin 2 x
0
S 3S 5S 7S S
3S
,
,
,
, , S,
4 4 4 4 2
2
Critical numbers: x
S
S
4
4
S
2
2
x 3S
4
3S
x S
4
0 x Sign of f c( x):
fc 0
fc ! 0
fc 0
fc ! 0
Conclusion:
Decreasing
Increasing
Decreasing
Increasing
Test intervals:
S x 5S
3S
x 4
2
3S
7S
x 2
4
7S
x 2S
4
Sign of f c( x):
fc 0
fc ! 0
fc 0
fc ! 0
Conclusion:
Decreasing
Increasing
Decreasing
Increasing
5S
4
x S
Test intervals:
§ S S · § 3S · § 5S 3S · § 7S
·
Increasing on: ¨ , ¸, ¨ , S ¸, ¨ ,
¸, ¨ , 2S ¸
4
2
4
4
2
4
©
¹ ©
¹ ©
¹ ©
¹
§ S · § S 3S · § 5S · § 3S 7S ·
Decreasing on: ¨ 0, ¸, ¨ ,
¸, ¨ S ,
¸, ¨ ,
¸
4 ¹ © 2 4 ¹
© 4¹ ©2 4 ¹ ©
§S ·
§ 3S ·
(b) Relative maxima: ¨ , 1¸, S , 1 , ¨ , 1¸
2
©
¹
© 2 ¹
§ S · § 3S · § 5S · § 7S ·
Relative minima: ¨ , 0 ¸, ¨ , 0 ¸, ¨ , 0 ¸, ¨ , 0 ¸
©4 ¹ © 4 ¹ © 4 ¹ © 4 ¹
(c)
3
2
0
−1
46. (a)
f x
sin x 3 cos x, 0 x 2S
fc x
cos x 3 sin x
tan x
1
3
0 Ÿ
3 sin x
cos x
3
3
Critical numbers: x
5S 11S
,
6 6
5S
11S
x 6
6
11S
x 2S
6
fc ! 0
fc 0
fc ! 0
Increasing
Decreasing
Increasing
Test intervals:
0 x Sign of f c( x):
Conclusion:
5S
6
§ 5S · § 11S
·
Increasing on: ¨ 0,
, 2S ¸
¸, ¨
© 6 ¹ © 6
¹
§ 5S 11S ·
Decreasing on: ¨ ,
¸
© 6 6 ¹
§ 5S ·
(b) Relative maximum: ¨ , 2 ¸
© 6 ¹
(c)
§ 11S
·
, 2¸
Relative minimum: ¨
6
©
¹
3
0
2
INSTRUCTOR USE ONLY
−3
3
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.3
Increasing and Decreasing Functions and the First De
Derivative Test
0 x 2S
f x
sin 2 x sin x,
fc x
2 sin x cos x cos x
47. (a)
Critical numbers: x
235
cos x 2 sin x 1
0
S 7S 3S 11S
2
,
,
6
,
2
Test intervals:
0 x Sign of f c( x):
fc ! 0
Conclusion:
Increasing
6
S
S
7S
3S
x 6
2
3S
11S
x 2
6
11S
x 2S
6
2
2
fc 0
fc ! 0
fc 0
fc ! 0
Decreasing
Increasing
Decreasing
Increasing
x 7S
6
§ S · § 7S 3S · § 11S
·
Increasing on: ¨ 0, ¸, ¨ ,
, 2S ¸
¸, ¨
© 2¹ © 6 2 ¹ © 6
¹
§ S 7S · § 3S 11S ·
Decreasing on: ¨ ,
¸, ¨ ,
¸
©2 6 ¹ © 2 6 ¹
§ 7S 1 · § 11S 1 ·
, ¸
(b) Relative minima: ¨ , ¸, ¨
4¹ © 6
4¹
© 6
§ S · § 3S ·
Relative maxima: ¨ , 2 ¸, ¨ , 0 ¸
©2 ¹ © 2 ¹
(c)
3
2
0
−1
f x
48. (a)
fc x
sin x
, 0 x 2S
1 cos 2 x
cos x 2 sin 2 x
1 cos 2 x
0
2
S 3S
Critical numbers: x
2
,
2
S
S
2
2
x 3S
2
3S
x 2S
2
Test intervals:
0 x Sign of f c( x):
fc ! 0
fc 0
fc ! 0
Conclusion:
Increasing
Decreasing
Increasing
§ S · § 3S
·
Increasing on: ¨ 0, ¸, ¨ , 2S ¸
© 2¹ © 2
¹
§ S 3S ·
Decreasing on: ¨ ,
¸
©2 2 ¹
§S ·
(b) Relative maximum: ¨ , 1¸
©2 ¹
§ 3S
·
Relative minimum: ¨ , 1¸
2
©
¹
2
(c)
0
2
INSTRUCTOR USE ONLY
−2
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236
NOT FOR SALE
Chapter 3
Applications
lications of Differentiation
2 x 9 x 2 , >3, 3@
49. f x
(a) f c t
2 9 2x2
(a) f c x
9 x
(b)
t 2 sin t , >0, 2S @
51. f t
2
(b)
y
f′
30
f
10
8
20
10
4
2
x
−1
1
π
2
− 10
2
−8
− 10
2 9 2 x2
r
3
2
r
t
2 tan t
3 2
2
0 or t
2
§ 3 2 3 2·
,
¨¨ ¸
2
2 ¸¹
©
fc x ! 0
§3 2 ·
, 3¸¸
¨¨
© 2
¹
fc x 0
Decreasing
Increasing
Decreasing
f is increasing when f c is positive and decreasing
when f c is negative.
x 2 3 x 16 , >0, 5@
10 5 5 2x 3
Critical numbers: t
2.2889, 5.0870
(d) Intervals:
§
3 2·
¨¨ 3,
¸
2 ¸¹
©
fc x 0
(b)
0
t | 2.2889, 5.0870 (graphing utility)
(d) Intervals:
(a) f c x
(c) t t cos t 2 sin t
t cot t
Critical numbers: x
50. f x
f
0
9 x2
t
2π
− 20
(c)
t t cos t 2 sin t
40
y
f′
t 2 cos t 2t sin t
0, 2.2889
2.2889, 5.0870
5.0870, 2S
fc t ! 0
fc t 0
fc t ! 0
Increasing
Decreasing
Increasing
f is increasing when f c is positive and decreasing
when f c is negative.
x
x
cos , >0, 4S @
2
2
52. f x
(b)
x 2 3 x 16
1
1
x
sin
2
2
2
(a) f c x
y
8
y
6
f
15
4
f
12
2
f′
6
3
−1
(c) f′
π
2π
3π
4π
x
x
1
3
4
−3
(c)
5 2x 3
x 2 3x 16
Critical number: x
0
3
2
(d) Intervals:
1
1
x
sin
2
2
2
x
sin
2
x
2
0
1
S
2
Critical number: x
§ 3·
§3 ·
¨ 0, ¸
¨ , 5¸
© 2¹
©2 ¹
fc x ! 0 fc x 0
S
(d) Intervals:
0, S
S , 4S
fc x ! 0
fc x ! 0
Increasing Decreasing
Increasing Increasing
f is increasing when f c is positive and decreasing
when f c is negative.
f is increasing when f c is positive.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.3
x
3 sin , >0, 6S @
3
x
cos
3
f x
53. (a)
fc x
(b)
Increasing and Decreasing Functions and the First De
Derivative Test
237
3S 9S
,
2 2
(c) Critical numbers: x
(d) Intervals:
§ 3S ·
¨ 0,
¸
© 2 ¹
fc 0
y
4
f
§ 3S 9S ·
¨ ,
¸
© 2 2 ¹
fc ! 0
§ 9S
·
¨ , 6S ¸
© 2
¹
fc 0
Decreasing Increasing Decreasing
2
f′
f is increasing when f c is positive and decreasing
when f c is negative.
x
2π
4π
−2
−4
54. (a)
f x
2 sin 3 x 4 cos 3 x, >0, S @
fc x
6 cos 3 x 12 sin 3x
(b)
(c) f c x
Critical numbers: x | 0.1545, 1.2017, 2.2489
y
(d) Intervals:
12
f'
8
f
4
x
π
−4
0, 0.1545
0.1545, 1.2017
1.2017, 2.2489
2.2489, S
fc ! 0
fc 0
fc ! 0
fc 0
Increasing
Decreasing
Increasing
Decreasing
f is increasing when f c is positive and decreasing when f c is
negative.
−8
− 12
55. f x
1
2
0 Ÿ tan 3 x
x 2 1 x3 3x
x5 4 x3 3x
x2 1
x2 1
x 3 3 x, x z r 1
y
x3 3 x for all x z r1.
f x
g x
fc x
3x 2 3
5
4
3
(−1, 2)
3 x 2 1 , x z r1 Ÿ f c x z 0
x
−4 −3
f symmetric about origin
zeros of f : 0, 0 , r
3, 0
−1
1 2 3 4 5
−2
−3
−4
−5
(1, −2)
g x is continuous on f, f and f x has holes at 1, 2 and 1, 2 .
56.
f t
cos 2 t sin 2 t
1 2 sin 2 t
fc t
4 sin t cos t
2 sin 2t
gt
c is constant Ÿ f c x
57. f x
4
f symmetric with respect to y-axis
zeros of f : r
0.
y
S
2
f′
4
−4
−2
2
x
4
−2
Relative maximum: (0, 1)
§ S
· §S
·
Relative minimum: ¨ , 1¸, ¨ , 1¸
2
2
©
¹ ©
¹
y
−4
58. f x is a line of slope | 2 Ÿ f c x
2
2.
6
1
−π
π
x
−6
−1
−2
6
−2
INSTRUCTOR USE ONLY
The ggraphs
p of f x and g x are the same.
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238
NOT FOR SALE
Chapter 3
Applications
lications of Differentiation
59. f is quadratic Ÿ f c is a line.
65.
y
4
fc x
g c 6
f c 6 0
66. g x
f x
−2
gc x
fc x
−4
gc 0
fc 0 ! 0
x
−2
2
4
60. f is a 4th degree polynomial Ÿ f c is a cubic polynomial.
y
6
f′
x
−6 −4 −2
2
y
4
f x 10
gc x
f c x 10
gc 0
f c 10 ! 0
68. g x
f x 10
gc x
f c x 10
gc 8
f c 2 0
­! 0,
x 4 Ÿ f is increasing on f, 4 .
°
69. f c x ®undefined, x 4
° 0,
x ! 4 Ÿ f is decreasing on 4 f .
¯
Two possibilities for f x are given below.
2
f′
−2
67. g x
6
4
61. f has positive, but decreasing slope.
−4
f x
gc x
f′
2
−4
g x
x
2
(a)
4
−2
y
6
−4
4
2
62. f has positive slope.
x
y
2
6
8
−2
4
3
2
(b)
f′
2
x
−3 −2 −1
1
2
y
3
1
x
−2
1
3
4
5
−1
In Exercises 63–68, f c x > 0 on f , 4 , f c x < 0 on
(– 4, 6) and f c x > 0 on 6, f .
63. g x
64.
−3
f x 5
gc x
fc x
gc 0
fc 0 0
g x
3f x 3
gc x
3fc x
g c 5
3 f c 5 ! 0
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.3
70.
Increasing and Decreasing Functions and the First De
Derivative Test
2 Because f c 2
(i) (a) Critical number: x
0
fc 4
(b) f increasing on
2, f
4.
3 Ÿ f is increasing at x
6.
5, f 5 is a relative minimum.
Because f c 0 on f, 2
(c) f has a relative minimum at x
(ii) (a) Critical numbers:
x
0, 1 Because f c 1
72. Critical number: x
2.
0
fc1
2 Ÿ f is decreasing at x
1.
fc 3
6 Ÿ f is increasing at x
3.
In Exercises 73 and 74, answers will vary.
Because f c ! 0 on these intervals
Sample answers:
73. (a)
f decreasing on
y
0, 1 Because f c 0 on 0, 1
1
(c) f has a relative maximum at x
relative minimum at x 1.
fc 0
f
0, and a
x
−1
1
1, 0, 1
(iii) (a) Critical numbers: x
2
2, f 2 is not a relative extremum.
(b) f increasing on f, 0 and 1, f
Because f c 1
5
2.5 Ÿ f is decreasing at x
fc 6
Because f c ! 0 on 2, f
f decreasing on
f, 2
71. Critical number: x
239
−1
fc1
0
(b) The critical numbers are in intervals
0.50, 0.25 and 0.25, 0.50 because the sign of
(b) f increasing on f, 1 and 0, 1
Because f c ! 0 on these intervals
f c changes in these intervals. f is decreasing on
f decreasing on 1, 0 and 1, f
approximately 1, 0.40 , 0.48, 1 , and increasing
Because f c 0 on these intervals
on 0.40, 0.48 .
1 and
(c) f has a relative maximum at x
x
1. f has a relative minimum at x
(iv) (a) Critical numbers: x
Because f c 3
(c) Relative minimum when x | 0.40: 0.40, 0.75
Relative maximum when x | 0.48: 0.48, 1.25
0.
3, 1, 5
fc1
74. (a)
fc s
0
(b) f increasing on 3, 1 and 1, 5
y
2
1
Because f c ! 0 on these intervals . In fact,
π
2
x
f is increasing on 3, 5 .
f decreasing on f, 3 and 5, f
Because f c 0 on these intervals
(c) f has a relative minimum at x
relative maximum at x
x
5.
1 is not a relative extremum.
3, and a
−2
(b) The critical numbers are in the intervals
§ S · §S S ·
§ 3S 5S ·
¨ 0, ¸, ¨ , ¸, and ¨ ,
¸ because the sign of
© 6¹ ©3 2¹
© 4 6 ¹
f c changes in these intervals. f is increasing on
§ S·
§ 3S 6S ·
approximately ¨ 0, ¸ and ¨ ,
¸ and decreasing
© 7¹
© 7 7 ¹
§ S 3S ·
§ 6S ·
on ¨ ,
¸ and ¨ , S ¸.
7
7
©
¹
© 7
¹
(c) Relative minima when x |
Relative maxima when x |
3S
,S
7
S 6S
7
,
7
INSTRUCTOR USE ONLY
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240
NOT FOR SALE
Chapter 3
Applications
lications of Differentiation
4.9 sin T t 2
75. s t
(a) sc t
4.9 sin T 2t
speed
sc t
T
0
sc t
0
9.8 sin T t
9.8 sin T t
(b)
S
S
S
4
3
2
2S
3
3S
4
S
4.9 2t
4.9 3t
9.8t
4.9 3t
4.9 2t
0
The speed is maximum for T
76. (a) M
(b)
S
2
.
0.06803t 4 3.7162t 3 76.281t 2 716.56t 2393.0
350
9
100
20
(c) Using a graphing utility, the maximum is approximately 17.7, 322.0 , which compares well with the actual maximum in
2007: 17, 326.0 .
3t
,t t 0
27 t 3
77. C
(a)
t
0
0.5
1
1.5
2
2.5
3
C(t)
0
0.055
0.107
0.148
0.171
0.176
0.167
The concentration seems greatest near t
(b)
2.5 hours.
0.25
0
3
0
The concentration is greatest when t | 2.38 hours.
(c) C c
27 t 3 3 3t 3t 2
27 t 3
2
3 27 2t 3
27 t 3
Cc
0 when t
2
3 3 2 | 2.38 hours.
By the First Derivative Test, this is a maximum.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.3
78. f x
241
sin x, 0 x S
x, g x
(a)
Increasing and Decreasing Functions and the First De
Derivative Test
x
0.5
1
1.5
2
2.5
3
f(x)
0.5
1
1.5
2
2.5
3
g(x)
0.479
0.841
0.997
0.909
0.598
0.141
f x seems greater than g x on 0, S .
(b)
5
f
g
0
−2
x ! sin x on 0, S so, f x ! g x .
f x g x
(c) Let h x
hc x
x sin x
1 cos x ! 0 on 0, S .
Therefore, h x is increasing on 0, S . Because h 0
0 and hc x ! 0 on 0, S ,
h x ! 0
x sin x ! 0
x ! sin x
f x ! g x on 0, S
79. v
k R r r2
k Rr 2 r 3
vc
k 2 Rr 3r 2
kr 2 R 3r
r
81. (a) s t
vt
(b) v t
0
2 R.
3
(a) Rc
2 0.001T 4 4T 100
0
10q
Minimum resistance: R | 8.3666 ohms
125
(b)
− 100
3.
(d) The particle changes direction at t
0.004T 3 4
Critical number: T
0 when t
(c) Moving in negative direction when t ! 3.
0.001T 4 4T 100
80. R
6 2t
Moving in positive direction for 0 d t 3 because
v t ! 0 on 0 d t 3.
0 or 23 R
Maximum when r
6t t 2 , t t 0
82. (a) s t
vt
(b) v t
3.
t 2 7t 10, t t 0
2t 7
0 when t
7
2
Particle moving in positive direction for
t ! 72 because vc t ! 0 on
7
,f.
2
(c) Particle moving in negative direction on ª¬0, 72 .
100
(d) The particle changes direction at t
− 25
7
.
2
The minimum resistance is approximately
R | 8.37 ohms at T
10q.
INSTRUCTOR USE ONLY
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242
NOT FOR SALE
Chapter 3
83. (a) s t
vt
(b) v t
Applications
lications of Differentiation
t 3 5t 2 4t , t t 0
3t 2 10t 4
10 r 100 48
6
0 for t
5 r 13
3
Particle is moving in a positive direction on
ª 5 13 ·
§ 5 13 ·
, f ¸¸ | 2.8685, f because v ! 0 on these intervals.
¸¸ | >0, 0.4648 and ¨¨
«0,
3
3
«¬
¹
©
¹
(c) Particle is moving in a negative direction on
§ 5 13 5 13 ·
,
¨¨
¸¸ | 0.4648, 2.8685
3
3
©
¹
(d) The particle changes direction at t
5 r 13
.
3
t 3 20t 2 128t 280
85. Answers will vary.
vt
3t 2 40t 128
86. Answers will vary.
(b) v t
3t 16 t 8
84. (a) s t
vt
16
,8
3
0 when t
and 8, f
v t ! 0 for ª¬0, 16
3
(c) v t 0 for
16
,8
3
(d) The particle changes direction at t
16
and 8.
3
87. (a) Use a cubic polynomial
f x
a3 x3 a2 x 2 a1 x a0
(b) f c x
3a3 x 2 2a2 x a1.
f 0
0: a3 0
fc 0
3
3a3 0
0:
f 2
2: a3 2
fc 2
3
(c) The solution is a0
a1
a1 0 a0
0 Ÿ
a0
0
2a2 0 a1
0 Ÿ
a1
0
a1 2 a0
2 Ÿ
8a3 4a2
2
2a2 2 a1
0 Ÿ 12a3 4a2
0
2
2
a2 2
3a3 2
0:
2
a2 0
2
0, a2
3
,a
2 3
12 :
12 x3 32 x 2 .
f x
(d)
4
(2, 2)
−2
(0, 0)
4
−4
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section 3.3
Increasing and Decreasing Functions and the First De
Derivative Test
243
88. (a) Use a cubic polynomial
3a3 x3 a2 x 2 a1 x a0
f x
(b) f c x
3a3 x 2 2a2 x a1
f 0
fc 0
f 4
a3 0
0:
3
1000: a3 4
fc 4
2
3a3 0
0:
3
2
a2 4
a1
a1 0 a0
0 Ÿ
a0
0
2a2 0 a1
0 Ÿ
a1
0
a1 4 a0
2
3a3 4
0:
(c) The solution is a0
2
a2 0
1000 Ÿ 64a3 16a2
2a2 4 a1
125
4
375
, a3
2
0, a2
0 Ÿ
100
48a3 8a2
0
1200
(d)
(4, 1000)
125
x3 375
x2.
4
2
f x
−3
8
(0, 0)
−400
89. (a) Use a fourth degree polynomial
a4 x 4 a3 x3 a2 x 2 a1 x a0 .
f x
(b) f c x
4a4 x3 3a3 x 2 2a 2 x a1
f 0
0: a4 0
fc 0
4
0: a4 4
fc 4
4: a4 2
fc 2
4
4
3
4a4 2
0:
(c) The solution is a0
a1
a2 0
3
3
2
a1 0 a0
0 Ÿ
a0
0
2a2 0 a1
0 Ÿ
a1
0
2
a1 4 a0
0 Ÿ 256a4 64a3 16a2
0
2a2 4 a1
0 Ÿ
256a4 48a3 8a2
0
2
a1 2 a0
4 Ÿ
16a4 8a3 4a2
4
2a2 2 a1
0 Ÿ
32a4 12a3 4a2
0
2
a2 4
3a3 4
a3 2
3
3
3a3 0
a3 4
4a4 4
0:
f 2
3
4a4 0
0:
f 4
a3 0
2
a2 2
3a3 2
2
0, a2
4, a3
2,
a4
1.
4
1 x 4 2 x3 4 x 2
4
f x
(d)
5
(2, 4)
−2
(0, 0)
5
(4, 0)
−1
90. (a) Use a fourth-degree polynomial
f x
(b) f c x
f 1
fc1
f 1
f c 1
f 3
fc 3
a4 x 4 a3 x3 a2 x 2 a1 x a0 .
4a4 x3 3a3 x 2 2a 2 x a1
4
a4 1
2:
4: a4 1
4:
0:
3
4a4 1
0:
0:
a3 1
4
a3 1
4a4 1
a4 3
4
3
3
a2 1
3
2
a1 1 a0
2 Ÿ
a4 a3 a2 a1 a0
2
3a3 1
2
2a2 1 a1
0 Ÿ
4a4 3a3 2a2 a1
0
a2 1
2
a1 1 a0
4 Ÿ
a4 a3 a2 a1 a0
4
2a2 1 a1
0 Ÿ
4a4 3a3 2a2 a1
0
2
a1 3 a0
4 Ÿ 81a4 27 a3 9a2 a1 a0
4
2a2 3 a1
0 Ÿ
0
3a3 1
a3 3
4a4 3
3
3
2
a2 3
3a3 3
2
108a4 27 a3 6a2 a1
INSTRUCTOR USE ONLY
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244
NOT FOR SALE
Chapter 3
Applications
lications of Differentiation
(c) The solution is a0
23 , a2
23
, a1
8
1, a
4 3
1, a
2 4
81
18 x 4 12 x3 14 x 2 23 x 23
.
8
f x
(d)
6
(−1 , 4)
(3, 4)
(1, 2)
−4
6
−2
97. Assume that f c x 0 for all x in the interval (a, b) and
91. True.
f x g x where f and g are increasing.
Let h x
Then hc x
f c x g c x ! 0 because
f c x ! 0 and g c x ! 0.
fc c
92. False.
Let h x
f x g x where f x
h x
x is decreasing on f, 0 .
g x
x. Then
2
x3 , then f c x
3 x 2 and f only has one
x3 3 x 1, then
critical number. Or, let f x
fc x
f x2 f x1
x2 x1
Because f c c 0 and x2 x1 ! 0, then
f x2 f x1 0, which implies that
f x2 f x1 . So, f is decreasing on the interval.
93. False.
Let f x
let x1 x2 be any two points in the interval. By the
Mean Value Theorem, you know there exists a number c
such that x1 c x2 , and
3 x 2 1 has no critical numbers.
Then there exists a and b in I such that f c x ! 0 for all
x in (a, c) and f c x 0 for all x in (c, b). By Theorem
3.5, f is increasing on (a, c) and decreasing on (c, b).
Therefore, f c is a maximum of f on (a, b) and so, a
94. True.
If f x is an nth-degree polynomial, then the degree of
f c x is n 1.
95. False. For example, f x
98. Suppose f c x changes from positive to negative at c.
relative maximum of f.
99. Let x1 and x2 be two real numbers, x1 x2 . Then
x3 does not have a relative
extrema at the critical number x
0.
96. False. The function might not be continuous on the
interval.
x13 x23 Ÿ f x1 f x2 . So f is increasing on
f, f .
100. Let x1 and x2 be two positive real numbers,
0 x1 x2 . Then
1
1
!
x1
x2
f x1 ! f x2
So, f is decreasing on 0, f .
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.4
Concavity and the Second De
Derivative Test
245
101. First observe that
sin x
cos x
1
1
cos x
sin x
cos x sin x
tan x cot x sec x csc x
sin 2 x cos 2 x sin x cos x
sin x cos x
1 sin x cos x § sin x cos x 1 ·
¨
¸
sin x cos x © sin x cos x 1 ¹
2
sin x cos x 1
sin x cos x sin x cos x 1
2 sin x cos x
sin x cos x sin x cos x 1
2
sin x cos x 1
Let t
sin x cos x 1. The expression inside the absolute value sign is
f t
sin x cos x 2
sin x cos x 1
sin x cos x 1 1 t 1
2
sin x cos x 1
2
t
S·
§
Because sin ¨ x ¸
4¹
©
sin x cos
S
4
cos x sin
S
4
2
sin x cos x ,
2
sin x cos x  ª¬
t
2, 2 º¼ and
sin x cos x 1  ª¬1 f 1 fc t
1
2
t2
t2 2
t2
2
1 2 1
4 2§
¨
2 1 ¨©
2 º¼.
2, 1 t 2 t t
2
1 2 1·
¸
2 1 ¸¹
2
2 1
2 2
4 2 2 4
1
For t ! 0, f is decreasing and f t ! f 1 For t 0, f is increasing on 2
2
2 1, 2
2
2 3 2
2 3 2
2 , then decreasing on 2, 0 . So f t f 2
1 2 2.
Finally, f t t 2 2 1.
(You can verify this easily with a graphing utility.)
Section 3.4 Concavity and the Second Derivative Test
1. The graph of f is increasing and concave downward:
f c ! 0, f cc 0
2. The graph of f is decreasing and concave upward:
f c 0, f cc ! 0
INSTRUCTOR USE ONLY
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Chapter 3
246
3.
y
x2 x 2
yc
2x 1
ycc
2
NOT FOR SALE
Applications
pplications of Differentiation
ycc ! 0 for all x.
Concave upward: f, f
4.
g x
3x 2 x3
gc x
6 x 3x 2
g cc x
6 6x
g cc x
0 when x
1.
Intervals:
f x 1
1 x f
Sign of g cc :
g cc ! 0
g cc 0
Conclusion:
Concave upward
Concave downward
Concave upward: f, 1
Concave downward: 1, f
5.
f x
x3 6 x 2 9 x 1
fc x
3x 2 12 x 9
f cc x
6 x 12
6 x 2
f cc x
0 when x
2.
Intervals:
f x 2
2 x f
Sign of f cc :
f cc ! 0
f cc 0
Conclusion:
Concave upward
Concave downward
Concave upward: f, 2
Concave downward: 2, f
6.
h x
x5 5 x 2
hc x
5x4 5
hcc x
20 x3
hcc x
0 when x
0.
Intervals:
f x 0
0 x f
Sign of hcc :
hcc 0
hcc ! 0
Conclusion:
Concave downward
Concave upward
Intervals:
f x 2
2 x 2
2 x f
Sign of f cc :
f cc ! 0
f cc 0
f cc ! 0
Conclusion:
Concave upward
Concave downward
Concave upward
Concave upward: 0, f
Concave downward: f, 0
7.
f x
fc x
f cc x
f cc x
24
x 2 12
48 x
x 2 12
2
144 4 x 2
x 2 12
0 when x
3
r 2.
Concave upward: f, 2 , 2, f
Concave downward: 2, 2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.4
8.
f x
fc x
f cc x
f cc x
Concavity and the Second De
Derivative Test
247
2x2
3x 2 1
4x
3x 2 1
2
4 3x 1 3x 1
3x 2 1
0 when x
3
1
r .
3
Intervals:
f x 13
13 x 13
1
3
x f
Sign of f cc :
f cc 0
f cc ! 0
f cc 0
Conclusion:
Concave downward
Concave upward
Concave downward
§ 1 1·
Concave upward: ¨ , ¸
© 3 3¹
1 ·§ 1 ·
§
Concave downward: ¨ f, ¸¨ , f ¸
3 ¹© 3 ¹
©
9. f x
fc
f cc
x2 1
x2 1
4 x
x2 1
2
4 3x 2 1
x2 1
3
f is not continuous at x
r 1.
Intervals:
f x 1
1 x 1
1 x f
Sign of f cc :
f cc ! 0
f cc 0
f cc ! 0
Conclusion:
Concave upward
Concave downward
Concave upward
Concave upward: f, 1 , 1, f
Concave downward: 1, 1
10.
y
1 3 x5 40 x 3 135 x
270
yc
1 15 x 4 120 x 2 135
270
ycc
92 x x 2 x 2
ycc
0 when x
0, r 2.
Intervals:
f x 2
2 x 0
0 x 2
2 x f
Sign of ycc :
ycc ! 0
ycc 0
ycc ! 0
ycc 0
Conclusion:
Concave upward
Concave downward
Concave upward
Concave downward
Concave upward: f, 2 , 0, 2
Concave downward: 2, 0 , 2, f
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Chapter 3
248
11.
g x
gc x
g cc x
Applications
pplications of Differentiation
x2 4
4 x2
16 x
4 x2
2
16 3x 2 4
16 3x 2 4
3
4 x2
2 x
f is not continuous at x
3
2 x
3
r 2.
Intervals:
f x 2
2 x 2
2 x f
Sign of g cc :
g cc 0
g cc ! 0
g cc 0
Conclusion:
Concave downward
Concave upward
Concave downward
Concave upward: 2, 2
Concave downward: f, 2 , 2, f
12.
h x
x2 1
2x 1
2x x 1
2
hc x
hcc x
2x 1
2
6
2x 1
Intervals:
f x 12
1
2
x f
Sign of hcc :
hcc ! 0
hcc 0
Conclusion:
Concave upward
Concave downward
Intervals:
Sign of ycc :
ycc ! 0
ycc 0
Conclusion:
Concave upward
Concave downward
3
f cc is not continuous at x
1
.
2
1·
§
Concave upward: ¨ f, ¸
2¹
©
§1 ·
Concave downward: ¨ , f ¸
©2 ¹
13.
y
§ S S·
2 x tan x, ¨ , ¸
© 2 2¹
yc
2 sec x
ycc
2 sec 2 x tan x
ycc
0 when x
2
0.
S
2
x 0
0 x S
2
§ S ·
Concave upward: ¨ , 0 ¸
© 2 ¹
§ S·
Concave downward: ¨ 0, ¸
© 2¹
14.
S , S
y
x 2 csc x,
yc
1 2 csc x cot x
ycc
2 csc x csc 2 x 2 cot x csc x cot x
2 csc3 x csc x cot 2 x
ycc
0 when x
Intervals:
S x 0
0 x S
Sign of ycc :
ycc 0
ycc ! 0
Conclusion:
Concave downward
Concave upward
0.
Concave upward: 0, S
INSTRUCTOR USE ONLY
Concave downward:
downward S , 0
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.4
15.
f x
x3 6 x 2 12 x
fc x
3 x 2 12 x 12
f cc x
6x 2
0 when x
2.
Concave upward: 2, f
Concavity and the Second Derivative
De
Test
Intervals:
f x 2
2 x f
Sign of f cc :
f cc 0
f cc ! 0
Conclusion:
Concave downward
Concave upward
249
Concave downward: f, 2
Point of inflection: 2, 8
16.
f x
x3 6 x 2 5
fc x
3 x 2 12 x
f cc x
6 x 12
6 x 2
0 when x
2.
Concave upward: f, 2
Intervals:
f x 2
2 x f
Sign of f cc :
f cc ! 0
f cc 0
Conclusion:
Concave upward
Concave downward
Concave downward: 2, f
Point of inflection: 2, 11
17.
f x
1 4
x 2 x3
2
fc x
2 x3 6 x 2
f cc x
6 x 2 12 x
6x x 2
f cc x
0 when x
0, 2
Intervals:
f x 2
2 x 2
0 x f
Sign of f cc :
f cc ! 0
f cc 0
f cc ! 0
Conclusion:
Concave upward
Concave downward
Concave upward
Intervals:
f x 0
0 x f
Sign of f cc :
f cc 0
f cc 0
Conclusion:
Concave downward
Concave downward
Concave upward: f, 2 , 0, f
Concave downward: 2, 0
Points of inflection: 2, 8 and 0, 0
18.
f x
4 x 3x 4
fc x
1 12 x3
f cc x
36 x 2
0 when x
0.
Concave downward: f, f
No points of inflection
19.
3
f x
x x4
fc x
2
3
x ª3 x 4 º x 4
¬
¼
f cc x
4 x 1 ª¬2 x 4 º¼ 4 x 4
f cc x
12 x 4 x 2
x4
0 when x
2
2
4x 4
4 x 4 ª¬2 x 1 x 4 º¼
4 x 4 3x 6
12 x 4 x 2
2, 4.
Intervals:
f x 2
2 x 4
4 x f
Sign of f cc x :
f cc x ! 0
f cc x 0
f cc x ! 0
Conclusion:
Concave upward
Concave downward
Concave upward
Concave upward: f, 2 , 4, f
Concave downward: 2, 4
Points of inflection: 2, 16 , 4, 0
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 3
250
Applications
pplications of Differentiation
x 2
3
x 1
fc x
x 2
2
4x 5
f cc x
6 x 2 2x 3
f cc x
0 when x
f x
20.
3
, 2.
2
Intervals:
f x 32
3
2
x 2
Sign of f cc:
f cc ! 0
f cc 0
f cc ! 0
Conclusion:
Concave upward
Concave downward
Concave upward
2 x f
3·
§
Concave upward: ¨ f, ¸, 2, f
2¹
©
§3 ·
Concave downward: ¨ , 2 ¸
©2 ¹
1·
§3
Points of inflection: ¨ , ¸, 2, 0
16 ¹
©2
x 3, Domain: >3, f
f x
x
fc x
1 2
§1·
x¨ ¸ x 3
2
© ¹
21.
6
f cc x
3x 2
x 3
2
x 3 3x 2 x 3
x 3
1 2
4x 3
3x 4
4x 3
32
0 when x
f x
22.
4.
fc x
f cc x
x
x
x 9 x , Domain: x d 9
36 x
2 9 x
3 x 12
49 x
32
0 when x
12 is not in the domain. f cc is not continuous at
9.
x
4 is not in the domain. f cc is not continuous at
Interval:
f x 9
x
3.
Sign of f cc:
f cc 0
Conclusion:
Concave downward
Interval:
3 x f
Sign of f cc:
f cc ! 0
Conclusion:
Concave upward
12.
Concave downward: f, 9
No point of inflection
Concave upward: 3, f
There are no points of inflection.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.4
23.
f x
fc x
f cc x
f cc x
Concavity and the Second De
Derivative Test
251
4
x 1
8 x
2
2
x2 1
8 3x 2 1
x2 1
3
3
3
r
0 for x
3
3
3
x 3
3
3
3
x f
3
Intervals:
f x Sign of f cc:
f cc ! 0
f cc 0
f cc ! 0
Conclusion:
Concave upward
Concave downward
Concave upward
§
3· § 3 ·
, f ¸¸
Concave upward: ¨¨ f, ¸, ¨
3 ¸¹ ¨© 3
©
¹
§
3
3·
Concave downward: ¨¨ ,
¸¸
3
3
©
¹
§
§ 3 ·
3 ·
Points of inflection: ¨¨ , 3¸¸ and ¨¨
, 3¸¸
© 3
¹
© 3
¹
24.
f x
fc x
f cc x
x 3
, Domain: x ! 0
x
x 3
2 x3 2
9 x
0 when x
9
4 x5 2
Intervals:
0 x 9
9 x f
Sign of f cc:
f cc ! 0
f cc 0
Conclusion:
Concave upward
Concave downward
Concave upward: 0, 9
Concave downward: 9, f
Points of inflection: 9, 4
25.
f x
fc x
x
sin , 0 d x d 4S
2
1
§ x·
cos¨ ¸
2
© 2¹
1
§ x·
sin ¨ ¸
4
© 2¹
f cc x
f cc x
0 when x
Intervals:
0 x 2S
2S x 4S
Sign of f cc:
f cc 0
f cc ! 0
Conclusion:
Concave downward
Concave upward
0, 2S , 4S .
Concave upward: 2S , 4S
Concave downward: 0, 2S
INSTRUCTOR USE ONLY
Point of inflection: 2S , 0
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© Cengage Learning. All Rights Reserved.
252
26.
Chapter 3
f x
fc x
f cc x
NOT FOR SALE
Applications
pplications of Differentiation
3x
, 0 x 2S
2
3x
3x
3 csc
cot
2
2
9 § 3 3x
3x
3x ·
cot 2 ¸ z 0 for any x in the domain of f .
csc
¨ csc
2©
2
2
2¹
2 csc
f cc is not continuous at x
2S
and x
3
4S
.
3
2S
3
2S
4S
x 3
3
4S
x 2S
3
Intervals:
0 x Sign of f cc x :
f cc ! 0
f cc 0
f cc ! 0
Conclusion:
Concave upward
Concave downward
Concave upward
§ 2S · § 4S
·
Concave upward: ¨ 0,
¸, ¨ , 2S ¸
© 3 ¹ © 3
¹
§ 2S 4S ·
Concave downward: ¨ ,
¸
© 3 3 ¹
No point of inflection
27.
f x
S·
§
sec¨ x ¸, 0 x 4S
2¹
©
fc x
S· §
S·
§
sec¨ x ¸ tan ¨ x ¸
2
2¹
©
¹
©
f cc x
S·
S·
S·
§
§
§
sec3 ¨ x ¸ sec¨ x ¸ tan 2 ¨ x ¸ z 0 for any x in the domain of f .
2
2
2¹
©
¹
©
¹
©
f cc is not continuous at x
S, x
2S , and x
3S .
Intervals:
0 x S
S x 2S
2S x 3S
3S x 4S
Sign of f cc:
f cc ! 0
f cc 0
f cc ! 0
f cc 0
Conclusion:
Concave upward
Concave downward
Concave upward
Concave upward
Concave upward: 0, S , 2S , 3S
Concave downward: S , 2S , 3S , 4S
No point of inflection
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.4
28.
f x
sin x cos x, 0 d x d 2S
fc x
cos x sin x
f cc x
sin x cos x
f cc x
0 when x
Concavity and the Second De
Derivative Test
253
3S 7S
,
.
4 4
3S
7S
x 4
4
7S
x 2S
4
f cc x 0
f cc x ! 0
f cc x 0
Concave downward
Concave upward
Concave downward
Intervals:
0 x Sign of f cc:
Conclusion:
3S
4
§ 3S 7S ·
Concave upward: ¨ ,
¸
© 4 4 ¹
§ 3S · § 7S
·
Concave downward: ¨ 0,
¸, ¨ , 2S ¸
© 4 ¹ © 4
¹
§ 3S · § 7S ·
Points of inflection: ¨ , 0 ¸, ¨ , 0 ¸
© 4 ¹ © 4 ¹
29.
f x
2 sin x sin 2 x, 0 d x d 2S
fc x
2 cos x 2 cos 2 x
f cc x
2 sin x 4 sin 2 x
f cc x
0 when x
2 sin x 1 4 cos x
0, 1.823, S , 4.460.
Intervals:
0 x 1.823
1.823 x S
S x 4.460
4.460 x 2S
Sign of f cc:
f cc 0
f cc ! 0
f cc 0
f cc ! 0
Conclusion:
Concave downward
Concave upward
Concave downward
Concave upward
Concave upward: 1.823, S , 4.460, 2S
Concave downward: 0, 1.823 , S , 4.460
Points of inflection: 1.823, 1.452 , S , 0 , 4.46, 1.452
30.
f x
x 2 cos x, >0, 2S @
fc x
1 2 sin x
f cc x
2 cos x
f cc x
0 when x
S 3S
2
,
2
.
S
S
2
2
x 3S
2
3S
x 2S
2
Intervals:
0 x Sign of f cc:
f cc 0
f cc ! 0
f cc 0
Conclusion:
Concave downward
Concave upward
Concave downward
§ S 3S ·
Concave upward: ¨ ,
¸
©2 2 ¹
§ S · § 3S
·
Concave downward: ¨ 0, ¸, ¨ , 2S ¸
© 2¹ © 2
¹
§ S S · § 3S 3S ·
Points of inflection: ¨ , ¸, ¨ ,
¸
©2 2¹ © 2 2 ¹
INSTRUCTOR USE ONLY
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254
31.
Chapter 3
NOT FOR SALE
Applications
pplications of Differentiation
f x
x 4 4 x3 2
6 2x
fc x
4 x 3 12 x 2
4x2 x 3
2
f cc x
12 x 2 24 x
12 x x 2
f x
6 x x2
fc x
f cc x
35.
Critical number: x
f cc 3
3
Critical numbers: x
2 0
However, f cc 0
f x
x 2 3x 8
fc x
2x 3
f cc x
2
33.
3, 25 is a relative minimum.
32
f cc 32
2 ! 0
Therefore,
32 , 41
4
is a relative minimum.
x3 3x 2 3
fc x
3x 2 6 x
f cc x
6x 6
f cc 0
4 x x 4 x 1
f cc x
12 x 2 24 x 16
4 3 x 2 6 x 4
0, x
x 7 x 15 x
fc x
3 x 2 14 x 15
f cc x
6 x 14
f cc 4
2
37.
fc x
2
3 x1 3
2
43
9x
Critical number: x
0
However, f cc 0 is undefined, so you must use the First
2 3x 7
Derivative Test. Because f c x 0 on f, 0 and
3, 53
f c x ! 0 on 0, f , 0, 3 is a relative minimum.
38.
f x
Therefore, 3, 9 is a relative maximum.
5
, 275
3
27
80 0
x2 3 3
f cc x
x 3 3x 5
4 0
Therefore,
16 ! 0
f x
2
Critical numbers: x
4 ! 0
20 0
Therefore, 4, 128 is a relative maximum.
f x
f cc 53
1, 0, 4
Therefore, 0, 0 is a relative minimum.
6 x 1
6 ! 0
f cc 3
4 x3 12 x 2 16 x
f cc 0
3x x 2
Therefore, 2, 1 is a relative minimum.
34.
fc x
Therefore 1, 3 is a relative maximum.
Therefore, 0, 3 is a relative maximum.
3
x 4 4 x3 8 x 2
f cc 1
6 0
f cc 2
f x
Critical numbers: x
f x
Critical numbers: x
0, so you must use the First
0, 3 ; so, 0, 2 is not an extremum. f cc 3 ! 0 so
36.
Critical number: x
3
Derivative Test. f c x 0 on the intervals f, 0 and
Therefore, 3, 9 is a relative maximum.
32.
0, x
fc x
is a relative minimum.
f cc x
x2 1
x
x 1
1
2
x2 1
32
Critical number: x
f cc 0
0
1 ! 0
Therefore, 0, 1 is a relative minimum.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.4
39.
4
x
f x
x fc x
4
1 2
x
8
x3
f cc x
(a)
x2 4
x2
3
2
fc x
0.2 x 5 x 6 x 3
f cc x
x 3 4 x 2 9.6 x 3.6
0.4 x 3 10 x 2 24 x 9
(b) f cc 0 0 Ÿ 0, 0 is a relative maximum.
r2
f cc 56 ! 0 Ÿ 1.2, –1.6796 is a relative minimum.
1 0
Points of inflection:
3, 0 , 0.4652, 0.7048 , 1.9348, 0.9049
Therefore, 2, 4 is a relative maximum.
f cc 2
255
0.2 x 2 x 3 , >1, 4@
43. f x
Critical numbers: x
f cc 2
Concavity and the Second De
Derivative Test
1! 0
(c)
y
Therefore, 2, 4 is a relative minimum.
f″
f′
2
40.
f x
fc x
x
x 1
1
x 1
1
x
−2 −1
f
2
f is increasing when f c ! 0 and decreasing when
f c 0. f is concave upward when f cc ! 0 and
concave downward when f cc 0.
There are no critical numbers and x 1 is not in the
domain. There are no relative extrema.
41.
f x
cos x x, 0 d x d 4S
fc x
sin x 1 d 0
Therefore, f is non-increasing and there are no relative
extrema.
42. f x
fc x
2 cos x 1 2 sin x
f cc x
§S ·
f cc¨ ¸
©6¹
x2
(a)
fc x
fc x
2 cos x 4 sin x cos x
0 when x
S S 5S 3S
6 x 2 , ª¬
f cc x
6 x2
0 when x
f cc x
r 2.
0, x
6 x 9 x 12
2
32
6 x2
2 sin x 4 cos 2 x
3 0
6 º¼
6,
3x 4 x 2
4
, , , .
6 2 6 2
§S 3·
Therefore, ¨ , ¸ is a relative maximum.
© 6 2¹
§S ·
f cc¨ ¸
©2¹
44. f x
2 sin x cos 2 x, 0 d x d 2S
2 cos x 2 sin 2 x
4
9
r
0 when x
33
2
.
(b) f cc 0 ! 0 Ÿ 0, 0 is a relative minimum.
f cc r2 0 Ÿ r2, 4 2 are relative maxima.
2 ! 0
Points of inflection: r1.2758, 3.4035
§S ·
Therefore, ¨ , 1¸ is a relative minimum.
©2 ¹
§ 5S ·
3 0
f cc¨ ¸
© 6 ¹
§ 5S 3 ·
Therefore, ¨ , ¸ is a relative maximum.
© 6 2¹
§ 3S ·
f cc¨ ¸
6 ! 0
© 2 ¹
§ 3S
·
Therefore, ¨ , 3¸ is a relative minimum.
© 2
¹
(c)
y
6
f
x
−3
3
f ''
f'
−6
The graph of f is increasing when f c ! 0 and
decreasing when f c 0. f is concave upward
when f cc ! 0 and concave downward when
f cc 0.
INSTRUCTOR USE ONLY
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Chapter 3
256
sin x 45. f x
(a)
NOT FOR SALE
Applications
pplications of Differentiation
1
1
sin 3x sin 5 x,
3
5
fc x
cos x cos 3x cos 5 x
fc x
0 when x
S
,x
S
,x
>0, S @
f cc x
6
2
sin x 3 sin 3 x 5 sin 5 x
f cc x
0 when x
S
6
x | 1.1731, x | 1.9685
,x
2 x sin x, >0, 2S @
46. f x
(a) f c x
5S
.
6
2 x cos x Critical numbers: x | 1.84, 4.82
f cc x
4 x 2 1 sin x
2 cos x
2x
2x 2x
4 x cos x 4 x 2 1 sin x
2x 2x
(b) Relative maximum: 1.84, 1.85
§S
·
Points of inflection: ¨ , 0.2667 ¸, 1.1731, 0.9638 ,
©6
¹
§ 5S
·
1.9685, 0.9637 , ¨ , 0.2667 ¸
6
©
¹
Relative minimum: 4.82, 3.09
Points of inflection: 0.75, 0.83 , 3.42, 0.72
(c)
y
Note: 0, 0 and S , 0 are not points of inflection
4
f′
because they are endpoints.
2
f
(c)
y
x
π
2
4
−2
π
4
π
2
f′
π
x
−4
f is increasing when f c ! 0 and decreasing when
f c 0. f is concave upward
when f cc ! 0 and concave downward when
f cc 0.
−4
−6
−8
f ''
−2
f
2
cos x
cos x
sin x
2x
2x
2x 2x
2 x sin x 5S
,
6
§S ·
§S
·
(b) f cc¨ ¸ 0 Ÿ ¨ , 1.53333¸ is a relative
2
2
© ¹
©
¹
maximum.
sin x
2x
f″
The graph of f is increasing when f c ! 0 and
decreasing when f c 0. f is concave upward
when f cc ! 0 and concave downward when
f cc 0.
47. (a)
y
4
3
2
1
x
1
2
3
4
f c 0 means f decreasing
f c increasing means concave upward
(b)
y
4
3
2
1
x
1
2
3
4
f c ! 0 means f increasing
f c increasing means concave upward
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.4
48. (a)
Concavity and the Second De
Derivative Test
y
51.
y
f
4
257
f'
3
f ''
3
2
x
−2
1
−1
3
−1
x
1
2
3
4
f c 0 means f decreasing
f c decreasing means concave downward
y
52.
f″
f′
(b)
f
4
y
4
x
−2
3
2
−2
2
−4
1
53.
x
1
2
3
4
y
f c ! 0 means f increasing
4
f c decreasing means concave downward
2
(2, 0) (4, 0)
49. Answers will vary. Sample answer:
x
2
f x
x .
f cc x
12 x 2
Let
f cc 0
4
6
4
0, but 0, 0 is not a point of inflection.
54.
y
y
2
6
1
5
(0, 0)
4
(2, 0)
x
−1
1
3
3
2
1
x
−3
−2
−1
1
2
3
55.
50. (a) The rate of change of sales is increasing.
y
3
S cc ! 0
2
1
(b) The rate of change of sales is decreasing.
(2, 0)
(4, 0)
x
S c ! 0, S cc 0
1
2
3
4
5
(c) The rate of change of sales is constant.
Sc
C , S cc
0
(d) Sales are steady.
S
C , S c 0, S cc
56.
y
0
3
(e) Sales are declining, but at a lower rate.
2
S c 0, S cc ! 0
(0, 0)
(f ) Sales have bottomed out and have started to rise.
S c ! 0, S cc ! 0 Answers will vary.
(2, 0)
x
−1
1
3
−1
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 3
258
Applications
pplications of Differentiation
y
57.
f
x
−4
8
12
f″
−8
f cc is linear.
f c is quadratic.
f is cubic.
f concave upward on f, 3 , downward on 3, f .
58. (a)
d
12
t
10
(b) Because the depth d is always increasing, there are no relative extrema. f c x ! 0
(c) The rate of change of d is decreasing until you reach the widest point of the jug, then the rate increases until you reach the
narrowest part of the jug's neck, then the rate decreases until you reach the top of the jug.
59. (a) n
n
1:
n
2:
f x
x 2
f x
x 2
fc x
1
fc x
f cc x
0
f cc x
No point of inflection
2
n
3:
3
f x
x 2
2 x 2
fc x
3x 2
2
f cc x
6 x 2
2
Point of inflection: 2, 0
No point of inflection
4
f x
x 2
fc x
4x 2
f cc x
12 x 2
3
2
No point of inflection
Relative minimum: 2, 0
Relative minimum: 2, 0
6
6
6
4:
6
f(x) = (x − 2)3
−9
9
−9
−9
9
9
f(x) = (x − 2)2
f(x) = x − 2
Point of
inflection
−6
−6
−6
−9
9
f(x) = (x − 2)4
−6
Conclusion: If n t 3 and n is odd, then 2, 0 is point of inflection. If n t 2 and n is even, then 2, 0 is a relative minimum.
(b) Let f x
n
x 2 , fc x
n x 2
n 1
, f cc x
nn 1 x 2
For n t 3 and odd, n 2 is also odd and the concavity changes at x
n2
.
2.
For n t 4 and even, n 2 is also even and the concavity does not change at x
So, x
2.
2 is point of inflection if and only if n t 3 is odd.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.4
60. (a)
f x
3
fc x
1 2 3
x
3
f cc x
92 x 5 3
Concavity and the Second De
Derivative Test
259
x
Point of inflection: 0, 0
(b) f cc x does not exist at x
0.
y
3
2
1
(0, 0)
−6
−4
x
−2
2
4
6
−2
−3
61. f x
ax3 bx 2 cx d
Relative maximum: 3, 3
Relative minimum: 5, 1
Point of inflection: 4, 2
fc x
3ax 2 2bx c, f cc x
f 3
3 ½°
¾ 98a 16b 2c
125a 25b 5c d
1°¿
27 a 6b c
0, f cc 4
24a 2b
0
f 5
fc 3
27 a 9b 3c d
49a 8b c
1
24a 2b
0
0
22a 2b
1
2a
1
27 a 6b c
22a 2b
a
f x
62. f x
6ax 2b
1, b
2
1
6, c
2 Ÿ 49a 8b c
1
2 Ÿ 28a 6b c
1
0, f cc 3
0
24
45
,d
2
1 x 3 6 x 2 45 x 24
2
2
ax 3 bx 2 cx d
Relative maximum: 2, 4
Relative minimum: 4, 2
Point of inflection: 3, 3
fc x
3ax 2 2bx c, f cc x
f 2
8a 4b 2c d
6ax 2b
f 4
4 °½
¾ 56a 12b 2c
64a 16b 4c d
2°¿
fc 2
12a 4b c
0, f c 4
48a 8b c
28a 6b c
1
18a 2b
0
12a 4b c
16a 2b
0
1
16a 2b
2a
1
1
a
f x
1, b
2
92 , c
12, d
18a 2b
6
1 x3 9 x 2 12 x 6
2
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 3
260
Applications
pplications of Differentiation
ax3 bx 2 cx d
63. f x
Maximum: 4, 1
Minimum: 0, 0
(a) f c x
f cc x
3ax 2 2bx c,
f 0
0 Ÿ d
f 4
f c 4
fc 0
1Ÿ
0 Ÿ
0 Ÿ
0
64a 16b 4c
48a 8b c
c
1
0
0
1 and b
32
Solving this system yields a
f x
6ax 2b
6a
3
.
16
1 x3 3 x 2
32
16
(b) The plane would be descending at the greatest rate at the point of inflection.
f cc x
6ax 2b
3
x 83
16
0 Ÿ x
2.
Two miles from touchdown.
64. (a) line OA : y
line CB : y
0.06 x
slope: 0.06
0.04 x 50
slope: 0.04
f x
ax3 bx 2 cx d
fc x
3ax 2bx c
y
150
2
1000, 60 :
60
0.06
1000, 90 :
90
0.04
2
1000 a 1000 b 1000c d
1000
2
3
2
(b) y
C
(0, 50)
2
−1000
O
x
1000
3a 2000b c
The solution to this system of four equations is a
8 3
(−1000, 60)
A
3a 2000b c
1000 a 1000 b 1000c d
1000
(1000, 90)
B
100
3
1.25 u 108 , b
0.000025, c
0.0275, and d
50.
1.25 u 10 x 0.000025 x 0.0275 x 50
2
100
−1100
1100
−10
(c)
0.1
−1100
1100
− 0.1
(d) The steepest part of the road is 6% at the point A.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section 3.4
C
0.5 x 2 15 x 5000
C
C
x
C
average cost per unit
dC
dx
0.5 65.
0.5 x 15 5000
x2
Concavity and the Second De
Derivative Test
261
5000
x
0 when x
100
By the First Derivative Test, C is minimized when x
100 units.
5.755 3 8.521 2
6.540
T T T 0.99987, 0 T 25
108
106
105
66. S
17.265 2 17.042
6.540
T T 108
106
105
34.53
17.042
S cc
T 0 when T | 49.4, which is not in the domain
108
106
S cc 0 for 0 T 25 Ÿ Concave downward.
(a) S c
(b) The maximum is approximately 4, 1 .
(c)
1.001
0
0.996
25
20, S | 0.998.
(d) When t
5000t 2
,0 d t d 3
8 t2
67. S
(a)
t
0.5
1
1.5
2
2.5
3
S
151.5
555.6
1097.6
1666.7
2193.0
2647.1
Increasing at greatest rate when 1.5 t 2
(b)
3000
0
3
0
Increasing at greatest rate when t | 1.5.
(c)
S
Sc t
S cc t
S cc t
5000t 2
8 t2
80,000t
8 t2
2
80,000 8 3t 2
8 t2
0 for t
r
3
8
. So, t
3
2 6
| 1.633 yrs.
3
INSTRUCTOR USE ONLY
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Chapter 3
262
NOT FOR SALE
Applications
pplications of Differentiation
100t 2
,t ! 0
65 t 2
68. S
(a)
100
0
35
0
13,000t
(b) S c t
65 t 2
2
13,000 65 3t 2
S cc t
65 t 2
0 Ÿ t
3
4.65
S is concave upwards on 0, 4.65 , concave downwards on 4.65, 30 .
(c) S c t ! 0 for t ! 0.
As t increases, the speed increases, but at a slower rate.
69.
f x
2 sin x cos x ,
§S ·
f¨ ¸
©4¹
2 2
fc x
2 cos x sin x ,
§S ·
f c¨ ¸
©4¹
0
f cc x
§S ·
2 sin x cos x , f cc¨ ¸
©4¹
P1 x
S·
§
2 2 0¨ x ¸
4¹
©
P1c x
0
P2 x
S· 1
S·
§
§
2 2 0¨ x ¸ 2 2 ¨ x ¸
4¹ 2
4¹
©
©
2 2
2 2
2
2 2 S·
§
2¨ x ¸
4¹
©
2
4
P1
S·
P2c x
§
2 2 ¨ x ¸
4¹
©
P2cc x
2 2
− 2
2
f
P2
−4
S 4. The values of the second derivatives of f and P2 are
S 4. The approximations worsen as you move away from x S 4.
The values of f , P1 , P2 , and their first derivatives are equal at x
equal at x
70.
f x
2 sin x cos x ,
f 0
2
fc x
2 cos x sin x ,
fc 0
2
f cc x
2 sin x cos x ,
f cc 0
2
P1 x
2 2 x 0
P1c x
2
21 x
4
P2 x
2 2 x 0
P2c x
2 2x
P2cc x
2
12 2
x 0
2
2 2x x
2
−6
6
P1
−4
The values of f , P1 , P2 , and their first derivatives are equal at x
equal at x
P2
f
0. The values of the second derivatives of f and P2 are
0. The approximations worsen as you move away from x
0.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.4
f x
71.
fc x
f cc x
1 x,
f 0
1
1
,
2 1 x
1
fc 0
1
2
f cc 0
1
4
41 x
32
,
x
§ 1·
1 ¨ ¸ x 0
1
2
© 2¹
1
2
1§ 1 ·
2
§ 1·
1 ¨ ¸ x 0 ¨ ¸ x 0
2
2
4
© ¹
© ¹
1
x
2
4
1
4
P1 x
P1c x
P2 x
P2c x
P2cc x
Concavity and the Second De
Derivative Test
5
x
x2
1 2
8
P1
f
−8
−3
f 2
2
,
fc 2
3x 6 x 1
,
3
4 x3 2 x 1
f cc 2
23
8 2
x 1
fc x
x x 1
2
2
2
f cc x
§ 3 2·
2 ¨¨ ¸ x 2
4 ¸¹
©
P1 x
P1c x
3
2 2
0.
3 2
4
23 2
16
3 2
5 2
x 4
2
3 2
4
§ 3 2·
1 § 23 2 ·
2
2 ¨¨ ¸¸ x 2 ¨¨
¸¸ x 2
4
2
16
©
¹
©
¹
P2 x
2 3 2
23 2
2
x 2 x 2
4
32
3 2
23 2
x 2
4
16
P2c x
P2cc x
23 2
16
The values of f , P1 , P2 and their first derivatives are equal at x
at x
0. The values of the second derivatives of f and P2 are
0. The approximations worsen as you move away from x
x
,
x 1
f x
72.
4
P2
The values of f , P1 , P2 , and their first derivatives are equal at x
equal at x
263
2. The approximations worsen as you move away from x
2. The values of the second derivatives of f and P2 are equal
2.
3
P1
P2
f
−1
5
−1
INSTRUCTOR USE ONLY
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Chapter 3
264
Applications
pplications of Differentiation
f x
§1·
x sin ¨ ¸
© x¹
fc x
ª 1
§ 1 ·º
§1·
x « 2 cos¨ ¸» sin ¨ ¸
© x ¹¼
© x¹
¬ x
73.
f cc x
x
1
§1·
§1·
cos¨ ¸ sin ¨ ¸
x
© x¹
© x¹
1ª 1
1
1
§ 1 ·º
§1·
§1·
« 2 sin ¨ ¸» 2 cos¨ ¸ 2 cos¨ ¸
x¬x
x
© x ¹¼
© x¹ x
© x¹
1
1
§1·
sin ¨ ¸
x3
© x¹
0
S
§1 ·
Point of inflection: ¨ , 0 ¸
©S ¹
When x ! 1 S , f cc 0, so the graph is concave downward.
1
−1
1
( π1 , 0(
−1
74.
2
f x
x x 6
x3 12 x 2 36 x
fc x
3 x 2 24 x 36
f cc x
6 x 24
3x 2 x 6
6x 4
0
0
Relative extrema: 2, 32 and 6, 0
Point of inflection 4, 16 is midway between the relative extrema of f.
75. True. Let y
ycc
ax3 bx 2 cx d , a z 0. Then
6ax 2b
0 when x
b 3a , and the
1 x has a discontinuity at x
0.
77. False. Concavity is determined by f cc. For example, let
f x
x and c
2. f c c
f c 2 ! 0, but f is not
concave upward at c
g c are increasing on a, b , and f cc ! 0 and g cc ! 0 .
So, f g cc ! 0 Ÿ f g is concave upward on
concavity changes at this point.
76. False. f x
79. f and g are concave upward on a, b implies that f c and
a, b by Theorem 3.7.
80. f, g are positive, increasing, and concave upward on
a, b Ÿ f x ! 0, f c x t 0 and f cc x ! 0, and
g x ! 0, g c x t 0 and g cc x ! 0 on a, b . For
2.
4
x 2 .
78. False. For example, let f x
x  a, b ,
fg c x
f c x g x f x gc x
fg cc x
f cc x g x 2 f c x g c x f x g cc x ! 0
So, fg is concave upward on a, b .
Section 3.5 Limits at Infinity
1. f x
2x2
x2 2
2. f x
No vertical asymptotes
Horizontal asymptote: y
Matches (f ).
2x
x2 2
No vertical asymptotes
2
Horizontal asymptotes: y
r2
Matches (c).
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.5
3. f x
x
x 2
5. f x
2
No vertical asymptotes
Horizontal asymptote: y
4 sin x
x2 1
Horizontal asymptote: y
0
f 1 !1
Matches (d).
Matches (b).
2
x
x4 1
2
6. f x
No vertical asymptotes
Horizontal asymptote: y
2 x 2 3x 5
x2 1
No vertical asymptotes
2
Horizontal asymptote: y
Matches (a).
7. f x
265
No vertical asymptotes
0
f 1 1
4. f x
Limits
Limi at Infinity
Lim
2
Matches (e).
4x 3
2x 1
x
100
101
102
103
104
105
106
f(x)
7
2.26
2.025
2.0025
2.0003
2
2
lim f x
2
xof
10
− 10
10
− 10
8. f x
2x2
x 1
x
100
101
102
103
104
105
106
f(x)
1
18.18
198.02
1998.02
19,998
199,998
1,999,998
lim f x
xof
f
Limit does not exist
20
0
10
−2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 3
266
Applications
pplications of Differentiation
6 x
9. f x
4x2 5
x
100
101
102
103
104
105
106
f(x)
–2
–2.98
–2.9998
–3
–3
–3
–3
3
lim f x
xof
10
− 10
10
− 10
10
10. f x
2x2 1
x
100
101
102
103
104
105
106
f(x)
10.0
0.7089
0.0707
0.0071
0.0007
0.00007
0.000007
lim f x
0
xof
10
−9
9
−2
11. f x
5
1
x2 1
x
100
101
102
103
104
105
106
f(x)
4.5
4.99
4.9999
4.999999
5
5
5
lim f x
5
xof
6
−1
8
0
12. f x
4
3
x2 2
x
100
101
102
103
104
105
106
f(x)
5
4.03
4.0003
4.0
4.0
4
4
lim f x
xof
4
10
INSTRUCTOR USE ONLY
0
15
1
5
0
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.5
5 x3 3 x 2 10 x
10
5x 3 x2
x
Limit does not exist
f x
13. (a) h x
x
2
f
lim h x
xof
5 x3 3x 2 10 x
x3
x
5
5 x3 3x 2 10 x
x4
5
3
10
3
x x2
x
3
lim h x
f x
x4
lim h x
0
x of
4 x 2 2 x 5
5
14. (a) h x
4 x 2 x
x
x
lim h x
f
Limit does not exist
f x
xof
4 x 2 2 x 5
x2
f x
(b) h x
x
2
xof
4 x 2 2 x 5
x3
f x
(c) h x
x
3
lim h x
x of
5
4
5 x3 2
xof 4
x 1
f
2
5
2
x
x
4
2
5
2 3
x
x
x
3·
§
19. lim ¨ 4 ¸
x o f©
x¹
4 0
x·
§5
20. lim ¨ ¸
x o f© x
3¹
f
21. lim
2x 1
lim
Limit does not exist
2 1x
xof 3 4x2 5
x o f x 2 3
lim
23. lim
x o f x2
x
1
0
1x
0
1
2
2
3
4
3 x2
1x
xof 1 5 x3 1
x of 10 x 3 3 x 2 7
(b) lim
x 2
x2 1
1
2
x2 2
(c) lim
xof x 1
f
3 2x
x o f 3x3 1
0
16. (a) lim
Limit does not exist
0
3 2 x2
x o f 3x 1
f
x3 4
x o f x 2 1
2
3
1
2
5x
3x
f
lim
x o f
x 4 x2
f
1 1 x2
Limit does not exist.
Limit does not exist
27. lim
x o f
0
5 2 x3 2
x o f 3x3 2 4
5 2 x3 2
x o f 3x 4
f
(b) lim
lim
50
10 0
x o –f 1 26. lim
5 2 x3 2
x o f 3x 2 4
17. (a) lim
5 1 x3
x of 10 3 x 7 x 3
lim
Limit does not exist.
(c) lim
5x2
x o f x 3
25. lim
3 2x
x o f 3x 1
(b) lim
(c) lim
4 5 x2
x o f 1 lim
2 0
3 0
2 x
24. lim
x2 2
x3 1
xof
4
x o f 3x 2
22. lim
Limit does not exist
0
15. (a) lim
xof
4 4
lim h x
5 x3 2
x o f 4 x3 2 1
(c) lim
5
xof
(c) h x
3 10
2
x
x
267
0
x o f 4x2
(b) lim
f x
(b) h x
5 x3 2
1
18. (a) lim
Limits
Limi at Infinity
Lim
2
3
Limit does not exist
x
x x
1
lim
x o f §
2
x x·
¨
¸
¨ x2 ¸
©
¹
1
lim
x o f
1 1x
2
1, for x 0 we have x
x2
INSTRUCTOR USE ONLY
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Chapter 3
268
NOT FOR SALE
Applications
pplications of Differentiation
x
28. lim
x o f
x 1
1
lim
x o f §
2
x 1·
¨
¸
¨ x2 ¸
©
¹
1
lim
x o f
1 1 x2
x 1
33. lim
2
xof
x2 1
xof
lim
xof
x
2x
34. lim
2
23
1 1 x2
13
f
x o f
x6 1
13
x o f
x6 1
x x
1
x
2
x x·
¸
x 2 ¸¹
13
1 x6
2 x
lim
x o f
2
1 x2
2x
lim
2x 1
x o f
x1 3 1 x
Limit does not exist.
1, for x 0 we have x
29. lim
§
·
x 1 ¨ 1 x2 3 ¸
13¨
13¸
x 2 1 ¨© 1 x 2 ¸¹
lim
13
11x
6
13
13
0
2
lim
x o f §
¨
¨
©
1
35. lim
0
x o f 2 x sin x
§1·
36. lim cos¨ ¸
xof
© x¹
§1·
2 ¨ ¸
© x¹
lim
x o f
1
1
x
cos 0
1
37. Because 1 x d sin 2 x x d 1 x for all x z 0, you
2, for x 0, x
have by the Squeeze Theorem,
x2
1
sin 2 x
1
d lim
d lim
x
o
f
x
o
f
x
x
x
sin 2 x
0 d lim
d 0.
xof
x
lim 30. lim
xof
5x2 2
xof
x2 3
lim
xof
lim
xof
5x2 2
xof
5x2 2 x
38. lim
1 3 x2
xof
f
Limit does not exist.
31. lim
xof
x2 1
2x 1
x2 1
x2
2 1x
lim
1 1 x2
2 1x
xof
32. lim
x o f
x4 1
x3 1
lim
x o f
lim
x o f
for x 0, we have x cos x
x
0.
cos x ·
§
lim ¨1 ¸
x ¹
10 1
x o f©
Note:
lim
xof
sin 2 x
x
Therefore, lim
x 1 3 x2
cos x
0 by the Squeeze Theorem because
x
1
cos x
1
d
d .
x
x
x
lim
xof
1
2
6 ·
§
x4 1 ¨1 x ¸
x3 1 ¨¨ 1 x3 ¸¸
©
¹
1 x2 1 x
1 1 x3
x6
x3
6
0,
x
39. f x
lim
x 1
x
1
xof x 1
lim
x
1
x o f x 1
Therefore, y
asymptotes.
1 and y
1 are both horizontal
4
y=1
y = −1
−6
6
INSTRUCTOR USE ONLY
−4
−4
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.5
3x 2
40. f x
y
3 is a horizontal asymptote (to the right).
y
3 is a horizontal asymptote (to the left).
8
y
3
is a horizontal asymptote (to the right).
2
y
y=3
− 10
269
9 x2 2
2x 1
42. f x
x 2
Limits
Limi at Infinity
Lim
3
is a horizontal asymptote (to the left).
2
6
10
y=3
y = −3
2
y = −3
2
−9
−6
9
3x
41. f x
−6
x2 2
lim f x
3
xof
43. lim x sin
xof
1
x
(Let x
1 t. )
3
lim f x
x o f
Therefore, y
asymptotes.
3 and y
lim
sin t
t
lim
tan t
t
t o 0
3 are both horizontal
1
x
44. lim x tan
xof
6
x o 0
1
ª sin t
1 º
lim «
˜
»
t
cos t ¼
x o 0 ¬
1 1
y=3
−9
(Let x
9
1
1 t. )
y = −3
−6
45. lim x x2 3
46. lim x x2 x
x o f
x of
47. lim 3x x o f
9 x2 x
ª
lim « x «¬
x2 3 ˜
x o f
ª
lim « x x of
¬«
ª
lim « 3 x x o f
«¬
3x lim
x o f
3
x o f 3 48. lim 4 x xof
16 x 2 x
x2 x º
»
x 2 x ¼»
x 9x2 x ˜
lim
x o f
lim
x of
3
x x2 3
x
x
x x
2
0
lim
x of 1 1
1 1x
1
2
9x2 x º
»
9 x 2 x »¼
3x 3x x
lim
x o f
lim
x x x2 x ˜
x2 3 º
»
x 2 3 »¼
x 9 x2 x
1
9x2 x
for x 0 you have x
x2
1
16 x 2 x
4x 16 x 2 x
lim
xof
lim
xof
x2
1
6
9 1x
4x 16 x 2 16 x 2 x
4x 16 x 2 x
x
4x 16 x 2 x
1
16 1 x
lim
xof 4 1
4 4
1
8
INSTRUCTOR USE ONLY
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Chapter 3
270
49.
NOT FOR SALE
Applications
pplications of Differentiation
x
100
101
102
103
104
105
106
f(x)
1
0.513
0.501
0.500
0.500
0.500
0.500
lim x x x 1
lim
x of
x x of
x2 x x ˜
1
x x2 x
x2 x
lim
x of
x
x
x2 x
lim
x of 1 1
1 1x
1
2
2
−1
8
−2
50.
x
100
101
102
103
104
105
106
f(x)
1.0
5.1
50.1
500.1
5000.1
50,000.1
500,000.1
x2 x
x2 x
lim
xof
˜
1
x2 x
x2 x
x2 x
x2 x
xof
lim
x3
x2 x
x2 x
f
Limit does not exist.
30
0
50
0
51.
x
100
101
102
103
104
105
106
f(x)
0.479
0.500
0.500
0.500
0.500
0.500
0.500
Let x
1 t.
§ 1 ·
lim x sin ¨ ¸
© 2x ¹
xof
lim
sin t 2
t o 0
t
lim
1 sin t 2
t 2
t o 0 2
1
2
1
−2
2
−1
52.
x
100
101
102
103
104
105
106
f(x)
2.000
0.348
0.101
0.032
0.010
0.003
0.001
lim
x 1
x
xof x
0
3
0
25
−1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.5
53. lim f x
4 means that f x approaches 4 as x
xof
Limits
Limi at Infinity
Lim
271
x
1 x
59. y
becomes large.
Intercept: 0, 0
54. lim f x
2 means that f x approaches 2 as x
x o f
becomes very large (in absolute value) and negative.
55. x
2 is a critical number.
Symmetry: none
1
Horizontal asymptote: y
Vertical asymptote: x
f c x 0 for x 2.
1
y
4
f c x ! 0 for x ! 2.
3
2
lim f x
lim f x
x o f
6
xof
1
x
6
For example, let f x
0.1 x 2
1
−1
2
1
2
3
4
5
−2
6.
−3
−4
y
8
x 4
x 3
60. y
Intercepts: 0, 4 3 , 4, 0
4
Symmetry: none
x
−2
2
4
Horizontal asymptote: y
6
Vertical asymptote: x
6 x 2
56. Yes. For example, let f x
x 2
2
1
.
1
3
y
5
4
y
3
2
8
−1
4
x
1
3
2
4
5
6
7
−2
2
−3
x
−4
−2
2
4
6
−2
61. y
57. (a) The function is even: lim f x
5
(b) The function is odd: lim f x
5
x o f
x o f
58. (a)
y
4
f
3
x 1
x2 4
Intercepts: 0, 1 4 , 1, 0
Symmetry: none
Horizontal asymptote: y
0
Vertical asymptotes: x
r2
y
2
f′
4
x
−4
1
2
3
3
4
2
1
−3
−4
(b) lim f x
−4
3
xof
(c) Because lim f x
xof
lim f c x
xof
0
−1
x
2
3
4
−2
−3
−4
3, the graph approaches that
of a horizontal line, lim f c x
xof
0.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
272
Chapter 3
62. y
2x
9 x2
Applications
pplications of Differentiation
65. xy 2
Domain: x ! 0
Intercept: 0, 0
Intercepts: none
Symmetry: origin
Symmetry: x-axis
Horizontal asymptote: y
0
Vertical asymptotes: x
r3
3
x
r
y
Horizontal asymptote: y
y
6
5
4
3
2
1
Vertical asymptote: x
0
0
y
4
x
−5−4
9
−1
−2
−3
−4
−5
−6
1 2
3
6
2
1
x
−1
1
2
3
4
5
6
7
−2
−3
2
x
x 16
63. y
−4
2
Intercept: 0, 0
Symmetry: y-axis
2
Horizontal asymptote: y
Intercepts: none
1
1
Symmetry: y-axis
32 x
yc
x 2 16
9
x2
9 Ÿ y
66. x 2 y
y
−8 −6 −4 −2
2
Relative minimum: 0, 0
x
2
4
6
8
Horizontal asymptote: y
−1
Vertical asymptote: x
−2
y
0
0
2x2
x2 4
64. y
Intercept: 0, 0
2
Symmetry: y-axis
2
Vertical asymptotes: x
r2
4x
x 4
2
ycc
4
6
8
3x
x 1
0, 0
Symmetry: none
16 x 2 4
ycc
67. y
Intercept:
2
x 4
2
2
−2
−4
Horizontal asymptote: y
yc
x
−8 −6 −4
Horizontal asymptote: y
3
3
Vertical asymptote: x
0 0
1
y
Relative maximum: 0, 0
y
8
7
6
5
4
3
2
6
−4 − 3 − 2 − 1
4
x
1 2 3 4 5 6
−2
2
x
−4
−2
2
4
6
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.5
68. y
3x
1 x2
Intercept: 0, 0
Intercept:
Symmetry: origin
y
Horizontal asymptote: y
0
Vertical asymptotes: x
r1
y
3
0
2
2
Ÿ
x
x
3 Ÿ x
2 § 2 ·
; ¨ , 0¸
3 © 3 ¹
Symmetry: none
Horizontal asymptote: y
Vertical asymptote: x
8
273
2
x
3
71. y
Limits
Limi at Infinity
Lim
6
3
0
y
4
8
7
6
5
4
3
2
2
x
−4 −3 −2 −1
1
x
69. y
−4 −3 −2 −1
2x2 3
x2
3
2 2
x
§
Intercepts: ¨¨ r
©
3 ·
, 0¸
2 ¸¹
1 2 3 4 5
4 x2
x2
4
1
x2
72. y
Intercept: none
Symmetry: y-axis
Horizontal asymptote: y
Vertical asymptote: x
2
0
Symmetry: none
Horizontal asymptote: y
Vertical asymptote: x
1
0
y
y
4
3
2
4
1
x
− 4 − 3 −2
2
3
4
2
−4
x
−2
2
4
−2
70. y
1
x 1
x
1
x
73. y
Intercept: 1, 0
x3
x 4
2
Domain: f, 2 , 2, f
Symmetry: none
Horizontal asymptote: y
Vertical asymptote: x
1
0
Intercepts: none
Symmetry: origin
Horizontal asymptote: none
y
r 2 (discontinuities)
Vertical asymptotes: x
6
5
y
−4 −3 −2 −1
−2
−3
−4
−5
−6
x
1
2
3
4
20
16
12
8
4
x
− 5 − 4 − 3 −2 − 1
1 2 3 4 5
−8
− 12
− 16
− 20
INSTRUCTOR USE ONLY
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Chapter 3
274
Applications
pplications of Differentiation
x
74. y
x 4
5
x2
9 75. f x
2
Domain: f, 2 , 2, f
Domain: all x z 0
Intercepts: none
fc x
10
Ÿ No relative extrema
x3
f cc x
Symmetry: origin
r1 because
Horizontal asymptotes: y
x
lim
1, lim
x2 4
xof
x
x o f
x2 4
1.
Vertical asymptote: x
r 2 (discontinuities)
Vertical asymptotes: x
30
Ÿ No points of inflection
x4
Horizontal asymptote: y
y=9
x=0
5
4
3
2
−6
1
6
−2
x
−1
9
12
y
−5 −4 −3
0
3 4 5
−2
−3
−4
−5
76.
1
x2 x 2
f x
1
x 1 x 2
2x 1
fc x
x x 2
2
x2 x 2
2
2
f cc x
0 when x
1
.
2
2 2 x 1 2 x 2 x 2 2 x 1
6 x2 x 1
4
x2 x 2
x2 x 2
3
§1·
§1 4·
Because f cc¨ ¸ 0, ¨ , ¸ is a relative maximum.
© 2¹
©2 9¹
Because f cc x z 0, and it is undefined in the domain of f, there are no points of inflection.
Vertical asymptotes: x
1, x
Horizontal asymptote: y
0
x = −1
2
2
x=2
( 12 , − 94(
−3
3
y=0
−2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.5
77.
x 2
x 4x 3
f x
x2 4x 3 x 2 2x 4
x2 4x 3
2
x2 4x 3
f cc x
275
x 2
x 1 x 3
2
fc x
Limits
Limi at Infinity
Lim
x2 4x 5
2
x2 4 x 3
2
z 0
2 x 4 x 2 4 x 5 2 x 2 4 x 3 2 x 4
x2 4x 3
2 x3 6 x 2 15 x 14
x2 4 x 3
4
2 x 2 x2 4x 7
3
x2 4 x 3
3
0 when x
2.
Because f cc x ! 0 on 1, 2 and f cc x 0 on 2, 3 , then 2, 0 is a point of inflection.
Vertical asymptotes: x
1, x
Horizontal asymptote: y
0
3
2
x=3
−1
5
y=0
x=1
−2
78.
f x
fc x
f cc x
x 1
x2 x 1
x x 2
x2 x 1
0 when x
2
2 x3 3x 2 1
x2 x 1
4x2 1
3
fc x
4 x2 1
when x | 0.5321, 0.6527, 2.8794.
f cc 0 0
Ÿ No relative extrema
32
36 x
f cc x
0
3
0, 2.
3x
f x
79.
4 x2 1
0 when x
52
Point of inflection: 0, 0
Horizontal asymptotes: y
Therefore, 0, 1 is a relative maximum.
0.
r
3
2
No vertical asymptotes
f cc 2 ! 0
2
Therefore,
y= 3
1·
§
¨ 2, ¸
3¹
©
−3
2
3
y= −3
2
−2
is a relative minimum.
Points of inflection:
0.5321, 0.8440 , 0.6527, 0.4491 and
2.8794, 0.2931
Horizontal asymptote: y
(− 0.6527, 0.4491)
2
0
(0.5321, 0.8440)
−3
3
(−2, − 13(
(0, 1)
y=0
−2
(− 2.8794, − 0.2931)
INSTRUCTOR USE ONLY
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Chapter 3
276
2x
g x
80.
NOT FOR SALE
Applications
pplications of Differentiation
3x 1
2
gc x
3x 1
2
x3 3x 2 2
,g x
x x 3
83. f x
2
f=g
18 x
g cc x
3x 2 1
−4
52
8
−2
No relative extrema. Point of inflection: 0, 0 .
Horizontal asymptotes: y
x3 3x 2 2
x x 3
(b) f x
2
r
3
x2 x 3
No vertical asymptotes
4
2
x x 3
8
(a)
32
x y= 2
3
−6
2
x x 3
2
x x 3
g x
x x 3
x 6
(c)
y=− 2
70
3 −4
81. g x
gc x
− 80
§ x ·
sin ¨
¸, 3 x f
© x 2¹
− 70
§ x ·
2cos¨
¸
© x 2¹
2
x 2
Horizontal asymptote: y
Relative maximum:
x
S
Ÿ x
x 2
2
80
The graph appears as the slant asymptote y
84. f x
sin 1
(a)
2S
| 5.5039
S 2
x3 2 x 2 2
,g x
2x2
f=g
x.
1
1
x 1 2
2
x
4
−6
6
No vertical asymptotes
1.2
−4
( π 2−π 2 , 1(
(b) f x
y = sin(1)
3
12
f x
fc x
2 sin 2 x
; Hole at 0, 4
x
4 x cos 2 x 2 sin 2 x
x2
(c)
70
−80
There are an infinite number of relative extrema. In the
interval 2S , 2S , you obtain the following.
80
−70
Relative minima: r 2.25, 0.869 , r 5.45, 0.365
The graph appears as the slant asymptote
1
x 1.
y
2
Relative maxima: r 3.87, 0.513
Horizontal asymptote: y
x3 2 x 2 2
2x2
ª x3
2 x2
2 º
« 2 »
2 x2
2x2 ¼
¬2x
1
1
x 1 2
g x
x
2
0
82.
0
6
No vertical asymptotes
85.
y=0
(−3.87, 0.513)
− 2
(3.87, 0.513)
2
−2
(−5.45, −0.365)
((−2.25,
− 2.25,
2.
−0.869)
− 0.869)
ª
º
1
»
lim 100 «1 c
v1 v 2 o f
«
»
v
v
1
2
¬
¼
100>1 0@
100%
INSTRUCTOR
CT
USE ONLY
(5.45, −0.365)
(2.25, −0.869)
− 0.869)
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NOT FOR SALE
Section 3.5
0.5 x 500
86. C
2
(a) lim f x
87. lim N t
f
lim E t
c
t of
t of
x12
H
2
(b) f x1 H
0.5
t o 0
2 x12 H x12 2H
2 x12 4
x12H
4 2H
4 2H
(c) Let M
x1 by symmetry
! 0. For x ! M :
H
4 2H
x !
This is the temperature of the room.
(c) No. y
H
x2
72q
t of
4 2H
x1
This is the temperature of the kiln.
(b) lim T
2
x12
1700q
88. (a) lim T
L
2
xof
500
C
0.5 x
500 ·
§
lim ¨ 0.5 ¸
x o f©
x ¹
277
2x2
x 2
91. f x
C
x
C
Limits
Limi at Infinity
Lim
H
x H ! 4 2H
2
72 is the horizontal asymptote.
2 x 2 x 2H 2H ! 2 x 2 4
100t 2
,t ! 0
65 t 2
89. S
(a)
2 x2
H ! 2
x2 2
2x2
2 ! H
x 2
120
2
H
f x L ! H
5
30
0
100
1
(b) Yes. lim S
t of
(b)
T1
x2 2
lim f x
6
lim f x
6
xof
x o f
(b) f x1 H
130
− 10
(c)
L
K
6 x1
x12 2
90
T2
H
6 H
36 x12
x12 2 6 H
1451 86t
58 t
(d) T1 0 | 26.6q
26H
2
x12 ª¬36 36 12H H 2 º¼
26H
2
26H
2
T2
x12
x1
T2 0 | 25.0q
(e) lim T2
t of
86
1
x2
86
(f ) No. The limiting temperature is 86q.
T1 has no horizontal asymptote.
x12 2
36 x12 6 H x12
120
− 10
6
6 x1
2
− 10
.
6x
92. f x
(a)
90
− 10
H
100
0.003t 2 0.68t 26.6
90. (a) T1 t
4 2H
(d) Similarly, N
(c) M
x1
2
12H H 2
2
12H H 2
x1 by symmetry
6H
6H
2
12H H 2
H 6
2
122H H 2
INSTRUCTOR USE ONLY
(d)
d) N
x2
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NOT FOR SALE
Chapter 3
278
Applications
pplications of Differentiation
3x
93. lim
3
x 3
xof
2
95. lim
3 x1
f x1 H
x12 3
9 x12
3
3 x1
3H
9 x12
3H
2
x12
H H
2
3H
x12
(a) When H
0.5:
(b) When H
x o f
3
3 x1
3
6H H 2
5 33
11
H
.
1
H 3
9 x12
H 3
9 x12 H 3 x12
2
3H 3
2
x12 9 H 2 6H 9
3H 3
2
3H 3
2
H 3
x12 3
2
x12 3
H
3
6H H 2
3
6H H 2
3
6 0.5 0.5
2
2
0
x
f x L
2
H whenever
x
x ! M.
x
1
4
!
Ÿ x ! 2
2
H
H
2
H Ÿ
x
4H .
2
Let M
4 H , you have
2
2
H Ÿ f x L H.
x
97. lim
1
0. Let H ! 0. You need N 0 such that
x o f x 3
1
0
x3
f x L
1
H whenever x N .
x3
5 33
11
1
3
For x N
H
.
1
3
H
,
1
! 3 H
x
1
3 H
x
1
3 H
x
Ÿ f x L H.
0.1:
0.1 3
1
H Ÿ f x L H.
x2
2
0. Let H ! 0 be given. You need
x
M ! 0 such that
Let N
0.5:
0.5 3
H
Ÿ
1
1
1
H Ÿ x3 !
Ÿ x 13
x3
H
H
6H H 2
H 3
1
Ÿ x2 !
x ! 2H Ÿ
3 x1
(b) When H
1
H
xof
3
N
Ÿ x !
96. lim
H
x12 3
(a) When H
N
x !
29 177
59
2
1
H
1
H whenever x ! M .
x2
For x ! M , you have
3
x1
N
2
6 0.1 0.1
1
x2 !
Let M
3
6H H 2
x12
Let x1
33H
2
1
0
x2
For x ! M
x2 3
f x1 H
33H
2
0.1:
3x
94. lim
33H
6 0.5 0.5
3 0.1
M
x12 3
2
3
3 0.5
M
2
3H
3H
f x L
x12 3
6H H 2
x1
x1
0. Let H ! 0 be given. You need
M ! 0 such that
H
x12 9 9 6
Let M
1
x o f x2
3
6 0.1 0.1
2
29 177
59
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.5
1
98. lim
Limits
Limi at Infinity
Lim
279
0. Let H ! 0 be given.
x o f x 2
1
0
x 2
You need N 0 such that f x L
1
H whenever x N .
x 2
1
1
1
H Ÿ x 2 Ÿ x 2
x 2
H
H
2 Let N
x 2 1
H
. For x N
2
1
H
,
1
H
1
H
x 2
Ÿ f x L H.
99. line: mx y 4
100. line: y 2
0
y
m x 0 Ÿ mx y 2
0
y
5
4
(4, 2)
y = mx + 4
2
3
d
2
x
−4
(3, 1)
1
(0, −2)
x
−2 −1
−1
(a) d
1
2
3
4
4
−4
Ax1 By1 C
m 3 11 4
A2 B 2
m2 1
(a) d
3m 3
m 4 12 2
A2 B 2
m2 1
4m 4
m2 1
(b)
Ax1 By1 C
m2 1
6
(b)
− 12
12
−5
−2
(c) lim d m
mof
10
3
lim d m
5
−3
m o f
(c) lim d m
The line approaches the vertical line x
distance from 3, 1 approaches 3.
0. So, the
mof
4; lim d m
m o f
4
The line approaches the vertical line x
distance from 4, 2 approaches 4.
0. So, the
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
280
NOT FOR SALE
Chapter 3
Applications
pplications of Differentiation
p x
an x n " a1 x a0
" b1 x b0
101. lim
lim
x o f b xm
m
xof q x
Divide p x and q x by x m .
an
a
a
" m11 m0
mn
0 " 0 0
0
x
x
x
lim
b1
b0
x of
0
0
b
b
"
m
m
bm " m 1 m
x
x
a
a
an " m11 m0
an " 0 0
an
x
x
lim
.
b1
b0
x of
b
b
0
0
"
m
m
bm " m 1 m
x
x
a
a
an x n m " m11 m0
rf " 0
x
x
rf.
lim
b1
b0
x of
bm " 0
bm " m 1 m
x
x
p x
Case 1: If n m: lim
x of q x
Case 2: If m
n: lim
p x
x of q x
Case 3: If n ! m: lim
p x
x of q x
102. lim x3
xof
f. Let M ! 0 be given. You need N ! 0 such that f x
x3 ! M Ÿ x ! M 1 3. Let N
103. False. Let f x
pcc 0
x3 ! M whenever x ! N .
M 1 3 , x ! M 1 3 Ÿ x3 ! M Ÿ f x ! M .
. (See Exercise 54.)
x 1, then y1 0
104. False. Let y1
y1cc 0
2x
x2 2
M 1 3 . For x ! N
0.
1. So y1c
1 2
1
ax 2 bx 1, then p 0
. Let p
4
1
1
Ÿ a
. Therefore,
4
8
x 1 and y1c 0
1. So, pc
f x
2
°­ 1 8 x 1 2 x 1, x 0
and f 0
®
x t 0
°̄ x 1,
fc x
­° 1 2 1 4 x, x 0
and f c 0
®
°̄1 2 x 1 , x ! 0
1
, and
2
f cc x
­
1 4 , x 0
°
and f cc 0
®
32
, x ! 0
°̄1 4 x 1
1
.
4
1 2. Finally, y1cc
2ax b and pc 0
1
Ÿ b
2
1
4 x 1
32
and
1
. Finally, pcc
2
2a and
1,
f cc x 0 for all real x, but f x increases without bound.
Section 3.6 A Summary of Curve Sketching
1. f has constant negative slope. Matches (d)
3. The slope is periodic, and zero at x
2. The slope of f approaches f as x o 0 , and approaches
4. The slope is positive up to approximately
x 1.5. Matches (b)
f as x o 0 . Matches (c)
0. Matches (a)
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section 3.6
5.
1
3
x 2
1
Ÿ undefined when x
2
x 2
y
yc
2
ycc
x 2
3
2
7·
§7 · §
Intercepts: ¨ , 0 ¸, ¨ 0, ¸
2¹
©3 ¹ ©
y
2
x=2
( 73 , 0 (
3
Horizontal asymptote: y
281
2
Ÿ undefined when x
Vertical asymptote: x
A Summary of Cu
Curve
Cur Sketching
x
4
yc
ycc
Conclusion
f x 2
Decreasing, concave down
2 x f
Decreasing, concave up
y
−2
−4
( 0, − 72 ( y = −3
No relative extrema, no points of inflection
x
x 1
y
6.
2
1 x x 1
1 x2
yc
x 1
2
ycc
2
2 x 3 x2
x2 1
Horizontal asymptote: y
y
f x 3
x
3 x 1
x
1
3
4
3
1
2
1 x 0
x
0
0
0 x 1
x
1
2
1
1 x x
3
3
3 x f
3
4
r1.
0, r
3.
yc
ycc
Conclusion
–
–
Decreasing, concave down
–
0
Point of inflection
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
+
0
Point of inflection
+
–
Increasing, concave down
0
–
Relative maximum
–
–
Decreasing, concave down
–
0
Point of inflection
0 when x
3
0 when x
2
x2 1
0
y
1
(1, 12 )
(−1, − 12 )
x
1
(0, 0)
–
+
Decreasing, concave up
( 3, 43 )
2
y=0
(− 3, − 43 )
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
7.
NOT FOR SALE
Chapter 3
282
Applications
pplications of Differentiation
x2
x 3
6x
y
2
yc
2
x2 3
18 1 x 2
ycc
x2 3
3
0 when x
0.
0 when x
r1.
Horizontal asymptote: y
y
f x 1
1
4
1
x
1 x 0
1
yc
ycc
Conclusion
–
–
Decreasing, concave down
–
0
Point of inflection
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
+
0
Point of inflection
+
–
Increasing, concave down
y
x
0
0
0 x 1
y=1
1
x
1
4
1
1 x f
8.
x2 1
x2 4
10 x
y
yc
0 when x
2
x2 4
10 3x 2 4
ycc
x2 4
3
0 when x
0 and undefined when x
(− 1, 14 (
−4
(1, 14 (
(0, 0))
2
x
4
r 2.
0.
Intercept: 0, 1 4
Symmetric about y-axis
Vertical asymptotes: x
r2
Horizontal asymptote: y
1
y
yc
ycc
Conclusion
f x 2
Increasing, concave up
2 x 0
Increasing, concave down
x
14
0
Relative maximum
0 x 2
Decreasing, concave down
2 x f
Decreasing, concave up
y
8
6
(0, − 14 ) 4
2
−8 −6 −4 −2
x = −2
x
2
4
y=1
INSTRUCTOR
ST
USE ONLY
x=2
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.6
9.
A Summary of Cu
Curve
Cur Sketching
283
3x
x 1
y
2
3 x 2 1
yc
x2 1
2
undefined when x
6x x 3
2
ycc
x2 1
3
r1
yc
ycc
Conclusion
f x 1
Decreasing, concave down
1 x 0
Decreasing, concave up
3
0
Point of inflection
0 x 1
Decreasing, concave down
1 x f
Decreasing, concave up
y
x
Intercept: 0, 0
Symmetry with respect to origin
Vertical asymptotes: x
r1
Horizontal asymptote: y
0
0
0
x = −1 y x = 1
y=0
1
(0, 0)
x
−3 −2 −1
10.
1
2
3
4
x 3
3
1
x
x
3
undefined when x
x2
6
3 z 0
x
f x
fc x
f cc x
Vertical asymptote: x
0
0
Intercept: 3, 0
Horizontal asymptote: y
1
yc
ycc
Conclusion
f x 0
Increasing, concave up
0 x f
Increasing, concave down
y
y
4
3
y=1
2
1
−4 −3 −2 −1
x=0
−2
x
1
3
4
(3, 0)
−3
−4
INSTRUCTOR USE ONLY
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284
11.
Chapter 3
f x
x 32
x2
fc x
1
64
x3
f cc x
192
x4
Applications
pplications of Differentiation
x 4 x 2 4 x 16
0 when x
x3
4 and undefined when x
0.
Intercept: 2 3 4, 0
Vertical asymptote: x
Slant asymptote: y
0
x
y
yc
ycc
Conclusion
f x 0
Increasing, concave up
0 x 4
Decreasing, concave up
0
Relative minimum
Increasing, concave up
y
x
4
6
4 x f
x3
x 9
12. f x
x 2
x 2 x 2 27
fc x
x2 9
2
x2 9
(4, 6)
8
6
4
y=x
2
x
−8 −6
2
4
6
8
x=0
−4
−6
9x
x2 9
0 when x
0, r 3 3 and is undefined when x
0 when x
0
18 x x 2 27
f cc x
3
(−2 4, 0)
3
r 3.
Intercept: 0, 0
Symmetry: origin
r3
Vertical asymptotes: x
Slant asymptote: y
x
yc
ycc
Conclusion
Increasing, concave down
0
Relative maximum
3 3 x 3
Decreasing, concave down
3 x 0
Decreasing, concave up
0
0
Point of inflection
0 x 3
Decreasing, concave down
3 x 3 3
+
Decreasing, concave up
0
+
Relative minimum
+
Increasing, concave up
y
f x 3 3
3 3
x
x
x
0
3 3
3 3 x f
9 3
2
0
9 3
2
(
y
3 3, 9 3
2
)
9
x = −3
y=x
x
−12 −9 −6 −3
3
−6
(
−3 3, − 9 3
2
6
9 12
x=3
)
INSTRUCTOR USE ONLY
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Section 3.6
13.
x 2 6 x 12
x 4
4
1
2
x 4
y
yc
x 2
x 2 x 6
285
4
x 4
0 when x
2
x 4
A Summary of Cu
Curve
Cur Sketching
2, 6 and is undefined when x
4.
8
ycc
3
x 4
Vertical asymptote: x
4
Slant asymptote: y
x 2
yc
ycc
Conclusion
Increasing, concave down
0
Relative maximum
2 x 4
Decreasing, concave down
4 x 6
Decreasing, concave up
0
Relative minimum
Increasing, concave up
y
f x 2
x
2
2
x
6
6
6 x f
x=4
y
8
6
(6, 6)
4
y=x−2
2
x
(0, − 3)
6
8
10
(2, −2)
14.
y
yc
ycc
x2 4x 7
x 3
x2 6x 5
2
x 3
4
x 3
x 1 x 5
x 1 x 3
2
0 when x
1, 5 and is undefined when x
3.
8
x 3
3
Intercept: 0, 73
Vertical asymptote: x
3
Slant asymptote: y
x 1
y
12
10
8
6
4
2
−10 −8 −6 −4 −2
−4
−6
−8
−10
ycc
Conclusion
Decreasing, concave up
0
Relative minimum
5 x 3
Increasing, concave up
3 x 1
Increasing, concave down
0
Relative maximum
Decreasing, concave down
f x 5
No symmetry
(−5, 6)
yc
y
(−1, −2)
x
2
x
x
5
1
1 x f
6
2
(0, − 73)
INSTRUCTOR
ST
USE ONLY
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286
Chapter 3
Applications
pplications of Differentiation
15. y
x 4 x , Domain: f, 4@
8 3x
2 4 x
yc
3x 16
ycc
44 x
(
4
0 when x
32
8 16 3
,
3
9
(
(4, 0)
x
−2
16.
2
4.
y
2
(0, 0)
4.
16
and undefined when x
3
16
is not in the domain.
3
Note: x
y
8
and undefined when x
3
0 when x
4
8
f x 3
8
x
3
8
x 4
3
16
3 3
x
0
4
yc
ycc
Conclusion
+
–
Increasing, concave down
0
–
Relative maximum
–
–
Decreasing, concave down
Undefined
Undefined
Endpoint
x 9 x 2 , Domain: 3 d x d 3
h x
9 2 x2
hc x
x 2 x 2 27
hcc x
9 x2
r
0 when x
9 x2
0 when x
32
3
2
r
3 2
and undefined when x
2
0 and undefined when x
r 3.
r 3.
Intercepts: 0, 0 , r3, 0
Symmetric with respect to the origin
3
x
yc
ycc
Conclusion
0
Undefined
Undefined
Endpoint
Decreasing, concave up
0
Relative minimum
Increasing, concave up
3
0
Point of inflection
Increasing, concave down
3
2
3 x x
y
3
2
9
2
3
x 0
2
x
0
0
3
2
0 x 3
2
x
9
2
3
x 3
2
x
3
0
Decreasing, concave down
Undefined
Undefined
Endpoint
y
5
4
3
2
1
Relative maximum
(− 3, 0)
−5 −4
−2 −1
(
(0, 0)
)
(3, 0)
x
1 2 3 4 5
(
−
0
3 2, 9
2
2
3 2, 9
−
2
2
)
−5
INSTRUCTOR USE ONLY
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Section 3.6
17.
y
3x 2 3 2 x
yc
2 x 1 3 2
A Summary of Curve
Cur
Cu
Sketching
2 1 x1 3
x1 3
1 and undefined when x
0 when x
0.
2
0 when x z 0.
3x 4 3
ycc
y
f x 0
x
0
0
0 x 1
yc
ycc
Conclusion
–
–
Decreasing, concave down
Undefined
Undefined
Relative minimum
+
–
Increasing, concave down
0
–
Relative maximum
y
5
(1, 1)
x
18.
287
1
1
x
(0, 0)
1 x f
–
2
–
y
x 1
3x 1
23
yc
2 x 1 2 x 1
1 3
ycc
2
2
4 3
x 1
3
Decreasing, concave down
2 x 1
43
2
0, 2 and undefined when x
yc
ycc
Conclusion
–
Decreasing, concave up
0
Relative minimum
+
Increasing, concave up
Undefined
Relative maximum
–
Decreasing, concave up
0
Relative minimum
Increasing, concave up
6x 1
43
3x 1
1
2
3
5
−2
0 when x
x 1
13
( 278 , 0 )
1.
2
43
1, 0 , r 33 4 1, 0
Intercepts:
y
f x 2
2
2
x
2 x 1
1
x
0
1 x 0
x
0
0
0 x f
y
(−33/4 − 1, 0)
8
6
(−2, 0) 4
(−1, 0) (0, 0)
2
−5 −4
x
−2
2
−4
3
(33/4 − 1, 0)
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Chapter 3
288
Applications
pplications of Differentiation
2 x x3
19. y
yc
1 3 x 2
No critical numbers
ycc
6 x
0 when x
0.
yc
ycc
Conclusion
–
+
Decreasing, concave up
–
0
Point of inflection
–
–
Decreasing, concave down
y
f x 0
x
2
0
0 x f
y
5
4
(0, 2)
1
20.
(1, 0)
x
−3 −2 − 1
2
y
13 x3 3 x 2
yc
x2 1
ycc
2 x
3
y
f x 1
43
1
x
1 x 0
x
23
0
0 x 1
x
r1.
0 when x
0 when x
0
1
1 x f
0.
yc
ycc
Conclusion
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
+
0
Point of inflection
+
–
Increasing, concave down
0
–
Relative maximum
–
–
Decreasing, concave down
y
2
1
(−2, 0)
(1, 0)
x
−1
1
0, − 2
3
(
2
(
(−1, − 43 ( −2
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Section 3.6
21.
y
3x 4 4 x3
yc
12 x3 12 x 2
12 x 2 x 1
0 when x
0, x
1.
ycc
36 x 24 x
12 x 3 x 2
0 when x
0, x
23 .
2
y
f x 1
1
x
–1
1 x 23
23
x
16
27
23 x 0
x
0
0
0 x f
yc
ycc
Conclusion
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
+
0
Point of inflection
+
–
Increasing, concave down
0
0
Point of inflection
+
+
Increasing, concave up
A Summary of Curve
Cur Sketching
Cu
289
y
2
1
(− 43 , 0 (
(0, 0)
x
−2
(
(− 1, − 1)
22.
1
16
− 2 , − 27
3
(
y
2 x 4 3x 2
yc
8x 6x
0 when x
ycc
24 x 6
0 when x
y
3
2
3
0, r
.
2
1
r .
2
y
(
) (12, 58)
( 23 , 98 )
( )
( )
( 26 , 0) x
− 1, 5
2 8
2
− 3, 9
2 8
1
− 6, 0
2
(0, 0)1
−2
−1
3
2
x
Symmetry: y-axis
§
6 ·
, 0 ¸¸
Intercepts: ¨¨ r
2
©
¹
3
2
f x 3
1
x 2
2
1
x
2
1
x 0
2
x
0 x −2
x
x
1
2
5
8
1
2
1
x 2
5
8
0
0
2
9
8
3
2
3
2
3
x f
2
9
8
yc
ycc
Conclusion
Increasing, concave down
0
Relative maximum
Decreasing, concave down
2
0
Point of inflection
Decreasing, concave up
0
Relative minimum
+
+
Increasing, concave up
2
0
Point of inflection
+
Increasing, concave down
0
Relative maximum
Decreasing, concave down
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23.
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Chapter 3
290
y
x5 5 x
yc
5x4 5
ycc
3
20 x
Applications
pplications of Differentiation
5 x4 1
0 when x
0 when x
r1.
0.
y
y
yc
ycc
Conclusion
+
–
Increasing, concave down
0
–
Relative maximum
)− 4 5, 0 )
6
f x 1
x
1
4
–
Decreasing, concave down
–
0
Point of inflection
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
0 when x
1.
0
0
–4
1
1 x f
x 1
yc
5 x 1
4
20 x 1
3
0 when x
y
f x 1
x
1
0
f x
fc x
f cc x
−6
x
1
−2
−4
2
(1, −4)
ycc
Conclusion
Increasing, concave down
Point of inflection
Increasing, concave up
19 x 2 1
x x2 1
x x 1
2
3
(1, 0)
0 for x | r1.10
0 for x | r1.84
x
−1
1
2
3
−1
26.
2 19 x 6 63 x9 3 x 2 1
x3 x 2 1
2
1
0
19 x 4 22 x 2 1
2
y
yc
20 x
1
x2 1 x
2
1.
0
1 x f
25.
−1
5
24. y
ycc
) 4 5, 0 )
(0, 0)
–
0 x 1
x
4
−2
1 x 0
x
(−1, 4)
f x
fc x
f cc x
x
x4 2x2 8x 1
x2 1
8 3x 2 1
x 1
2
3
Slant asymptote: y
Vertical asymptote: x
0
Horizontal asymptote: y
x3 x 4
x2 1
4
x 1
2
2
0 for x
0 for x | 1.379
0 for x | 1.608, x | 0.129
r
1
| r 0.577
3
x
Points of inflection: 0.577, 2.423 , 0.577, 3.577
0
Relative maximum: 0.129, 4.064
Minimum: 1.10, 9.05
Relative minimum: 1.608, 2.724
Maximum: 1.10, 9.05
5
Points of inflection: 1.84, 7.86 , 1.84, 7.86
10
−6
− 15
15
6
−3
−10
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.6
f x
27.
fc x
f cc x
2 x
x2 7
0
32
42 x
x2 7
0 at x
52
Point of inflection: 0, 0
4
−6
6
fc x
f cc x
1 2sin x
f cc x
2cos x
f x
0 Ÿ x | 1.0299
fc x
0 Ÿ sin x
f cc x
0 Ÿ x
291
7S 11S
,
6
6
1
Ÿ x
2
S 3S
2
,
2
§ 7S
Relative minimum: ¨ , © 6
3 7S ·
¸
6 ¹
§ 11S
,
Relative maximum: ¨
© 6
3 11S ·
¸
6 ¹
§ S S · § 3S 3S ·
Points of inflection: ¨ , ¸, ¨ , ¸
2 ¹
©2 2¹ © 2
−4
f x
x 2cos x, 0 d x d 2S
fc x
0
r2
Horizontal asymptotes: y
28.
f x
30.
x2 7
14
A Summary of Cu
Curve
Cur Sketching
4x
5
x 15
60
2
x 2 15
−6
180 x
x 2 15
2
0
! 0
32
0 at x
52
0
y
1
cos 2 x, 0 d x d 2S
4
0 at x | 1.797, 4.486
y
sin x cos x 31. y
Horizontal asymptotes: y
r4
Point of inflection: 0, 0
6
1
sin 2 x
2
sin x sin x cos x
sin x cos x 1
−8
ycc
8
cos x 2cos 2 x 1
2cos x 1 cos x 1
−6
f x
2 x 4sin x, 0 d x d 2S
fc x
2 4cos x
fc x
4sin x
fc x
0 Ÿ cos x
f cc x
0 Ÿ x
29.
cos x cos 2 x
1
Ÿ x
2
0, S , 2S
S 5S
3
,
3
yc
0 Ÿ x
0, S , 2S
ycc
0 Ÿ x
2S 4S
,
, 0, 2S
3 3
5·
§
Relative minimum: ¨S , ¸
4¹
©
§ 2S 3 · § 4S 3 ·
Points of inflection: ¨ , ¸, ¨ , ¸
8¹ © 3
8¹
© 3
§ S 2S
·
2 3¸
Relative minimum: ¨ ,
3
3
©
¹
§ 5S 10S
·
2 3¸
Relative maximum: ¨ ,
3
© 3
¹
Points of inflection: 0, 0 , S , 2S , 2S , 4S
2
0
2p
−2
16
0
2p
−4
INSTRUCTOR USE ONLY
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Chapter 3
292
NOT FOR SALE
Applications
pplications of Differentiation
y
2 x tan x, yc
2 sec x
ycc
2sec x tan x
32.
2
S
x 2
S
35. Because the slope is negative, the function is decreasing
on 2, 8 , and so f 3 ! f 5 .
2
S
r .
4
0 when x
0.
0 when x
2
r
Vertical asymptotes: x
2 in >5, 5@, then f x
36. If f c x
f 2
7 is the least possible value of f 2 . If
S
fc x
4 in >5, 5@, then f x
2
f 2
11 is the greatest possible value of f 2 .
S·
§ S
Relative minimum: ¨ , 1 ¸
4
2¹
©
37. f is cubic.
§S S
·
Relative maximum: ¨ , 1¸
©4 2
¹
f cc is linear.
2
−3
33. y
yc
38. f cc is constant.
2 csc x sec x , 0 x S
f c is linear.
2
f is quadratic.
2 sec x tan x csc x cot x
0 Ÿ x
§S
·
Relative minimum: ¨ , 4 2 ¸
©4
¹
Vertical asymptotes: x
0,
S
2
S
4
39. f x
−4
2
−1
f′
−2
y
f ''
x
f'
2
x2 4x 5
Vertical asymptote: none
Horizontal asymptote: y
p
2
0
4 x 1
x
−2
f
The zero of f c corresponds
to the points where the graph
of f has a horizontal tangent.
There are no zeros on of f cc,
which means the graph of f c
has no horizontal tangent.
16
f″
f
The zeros of f c correspond
to the points where the graph
of f has horizontal tangents.
The zero of f cc corresponds to
the point where the graph of f c
has a horizontal tangent.
3
4 x 3 and
y
f c is quadratic.
Point of inflection: 0, 0
−
2
2 x 3 and
4
9
34. g x
gc x
x cot x, 2S x 2S
sin x cos x x
sin 2 x
−6
g 0 does not exist, but lim x cot x
xof
g cc x
1.
9
−1
The graph crosses the horizontal asymptote y
4. If a
c, the graph
function has a vertical asymptote at x
2 x cos x sin x
sin 3 x
would not cross it because f c is undefined.
r 2S , r S
Vertical asymptotes: x
§ 3S · § S ·
Intercepts: ¨ r , 0 ¸, ¨ r , 0 ¸
© 2 ¹ © 2 ¹
Symmetric with respect to y-axis
4
−2
2
−4
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.6
3x 4 5 x 3
x4 1
Vertical asymptote: none
Horizontal asymptote: y
40. g x
44. g x
A Summary of Cur
Curve
Cu
Sketching
x2 x 2
x 1
if x z 1
­x 2,
®
¯Undefined, if x 1
x 2 x 1
3
x 1
7
293
The rational function is not reduced to lowest terms.
4
−6
6
−8
4
−1
The graph crosses the horizontal asymptote y
3. If a
function has a vertical asymptote at x
c, the graph
would not cross it because f c is undefined.
sin 2 x
41. h x
x
Vertical asymptote: none
Horizontal asymptote: y
−4
There is a hole at 1, 3 .
45. f x
x 2 3x 1
x 2
x 1 3
x 2
3
0
−3
3
6
−3
The graph appears to approach the slant asymptote
x 1.
y
2
− 2
−1
Yes, it is possible for a graph to cross its horizontal
asymptote.
It is not possible to cross a vertical asymptote because
the function is not continuous there.
cos 3 x
4x
Vertical asymptote: x
0
Horizontal asymptote: y
0
46. g x
2 x 2 8 x 15
x 5
2x 2 5
x 5
18
42. f x
−10
The graph appears to approach the slant asymptote
y
2 x 2.
2
− 2
2
47. f x
2 x3
x 1
2
2x 2x
x2 1
4
−2
Yes, it is possible for a graph to cross its horizontal
asymptote.
It is not possible to cross a vertical asymptote because
the function is not continuous there.
43. h x
20
−2
6 2x
3 x
−6
6
−4
The graph appears to approach the slant asymptote
y
2 x.
if x z 3
­2,
®
Undefined,
if x
3
¯
The rational function is not reduced to lowest terms.
23 x
3 x
48. h x
x3 x 2 4
x2
x 1 4
x2
10
3
−10
−2
10
4
−10
−1
The graph appears to approach the slant asymptote
x 1.
y
1
INSTRUCTOR USE ONLY
There is a hole at 3, 2 .
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Chapter 3
294
49.
Applications
pplications of Differentiation
y
4
−2
2
4
−4
x
−2
2
4
−4
(or any vertical translation of f )
50.
x
4
−2
−2
−4
−4
52.
y
y
120
y
2
100
10
f
80
1
8
60
x
−2
2
x
3
6
9
cos 2 S x
x2 1
1
2
x
−2
−1
1
−1
−1
−2
−2
2
x
−4 −2
−2
12 15
2
4
6
8
10
(or any vertical translation of f )
53. f x
1
f ′′
6
4
(a)
f″
−8
8
(or any vertical translation of f )
y
−6 −3
2
x
−4
−2
−4
f
2
2
x
−4
4
4
f″
4
f
y
y
51.
y
(or any vertical translation of the 3 segments of f )
, 0, 4
1.5
4
0
−0.5
On 0, 4 there seem to be 7 critical numbers: 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5
cos S x x cos S x 2S x 2 1 sin S x
(b) f c x
x2 1
Critical numbers |
32
0
1
3
5
7
, 0.97, , 1.98, , 2.98, .
2
2
2
2
The critical numbers where maxima occur appear to be integers in part (a), but approximating them using f c shows that
they are not integers.
tan sin S x
54. f x
(a)
3
−2
2
−3
(b) f x
tan sin S x
tan sin S x
tan sin S x
f x
Symmetry with respect to the origin
(c) Periodic with period 2
(d) On 1, 1 , there is a relative maximum at
1
, tan 1 and a relative minimum at
2
12 , tan 1 .
(e) On 0, 1 , the graph of f is concave downward.
INSTRUCTOR USE ONLY
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Section 3.6
55. Vertical asymptote: x
0
(a) Let y
1
x 3
y
Slant asymptote: y
3x 2
3x 2 7 x 5
x 3
1
3x 2 x 3
58. Vertical asymptote: x
y
x 59. (a) f c x
(b) f cc x
1
x 2
x0 f c x0 y0
x0 y-intercept:
x2 2x 1
x 2
y0 f c x0
x-intercept:
(d) Let x
2 (relative minimum).
x0 y0 f c x0
(e)
0: y y0
1
f c x0
x0
y
y0 x0
f c x0
x0 BC
(f ) PC
2
f cc is positive for x ! 0 (concave upward).
f cc is negative for x 0 (concave downward).
PC
(c) f c is increasing on 0, f . f cc ! 0
(d) f c x is minimum at x
0. The rate of change of f
at x
0 is less than the rate of change of f for all
other values of x.
x0 f x0 f c x0
x0 f x0 f c x0 , 0
f c is positive for x ! 2 and x 2 (increasing).
0 (point of inflection).
1
x x0
f c x0
§
x0 ·
y-intercept: ¨ 0, y0 ¸
¨
c
f x0 ¸¹
©
2 (relative maximum) and
f c is negative for 2 x 2 (decreasing).
1
x x0
c
f x0
x x0
x
0 at x2 and x3 (point of inflection).
0 at x
f x0 x0 f c x0
0, f x0 x0 f c x0
(e) f has a point of inflection at x2 and x3 (change in
concavity).
(b) f cc x
y
0: y0
Let y
(d) f has a relative maximum at x1 .
x
y0 x0 f c x0
x
0 at x0 , x2 and x4 (horizontal tangent).
0 for x
f x0
f c x0
x0 y
(c) Normal line: y y0
2
(c) f c x does not exist at x1 (sharp corner).
60. (a) f c x
y0
f c x0
f c x0 x0
0: y y0
(b) Let x
3
Slant asymptote: y
f c x0 x
§
·
f x0
x-intercept: ¨ x0 , 0¸
¨
¸
f c x0
©
¹
x2
x 5
57. Vertical asymptote: x
y
f c x0 x x0
x
Horizontal asymptote: none
y
0: y0
5
56. Vertical asymptote: x
295
f c x0 x x0
61. Tangent line at P : y y0
3
Horizontal asymptote: y
A Summary of Cu
Cur
Curve Sketching
(g)
AB
(h)
AP
f x0
f c x0
f x0
x0
§ f x0 ·
y02 ¨¨
¸¸
© f c x0 ¹
f c x0
AP
2
f x0
2
2
2
f c x0
x0 x0 f x0 f c x0
2
f c x0
f c x0
1 ª¬ f c x0 º¼
f x0
2
f x0
f x0
f x0
2
f c x0
2
f x0 f c x0
y02
1 ª¬ f c x0 º¼
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 3
296
Applications
pplications of Differentiation
2 xn
x 1
62. f x
4
(a) For n even, f is symmetric about the y-axis. For n odd, f is symmetric about the origin.
(b) The x-axis will be the horizontal asymptote if the degree of the numerator is less than 4.
0, 1, 2, 3.
That is, n
(c) n
4 gives y
2 as the horizontal asymptote.
(d) There is a slant asymptote y
(e)
2 x if n
5:
2 x5
x4 1
2x 2x
.
x4 1
2.5
n=0
n=2
−3
3
n=1
− 1.5
2.5
n=5
n=4
−3
3
n=3
− 1.5
n
0
1
2
3
4
5
M
1
2
3
2
1
0
N
2
3
4
5
2
3
ax
63. f x
x b
2
Answers will vary. Sample answer: The graph has a vertical asymptote at x
b. If a and b are both positive, or both negative,
– b. If a and b have opposite signs,
then the graph of f approaches f as x approaches b, and the graph has a minimum at x
– b.
then the graph of f approaches f as x approaches b, and the graph has a maximum at x
64.
f x
1
2
ax ax
2
1
ax ax 2 , a z 0
2
1
a ax 1
0 when x
.
a
fc x
a2 x a
f cc x
a 2 ! 0 for all x.
§2 ·
(a) Intercepts: 0, 0 , ¨ , 0 ¸
©a ¹
§1 1·
Relative minimum: ¨ , ¸
©a 2¹
Points of inflection: none
(b)
y
a=2
a = −2
5
4
a=1
x
−3
a = −1
−1
2
3
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
Section 3.6
4 16 x 2
65. y
66. y
As x o f, y o 4 x. As x o f, y o 4 x.
x2 6x
x 3
2
297
9
y o x 3 as x o f, and y o x 3 as x o f.
r 4x
Slant asymptotes: y
A Summary of Cu
Curve Sketching
Cur
x 3, y
Slant asymptotes: y
y
x 3
y
12
15
10
12
8
9
6
2
−8 −6 − 4 − 2
3
x
2
4
6
8
− 9 − 6 −3
3
6
f x f a
f b f a
x a
b a
, a x b.
x b
67. Let O
O xb
O xb xa
f x
f x f a
x a
f x f a f b f a
b a
f b f a
x a
b a
f b f a
f a x a O x b x a
b a
f b f a
­°
½°
f t ®f a t a O t a t b ¾.
b a
¯°
¿°
Let h t
ha
x
−3
0, h b
0, h x
0
By Rolle’s Theorem, there exist numbers D1 and D 2 such that a D1 x D 2 b and hc D1
By Rolle’s Theorem, there exists E in a, b such that hcc E
hc D 2
0.
0.
Finally,
0
hcc E
f cc E ^2O` Ÿ O
1 f cc E .
2
INSTRUCTOR USE ONLY
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Chapter 3
298
NOT FOR SALE
Applications
pplications of Differentiation
Section 3.7 Optimization Problems
1. (a)
(b)
First Number, x
Second Number
Product, P
10
110 10
10(110 10) = 1000
20
110 20
20(110 20) = 1800
30
110 30
30(110 30) = 2400
40
110 40
40(110 40) = 2800
50
110 50
50(110 50) = 3000
60
110 60
60(110 60) = 3000
First Number, x
Second Number
Product, P
10
110 10
10(110 10) = 1000
20
110 20
20(110 20) = 1800
30
110 30
30(110 30) = 2400
40
110 40
40(110 40) = 2800
50
110 50
50(110 50) = 3000
60
110 60
60(110 60) = 3000
70
110 70
70(110 70) = 2800
80
110 80
80(110 80) = 2400
90
110 90
90(110 90) = 1800
100
110 100
100(110 100) = 1000
The maximum is attained near x
(c) P
(d)
x 110 x
110 x x
50 and 60.
2
3500
(55, 3025)
0
120
0
The solution appears to be x
55.
dP
55.
(e)
110 2 x
0 when x
dx
d 2P
dx 2
2 0
P is a maximum when x
110 x
55. The two numbers are 55 and 55.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.7
2. (a)
Height, x
Length & Width
Volume
1
24 2(1)
1[24 2(1)]2 = 484
2
24 2(2)
2[24 2(2)]2 = 800
3
24 2(3)
3[24 2(3)]2 = 972
4
24 2(4)
4[24 2(4)]2 = 1024
5
24 2(5)
5[24 2(5)]2 = 980
6
24 2(6)
6[24 2(6)]2 = 864
The maximum is attained near x
4.
x 24 2 x , 0 x 12
dV
dx
2 x 24 2 x 2 24 2 x
12 12 x 4 x
2
d V
dx 2
2
0 when x
24 2 x 24 6 x
12, 4 12 is not in the domain .
12 2 x 16
d 2V
0 when x
dx 2
When x
(d)
299
2
(b) V
(c)
Optimizat
Optimization Problems
Optimizati
4.
4, V
1024 is maximum.
1200
0
12
0
The maximum volume seems to be 1024.
3. Let x and y be two positive numbers such that
x y
S.
xS x
P
xy
dP
dx
S 2x
2
d P
dx 2
Sx x 2
0 when x
2 0 when x
S
.
2
P is a maximum when x
y
S
.
2
S 2.
5. Let x and y be two positive numbers such that
xy 147.
S
x 3y
147
3y
y
dS
dy
3
147
y2
0 when y
d 2S
dy 2
294
! 0 when y
y3
S is minimum when y
4. Let x and y be two positive numbers such that
xy 185.
x y
dS
dx
1
d 2S
dx 2
370
! 0 when x
x3
185
x2
x 185
x
S
S is a minimum when x
185.
185
y
185.
7.
7 and x
21.
6. Let x be a positive number.
dS
dx
1
x
1
1 2
x
d 2S
dx 2
2
! 0 when x
x3
S
0 when x
7.
x 0 when x
1.
1.
The sum is a minimum when x
1 and 1 x
1.
INSTRUCTOR USE ONLY
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Chapter 3
300
NOT FOR SALE
Applications
pplications of Differentiation
7. Let x and y be two positive numbers such that
x 2 y 108.
y 108 2 y
P
xy
dP
dy
108 4 y
2
d P
dy 2
11. Let x be the length and y the width of the rectangle.
0 when y
4 0 when y
54 and y
x 54 x 2
dP
dx
54 3 x 2
2
d P
dx 2
27.
0 when x
P
2x 2 y
dP
dx
2
d 2P
dx 2
128
! 0 when x
x3
9. Let x be the length and y the width of the rectangle.
80
y
40 x
A
xy
dA
dx
40 2 x
d2A
dx 2
2 0 when x
x 40 x
40 x x 2
0 when x
y
P 2x
2
A
xy
y
A
x
P
2x 2 y
36.
dP
dx
d 2P
dx 2
2
dA
dx
P
2x
2
d2A
dx 2
A is maximum when x
y
64
x
4 2.
4 2.
y
4 2 ft.
§ A·
2 x 2¨ ¸
©x¹
2x 0 when x
A.
4A
! 0 when x
x3
P is minimum when x
2A
x
A.
y
A cm. (A square!)
y
20 m.
x
x 2
13. d
2
ª¬ x 2 1 2 º¼
2
x 4 4 x 17 4
P
x x2
2
P
.
4
0 when x
2 0 when x
2A
x2
20.
P
x
2
§P
·
x¨ x ¸
2
©
¹
2x 20.
P
y
0 when x
A
10. Let x be the length and y the width of the rectangle.
2x 2 y
64
x2
xy
3 2.
3 2 and y
§ 32 ·
2 x 2¨ ¸
© x¹
12. Let x be the length and y the width of the rectangle.
3 2.
The product is a maximum when x
A is maximum when x
32
x
P is minimum when x
54 x x3
6 x 0 when x
2x 2 y
y
27.
8. Let x and y be two positive numbers such that
x2 y
54.
xy
32
27.
P is a maximum when x
P
xy
108 y 2 y 2
P
.
4
Because d is smallest when the expression inside the
radical is smallest, you need only find the critical
numbers of
f x
x 4 4 x 17
.
4
fc x
4 x3 4
x
P 4 units. (A square!)
0
1
By the First Derivative Test, the point nearest to 2, 12 is
1, 1 .
y
y
4
3
2
x
( x, x 2 )
(2, 12(
d
1
INSTRUCTOR USE
S ONLY
x
−2
−
2
−1
−
1
1
2
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.7
2
x 1 , 5, 3
14. f x
x 5
d
2
2
3º
¼
2
x 2 10 x 25 x 2 2 x 2
x 12
d
gc x
4 x3 12 x 2 2 x 18
2 x 1 2x 8x 9
2
x 4
2
x 0
1 yields a minimum.
2
x 2 23 x 136
gc x
2 x 23
0 when x
g cc x
2 ! 0 at x
23
2
Because d is smallest when the expression inside the
radical is smallest, you need only find the critical
numbers of
f x
x 2 7 x 16.
fc x
2x 7
§ 23
¨¨ ,
© 2
§ 23 § 23 · ·
¨ , f ¨ ¸¸
© 2 ¹¹
©2
17. xy
x 2 7 x 16
23
2
14 ·
¸
2 ¸¹
30
x
30 Ÿ y
A
§ 30
·
x 2¨
2 ¸ (see figure)
© x
¹
dA
dx
§ 30 · § 30
·
x 2¨ 2 ¸¨
2¸
© x ¹ © x
¹
2 x 2 30
0 when x
x2
0
y
7
2
x
2
The point nearest to 12, 0 is
By the First Derivative Test, x
So, 1, 4 is closest to 5, 3 .
15. d
g x
0
1
x
x 8 0
Because d is smallest when the expression inside the radical
is smallest, you need to find the critical numbers of
Because d is smallest when the expression inside the radical
is smallest, you need to find the critical numbers of
x 4 4 x3 x 2 18 x 29
x 2 23x 136
x 2 10 x 25 x 4 4 x3 8 x 4
g x
2
x 2 24 x 144 x 8
2
x 4 4 x3 x 2 18 x 29
301
x 8, 12, 0
16. f x
ªx 1
¬
Optimizat
Optimization
Optimizati
Problems
30
30
30.
30
By the First Derivative Test, the point nearest to 4, 0 is
By the First Derivative Test, the dimensions x 2 by
7 2,
y 2 are 2 7 2.
30 by 2 30 (approximately
7.477 by 7.477). These dimensions yield a minimum
area.
y
4
x+2
3
x
( x, x )
2
d
1
y
y+2
x
1
2
3
(4, 0)
INSTRUCTOR USE ONLY
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Chapter 3
302
NOT FOR SALE
Applications
pplications of Differentiation
36
x
36 Ÿ y
18. xy
19.
36 x 2y
490,000 ·
§
¨x ¸ where S is the length
x
©
¹
of fence needed.
108
3x 9
x
108
3
x2
dA
dx
245,000 see figure
S
§ 36
·
3¸
x 3¨
© x
¹
x 3 y 3
A
xy
0 Ÿ 3x 2
108 Ÿ x
6, y
dS
dx
1
d 2S
dx 2
980,000
! 0 when x
x3
6
Dimensions: 9 u 9
x+3
490,000
x2
S is a minimum when x
0 when x
700.
700.
700 m and y
350 m.
x
y
y
y+3
x
20.
S
2 x 2 4 xy
y
337.5 2 x
4x
V
dV
dx
d 2V
dx 2
337.5
y
2
ª 337.5 2 x 2 º
1
x2 «
84.375 x x3
»
4x
2
¬
¼
3 2
84.375 x
0 Ÿ x2
56.25 Ÿ x
7.5 and y
2
3x 0 for x
16
32
7.5.
y
7.5 cm.
§ x·
2y x S ¨ ¸
© 2¹
4 y 2x S x
y
32 2 x S x
4
A
xy dA
dx
8 x
d2A
dx 2
S·
§
¨1 ¸ 0 when x
4¹
©
y
x
7.5.
The maximum value occurs when x
21.
x
x2 y
S § x·
2
¨ ¸
2© 2¹
S
2
x S x2
§ 32 2 x S x ·
¨
¸x 4
8
©
¹
S
4
x
S·
§
8 x¨1 ¸
4¹
©
0 when x
1 2 S 2 S 2
x x x
2
4
8
8
1 S 4
32
.
4S
32
.
4S
32 2 ª¬32 4 S º¼ S ª¬32 4 S ¼º
4
The area is maximum when y
8x 16
ft and x
4S
16
4S
32
ft.
4S
x
2
y
x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.7
22. You can see from the figure that A
Optimizat
Optimizati
Optimization Problems
303
6 x
.
2
xy and y
y
6
5
y=
4
6−x
2
3
2
( x, y )
1
x
1
2
4
3
5
6
§6 x·
x¨
¸
© 2 ¹
1
6 2x
2
A
dA
dx
d2A
dx 2
1
6 x x2 .
2
0 when x
1 0 when x
3.
A is a maximum when x
23 (a)
y 2
0 1
3 and y
3 2.
02
x 1
2
x 1
2
y
2 ·
§
x2 ¨ 2 ¸
x 1¹
©
x2 y2
L
(b)
3.
2
x2 4 8
4
,
2
x 1
x 1
x !1
10
(2.587, 4.162)
0
10
0
L is minimum when x | 2.587 and L | 4.162.
1
1 §
2 ·
x
Ax
xy
x¨ 2 x (c) Area
¸
2
2 ©
x 1¹
x 1
Ac x
1
2
x 1
x 1
x
They y
x 1 x
x 1
2
1
1
x 1
2
0
1
r1
0, 2 select x
4 and A
2
4.
Vertices: 0, 0 , 2, 0 , 0, 4
INSTRUCTOR USE ONLY
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Chapter 3
304
24. (a)
A
dA
dh
NOT FOR SALE
Applications
pplications of Differentiation
1
1
base u height
2 36 h 2 6 h
2
2
1/2
1/2
1
2h 6 h 36 h 2
36 h 2
2
1/2
36 h 2
dA
dh
ª h 6 h 36 h 2 º
¬
¼
36 h 2 6 h
2 h 2 3h 18
2 h 6 h 3
36 h 2
36 h 2
3, which is a maximum by the First Derivative Test. So, the sides are 2 36 h 2
0 when h
equilateral triangle. Area
6 3, an
27 3 sq. units.
6
6
h
36 − h 2
6 h
2 3 6 h
cos D
(b)
tan D
Area
6 h
2 3
36 h 2
6 h
§1·
2¨ ¸ 36 h 2 6 h
© 2¹
6 h
2
tan D
144 cos 4 D tan D
Ac D
144 ª¬cos 4 D sec2 D 4 cos3 sin D tan D º¼
Ÿ cos 4 D sec2 D
4 cos3 D sin D tan D
1
4 cos D sin D tan D
1
4
sin 2 D
sin D
1
Ÿ D
2
2
α
3
30q and A
0
27 3.
6+h
6
h
6
36 − h 2
(c) Equilateral triangle
25.
A
2 xy
dA
dx
§ 1 ·§
2 x¨ ¸¨
© 2 ¹©
2 x 25 x 2 see figure
·
2
¸ 2 25 x
2
25 x ¹
2 x
§ 25 2 x 2 ·
2¨
¸
© 25 x 2 ¹
0 when x
y
5 2
| 3.54.
2
By the First Derivative Test, the inscribed rectangle of maximum area has vertices
§ 5 2 · § 5 2 5 2·
, 0 ¸¸, ¨¨ r
,
¨¨ r
¸.
2
2
2 ¸¹
©
¹ ©
y
8
6
( x,
25 − x 2
(
5 2
Width:
; Length: 5 2
2
x
−6 −4 −2
−2
2
4
6
INSTRUCTOR
O USE ONLY
−4
4
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.7
26.
Optimization
Optimizati
Optimizat
Problems
305
2 x r 2 x 2 see figure
A
2 xy
dA
dx
2 r 2 2x2
2r
.
2
0 when x
r 2 x2
2r by
By the First Derivative Test, A is maximum when the rectangle has dimensions
2r 2.
y
(−x,
r 2 − x2
( ( x,
r 2 − x2
(
x
(−r, 0)
(r, 0)
27. (a) P
2 x 2S r
§ y·
2 x 2S ¨ ¸
©2¹
2x S y
200
Ÿ y
200 2 x
2
S
S
y
2
100 x
y
x
(b)
Length, x
Width, y
10
S
20
S
2
2
2
30
S
40
S
50
S
2
2
2
60
S
Area, xy
2
100 10
10
S
100 20
20
S
100 30
30
S
100 40
40
S
100 50
50
S
100 60
60
2
2
2
2
2
S
100 10 | 573
100 20 | 1019
100 30 | 1337
100 40 | 1528
100 50 | 1592
100 60 | 1528
The maximum area of the rectangle is approximately 1592 m2.
2
2
(c) A
100 x x 2
xy
x 100 x
S
(d) Ac
2
S
S
100 2 x . Ac
0 when x
50.
Maximum value is approximately 1592 when length
(e)
50 m and width
100
S
.
2000
(50, 1591.6)
0
100
0
Maximum area is approximately
1591.55 m 2 x
50 m .
INSTRUCTOR USE ONLY
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306
Chapter 3
28. V
S r 2h
(a)
NOT FOR SALE
Applications
pplications of Differentiation
22 cubic inches or h
Radius, r
Height
Surface Area
2
ª
22 º
» | 220.3
2S 0.2 «0.2 2
«¬
S 0.2 »¼
2
ª
22 º
» | 111.0
2S 0.4 «0.4 2
«¬
S 0.4 »¼
2
ª
22 º
» | 75.6
2S 0.6 «0.6 2
S 0.6 ¼»
¬«
2
ª
22 º
» | 59.0
2S 0.8 «0.8 2
«¬
S 0.8 »¼
22
0.2
S 0.2
22
0.4
S 0.4
22
0.6
S 0.6
22
0.8
S 0.8
Radius, r
Height
22
Sr2
(b)
Surface Area
2
ª
22 º
» | 220.3
2S 0.2 «0.2 2
«¬
S 0.2 »¼
2
ª
22 º
» | 111.0
2S 0.4 «0.4 2
S 0.4 ¼»
¬«
2
ª
22 º
» | 75.6
2S 0.6 «0.6 2
«¬
S 0.6 »¼
2
ª
22 º
» | 59.0
2S 0.8 «0.8 2
«¬
S 0.8 »¼
2
ª
22 º
» | 50.3
2S 1.0 «1.0 2
«¬
S 1.0 »¼
2
ª
22 º
» | 45.7
2S 1.2 «1.2 2
«¬
S 1.2 »¼
2
ª
22 º
» | 43.7
2S 1.4 «1.4 2
«¬
S 1.4 »¼
2
ª
22 º
» | 43.6
2S 1.6 «1.6 2
«¬
S 1.6 »¼
2
ª
22 º
» | 44.8
2S 1.8 «1.8 2
«¬
S 1.8 »¼
2
ª
22 º
» | 47.1
2S 2.0 «2.0 2
S 2.0 ¼»
¬«
22
0.2
S 0.2
22
0.4
S 0.4
22
0.6
S 0.6
22
0.8
S 0.8
22
1.0
S 1.0
22
1.2
S 1.2
1.4
S 1.4
22
22
1.6
S 1.6
22
1.8
S 1.8
22
2.0
S 2.0
The minimum seems to the about 43.6 for r
(c) S
1.6.
2S r 2 2S rh
2S r r h
22 º
ª
2S r «r S r 2 »¼
¬
2S r 2 44
r
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.7
(d)
Optimization Problems
Optimizati
Optimizat
307
100
(1.52, 43.46)
−1
4
−10
The minimum seems to be 43.46 for r | 1.52.
dS
44
3 11 S | 1.52 in.
4S r 2
0 when r
(e)
dr
r
22
| 3.04 in.
h
S r2
Note: Notice that h
22
Sr2
22
S 11 S
23
§ 111 3 ·
2¨ 1 3 ¸
©S ¹
2r.
29. Let x be the sides of the square ends and y the length of
the package.
P
4x y
V
x2 y
dV
dx
216 x 12 x 2
2
d V
dx 2
108 Ÿ y
x 2S r
108 4 x
x 2 108 4 x
108 x 2 4 x3
12 x 18 x
0 when x
216 24 x
216 0 when x
S r2x
V
30.
108 Ÿ x
V
S r 2 108 2S r
S 108r 2 2S r 3
dV
dr
S 216r 6S r 2
6S r 36 S r
18.
The volume is maximum when x
y 108 4 18
36 in.
0 when r
d 2V
dr 2
18.
108 2S r see figure
36
S
and x
36.
S 216 12S r 0 when r
18 in. and
r
36
S
.
x
Volume is maximum when x
36 in. and
36 S | 11.459 in.
r
31. No. The volume will change because the shape of the container changes when squeezed.
32. No, there is no minimum area. If the sides are x and y, then 2 x 2 y
Ax
33.
V
h
x 10 x
14
20 Ÿ y
10 x. The area is
10 x x . This can be made arbitrarily small by selecting x | 0.
2
4 3
S r S r 2h
3
14 4 3 S r 3
14
Sr2
Sr2
4
r
3
4 ·
§ 14
4S r 2 2S r ¨ 2 r ¸
3 ¹
©Sr
S
4S r 2 2S rh
dS
dr
8
28
Sr 2
3
r
d 2S
dr 2
8
56
S 3 ! 0 when r
3
r
0 when r
3
3
The surface area is minimum when r
4S r 2 28 8 2
Sr
3
r
4 2
28
Sr 3
r
21
| 1.495 cm.
2S
21
.
2S
r
3
21
cm and h
2S
h
0.
The resulting solid is a sphere of radius r | 1.495 cm.
INSTRUCTOR USE ONLY
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Chapter 3
308
34. V
4 3
S r S r 2h
3
4000
4
r
Sr2
3
4000
h
Let k
NOT FOR SALE
Applications
pplications of Differentiation
cost per square foot of the surface area of the sides, then 2k
C
2k 4S r 2 k 2S rh
dC
dr
8000 º
ª 32
k« Sr 2 »
3
r ¼
¬
ª
4 ·º
§ 4000
k «8S r 2 2S r ¨ 2 r ¸»
3 ¹¼
© Sr
¬
0 when r
3
By the Second Derivative Test, you have
The cost is minimum when r
750
3
S
750
S
cost per square foot of the hemispherical ends.
8000 º
ª16
k « Sr2 r ¼»
¬3
| 6.204 ft and h | 24.814 ft.
12,000 º
ª 32
! 0 when r
k« S 3
r 3 »¼
¬
d 2C
dr 2
3
750
S
.
ft and h | 24.814 ft.
35. Let x be the length of a side of the square and y the length of a side of the triangle.
4x 3y
A
10
x2 1 § 3 ·
y¨
y¸
2 ¨© 2 ¸¹
10 3 y
2
16
dA
dy
30 9 y 4 3 y
y
d2A
dy 2
A is minimum when y
3 2
y
4
1
3
10 3 y 3 y
8
2
0
0
30
9 4 3
9 4 3
! 0
8
30
and x
9 4 3
10 3
.
9 4 3
36. (a) Let x be the side of the triangle and y the side of the square.
A
3§
S · 2 4§
S· 2
¨ cot ¸ x ¨ cot ¸ y where 3 x 4 y
4©
3¹
4©
4¹
20
2
3 2 §
3 ·
20
.
x ¨5 x¸ , 0 d x d
4
4 ¹
3
©
Ac
3
3 ·§ 3 ·
§
x 2¨ 5 x ¸¨ ¸
2
4 ¹© 4 ¹
©
x
60
4 3 9
When x
0, A
25, when x
0
60 4 3 9 , A | 10.847, and when x
20 3, A | 19.245. Area is maximum
when all 20 feet are used on the square.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.7
Optimizat
Optimization
Optimizati
Problems
309
(b) Let x be the side of the square and y the side of the pentagon.
A
4§
S · 2 5§
S· 2
¨ cot ¸ x ¨ cot ¸ y where 4 x 5 y
4©
4¹
4©
5¹
20
2
4 ·
§
x 2 1.7204774¨ 4 x ¸ , 0 d x d 5.
5 ¹
©
4 ·
§
2 x 2.75276384¨ 4 x ¸
5 ¹
©
x | 2.62
Ac
When x
0
0, A | 27.528, when x | 2.62, A | 13.102, and when x
5, A | 25. Area is maximum when all 20 feet
are used on the pentagon.
(c) Let x be the side of the pentagon and y the side of the hexagon.
A
5§
S · 2 6§
S· 2
¨ cot ¸ x ¨ cot ¸ y where 5 x 6 y
4©
5¹
4©
6¹
2
S· 2 3
5§
¨ cot ¸ x 4©
5¹
2
§ 20 5 x ·
3¨
¸ , 0 d x d 4.
6
©
¹
S·
5§
§ 5 ·§ 20 5 x ·
¨ cot ¸ x 3 3 ¨ ¸¨
¸
2©
5¹
6
© 6 ¹©
¹
x | 2.0475
Ac
When x
20
0
0, A | 28.868, when x | 2.0475, A | 14.091, and when x
4, A | 27.528. Area is maximum when all 20
feet are used on the hexagon
(d) Let x be the side of the hexagon and r the radius of the circle.
A
6§
S· 2
2
¨ cot ¸ x S r where 6 x 2S r
4©
6¹
20
2
3 3 2
10
§ 10 x ·
x S¨
¸ ,0 d x d .
S ¹
2
3
©S
§ 10 x ·
3 3 6¨
S ¸¹
©S
x | 1.748
Ac
When x
0
0, A | 31.831, when x | 1.748, A | 15.138, and when x
10 3, A | 28.868. Area is maximum when all
20 feet are used on the circle.
In general, using all of the wire for the figure with more sides will enclose the most area.
37. Let S be the strength and k the constant of
proportionality. Given
h 2 w2
202 , h 2
202 w2 ,
S
kwh 2
S
kw 400 w2
k 400 w w3
dS
dw
k 400 3w2
and h
20 6
in.
3
d 2S
dw2
6kw 0 when w
0 when w
20 3
.
3
20 3
in.
3
38. Let A be the amount of the power line.
A
h y 2
dA
dy
1 d2A
dy 2
x2 y 2
2y
x2 y2
2 x2
x y
2
2
32
0 when y
! 0 for y
x
.
3
x
.
3
The amount of power line is minimum when
y
x 3.
y
(0, h)
These values yield a maximum.
h−y
y
INSTRUCTOR USE
S ONLY
(−x, 0)
x
(x, 0)
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Chapter 3
310
39. C x
NOT FOR SALE
Applications
pplications of Differentiation
x2 4 k 4 x
2k
2 xk
Cc x
x2 4
k
40. sin D
h
Ÿ s
s
S
h
,0 D sin D
2
tan D
h
Ÿ h
2
2 tan D Ÿ s
I
k sin D
s2
k sin D
4 sec 2 D
0
x 4
2
2x
4x2
x2 4
3x 2
4
dI
dD
2
3
x
Or, use Exercise 50(d): sin T
So, x
2
.
3
θ
x2 + 4
C2
C1
1
ŸT
2
30q.
2 tan D
sin D
2 sec D
k
sin D cos 2 D
4
k
ªsin D 2 sin D cos D cos 2 D cos D º¼
4¬
k
cos D ª¬cos 2 D 2 sin 2 D º¼
4
k
cos D ª¬1 3 sin 2 D º¼
4
S 3S
1
r
0 when D
,
, or when sin D
.
2 2
3
Because D is acute, you have
2
1
Ÿ h
3
sin D
4−x
x
Because
d 2 I dD 2
sin D
1
2 tan D
§ 1 ·
2¨
¸
© 2¹
2 ft.
k 4 sin D 9 sin 2 D 7 0 when
3, this yields a maximum.
h
α
s
α
4 ft
x 2 4, L
S
41.
Time
dT
dx
x2
x2 4
x 4 6 x3 9 x 2 8 x 12
S=
2
x
x2 4
2
T
x
2
1 3 x
x2 4
4
2
x 2 6 x 10
4
x 3
x 2 6 x 10
0
9 6 x x2
4 x 2 6 x 10
0
x2 + 4
3−x
1
L=
1 + (3 − x( 2
Q
You need to find the roots of this equation in the interval >0, 3@. By using a computer or graphing utility you can determine that
this equation has only one root in this interval x
1 . Testing at this value and at the endpoints, you see that x
1 yields the
minimum time. So, the man should row to a point 1 mile from the nearest point on the coast.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.7
42.
x2 4
v1
T
dT
dx
x
v1
x 2 6 x 10
v2
x2 4
x 3
dT
dx
0
x 2 6 x 10
v2
Because
x
sin T 2
x 2 6 x 10
x
sin T1
v1
0 Ÿ
sin T 2
.
v2
4
v1 x 2 4
32
x a
d2 a x
2
v2
0
2
sin T1
sin T 2
v1
v2
1
v2 x 2 6 x 10
x a
d2 a x
2
sin T 2
2
you have
Because
d 2T
dx 2
2
v2
sin T1 and
x 2 d12
you have
sin T1
sin T 2
v1
v2
x
x 2 d12
v1
311
Because
x 3
sin T1 and
x2 4
d22 a x
x 2 d12
v1
T
44.
Optimizat
Optimization
Optimizati
Problems
sin T 2
.
v2
Because
! 0
32
sin T1
v1
0 Ÿ
d 2T
dx 2
this condition yields a minimum time.
d12
v1 x 2 d12
32
d22
2
v2 ªd 2 2 a x º
¬
¼
32
! 0
this condition yields a minimum time.
θ1
2
V
45.
3−x
x
dV
dr
1
θ2
Q
1 2
S r 144 r 2
3
1 2
º
1 ª 2§ 1 ·
S r ¨ ¸ 144 r 2
2r 2r 144 r 2 »
3 «¬ © 2 ¹
¼
1 ª 288r 3r 3 º
S«
»
3 ¬ 144 r 2 ¼
2 2 sin x
43. f x
1 2
Sr h
3
ª r 96 r 2 º
»
«¬ 144 r 2 »¼
S«
y
3
0 when r
0, 4 6.
By the First Derivative Test, V is maximum when
r
4 6 and h
4 3.
2
1
−π
4
−1
π
4
Area of circle: A
x
π
2
S
Distance from origin to x-intercept is S 2 | 1.57.
x2 y 2
x 2 2 2 sin x
2
3
144S
2
S 4 6
4 6
2
4 3
2
48 6S
Area of sector:
144S 48 6S
T
(0.7967, 0.9795)
−
4
2
Lateral surface area of cone:
(a) Distance from origin to y-intercept is 2.
(b) d
S 12
1 2
Tr
2
72T
144S 48 6S
72
2S
3 6 | 1.153 radians or 66q
3
−1
Minimum distance
d2 x
(c) Let f x
fc x
0.9795 at x
0.7967.
2
x 2 2 2 sin x .
2 x 2 2 2 sin x 2 cos x
Setting f c x
corresponds to d
0, you obtain x | 0.7967, which
0.9795.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Chapter 3
312
46. (a)
NOT FOR SALE
Applications
pplications of Differentiation
47. Let d be the amount deposited in the bank, i be the
interest rate paid by the bank, and P be the profit.
Base 1
Base 2
Altitude
Area
8
8 16 cos 10q
8 sin 10q
| 22.1
P
0.12 d id
8
8 16 cos 20q
8 sin 20q
| 42.5
d
ki 2 because d is proportional to i 2
8
8 16 cos 30q
8 sin 30q
| 59.7
P
0.12 ki 2 i ki 2
8
8 16 cos 40q
8 sin 40q
| 72.7
k 0.24i 3i 2
8
8 16 cos 50q
8 sin 50q
| 80.5
dP
di
8
8 16 cos 60q
8 sin 60q
| 83.1
d 2P
di 2
k 0.24 6i 0 when i
Base 1
Base 2
Altitude
Area
8
8 16 cos 10q
8 sin 10q
| 22.1
8
8 16 cos 20q
8 sin 20q
| 42.5
8
8 16 cos 30q
8 sin 30q
| 59.7
8
8 16 cos 40q
8 sin 40q
| 72.7
8
8 16 cos 50q
8 sin 50q
| 80.5
8
8 16 cos 60q
8 sin 60q
| 83.1
8
8 16 cos 70q
8 sin 70q
| 80.7
8
8 16 cos 80q
8 sin 80q
| 74.0
8
8 16 cos 90q
8 sin 90q
| 64.0
The maximum cross-sectional area is approximately
83.1 ft2.
(c) A
8 sin T
2
64 1 cos T sin T , 0q T 90q
64 1 cos T cos T 64 sin T sin T
64 cos T cos 2 T sin 2 T
(d) The point of diminishing returns is the point where
the concavity changes, which in this case is
x
20 thousand dollars.
5m 6
2
dS1
dm
2 4m 1 4 2 5m 6 5 2 10m 3 10
282m 128
Line: y
10m 3
2
4m 1
0 when m
64
.
141
64
x
141
§ 64 ·
§ 64 ·
§ 64 ·
4¨
¸ 1 5¨
¸ 6 10¨
¸3
141
141
©
¹
©
¹
© 141 ¹
256
320
640
1 6 3
141
141
141
50. S 2
858
| 6.1 mi
141
4m 1 5m 6 10m 3
Using a graphing utility, you can see that the minimum
occurs when m
0.3.
Line y
0.3x
4 0.3 1 5 0.3 6 10 0.3 3
4.7 mi.
S2
60q, 180q, 300q.
The maximum occurs when T
2
S1
49.
64 2 cos T 1 cos T 1
0 when T
8%.
(c) In order to yield a maximum profit, the company
should spend about $40 thousand.
S2
64 2 cos 2 T cos T 1
(e)
0.08 Note: k ! 0 .
(b) The profit is decreasing on 40, 60 .
ª¬8 8 16 cos T º¼
dA
dT
0.08.
48. (a) The profit is increasing on 0, 40 .
S
h
a b
2
0.24
3
0 when i
The profit is a maximum when i
(b)
(d)
k 0.12i 2 i 3
60q.
30
20
10
100
(60°, 83.1)
(0.3, 4.7)
m
1
0
2
3
90
0
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.7
4m 1
51. S3
m2 1
5m 6
m2 1
Optimization Problems
Optimizati
Optimizat
313
10m 3
m2 1
Using a graphing utility, you can see that the minimum occurs when x | 0.3.
Line: y | 0.3x
4 0.3 1 5 0.3 6 10 0.3 3
S3
0.3
2
| 4.5 mi.
1
S3
30
20
10
(0.3, 4.5)
m
1
2
3
x2 h2 .
52. (a) Label the figure so that r 2
Then, the area A is 8 times the area of the region given by OPQR:
ª1
º
8 « h 2 x h h»
¬2
¼
A
Ac x
8x2
r 2 x2
x2
ª1
8« r 2 x 2 x ¬2
8 r 2 x2 4 x4 4 x2r 2 r 4
x2 r 2 x2
5x4 5x2r 2 r 4
0
25r 4 20r 4
10
Take positive value.
r2 ª
5r
10 ¬
cos T
tan T
T
θ
2
5 ¼º.
R
O
x
h
P r
T
h
and cos
2
r
2
square in the middle.
Ac T
h
Q
5 5
| 0.85065r Critical number
10
(b) Note that sin
A
0
Quadratic in x 2 .
5r 2 r
r
8x
8 x r 2 x 2 4 x 2 4r 2
x r 2 x2 r 2 x2
x r 2 x2
x
r 2 x2
º
r 2 x2 »
¼
8x 8 r 2 x2
2x2 r 2
x2
8x2
r 2 x2
T
2 2 x 2h 4h 2
x
. The area A of the cross equals the sum of two large rectangles minus the common
r
8 xh 4h 2
T
T·
§
4r 2 ¨ cos T sin cos ¸
2
2¹
©
T
T
1
sin cos
sin T
2
2
2
2
arctan 2 | 1.10715
or
8r 2 sin
T
2
cos
T
2
4r 2 sin 2
T
2
T·
§
4r 2 ¨ sin T sin 2 ¸
2¹
©
0
63.4q
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 3
314
Applications
pplications of Differentiation
r2
5
10
(c) Note that x 2
ªr 2
8« 5 ¬10
5
12
º
5»
¼
r2
5
10
12
ªr 4
º
8« 20 »
¬10
¼
2r 2 2
5
8 2
r
5
5 2r 2 2 5 2
r
5
ª4
2r 2 «
«¬ 5
5 1
5º
»
5 »¼
r2
4 5
10
So, A T
2r 2
T·
§
4r 2 ¨ sin T sin 2 ¸
2¹
©
5 1
2, sin T
x 13 x 36
2
§T ·
and sin 2 ¨ ¸
5
© 2¹
§ 2
1§
1 ··
4r 2 ¨
¨1 ¸¸
2
5
5 ¹¹
©
©
x3 3 x; x 4 36 d 13 x 2
2
5 4r 2
5r 2 4 r 2
Using the angle approach, note that tan T
4
5.
8 x r 2 x 2 4 x 2 4r 2
Ax
53. f x
r2
5
10
5 and r 2 x 2
5 1
4r 2
3
54. Let a
2
x 3 x 2 x 2 x 3 d 0
3x 2 3
1
, x ! 0.
x3
1·
1·
§
§ 3
¨x ¸ ¨x 3 ¸
x¹
x ¹
©
©
2
6
So, f is increasing on >3, 2@ and >2, 3@.
2, f 3
x3 1·
1
§
§ 6
·
¨ x ¸ ¨ x 6 2¸
x¹
x
©
©
¹
3x 1 x 1
f is increasing on f, 1 and 1, f .
f 2
1·
§
¨ x ¸ and b
x¹
©
6
a 2 b2
So, 3 d x d 2 or 2 d x d 3.
fc x
5 1
2r 2
2
x 9 x 4
2
1§
1 ·
¨1 ¸.
2©
5¹
1
1 cos T
2
Let f x
x 1x
6
6
x6 1 x 2
3
x 1x
a 2 b2
a b
18. The maximum value of f is 18.
x3 1 x3
a b
3
1· § 3
1·
§ 3
¨ x 3x 3 ¸ ¨ x 3 ¸
x
x ¹ ©
x ¹
©
3x Let g x
g cc x
x
x 3
x
1·
§
3¨ x ¸.
x¹
©
1
, gc x
x
2
and g cc 1
x3
1: g 1
2.
1
1
x2
0 Ÿ x
1.
2 ! 0. So g is a minimum at
Finally, f is a minimum of 3 2
6.
INSTRUCTOR USE ONLY
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Section 3.8
Newt
New
Newton’s Method
315
Section 3.8 Newton's Method
The following solutions may vary depending on the software or calculator used, and on rounding.
1.
2.
3.
4.
f x
x2 5
fc x
2x
x1
2.2
n
xn
f xn
f c xn
1
2.2000
–0.1600
4.4000
–0.0364
2.2364
2
2.2364
0.0013
4.4727
0.0003
2.2361
f xn
f c xn
f xn
xn f c xn
f x
x3 3
fc x
3x 2
x1
1.4
n
xn
f xn
f c xn
f xn
f c xn
1
1.4000
–0.2560
5.8800
–0.0435
1.4435
2
1.4435
0.0080
6.2514
0.0013
1.4423
f x
cos x
fc x
sin x
f xn
f c xn
xn x1
1.6
n
xn
f xn
f c xn
1
1.6000
–0.0292
–0.9996
0.0292
1.5708
2
1.5708
0.0000
–1.0000
0.0000
1.5708
f x
tan x
fc x
sec2 x
f xn
f c xn
xn f xn
f c xn
x1
0.1
n
xn
f xn
f c xn
f xn
f c xn
1
0.1000
0.1003
1.0101
0.0993
0.0007
2
0.0007
0.0007
1.0000
0.0007
0.0000
xn f xn
f c xn
INSTRUCTOR USE ONLY
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5.
NOT FOR SALE
Chapter 3
316
Applications
pplications of Differentiation
f x
x3 4
fc x
3x 2
x1
2
n
xn
f xn
f c xn
f xn
f c xn
1
–2.0000
–4.0000
12.0000
–0.3333
–1.6667
2
–1.6667
–0.6296
8.3333
–0.0756
–1.5911
3
–1.5911
–0.0281
7.5949
–0.0037
–1.5874
4
–1.5874
–0.0000
7.5596
0.0000
–1.5874
xn f xn
f c xn
Approximation of the zero of f is –1.587.
6.
f x
2 x3
fc x
3 x 2
x1
1.0
n
xn
f xn
f c xn
f xn
f c xn
1
1.0000
1.0000
–3.0000
–0.3333
1.3333
2
1.3333
–0.3704
–5.3333
0.0694
1.2639
3
1.2639
–0.0190
–4.7922
0.0040
1.2599
4
1.2599
0.0001
–4.7623
0.0000
1.2599
xn f xn
f c xn
Approximation of the zero of f is 1.260.
7.
f x
x3 x 1
fc x
3x 2 1
n
xn
f xn
f c xn
1
0.5000
–0.3750
1.7500
–0.2143
0.7143
2
0.7143
0.0788
2.5307
0.0311
0.6832
3
0.6832
0.0021
2.4003
0.0009
0.6823
f xn
f c xn
xn f xn
f c xn
Approximation of the zero of f is 0.682.
8.
f x
x5 x 1
fc x
5x4 1
n
xn
f xn
f c xn
f xn
f c xn
1
0.5000
–0.4688
1.3125
–0.3571
0.8571
2
0.8571
0.3196
3.6983
0.0864
0.7707
3
0.7707
0.0426
2.7641
0.0154
0.7553
4
0.7553
0.0011
2.6272
0.0004
0.7549
xn f xn
f c xn
INSTRUCTOR USE ONLY
Approximation of the zero off f is 0.755.
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NOT FOR SALE
Section 3.8
9.
f x
fc x
5
x 1 2x
2
5
2
x 1
From the graph you see that these are two zeros. Begin with x
f xn
xn
f xn
f c xn
1
1.2000
–0.1639
3.5902
–0.0457
1.2457
2
1.2457
–0.0131
3.0440
–0.0043
1.2500
3
1.2500
–0.0001
3.0003
–0.0003
1.2500
xn 317
1.2.
f xn
n
f c xn
Newt
Newton’s Method
New
f c xn
Approximation of the zero of f is 1.250.
Similarly, the other zero is approximately 5.000.
(Note: These answers are exact)
10.
f x
x 2
x 1
fc x
1
1
x 1
n
xn
f xn
f c xn
f xn
f c xn
1
5.0000
0.1010
0.5918
0.1707
4.8293
2
4.8293
0.0005
0.5858
0.00085
4.8284
xn f xn
f c xn
Approximation of the zero of f is 4.8284.
11.
f x
x3 3.9 x 2 4.79 x 1.881
fc x
3x 2 7.8 x 4.79
n
xn
f xn
f c xn
f xn
f c xn
1
0.5000
–0.3360
1.6400
–0.2049
0.7049
2
0.7049
–0.0921
0.7824
–0.1177
0.8226
3
0.8226
–0.0231
0.4037
–0.0573
0.8799
4
0.8799
–0.0045
0.2495
–0.0181
0.8980
5
0.8980
–0.0004
0.2048
–0.0020
0.9000
6
0.9000
0.0000
0.2000
0.0000
0.9000
xn f xn
f c xn
Approximation of the zero of f is 0.900.
n
xn
f xn
f c xn
f xn
f c xn
1
1.1
0.0000
–0.1600
–0.0000
xn f xn
f c xn
1.1000
Approximation of the zero of f is 1.100.
n
xn
f xn
f c xn
f xn
f c xn
1
1.9
0.0000
0.8000
0.0000
xn f xn
f c xn
1.9000
INSTRUCTOR USE ONLY
Approximation of the zero off f is 1.900.
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Chapter 3
318
12.
Applications
pplications of Differentiation
f x
x 4 x3 1
fc x
4 x3 3x 2
From the graph you see that these are two zeros. Begin with x1
n
xn
f xn
f c xn
f xn
f c xn
1
1.0000
1.0000
7.0000
0.1429
0.8571
2
0.8571
0.1695
4.7230
0.0359
0.8213
3
0.8213
0.0088
4.2390
0.0021
0.8192
4
0.8192
0.0003
4.2120
0.0000
0.8192
xn 1.0
f xn
f c xn
Approximation of the zero of f is 0.819.
Similarly, the other zero is approximately –1.380.
13.
f x
1 x sin x
fc x
1 cos x
x1
2
n
xn
f xn
f c xn
1
2.0000
–0.0907
–1.4161
0.0640
1.9360
2
1.9360
–0.0019
–1.3571
0.0014
1.9346
3
1.9346
0.0000
–1.3558
0.0000
1.9346
f xn
f c xn
xn f xn
f c xn
Approximate zero: x | 1.935
14.
f x
x3 cos x
fc x
3x 2 sin x
n
xn
f xn
f c xn
f xn
f c xn
1
0.9000
0.1074
3.2133
0.0334
0.8666
2
0.8666
0.0034
3.0151
0.0011
0.8655
3
0.8655
0.0000
3.0087
0.0000
0.8655
xn f xn
f c xn
Approximation of the zero of f is 0.866.
15. h x
f x g x
2x 1 x 4
1
x 4
hc x
2
n
xn
h xn
hc xn
h xn
hc xn
1
0.6000
0.0552
1.7669
0.0313
0.5687
2
0.5687
0.0000
1.7661
0.0000
0.5687
2
xn h xn
hc xn
Point of intersection of the graphs of f and g occurs when x | 0.569.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.8
16. h x
f x g x
Newt
Newton’s Method
New
319
1
x 1
3 x 2
2x
hc x
1 n
xn
h xn
hc xn
h xn
hc xn
1
2.9000
–0.0063
–0.9345
0.0067
2.8933
2
2.8933
0.0000
–0.9341
0.0000
2.8933
x 1
2
2
xn h xn
hc xn
Point of intersection of the graphs of f and g occurs when x | 2.893.
17. h x
f x g x
x tan x
hc x
1 sec 2 x
n
xn
h xn
hc xn
h xn
hc xn
1
4.5000
–0.1373
–21.5048
0.0064
4.4936
2
4.4936
–0.0039
–20.2271
0.0002
4.4934
h xn
hc xn
xn Point of intersection of the graphs of f and g occurs when x | 4.493.
Note: f x
x and g x
tan x intersect infinitely often.
18. h x
f x g x
x 2 cos x
hc x
2 x sin x
n
xn
h xn
hc xn
h xn
hc xn
1
0.8000
–0.0567
2.3174
–0.0245
0.8245
2
0.8245
0.0009
2.3832
0.0004
0.8241
xn h xn
hc xn
One point of intersection of the graphs of f and g occurs when x | 0.824.
Because f x
x 2 and g x
cos x are both symmetric with respect to the y-axis,
the other point of intersection occurs when x | 0.824.
INSTRUCTOR USE ONLY
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NOT FOR SALE
320
Chapter 3
Applications
pplications of Differentiation
19. (a)
f x
x 2 a, a ! 0
fc x
2x
xn 1
xn 5: xn 1
1§
5·
¨ xn ¸, x1
xn ¹
2©
2
f xn
n
1
2
3
4
f c xn
xn
2
2.25
2.2361
2.2361
(b)
x2 a
xn n
2 xn
For example, given x1
2,
a·
1§
¨ xn ¸
xn ¹
2©
x2
1§
5·
¨2 ¸
2©
2¹
2.25.
9
4
5 | 2.236
7 : xn 1
1§
7·
¨ xn ¸, x1
2©
xn ¹
2
n
1
2
3
4
5
xn
2
2.75
2.6477
2.6458
2.6458
7 | 2.646
x n a, a ! 0
20. (a) f x
fc x
nx
n 1
xi xi 1
21. y
f xi
f c xi
xi xi a
nxi n 1
3 xi 6
, x1
4 xi 3
yc
6 x 12 x 6
n 1 xi a
x1
1
nxi n 1
fc x
n
n
2 x3 6 x 2 6 x 1
2
f x
fc x
0; therefore, the method fails.
4
(b)
4
4
3
6 : xi 1
i
1
2
3
4
xi
1.5
1.5694
1.5651
1.5651
2 xi 3 15
, x1
3 xi 2
15: xi 1
1
2.5
xi
2
2.4667
2.5
3
4
2.4662
n
xn
f xn
f c xn
1
1
1
0
22. y
6 | 1.565
i
3
1.5
x3 2 x 2, x1
yc
3x 2
x1
0
x2
1
x3
0
x4
1
0
2
and so on.
Fails to converge
2.4662
15 | 2.466
f x x
23. Let g x
gc x
cos x x
sin x 1.
g xn
g xn
n
xn
g xn
g c xn
1
1.0000
–0.4597
–1.8415
0.2496
0.7504
2
0.7504
–0.0190
–1.6819
0.0113
0.7391
3
0.7391
0.0000
–1.6736
0.0000
0.7391
g c xn
xn g c xn
The fixed point is approximately 0.74.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.8
f x x
24. Let g x
gc x
Newt
Newton’s Method
New
321
cot x x
csc 2 x 1.
g xn
g xn
n
xn
g xn
g c xn
1
1.0000
–0.3579
–2.4123
0.1484
0.8516
2
0.8516
0.0240
–2.7668
–0.0087
0.8603
3
0.8603
0.0001
–2.7403
0.0000
0.8603
xn g c xn
g c xn
The fixed point is approximately 0.86.
25.
1
a
x
1
2
x
f x
fc x
§1
·
xn xn 2 ¨
a¸
x
© n
¹
2 xn xn 2 a
xn 2 axn
xn 2 11xn
(b) xn 1
i
1
2
3
4
i
1
2
3
4
xi
0.3000
0.3300
0.3333
0.3333
xi
0.1000
0.0900
0.0909
0.0909
| 0.333
1 | 0.091
11
x3 3x 2 3, f c x
27. f x
xn xn xn 2 a
xn 2 3xn
26. (a) xn 1
1
3
1 xn a
1 xn 2
xn xn 1
0
3x 2 6 x
4
(a)
(d)
y = −3x + 4
y
f
−4
5
3
−2
(b) x1
x2
1
f x1
| 1.333
x1 f c x1
Continuing, the zero is 1.347.
1
(c) x1
4
f x1
| 2.405
x2
x1 f c x1
Continuing, the zero is 2.532.
x
−2
1
4
5
y = −1.313x + 3.156
The x-intercept of y
y
3x 4 is 43 . The x-intercept of
1.313 x 3.156 is approximately 2.405.
The x-intercepts correspond to the values resulting from the
first iteration of Newton's Method.
(e) If the initial guess x1 is not "close to" the desired zero of the
function, the x-intercept of the tangent line may approximate
another zero of the function.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 3
322
Applications
pplications of Differentiation
sin x, f c x
28. f x
(a)
29. Answers will vary. See page 229.
cos x
If f is a function continuous on >a, b@ and differentiable
2
on a, b where c  >a, b@ and f c
−
First, estimate an initial x1 close to c (see graph).
−2
(b) x1
1.8
x2
x1 (c) x1
(d)
y
f x1
f c x1
| 6.086
1
−1
y
c
2 b
x
−1
f x1
| 3.143
f c x1
x1 x
a 3
f(x)
x1
x2
3
x2
0, Newton's
Method uses tangent lines to approximate c such that
f c
0.
−2
y = 0.99x + 3.111
2
y = −0.227x + 1.383
x1 Then determine x2 by x2
1
π
2
x
π
Calculate a third estimate by x3
f x1
f c x1
x2 −1
.
f x2
.
f c x2
Continue this process until xn xn 1 is within the
−2
desired accuracy.
The x-intercept of y
0.227 x 1.383 is
approximately 6.086. The x-intercept of
y
0.99 x 3.111 is approximately 3.143.
Let xn 1 be the final approximation of c.
3 and x
30. At x
The x-intercepts correspond to the values resulting
from the first iteration of Newton's Method.
2, the tangent lines to the curve are
horizontal. Hence, Newton’s Method will not converge
for these initial approximations.
(e) If the initial guess x1 is not "close to" the desired
zero of the function, the x-intercept of the tangent
line may approximate another zero of the function.
31. y
4 x 2 , 1, 0
f x
x 1
d
2
y 0
d is minimized when D
g x
Dc
gc x
12 x 2 14
2
x 1
2
4 x2
2
x 4 7 x 2 2 x 17
x 4 7 x 2 2 x 17 is a minimum.
4 x3 14 x 2
n
xn
g xn
g c xn
g xn
g c xn
1
2.0000
2.0000
34.0000
0.0588
xn y
g xn
g c xn
5
1.9412
(1.939, 0.240)
3
2
2
1.9412
0.0830
31.2191
0.0027
1.9385
3
1.9385
–0.0012
31.0934
0.0000
1.9385
1
−3
−1
−1
(1, 0)
1
x
3
x | 1.939
Point closest to 1, 0 is | 1.939, 0.240 .
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.8
32. Maximize: C
Cc
Newt
Newton’s Method
New
323
3t 2 t
50 t 3
3t 4 2t 3 300t 50
0
2
50 t 3
n
xn
f xn
f c xn
f xn
f c xn
xn f xn
f c xn
1
4.5000
12.4375
915.0000
0.0136
4.4864
2
4.4864
0.0658
904.3822
0.0001
4.4863
3t 4 2t 3 300t 50
Let f x
fc x
12t 3 6t 2 300.
Because f 4
354 and f 5
575, the solution is in the interval 4, 5 .
Approximation: t | 4.486 hours
Distance rowed
Distance walked
Rate rowed
Rate walked
Minimize: T
33.
x2 4
x 2 6 x 10
3
4
x
x 3
3 x2 4
4 x 2 6 x 10
T
Tc
x 2 6 x 10
3 x 3
16 x 2 x 2 6 x 10
9x 3
4x
7 x 4 42 x3 43x 2 216 x 324
2
x2 4
x2 4
0
7 x 4 42 x3 43 x 2 216 x 324 and f c x
Let f x
0
28 x3 126 x 2 86 x 216. Becasuse f 1
f 2
56, the solution is in the interval 1, 2 .
n
xn
f xn
f c xn
f xn
f c xn
xn 1
1.7000
19.5887
135.6240
0.1444
1.5556
2
1.5556
–1.0480
150.2780
–0.0070
1.5626
3
1.5626
0.0014
49.5591
0.0000
1.5626
100 and
f xn
f c xn
Approximation: x | 1.563 mi
34. Set T
300 and obtain the following equation.
0.2988 x 22.625 x3 628.49 x 2 7565.9 x 33,478
300
0.2988 x 22.625 x 628.49 x 7565.9 x 33,178
0
4
4
From the graph, T
3
2
300 when x | 17,and x | 22.
Using Newton’s Method with x1
17, you obtain x
17.2 years.
Using Newton’s Method with x1 | 22, you obtain x | 22.1 years.
35. False. Let f x
f x
x2 1
x 1. x
1 is a discontinuity. It is not a zero of f x . This statement would be true if
p x q x was given in reduced form.
INSTRUCTOR USE ONLY
36. True
3 True
37.
rue
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NOT FOR SALE
Chapter 3
324
Applications
pplications of Differentiation
38. True
40. Let x1 , y1 be the point of tangency.
cos x, f c x
sin x, f c x1
39. f x
sin x
f x
fc x
cos x
At the point of tangency,
y1 0
f c x1
x1 0
x0 , sin x0
Let x0 , y1
be a point on the graph of
f. If x0 , y0 is a point of tangency, then
y0 0
x0 0
cos x0
So, x0
sin x0
y0
x0
x0
sin x1
cos x1 x1
cos x1 x1 sin x1
.
sin x1 .
0
Using Newton's method with initial guess 3, you obtain
x1 | 2.798 and y1 | 0.942.
tan x0 .
x0 | 4.4934
Slope
cos x0 | 0.217
sin x and
You can verify this answer by graphing y1
the tangent line y2
0.217 x.
2
−1
5
−2
Section 3.9 Differentials
1.
f x
x2
fc x
2x
fc 2 x 2
Tangent line at (2, 4): y f 2
x
2.
2
f x
x
T x
4x 4
f x
6
x2
fc x
12 x 3
y 4
4 x 2
y
4x 4
1.9
1.99
2
2.01
2.1
3.6100
3.9601
4
4.0401
4.4100
3.6000
3.9600
4
4.0400
4.4000
6 x 2
12
x3
§ 3·
Tangent line at ¨ 2, ¸ :
© 2¹
y 3
2
y
12
x 2
8
3
9
x 2
2
x
f x
T x
6
x2
3
9
x 2
2
3
x 2
2
1.9
1.99
2
2.01
2.1
1.6620
1.5151
1.5
1.4851
1.3605
1.65
1.515
1.5
1.485
1.35
INSTRUCTOR
NSTR
RUCTOR USE ONLY
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NOT FOR SALE
Section 3.9
3.
f x
x5
fc x
5x4
D
Differentials
325
Tangent line at 2, 32 :
y f 2
fc 2 x 2
y 32
80 x 2
y
80 x 128
x
4.
1.9
1.99
2
2.01
2.1
f x
x5
24.7610
31.2080
32
32.8080
40.8410
T x
80 x 128
24.0000
31.2000
32
32.8000
40.0000
f x
x
fc x
1
1.9
1.99
2
2.01
2.1
1.3784
1.4107
1.4142
1.4177
1.4491
1.3789
1.4107
1.4142
1.4177
1.4496
1.9
1.99
2
2.01
2.1
x
2
Tangent line at 2,
2:
y f 2
fc 2 x 2
y 1
x 2
2 2
1
x
2 2
2
2
y
x
f x
x
x
T x
5.
2
2
f x
sin x
fc x
cos x
1
2
Tangent line at 2, sin 2 :
y f 2
fc 2 x 2
y sin 2
cos 2 x 2
y
cos 2 x 2 sin 2
x
f x
sin x
0.9463
0.9134
0.9093
0.9051
0.8632
T x
cos 2 x 2 sin 2
0.9509
0.9135
0.9093
0.9051
0.8677
INSTRUCTOR USE ONLY
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6.
NOT FOR SALE
Chapter 3
326
Applications
pplications of Differentiation
f x
csc x
fc x
csc x cot x
Tangent line at 2, csc 2 :
y f 2
fc 2 x 2
y csc 2
csc 2 cot 2 x 2
csc 2 cot 2 x 2 csc 2
y
x
7. y
'y
1.9
1.99
2
2.01
2.1
f x
csc x
1.0567
1.0948
1.0998
1.1049
1.1585
T x
csc 2 cot 2 x 2 csc 2
1.0494
1.0947
1.0998
1.1048
1.1501
f x
x3 , f c x
1, 'x
3x 2 , x
f x 'x f x
dx
0.1
f c x dx
dy
f 1.1 f 1
f c 1 0.1
0.331
3 0.1
0.3
8. y
'y
f x
6 2 x2 , f c x
4 x, x
f x 'x f x
2, 'x
dy
f 1.9 f 2
6 2 1.9
2
9. y
'y
f x
f c x dx
0.8
2
0.78
x 4 1, f c x
4 x3 , x
1, 'x
f x 'x f x
dx
0.01
dy
f c x dx
f c 1 0.01
f 0.99 f 1
ª 0.99
¬
10. y
'y
f x
4
0.1
4 2 0.1
6 2 2
1.22 2
dx
1º ª 1
¼ ¬
2 x4 , f c x
4
4 x3 , x
f x 'x f x
2, 'x
dy
f 2.01 f 2
| 14.3224 14
4 0.01
1º | 0.0394
¼
dx
0.04
0.01
f c x dx
4 x3 dx
0.3224
4 2
3
0.01
0.32
11.
12.
y
3x 2 4
dy
6 x dx
y
3x 2 3
dy
2 x 1 3 dx
13.
14.
2
dx
x1 3
y
x tan x
dy
x sec2 x tan x dx
y
csc 2 x
dy
2csc 2 x cot 2 x dx
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 3.9
15.
y
dy
x 1
2x 1
3
dx
2
2x 1
D
Differentials
g 3 0.07 | g 3 g c 3 0.07
24. (a) g 2.93
| 8 3 0.07
| 8 3 0.1
16.
y
dy
17.
y
dy
18.
19.
x 25. x
x 1
dx
2x x
1 ·
§ 1
¨
¸ dx
x
x
x¹
2
2
©
10 in., 'x
dA
1 2
1
9 x2
2 x dx
2
9 x2
x 1 x2
dy
§
x
¨x
1 x2
©
y
3x sin 2 x
dy
3 2 sin x cos x dx
·
1 x 2 ¸ dx
¹
dx
1 2x2
dx
1 x2
26. r
58
100
sec x
x2 1
dy
ª x 2 1 2 sec 2 x tan x sec 2 x 2 x º
«
» dx
2
«
»
2
1
x
¬
¼
dA
A
(a)
0.9
f 2 0.04 | f 2 f c 2 0.04
| 1 1 0.04
1.04
1.05
1
in.
4
2S r dr
§ 1·
2S 16 ¨ r ¸
© 4¹
r 8S in.2
8S
S 16
2
1
32
0.03125
3.125%
1
bh
2
1
1
dA
b dh h db
2
2
1
1
36 r 0.25 50 r 0.25
'A | dA
2
2
r10.75 cm 2
A
dA
A
10.75
| 0.011944
50
1 36
2
1.19%
f 2 0.04 | f 2 f c 2 0.04
| 1 12 0.04
0.98
g 3 0.07 | g 3 g c 3 0.07
| 8 12 0.07
(b) g 3.1
0.625%
(b) Percent error:
f 2 0.1 | f 2 f c 2 0.1
| 1 12 0.1
23. (a) g 2.93
r
0.00625
27. b
36 cm, h
50 cm,
'b
'h
db
dh
r 0.25 cm
f 2 0.1 | f 2 f c 2 0.1
| 1 1 0.1
(b) f 2.04
dr
1
100
(b) Percent error:
ª 2 sec 2 x x 2 tan x tan x x º
«
» dx
2
«
»
2
1
x
¬
¼
22. (a) f 1.9
5 2
in.
8
Sr2
dA
3 sin 2 x dx
5
800
16 in., 'r
(a) A
y
(b) f 2.04
r
(b) Percent error:
2
21. (a) f 1.9
1
in.
32
§ 1·
2 10 ¨ r ¸
© 32 ¹
'A | dA
20.
r
2 xdx
'A | dA
x
dA
A
y
dx
8.3
x2
(a) A
9 x2
7.79
g 3 0.1 | g 3 g c 3 0.1
(b) g 3.1
1
x
327
8.035
g 3 0.1 | g 3 g c 3 0.1
| 8 12 0.1
7.95
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 3
328
28. (a)
Applications
pplications of Differentiation
C
64 cm
'C
dC
C
2S r Ÿ r
A
Sr2
31.
r0.9 cm
C
2S
T
2.5 x 0.5 x 2 , 'x
dT
2.5 x dx
dT
T
Percentage change
2
dA
dA
A
(b)
dA
A
1 2
§C ·
C
¸
4S
© 2S ¹
1
1
C dC
64 r0.9
2S
2S
28.8 S
| 0.028125
2
ª¬1 4S º¼ 64
S¨
ª¬1 2S º¼ C dC
ª¬1 4S º¼ C 2
dC
0.03
d
C
2
29. x
(a)
0.015
15 in., 'x
x3
dV
3 x 2 dx
'V | dV
(b)
6x2
dS
12 x dx
'S | dS
2
3 15
S
S
2.8%
(a)
'r
V
4 3
Sr
3
dV
4S r 2 dr
'V | dV
(b)
S
4S r 2
dS
8S r dr
'S | dS
L g
dL
S dL
g
2S
r 5.4 in.2
(b)
2
E
IR
R
E
I
dR
34.
dR
R
r 5.12S in.3
r 0.02
L g
dT
100
T
0.0025 3600 24
dR
R
4S 8
L g
dL
2L
1
relative error in L
2
1
0.005
0.0025
2
r 20.25 in.3
r 0.02 in.
27.5
| 7.3%
375
Relative error:
dT
T
Percent error of surface area:
dS
5.4
0.004 or 0.4%
2
S
6 15
8 in., dr
g
Percentage error:
12 15 r 0.03
25
L g
S
dT
1.5%
(c) Percent error of volume:
dV
20.25
0.006 or 0.6%
V
153
30. r
2S
T
33. (a)
2 dC
d 0.03
C
r 0.03
1, x
27.5 mi
32. Because the slope of the tangent line is greater at
400, the change in profit is greater
x
900 than at x
at x
900 units.
r 0.03 in.
dx
V
r 28.8
26 25
dx
2.5 25 1
35. R
v0
0.25%
216 sec
1
%
4
3.6 min
E
dI
I2
2
E I dI
E I
dI
I
dI
I
dI
I
v0 2
sin 2T
32
2500 ft/sec
T changes from 10q to 11q.
8S 8 r 0.02
r1.28S in.2
(c) Percent error of volume:
dV
5.12S
0.0075 or 0.75%
4S 8 2
V
dR
T
3
Percent error of surface area:
dS
1.28S
0.005 or 0.5%
2
S
4S 8
dT
v0 2
2 cos 2T dT
32
§ S ·
10¨
¸
© 180 ¹
11 10
2500
16
2
cos 2T dT
S
180
'R | dR
2500
2
16
| 6407 ft
§ 20S ·§ S ·
cos¨
¸¨
¸
© 180 ¹© 180 ¹
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 3.9
50 tan T
36. h
x3 , x
40. Let f x
D
Differentials
0.01.
3, dx
f x 'x | f x f c x dx
f x 'x
3
2.99
Using a calculator: 2.99
θ
50 ft
T
71.5q
dh
50 sec 2T ˜ dT
50 tan 1.2479
dT d 0.06
0.6.
100, dx
f x 'x
2
x
y
100 Using a calculator:
dx
3
x, x
26 | 3 27 6
42.
1
0.6
2 100
9.97
99.4 | 9.96995
1.
27, dx
1
3 3 27 2
tan x
fc x
sec 2 x
f 0
0
fc 0
1
Tangent line at 0, 0 : y 0
1
x
3
f x
dx
3 3 x2
1
3
| 2.9630
27
1
y
f
(0, 0)
Using a calculator, 3 26 | 2.9625
625, dx
1.
f x 'x | f x f c x dx
4
f x 'x
4
x, x
4
4
1
5
500
x 1
4 4 x3
4
625
3
1
dx
y
43. In general, when 'x o 0, dy approaches 'y.
44. Propagated error
f x 'x f x ,
relative error
dy
, and the percent error
y
4.998
Using a calculator, 4 624 | 4.9980.
x
−4
1
624 | 4 625 x 0
4
−
39. Let f x
f
−2
f x 'x | f x f c x dx
3
(0, 2)
−6
99.4
|
38. Let f x
1
1
4
6
f x 'x | f x f c x dx
x | 26.7309
1
x 0
4
1
x 2
4
Tangent line: y 2
y
x, x
26.73
3
2, f c 0
dT d 0.018
37. Let f x
0.01
1
x 4
2
At 0, 2 , f 0
9.9316
dT d 0.06
2.9886
2
x 4
f x
fc x
50 sec2 1.2479
dh
h
41.
1.2479 radians
x3 3 x 2 dx
| 33 3 3
27 0.27
h
329
dy
u 100.
y
INSTRUCTOR USE ONLY
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330
NOT FOR SALE
Chapter 3
Applications
lications of Differentials
45. (a) Let f x
fc x
x, x
4, dx
46. Yes. y
0.02,
x.
1 2
Then
f 4.02 | f 4 f c 4 dx
4.02 |
1
4 2 4
2
0.02
x is the tangent line approximation to
f x
sin x at 0, 0 .
fc x
cos x
fc 0
1
Tangent line: y 0
1
0.02 .
4
1x 0
y
x
47. True
(b) Let
f x
tan x, x
0.05, f c x
0, dx
sec 2 x.
48. True,
Then
f 0.05 | f 0 f c 0 dx
tan 0.05 | tan 0 sec 2 0 0.05
0 1 0.05 .
'y
'x
dy
dx
a
49. True
50. False
Let f x
1, and 'x
dx
3. Then
f x 'x f x
f 4 f 1
1
1
3
2 1
So, dy ! 'y in this example.
3
.
2
'y
x, x
f c x dx
and dy
Review Exercises for Chapter 3
1.
f x
x 2 5 x,
fc x
2x 5
>4, 0@
4. h x
5 2
0 when x
Critical number: x
hc x
5 2
2.
Critical number: 5 2, 25 4
Minimum
Right endpoint: 0, 0
Maximum
f x
x 6 x , > 6, 1@
fc x
3x 2 12 x
3x x 4
0 when x
0, 4
Left endpoint: 6, 0
Minimum
Critical number: 0, 0
Minimum
Critical number: 4, 32
Maximum
Right endpoint: 1, 7
f x
fc x
2
2
x
>0, 9@
1
0 Ÿ 2
3 Ÿ x
x
94
94
Left endpoint: 0, 0
Minimum
Critical number: 9 4, 9 4
Maximum
Right endpoint: 9, 0
Minimum
2
Critical numbers: x
3.
3
Critical number: x
Left endpoint: 4, 4
3
3 x x,
0, 4
5.
f x
fc x
4x
, > 4, 4@
x2 9
x2 9 4 4x 2x
x 9
2
2
0 Ÿ 36 4 x 2
Critical numbers: x
36 4 x 2
x2 9
0 Ÿ x
2
r3
r3
Left endpoint: 4, 16
25
x 2, >0, 4@
Critical number: 3, 23
Minimum
1
Critical number: 3, 23
Maximum
x
Right endpoint: 4, 16
25
No critical numbers on 0, 4
Left endpoint: 0, 2
Minimum
Right endpoint: 4, 0
Maximum
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 3
6.
x
f x
fc x
x 1
2
12. f x
, >0, 2@
x2 1
fc x
2 cos 2 x
3S
S
,r .
4
4
S 3S
c-values: r , r
4
4
Minimum
Right endpoint: 2, 2
5
13. f x
Maximum
2 1/3
x
3
f b f a
4 1
81
b a
2 1 3
3
fc c
c
3
7
2 x 5 cos x, >0, 2S @
2 5 sin x
2.
5
0 when sin x
x2 3, 1 d x d 8
fc x
Critical numbers: x | 0.41, x | 2.73
Left endpoint: 0, 5
Critical number: 2.73, 0.88
Minimum
Right endpoint: 2S , 17.57
Maximum
f x
sin 2 x, >0, 2S @
fc x
2 cos 2 x
§ 14 ·
¨ ¸
©9¹
c
Critical number: 0.41, 5.41
8.
r
0 for x
32
Left endpoint: 0, 0
gc x
0. f is continuous on > S , S @
f S
and differentiable on S , S .
No critical numbers
7. g x
sin 2 x, > S , S @
Yes. f S
3 2
1 2
ª 1
º
x « x 2 1
2 x » x2 1
¬ 2
¼
1
331
0 when x
,
4
,
3
2744
| 3.764
729
1
,1 d x d 4
x
1
2
fc x
x
f b f a
14 1
14. f x
S 3S 5S 7S
4
3
7
4
,
4
.
Left endpoint: 0, 0
b a
1
fc c
c2
c
3 4
3
4 1
1
4
1
4
§S ·
Critical number: ¨ , 1¸
©4 ¹
Maximum
§ 3S
·
Critical number: ¨ , 1¸
4
©
¹
Minimum
15. The Mean Value Theorem cannot be applied. f is not
differentiable at x
5 in >2, 6@.
§ 5S ·
Critical number: ¨ , 1¸
© 4 ¹
Maximum
16. The Mean Value Theorem cannot be applied. f is not
defined for x 0.
§ 7S
·
Critical number: ¨ , 1¸
© 4
¹
Minimum
17. f x
Right endpoint: 2S , 0
fc x
9. No, Rolle's Theorem cannot be applied.
f 0
7 z 25
f 4
10. Yes. f 3
f 2
0. f is continuous on >3, 2@,
2
x cos x, x 3 3x 1
0 for x
1.
3
1 sin c
1
c
0
x2
is not continuous on >2, 2@. f 1
1 x2
is not defined.
11. No. f x
1
2
x
2
1
2
f b f a
b a
fc c
S
x 2 x, 0 d x d 4
18. f x
fc x
c-value: 13
d x d
S 2 S 2
S 2 S 2
b a
differentiable on 3, 2 .
fc x
2
1 sin x
f b f a
fc c
S
6 0
40
1
2
2 c
c
3
2
3
2
1
INSTRUCTOR USE ONLY
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332
NOT FOR SALE
Chapter 3
Applications
lications of Differentials
19. No; the function is discontinuous at x
the interval >2, 1@.
0 which is in
21.
Ax 2 Bx C
20. (a) f x
fc x
f x2 f x1
A x22 x12 B x2 x1
x2 x1
x2 x1
A x1 x2 B
2 Ac B
A x1 x2
32 x f
Sign of f c x :
fc x 0
fc x ! 0
Conclusion:
Decreasing
Increasing
x 2
13
21 1
40
b a
4c 3
5
c
2
2
x 1
5
2
Intervals:
f, 2
2, f
Sign of hc x :
hc x ! 0
hc x ! 0
Conclusion:
Increasing
Increasing
1 and x
x 1 3x 7
7
3
Intervals:
f x 1
1 x 73
7
3
Sign of f c x :
fc x ! 0
fc x 0
fc x ! 0
Conclusion:
Increasing
Decreasing
Increasing
3
3x 1
23
x3
Critical numbers: x
gc x
3x 2
Midpoint of >0, 4@
x 1 1 x 3 2 x 1
x 1
1
h is increasing on f, f .
2
24. g x
8
1
2 3
x 2
3
Critical number: x
4x 3
f b f a
fc x
f x 32
hc x
2 x 3x 1
fc x
32
Intervals:
2
(b) f x
f x
2x 3
22. h x
x1 x2
2
Midpoint of > x1, x2 @
c
23.
fc x
A x1 x2 B
2 Ac
fc c
x 2 3x 12
Critical number: x
2 Ax B
fc c
f x
x f
25. h x
2
Critical number: x
x x 3
x3 2 3 x1 2
Domain: 0, f
1
hc x
Intervals:
f x 1
1 x f
Sign of g c x :
gc x ! 0
gc x ! 0
Conclusion:
Increasing
Increasing
3 3 2 3 1 2
x x
2
2
Critical number: x
3 1 2
x
x 1
2
3 x 1
2
x
1
Intervals:
0 x 1
1 x f
Sign of hc x :
hc x 0
hc x ! 0
Conclusion:
Decreasing
Increasing
INSTRUCTOR USE ONLY
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Review Exercises ffor Chapter 3
26. f x
sin x cos x,
fc x
cos x sin x
0 d x d 2S
S 5S
Critical numbers: x
4
,
4
S
S
4
4
x 5S
4
5S
x 2S
4
Intervals:
0 x Sign of f c x :
fc x ! 0
fc x 0
fc x ! 0
Conclusion:
Increasing
Decreasing
Increasing
x2 6x 5
27. (a) f x
fc x
(b)
333
2x 6
0 when x
3.
Intervals:
f x 3
3 x f
Sign of f c x :
fc x 0
fc x ! 0
Conclusion:
Decreasing
Increasing
(c) Relative minimum: 3, 4
(d)
3
−3
9
−5
28. (a) f x
4 x3 5 x
fc x
12 x 2 5
0 when x
r
5
12
Intervals:
f x 15
6
Sign of f c x :
fc x ! 0
fc x 0
fc x ! 0
Conclusion:
Increasing
Decreasing
Increasing
(b)
r
15
.
6
15
x 6
15
6
15
x f
6
§
15 5 15 ·
(c) Relative maximum: ¨¨ ,
¸
6
9 ¸¹
©
§ 15
5 15 ·
Relative minimum: ¨¨
, ¸
9 ¸¹
© 6
(d)
4
−6
6
−4
INSTRUCTOR USE ONLY
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334
Chapter 3
29. (a) h t
hc t
(b)
Applications
lications of Differentials
1 t 4 8t
4
t 8
3
(c) Relative minimum: 2, 12
0 when t
2.
Intervals:
f t 2
2 t f
Sign of hc t :
hc t 0
hc t ! 0
Conclusion:
Decreasing
Increasing
(d)
10
−2
6
−15
30. (a) g x
1 3
x 8x
4
gc x
3 2
x 2
4
0 Ÿ x2
8
Ÿ x
3
r
Intervals:
f x 2 6
3
Sign of g c x :
Conclusion:
(b)
2 6
3
2 6
2 6
x 3
3
2 6
x f
3
gc x ! 0
gc x 0
gc x ! 0
Increasing
Decreasing
Increasing
§ 2 6 8 6·
(c) Relative maximum: ¨¨ ,
¸
3
9 ¸¹
©
§2 6
8 6·
Relative minimum: ¨¨
, ¸
3
9 ¸¹
©
(d)
6
−9
9
−6
31. (a) f x
fc x
fc x
(b)
x 4
x2
x2 1 x 4 2x
x
4
0 when x
8.
Discountinuity at: x
0
x2 8x
x4
x 8
x3
Intervals:
f x 8
8 x 0
0 x f
Sign of f c x :
fc x 0
fc x ! 0
fc x 0
Conclusion:
Decreasing
Increasing
Decreasing
1
(c) Relative minimum: 8, 16
(d)
8
− 10
5
−2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Review Exercises ffor Chapter 3
32. (a)
f x
fc x
335
x 2 3x 4
x 2
x 2 2 x 3 x 2 3x 4 1
2
x 2
2 x 2 7 x 6 x 2 3x 4
x 2
2
x 2 4 x 10
x 2
2
f c x z 0 since x 2 4 x 10
Discountinuity at: x
(b)
0 has no real roots.
2
Intervals:
f x 2
2 x f
Sign of f c x :
fc x ! 0
fc x ! 0
Conclusion:
Increasing
Increasing
(c) No relative extrema
(d)
6
−8
10
−6
f x
cos x sin x, 0, 2S
fc x
sin x cos x
33. (a)
1
3S
7S
x 4
4
7S
x 2S
4
fc x 0
fc x ! 0
fc x 0
Decreasing
Increasing
Decreasing
Intervals:
0 x Sign of f c x :
Conclusion:
§ 3S
(c) Relative minimum: ¨ , © 4
§ 7S
Relative maximum: ¨ ,
© 4
(d)
sin x Ÿ tan x
3S 7S
,
4 4
Critical numbers: x
(b)
0 Ÿ cos x
3S
4
·
2¸
¹
·
2¸
¹
2
0
2p
−2
INSTRUCTOR USE ONLY
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336
NOT FOR SALE
Chapter 3
Applications
lications of Differentials
g x
3
§S x
·
sin ¨
1¸,
2
© 2
¹
gc x
3§ S ·
§S x
·
1¸
¨ ¸ cos¨
2© 2 ¹
© 2
¹
34. (a)
(b)
>0, 4@
0 x 1
Intervals:
1
0 when x
2
1
S
2
S
2
S
,3 2
S
.
x 3
2
2
3
S
S
x 4
Sign of g c x :
gc x ! 0
gc x 0
gc x ! 0
Conclusion:
Increasing
Decreasing
Increasing
2 3·
§
(c) Relative maximum: ¨1 , ¸
S 2¹
©
2 3·
§
Relative minimum: ¨ 3 , ¸
S 2¹
©
2
(d)
0
4
−2
35.
f x
x3 9 x 2
fc x
3 x 2 18 x
f cc x
6 x 18
0 when x
3.
Intervals:
f x 3
3 x f
Sign of f cc x :
f cc x 0
f cc x ! 0
Conclusion:
Concave downward
Concave upward
Point of inflection: 3, 54
36.
f x
6x4 x2
fc x
24 x3 2 x
f cc x
72 x 2 2
0 Ÿ x2
Ÿ x
r 16
Intervals:
f x 16
16 x 16
1
6
Sign of f cc x :
f cc x ! 0
f cc x 0
f cc x ! 0
Conclusion:
Concave upward
Concave downward
Concave upward
5
Points of inflection: 16 , 216
,
37.
1
36
5
1
, 216
6
x 5, Domain: x t 5
g x
x
gc x
1 2
12
§1·
x 5
x¨ ¸ x 5
© 2¹
g cc x
2
x f
x 5 3 3 x 10 x 5
4x 5
1
1 2
x 5
x 2x 5
2
1 2
6 x 5 3 x 10
4x 5
32
3x 10
2 x 5
3x 20
4x 5
32
! 0 on 5, f .
Concave upward on 5, f
INSTRUCTOR USE ONLY
No point of inflection
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Review Exercises ffor Chapter 3
38.
f x
3x 5 x3
fc x
3 15 x 2
f cc x
30 x
0 when x
337
0.
Intervals:
f x 0
0 x f
Sign of f cc x :
f cc x ! 0
f cc x 0
Conclusion:
Concave upward
Concave downward
Point of inflection: 0, 0
39.
f x
x cos x, 0 d x d 2S
fc x
1 sin x
f cc x
cos x
S 3S
0 when x
, .
2 2
S
S
2
2
x 3S
2
3S
x 2S
2
Intervals:
0 x Sign of f cc x :
f cc x 0
f cc x ! 0
f cc x 0
Conclusion:
Concave downward
Concave upward
Concave downward
§ S S · § 3S 3S ·
Points of inflection: ¨ , ¸, ¨ ,
¸
©2 2¹ © 2 2 ¹
40.
41.
42.
x
tan , 0, 2S
4
x
1
sec 2
fc x
4
4
x ·§ 1 ·
1 § 2 x
2 ¨ sec
tan ¸¨ ¸
f cc x
4 ©
4
4 ¹© 4 ¹
x
1
2 x
sec
tan ! 0 on 0, 2S .
8
4
4
Concave upward on 0, 2S
43.
No point of inflection
44.
f x
x 9
fc x
2x 9
f cc x
2 ! 0 Ÿ 9, 0 is a relative minimum.
f x
2 x3 11x 2 8 x 12
fc x
6x2
22 x 8
Critical numbers: x
f cc x
0 Ÿ x
9
4 x 2 x 2 1
g cc x
4 24 x 2
g cc 0
4 ! 0
0 Ÿ x
0, r
1
2
0, 0 is a relative minimum.
§ 1 1·
, ¸ are relative maxima.
8 0 ¨ r
2 2¹
©
ht
t 4 t 1, Domain: >1, f@
hc t
1
hcc t
45.
4, 13
2
t 1
0 Ÿ t
3
1
t 1
hcc 3
1
! 0
8
f x
2x fc x
f cc 4 0 Ÿ 4, 68 is a relative maximum.
1 , 361
3
27
gc x
3/ 2
3, 5 is a relative minimum.
2 x 4 3x 1
12 x 22
f cc 13 ! 0 Ÿ
2 x2 1 x2
§ 1 ·
g cc¨ r
¸
2¹
©
2
f x
g x
is a relative minimum.
18
x
18
2 2
x
Critical numbers: x
0 Ÿ 2 x2
18 Ÿ x
r3
r3
36
x3
f cc 3 0 Ÿ 3, 12 is a relative maximum.
f cc x
INSTRUCTOR USE ONLY
f cc 3 ! 0 Ÿ 3, 12 is a relative minimum.
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338
NOT FOR SALE
Chapter 3
Applications
lications of Differentials
46. h x
x 2 cos x,
hc x
1 2 sin x
Critical numbers: x
hcc x
1
2
7S 11S 19S 23S
,
,
,
6 6
6
6
0 Ÿ sin x
2 cos x
§ 7S ·
hcc¨ ¸
© 6 ¹
§ 11S ·
hcc¨
¸
© 6 ¹
§ 19S ·
hcc¨
¸
© 6 ¹
§ 23S ·
hcc¨
¸
© 6 ¹
47.
>0, 4S @
§ 7S 7S
3 0 Ÿ ¨ ,
© 6 6
§ 11S 11S
3 ! 0 Ÿ ¨
,
6
© 6
·
3 ¸ | 3.665, 5.397 is a relative maximum
¹
·
3 ¸ | 5.760, 4.028 is a relative minimum.
¹
§ 19S 19S
·
3 0 Ÿ ¨
3 ¸ | 9.948, 11.680 is a relative maximum
,
6
© 6
¹
§ 23S 23S
·
3 ¸ | 12.043, 10.311 is a relative minimum.
3 ! 0 Ÿ ¨
,
6
© 6
¹
51. (a)
y
7
0.00188t 4 0.1273t 3 2.672t 2 7.81t 77.1,
D
6
(5, f(5))
5
0 d t d 40
4
(3, f(3))
3
(b)
2
1
800
(6, 0)
x
−1
(0, 0) 2 3 4 5
7
y
48.
0
40
0
7
6
(c) Maximum occurs at t
5
3
(d) Dc t is greatest at t
2
1
40 (2010).
x
−1
1
2
3
4
5
6
7
0.1222t 3 3.199t 2 23.73t 58.8,
52. (a) S
49. The first derivative is positive and the second derivative
is negative. The graph is increasing and is concave
down.
50.
40 (2010).
Minimum occurs at t | 1.6 (1970).
4
C
dC
dx
Qs
x2
x2
x
§Q·
§ x·
¨ ¸ s ¨ ¸r
©x¹
©2¹
Qs
r
2 0
x
2
r
2
2Qs
r
2Qs
r
6 d t d 12
(b)
25
6
12
0
(c) S c t
S cc t
0.3666t 2 6.398t 23.73
0.7332t 6.398
0
when t | 8.7 year 2008 .
(d) No. The coefficient of t 3 is negative and therefore
eventually falls to the right.
·
§
53. lim ¨ 8 ¸
x o f©
x¹
54. lim
8 0
1 4x
x o f x 1
lim
8
1x 4
x o f 1 4 x
4
INSTRUCTOR USE ONLY
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NOT FOR SALE
Review Exercises ffor Chapter 3
55. lim
2 x2
5
2
lim
x o f 3x 2
x o f 3 5 x2
4 x3
56. lim 4
xof x 3
4 x
2
3
2x 3
x 4
65. h x
Discontinuity: x
lim
3x 2
x o f x 5
x o f
lim
2 3x
xof 1 4
Horizontal asymptote: y
12
12
0, because 5 cos x d 5.
−4
x3
lim
xof
x 2
2
x3
3x
66. f x
x 1 2 x
lim
x2
1 2 x2
x2 2
3x
2
lim
f
xof
6x
x 2
x
x o f 2 sin x
63. f x
x 2
2
3x
lim
x o f
x 2
lim
x o f
Horizontal asymptotes: y
3
3x x
lim
does not exist.
3
2
x
1 2 x2
x o f
2
x2
3
lim
xof
6
x o f x cos x
3x x
lim
xof
2
Limit does not exist.
62. lim
2
y=2
xof
61. lim
2
4 x
Vertical asymptote: x
−6
5 cos x
xof
x
xof
2x 3
4
8
x2 x
2 x
59. lim
60. lim
lim
xof x 4
f
57. lim
58. lim
0
x o f1 3 x 4
339
x 2 2
3
1 2 x2
x2
3
r3
4
Discontinuity: x
0
y=3
§3
·
lim ¨ 2 ¸
x o f© x
¹
−6
2
Vertical asymptote: x
y = −3
−4
0
Horizontal asymptote: y
6
2
4 x x2
67. f x
3
x4 x
Domain: f, f ; Range: (f, 4]
−5
5
y = −2
fc x
4 2x
f cc x
2
0 when x
2.
−7
64. g x
Therefore, 2, 4 is a relative maximum.
5x2
2
x 2
5x2
xof x 2
lim
2
Intercepts: 0, 0 , 4, 0
lim
x o f1 5
2 x2
y
5
5
(2, 4)
Horizontal asymptote: y
5
4
3
10
2
y=5
1
(0, 0)
(4, 0)
x
1
−9
2
3
5
9
INSTRUCTOR USE ONLY
−2
−
2
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340
NOT FOR SALE
Chapter 3
Applications
lications of Differentials
4 x3 x 4
68. f x
x3 4 x
Domain: f, f ; Range: f, 27
fc x
x
12 x 2 4 x3
2
x2 4
70. f x
Domain: f, f ; Range: [0, f)
4x2 3 x
0 when
12 x 2 x
0 when
fc x
4x x2 4
0 when x
0, r 2.
f cc x
4 3x 2 4
0 when x
r
0, 3.
f cc x
x
24 x 12 x 2
2 3
.
3
f cc 0 0
0, 2.
f cc 3 0
Therefore, 0, 16 are relative maximum.
Therefore, 3, 27 is a relative maximum.
f cc r 2 ! 0
Points of inflection: 0, 0 , 2, 16
Therefore, r 2, 0 are relative minima.
Intercepts: 0, 0 , 4, 0
Points of inflection: r 2 3 3, 64 9
Intercepts: 2, 0 , 0, 16 , 2, 0
y
30
(3, 27)
Symmetry with respect to y-axis
25
20
y
(2, 16)
15
24
10
20
(0, 0) 5
(4, 0)
(0, 16)
x
−2
1
2
3
5
(−2, 0)
(2, 0)
8
4
x 16 x 2
69. f x
x
−3 −2 −1
Domain: >4, 4@; Range: >8, 8@
fc x
0 when x
16 x 2
r 4.
undefined when x
r 2 2 and
f cc x
16 x 2
32
3
23
Domain: f, f ; Range: f, f
x 1
fc x
2 x x 2 24
2
x1 3 x 3
71. f x
16 2 x 2
1
x
2
f cc x
Therefore, 2 2, 8 is a relative minimum.
f cc 2 2 0
x
53
x 3
43
is undefined when x
0, 3.
By the First Derivative Test 3, 0 is a relative
maximum and 1, 3 4 is a relative minimum. (0, 0)
Therefore, 2 2, 8 is a relative maximum.
is a point of inflection.
Point of inflection: 0, 0
Intercepts: 3, 0 , 0, 0
Intercepts: 4, 0 , 0, 0 , 4, 0
y
4
Symmetry with respect to origin
3
y
8
1 and
3, 0.
undefined when x
f cc 2 2 ! 0
0 when x
13 23
x 3
2
) 2 2, 8 )
1
(− 3, 0)
6
−5 −4
(0, 0)
−2 −1
x
1
2
4
2
(− 4, 0)
−8 −6
−2
(− 1, − 1.59)
(4, 0)
2
4
6
x
8
−3
(0, 0)
−8
)−2 2, −8 )
INSTRUCTOR USE ONLY
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NOT FOR SALE
Review Exercises ffor Chapter 3
x 3 x 2
72. f x
3
16,875
fc x
x 3 3 x 2
4x 7 x 2
f cc x
2
x 2
2
Domain: f, f ; Range: >1, 1@
3
2, 74 .
0 when x
4x 7 2 x 2 x 2
6 2x 1 x 2
0 when x
2
4
2, 12 .
16,875
7
, 256
4
is a relative minimum.
Points of inflection: 2, 0 , 12 , 625
16
Intercepts: 2, 0 , 0, 24, 3, 0
(3, 0)
(−2, 0)
−2
2
( 74 , − 16.875
256 )
3 2
0
x 2
x
−1
( 53 , 0 (
x
2
3
4
5
x3 x 4
x
Domain: f, 0 , 0, f ; Range: (f, 6], [6, f )
fc x
3x2 1 2
x=2
1
3
−2
75. f x
3
5·
§5 · §
Intercepts: ¨ , 0 ¸, ¨ 0, ¸
3
2¹
©
¹ ©
2
2
−3
Horizontal asymptote: y
y
1
(−1, −1)
! 0 for all x z 2
2
2
x 2
(1, 1)
1
Vertical asymptote: x
−6
3,
2
5 3x
x 2
1
1
3 2 , 0, 0 ,
3
Concave downward on 2, f
−5
Therefore, 1, 1 is a relative minimum.
Symmetric with respect to the origin
Concave upward on f, 2
−4
f cc 1 ! 0
y
f cc x
−2
Therefore, (1, 1) is a relative maximum.
Horizontal asymptote: y
− 60
−2 −1
3.
Intercept: (0, 0)
− 40
fc x
0, r
4
(0,−24)
f x
0 when x
3
1 x2
3, x
− 20
73.
4 x 3 x 2
f cc x
r1.
0 when x
2
1 x2
Points of inflection:
y
−4
21 x 1 x
fc x
f cc 1 0
f cc 74 ! 0
Therefore,
2x
1 x2
74. f x
Domain: f, f ; Range: ª 256 , f
¬
341
4
x2
3x 2 4 x 2 1
3x 4 x 2 4
x2
3
when x
r1.
f cc x
6x 8
x3
0
x2
6x4 8
z 0
x3
f cc 1 0
6
(0, − 52 ( y = −3
Therefore, 1, 6 is a relative maximum.
f cc 1 ! 0
y
Therefore, (1, 6) is a relative
minimum.
Vertical asymptote: x
10
5
(1, 6)
0
Symmetric with respect to
origin
−2
−1
(−1, −6) − 5
x
1
2
x=0
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 3
Applications
lications of Differentials
x3 1
x
1
x
x2 76. f x
x2
y2
144 16
78. Ellipse:
Domain: f, 0 , 0, f ; Range: f, f
fc x
2x 1
x2
2 x3 1
x2
f cc x
2
2
x3
2 x3 1
x3
0 when x
3
0 when x
1
.
2
A
§2
·
2x ¨
144 x 2 ¸
©3
¹
dA
dx
x2
4ª
«
3 ¬ 144 x 2
º
144 x 2 »
¼
4 ª 144 2 x 2 º
«
»
3 ¬ 144 x 2 ¼
0 when x
1.
§ 1 ·
f cc¨ 3 ¸ ! 0
© 2¹
3 ·
§ 1
Therefore, ¨ 3 , 3 ¸ is a relative minimum.
4¹
© 2
1
144 x 2
3
1, y
4
x 144 x 2
3
The dimensions of the rectangle are 2 x
2
y
144 72
4 2.
3
72
6 2.
12 2 by
y
Point of inflection: 1, 0
12
8
Intercept: 1, 0
( x,
1
3
144 − x 2
(
x
Vertical asymptote: x
−12
0
12
−8
y
−12
3
2
1
(−1, 0)
( 12 , 34 )
3
79. You have points 0, y , x, 0 , and (1, 8). So,
3
x
−3 −2
1
2
y 8
0 1
3
m
0 8
or y
x 1
8x
.
x 1
2
L2
§ 8x ·
x2 ¨
¸ .
© x 1¹
fc x
§ x ·ª x 1 x º
2 x 128¨
»
¸«
2
© x 1 ¹«¬ x 1 »¼
Let f x
77. 4 x 3 y
400 is the perimeter.
§ 400 4 x ·
2 x¨
¸
3
©
¹
A
2 xy
dA
dx
8
100 2 x
3
d2A
dx 2
0 when x
16
0 when x
3
A is a maximum when x
8
100 x x 2
3
200
ft.
3
y
0 when x
0, 5 (minimum).
y
8
x
0
3
Vertices of triangle: (0, 0), (5, 0), (0, 10)
10
x
x 1
3
x ª x 1 64º
¬
¼
50.
50 ft and y
64 x
x
50.
0
(0, y)
(1, 8)
6
4
2
(x, 0)
x
2
4
6
8
10
INSTRUCTOR USE ONLY
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NOT FOR SALE
Review Exercises ffor Chapter 3
343
80. You have points 0, y , x, 0 , and (4, 5). So,
y 5
0 4
m
50
or y
4 x
§ 5x ·
x2 ¨
¸
© x 4¹
L2
Let f x
fc x
100 x
x xª x 4
¬
x 4
3
x2 L
5x
.
x 4
3
2
(0, y)
0
5
(x, 0)
4
0 when x
25 x 2
x
x 4
x 4
(4, 5)
0
100º
¼
2
L
§ x ·ª x 4 x º
2 x 50¨
»
¸«
2
© x 4 ¹«¬ x 4 »¼
4 3 100.
0 or x
x 4
2
25
3
100 4
100 2 3 25 | 12.7 ft
3
100
81. You can form a right triangle with vertices (0, 0), ( x , 0)
and (0, y ). Assume that the hypotenuse of length L
passes through (4, 6).
82.
csc T
sec T
(0, y)
L
dL
dT
(4, 6)
L
(0, 0)
m
60
or y
4 x
6x
x 4
3
tan 3 T
2
Ÿ tan T
3
sec T
1 tan 2 T
csc T
2
fc x
6 csc T cot T 9 sec T tan T
3
(x, 0)
y 6
0 4
Let f x
L1
or L1 6 csc T
see figure
6
L2
or L2 9 sec T
9
L1 L2 6 csc T 9 sec T
L2
x2 y 2
§ 6x ·
x2 ¨
¸ .
© x 4¹
§ x · ª 4 º
»
2 x 72¨
¸«
2
© x 4 ¹«¬ x 4 ¼»
3
x ª x 4 144º 0 when x
¬
¼
L | 14.05 ft
0 or x
L
0
4
3
6
sec T
tan T
32 3 22 3
2
3
§2·
1 ¨ ¸
©3¹
23
32 3 22 3
31 3
32 3 22 3
21 3
12
9
21 3
3 32 3 22 3
0
32
32 3 22 3
12
31 3
ft | 21.07 ft
144.
L1
θ
L2
θ
9
6
(π2 − θ(
INSTRUCTOR USE ONLY
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NOT FOR SALE
344
Chapter 3
Applications
lications of Differentials
83.
V
1 2
Sx h
3
dV
dx
1 ª
S«
3 ¬
1 2
Sx r 3
x3
3 r 2 x2
2r 2 2r
2x r r 2 x2
Sx
2 r 2 2r
r 2 x 2 3x 2
2r
r 2 x2
r x
2
2
4r 2 r 2 x 2
84.
see figure
º
r 2 x2 »
¼
r 2 x 2 3x 2
V
S x2 2 r 2 x2
2S x 2
r 2 x 2 see figure
1 2
ª §1·
º
2S « x 2 ¨ ¸ r 2 x 2
2 x 2 x r 2 x 2 »
¬ © 2¹
¼
dV
dx
0
S x 2h
2S x
r 2 x2
0
3 x 2 2r 2
0 when x
9 x 4 12 x 2 r 2 4r 4
0
9x 8x r
x
2 2r
0,
3
4
x 9 x 8r
2 2
2
2
2r 2 3 x 2
0 and x 2
(x,
r2 − x2
(
(x, −
r2 − x2
(
2r 2
Ÿ x
3
6r
.
3
2
x
r
h
(0, r)
h
x
By the First Derivative Test, the volume is a maximum
6r
2r
and h
.
when x
3
3
r
(x, −
r 2 − x2
(
By the First Derivative Test, the volume is a maximum
when
x
2 2r
and h
3
r r 2 x2
Thus, the maximum volume is
4S r 3
§ 2 ·§ 2r ·
S ¨ r 2 ¸¨
V
.
¸
3 3
© 3 ¹© 3 ¹
4r
.
3
Thus, the maximum volume is
V
1 § 8r 2 ·§ 4r ·
S¨
¸¨ ¸
3 © 9 ¹© 3 ¹
32S r 3
cubic units.
81
INSTRUCTOR USE ONLY
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Review Exercises ffor Chapter 3
345
x3 3x 1
85. f x
From the graph you can see that f x has three real zeros.
fc x
3x 2 3
n
xn
f xn
f c xn
1
–1.5000
0.1250
3.7500
0.0333
–1.5333
2
–1.5333
–0.0049
4.0530
–0.0012
–1.5321
n
xn
f xn
f c xn
f xn
xn 1
–0.5000
0.3750
–2.2500
–0.1667
–0.3333
2
–0.3333
–0.0371
–2.6667
0.0139
–0.3472
3
–0.3472
–0.0003
–2.6384
0.0001
–0.3473
n
xn
f xn
f c xn
1
1.9000
0.1590
7.8300
0.0203
1.8797
2
1.8797
0.0024
7.5998
0.0003
1.8794
f xn
xn f c xn
f c xn
f xn
f c xn
xn f xn
f c xn
f xn
f c xn
f xn
f c xn
The three real zeros of f x are x | 1.532, x | 0.347, and x | 1.879.
x3 2 x 1
86. f x
From the graph, you can see that f x has one real zero.
fc x
3x 2 2
f changes sign in >1, 0@.
f xn
f xn
n
xn
f xn
f c xn
1
–0.5000
–0.1250
2.7500
–0.0455
–0.4545
2
–0.4545
–0.0029
2.6197
–0.0011
–0.4534
f c xn
xn f c xn
The real zero of f x is: x | 0.453.
INSTRUCTOR USE ONLY
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346
NOT FOR SALE
Chapter 3
87. f x
Applications
lications of Differentials
x 4 x3 3x 2 2
From the graph you can see that f x has two real zeros.
fc x
4 x3 3x 2 6 x
n
xn
f xn
f c xn
1
2.0
2.0
8.0
0.25
2.25
2
2.25
1.0508
16.875
0.0623
2.1877
3
2.1877
0.0776
14.3973
0.0054
2.1823
4
2.1823
0.0004
14.3911
0.00003
2.1873
n
xn
f xn
f c xn
f xn
f c xn
1
1.0
1.0
5.0
0.2
0.8
2
0.8
0.0224
4.6720
0.0048
0.7952
3
0.7952
0.00001
4.6569
0.0000
0.7952
f xn
f xn
xn f c xn
xn f c xn
f xn
f c xn
The two zeros of f x are x | 2.1823 and x | 0.7952.
88. f x
3 x 1 x
From the graph you can see that f x has two real zeros.
2
3
1
x 1
n
xn
f xn
f c xn
1
1.1
0.1513
3.7434
0.0404
1.1404
2
1.1404
0.0163
3.0032
0.0054
1.1458
3
1.1458
0.0003
2.9284
0.0000
1.1459
n
xn
f xn
f c xn
1
8.0
0.0627
0.4331
0.1449
7.8551
2
7.8551
0.0004
0.4271
0.0010
7.8541
fc x
f xn
f c xn
f xn
f c xn
xn xn f xn
f c xn
f xn
f c xn
The two zeros of f x are x | 1.1459 and x | 7.8541.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Review Exercises ffor Chapter 3
89. h x
347
f x g x
1 x x5 2
x5 x 1
hc x
5x4 1
From the graph you can see that there is one point of intersection. That is, h x has one real zero.
n
xn
f xn
f c xn
f xn
f c xn
1
1.0
1.0
6.0
0.1667
0.8333
2
0.8333
0.2351
3.4109
0.0689
0.7644
3
0.7644
0.0254
2.7071
0.0094
0.7550
4
0.7550
0.0003
2.6246
0.0001
0.7549
xn f xn
f c xn
The point of intersection is x | 0.7549.
90. h x
f x g x
sin x x 2 2 x 1
sin x x 2 2 x 1
From the graph you can see that there are two points of intersection. That is, h x has two real zeros.
hc x
cos x 2 x 2
n
xn
f xn
f c xn
f xn
f c xn
1
0.3
0.1945
2.3553
0.0826
0.3826
2
0.3826
0.0078
2.1625
0.0036
0.3862
n
xn
f xn
f c xn
1
2.0
0.0907
2.4161
0.0375
1.9625
2
1.9625
0.0021
2.3068
0.0009
1.9616
xn f xn
f c xn
xn f xn
f c xn
f xn
f c xn
The two points of intersection are x | 0.3862 and x | 1.9616.
91. y
'y
0.5 x 2 , f c x
x, x
3, 'x
f x 'x f x
dy
f c x dx
f x
f 3.01 f 3
f c 3 dx
4.53005 4.5
3 0.01
0.03005
0.03
dx
0.01
INSTRUCTOR USE ONLY
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348
NOT FOR SALE
Chapter 3
92. y
Applications
lications of Differentials
x 3 6 x, f c x
f x
'y
3 x 2 6, x
f x 'x f x
94.
dx
0.1
f c x dx
dy
f c 2 dx
f 2.1 f 2
93.
2, 'x
3.339 4
6 0.01
0.661
0.06
y
x 1 cos x
x x cos x
dy
dx
dy
1 x sin x cos x
95. r
(a)
1 x sin x cos x dx
x
dy
36 x 2
(b)
x
1 2
1
36 x 2
2 x
2
dy
dx
V
4 3
Sr
3
dV
4S r 2 dr
4S 9
'V | dV
36 x 2
y
'r
9 cm, dr
36 x 2
S
4S r 2
dS
8S r dr
2
r 0.025
8S 9 r 0.025
'S | dS
dx
r 0.025
r 8.1S cm3
r1.8S cm 2
(c) Percent error of volume:
dV
8.1S
0.0083, or 0.83%
4S 9 3
V
3
Percent error of surface area:
dS
1.8S
0.0056, or 0.56%
2
S
4S 9
96.
p
75 14 x
'p
p8 p7
75 84 75 74
dp
14 dx
14
>'p
dp because p is linear.@
14 1
14
Problem Solving for Chapter 3
x 4 ax 2 1
1. p x
(a)
pc x
4 x3 2ax
pcc x
12 x 2 2a
2 x 2 x2 a
For a t 0, there is one relative minimum at 0, 1 .
(b) For a 0, there is a relative maximum at 0, 1 .
(c) For a 0, there are two relative minima at x
r
a
.
2
(d) If a 0, there are three critical points; if a ! 0, there is only one critical point.
a=1 a=3 y
a=2
a=0
8
7
6
5
4
3
2
a = −1
a = −2
a = −3
−2
−1
−2
x
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Problem Solving ffor Chapter 3
3, 2, 1, 0, p has a relative maximum at 0, 0 .
2. (a) For a
For a
1, 2, 3, p has a relative maximum at 0, 0 and 2 relative minima.
pc x
4ax3 12 x
4 x ax 2 3
pcc x
12ax 2 12
12 ax 2 1
For x
0, pcc 0
(b)
(c) If a ! 0, x
§
pcc¨¨ r
©
r
3·
¸
a ¸¹
Let x
r
So, y
§
3¨¨ r
©
9
a
3·
¸
a ¸¹
9 18
a
a
9
.
a
2
3 x 2 is satisfied by all the relative extrema of p.
a=1
x
1
a = −1
a = −3
−8
2
3
a = −2
a=0
f x
c
x2
x
fc x
If c
2
§ 3·
§3·
a¨ ¸ 6¨ ¸
©a¹
©a¹
a=3
−3
f cc x
24 ! 0 Ÿ p has relative minima for a ! 0.
3x 2 .
2
3.
3
a
0, r
3
are the remaining critical numbers.
a
3
. Then p x
a
y
a=2
0 Ÿ x
12 0 Ÿ p has a relative maximum at 0, 0 .
§ 3·
12a¨ ¸ 12
©a¹
0, 0 lies on y
(d)
349
c
2x
x2
2c
2
x3
0 Ÿ
c
x2
2 x Ÿ x3
c
Ÿ x
2
3
c
2
x 2 has a relative minimum, but no relative maximum.
0, f x
If c ! 0, x
3
§ c·
c
is a relative minimum, because f cc¨ 3 ¸ ! 0.
¨ 2¸
2
©
¹
If c 0, x
3
c
is a relative minimum, too.
2
Answer: All c.
INSTRUCTOR USE ONLY
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350
Chapter 3
NOT FOR SALE
Applications
lications of Differentials
f x
ax 2 bx c, a z 0
fc x
2ax
f cc x
2a z 0
4. (a)
No point of inflection
f x
ax3 bx 2 cx d , a z 0
fc x
3ax 2 2bx c
f cc x
6ax 2b
(b)
b
3a
0 Ÿ x
One point of inflection
y·
§
ky¨1 ¸
L¹
©
yc
(c)
ky 2k
yyc
L
ycc
kyc If y
L
, then ycc
2
5. Set
k 2
y
L
2 ·
§
kyc¨1 y ¸
L ¹
©
L
2
2
k.
2
f x f a fc a x a k x a .
Define F x
F a
y=
0, and this is a point of inflection because of the analysis above.
f b f a fc a b a
b a
+ + + + + + − − − − −
y″
0, F b
f b f a fc a b a k b a
2
0
F is continuous on >a, b@ and differentiable on a, b .
There exists c1 , a c1 b, satisfying F c c1
0.
Fc x
f c x f c a 2k x a satisfies the hypothesis of Rolle’s Theorem on >a, c1@:
Fc a
0, F c c1
0.
There exists c2 , a c2 c1 satisfying F cc c2
Finally, F cc x
k
So, k
f cc c2
2
f cc x 2k and F cc c2
0.
0 implies that
.
f b f a fc a b a
b a
2
f cc c2
2
Ÿ f b
f a fc a b a 1
2
f cc c2 b a .
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Problem Solving ffor Chapter 3
351
6.
d
5
13
θ
12
x
.
d
132 x 2 , sin T
d
Let A be the amount of illumination at one of the corners, as indicated in the figure. Then
kI
sin T
132 x 2
A
x 2 169
Ac x
32
kI
kIx
13 x 2
2
12
§ 3·
1 x¨ ¸ x 2 169
2x
© 2¹
3
169 x 2
Ÿ x 2 169
32
3x 2 x 2 169
x 2 169
3x 2
2x2
169
32
0
12
13
| 9.19 ft
2
x
By the First Derivative Test, this is a maximum.
42 x 2 7. Distance
x
fc x
x
4 x
4 x
2
2
2
42
x 2 ª¬16 8 x x 2 16º¼
32 x 2 8 x3 x 4
128 x
x
4 x
2
42
f x
4 x
4 x
x 4
2
0
42
42 x 2
x 2 8 x 16 16 x 2
x 4 8 x3 32 x 2 128 x 256
256
2
The bug should head towards the midpoint of the opposite side.
Without Calculus: Imagine opening up the cube:
P
x
Q
The shortest distance is the line PQ, passing through the midpoint.
INSTRUCTOR USE ONLY
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352
Chapter 3
NOT FOR SALE
Applications
lications of Differentials
8. Let T be the intersection of PQ and RS. Let MN be the perpendicular to SQ and PR passing through T.
Let TM
x and TN
b x.
MR
b x
Ÿ SN
MR
x
x
PM
b x
Ÿ NQ
PM
x
x
b x
b x
MR PM
d
x
x
SN
b x
NQ
b x
SQ
Ax
2
b x º
1 ª
»
d «x 2 «
x
»¼
¬
1
1§ b x ·
dx ¨
d¸ b x
2
2© x
¹
Area
1 ª 2 x 2 2bx b 2 º
d«
»
2 ¬
x
¼
2
2
1 ª x 4 x 2b 2 x 2bx b º
»
d«
x2
2 «
»¼
¬
0 Ÿ 4 x 2 2 xb
2 x 2 2bx b 2
Ac x
Ac x
2x2
b2
x
b
2
b b
b x
d
x
So, you have SQ
b
2
2
d
2 1 d.
Using the Second Derivative Test, this is a minimum. There is no maximum.
S
Q
N
b−x
b
T
x
P
d
M
R
9. f continuous at x
10. f continuous at x
1: a
f continuous at x
0: 2
f continuous at x
1: b 2
0
f differentiable at x
0: 0
0
0 x d1
f differentiable at x
1: 2b
d
1 x d 3
So, b
b
0: 1
f continuous at x
1: a 1
5 c
f differentiable at x
1: a
2 4
­1,
°
®6 x 1,
°x 2 4 x 2,
¯
­6 x 1,
® 2
¯x 4 x 2,
f x
11. Let h x
x
6. So, c
2.
0 d x d1
2 and d
2
c
d 4 Ÿ b
d 2
4.
1 x d 3
g x f x , which is continuous on >a, b@ and differentiable on a, b . h a
0 and h b
g b f b.
By the Mean Value Theorem, there exists c in a, b such that
hb ha
hc c
b a
Because hc c
g b f b
.
b a
g c c f c c ! 0 and b a ! 0, g b f b ! 0 Ÿ g b ! f b .
y
g
f
INSTRUCTOR
NST
USE ONLY
x
a
b
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffor Chapter 3
12. (a) Let M ! 0 be given. Take N
M . Then whenever x ! N
353
M , you have
x ! M.
2
f x
1
(b) Let H ! 0 be given. Let M
H
. Then whenever x ! M
1
H
, you have x 2 !
1
H
Ÿ
1
1
H Ÿ 2 0 H.
2
x
x
(c) Let H ! 0 be given. There exists N ! 0 such that f x L H whenever x ! N .
1
1
1
Ÿ x ! N and f x L
, then N
x
N
If 0 y G
13.
1 x2
y
§1·
f ¨ ¸ L H.
© y¹
1
2 x
yc
1 x2
2
2 3x 2 1
ycc
y :
1
.
y
1
. Let x
N
Let G
x 1
2
0 Ÿ x
3
1
3
r
r
3
3
+++ −−−− −−−− +++
0
3
3
−
3
3
§
§ 3 3·
3 3·
The tangent line has greatest slope at ¨ ¨ 3 , 4 ¸¸ and least slope at ¨¨ 3 , 4 ¸¸.
©
¹
©
¹
v
14. (a) s
km §
m·
¨1000
¸
h ©
km ¹
sec ·
§
¨ 3600 ¸
h ¹
©
5
v
18
v
20
40
60
80
100
s
5.56
11.11
16.67
22.22
27.78
d
5.1
13.7
27.2
44.2
66.4
d s
0.071s 2 0.389 s 0.727
(b) The distance between the back of the first vehicle and the front of the second vehicle is d s , the safe stopping distance.
The first vehicle passes the given point in 5.5/s seconds, and the second vehicle takes d s s more seconds. So,
T
d s
s
5.5
.
s
10
(c)
1
5.5
0.071s 2 0.389 s 0.727 s
s
The minimum is attained when s | 9.365 m/sec.
T
0
30
0
(d) T s
Tc s
0.071s 0.389 0.071 6.227
s
6.227
Ÿ s2
s2
(e) d 9.365
10.597 m
6.227
Ÿ s | 9.365 m/sec
0.071
T 9.365 | 1.719 seconds
3600
1000
INSTRUCTOR USE ONLY
9.365 m/sec ˜
33.7 km
km/h
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354
NOT FOR SALE
Chapter 3
Applications
lications of Differentials
f x d x a . g is continuous on >a, b@ and therefore has a minimum c, g c on
15. Assume y1 d y2 . Let g x
>a, b@. The point c cannot be an endpoint of >a, b@ because
gc a
fc a d
y1 d 0
gc b
fc b d
y2 d ! 0.
So, a c b and g c c
16. The line has equation
0 Ÿ fc c
x
y
3
4
1 or y
d.
4
x 4.
3
Rectangle:
Area
A
Ac x
8
x 4
3
4
§ 4
·
x 2 4 x.
x¨ x 4 ¸
3
© 3
¹
8
3
0 Ÿ x
4 Ÿ x
3
2
xy
3
u2
2
Dimensions:
Calculus was helpful.
Circle: The distance from the center r , r to the line
r
r
r
1
3
4
1
1
9 16
12 7 r 12
5
12
5r
7 r 12 Ÿ r
1 or r
Clearly, r
17.
5
6.
1.
r
r
3
4
1Ÿ
7
r
12
1Ÿ r
p x
ax3 bx 2 cx d
pc x
3ax 2 2bx c
pcc x
6ax 2b
6ax 2b
0
x
y0
x3 3 x 2 2, a
3
27 1
y
r.
12
. No calculus necessary.
7
b 3a. Therefore, b 3a, p b 3a is a point of inflection.
1, b
3, c
2b3
bc
d
27 a 2
3a
0, and d
2.
1
31
2 3
1 and satisfies x
§ b3 ·
§ b2 ·
§ b·
a¨ b
¸
¨ 2 ¸ c¨ ¸ d
3
a
a
27
9
© 3a ¹
©
¹
©
¹
When p x
x0
x
y
3
4
b
3a
The sign of pcc x changes at x
§ b·
p¨ ¸
© 3a ¹
0 must be r:
7 r 12
Semicircle: The center lies on the line
So
x
y
1
3
4
3
2
3 0
31
2
The point of inflection of p x
2 0 2
0
x3 3 x 2 2 is x0 , y0
1, 0 .
INSTRUCTOR USE ONLY
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NOT FOR SALE
Problem Solving ffor Chapter 3
18. (a)
T
355
R
PQ − TR
x
8.5 − x
PQ
8.5 in.
x
C
P
Q
2
x PQ
2
C 2 Ÿ PQ 2
C 2 x2
TR 2 8.5 x
2
x 2 Ÿ TR 2
17 x 8.52
PQ TR
2
8.52
PQ 2 Ÿ 2 PQ TR
TR 2 8.52
17 x 8.52 8.52.
So, 2 PQ TR
8.5 x
C 2 x2
PQ TR
17 x 8.52
2
8.5 x
17 x 8.52
C 2 x2
2
17 x3
17 x 8.52
8.5 x
17 x 8.52
C2
x2 C2
2 x3
2 x 8.5
(b) Domain: 4.25 x 8.5
C2 :
(c) To minimize C, minimize f x
fc x
2 x 8.5 6 x 2 2 x3 2
2 x 8.5
8 x3 51x 2
2
2 x 8.5
0
2
51
6.375
8
By the First Derivative Test, x
6.375 is a minimum.
6.375, C | 11.0418 in.
(d) For x
x
19. (a) f x
x
, fc x
x 1
1
x 1
P0
f 0:
c0
0
Pc 0
f c 0 : c1
1
Pcc 0
f cc 0 : 2c2
P x
x x
2
, f cc x
2 Ÿ c2
2
x 1
3
1
2
5
(b)
−3
P(x)
(0, 0)
3
f(x)
−3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
C H A P T E R
Integration
4
Section 4.1
Antiderivatives and Indefinite Integration.........................................357
Section 4.2
Area .....................................................................................................366
Section 4.3
Riemann Sums and Definite Integrals...............................................382
Section 4.4
The Fundamental Theorem of Calculus ............................................391
Section 4.5
Integration by Substitution.................................................................404
Section 4.6
Numerical Integration.........................................................................418
Review Exercises ........................................................................................................426
Problem Solving .........................................................................................................436
INSTRUCTOR USE ONLY
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NOT FOR SALE
C H A P T E R
Integration
4
Section 4.1 Antiderivatives and Indefinite Integration
1.
d§2
·
C¸
¨
dx © x3
¹
d
2 x 3 C
dx
1
d§ 4
·
C¸
2.
¨ 2x 2x
dx ©
¹
dy
dt
9t 2
y
3t 3 C
6
x4
4.
d § 4 1 1
·
¨ 2x x C ¸
2
dx ©
¹
8 x3 3.
6 x 4
1 2
x
2
8 x3 dy
dt
y
5
5t C
Check:
1
2x2
5.
dy
dx
x3 2
y
2 52
x C
5
Check:
d 3
ª3t C º¼
Check:
dt ¬
9t
2
6.
Rewrite
2 x 3
y
2 x 2
C
2
3 43
x C
4
1 2
x dx
4³
1 x 1
C
4 1
1
C
4x
dx
³x
x 1 2
C
1 2
2
C
x
dx
1 2
x dx
9³
1 § x 1 ·
¨
¸C
9 © 1 ¹
1
C
9x
1
dx
4x2
9. ³
1
10. ³
1
3x
2
11. ³ x 7 dx
Check:
13
3 2
dx
x2
7x C
2
º
d ª x2
« 7 x C»
dx ¬ 2
¼
12. ³ 13 x dx
Check:
³x
2 x 3
Simplify
x
C
43
8. ³
x
43
x3 2
1
C
x2
d ª 1
º
C»
dx «¬ x 2
¼
dx
7. ³ 3 x dx
x
Integrate
5
d ª2 5 2
º
x C»
dx «¬ 5
¼
dy
dx
Check:
Given
d
>5t C@
dt
13x x7
x2
C
2
º
d ª
x2
C»
«13 x dx ¬
2
¼
13. ³ x 5 1 dx
13 x
Check:
x6
xC
6
·
d § x6
x C¸
¨
dx © 6
¹
x5 1
INSTRUCTOR USE ONLY
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357
NOT FOR SALE
Chapter 4
358
Integration
egration
14. ³ 8 x3 9 x 2 4 dx
Check:
2 x 4 3x3 4 x C
d
2 x 4 3x3 4 x C
dx
§
16. ³ ¨
©
1 ·
¸ dx
2 x¹
x3 2 2 x 1
1 1 2 ·
§ 12
³ ¨© x 2 x ¸¹ dx
d §2 3 2
·
12
¨ x x C¸
dx © 3
¹
Check:
d §2 3 2
·
12
¨ x 12 x C ¸
dx © 3
¹
2§ 3 1 2 ·
§ 1 1 2 ·
¨ x ¸ 12¨ x ¸
3© 2
¹
©2
¹
³x
23
d §3 5 3
·
Check:
¨ x C¸
dx © 5
¹
18. ³
4
x3 1 dx
Check:
³ x
³ x dx
Check:
d § 1
·
C¸
¨
dx © 4 x 4
¹
34
5
3
dx
x7
Check:
3
d§ 1
·
C¸
¨
dx © 2 x 6
¹
x 4 3x 2 5
dx
x4
³ 1 3x
x2
x3 4 1
4
x 3
5
3 C
3x
x
Check:
d ª
3
5
º
x 3 C»
«
dx ¬
x
3x
¼
d ª
5
º
x 3 x 1 x 3 C »
«
dx ¬
3
¼
1 3x 2 5 x 4
1
23. ³ x 1 3 x 2 dx
3 x 6
C
6
Check:
1
C
2 x6
d § 1 6
·
¨ x C ¸
dx © 2
¹
§ 1·
7
¨ ¸ 6 x
© 2¹
³ 3x x 2 dx
2
x3 1
x5
3
x7
3
5
4
x2
x
x 4 3x 5
x4
d § 1 4
·
¨ x C¸
dx © 4
¹
1
4 x 5
4
5 x 4 dx
3x 1
5x 3
C
1
3
x3 1
1
C
4x4
2
x 4 74
x xC
7
x 4
C
4
7
³ 3x dx
22. ³
x
2
3 53
x C
5
1 dx
20. ³
x 1
x2 3
d §4 7 4
·
¨ x x C¸
dx © 7
¹
1
dx
x5
19. ³
1 1 2
x
2
x
C
53
dx
x6
x
x1 2 6 x 1 2
x1 2 53
6 x 1 2 dx
2 32
x 12 x1 2 C
3
2 12
x x 18 C
3
2 32
x x1 2 C
3
17. ³ 3 x 2 dx
12
x3 2
x1 2
6
C
32
12
12
x3 2
1§ x ·
¸ C
¨
32
2 ¨© 1 2 ¸¹
Check:
³ x
8 x3 9 x 2 4
d §2 5 2
·
2
¨ x x x C¸
dx © 5
¹
x x 6
dx
x
2 52
x x2 x C
5
15. ³ x3 2 2 x 1 dx
Check:
21. ³
1 2
x 2x C
2
d§ 3 1 2
·
¨ x x 2x C ¸
dx ©
2
¹
3x 2 x 2
x 1 3x 2
24. ³ 4t 2 3
2
dt
³ 16t 24t 9 dt
4
2
16t 5
8t 3 9t C
5
Check:
·
d § 16t 5
8t 3 9t C ¸
¨
dt © 5
¹
16t 4 24t 2 9
4t 2 3
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 4.1
25. ³ 5 cos x 4 sin x dx
5 sin x 4 cos x C
Antiderivatives and Indefinite
Indefinit Integration
34. f c x
x
x2
C
2
f x
Check:
d
5 sin x 4 cos x C
dx
5 cos x 4 sin x
2
f ( x) = x + 2
y
2
x2
8
t3
sin t C
3
26. ³ t cos t dt
2
Check:
6
1 csc t cot t
1 3
T tan T C
3
d §1 3
·
Check:
¨ T tan T C ¸
dT © 3
¹
29. ³ sec 2 T sin T dT
T 2 sec 2 T
30. ³ sec y tan y sec y dy
³ sec y tan y sec y dy
³ sec y dy
32. ³ 4 x csc 2 x dx
Check:
33. f c x
f x
f 0
8
f x
3x 8
4x C
f(x) = 4x + 2
5
3
2
f(x) = 4x
x
30
2
C Ÿ C
3
g x
³ 4 x dx
4 3
x C
3
g 1
4
3
4 3 13
x 3
3
C Ÿ C
g x
8
2
4 x 2 , g 1
2
3
13
3
37. hc t
8t 3 5, h 1
4
sec y tan y sec y
ht
³ 8t 5 dt
2t 4 5t C
h1
4
3
25C Ÿ C
11
tan y C
ht
2t 5t 11
sec 2 y
tan 2 y 1
38. f c s
10 s 12 s 3 , f 3
2
f s
³ 10s 12s ds
5s 2 3s 4 C
f 3
2
f s
5s 3s 200
4 x csc 2 x
39. f cc x
4
3x 2 C
2
d
2 x 2 cot x C
dx
y
−3 −2 −1
³ 6 x dx
8
sec y tan y sec 2 y
2 x 2 cot x C
Answers will vary.
f′
f x
2
d
sec y tan y C
dy
d
tan y C
dy
6 x, f 0
sec 2 T sin T
sec y tan y C
Check:
4
35. f c x
36. g c x
tan T cos T C
d
Check:
tan T cos T C
dT
31. ³ tan y 1 dy
2
Answers will vary.
28. ³ T 2 sec2 T dT
2
x
−2
t csc t C
d
Check:
t csc t C
dt
Check:
f′
t 2 cos t
−4
27. ³ 1 csc t cot t dt
f ( x) = 2
4
·
d § t3
¨ sin t C ¸
dt © 3
¹
359
4
3
2
4
53 33 C
2
45 243 C Ÿ C
200
4
2
fc 2
5
f 2
10
fc x
2 x C1
fc 2
³ 2 dx
4 C1
5 Ÿ C1
fc x
2x 1
f x
³ 2 x 1 dx
f 2
6 C2
f x
x x 4
1
x 2 x C2
10 Ÿ C2
4
2
INSTRUCTOR
S
USE ONLY
1
2
3
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copied or duplicated, or posted to a publicly accessible website,
website, in whole or in part.
© Cengage Learning. All Rights Reserved.
360
NOT FOR SALE
Chapter 4
Integration
egration
40. f cc x
x2
fc 0
8
f 0
4
fc x
³ x dx
43. (a) Answers will vary. Sample answer.
y
5
1 3
x C1
3
8 Ÿ C1
8
2
fc 0
0 C1
fc x
1 3
x 8
3
§1 3
·
³ ¨© 3 x 8¸¹ dx
f x
1 4
x 8 x C2
12
0 0 C2
f x
1 4
x 8x 4
12
x
fc 4
2
4 Ÿ C2
x 2 1, 1, 3
y
x3
xC
3
4
1
3
7
3
C
3
1 C
5
(−1, 3)
−4
4
3
fc x
³x
fc 4
2
C1
2 Ÿ C1
2
2
3
x
7
x
x
3
3
y
3 2
³ 2 x
2 x 1 2 C1
dx
1 2
3 dx
f x
4 x
4
y
x
1
0
(b)
1
f 0
6
fc x
³ sin x dx
fc 0
1 C1
fc x
cos x 2
f x
³ cos x 2 dx
f 0
0 0 C2
f x
sin x 2 x 6
7
x 3x
sin x
fc 0
−5
44. (a) Answers will vary. Sample answer:
4 x1 2 3x C2
0 Ÿ C2
3x
2
C1
x
(1, 3)
0 0 C2
12
3
f 0
42. f cc x
dy
dx
3 2
0
f x
(b)
3
f 0
fc x
4
−5
f 0
41. f cc x
x
−4
dy
dx
1
, x ! 0, 1, 3
x2
y
³ x 2 dx ³ x
cos x C1
1 Ÿ C1
1
1
C ŸC
1
1
2
x
3
2
y
sin x 2 x C2
6 Ÿ C2
2
dx
x 1
C
1
1
C
x
2
5
6
−1
8
−1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.1
45. (a)
361
49. Because f cc is negative on f, 0 , f c is decreasing on
9
f, 0 . Because f cc is positive on 0, f , f c is
−3
3
increasing on 0, f . f c has a relative minimum at
(0, 0). Because f c is positive on f, f , f is
−9
(b)
Antiderivatives and Indefinite
Indefinit Integration
dy
dx
2 x, 2, 2
y
³ 2 x dx
2
2
y
x2 6
increasing on f, f .
y
2
x2 C
C
3
4C Ÿ C
f′
6
2
f″
1
x
−3
−2
1
2
3
12
(c)
−2
f
− 15
−3
15
4. Graph of f c is given.
50. f 0
−8
46. (a)
(a) f c 4 | 1.0
20
(b) No. The slopes of the tangent lines are greater than 2
on >0, 2@. Therefore, f must increase more than 4
0
units on >0, 4@.
6
(c) No, f 5 f 4 because f is decreasing on >4, 5@.
0
(b)
dy
dx
2
y
³ 2x
12
y
(c)
x , 4, 12
(d) f is a maximum at x
4 32
x C
3
4 32
4
C
4
8 C
3
3
4 32 4
x 3
3
12
3.5 because f c 3.5 | 0
and the First Derivative Test.
dx
(e) f is concave upward when f c is increasing on
32
C ŸC
3
4
3
f, 1 and 5, f . f is concave downward on
(1, 5). Points of inflection at x
(f ) f cc is a minimum at x
1, 5.
3.
(g) NEED NEW ART
20
0
6
0
47. They are the same. In both cases you are finding a
function F x such that F c x
f x.
48. f x
tan 2 x Ÿ f c x
g x
sec x Ÿ g c x
2
51. (a) h t
2 tan x ˜ sec 2 x
2 sec x ˜ sec x tan x
fc x
The derivatives are the same, so f and g differ by a
constant. In fact, tan 2 x 1 sec 2 x.
³ 1.5t 5 dt
0.75t 2 5t C
h0
00C
ht
0.75t 5t 12
(b) h 6
12 Ÿ C
12
2
0.75 6
2
5 6 12
69 cm
INSTRUCTOR USE ONLY
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copied or duplicated, or posted to a publicly accessible website,
website, in whole or in part.
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362
Chapter 4
52.
dP
dt
Pt
P0
P1
Pt
P7
53. a t
NOT FOR SALE
Integration
egration
55. v0
t , 0 d t d 10
k
s0
2 32
³ kt dt 3 kt C
0C
500 Ÿ C
500
12
32
16 t t 4
t
60
st
³ 32t 60 dt
s0
6
st
16t 2 60t 6, Position function
32t C1
C1
60
0
t
15 seconds.
8
s 15
8
16 15
8
2
2
62.25 feet
³ 32 dt
v0
0 C1
sc t
32t V0
st
³ 32t V0 dt
s0
0 0 C2
st
16t 2 V0t S0
sc t
32t v0
sc t
§ 1 17 ·
32¨¨
¸¸ 16
2
©
¹
v0
v0
f t
³ 9.8t v0 dt
4.9t 2 v0t C2
f 0
s0
C2 Ÿ f t
4.9t 2 v0t s0
vt
9.8t C1
C1 Ÿ v t
9.8t v0
4.9t 2 10t 2.
9.8t 10
0 Maximum height when v
9.8t
10
t
10
9.8
0.
§ 10 ·
f ¨ ¸ | 7.1 m
© 9.8 ¹
V0
16t 2 V0t C2
S 0 Ÿ C2
32t 16
³ 9.8 dt
32t C1
V0 Ÿ C1
| 2.562 seconds.
vt
32 ft sec2
vt
2
9.8
So, f t
60 15
6
8
17
16 17 | 65.970 ft sec
56. a t
The ball reaches its maximum height when
32t
17
vt
16t 2 60t C2
C2
32t 60
1
§ 1 17 ·
v¨¨
¸¸
2
©
¹
v0
1r
Choosing the positive value,
(b)
³ 32 dt
0
0
t
500 | 2352 bacteria
32 ft sec 2
vt
16t 2 16t 64
st
2
vt
54. a t
64 ft
(a)
2
k 500
600 Ÿ k 150
3
2
150 t 3 2 500 100t 3 2 500
3
100 7
16 ft sec
f t
S0
4.9t 2 v0t 2. If
57. From Exercise 56, f t
4.9t 2 v0t 2,
200
then
0 when t
v0
32
time to reach
vt
9.8t v0
0
for this t value. So, t
maximum height.
v0 9.8 and you solve
2
§v ·
s¨ 0 ¸
© 32 ¹
2
§v ·
§v ·
16¨ 0 ¸ v0 ¨ 0 ¸
© 32 ¹
© 32 ¹
550
v2
v2
0 0
64
32
550
v0 2
§v ·
§v ·
4.9¨ 0 ¸ v0 ¨ 0 ¸ 2
9.8
© ¹
© 9.8 ¹
200
§v 2 ·
¨ 0 ¸
© 9.8 ¹
198
4.9v0 2
35,200
9.8
2
2
4.9v0 2 9.8v0 2
9.8 198
4.9v0 2
9.8 198
v0 2
3880.8
v0 | 187.617 ft sec
2
v0 | 62.3 m sec.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.1
4.9t 2 1800. (Using the
58. From Exercise 56, f t
Antiderivatives and Indefinite
Indefinit Integration
t 1 t 3
62. x t
4.9t 2 1800
0
2
1800
t2
1800
Ÿ t | 9.2 sec
4.9
4.9t
(a) v t
xc t
3t 2 14t 15
at
vc t
6t 14
6t 14
(c) a t
3 73 5
v 73
³ 1.6 dt
vt
1.6t v0
stone was dropped, v0
1.6t , because the
³ 1.6t dt
s 20
0 Ÿ 0.8 20
0.8t 2 s0
xt
2
x1
s0
0
s0
320
1.6t
v 20
32 m sec
GM ³
4
GM
C
y
When y
R, v
1 2
v0
2
C
1 2
v
2
v
v
2
2
64. (a) a t
v0 .
GM
C
R
1 2 GM
v0 R
2
GM
1 2 GM
v0 y
R
2
2GM
2GM
v0 2 y
R
v0
2
65. (a)
§1
1·
2GM ¨ ¸
R¹
©y
(a) v t
xc t
at
3t 1 t 3
vc t
6t 2
6t 12
(b) v t ! 0 when 0 t 1 or 3 t 5.
(c) a t
6t 2
0 when t
v2
3 1 1
3
21 C Ÿ C
2t
2
(b)
2
12
1
2t 3 2
1
t 3 2
2
vc t
³ a t dt
³ cos t dt
sin t C1
sin t
f t
³ v t dt
³ sin t dt
f 0
3
f t
cos t 4
because v0
cos 0 C2
1 C2 Ÿ C2
v0
25 km h
25 ˜ 1000
3600
v 13
80 km h
80 ˜ 1000
3600
at
a constant acceleration
vt
at C
v0
v 13
250
36
800
36
550
36
13a
a
550
468
Ÿ vt
0
cos t C2
kS , k
sin t for t
0
3t 2 12t 9
3 t 2 4t 3
43
cos t
(b) v t
t 3 6t 2 9t 2, 0 d t d 5
61. x t
2 23
Acceleration function: a t
1
dy
y2
1 2
v
2
7 3
3
t 1 2 t ! 0
t
12
³ v t dt 2t C
vt
60. ³ v dv
7.
3
0 when t
Position function: x t
So, the height of the cliff is 320 meters.
vt
3t 5 t 3
1
63. v t
0.
st
2
(b) v t ! 0 when 0 t 53 and 3 t 5.
1.6
59. a
0 d t d 5
t 7t 15t 9
3
canyon floor as position 0.)
f t
2
363
4
0, 1, 2, !
250 m sec
36
800 m sec
36
at 250
36
13a 250
36
275
234
| 1.175 m sec 2
t2
250
t s0
2
36
st
a
s 13
275 13
234 2
2
0
250
13 | 189.58 m
36
2.
INSTRUCTOR USE ONLY
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copied or duplicated, or posted to a publicly accessible website,
website, in whole or in part.
© Cengage Learning. All Rights Reserved.
364
NOT FOR SALE
Chapter 4
66. v 0
Integration
egration
45 mi h
66 ft sec
67. Truck: v t
30 mi h
44 ft sec
st
15 mi h
22 ft sec
at
a
vt
at 66
vt
at 66
66
.
a
a § 66 ·
§ 66 ·
¨ ¸ 66¨ ¸
2© a ¹
©a¹
33
132 when a
16.5.
2
at
16.5
vt
16.5t 66
st
8.25t 2 66t
(a) 16.5t 66
t
30t
0
3t 2 30t
0
3t t 10 when t
(a) s 10
3 10
(b) v 10
6 10
k
vt
kt
st
k 2
t because v 0
2
§
v¨¨
©
44
s 16.5
| 117.33 ft
0.
160 and k 2 t 2
0.7.
0.7,
1.4
k
1.4 ·
¸
k ¸¹
k
1.4k
1602 Ÿ k
1.4
k
160
1602
1.4
=4
| 7.45 ft sec2 .
15
/h
mi
30
s0
| 18,285.714 mi h 2
4f
t/se
c
mi
0 m /h =
i/h 22 f
t/se
c
c
t/se
6f
=6
/h
mi
10 sec.
60 ft sec | 41 mi h
68. a t
t
| 2.667
0.
300 ft
Because k 2 t 2
t
45
2
22
| 1.333
16.5
44
16.5
0.
3t 2
At the time of lift-off, kt
22
69. False. f has an infinite number of antiderivatives, each
differing by a constant.
132
73.33
feet
Let s 0
44
(b) 16.5t 66
0
3t
2
At the point where the automobile overtakes the truck:
§ 22 ·
s¨
¸ | 73.33 ft
© 16.5 ¹
(c)
6t Let v 0
st
0 when t
0.
6
vt
2
§ 66 ·
s¨ ¸
©a¹
30t Let s 0
Automobile: a t
a
t 2 66t Let s 0
0.
2
0 after car moves 132 ft.
st
30
117.33
feet
70. True
It takes 1.333 seconds to reduce the speed from 45
mi h to 30 mi h, 1.333 seconds to reduce the speed
from 30 mi h to 15 mi h, and 1.333 seconds to
reduce the speed from 15 mi h to 0 mi h. Each
time, less distance is needed to reach the next speed
reduction.
71. True
72. True
73. True
74. False. For example,
³ x ˜ x dx z ³ x dx ˜ ³ x dx because
§ x2
·§ x 2
·
x3
C z ¨
C1 ¸¨
C2 ¸.
3
2
2
©
¹©
¹
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.1
75. f cc x
Antiderivatives and Indefinite
Indefinit Integration
­1, 0 d x 2
°
®2, 2 x 3
°
¯0, 3 x d 4
2x
76. f c x
fc x
x2 C
fc 2
0 Ÿ 4C
0 Ÿ C
4
f x
f 2
x
4 x C1
3
8
0 Ÿ 8 C1
3
f x
x3
16
4x 3
3
2
1
f x
f 0
1 Ÿ C1
16
3
0 Ÿ C1
y
­ x C1 , 0 d x 2
°
®2 x C2 , 2 x 3
°C ,
3 x d 4
¯ 3
3
x
1
2
3
4
1
f continuous at
x
2 Ÿ 2 1
4 C2 Ÿ C2
f continuous at x
3 Ÿ 65
5
C3
1
­ x 1, 0 d x 2
°
®2 x 5, 2 d x 3
°1,
3 d x d 4
¯
f x
2
2
d ª
ªs x º¼ ª¬c x º¼ º»
¼
dx ¬«¬
77.
365
2 s x sc x 2c x cc x
2 s x c x 2c x s x
2
So, ¬ªs x ¼º ¬ªc x ¼º
Because, s 0
2
k for some constant k.
0 and c 0
2
Therefore, ª¬s x ¼º ¬ªc x ¼º
[Note that s x
0
1, k
2
sin x and c x
1.
1.
cos x satisfy these
properties.]
78. f x y
f x f y g x g y
g x y
f x g y g x f y
fc 0
0
[Note: f x
cos x and g x
sin x satisfy these conditions]
fc x y
f x f c y g x g c y (Differentiate with respect to y)
gc x y
f x g c y g x f c y (Differentiate with respect to y)
Letting y
0, f c x
f x f c 0 g x gc 0
g x gc 0
gc x
f x gc 0 g x f c 0
f x gc 0
So, 2 f x f c x
2 f x g x g c 0
2g x gc x
2 g x f x gc 0 .
Adding, 2 f x f c x 2 g x g c x
Integrating, f x
2
g x
Clearly C z 0, for if C
Now, C
f x y
2
2
0.
C.
2
0, then f x
g x y
2
g x
2
Ÿ f x
f x f y g x g y
f x
2
f y
2
2
g x
2
g x g y
0, which contradicts that f, g are nonconstant.
f x g y g x f y
2
2
f x g y
ª f x 2 g x 2 ºª f y 2 g y 2 º
¬
¼¬
¼
2
g x
2
2
f y
2
C2
INSTRUCTOR USE ONLY
So, C
1 and you have f x
2
g x
2
1.
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied
copied or duplicated, or posted to a publicly accessible website,
website, in whole or in part.
© Cengage Learning. All Rights Reserved.
366
NOT FOR SALE
Chapter 4
Integration
egration
Section 4.2 Area
6
1. ¦ 3i 2
6
6
i 1
i 1
3¦i ¦ 2
i 1
9
2. ¦ k 2 1
3 1 2 3 4 5 6 12
32 1 42 1 ! 92 1
75
287
k 3
4
1
2
k
1
k 0
3. ¦
6
3
j 4 j
3
3
3
4
5
6
37
20
cccc
4c
4. ¦
4
5. ¦c
1 1
1
1
2 5 10 17
1
158
85
k 1
4
6. ¦ ª i 1
¬
i 1
2
i 1 º
¼
3
0 8 1 27 4 64 9 125
11
238
24
1
5
i 1 i
7. ¦
15. ¦ 4i
i 1
14
9
1
i
i 1
8. ¦
i 1
16
5¦ i 4 16
i 1
20
17. ¦ i 1
i 1
2
i 1
4 ª
§ j· º
10. ¦ «1 ¨ ¸ »
© 4 ¹ ¼»
j 1 ¬
«
1200
ª16 17 º
5«
» 64
¬ 2 ¼
16
16. ¦ 5i 4
6
ª § j·
º
9. ¦ «7¨ ¸ 5»
¼
j 1 ¬ ©6¹
ª 24 25 º
4«
»
¬ 2 ¼
24
4¦i
¦ i2
i 1
ª19 20 39 º
«
»
6
¬
¼
10
10
i 1
i 1
19
616
2470
2
10
18. ¦ i 2 1
i 1
11.
12.
3
2 n ª§ 2i ·
§ 2i ·º
«¨ ¸ ¨ ¸»
¦
n i 1 ¬«© n ¹
© n ¹¼»
n ª
15
19. ¦ i i 1
ª10 11 21 º
«
» 10
6
¬
¼
¦ i2 ¦1
2
i 1
3
3i · º
§
«2¨1 ¸ »
¦
n i 1 ¬« ©
n ¹ ¼»
15
15
i 1
i 1
¦ i 3 2¦ i 2 ¦ i
i 1
2
2
2
15
375
15 16
15 16 31
15 16
2
4
6
2
14,400 2480 120 12,040
12
13. ¦ 7
7 12
84
25
20. ¦ i 3 2i
i 1
i 1
30
14. ¦ 18
i 1
18 30
25
25
¦ i 3 2¦ i
i 1
540
25
i 1
2
26
2
2
4
105,625 650
25 26
2
104,975
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.2
Secti
Sectio
2i 1
n2
i 1
n
21. ¦
º
1ª nn 1
n»
«2
n2 ¬
2
¼
1 n
¦ 2i 1
n2 i 1
S 100
12
1.2
10
1.02
S 1000
1.002
S 10,000
1.0002
7j 4
n2
j 1
1 n
¦ 7j 4
n2 j 1
S 10
n
22. ¦
n 2
n
1
2
n
Area
367
S n
º
1ª nn 1
4 n»
«7
2
n2 ¬
¼
7n 2 7n
4n
2
2n 2
n
S 10
17
4
S 100
3.575
S 1000
3.5075
S 10,000
3.50075
6k k 1
n3
k 1
n
23. ¦
7 n 15
2n
4.25
nn1º
6 ª n n 1 2n 1
«
»
n3 ¬
6
2 ¼
6 n 2
¦k k
n3 k 1
6 ª 2n 2 3n 1 3n 3 º
«
»
n2 ¬
6
¼
S 10
1.98
S 100
1.9998
S 1000
1.999998
S 10,000
1.99999998
2i 3 3i
n4
i 1
n
24. ¦
S n
1
ª2n 2 2º¼
n2 ¬
2
2
n2
S n
1 n
¦ 2i3 3i
n4 i 1
1 ª n2 n 1
«2
4
n 4 «¬
n 1
2
2
2
3
3n 1
S 10
2n
0.5885
S 100
0.5098985
S 1000
0.5009989985
S 10,000
0.50009999
2n
3
nn 1º
»
2
»¼
n3 2n 2 2n 3
2n3
S n
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
368
NOT FOR SALE
Chapter 4
Integration
egration
y
25.
y
10
10
8
8
6
6
4
4
2
2
x
'x
1
2
20
4
1
2
x
1
2
Left endpoints: Area | 12 >5 6 7 8@
26
2
Right endpoints: Area | 12 >6 7 8 9@
13
30
2
15
13 Area 15
26.
y
y
10
10
8
8
6
6
4
4
2
2
x
1
2
3
42
6
'x
x
4
1
2
3
4
1
3
1ª
20 19
17 16 º
6
»
7
3 «¬
3
3
3
3¼
Left endpoints: Area |
Right endpoints: Area |
37
| 12.333
3
1 ª 20 19
17 16 15 º
6
»
3 «¬ 3
3
3
3
3¼
35
| 11.667
3
35
37
Area 3
3
27.
y
y
50
50
40
40
30
30
20
20
10
10
x
1
'x
2
52
6
3
4
5
x
1
2
3
4
5
1
2
Left endpoints: Area | 12 >5 9 14 20 27 35@
Right endpoints: Area | 12 >9 14 20 27 35 44@
55
149
2
74.5
55 Area 74.5
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
Section 4.2
Secti
Sectio
28.
y
Area
369
y
10
10
8
8
6
6
4
4
2
2
x
1
2
31
8
'x
x
3
1
2
3
1
4
41 13 65 5 97 29 137 º
Left endpoints: Area | 14 ª¬2 16
4
16
16
4
16 ¼
155
16
41 13 65 5 97 29 137 10º
Right endpoint: Area | 14 ª¬16
4
16
16
4
16
¼
11.6875
9.6875
9.6875 Area 11.6875
29.
y
y
1
1
π
4
S
2
'x
x
π
2
0
S
4
8
Left endpoints: Area |
π
4
π
2
x
Sª
§S ·
§S ·
§ 3S ·º
cos 0 cos¨ ¸ cos¨ ¸ cos¨ ¸» | 1.1835
8 «¬
8
4
© ¹
© ¹
© 8 ¹¼
Sª
§S ·
§S ·
§ 3S ·
§ S ·º
cos¨ ¸ cos¨ ¸ cos¨ ¸ cos¨ ¸» | 0.7908
8 «¬ © 8 ¹
©4¹
© 8 ¹
© 2 ¹¼
Right endpoints: Area |
0.7908 Area 1.1835
y
30.
y
1
1
x
π
2
'x
S 0
S
6
6
Left endpoints: Area |
π
2
x
Sª
Right endpoints: Area |
S
S
S
2S
5S º
sin
sin
sin
sin
| 1.9541
sin 0 sin
6 «¬
6
3
2
3
6 »¼
Sª
S
S
S
2S
S
º
sin
sin
sin
sin
sin S » | 1.9541
sin
6 «¬
6
3
2
3
6
¼
By symmetry, the answers are the same. The exact area (2) is larger.
31. S
ª3 4 92 5º 1
¬
¼
33
2
16.5
s
ª1 3 4 92 º 1
¬
¼
25
2
12.5
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
Chapter 4
370
32. S
s
NOT FOR SALE
Integration
egration
>5 5 4 2@ 1
>4 4 2 0@ 1
1§1·
¨ ¸
4© 4¹
33. S 4
§1·
0¨ ¸ © 4¹
s4
§
¨¨
©
34. S 8
16
10
1§1·
¨ ¸
2© 4¹
1§1·
¨ ¸
4© 4¹
·1 §
1
2 ¸¸ ¨¨
4
¹4 ©
3§1·
¨ ¸
4© 4¹
1§1·
¨ ¸
2© 4¹
1
3§1·
¨ ¸
4© 4¹
·1 §
1
2 ¸¸ ¨¨
2
¹4 ©
1
§1·
1¨ ¸
© 4¹
2 8
2 8
·1
3
2 ¸¸ 4
¹4
12
1§
1
2
3
5
6
7
1
¨16 4 ¨©
2
2
2
2
2
2
0 2
s8
1 §
¨
4 ¨©
·1 §
1
2 ¸¸ ¨¨
4
¹4 ©
3 2
3
| 0.768
| 0.518
1 §
¨
4 ¨©
·1 §
5
2 ¸¸ ¨¨
4
¹4 ©
·1 §
3
2 ¸¸ ¨¨
2
¹4 ©
·1
§
1
2 ¸¸ ! ¨¨
2
4
¹
©
1 1 1 1 1
| 0.746
5 6 7 8 9
s5
1 §1·
1 §1·
1 §1·
1 § 1 · 1§ 1 ·
¨ ¸
¨ ¸
¨ ¸
¨ ¸ ¨ ¸
6 5© 5 ¹ 7 5© 5 ¹ 8 5© 5 ¹ 9 5© 5 ¹ 2 © 5 ¹
1 1 1 1
1
| 0.646
6 7 8 9 10
2
§1· §1·
1¨ ¸ ¨ ¸ © 5¹ © 5¹
1ª
«1 5 ¬«
24
5
2
n
§ 24i ·
37. lim ¦ ¨ 2 ¸
n of
i 1© n ¹
2
§ 2· §1·
1¨ ¸ ¨ ¸ © 5¹ ©5¹
24 n
¦i
n of n 2
i 1
lim
24 § n n 1 ·
¨
¸
n of n 2
2
©
¹
2
lim
1 ª n 1 n 2n 1 º
«
»
6
¬
¼
n
1 n 1 2
¦i
n of n 3
i 1
n of n3
9 § n 1·
lim ¨
¸
n ¹
nof 2©
1 ª 2n3 3n 2 n º
«
»
n of 6
n3
¬
¼
lim
2
n
2i · § 2 ·
§
40. lim ¦ ¨1 ¸ ¨ ¸
n of
n ¹ ©n¹
©
i 1
§ 4· §1·
1 ¨ ¸ ¨ ¸ 0 | 0.659
© 5¹ © 5¹
ª § n 2 n ·º
lim «12¨
¸»
2
n of
¹¼
¬ © n
9 ªn n 1 º
«
»
2
¬
¼
n o f n2
lim
2
lim
9 n
¦i
n o f n2
i 1
lim
1
2
i 1
3
n of
n
i 1
2
§ 4· §1·
1¨ ¸ ¨ ¸
© 5¹ © 5¹
§ 3· § 1·
1¨ ¸ ¨ ¸ © 5¹ © 5¹
lim
n
§ 3i ·§ 3 ·
38. lim ¦ ¨ ¸¨ ¸
nof
i 1 © n ¹© n ¹
39. lim ¦
2
§ 3· § 1·
1¨ ¸ ¨ ¸ © 5¹ © 5¹
16
9º
» | 0.859
5
5 »¼
21
5
§1· §1·
1¨ ¸ ¨ ¸ © 5¹ ©5¹
s5
2
§ 2· §1·
1¨ ¸ ¨ ¸ © 5¹ ©5¹
1
4
·1
7
2 ¸¸ | 5.685
4
¹4
1 §1·
1 §1·
1 §1·
1 §1·
§1·
1¨ ¸ ¨ ¸
¨ ¸
¨ ¸
¨ ¸
© 5 ¹ 6 5© 5 ¹ 7 5© 5 ¹ 8 5© 5 ¹ 9 5© 5 ¹
§1·
1¨ ¸ ©5¹
22
·
2 ¸¸ | 6.038
¹
35. S 5
36. S 5
·1
7
2 ¸¸ 4
¹4
1·
§
12 lim ¨1 ¸
n of ©
n¹
12
9
2
ª 1 § 2 3 n 1 n 2 ·º
¸»
lim « ¨
n of « 6 ¨
¸»
1
¹¼
¬ ©
1
3
2 n
2
¦ n 2i
n of n 3
i 1
lim
n
n
º
2ªn 2
n 4n ¦ i 4 ¦ i 2 »
¦
«
3
n of n
i 1
i 1 ¼
¬i 1
lim
§ n n 1 · 4 n n 1 2n 1 º
2ª 3
n 4n ¨
«
»
¸
3
n of n «
2
6
»¼
©
¹
¬
lim
2 4 2
2 º
ª
2 lim «1 2 2 »
n of ¬
3 n 3n ¼
n
4·
§
2¨1 2 ¸
3
©
¹
26
3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.2
Secti
Sectio
1ª
1 § n n 1 ·º
«n ¨
¸»
n of n «
n©
2
¹»¼
¬
1ª n
1 n º
2 lim «¦ 1 ¦ i»
n of n
ni 1 ¼
¬i 1
n
i ·§ 2 ·
§
41. lim ¦ ¨1 ¸¨ ¸
n of
n
¹© n ¹
i 1©
3
n
3i · § 3 ·
§
42. lim ¦ ¨ 2 ¸ ¨ ¸
nof
n ¹ ©n¹
i 1©
3 n ª 2n 3i º
¦ ¬« n ¼»
nof n
i 1
2 lim
ª
n2 n º
2 lim «1 »
n of
2n 2 ¼
¬
Area
1·
§
2¨1 ¸
2¹
©
371
3
3
lim
3 n
¦ 8n3 36n 2i 54ni 2 27i3
n o f n4
i 1
lim
nn 1
n n 1 2n 1
n2 n 1
3 § 4
¨ 8n 36n 2
54n
27
4
nof n ¨
2
6
4
©
lim
2
·
¸
¸
¹
2
§
9 n 1 2n 1
n 1
27 n 1 ·
¸
lim 3 ¨ 8 18
˜
nof ¨
4
n
n2
n2 ¸
©
¹
27 ·
§
3 ¨ 8 18 18 ¸
4¹
©
43. (a)
609
4
152.25
y
3
2
1
x
(b) 'x
1
3
20
n
2
n
§ 2·
§ 2·
§ 2·
§ 2·
Endpoints: 0 1¨ ¸ 2¨ ¸ ! n 1 ¨ ¸ n¨ ¸
n
n
n
© ¹
© ¹
© ¹
©n¹
x is increasing, f mi
(c) Because y
n
sn
¦ f ¨©
i 1
n
¦ f xi 'x
i 1
(e)
i 1
ª
n
§ 2 ·º§ 2 ·
¦ «¬ i 1 ¨© n ¸¹»¼¨© n ¸¹
i 1
f xi on > xi 1 , xi @
(d) f M i
S n
f xi 1 on > xi 1 , xi @.
§ 2i 2 ·§ 2 ·
¸¨ ¸
n ¹© n ¹
n
¦ f xi 1 'x
2
n
§ 2i · 2
¦ f ¨© n ¸¹ n
i 1
x
5
10
50
100
sn
1.6
1.8
1.96
1.98
S n
2.4
2.2
2.04
2.02
n
ª
§ 2 ·º§ 2 ·
(f ) lim ¦ « i 1 ¨ ¸»¨ ¸
n of
© n ¹¼© n ¹
i 1 ¬
n
ª § 2 ·º§ 2 ·
lim ¦ «i¨ ¸»¨ ¸
n of
i 1 ¬ © n ¹¼ © n ¹
n
ª § 2 ·º§ 2 ·
¦ «¬i¨© n ¸¹»¼¨© n ¸¹
i 1
4 n
¦ i 1
n of n 2
i 1
lim
4 n
¦i
n of n 2
i 1
lim
lim
º
4 ªn n 1
n»
«
2
¬
¼
n of n 2
§ 4 ·n n 1
lim ¨ ¸
2
n of © n 2 ¹
lim
n of
ª2 n 1
4º
lim «
»
n¼
¬ n
n of
2n 1
n
2
2
INSTRUCTOR USE ONLY
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372
Chapter 4
44. (a)
Integration
egration
y
4
3
2
1
x
2
4
31
n
(b) ' x
2
n
Endpoints:
2
4
2n
1
" 1
3
n
n
n
§2·
§ 2·
§ 2·
§ ·
1 1 1¨ ¸ 1 2¨ ¸ " 1 n 1 ¨ ¸ 1 n¨ ¸
©n¹
©n¹
©n¹
©n¹
1 1
x is increasing, f mi
(c) Because y
n
§ 2 ·º § 2 ·
ª
§ 2 ·º§ 2 ·
¦ «¬1 i 1 ¨© n ¸¹»¼¨© n ¸¹
i 1
i 1
f xi on > xi 1 , xi @
(d) f M i
n
n
ª
§ 2 ·º§ 2 ·
¦ f xi 'x
¦ f «¬1 i¨© n ¸¹»¼¨© n ¸¹
x
5
10
50
100
sn
3.6
3.8
3.96
3.98
S n
4.4
4.2
4.04
4.02
S n
n
¦ f «¬1 i 1 ¨© n ¸¹»¼ ¨© n ¸¹
i 1
i 1
(e)
ª
n
¦ f xi 1 'x
sn
f xi 1 on > xi 1 , xi @.
i 1
n
ª
§ 2 ·º§ 2 ·
(f ) lim ¦ «1 i 1 ¨ ¸»¨ ¸
n of
© n ¹¼© n ¹
¬
i 1
n
ª
§ 2 ·º§ 2 ·
¦ «¬1 i¨© n ¸¹»¼¨© n ¸¹
i 1
·º
2§ n n 1
§ 2 ·ª
n ¸»
lim ¨ ¸ «n ¨
n of© n ¹ «
n©
2
¹»¼
¬
2n 2 4 º
ª
»
lim 2 n
n¼
2º
ª
lim 4 »
n¼
n of «
¬
n
ª
§ 2 ·º§ 2 ·
lim ¦ «1 i¨ ¸»¨ ¸
n of
© n ¹¼© n ¹
i 1 ¬
4
n of «
¬
2ª
§ 2·n n 1 º
lim «n ¨ ¸
»
2 ¼
©n¹
¬
n of n
2n 1º
ª
lim «2 »
n ¼
¬
n of
2º
ª
lim 4 »
n¼
n of «
¬
4
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.2
Secti
Sectio
§
4 x 5 on [0, 1]. ¨ Note: 'x
©
45. y
§ i ·§ 1 ·
n
n
¦ f ¨© n ¸¨
¸
¹© n ¹
sn
ª
1·
¸
n¹
º§ 1 ·
§i·
i 1
i 1
ª§ i · 2
º§ 1 ·
«
¦ ¨© n ¸¹ 2»¨© n ¸¹
»¼
i 1«
¬
n
4 nn 1
5
n2
2
1·
§
2¨1 ¸ 5
n¹
©
ª 1 n 2º
« 3¦i » 2
¬n i 1 ¼
lim s n
n n 1 2n 1
6n 3
3
nof
1·
¸
n¹
¦ f ¨© n ¸¨
¸
¹© n ¹
S n
4 n
2¦ i 5
n i 1
Area
373
§ i ·§ 1 ·
n
¦ «¬4¨© n ¸¹ 5»¼¨© n ¸¹
i 1
§
x 2 2 on [0, 1]. ¨ Note: 'x
©
47. y
Area
7
3
lim S n
Area
n of
y
1§
3
1·
¨2 2 ¸ 2
n
n ¹
6©
2
y
5
4
3
3
2
1
1
−2
x
−1
1
2
3
x
1
52
n
§
3 x 2 on [2, 5]. ¨ Note: 'x
©
46. y
§
n
3·
¸
n¹
ª §
3i ·
S n
º§ 3 ·
¦ «¬3¨© 2 n ¸¹ 2»¼¨© n ¸¹
2 n
§ 3·
18 3¨ ¸ ¦ i 6
©n¹ i 1
27 § n 1 n ·
¨
¸
n2 ©
2
¹
12 27
2
51
2
12 lim S n
Area
n of
2·
¸
n¹
n
24 n 2
2 n
i ¦1
3 ¦
n i 1
ni 1
i 1
12 2 0
n
ª § 2i · 2
º§ 2 ·
«
¦ 3¨© n ¸¹ 1»¨© n ¸¹
»¼
i 1«
¬
§ 2i ·§ 2 ·
¦ f ¨© n ¸¨
¸
¹© n ¹
i 1
n
i 1
n
3
§
3 x 2 1 on [0, 2]. ¨ Note: 'x
©
48. y
3i ·§ 3 ·
¦ f ¨© 2 n ¸¨
¸
¹© n ¹
S n
2
24 § n n 1 2n 1 · 2
¨
¸ n
6
n3 ©
¹ n
27 §
1·
¨1 ¸
n¹
2©
4 n 1 2n 1
n2
8 2
lim S n
Area
nof
y
2
10
y
15
3
10
2
5
x
1
2
3
4
5
6
x
1
2
3
INSTRUCTOR USE ONLY
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374
NOT FOR SALE
Chapter 4
Integration
egration
§
25 x 2 on [1, 4]. ¨ Note: 'x
©
49. y
§
n
2
ª
3i · º§ 3 ·
§
25
1
«
¦
¨
¸ »¨ ¸
n ¹ ¼»© n ¹
©
«
i 1¬
3i ·§ 3 ·
i 1
4 x 2 on >2, 2@. Find area of region over the
50. y
n
¦ f ¨©1 n ¸¨
¸
¹© n ¹
sn
3·
¸
n¹
3 n ª
9i 2
6i º
¦
«24 2 »
ni 1¬
n
n¼
§
interval [0, 2]. ¨ Note: 'x
©
i 1
72 9 9
n of
ª
n
§ 2i · º§ 2 ·
2
¦ «4 ¨© n ¸¹ »¨© n ¸¹
i 1¬
«
9
9
n 1 2n 1 n 1
72 n
2n 2
lim s n
§ 2i ·§ 2 ·
n
¦ f ¨© n ¸¨
¸
¹© n ¹
sn
3ª
9 n n 1 2n 1
6n n 1º
«24n 2
»
n¬
n
n
6
2 ¼
Area
2·
¸
n¹
54
8
8
¦ i2
n3 i 1
8
8n n 1 2n 1
6n3
1
Area
2
y
20
8
lim s n
n of
8
3
8
4§
3
1·
¨2 2 ¸
n
n ¹
3©
16
3
32
3
Area
15
¼»
n
10
y
5
−1
x
−5
1
2
3
4
5
3
2
1
x
−1
51. y
§
27 x3 on [1, 3]. ¨ Note: 'x
©
2·
¸
n¹
3
ª
2i · º§ 2 ·
§
27
1
«
¦
¨
¸ »¨ ¸
n ¹ »¼© n ¹
©
i 1«
¬
2i ·§ 2 ·
§
¦ f ¨©1 n ¸¨
¸
¹© n ¹
i 1
n
sn
31
n
n
y
2 n ª
8i 3 12i 2
6i º
¦
«26 3 2 »
ni 1¬
n
n
n¼
2ª
8 n n 1
«26n 3
4
n «¬
n
2
30
24
12 n n 1 2n 1
6n n 1º
»
2
6
2 »¼
n
n
2
4
4
6n 1
2
n 1 2 n 1 2n 1 n2
n
n
lim s n
52 4 8 6 34
52 Area
1
18
12
6
x
−2 −1
−6
1
2
4
5
n of
10
1·
§
2 x x3 on [0, 1]. ¨ Note: 'x
¸
n
n¹
©
Because y both increases and decreases on [0, 1], T(n) is neither an upper nor lower sum.
52. y
n
T n
§ i ·§ 1 ·
¦ f ¨© n ¸¨
¸
¹© n ¹
i 1
n
lim T n
n of
1
§ i · º§ 1 ·
¦ «2¨© n ¸¹ ¨© n ¸¹ »¨© n ¸¹
«
i 1¬
2 n
1 n
i 4 ¦ i3
2¦
n i 1
n i 1
Area
ª §i·
1
4
y
3
2.0
¼»
2
nn 1
1 ª n2 n 1 º
«
»
4
n2
n 4 «¬
»¼
3
4
1.5
1 1
2
1
1 n 4 4 n 4n 2
1.0
0.5
x
0.5
1.0
2.0
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.2
Secti
Sectio
§
x 2 x3 on >1, 1@. ¨ Note: 'x
©
53. y
1 1
n
Area
375
2·
¸
n¹
Because y both increases and decreases on >1, 1@, T(n) is neither an upper nor a lower sum.
§
2i ·§ 2 ·
ª§
4i ·
§
¹
©
n
2
2i · º§ 2 ·
3
§
«
i 1¬
2
4i
2i ·
¦ «¨© 1 n ¸¹ ¨© 1 n ¸¹ »¨© n ¸¹
i 1
n
ª§
n
¦ f ¨© 1 n ¸¨
¸
¹© n ¹
T n
¼»
6i
12i
2
8i ·º § 2 ·
3
¦ «¨1 n n2 ¸ ¨ 1 n n2 n3 ¸»¨© n ¸¹
i 1 ©
¹¼
ª
10i 16i 2
8i 3 º§ 2 ·
¦ «2 n n2 n3 »¨© n ¸¹
i 1¬
¼
n
4 n
20 n
32 n
16 n
1 2 ¦ i 3 ¦i 2 4 ¦ i 3
¦
ni 1
n i 1
n i 1
n i 1
4
20 n n 1
32 n n 1 2n 1
16 n 2 n 1
n 2 ˜
3 ˜
4 ˜
n
n
n
n
2
6
4
1 · 16 §
3
1·
2
1·
§
§
4 10¨1 ¸ ¨ 2 2 ¸ 4¨1 2 ¸
n¹
n
n ¹
n
n ¹
3©
©
©
Area
4 10 lim T n
n of
32
4
3
2
2
3
y
2
1
x
−1
1
2 1
n
§
2 x3 x 2 on >1, 2@. ¨ Note: 'x
©
54. y
3
2
ª §
i·
i · º§ 1 ·
§
2
1
1
«
¦ ¨© n ¸¹ ¨© n ¸¹ »¨© n ¸¹
»¼
i 1«
¬
i ·§ 1 ·
§
¦ f ¨©1 n ¸¨
¸
¹© n ¹
i 1
n
sn
n
§ 2i 3
5i 2
1·
¸
n¹
n
4i
·§ 1 ·
¦ ¨ n3 n 2 n 1¸¨© n ¸¹
i 1©
¹
2 n2 n 1
˜
n4
4
Area
2
5 n n 1 2n 1
4 nn 1
˜
2 ˜
1
3
n
n
6
2
1
5
21
2
3
lim sn
nof
31
6
y
12
10
8
6
4
2
−3
−2
−1
x
1
2
3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
376
NOT FOR SALE
Chapter 4
Integration
egration
§
4 y, 0 d y d 2 ¨ Note: 'y
©
55. f y
20
n
2·
¸
n¹
2i ·§ 2 ·
1§
2i ·§ 2 ·
i 1
¦ g ¨© 2 n ¸¨
¸
¹© n ¹
§ 2i ·§ 2 ·
¦ f ¨© n ¸¨
¸
¹© n ¹
i 1
¦ 2 ¨© 2 n ¸¨
¸
¹© n ¹
i·
2 n §
¦ ¨1 n ¸¹
n i 1©
§ 2i ·§ 2 ·
¦ 4¨© n ¸¨
¸
¹© n ¹
i 1
2ª
1n n 1º
«n »
2
n¬
n
¼
2
S n
n
i 1
n
16 n
¦i
n2 i 1
§ 16 · n n 1
¨ 2¸ ˜
2
©n ¹
lim S n
n of
n 1
n
3
y
8n1
n
8·
§
lim ¨ 8 ¸
n of©
n¹
21
lim S n
Area
n of
2·
¸
n¹
i 1
n
Area
§
n
n
¦ f mi 'y
S n
42
n
1
§
y, 2 d y d 4. ¨ Note: 'y
2
©
56. g y
8
8
n
5
4
8
3
2
1
y
x
4
2
1
3
4
5
3
2
§
y 2 , 0 d y d 5 ¨ Note: 'y
©
57. f y
1
x
2
−1
4
6
8
n
50
n
5·
¸
n¹
§ 5i ·§ 5 ·
¦ f ¨© n ¸¨
¸
¹© n ¹
S n
i 1
n
2
§ 5i · § 5 ·
¦ ¨© n ¸¹ ¨© n ¸¹
i 1
125 n 2
¦i
n3 i 1
125 n n 1 2n 1
˜
6
n3
125 § 2n 2 3n 1 ·
¨
¸
6
n2 ©
¹
125 125 125
3
2n
6n 2
§ 125 125 125 ·
lim ¨
¸
3
2n
6n 2 ¹
Area lim S n
n of ©
n of
125
3
y
6
4
2
x
−5
−2
5
10
15
20
25
−4
−6
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.2
Secti
Sectio
2 1
n
§
4 y y 2 , 1 d y d 2. ¨ Note: 'y
©
58. f y
§
n
Area
377
1·
¸
n¹
i ·§ 1 ·
¦ f ¨©1 n ¸¨
¸
¹© n ¹
S n
i 1
2
1 n ª §
i· §
i· º
4
1
1
«
¦ ¨ n ¸¹ ¨© n ¸¹ »
n i 1 «¬ ©
»¼
1 n §
4i
2i
i2 ·
1
2¸
¨4 ¦
ni 1©
n
n
n ¹
1 n §
2i
i2 ·
2¸
¨3 ¦
ni 1©
n
n ¹
1ª
2n n 1
1 n n 1 2n 1 º
2
«3n »
2
6
n¬
n
n
¼
n 1 2n 1
n 1
6
n
3
Area
31
lim S n
n of
1
3
11
3
y
5
3
2
1
x
1
2
3
4
5
31
n
§
4 y 2 y 3 , 1 d y d 3. ¨ Note: 'y
©
59. g y
§
2i ·§ 2 ·
ª §
2i ·
n
2·
¸
n¹
¦ g ¨©1 n ¸¨
¸
¹© n ¹
S n
i 1
n
2
2i · º 2
3
§
¦ «4¨©1 n ¸¹ ¨©1 n ¸¹ » n
«
i 1¬
¼»
ª
2
4i
4i º ª
6i 12i 2
8i 3 º
4 «1 2 » «1 2 3»
¦
ni 1 ¬
n
n ¼ ¬
n
n
n ¼
2
n
2 n ª
10i
4i 2
8i 3 º
2 3»
¦
«3 ni 1¬
n
n
n ¼
2ª
10 n n 1
4 n n 1 2n 1
8 n2 n 1
«3n 2
2
2
6
4
n«
n
n
n
¬
Area
lim S n
n of
6 10 8
4
3
2
º
»
»¼
44
3
y
10
8
6
2
−4 −2
−2
x
−4
INSTRUCTOR USE ONLY
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378
Chapter 4
NOT FOR SALE
Integration
egration
§
y 3 1, 1 d y d 2 ¨ Note: 'y
©
60. h y
n
§
1·
¸
n¹
i ·§ 1 ·
¦ h¨©1 n ¸¨
¸
¹© n ¹
S n
i 1
3
ª§
º1
i·
1
«
¦ ¨© n ¸¹ 1» n
»¼
i 1«
¬
n
y
1 n §
3i 2
3i ·
i3
¨2 3 2 ¸
¦
ni 1©
n
n
n¹
5
4
2
1ª
1 n2 n 1
3 n n 1 2n 1
3 3n n 1 º
«2n 3
2
»
4
6
2n »
n «¬
n
n
n
¼
Area
n 1
2
n2 4
2
lim S n
2
n of
1
, c1
2
1
, c2
4
1
x
2
4
6
8
10
19
4
4
3
, c3
4
n
4
Area | ¦ f ci 'x
5
, c4
4
7
4
§1·
1 ª§ 1
· §9
· § 25
· § 49
·º
3¸ ¨
3¸ ¨
3¸ ¨
3¸»
¨
2 «¬© 16
¹ © 16
¹ © 16
¹ © 16
¹¼
¦ ¬ªci2 3¼º¨© 2 ¸¹
i 1
i 1
x 2 4 x, 0 d x d 4, n
62. f x
69
8
4
xi xi 1
.
2
Let ci
1, c1
1
, c2
2
3
, c3
2
n
¦ ¬ªci2 4ci ¼º 1
i 1
i 1
tan x, 0 d x d
63. f x
5
, c4
2
4
Area | ¦ f ci 'x
S
4
,n
7
2
ª§ 1
· §9
· § 25
· § 49
·º
«¨ 4 2 ¸ ¨ 4 6 ¸ ¨ 4 10 ¸ ¨ 4 14 ¸»
©
¹
©
¹
©
¹
©
¹
¬
¼
53
4
xi xi 1
.
2
Let ci
'x
2
xi xi 1
.
2
Let ci
'x
1
3
1
4
2
x 2 3, 0 d x d 2, n
61. f x
'x
3n1
1 n 1 2n 1
2
2n
n2
3
S
16
, c1
n
S
32
Area | ¦ f ci 'x
i 1
3S
, c3
32
, c2
4
5S
, c4
32
§S ·
¦ tan ci ¨© 16 ¸¹
i 1
7S
32
S§
S
3S
5S
7S ·
tan
tan
tan ¸ | 0.345
¨ tan
16 ©
32
32
32
32 ¹
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.2
Secti
Sectio
S
'x
2
, n
379
4
xi xi 1
.
2
Let ci
8
S
, c1
16
3S
, c3
16
, c2
n
5S
, c4
§S ·
4
Area | ¦ f ci 'x
¦ cos ci ¨© 8 ¸¹
i 1
65.
S
cos x, 0 d x d
64. f x
Area
i 1
7S
16
S§
S
3S
5S
7S ·
cos
cos
cos
¨ cos
¸ | 1.006
8©
16
16
16
16 ¹
67. You can use the line y
x bounded by x
a and
x
b. The sum of the areas of these inscribed
rectangles is the lower sum.
y
4
y
3
2
1
1
2
3
x
4
(b) A | 6 square units
x
a
b
y
66.
The sum of the areas of these circumscribed rectangles is
the upper sum.
4
3
y
2
1
1
2
3
x
4
(a) A | 3 square units
x
a
b
You can see that the rectangles do not contain all of the
area in the first graph and the rectangles in the second
graph cover more than the area of the region. The exact
value of the area lies between these two sums.
68. See the definition of area, page 260.
69. (a)
y
8
6
4
2
x
1
2
3
4
0 4 5 13 6
Lower sum: s 4
(b)
15 13
46
3
| 15.333
y
8
6
4
2
x
1
2
3
4
INSTRUCTOR USE ONLY
Upper sum: S 4
4 5 13 6 6 52
111
1
2115
326
32
15
| 21.733
21.73
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380
NOT FOR SALE
Chapter 4
Integration
egration
y
(c)
8
6
4
2
x
1
2
3
4
Midpoint Rule: M 4
(d) In each case, 'x
2 23 4 54 5 75 6 92
6112
315
| 19.403
4 n. The lower sum uses left end-points, i 1 4 n . The upper sum uses right endpoints,
i 4 n . The Midpoint Rule uses midpoints, i 12 4 n .
(e)
N
4
8
20
100
200
s(n)
15.333
17.368
18.459
18.995
19.06
S(n)
21.733
20.568
19.739
19.251
19.188
M(n)
19.403
19.201
19.137
19.125
19.125
(f ) s (n) increases because the lower sum approaches the exact value as n increases. S(n)decreases because the upper sum
approaches the exact value as n increases. Because of the shape of the graph, the lower sum is always smaller than the
exact value, whereas the upper sum is always larger.
70. (a) Left endpoint of first subinterval is 1.
Left endpoint of last subinterval is 4 14
15 .
4
(b) Right endpoint of first subinterval is 1 14
Right endpoint of second subinterval is 1 12
5.
4
3
.
2
(c) The rectangles lie above the graph.
(d) The heights would be equal to that constant.
71. True. (Theorem 4.2 (2))
2S
n
74. (a) T
72. True. (Theorem 4.3)
73. Suppose there are n rows and n 1 columns in the
figure. The stars on the left total 1 2 " n, as do
(b) sin T
h
the stars on the right. There are n n 1 stars in total, so
2>1 2 " n@
1 2" n
nn 1
1 n
2
n 1.
A
(c) An
r
h
r
r sin T
1
bh
2
r
1
r r sin T
2
1 2
r sin T
2
2S ·
§1
n¨ r 2 sin
¸
n ¹
©2
r 2n
2S
sin
2
n
Let x
lim An
n of
h
θ
§ sin 2S n ·
¸
© 2S n ¹
Sr2¨
2S n. As n o f, x o 0.
§ sin x ·
lim S r 2 ¨
¸
© x ¹
x o0
Sr2 1
Sr2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.2
Secti
Sectio
Area
381
75. For n odd,
n
1, 1 row,
1 block
n
3, 2 rows,
4 blocks
n
5, 3 rows,
9 blocks
2
n 1
§ n 1·
rows, ¨
¸ blocks,
2
© 2 ¹
n,
For n even,
n
2, 1 row,
2 block
n
4, 2 rows,
6 blocks
n
6, 3 rows,
12 blocks
n,
n
rows,
2
n
76. (a)
¦ 2i
n 2 2n
blocks,
4
nn 1
i 1
The formula is true for n
1: 2
11 1
2.
k
k : ¦ 2i
Assume that the formula is true for n
k k 1.
i 1
k 1
Then you have ¦ 2i
i 1
k
¦ 2i 2 k 1
k k 1 2k 1
which shows that the formula is true for n
n
(b) ¦ i 3
i 1
n2 n 1
4
k 1.
2
1 because 13
The formula is true for n
12 1 1
4
2
k
k2 k 1
.
4
k : ¦ i3
Assume that the formula is true for n
i 1
k 1
Then you have ¦ i 3
i 1
k 1 k 2
i 1
k
¦ i3 k 1
3
i 1
which shows that the formula is true for n
k2 k 1
4
4
4
1.
2
2
2
k 1
3
k 1
ªk 2 4 k 1 º¼
4 ¬
k 1
4
2
k 2
2
k 1.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
382
Chapter 4
Integration
egration
77. Assume that the dartboard has corners at r1, r1 .
A point (x, y) in the square is closer to the center than the top edge if
x y
2
1
x y d 1 2y y
2
y
d1 y
2
2
(x, 1)
2
(x, y)
y d 12 1 x 2 .
−1
By symmetry, a point (x, y) in the square is closer to the center than the right edge if
(0, 0)
1
x
−1
x d 12 1 y 2 .
1 1 x2
2
In the first quadrant, the parabolas y
1 1 y2
2
and x
intersect at
2 1,
2 1 . There are 8 equal
regions that make up the total region, as indicated in the figure.
y
1
( 2 − 1,
2 − 1(
(1, 1)
−1
x
1
−1
Probability
2 1 ª 1
³0
Area of shaded region S
8S
Area square
«2 1 x
¬
2
ª2 2
5º
2«
»
3
6
¬
¼
º
x» dx
¼
2 2
5
3
6
4 2
5
3
3
Section 4.3 Riemann Sums and Definite Integrals
0, x
1. f x
x, y
'xi
3i 1
3i 2
2
n
n2
n
lim ¦ f ci 'xi
n of
i 1
0, x
2
3i 2
n2
3
2i 1
n2
n
lim ¦
n of
3, ci
i 1
3i 2 3
2i 1
n2 n2
3 3 n
lim 3 ¦ 2i 2 i
n of n
i 1
lim
n of
nn 1º
3 3 ª n n 1 2n 1
«2
»
n3 ¬
6
2 ¼
ª n 1 2n 1
n 1º
lim 3 3 «
»
2
n
3
2n 2 ¼
¬
n of
ª2
º
3 3 « 0»
¬3
¼
2 3 | 3.464
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.3
3
2. f x
'xi
x, y
0, x
i 1
i3
3
n
n3
3
n
i3
n3
1, ci
i 3 ª 3i 2 3i 1º
«
»
n3 ¬
n3
¼
lim ¦ 3
n of
i 1
383
3i 2 3i 1
n3
n
lim ¦ f ci 'xi
n of
0, x
Riemann Sums and Definite
Defin
Defi
Integrals
lim
i 1
n
1
n of n 4
¦ 3i3 3i 2 i
i 1
1 ª § n2 n 1
«3¨
n of n 4 « ¨
4
¬ ©
2
lim
·
§ n n 1 2n 1 · n n 1 º
»
¸ 3¨
¸
¸
6
2 »
©
¹
¹
¼
1 ª 3n 4 6n3 3n 2
2n3 3n 2 n
n2 n º
«
»
4
n of n
4
2
2 ¼
¬
lim
1 ª 3n 4
n3
n2 º
«
»
n of n 4
2
4¼
¬ 4
lim
3. y
n
¦ f ci 'xi
i 1
6
³ 2 8 dx
4. y
62
n
§
8 on >2, 6@. ¨ Note: 'x
©
§
n
4i ·§ 4 ·
lim 32
n
i 1
§
n
i 1
3 2
n
n
i 1
32
i 1
1 n
¦ 32
ni 1
1
32n
n
32
5i ·§ 5 ·
§
5i ·§ 5 ·
25 ·
§5
lim ¨ ¸
2n ¹
10 §
n
25 n
¦i
n2 i 1
§ 25 · n n 1
10 ¨ 2 ¸
2
©n ¹
10 25 §
1·
¨1 ¸
2©
n¹
5
25
2 2n
5
2
n of © 2
§
x3 on >1, 1@. ¨ Note: 'x
©
¦ f ci 'xi
n
¦ n
·
5
, ' o 0 as n o f ¸
n
¹
i 1
5. y
3
4
i 1
n
³ 2 x dx
1
1 º
2n 4n 2 »¼
¦ f ¨© 2 n ¸¨
¸
¹© n ¹
¦ ©¨ 2 n ¹©
¸¨ ¸
n¹
3
32
n of
§
x on >2, 3@. ¨ Note: 'x
©
¦ f ci 'xi
§ 4·
¦ 8¨© n ¸¹
i 1
ª3
4
·
, ' o 0 as n o f ¸
n
¹
n
¦ f ¨© 2 n ¸¨
¸
¹© n ¹
lim
n of «
¬4
1 1
n
·
2
, ' o 0 as n o f ¸
n
¹
2i ·§ 2 ·
¦ f ¨© 1 n ¸¨
¸
¹© n ¹
i 1
n
3
§
2i · § 2 ·
ª
6i
¦ ¨© 1 n ¸¹ ¨© n ¸¹
i 1
n
8i 3 º § 2 ·
12i 2
¦ «1 n n 2 n3 »¨© n ¸¹
i 1¬
2 1
³ 1 x dx
3
¼
n
n
12
24
16 n
i 3 ¦ i 2 4 ¦ i3
2¦
n i 1
n i 1
n i 1
1·
3
1·
2
1·
§
§
§
2 6¨1 ¸ 4¨ 2 2 ¸ 4¨1 2 ¸
n¹
n
n ¹
n
n ¹
©
©
©
2
lim
0
n of n
2
n
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
384
Chapter 4
NOT FOR SALE
Integration
egration
4 1
n
§
4 x 2 on >1, 4@. ¨ Note: 'x
©
6. y
n
¦ f ci 'xi
i 1
n
§
3i ·§ 3 ·
§
3i · § 3 ·
3
·
, ' o 0 as n o f ¸
n
¹
¦ f ¨©1 n ¸¨
¸
¹© n ¹
i 1
n
2
¦ 4¨©1 n ¸¹ ¨© n ¸¹
i 1
12 n §
6i 9i 2 ·
2¸
¨1 ¦
n i 1©
n
n ¹
12 ª
6n n 1
9 n n 1 2n 1 º
2
«n »
n¬
n
n
2
6
¼
12 36
4
³1 4 x dx
2
n 1 2n 1
n1
18
n
n2
ª
36 n 1
18 n 1 2n 1 º
lim «12 »
n
n2
¬
¼
12 36 36 84
n of
2 1
n
§
x 2 1 on [1, 2]. ¨ Note: 'x
©
7. y
n
§
n
¦ f ci 'xi
1
·
, ' o 0 as n o f ¸
n
¹
i ·§ 1 ·
¦ f ¨©1 n ¸¨
¸
¹© n ¹
i 1
i 1
2
ª§
º§ 1 ·
i·
1
«
¦ ¨© n ¸¹ 1»¨© n ¸¹
»¼
i 1 «
¬
n
n
ª
2i
º§ 1 ·
i2
¦ «1 n n2 1»¨© n ¸¹
i 1¬
2
2
¼
n
n
2
1
¦ i n3 ¦ i 2
n2 i 1
i 1
3
1 ·
§ 10
lim ¨
¸
2n 6n 2 ¹
³ 1 x 1 dx
2
n of © 3
§
2 x 2 3 on >2, 1@. ¨ Note: 'x
©
8. y
n
¦ f ci 'xi
i 1
n
§
1 · 1§
3
1·
§
2 ¨1 ¸ ¨ 2 2 ¸
n ¹ 6©
n
n ¹
©
10
3
1
3
2n 6n 2
10
3
1 2
n
·
3
, ' o 0 as n o f ¸
n
¹
3i ·§ 3 ·
¦ f ¨© 2 n ¸¨
¸
¹© n ¹
i 1
2
ª §
º§ 3 ·
3i ·
2
2
«
¦ ¨©
¸ 3» ¨ ¸
n¹
»¼ © n ¹
i 1 «
¬
n
º
3 n ª §
12i 9i 2 ·
2 ¸ 3»
«2¨ 4 ¦
ni 1 ¬ ©
n
n ¹
¼
3 n ª
24i 18i 2 º
2 »
¦
«11 ni 1 ¬
n
n ¼
3ª
24 n n 1
18 n n 1 2n 1 º
2
«11n »
n¬
n
n
2
6
¼
1
³ 2 2 x 3 dx
2
ª
n 1 2n 1 º
n 1
lim «33 36
9
»
n
n2
¬
¼
n of
33 36
n 1 2n 1
n 1
9
n
n2
33 36 18
15
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.3
n
9. lim ¦ 3ci 10 'xi
' o0
i 1
n
' o0
2
A
bh
4
³0
i 1
10 6
6
³ 4 6 dx
A
'xi
385
24. Rectangle
5
³ 1 3x 10 dx
on the interval >1, 5@.
10. lim ¦ 6ci 4 ci
Riemann Sums and Definite
Defin
Defi
Integrals
60
60
y
6x 4 x
2
dx
on the interval [0, 4].
4
Rectangle
n
11. lim ¦
' o0
ci2
3
³0
4 'xi
i 1
2
x 4 dx
2
x
−4
−2
2
4
6
1 4
2
4
on the interval [0, 3].
n
§3·
12. lim ¦ ¨ 2 ¸ 'xi
' o0
i 1 © ci ¹
3
³ 1 x 2 dx
3
on the interval [1, 3].
25. Triangle
A
1 bh
2
A
³ 0 x dx
4
13. ³ 5 dx
4
8
8
y
0
14. ³
15. ³
2
0
4
6 3x dx
Triangle
2
4
4 x dx
4
x
2
4
2
16. ³ x 2 dx
26. Triangle
0
5
17. ³
5
18. ³
1 x 2
19. ³
0
20. ³
0
21. ³
25 x 2 dx
A
4
dx
2
A
1
S 2
2
0
8
y
cos x dx
S 4
1
1
bh
8 2
2
2
8 x
³ 0 4 dx 8
4
2
tan x dx
Triangle
x
2
y 3 dy
4
6
8
27. Trapezoid
22. ³
2
0
2
y 2 dy
A
b1 b2
h
2
A
³ 0 3x 4 dx
23. Rectangle
A
bh
A
³ 0 4 dx
34
§ 4 10 ·
¨
¸2
© 2 ¹
2
14
14
y
3
12
12
y
8
Trapezoid
5
4
3
x
−1
Rectangle
2
1
2
3
−4
1
INSTRUCTOR
N
USE ONLY
x
1
2
3
4
5
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 4
386
Integration
egration
32. Semicircle
28. Trapezoid
b1 b 2
A
2
3
³0
A
8 2
3
2
h
8 2 x dx
15
A
1S r2
2
A
³ r
r
15
1S r 2
2
r 2 x 2 dx
y
y
Semicircle
r
8
6
Trapezoid
−r
x
r
4
2
−r
x
1
2
3
4
29. Triangle
4
1
1
³ 2 dx
³ 1 1 x dx
1
33. ³ x dx
1 bh
2
A
1 2
2
1
A
2
y
2
34. ³ x3 dx
2
Triangle
1
4
35. ³ 8 x dx
x
−1
2
1
4
36. ³ 25 dx
30. Triangle
A
2
1 bh
2
1 2a
2
a2
a
a
³ a a x dx
a
2
y
a
Triangle
−a
2
4
37. ³
2
38. ³
2
39. ³
2
4
4
4
³ x dx
4
2
x dx
6,
6
2
0
8³
4
x dx
2
4
25 ³ dx
2
86
48
25 2
50
4
4
692
12
4
60 4 2
68
x 9 dx
³ 2 x dx 9 ³ 2 dx
x3 4 dx
³ 2 x dx 4³ 2 dx
4
3
³
1 x 3 3 x 2 dx
2
4
³
4
³
36 22
1 60
2
4
40. ³ 10 4 x 3x3 dx
16
4
4
4
2
2
2
10³ dx 4³ x dx 3³ x3 dx
2
10 2 4 6 3 60
31. Semicircle
A
A
1 2
Sr
2
1
2
S 7
2
7
49 x dx
³ 7
2
y
49S
2
49S
2
41. (a)
8
Semicircle
5
0
³
7
³ 0 f x dx ³ 0 f x dx ³ 5 f x dx
(b)
³ 5 f x dx
(c)
³ 5 f x dx
(d)
³ 0 3 f x dx
12
10
7
5
5
4
1
x3 dx 3 x dx 2 dx
2 2
2
2
x
a
60, ³
x 3 dx
2
4
A
4
In Exercises 33 – 40, ³
5
0
f x dx
10
f x dx
3 10
136
10 3
13
0
3³
5
0
30
6
4
2
x
−8 −6 −4 −2
2
4
6
8
−4
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.3
42. (a)
(b)
6
3
3
³6
3
f x dx
³
f x dx
0
(c)
³3
(d)
³ 3 5 f x dx
43. (a)
6
³ 0 f x dx ³ 0 f x dx ³ 3 f x dx
6
1
f x dx
3
4 1
1
6
5 ³
6
3
6
6
³ 2 ª¬g x f x º¼ dx
³2
(d)
44. (a)
2³ g x dx
2
³ 2 3 f x dx
0
³ 1 f x dx
6
3³
f x dx
2
1
6
2
f x dx
12
1 4
2
2
4
0
(c)
³ 1 3 f x dx
(d)
³ 0 3 f x dx
1
3³
1
³ 3 3 f x dx
(c)
³ 0 f x dx
(d)
³ 5 f x dx
10
(e)
³ 0 f x dx
(f ) ³
4
3 10
5 5
30
30
0
3³ f x dx
35
15
1
0
1
(b)
2 2
f x dx
1
1 2S
(f ) Answers to (d) plus
2 10
20: 3 2S 20
³ 0 f x dx
5
³ 0 f x dx ³1 f x dx
2
3 2S
(e) Sum of absolute values of (b) and (c):
4 1 2S
5 2S
48. (a)
³ 1 f x dx ³ 0 f x dx
(b)
1
S
12 2 1 12 S 2
1
05
1
8
g x dx ³
6
6
2
14 S 2
(d) Sum of parts (b) and (c): 4 1 2S
³ 2 f x dx ³ 2 g x dx
6
³ 2 2 g x dx
14 S r 2
5
6
2 10
(c)
47. (a) Quarter circle below x-axis:
(b) Triangle: 12 bh
5 1
f x dx
6
³ 2 ª¬ f x g x º¼ dx
6
387
(c) Triangle Semicircle below x -axis:
10 2
(b)
3
Riemann Sums and Definite
Defin
Defi
Integrals
1
³ f x dx
1
2
0
4
7
23 2S
32
6
12 12 2 2 2 12 2 2 12
11
1 1 4
2
2
11
12 2 2 2 4 12
10
24
f x dx
4
2 12
5
3
2
2
y
2
(3, 2) (4, 2)
(11, 1)
45. Lower estimate: >24 12 4 20 36@ 2
48
Upper estimate: >32 24 12 4 20@ 2
88
46. (a) >6 8 30@ 2
1
236 right endpoint estimate
(c) >0 18 50@ 2
136 midpoint estimate
If f is increasing, then (a) is below the actual value and
(b) is above.
2
−1
4
49. (a)
8
12
(0, 1)
−2
64 left endpoint estimate
(b) >8 30 80@ 2
x
−2
(8, 2)
5
³ 0 f x dx ³ 0 2 dx
5
3
5
³ 0 ª¬ f x 2º¼ dx
4 10
(b)
³ 2 f x 2 dx ³ 0 f x dx
(c)
³ 5 f x dx
(d)
³ 5 f x dx
5
5
2³
0
5
0
f x dx
5
14
4 Let u
24
x 2.
8 f even
f odd
50. (a) The left endpoint approximation will be greater than
the actual area so,
n
¦ f xi 'x ! ³ 1 f x dx.
5
i 1
(b) The right endpoint approximation will be less than
the actual area so,
n
¦ f xi 'x ³ 1 f x dx.
5
INSTRUCTOR USE ONLY
i 1
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 4
388
Integration
egration
­4, x 4
®
¯ x, x t 4
51. f x
8
³ 0 f x dx
y
55.
2
4 4 4 4 12 4 4
3
2
40
1
y
1
2
(8, 8)
8
1
1
2
x
2
3
2
(d) A | 54 square units
4
56.
y
x
4
8
9
8
7
6
5
4
3
2
1
x ! 6
°­6,
® 1
x
9,
x d 6
°̄ 2
52. f x
y
10
x
1 2 3 4 5 6 7 8 9
12
(c) A | 27 square units
(0, 9)
8
6
1
x 4
57. f x
4
2
is not integrable on the interval [3, 5] because f has a
discontinuity at x
4.
x
2
4
6
8
12
³ 0 f x dx
10
12
6 6 12 6 3 6 6
81
x x is integrable on >1, 1@, but is not
58. f x
continuous on >1, 1@. There is discontinuity at
y
53.
36 9 36
x
0. To see that
4
1
x
³ 1 x dx
3
2
1
1
2
3
4
x
is integrable, sketch a graph of the region bounded by
f x
x x and the x-axis for 1 d x d 1. You see
that the integral equals 0.
(a) A | 5 square units
54.
y
2
y
1
4
x
−2
1
2
3
2
−2
1
x
1
4
1
2
3
4
1
(b) A | 43 square units
59. ³
1
2
2, b
a
60. ³
f x dx ³
3
3
5
1
5
³ 2 f x dx
f x dx
5
6
b
3
a
6
a
f x dx ³ f x dx ³ f x dx
6
³ 1 f x dx
6
³ 3 f x dx ³ b f x dx ³ 1 f x dx
a
3, b
1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.3
S, b
61. Answers will vary. Sample answer: a
2S
Riemann Sums and Defi
Defin
Definite Integrals
389
63. True
2S
³ S sin x dx 0
64. False
1
1
1
³ 0 x x dx z ³ 0 x dx ³ 0
y
x dx
1
65. True
π
2
x
3π
2
66. True
67. False
−1
2
62. Answers will vary. Sample answer: a
S
³ 0 cos x dx
0, b
³ 0 x dx
S
2
68. True. The limits of integration are the same.
0
y
π
4
π
2
3π
4
x
π
−1
x 2 3x, >0, 8@
69. f x
x0
0, x1
'x1
1, 'x2
c1
1, c2
1, x2
3, x3
2, 'x3
2, c3
4
¦ f ci 'x
7, x4
4, 'x4
5, c4
8
1
8
f 1 'x1 f 2 'x2 f 5 'x3 f 8 'x4
i 1
4 1 10 2 40 4 88 1
sin x, >0, 2S @
70. f x
x0
0, x1
S
'x1
c1
4
4
S
6
S
4
, x2
S
, 'x2
, c2
¦ f ci 'xi
i 1
272
12
S
3
, c3
S
3
, 'x3
S , x4
, x3
2S
, 'x4
3
2S
, c4
3
2S
S
3S
2
§S ·
§S ·
§ 2S ·
§ 3S ·
f ¨ ¸ 'x1 f ¨ ¸ 'x2 f ¨ ¸ 'x3 f ¨ ¸ 'x4
©6¹
©3¹
© 3 ¹
© 2 ¹
§ 1 ·§ S · § 3 ·§ S · § 3 ·§ 2S ·
¸¨ ¸ ¨
¸¨ ¸ 1 S | 0.708
¨ ¸¨ ¸ ¨¨
© 2 ¹© 4 ¹ © 2 ¸¹© 12 ¹ ¨© 2 ¸¹© 3 ¹
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
390
Chapter 4
71. 'x
NOT FOR SALE
Integration
egration
ba
, ci
n
§b a·
a i¨
¸
© n ¹
a i 'x
n
b
lim ¦ f ci 'x
³ 0 x dx
' o0
i 1
n
ª
§ b a ·º§ b a ·
lim ¦ «a i¨
¸»¨
¸
nof
© n ¹¼© n ¹
i 1¬
2
ª§ b a · n
§b a· n º
lim Ǭ
a
¸¦
¨
¸ ¦ i»
nof ©
© n ¹ i 1 »¼
«¬ n ¹ i 1
2
ªb a
§b a· n n 1 º
lim «
an ¨
»
¸
nof
2
© n ¹
¬« n
¼»
2
ª
b a n 1º
»
lim «a b a n of«
2 »
n
¬
¼
ab a b a
2
2
b aº
ª
b a «a 2 »¼
¬
b a a b
2
72. 'x
ba
, ci
n
b
³ a x dx
2
b2 a 2
2
§b a·
a i¨
¸
© n ¹
a i 'x
n
lim ¦ f ci 'x
' o0
i 1
2
n
ª
§ b a ·º § b a ·
lim ¦ «a i¨
¸» ¨
¸
n of
© n ¹¼ © n ¹
i 1 ¬
2 º
ª§ b a · n §
·
2ai b a
2
2§ b a ·
lim Ǭ
a
i
¨
¸»
¦
¸
¨
¸
¨
¸
n of «©
n ¹i 1 ©
n
© n ¹ ¹»¼
¬
2a b a n n 1
§ b a ·ª 2
§ b a · n n 1 2n 1 º
lim ¨
¨
»
¸ «na ¸
n of ©
n ¹ ¬«
n
2
6
© n ¹
¼»
2
2
3
ª
ab a n 1
b a n 1 2n 1 º
»
lim «a 2 b a n of «
n
n2
6
»¼
¬
1
1 3
2
3
a2 b a a b a b a
b a3
3
3
73. f x
­1, x is rational
®
¯0, x is irrational
is not integrable on the interval >0, 1@. As
' o 0, f ci
1 or f ci
0
­0, x
°
®1
° , 0 x d1
¯x
74. f x
0 in each subinterval
because there are an infinite number of both rational and
irrational numbers in any interval, no matter how small.
y
1
2
f(x) = x
The limit
1
n
lim ¦ f ci 'xi
' o0
i 1
x
1
2
does not exist.
This does not contradict Theorem 4.4
because f is not continuous on [0, 1].
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.4
1 and
75. The function f is nonnegative between x
x 1.
The Fundamental Theorem of Calculus
76. To find ³
y
So, ³
b
a
f(x) = 1 − x 2
3
2
is a maximum for
x
−2
1 and b
a
axb dx, use a geometric approach.
y
2
1 x 2 dx
2
0
391
1
1
2
−1
1.
x
−1
1
−2
x 2 , 0 d x d 1, and 'xi
n
¦ f ci 'xi
i 1
n
2
§i· 1
2
0
axb dx
12 1
1.
1 n. The appropriate Riemann Sum is
1 n 2
¦i .
n3 i 1
¦ ¨© n ¸¹ n
i 1
1 n 2n 1 n 1
˜
n of n 3
6
1 2
ª1 22 32 " n 2 ¼º
n of n 3 ¬
lim
1
3
−1
So, ³
77. Let f x
2
lim
1
2n 2 3n 1
n of
6n 2
lim
1
1 ·
§1
lim ¨ ¸
2n 6n 2 ¹
1
3
n of © 3
³ 0 x f x dx ³ 0 xf x dx.
78. I f J f
2
2
Observe that
x3
x·
§
x¨ f x ¸
4
2¹
©
2
§
x3
x2 ·
2
x¨ f x xf x ¸
4
4¹
©
1
1 ª x3
³ 0 ¬ªx f x xf x ¼º dx
So, I f J f
x
. Then I f
2
Furthermore, 6 f x
1
1
8 16
So I f J f
The maximum value is
§
x· º
2
x 2 f x xf x
1 x3
¬
1
¼
2§ x ·
³ 0 x ¨© 2 ¸¹ dx
1
and J f
8
§ x2 ·
1
³ 0 x¨© 4 ¸¹
2
1
16
³ 0 «« 4 x©¨ f x 2 ¹¸ »» dx d ³ 0 4 dx
2
2
x3
x3
2
xf x x 2 f x 4
4
1
16
1
16
1
.
16
Section 4.4 The Fundamental Theorem of Calculus
1. f x
S
³0
4
x2 1
5
3. f x
2
4
dx is positive.
x2 1
x
x2 1
³ 2 x x 1 dx
−5
2
5
0
−5
5
5
−5
−2
4. f x
x 2 x
5
2. f x
S
³ 2 x 2 x dx is negative.
2
cos x
³ 0 cos x dx
2
0
−2
0
2
−5
−2
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 4
392
2
2
5. ³ 6 x dx
ª¬3x 2 º¼
0
0
1
6. ³
3
7. ³
1
0
Integration
egration
32
>8t@3
1
8 dt
2
8. ³
0
8 1 8 3
2
1
7 3t dt
2
ª7t 3 t 2 º
2 ¼ 1
¬
ª7 2 3 4 º ª7 1 3 1 2 º
2
2
¬
¼ ¬
¼
32
14 6 7 32
33
2
0
ª¬ x 2 xº¼
1
2 x 1 dx
0
1
2
1
1
11
9. ³
2
1
1
t 2 2 dt
ªt 3
º
« 2t »
¬3
¼ 1
§1
· § 1
·
¨ 2¸ ¨ 2¸
©3
¹ © 3
¹
2
10. ³
1
11. ³
0
1
2
6 x 2 3x dx
ª2 x 3 3 x 2 º
2 ¼1
¬
2
³ 0 4t 4t 1 dt
2t 1
1
dt
3
ª¬ x 4 x3 º¼
1
1
14. ³
15. ³
2
1
8
16. ³
8
17. ³
1
18. ³
1
19. ³
1
8
u
2
dx
x
3
2
2t
21. ³
1
0
x
23 x
3ª 43
43
8 8 º
¼
4¬
1
8
2 ³ x 1 2 dx
1
dx
t dt
dx
3 2
4
8
ª 2 2 x1 2 º
¬
¼1
1 1
x x1 2 dx
3 ³0
2
³0
ª2
«3
¬
3
16 16
4
3 2
4
0
ª 3 t 4 3 3t5 3º
5
¬4
¼ 1
8
ª2 2 x º
¬
¼1
1
º ª2
º
4 4 » « 4»
3
¼ ¬
¼
2
3
82 2
1§ 1 2 ·
¨ ¸
3© 2 3 ¹
1
18
2
ªt t
º
20 6t »
«
15
¬
¼0
ª4 3 2 2 5 2º
«3t 5t »
¬
¼0
3 3
4
5
3
4
1 ª x2
2 3 2º
« x »
3¬ 2
3
¼0
0
4
0
2
2t1 2 t 3 2 dt
1
3
2
ª2 3 2
1 2º
« 3 u 4u »
¬
¼1
2u 1 2 du
19
2
1
2
1·
§1
· §
¨ 1¸ ¨ 2 ¸
2¹
©2
¹ ©
ª 3 t 4 3 2t º
¬4
¼ 1
t1 3 t 2 3 dt
1 x x 2
8
ª3 4 3º
«4 x »
¬
¼ 8
t 2 dt
0
0
12
8
3
4 2 1
3
4
4
³1 u
du
x1 3 dx
1 x 20. ³
22. ³
1
16 6 2 32
10
3
54
§ 3
·
¨ 2 ¸ 3 1
2
©
¹
ªu 2
1º
« »
2
u
¬
¼ 2
1·
u 2 ¸ du
2 ¨
u ¹
©
1
ª 4 t 3 2t 2 t º
¬3
¼0
81 27 1 1
ª 3
º
« x x»
¬
¼1
1 §
4 u 2
2
3
12. ³ 4 x 3 3 x 2 dx
2 § 3
·
13. ³ ¨ 2 1¸ dx
1
x
©
¹
ª2 8 3 4 º ª2 1 3 1 º
2
2 ¼
¬
¼ ¬
2 2
20 12
15
16 2
15
27
20
1 1 2 3
x x5 3 dx
2 ³ 8
1
1 ª3 5 3 3 8 3 º
x x »
2 «¬ 5
8
¼ 8
1
ª x5 3
º
24 15 x »
«
¬ 80
¼ 8
1
32
39 144
80
80
4569
80
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.4
23. ³
5
0
52
³0
2 x 5 dx
5 2 x dx ³
52
ª¬5 x x 2 º¼
0
Note: By Symmetry, ³
24. ³
4
ª¬ x 2 5 xº¼
2³
2 x 5 dx
0
2 x 5 dx split up the integral at the zero x
52
5
25 25
2
4
52
5
0 25 25 393
5
2
25 25
4
2
2 25
25
2
4
25
2
2 x 5 dx.
52
3
4
³1 ª¬3 x 3 º¼ dx ³ 3 ª¬3 x 3 º¼ dx
3 1x 31 dx
1
5
5
The Fundamental Theorem of Calculus
3
4
³1 x dx ³ 3 6 x dx
3
4
ª x2 º
ª
x2 º
« » «6 x »
2 ¼3
¬ 2 ¼1 ¬
9 ·º
§9 1· ª
§
¨ ¸ « 24 8 ¨18 ¸»
2 ¹¼
©2 2¹ ¬
©
4 16 18 25. ³
4
3
4
13
2
³ 0 9 x dx ³ 3 x 9 dx split up integral at the zero x
x 2 9 dx
0
9
2
2
3
2
4
ª
ª x3
º
x3 º
«9 x » « 9 x»
3 ¼0 ¬ 3
¬
¼3
26. ³
4
0
1
3
§ 64
·
36 ¸ 9 27
27 9 ¨
© 3
¹
3
4
64
3
³ 0 x 4 x 3 dx ³1 x 4 x 3 dx ³ 3 x 4 x 3 dx split up the integral at the zeros x
x 2 4 x 3 dx
2
2
1
2
3
1, 3
4
ª x3
º
ª x3
º
ª x3
º
2
2
2
« 2 x 3x» « 2 x 3 x» « 2 x 3 x»
¬3
¼0 ¬ 3
¼1 ¬ 3
¼3
§1
·
§1
· § 64
·
¨ 2 3¸ 9 18 9 ¨ 2 3¸ ¨ 32 12¸ 9 18 9
©3
¹
©3
¹ ©3
¹
4
4 4
0 0 4
3
3 3
S
27. ³
0
28. ³
0
29. ³
S
1 sin x dx
> x cos x@0
2 cos x dx
>2 x sin x@0
S
S 4 1 sin 2 T
0
S 4
30. ³
0
31. ³
S 6
32. ³
S 4
33. ³
S 3
S 6
S 2
S 3
cos 2 T
S
dT
sec 2 T
dT
tan 2 T 1
sec2 x dx
S 1 0 1
S 4
³ 0 dT
S 4
>T @0
S 4 sec 2 T
³0
sec 2 T
>tan x@ SS6 6
2S 0 0
2S
2S
S
4
S 4
S 4
dT
>T @0
3 §
3·
¨¨ ¸¸
3
3
©
¹
2 3
3
dT
2 csc 2 x dx
>2 x cot x@SS 24
4 sec T tan T dT
>4 sec T @SS 3
3
³0
§S
·
1¸
©2
¹
S 0 ¨
42 42
S
4
S
2
1
S 2
2
0
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 4
394
34. ³
S 2
S 2
Integration
egration
§S 2
· §S 2
·
1¸ ¨
1¸
¨
4
4
©
¹ ©
¹
S 2
ª¬t 2 sin t º¼
S 2
2t cos t dt
1
1
x x 2 dx
2
1
35. A
³0
36. A
³ 1 x 2 dx
37. A
³0
38. A
S 2
ª x2
x3 º
« »
3 ¼0
¬2
2
ª 1º
« x »
¬ ¼1
cos x dx
1
6
1
2
S 2
>sin x@0
1
S
S
³0
x sin x dx
2
ª x2
º
« cos x»
2
¬
¼0
S2
2
S2 4
2
2
39. Because y ! 0 on >0, 2@,
Area
2
³ 0 5 x 2 dx
2
2
ª 5 x 3 2 xº
¬3
¼0
40 4
3
52 .
3
40. Because y ! 0 on [0, 2],
2
Area
2
³0
ª x4
x2 º
« »
2 ¼0
¬4
x 3 x dx
4 2
6.
8
3
16
4
20.
32
8
3
8
.
3
41. Because y ! 0 on [0, 8],
Area
8
8
³0
1 x1 3 dx
3 4 3º
ª
«x 4 x »
¬
¼0
42. Because y t 0 on [0, 4],
Area
4
³ 0 2 x x dx
4
4
³0
2 x1 2 x dx
ª4 3 2
x2 º
« x »
2 ¼0
¬3
43. Because y ! 0 on >0, 4@,
4
Area
4
³ 0 x 4 x dx
2
ª x3
º
2x2 »
«
3
¬
¼0
64
32
3
32
.
3
44. Because y ! 0 on >1, 1@,
Area
1
³ 1 1 x dx
2³
4
1
0
1 x 4 dx
1
ª
x5 º
2«x »
5 ¼0
¬
1·
§
2¨1 ¸
5¹
©
8
.
5
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.4
3
ª x4 º
« »
¬ 4 ¼0
3
45. ³ x 3 dx
0
f c 30
f c
c
3
c
49. ³
81
4
81
4
27
4
27
4
3
3
4
The Fundamental Theorem of Calculus
S 4
>2 tan x@SS 4
4
2 sec 2 x dx
S 4
ªS
§ S ·º
f c « ¨ ¸»
¬ 4 © 4 ¹¼
8
S
4
sec 2 c
S
9
4
9
ª 2 x3 2 º
¬3
¼4
x dx
f c 94
38
3
f c
38
15
c
38
15
c
1444
225
r arccos
18
f c
3
c2
4
3
c
r 12
c
2 3
18
51.
9
dx
1 x3
ª 9 º
« 2 x 2 »
¬
¼1
f c 31
4
9
c3
2
c
3
c
3
3
1
3
3 3 ³ 3
3
1ª
1 º
9 x x3 »
6 «¬
3 ¼ 3
9 x 2 dx
1
ª 27 9 27 9 º¼
6¬
6
6
Average value
9 x
2
9 6 or x
6 when x 2
r 3 | r1.7321.
10
3
3
| r 0.4817
3 3
2S
c | r0.5971
2 3 is not in interval.
48. ³
S 3
>sin x@S 3
cos x dx
S
2
cos c
216
12
f c 6 0
S 3
S 3
ªS
§ S ·º
f c « ¨ ¸»
¬ 3 © 3 ¹¼
| 6.4178
6
S
§ 2 ·
r arcsec¨
¸
© S¹
38
3
50. ³
ª x3 º
« »
¬12 ¼ 0
x2
47. ³
dx
0 4
6
2 27 8
3
2
r
sec c
c
46. ³
4
4
2 sec 2 c
33
2 | 1.8899
2
2 1 2 1
395
1 9
2 2
(− 3, 6 )
( 3, 6 )
4
−4
4
0
52.
9
2
3
2
3 4 x 1
1
dx
³
x2
31 1
2³
3
1
1 x 2 dx
3
1º
ª
2«x »
x ¼1
¬
9
| 1.6510
2
1·
§
2¨ 3 ¸
3¹
©
16
3
10
( 3, )
16
3
4
0
0
Average value
4 x2 1
16
3
16
Ÿ x
3
3 on >1, 3@
INSTRUCTOR USE ONLY
O
x2
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© Cengage Learning. All Rights Reserved.
396
53.
NOT FOR SALE
Chapter 4
Integration
egration
1
1
1
x3 dx
1 0 ³0
ª x4 º
« »
¬ 4 ¼0
Average value
1
4
x3
1
4
x
1
3
4
3
5
58. The distance traveled is ³ v t dt. The area under the
1
4
0
curve from 0 d t d 5 is approximately
145 ft.
(29 squares) 5
1
59. (a)
(
2
| 0.6300
2
3
2, 1
2 4
7
³ 1 f x dx
Sum of the areas
A1 A2 A3 A4
)
1 31
2
1
0
12 1 2 12 2 1 3 1
8
0
y
54.
1
1
4 x3 3 x 2 dx
1 0 ³0
Average value
1
ª¬ x 4 x3 º¼
0
4
0
A1
3
A2
0
4 x3 3x 2
0
x 4x 3
0
2
A4
1
x
1
0, 34
x
A3
2
2
3
4
5
7
6
7
³ 1 f x dx
1
(b) Average value
0
( )
3
,0
4
(0, 0)
8 6 2
(c) A
8
6
4
3
20
1
−0.25
55.
7 1
20
6
Average value
Average value
sin x
y
S
ª 1
º
« S cos x»
¬
¼0
S
1
sin x dx
S 0 ³0
10
3
2
7
S
6
2
5
S
4
3
2
2
S
1
x | 0.690, 2.451
x
1
2
3
4
5
6
7
2
(0.690, π2 (
60. r t represents the weight in pounds of the dog at time t.
(2.451, π2 (
−
2
6
³ 2 rc t dt represents the net change in the weight of the dog
3
2
from year 2 to year 6.
−1
S 2
ª2
º
«S sin x»
¬
¼0
S 2
1
56.
cos x dx
S 2 0³0
Average value
cos x
2
61. (a) F x
2
S
1.5
S
(0.881, π2 (
2
S
0
x | 0.881
(b)
k sec 2 x
F 0
k
F x
500 sec 2 x
1
500
S 3
S 3 0³0
500 sec 2 x dx
1500
S
>tan x@S0 3
1500
2.71
3 0
S
−0.5
| 826.99 newtons
8
57. The distance traveled is ³ v t dt. The area under the
| 827 newtons
0
curve from 0 d t d 8 is approximately
(18 squares) 30 | 540 ft.
62.
R
1
k R 2 r 2 dr
R 0³0
R
kª 2
r3 º
«R r »
R¬
3 ¼0
2kR 2
3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section 4.4
63.
The Fundamental Theorem of Calculus
397
5
5
1
1
0.1729t 0.1522t 2 0.0374t 3 dt | ª¬0.08645t 2 0.05073t 3 0.00935t 4 º¼ | 0.5318 liter
0
5 0³0
5
64. (a)
1
0
24
−1
The area above the x-axis equals the area below the x-axis. So, the average value is zero.
(b)
10
0
24
0
The average value of S appears to be g.
0.00086t 3 0.0782t 2 0.208t 0.10
65. (a) v
(b)
90
− 10
70
− 10
60
(c)
60
³0
v t dt
ª 0.00086t 4
º
0.0782t 3
0.208t 2
0.10t » | 2476 meters
«
4
3
2
¬
¼0
66. (a) Because y 0 on [0, 2], ³
6
(b)
³ 2 f x dx
(c)
³ 0 f x dx
(d)
³ 0 2 f x dx
(e)
³ 0 ª¬2 f x º¼ dx
2
0
6
³
2
2
f x dx ³
0
2 ³
6
(f ) Average value
x
2
0
6
2
f x dx
6
6
³ f x dx
1.5 5.0
2 1.5
3.0
6
1 3.5
6
x
³ 0 4t 7 dt
ª¬2t 2 7t º¼
0
F 2
2 22 7 2
6
F 5
2 52 7 5
15
F 8
2 82 7 8
72
67. F x
2
f x dx
³ 0 2 dx ³ 0 f x dx
1
6 0
1.5.
³ 0 f x dx ³ 0 f x dx
area of region B
6
area of region A
f x dx
12 3.5
3.5 1.5
5.0
6.5
15.5
0.5833
2 x2 7 x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
398
NOT FOR SALE
Chapter 4
Integration
egration
x
68. F x
x
³2
0 ªNote: F 2
«¬
F 2
4 4 4 4
F 5
625
25 10 4
4
20
20
x
1·
§
20¨1 ¸
x¹
©
x
F 2
10
F 5
20 54
16
F 8
20 78
35
2
2
³ 2 t 3 dt
x
3
2
dv
x
1º
t 2 »¼ 2
³ 2t 3 dt
2
1 1
4 4
0
F 5
1
1
25 4
21
100
F 8
1
1
64 4
15
64
20 º
v »¼1
x
³ 0 f t dt
73. g x
0
³ 0 f t dt
(a) g 0
x
F 2
0º
»¼
1068
x
³1 20v
20 12
70. F x
2
x4
x2 2 x 4
4
167.25
x 20
³ 1 v 2 dv
§ x4
·
x2 2 x ¸ 4 4 4
¨
4
©
¹
³ 2 t 2t 2 dt
84
64 16 4
4
F 8
69. F x
ªt 4
º
2
« t 2t »
4
¬
¼2
t 3 2t 2 dt
1
1
4
x2
0
2
g 2
³ 0 f t dt | 4 2 1
g 4
³ 0 f t dt | 7 2
g 6
³ 0 f t dt | 9 1
g8
³ 0 f t dt | 8 3
4
7
9
6
8
8
5
(b) g increasing on (0, 4) and decreasing on (4, 8)
(c) g is a maximum of 9 at x
4.
(d)
y
10
0.21
8
6
4
2
71. F x
x
³ 1 cos T dT
x
sin T º
»¼1
F 2
sin 2 sin 1 | 0.0678
F 5
sin 5 sin 1 | 1.8004
F 8
sin 8 sin 1 | 0.1479
72. F x
x
³ 0 sin T dT
x
sin x sin 1
2
4
x
0
(a) g 0
³ 0 f t dt
g 2
³ 0 f t dt
g 4
³ 0 f t dt
g 6
³ 0 f t dt
g8
³ 0 f t dt
cos T º
»¼ 0
cos x cos 0
1 cos x
1 cos 2 | 1.4161
F 5
1 cos 5 | 0.7163
F 8
1 cos 8 | 1.1455
8
³ 0 f t dt
74. g x
x
F 2
6
0
2
12 2 4
4
4
12 4 4
8
6
8 2 4
8
2 6
2
4
(b) g decreasing on (0, 4) and increasing on (4, 8)
(c) g is a minimum of 8 at x
4.
(d)
y
4
2
−2
x
−2
2
4
8
10
−4
−6
INSTRUCTOR USE
S ONLY
−8
−8
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.4
x
75. (a)
x
³0
ªt 2
º
« 2t »
2
¬
¼0
t 2 dt
d ª1 2
º
x 2 x»
(b)
dx «¬ 2
¼
76. (a)
x
³ 0 t t 1 dt
2
1 2
x 2x
2
The Fundamental Theorem of Calculus
³1
4
Fc x
4
x
x 2
85. F x
x
Fc x
³ 0 t t dt
3
x
ª1 4 1 2 º
«4t 2t »
¬
¼0
d ª1 4 1 2 º
x x »
(b)
dx «¬ 4
2 ¼
77. (a)
(b)
78. (a)
x
³8
x x
d ª3 4 3
º
x 12»
dx «¬ 4
¼
x
³4
2
3 43
x 16
4
x1 3
3
Fc x
x
d ª 2 3 2 16 º
(b)
x »
3¼
dx «¬ 3
F x
Fc x
d
>tan x 1@
dx
sec 2 x
x
tan x 1
³ S 3 sec t tan t dt
Fc x
82. F x
Fc x
83. F x
Fc x
x
Fc x
sec x 2
Fc x
t2
Fc x
x2
x2 1
x
³ 1
t 4 1 dt
x
4t 1 dt ³
0
4t 1 dt
x2
4t 1 dt
0
4x 1 4 x 2 1
x
ªt 4 º
« »
¬ 4 ¼x
³ x t dt
8
3
0
0
x
³ x t dt
³
89. F x
x
x2
3
0
x2 2x
³ 1 t 2 1 dt
4t 1 dt
x
³ x t dt ³ 0 t dt
sec x tan x
t 2 2t dt
x2
³x
Alternate solution:
>sec t@S 3
x
³ 2
8
x
F x
x
4t 1 dt
x
88. F x
(b)
81. F x
x2
³x
³
x
>tan t@S 4
d
>sec x 2@
dx
sec3 x
³ x 4t 1 dt ³ 0
12
³ S 4 sec t dt
(b)
3
0
79. (a)
80. (a)
x
³ 0 sec t dt
Alternate solution:
ª2 3 2º
«3t »
¬
¼4
2
x cos x
ª2 x 2 2 x 2 º ª2 x 2 xº
¼
¬
¼ ¬
8 x 10
3 43
x 12
4
2 3 2 16
x 3
3
2 32
x 8
3
x
x
³ 0 t cos t dt
x2
x
t dt
t dt
ª¬2t 2 t º¼
x
ª3 4 3º
«4t »
¬
¼8
t dt
Fc x
87. F x
x x 1
3
x
3
86. F x
x2 2
x 2
4
1 4 1 2
x x
4
2
x
84. F x
399
3
x
0
3
x
t 3 dt ³ t 3 dt
x
0
3
sin x
³0
sin x
1 x3
t dt
12
0
sin x
ª2t3 2º
¬ 3 ¼0
cos x
2 sin x 3 2
3
cos x sin x
Alternate solution:
F x
Fc x
sin x
³0
sin x
t dt
d
sin x
dx
sin x cos x
x4 1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
400
NOT FOR SALE
Chapter 4
Integration
egration
x2
ª t 2 º
« »
¬ 2 ¼ 2
³2
90. F x
Fc x
x2
t 3 dt
x2
ª 1 º
« 2t 2 »
¬
¼2
1
1
2x4
8
Fc x
2 x 5
Alternate solution:
x2
F x
³2 t
Fc x
x2
3
3
x2
³ 0 sin T dT
92. F x
sin x 2
2 x 5
2 x sin x 4
2x
x
0, g 1 | 12 , g 2 | 1, g 3 | 12 , g 4
g 0
2x
2
³ 0 f t dt
93. g x
dt
2
g has a relative maximum at x
0
2.
y
x3
³ 0 sin t dt
91. F x
Fc x
2
sin x3
2
1
2
˜ 3x 2
f
g
3 x 2 sin x 6
x
1
2
3
4
−1
−2
4
94. (a) g t
4 2
t
lim g t
4
t of
Horizontal asymptote: y
x
x §
4·
³1 ¨© 4 t 2 ¸¹ dt
(b) A x
lim A x
x of
4
4º
ª
«4t t »
¬
¼1
4
§
·
lim ¨ 4 x 8¸
x
¹
f08
x of©
4x2 8x 4
x
4
8
x
4x 4 x 1
x
2
f
The graph of A x does not have a horizontal asymptote.
95. (a) v t
5t 7, 0 d t d 3
3
Displacement
3
³0
ª 5t 2
º
7t »
«
2
¬
¼0
5t 7 dt
(b) Total distance traveled
45
21
2
3
ft to the right
2
3
³ 0 5t 7 dt
75
³0
7 5t dt ³
75
ª
5t 2 º
«7t »
2 ¼0
¬
3
75
5t 7 dt
3
ª 5t 2
º
«
7t »
2
¬
¼7 5
2
2
§ 7 · 5§ 7 ·
§5
· § 5§ 7 ·
§ 7 ··
7¨ ¸ ¨ ¸ ¨ 9 21¸ ¨ ¨ ¸ 7¨ ¸ ¸
¨
© 5 ¹ 2© 2 ¹
©2
¹ © 2© 5 ¹
© 5 ¹ ¸¹
49 49 45
49 49
21 5
10
2
10
5
96. (a) v t
t 2 t 12
Displacement
5
113
ft
10
t 4 t 3,1 d t d 5
³1 t t 12 dt
2
5
ªt 3
º
t2
12t »
« 3
2
¬
¼1
§ 125 25
· §1 1
·
60 ¸ ¨ 12 ¸
¨
3
2
3
2
©
¹ ©
¹
56 § 56
·
ft to the left ¸
¨
3 ©3
¹
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.4
4
5
The Fundamental Theorem of Calculus
401
³1 t t 12 dt ³ 4 t t 12 dt
(b) Total distance traveled
2
2
4
5
ª t3
º
ªt 3
º
t2
t2
12t » « 12t »
« 3
2
3
2
¬
¼1
¬
¼4
§ 64
· § 1 1
· § 125 25
· § 64
·
8 48 ¸ ¨ 12 ¸ ¨
60 ¸ ¨
8 48 ¸
¨
2
© 3
¹ © 3 2
¹ © 3
¹ © 3
¹
104 73 § 185 · § 104 ·
¨
¸ ¨
¸
3
6
© 6 ¹ © 3 ¹
97. (a) v t
t 3 10t 2 27t 18
Displacement
7
79
ft
3
t 1 t 3 t 6,1d t d 7
³1 t 10t 27t 18 dt
3
2
7
ª t 4 10t 3
º
27t 2
18t »
« 3
2
¬4
¼1
ª 7 4 10 73
º ª 1 10 27
27 7 2
º
« 18 7 » « 18»
3
2
3
2
«¬ 4
»¼ ¬ 4
¼
91 § 91 ·
¨ ¸
12 © 12 ¹
0
7
³1 v t dt
(b) Total distance traveled
3
6
7
³1 t 10t 27t 18 dt ³ 3 t 10t 27t 18 dt ³ 6 t 10t 27t 18 dt
3
2
3
2
3
2
Evaluating each of these integrals, you obtain
Total distance
98. (a) v t
16 3
63
125
4
12
t 3 8t 2 15t
Displacement
5
63 ft
2
tt 3 t 5, 0 d t d 5
³ 0 t 8t 15t dt
3
2
5
ªt 4
8t 3 15t 2 º
« »
3
2 ¼0
¬4
625 8 125
375
4
3
2
(b) Total distance traveled
125
ft to the right
12
5
³ 0 v t dt
3
5
³ 0 t 8t 15t dt ³ 3 t 8t 15t dt
3
2
3
2
Evaluating each of these integrals, you obtain
Total distance
99. (a) v t
1
t
63 § 16 ·
¨ ¸
4
© 3¹
253
| 21.08 ft
12
,1d t d 4
Because v t ! 0,
Displacement
Total Distance
Displacement
³1 t
4
1 2
dt
4
ª¬2t1 2 º¼
1
42
2 ft to the right
INSTRUCTOR USE ONLY
((b)) Total distance
2 ft
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
402
Chapter 4
NOT FOR SALE
Integration
egration
cos t , 0 d t d 3S
100. (a) v t
3S
S
>sin t@30
³ 0 cos t dt
Displacement
S 2
³0
(b) Total distance
3S 2
cos t dt ³
S 2
S 2
3S 2
>sin t@0
0 ft
cos t dt ³
5S 2
3S 2
cos t dt ³
5S 2
3S
>sin t @S 2 >sin t @3S 2 >sin t@5S 2
3S
5S 2
cos t dt
1 2 2 1
6
t 3 6t 2 9t 2
101. x t
xc t
3t 2 12t 9
3 t 2 4t 3
3t 3 t 1
5
³ 0 xc t dt
Total distance
5
³ 0 3 t 3 t 1 dt
3³
1
t 1 t 3
102. x t
xc t
3
5
1
3
t 2 4t 3 dt 3 ³ t 2 4t 3 dt 3 ³
0
2
t 2 4t 3 dt
4 4 20
28 units
t 3 7t 2 15t 9
3t 2 14t 15
Using a graphing utility,
5
³ 0 xc t dt | 27.37 units.
Total distance
103. Let c(t) be the amount of water that is flowing out of the tank. Then cc t
18
³ 0 cc t dt
18
³ 0 500 5t dt
2 18
ª
5t º
«500t »
2 ¼0
¬
104. Let c t be the amount of oil leaking and t
9000 810
500 5t L min is the rate of flow.
8190 L
0 represent 1 p.m. Then cc t
4 0.75t gal min is the rate of flow.
(a) From 1 p.m. to 4 p.m. (3 hours):
3
³0
3
0.75 2 º
ª
«4t 2 t »
¬
¼0
4 0.75t dt
123
8
15.375 gal
(b) From 4 p.m. to 7 p.m. (3 hours)
6
³3
6
4 0.75t dt
0.75 2 º
ª
«4t 2 t »
¬
¼3
22.125 gal
(c) The second answer is larger because the rate of flow is increasing.
x 2 is not continuous on >1, 1@.
105. The function f x
1
³ 1 x
2
dx
0
³ 1 x
2
1
dx ³ x 2 dx
0
Each of these integrals is infinite. f x
1
2
0
0.
2
is not continuous on >2, 1@.
x3
106. The function f x
³ 2 x3 dx
x 2 has a nonremovable discontinuity at x
2
1
2
³ 2 x3 dx ³ 0 x3 dx
Each of these integrals is infinite. f x
2
has a nonremovable discontinuity at x
x3
0.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
Section 4.4
3S 4
S 2
3S 4
³ S 4 sec x dx ³ S 2 sec x dx
2
2
3S 2
S
³ S 2 csc x cot x dx
3S 2
³ S 2 csc x cot x dx ³ S
S 2
2
S 2
ª 2
º
« S cos T »
¬
¼0
sin T dT
csc x cot x dx
S.
csc x cot x has a nonremovable discontinuity at x
Each of these integrals is infinite f x
S ³0
S
ªS 3S º
csc x cot x is not continuous on « ,
».
¬2 2 ¼
108. The function f x
2
2
sec2 x has a nonremovable discontinuity at x
Each of these integrals is infinite. f x
109. P
403
ªS 3S º
sec2 x is not continuous on « ,
».
¬4 4 ¼
107. The function f x
³ S 4 sec x dx
The Fundamental Theorem of Calculus
2
S
0 1
2
S
| 63.7%
110. Let F t be an antiderivative of f t . Then,
³ u x f t dt
vx
ª¬F t º¼ u x
F v x
d ª vx
f t dt º
»¼
dx «¬³ u x
d
ªF v x
dx ¬
F u x
vx
F u x
F c v x vc x F c u x uc x
f v x vc x f u x uc x .
111. True
112. True
1
1x
x
1
³ 0 t 2 1 dt ³ 0 t 2 1 dt
113. f x
By the Second Fundamental Theorem of Calculus, you have f c x
Because f c x
114. ³
x
c
f t dt
x
³ c f t dt
x2 x 2
x
³ c 2t 1 dt
x2 x c2 c
1
§ 1·
¨ 2 ¸ 2
1
x
x
©
¹
1
1
1
2
x 1
1 x2
0.
2
c c 2
0
2
c 2 c 1
0 Ÿ c
2 x 1, and c
x
(a) G 0
x
ª¬t 2 t º¼
c
1, 2.
1 or c
2.
ª
º
s
³ 0 ¬«s ³ 0 f t dt ¼» ds
115. G x
x2 x 2
c 2 c
So, f x
1x
2
0, f x must be constant.
2t 1. Then
Let f t
1
0
ª
º
s
³ 0 ¬«s ³ 0 f t dt ¼» ds
s
(b) Let F s
s³
G x
³ 0 F s ds
Gc x
F x
Gc 0
0³
0
0
f t dt.
x
0
0
x³
x
0
f t dt
f t dt
0
x
(c) Gcc x
x˜ f x ³
0
(d) Gcc 0
0˜ f 0 ³
0
0
f t dt
f t dt
0
INSTRUCTOR USE ONLY
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Chapter 4
404
NOT FOR SALE
Integration
egration
Section 4.5 Integration by Substitution
³ f g x gc x dx
u
g x
gc x dx
du
1. ³ 8 x 2 1 16 x dx
2
8x2 1
16x dx
2. ³ x 2
x3 1 dx
x3 1
3x 2 dx
3. ³ tan 2 x sec2 x dx
tan x
sec 2 x dx
cos x
dx
sin 2 x
sin x
cos x dx
4. ³
5. ³ 1 6 x
Check:
7. ³
5
3 4 x 8 x dx
4
C
4 x2 9
10. ³ x 2 6 x3 dx
3
2x
32
2
25 x 2
C
3
C
32
12
2§ 3 ·
2
2 x
¨ ¸ 25 x
3© 2 ¹
³ 3 4x
2
13
25 x 2 2 x
3 4x2
8 x dx
43
3 x4 3
12
C
43
13
3§ 4 ·
2
8 x
¨ ¸ 3 4x
4© 3 ¹
4
1 x 3
4
3
3 4 x2
3
43
3
3 4 x2
C
4
13
x4 3
C
8 x
3
12
C
2
x4 3
4 x3
5
1
6 x3 3 x 2 dx
3³
6
3
ª
º
d « 6 x
C»
»
dx «
18
¬
¼
3
32
2
1
x 4 3 4 x3 dx
³
4
x2 9
2x
4
43
d ª3
º
C»
3 4x2
«
dx ¬ 4
¼
2
4
4
25 x 2
3
ª 4
º
d « x 3
Check:
C»
»
dx « 12
¬
¼
Check:
6 1 6x
32
d ª2
º
C»
25 x 2
«
dx ¬ 3
¼
9. ³ x x 3 dx
3
º
C»
»¼
4
ª 2
º
d « x 9
C»
»
4
dx «
¬
¼
2
Check:
C
4
25 x 2 x dx
3
5
x2 9
2 x dx
2
Check:
8. ³
3
1 6x
5
6 dx
d ª 1 6x
«
dx «
5
¬
6. ³ x 2 9
Check:
4
6 6 x3
18
5
2
3
1 6 x
˜
3
6
3x2
x3
6
C
x 2 6 x3
6 x3
18
6
C
5
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 4.5
3
5
ª 3
º
d « x 1
Check:
C»
»
dx « 15
¬
¼
5 x3 1
12. ³ x 5 x 4 dx
32
ª 2
º
d«t 2
Check:
C»
»
dt «
3
¬
¼
Check:
16. ³ u 2
13
dx
17. ³
x
1 x
Check:
2
dx
3
2
1 t 2
2
32
3 2 t2 2
12
4
C
40
2t
t2 2
4
1 2t 3
˜
8
32
12
2
3
1
2 1 x 2
2 x
4
C
12
43
C
5x 1 x2
13
32
43
15
1 x2
C
8
2 u3 2
C
5x 3 1 x2
32
C
9
u3 2
1 1 x
2
2
32
2t 4 3
C
5 1 x
˜
2
43
12
2 3 3
3u 2
˜ u 2
9 2
C
2t 4 3
t3
3
1 u 2
3
32
32
t
32
2
3
3
13
15 4
˜ 1 x2
2 x
8 3
3
1
1 x2
2 x dx
2³
ª
º
d «
1
»
C
»
dx « 4 1 x 2 2
¬
¼
5x2 4
t2 2
C
12
3 4
2t 3 8t 3
2
12
12
1
u3 2
3u 2 du
3³
32
3
13
5
1 x2
2 x dx
2³
32
ª 3
º
d «2 u 2
Check:
C»
»
du «
9
¬
¼
x 5x2 4
40
43
d ª 15
º
1 x2
C»
«
dx ¬ 8
¼
u 3 2 du
C
3
32
ª 4
º
d « 2t 3
C»
Check:
»
dt «
12
¬
¼
15. ³ 5 x 1 x 2
5
15
4
ª 2
º
1 « 5x 4 »
C
»
10 «
4
¬
¼
12
1
2t 4 3 8t 3 dt
8³
2t 4 3 dt
x3 1
405
4
x 2 x3 1
4 5 x 2 4 10 x
12
1
t2 2
2t dt
2³
t 2 2 dt
14. ³ t 3
3x 2
15
4
ª 2
º
d « 5x 4
Check:
C»
»
dx «
40
¬
¼
13. ³ t
4
3
1
5 x 2 4 10 x dx
³
10
3
2
5
ª 3
º
1« x 1 »
C
»
3«
5
¬
¼
4
1
x3 1 3x 2 dx
3³
4
11. ³ x x 1 dx
2
Integration by Substitution
12
u2
2
1
C
4 1 x2
2
C
x
1 x2
3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 4
406
18. ³
x3
1 x4
Check:
19. ³
21. ³
º
1
d ª
«
C»
3
dx « 3 1 x
»¼
¬
4x 9
3
dx
1 x2
dx
1 x
4
1 x3
2
C
3
4 4x 9
2
3
1
2 4 x3 9 12 x 2
4
C
6x2
4 x3 9
3
C
1 x2 C
1 x2
4
3
1
3
x
1 1 x
4
12
1 2
1 1
˜ 1 x4
4 x3
2 2
12
1 2
1
1 x2
2 x
2
º
d ª 1 x4
C»
«
dx ¬«
2
¼»
2
2
3
1 4x 9
˜
2
2
1 1 x
2
12
1 2
1
1 x4
4 x3 dx
³
4
1
C
3 1 x3
x2
2
1
3x 2
1 1 x3
3
1 2
1
³ 1 x2
2 x dx
2
dx
2
1 x4
2
d ª 1 3
º
4x 9
C»
dx «¬ 4
¼
12
d ª
1 x2
Cº
¼»
dx ¬«
x3
x3
1
3
ª
º
1« 1 x
» C
3«
1 »
¬
¼
3
1
4 x3 9 12 x 2 dx
³
2
ª
º
d «
1
»
C
»
dx « 4 4 x3 9 2
¬
¼
x
Check:
3
1
C
4 1 x4
1
1
1 x4
C
4
2
1
1 x4
4 x3
4
2
1
1 x3
3 x 2 dx
³
3
dx
6x2
Check:
22. ³
2
1 x3
Check:
2
1
1 x4
4 x3 dx
³
4
dx
º
d ª 1
«
C»
4
dx « 4 1 x
»¼
¬
x2
Check:
20. ³
2
Integration
egration
12
1 x4
C
2
C
x3
1 x4
4
1· § 1 ·
§
23. ³ ¨1 ¸ ¨ 2 ¸ dt
t ¹ ©t ¹
©
1· § 1 ·
§
³ ¨1 ¸ ¨ 2 ¸ dt
t¹ © t ¹
©
4
º
d ª« ª¬1 1 t º¼
Check:
C»
»
dt «
4
¬
¼
ª
1 º
» dx
24. ³ « x 2 2
«¬
3 x »¼
Check:
ª
§ 1 ·º
«1 ¨ t ¸»
© ¹¼
¬
C
4
3
1 §
1· § 1 ·
4 ¨1 ¸ ¨ 2 ¸
4 ©
t¹ © t ¹
§ 2 1 2 ·
³ ©¨ x 9 x ¹¸ dx
d ª1 3 1 1
º
x x C»
9
dx «¬ 3
¼
x2 1§
1·
1 ¸
2¨
t ©
t¹
x3
1 § x 1 ·
¨
¸C
3
9 © 1 ¹
1 2
x
9
x2 3
x3
1
C
3
9x
3x 4 1
C
9x
1
3x
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.5
1
dx
2x
25. ³
d ª
2 x C º¼
dx ¬
3
º
» C
»¼
407
2x C
1 x1 2
C
2 12
1
1 2
³ x dx
2
1
1 2
2x
2
2
1
x
dx ³ 3 x1 3 dx
5
5x2
Alternate Solution:
26. ³
12
1
dx
2x
Alternate Solution: ³
Check:
1 ª 2x
«
2 «¬ 1 2
1
1 2
2x
2 dx
³
2
Integration by Substitution
2x C
1
2x
1
3 43
˜ x
C
5 4
3
3 3
25 x 4 C
20
23
x
³ 3 5x2
Check:
d ª 1
3
º
˜ x4 3 C»
dx «¬ 3 5 4
¼
27. y
28. y
2
³ 5x
1 3
dx
ª
³ ¬«4 x ³
x dx
º
» dx
16 x 2 ¼
4x
10 x 2
1 x
3
5x2
1
˜
10
23
1 3
1
5x2
10 x dx
10 ³
1
3 4
˜ ˜ x1 3
5 4 3
3
23
3
C
5x2
20
C
x
3
4 ³ x dx 2³ 16 x 2
5x2
1 2
2 x dx
12º
ª
16 x 2 »
§ x2 ·
4¨ ¸ 2 «
C
«
»
©2¹
1
2
¬
¼
y
3
12º
ª
10 « 1 x 3 »
C
»
3« 12
¬
¼
20
1 x3 C
3
x
−2
2
−1
(b)
x 1
³ x 2 2 x 3 2 dx
dy
dx
x 4 x 2 , 2, 2
12
1
4 x2
2 x dx
2³
32
32
1 2
1
˜ 4 x2
C
4 x2
C
2 3
3
³ x 4 x dx
1
ª 2
º
1 « x 2x 3 »
C
»
2«
1
¬
¼
1
C
2
2 x 2x 3
³
x 4
dx
2, 2 : 2
y
32
1
C Ÿ C
4 22
3
2
32
1
2
4 x2
3
2
−2
x 8x 1
1 2
1
x2 8x 1
2 x 8 dx
2³
12º
ª 2
1 « x 8x 1 »
x2 8x 1 C
C
»
2«
12
¬
¼
2
2
y
2
1
x 2 2 x 3 2 x 2 dx
2³
30. y
2 x 2 4 16 x 2 C
31. (a) Answers will vary. Sample answer:
dx
1 2
10
1 x3
3x 2 dx
3³
29. y
3
1 43
˜
x C
4 3 5
2
−1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 4
408
Integration
egration
32. (a) Answers will vary: Sample answer:
y
2
33. ³ S sin S x dx
cos S x C
34. ³ sin 4 x dx
1
sin 4 x 4 dx
4³
1
cos 4 x C
4
35. ³ cos 8 x dx
1
cos 8 x 8 dx
8³
1
sin 8 x C
8
x
−2
2
−2
(b)
u
dy
dx
x 2 x3 1 , 1, 0
y
³ x x 1 dx
2
2
2
3
x 1
3
0
y
1
1§ 1 ·
³ cos ¨ 2 ¸ dT
T© T ¹
T2
cos
1
T
dT
1
sin x 2 2 x dx
2³
38. ³ x sin x 2 dx
3
1 x 1
C
3
3
C
3
1 3
x 1
9
§ x ·§ 1 ·
2 ³ csc 2 ¨ ¸¨ ¸ dx
© 2 ¹© 2 ¹
37. ³
2
1
x3 1 3 x 2 dx
³
3
3
§ x·
36. ³ csc 2 ¨ ¸ dx
© 2¹
3
1 3
x 1 C
9
§ x·
2 cot ¨ ¸ C
© 2¹
sin
1
T
C
1
cos x 2 C
2
2
−3
3
−2
39. ³ sin 2 x cos 2 x dx
1
cos 2 x 2 sin 2 x dx
2³
³ sin 2 x cos 2 x dx
³ sin 2 x cos 2 x dx
1
2 sin 2 x cos 2 x dx
2³
40. ³
41. ³
sin x
dx
cos3 x
32
³ cot x
42. ³
tan x
tan x sec 2 x dx
csc 2 x
dx
cot 3 x
1 sin 2 x
2
2
1
sin 2 x 2 cos 2 x dx
2³
cot x
2
3
C
3
C
C
2
1 cos 2 x
2
2
1
sin 4 x dx
2³
1
sin 2 2 x C OR
4
C1
1
cos 2 2 x C1 OR
4
1
cos 4 x C2
8
2
32
tan x
C
3
csc 2 x dx
2
³ cos x
32
2
1
C
2 cot 2 x
sin x dx
1
tan 2 x C
2
1
sec 2 x 1 C
2
43. f x
1
sec 2 x C1
2
x
³ sin 2 dx
2
cos x
C
2
1
1
sec 2 x C
C
2
2 cos x
2
Because f 0
f x
2 cos
6
2 cos
x
C
2
§0·
2 cos¨ ¸ C , C
© 2¹
4. So,
x
4.
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section 4.5
§S ·
sec 2 2 x , ¨ , 2 ¸
©2 ¹
44. f c x
47. u
f x
1
tan 2 x C
2
§S ·
f¨ ¸
©2¹
1
§ § S ··
tan ¨ 2¨ ¸ ¸ C
2
© © 2 ¹¹
45. f c x
f x
2
3
3
2 2
2x 5 8
3
f x
³x
2
C
18 C
3 x 4, x
³ x 3x 4 dx
10 Ÿ C
8
2 x 8 x 2 , 2, 7
2 8 x2
32
C
3
24
3
32
16
C
3
C
2 8 x2
1 x dx
³ u 6 u du
32
12
³ u 6u du
7 Ÿ C
5
3
³
u 4
˜
3
u ˜
1
du
3
1
u 3 2 4u1 2 du
9³
1§ 2 5 2
8 3 2·
¨ u u ¸ C
9© 5
3
¹
2
8
52
32
C
3x 4
3x 4
45
27
2
32
3x 4 ª¬3 3x 4 20º¼ C
135
2
32
3x 4
9x 8 C
135
32
3
1 x, x
49. u
3
48. u
3
3
C
du
2u 3 2
u 10 C
5
2
32
x 6 ª¬ x 6 10º¼ C
5
2
32
x6
x4 C
5
u 4
1
, dx
du
3
3
2
2 2 x2 5
C
12
f x
f 2
2
2
4 x 2 10
409
2 52
u 4u 3 2 C
5
2 x 4 x 2 10 , 2, 10
285
3
f x
u 6, dx
1
tan 2 x 2
2
f 2
46. f c x
x 6, x
³ x x 6 dx
1
0 C
2
C
f x
Integration by Substitution
5
3
1 u , dx
³ 1 u
du
2
u du
³ u1 2 2u 3 2 u 5 2 du
4
2
§2
·
¨ u 3 2 u 5 2 u 7 2 ¸ C
5
7
©3
¹
2u 3 2
35 42u 15u 2 C
105
2
32
2
1 x ª35 42 1 x 15 1 x º C
¬
¼
105
2
32
1 x
15 x 2 12 x 8 C
105
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Chapter 4
410
50. u
2 x, x
³ x 1
NOT FOR SALE
Integration
egration
2 u , dx
du
³ 3 u
2 x dx
u du
³ 3u1 2 u 3 2 du
2
§
·
¨ 2u 3 2 u 5 2 ¸ C
5
©
¹
2u 3 2
5u C
5
2
32
2 x ª¬5 2 x º¼ C
5
2
32
2 x
x 3 C
5
51. u
2 x 1, x
1
u 1 , dx
2
1
du
2
³
x2 1
dx
2x 1
ª¬ 1 2 u 1 º¼ 1 1
du
³
2
u
1 1 2 ª 2
u ¬ u 2u 1 4º¼ du
8³
1
u 3 2 2u1 2 3u 1 2 du
8³
1§ 2 5 2
4 32
1 2·
¨ u u 6u ¸ C
8© 5
3
¹
2
u1 2
3u 2 10u 45 C
60
2x 1 ª
2
3 2 x 1 10 2 x 1 45º C
¼
60 ¬
1
2 x 1 12 x 2 8 x 52 C
60
1
2 x 1 3 x 2 2 x 13 C
15
52. u
x 4, x
³
2x 1
dx
x 4
u 4, du
³
dx
2u 4 1
³ 2u
u
12
7u
53. u
du
1 2
du
x 1, x
u 1, dx
du
dx
³u
x
³ x 1 x 1
u 1
³
4 32
u 14u1 2 C
3
2 12
u 2u 21 C
3
2
x 4 ª¬2 x 4 21º¼ C
3
2
x 4 2 x 13 C
3
u
du
u 1
u 1
u 1
u
du
³ 1 u 1 2 du
u 2u1 2 C
u 2 u C
x 1 2
where C1
x 1C
x 2
x 1 1 C
x 2
x 1 C1
1 C.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.5
t 10, t
54. u
³ t t 10
13
u 10, du
Integration by Substitution
411
dt
³ u 10 u du
43
13
³ u 10u du
13
dt
3 7 3 15 4 3
u u C
7
2
3 43
u 2u 35 C
14
3
43
t 10 ª¬2 t 10 35º¼ C
14
3
43
2t 15 C
t 10
14
x 2 1, du
55. Let u
1
2 x dx.
1 1
2 1
3
³ 1 x x 1 dx
2
³
2 x 4 1, du
56. Let u
x2 1
3
1
ª1 x2 1 4 º
«¬ 8
»¼ 1
2 x dx
0
8 x3 dx.
1
1
1 1
8 0
2
³ 0 x 2 x 1 dx
3
4
57. Let u
³
2x 1
4
2
8x
3
dx
3
ª
2x4 1 º
«1 ˜
»
«8
»
3
¬
¼0
x3 1, du
3x 2 dx.
x 1 dx
12
1 2
2 ˜ ³ x3 1 3 x 2 dx
1
3
1 x 2 , du
2 x dx.
1 3
3 13
24
13
12
2
2
³1
2x
58. Let u
1
³0
2
3
x 1 x 2 dx
59. Let u
4
³0
60. Let u
2
2 x 1, du
1
dx
2x 1
x
1 2 x2
61. Let u
1
³1
1 4
1 2
2x 1
2 dx
2 ³0
x1
x , du
x
1
3 2º
ª 1
2
« 3 1 x
»
¬
¼0
0
1
3
4
ª27 2 2 º
¼
9¬
12 8
9
2
1
3
4
ª 2x 1 º
¬
¼0
9 1
2
4 x dx.
1 2
1 2
1 2 x2
4 x dx
³
0
4
dx
1
12
1 1
1 x2
2 x dx
2³0
32 2
4ª 3
x 1 º
¼»1
9 ¬«
2 dx.
1 2 x 2 , du
³0
9
3 2º
ª 3
« x 1 »
«
»
32
¬
¼1
dx
2
1
2
x
2³
9
1
2
ª1
2º
«2 1 2x »
¬
¼0
3 1
2 2
1
dx.
1
x
2 §
1 ·
¨
¸ dx
©2 x ¹
9
2 º
ª
«
»
¬ 1 x ¼1
1
1
2
1
2
INSTRUCTOR USE ONLY
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412
NOT FOR SALE
Chapter 4
2 x 1, du
62. Let u
When x
5
³1
Integration
egration
1, u
2 dx, x
1. When x
9 1 2 u 1
x
dx
2x 1
³1
u
1
u 1.
2
5, u
9.
1
du
2
1 9 12
u u 1 2 du
4 ³1
9
1 ª2 3 2
º
u 2u1 2 »
4 «¬ 3
¼1
1 ª§ 2
· §2
·º
¨ 27 2 3 ¸ ¨ 2 ¸»
4 «¬© 3
3
¹ ©
¹¼
16
3
63.
dy
dx
18 x 2 2 x3 1 , 0, 4
y
3³ 2 x 3 1
2
2x 1
C
3
4
13 C Ÿ C
y
2 x3 1
dy
dx
48
3
3
3x 5
6 x 2 dx u
2 x3 1
3
2 x3 1
3
8
3
dx
66. u
2
8
3x 5
2
u du
6, u
8.
u1 3 du
43
C
x 2, x
Area
u 2, dx
2, u
When x
16 3 x 5
C
2
8
C
2
3x 5
y
3
§ 384
· § 3 3·
12 ¸ ¨ ¸
¨
7
4¹
©
¹ ©7
1209
28
, 1, 3
2
8
³1 u 1
8.
8
3
8
7, u
ª3 7 3 3 4 3º
«7 u 4 u »
¬
¼1
3
48³ 3 x 5
3 1 5
du
1. When x
x 1 dx
3
³1 u
1
3
48 ³ 3x 5 3 dx
3
3
7
C
u 1, dx
0, u
³0 x
Area
3
y
y
x 1, x
When x
3
64.
65. u
2
6
³ 2 x
2 3
8
8
8
C Ÿ C
4
1
73
0. When x
x 2 dx
³0 u 2
³0 u
du
2 3
u du
4u 4 3 4u1 3 du
8
ª 3 10 3 12 7 3
º
u 3u 4 3 »
«10 u
7
¬
¼0
1
67. Area
4752
35
2§ x ·
2S 3
³ S 2 sec ¨© 2 ¸¹ dx
2³
2S 3
S 2
§ x ·§ 1 ·
sec 2 ¨ ¸¨ ¸ dx
© 2 ¹© 2 ¹
2S 3
ª
§ x ·º
«2 tan ¨ 2 ¸»
© ¹¼S 2
¬
2
3 1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.5
68. Let u
2 x, du
2 dx.
S 4
³
S 4
S 2
³ S 2 sin x cos x dx
1
csc 2 x cot 2 x 2
2 S 12
S 4
ª¬ 12 csc 2 xº¼
S 12
³ 2 x x 1 dx
2
2³
2
0
2
3
x x
4
2
72. f x
ª x5
x3 º
2« »
3 ¼0
¬5
dx
ª 32 8 º
2«
»
3¼
¬5
sin x cos x is odd.
S 2
³ S 2 sin x cos x dx
272
15
4
3
x x 2 1 is odd.
2
3
³ 2 x x 1 dx
2
S 4
³ S 4 sin x dx
(b)
³ S 4 cos x dx
(c)
³ S 2 cos x dx
(d)
³ S 2 sin x cos x dx
76. ³
S 2
3
S 2
S 2
0
4
(a)
³ 4 x dx
2
³ 0 x dx
(b)
³ 4 x dx
2 ³ x 2 dx
(c)
³ 0 x dx
(d)
³ 4 3x dx
4
0
4
2
4
2
0
2
128
3
4
³ x 2 dx
0
4
3³ x 2 dx
2
64
3
0
64
3
64
0 because sin x is symmetric to the origin.
S 4
3
64
; the function x 2 is an even
3
function.
0
74. (a)
0
ª x3 º
« »
¬ 3 ¼0
4
73. ³ x 2 dx
0
70. f x
sin 2 x cos x dx
ª sin 3 x º
2«
»
¬ 3 ¼0
1
2
2
S 2
0
S 2
dx
2
2
75. ³
2³
2
x 2 x 2 1 is even.
69. f x
413
sin 2 x cos x is even.
71. f x
³ S 12 csc 2 x cot 2 x dx
Area
Integration by Substitution
S 4
2³
0
2³
0
S 2
S 2
>2 sin x@0
cos x dx
>2 sin x@S0 2
2
3
sin x cos x and so, is symmetric to the origin.
3
³ 3 x 3x dx ³ 3 4 x 6 dx
0 2³
S 2
S 2
³ S 2
3
sin 4 x dx ³
S 2
S 2
2 x dx and ³ x 5 x 2
77. If u
5 x 2 , then du
78. f x
x x 2 1 is odd. So, ³
2
2 because cos x is symmetric to the y-axis.
0 because sin x cos x
x 3 4 x 2 3 x 6 dx
sin 4 x cos 4 x dx
S 4
cos x dx
2
2
2
x x 2 1 dx
3
2
cos 4 x dx
dx
0 2³
12 ³ 5 x 2
3
0
3
0
4 x 2 6 dx
2 ª¬ 43 x3 6 xº¼
3
0
36
S 2
cos 4 x dx
2 x dx
ª2
º
« 4 sin 4 x»
¬
¼0
0
12 ³ u 3 du.
0.
INSTRUCTOR USE ONLY
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414
NOT FOR SALE
Chapter 4
Integration
egration
x 3 1 and du
79. (a) The second integral is easier. Use substitution with u
³x
2
x3 1 dx
1
3
³ x 1
3
12
3 x 2 dx. The answer is
3 x 2 dx
2 x3 1 3 2 C.
9
(b) The first integral is easier. Use substitution with u
³ tan 3x sec 3x dx
1
3
2
2
³ 2 x 1 dx
1
2
³ 2 x 1 dx
³ 4 x 4 x 1 dx
2
80. (a)
³ tan 3x 3sec 3x dx
2
³ 2 x 1 2 dx
2
C1 16 .
tan 2 x
C1
2
³ tan x sec x dx
³ sec x sec x tan x dx
2
tan 2 x
C1
2
sec 2 x 1
C1
2
They differ by a constant: C 2
81.
dV
dt
4 x3 2 x 2 x 1 C
1
3
6
4 x3 2 x 2 x C
2
3
³ tan x sec x dx
2
3sec 2 3 x dx. The answer is
1 tan 2 3 x C .
6
1 2x 1 3 C
1
6
2
They differ by constant: C2
(b)
tan 3 x and du
sec 2 x
C2
2
sec 2 x
1
C1
2
2
1
C1 .
2
k
t 1
2
k
V t
³ t 1 2 dt
V 0
k C
V1
1
k C
2
k
C
t 1
500,000
400,000
200,000 and C
Solving this system yields k
When t
4, V 4
200,000
300,000.
t 1
300,000. So, V t
$340,000.
82. (a) The maximum flow is approximately R | 62 thousand gallons at 9:00 A.M. t | 9 .
(b) The volume of water used during the day is the area under the curve for 0 d t d 24. That is, V
24
³0 R t dt.
(c) The least amount of water is used approximately from 1 A.M. to 3 A.M. 1 d t d 3 .
83.
b ª
1
St º
74.50 43.75 sin » dt
b a ³ a «¬
6¼
b
1 ª
262.5
St º
74.50t cos »
b a «¬
6 ¼a
S
3
1§
262.5 ·
| 102.352 thousand units
¨ 223.5 3©
S ¹¸
6
1§
262.5
·
223.5¸ | 102.352 thousand units
¨ 447 3©
S
¹
(a)
1ª
262.5
St º
74.50t cos »
3 «¬
6 ¼0
S
(b)
1ª
262.5
St º
74.50t cos »
3 «¬
6 ¼3
S
(c)
1ª
262.5
St º
74.50t cos »
12 «¬
6 ¼0
S
12
1§
262.5 262.5 ·
¨ 894 12 ©
S
S ¹¸
74.5 thousand units
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.5
84.
Integration by Substitution
415
b
1 ª 1
1
º
cos 60S t sin 120S t »
120S
b a «¬ 30S
¼a
b
1
ª2 sin 60S t cos 120S t º¼ dt
b a³a ¬
1 60
(a)
1
1
ª 1
º
cos 60S t sin 120S t »
«
1 60 0 ¬ 30S
120S
¼0
(b)
1
1
ª 1
º
cos 60S t sin 120S t »
1 240 0 «¬ 30S
120S
¼0
ª§ 1
· § 1 ·º
60 Ǭ
0¸ ¨ ¸»
¹ © 30S ¹¼
© 30S
1 240
4
S
| 1.273 amps
ª§
1
1 · § 1 ·º
240 Ǭ 120
S ¸¹ ©¨ 30S ¹¸»¼
30
2
S
©
¬
2
5 2 2 | 1.382 amps
S
1 30
(c)
85. u
1
1
ª 1
º
cos 60S t sin 120S t »
1 30 0 «¬ 30S
120S
¼0
1 x, x
When x
1 u , dx
b 15
³ a 4 x 1 x dx
Pa , b
0 amp
du
1 a. When x
a, u
ª§ 1 · § 1 ·º
30 «¨ ¸ ¨
¸»
¬© 30S ¹ © 30S ¹¼
b, u
1 b.
15 1 b
1u
4 ³ 1 a
u du
1 b
15 ª 2 5 2 2 3 2 º
u u »
4 «¬ 5
3
¼1 a
15 1 b 3 2
u u1 2 du
4 ³ 1 a
1 b
º
15 ª 2u 3 2
3u 5 »
«
4 ¬ 15
¼1 a
b
ª 1 x 32
º
3x 2 »
«
2
¬«
¼» a
0.75
ª 1 x 32
º
«
3x 2 »
2
¬«
¼» 0.50
(a) P0.50, 0.75
b
ª 1 x 32
º
«
3x 2 »
2
«¬
»¼ 0
(b) P0, b
0.353
1b
2
35.3%
32
1b
3b 2 1
32
3b 2
0.5
1
b | 0.586
86. u
1 x, x
When x
Pa , b
1 u , dx
b 1155
du
1 a. When x
a, u
³ a 32 x 1 x
3
32
58.6%
dx
b, u
1 b.
1155 1 b
3
1 u u 3 2 du
32 ³ 1 a
1 b
1155 ª 2 11 2 2 9 2
6
2
º
u
u u7 2 u5 2 »
32 «¬11
3
7
5
¼1 a
1155 1 b 9 2
u 3u 7 2 3u 5 2 u 3 2 du
32 ³ 1 a
1 b
º
1155 ª 2u 5 2
105u 3 385u 2 495u 231 »
«
32 ¬ 1155
¼1 a
1 b
ªu5 2
º
105u 3 385u 2 495u 231 »
«
16
¬
¼1 a
0.75
(a) P0, 0.25
ªu5 2
º
105u 3 385u 2 495u 231 »
«
¬ 16
¼1
(b) P0.5, 1
ªu5 2
º
105u 3 385u 2 495u 231 » | 0.736
«
¬ 16
¼ 0.5
| 0.025
2.5%
0
73.6%
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
416
NOT FOR SALE
Chapter 4
87. (a)
Integration
egration
4
g
0
9.4
f
−4
(b) g is nonnegative because the graph of f is positive at the beginning, and generally has more positive sections than
negative ones.
(c) The points on g that correspond to the extrema of f are points of inflection of g.
S 2, do not correspond to an extrema of g. The graph of g continues to increase after
S 2 because f remains above the x-axis.
(d) No, some zeros of f , like x
x
(e) The graph of h is that of g shifted 2 units downward.
4
0
9.4
−4
i
,i
n
lim ¦
n of
i 1
f x dx ³
1, 2, !, n.
sin iS n
n
t
S 2
f x dx
2 ht.
Let u
S 2
n
lim ¦ f ci 'x
'x o 0
§S
·
cos¨ x ¸ and cos x
©2
¹
90. (a) sin x
1
and use righthand endpoints
n
Let 'x
n
³0
sin S x, 0 d x d 1.
88. Let f x
ci
S 2
t
³ 0 f x dx
g t
³0
S
2
x, du
sin 2 x dx
0
1
89. (a) Let u
1
1
1
S
³ 0 x 1 x dx
2
S 2
³0
S
dx, x
1u
1Ÿ u
0
1, x
5
2
(b) Let u
1 1
1 x, du
0 Ÿ u
x
S 2
³0
º
cos S x»
S
¼0
1
0
S
2
sin n x dx
³ 1 1 u u du
1
1
5
2
2
u:
§S
·
cos 2 ¨ x ¸ dx
©2
¹
2
cos 2 u du
S 2
³ 0 cos x dx
2
x as in part (a):
2 5
³0 u 1 u
S
³ S 2 cos u du
i 1
³ 0 sin S x dx
S 2
³0
dx, x
§S
·
sin ¨ x ¸
©2
¹
S 2
³0
0
§S
·
cos n ¨ x ¸ dx
©2
¹
³ S 2 cos u du
S 2
³0
n
cos n u du
S 2
³0
cos n x dx
du
³ 0 x 1 x dx
(b) Let u
1 x, du
0 Ÿ u
x
1
a
dx, x
1u
1Ÿ u
0
1, x
³ 0 x 1 x dx
b
2
5
0
³ 1 1 u u du
1
a
b
³ 0 u 1 u du
b
a
³0 x 1 x
a
1
b
dx
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.5
Integration by Substitution
417
91. False
³ 2 x 1 dx
2
1
2
³ 2 x 1 2 dx
2
1 2x 1 3 C
6
92. False
³ x x 1 dx
1
2
2
³ x 1 2 x dx
2
1 x2 1 2 C
4
93. True
10
10
³ 10 ax bx cx d dx
3
10
³ 10 ax cx dx ³ 10 bx d dx
2
3
Odd
0 2³
2
10
bx 2 d dx
0
Even
94. True
b
³ a sin x dx >cos x@a
b
cos b cos a
b 2S
³a
cos b 2S cos a
sin x dx
95. True
4 ³ sin x cos x dx
2 ³ sin 2 x dx
cos 2 x C
99. Because f is odd, f x
96. False
1
2
sin 2 x 2 cos 2 x dx
2³
³ sin 2 x cos 2 x dx
2
a
97. Let u
c³
b
a
cx, du
³ a f x dx ³ 0 f x dx
f cx dx
³
c dx:
Let x
u , dx
When x
0, u
1
³ a f x dx
du
c³ f u
ca
c
cb
cb
³ ca f u du
100. Let u
cb
³ ca f x dx
98. (a)
d
>sin u u cos u C@
du
cos u cos u u sin u
u sin u
So, ³ u sin u du
(b) Let u
S2
³ 0 sin
x , u2
x dx
sin u u cos u C.
x, 2u du
0
f x dx ³
a
0
f x dx.
du in the first integral.
a, u
0. When x
a
³
0
³
0
a
f u du ³
f u du ³
x h, then du
a, u
a h.
When x
b, u
b h. So,
b
³ a f x h dx
bh
a
0
a
0
a.
f x dx
f x dx
0
dx.
When x
³ a h f u du
bh
³ a h f x dx.
a0 a1x a2 x 2 " an x n .
101. Let f x
1
³0
³ 0 sin u 2u du
2³
a
0
1
dx.
S
S
a
0
³ a f x dx
3
1 sin 2 x
C
2
3
1
sin 3 2 x C
6
f x . Then
f x dx
ª
x2
x3
x n 1 º
a2
" an
«a0 x a1
»
2
3
n 1¼ 0
¬
a0 u sin u du
a1 a2
an
"
2
3
n 1
0 (Given)
2>sin u u cos u@0 (part (a))
By the Mean Value Theorem for Integrals, there exists c
in [0, 1] such that
2 ª
¬ S cos S º¼
³ 0 f x dx
S
2S
1
0
f c 10
f c.
So the equation has at least one real zero.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
418
NOT FOR SALE
Chapter 4
Integration
egration
1
102. D 2 ³ f x dx
D2 1
0
1
2D ³ f x x dx
2D D
0
1
³ 0 f x x dx
D2
2D 2
D2
2
Adding,
1
³ 0 ª¬D f x 2D xf x x f x º¼ dx
2
2
1
³ 0 f x D x dx
2
Because D x
2
0
0.
t 0, f
0. So, there are no such functions.
Section 4.6 Numerical Integration
2
1. Exact: ³ x 2 dx
0
2
ª 1 x3 º
¬3 ¼0
8
3
| 2.6667
Trapezoidal: ³ x 2 dx | 14 ª0 2 12
«¬
0
2
Simpson’s: ³ x 2 dx | 16 ª0 4 12
«¬
0
2
2
2
21
2
2 § x2
·
2. Exact: ³ ¨
1¸ dx
1
4
©
¹
21
ª x3
º
« x»
12
¬
¼1
2
2
2 32
4 23
2
2
2
2 º
»¼
2
2 º
»¼
11
4
8
3
2.7500
| 2.6667
19
| 1.5833
12
§ 542
·
§ 322
·
§ 742
· § 22
2 § x2
·
·
·º
1 ª§ 12
1¸ dx | «¨ 1¸ 2¨
1¸ 2¨
1¸ 2¨
1¸ ¨
1¸»
Trapezoidal: ³ ¨
1
¨ 4
¸
¨ 4
¸
¨ 4
¸ ©4
8 «© 4
©4
¹
¹
¹»¼
©
¹
©
¹
©
¹
¬
§ 542
·
§ 322
·
§ 742
· § 22
1 § x2
·
·
·º
1 ª§ 12
«¨ 1¸ 4¨
1¸ dx |
1¸ 2¨
1¸ 4¨
1¸ ¨
1¸»
Simpson’s: ³ ¨
0
¨
¸
¨
¸
¨
¸
12 «© 4
4
4
4
4
© 4
¹
¹
¹»¼
©
¹
©
¹
©
¹ ©
¬
203
| 1.5859
128
19
| 1.5833
12
2
2
3. Exact: ³ x 3 dx
0
ª x4 º
« »
¬ 4 ¼0
2
Trapezoidal: ³ x3 dx |
0
2
Simpson’s: ³ x3 dx |
0
4. Exact: ³
2
dx
2 x2
3
Trapezoidal: ³
Simpson’s: ³
3
2
4.0000
3
3
1ª
3
3º
§1·
§ 3·
«0 2¨ ¸ 2 1 2¨ ¸ 2 »
4 ¬«
© 2¹
© 2¹
¼»
3
3
1ª
3
3º
§1·
§ 3·
«0 4¨ ¸ 2 1 4¨ ¸ 2 »
6 ¬«
© 2¹
© 2¹
¼»
3
ª 2º
« x »
¬ ¼2
2
2
3
2
17
4
4.2500
24
6
4.0000
1
3
§ 2 ·
§ 2 ·
§ 2 ·
2
1ª 2
2º
¸ 2¨
¸ 2¨
¸ 2 » | 0.3352
dx | « 2 2¨
2
2
2
2
2 x
¨ 94 ¸
¨ 10 4 ¸
¨ 11 4 ¸ 3 »
8 «2
©
¹
©
¹
©
¹
¬
¼
3
§ 2 ·
§ 2 ·
§ 2 ·
2
1ª2
2º
« 2 4¨
¸ 2¨
¸ 4¨
¸ 2 » | 0.3334
dx |
2¸
2¸
2¸
2
¨
¨
¨
12 « 2
3 »
x
© 94 ¹
© 10 4 ¹
© 11 4 ¹
¬
¼
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.6
3
ª x4 º
« »
¬ 4 ¼1
3
5. Exact: ³ x3 dx
1
1
3
1
6. Exact: ³
8
0
3
8
Trapezoidal: ³
Simpson’s: ³
7. Exact: ³
9
4
8
3
0
8
3
0
9
Simpson’s: ³
4
9
4
12.0000
x dx | 13 ª¬0 4 2 3 2 4 3 3 2 3 4 4 3 5 2 3 6 4 3 7 2º¼ | 11.8632
9
ª 2 x3 2 º
¬3
¼4
18 16
3
38
3
x dx |
5ª
«2 2
16 ¬
37
2
8
x dx |
5ª
«2 4
24 ¬
37
8
4
8. Exact: ³
4
1
Simpson’s: ³
9. Exact: ³
ª
x3 º
«4 x »
3 ¼1
¬
4 x 2 dx
Trapezoidal: ³
4
4
4 x 2 dx |
1
2
0
x 2
Trapezoidal: ³
ª 2 º
«
»
¬« x 2 ¼» 0
dx
2
1
2
0
x 2
21
2
4
21 4
16 11
3
3
47
2
8
47
8
26
2
4
26 4
57
2
8
57
8
31
2
4
31 4
º
67
3» | 12.6640
8
¼
º
67
3» | 12.6667
8
¼
9
º
1ª
9·
25 ·
49 ·
§
§
§
3 4¨ 4 ¸ 0 4¨ 4 ¸ 10 4¨ 4 ¸ 12»
6 «¬
4¹
4¹
4¹
©
©
©
¼
1
1
| 12.6667
2
2
2
½°
ª
ª
ª
1 °­
§ 3· º
§5· º
§7· º
®3 2 «4 ¨ ¸ » 2 0 2 «4 ¨ ¸ » 2 5 2 «4 ¨ ¸ » 12¾ | 9.1250
4°
2
2
2
© ¹ ¼»
© ¹ ¼»
© ¹ ¼»
«¬
°¿
¬«
¬«
¯
4 x 2 dx |
1
20.0000
x dx | 12 ª¬0 2 2 3 2 2 3 3 2 3 4 2 3 5 2 3 6 2 3 7 2º¼ | 11.7296
x dx
Trapezoidal: ³
20
3
3
3
3
º
1ª
3
§ 4·
§5·
§7·
§8·
«1 4¨ ¸ 2¨ ¸ 4 2 2¨ ¸ 4¨ ¸ 27»
9 ¬«
© 3¹
© 3¹
© 3¹
© 3¹
¼»
ª 3 x4 3 º
¬4
¼0
x dx
419
3
3
3
3
º
1ª
3
§ 4·
§5·
§7·
§8·
«1 2¨ ¸ 2¨ ¸ 2 2 2¨ ¸ 2¨ ¸ 27» | 20.2222
6 ¬«
© 3¹
© 3¹
© 3¹
© 3¹
¼»
3
Trapezoidal: ³ x3 dx |
Simpson’s: ³ x3 dx |
81 1
4
4
Numerical
Numerica Integration
dx |
2
2 2
3
2
9
1
3
ª
§
·
§
·
§
· 2º
1 «1
2
2
2
¸ 2¨
¸ 2¨
¸ »
2¨
¨ 1 4 2 2¸
¨ 12 2 2¸
¨ 3 4 2 2 ¸ 9»
8 «2
©
¹
©
¹
©
¹
¬
¼
1 ª1
§ 32 ·
§8·
§ 32 · 2 º
2¨ ¸ 2¨ ¸ 2¨
¸ » | 0.3352
8 «¬ 2
© 81 ¹
© 25 ¹
© 121 ¹ 9 ¼
Simpson’s: ³
1
0
2
x 2
2
dx |
ª
§
·
§
·
§
· 2º
1 «1
2
2
2
¸ 2¨
¸ 4¨
¸ »
4¨
2
2
2
¨ 14 2 ¸
¨ 12 2 ¸
¨ 3 4 2 ¸ 9»
12 « 2
©
¹
©
¹
©
¹
¬
¼
1 ª1
§ 32 ·
§8·
§ 32 · 2 º
4¨ ¸ 2¨ ¸ 4¨
¸ » | 0.3334
«
12 ¬ 2
© 81 ¹
© 25 ¹
© 121 ¹ 9 ¼
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 4
420
Integration
egration
2
10. Exact: ³ x
0
2
1 53 2 1
3
1
2
1
2
1 x3 dx | 14 ª1 2 1 ¬«
1
8
0
0
º
»¼ 0
ª
x 2 1 dx | 16 «0 4 12
¬
2
Simpson’s: ³ x
11. Trapezoidal: ³
32 2
ª
x 2 1 dx | 14 «0 2 12
¬
2
Trapezoidal: ³ x
Simpson’s: ³
1 ª x2 1
3«
¬
x 2 1 dx
0
1 x3 dx | 16 ª1 4 1 ¬«
2
0
2
2
| 3.393
1 21
12 1 2 23
1 21
12 1 4 32
2 2 2 1
2 2 4 1
1
8
3
2
3
2
2
2
º
1 2 22 1» | 3.457
¼
º
1 2 22 1» | 3.392
¼
3º | 3.283
¼»
27
8
3º | 3.240
¼»
27
8
Graphing utility: 3.241
12. Trapezoidal: ³
Simpson’s: ³
2
1
0
1 x3
2
1
0
1 x3
ª
§
1«
1 2¨
¨¨
4«
©
¬«
·
¸ 2§
¨
3 ¸
©
1 1 2 ¸¹
§
·
2¨
¸
¨¨
1 13 ¹
©
º
·
¸ 1 » | 1.397
»
3 ¸
1 3 2 ¸¹ 3 »
¼
ª
§
1«
1 4¨
¨¨
6«
©
¬«
·
¸ 2§
¨
3 ¸
©
1 1 2 ¸¹
§
·
4¨
¸
¨¨
1 13 ¹
©
º
·
¸ 1 » | 1.405
»
3 ¸
1 3 2 ¸¹ 3 »
¼
dx |
dx |
1
1
1
1
1
1
Graphing utility: 1.402
13. ³
1
0
x
1
³0
1 x dx
Trapezoidal: ³
Simpson’s: ³
x 1 x dx | 18 ª0 2
«¬
1 1 1
4
4
2
1 1 1
2
2
2
3 1 3 º | 0.342
4
4 »
¼
1 ª0 4
x 1 x dx | 12
«¬
1 1 1
4
4
2
1 1 1
2
2
4
3 1 3
4
4
1
0
1
0
x 1 x dx
º | 0.372
»¼
Graphing utility: 0.393
14. Trapezoidal: ³
Simpson’s: ³
S
S 2
S
S 2
x sin x dx |
Sª S
1 2
«
16 ¬ 2
x sin x dx |
S ª S
4
«
24 ¬ 2
5S
§ 5S ·
sin ¨ ¸ 2
8
© 8 ¹
5S
§ 5S ·
sin ¨ ¸ 2
8
© 8 ¹
3S
§ 3S ·
sin ¨ ¸ 2
4
© 4 ¹
3S
§ 3S ·
sin ¨ ¸ 4
4
© 4 ¹
º
7S
§ 7S ·
sin ¨ ¸ 0» | 1.430
8
8
© ¹
¼
º
7S
§ 7S ·
sin ¨ ¸ 0» | 1.458
8
© 8 ¹
¼
Graphing utility: 1.458
15. Trapezoidal: ³
Simpson’s: ³
S 2
0
S 2
0
sin x
sin x
2
2
dx |
S 2 ǻ
§
sin 0 2 sin ¨
¨
«
©
¬
S 2·
dx |
S 2 ǻ
§
sin 0 4 sin ¨
¨
«
12
©
¬
S 2·
8
§
2 sin ¨
¸¸
¨
¹
©
4
4
2
2
§
¸¸ 2 sin ¨¨
¹
©
S 2·
2
§3 S 2 ·
§
2 sin ¨
sin ¨¨
¸¸
¨ 4 ¸¸
©
¹
©
¹
2
S 2·
2
2
2
2
§3 S 2 ·
§
¸¸ 4 sin ¨¨
¸¸ sin ¨¨
©
¹
© 4 ¹
S · »º
2
¸ | 0.550
2 ¸¹ »
¼
S · »º
2
¸ | 0.548
2 ¸¹ »
¼
Graphing utility: 0.549
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.6
16. Trapezoidal: ³
Simpson’s: ³
tan x 2 dx |
S 4 ǻ
§
tan 0 2 tan ¨
¨
«
©
¬
S 4·
tan x 2 dx |
S 4 ª«
§
tan 0 4 tan ¨
¨
12 «
©
¬
S 4·
S 4
0
S 4
0
8
2
§
¸¸ 2 tan ¨¨
¹
©
4
S 4·
2
4
S 4·
2
2
2
421
S · »º
2
2
§3 S 4 ·
§
¸¸ 2 tan ¨¨
¸¸ tan ¨¨
4
©
¹
©
¹
2
§
2 tan ¨
¸¸
¨
¹
©
Numerical
Numerica Integration
¸ | 0.271
4 ¸¹ »
¼
S · º»
2
2
§3 S 4 ·
§
4 tan ¨
tan ¨¨
¸¸
¨ 4 ¸¸
©
¹
©
¹
¸ | 0.257
4 ¹¸ »
¼
Graphing utility: 0.256
17. Trapezoidal: ³
Simpson’s: ³
3.1
ªcos 3
cos x 2 dx | 0.1
8 ¬
3
3.1
3
ªcos 3
cos x 2 dx | 0.1
12 ¬
2
2
2 cos 3.025
4 cos 3.025
2
2
2 cos 3.05
2 cos 3.05
2
2
2 cos 3.075
4 cos 3.075
2
2
cos 3.1 º | 0.098
¼
2
cos 3.1 º | 0.098
¼
2
Graphing utility: 0.098
18. Trapezoidal: ³
Simpson’s: ³
S 2
§S ·
§S ·
§ 3S ·
«1 2 1 sin 2 ¨ ¸ 2 1 sin 2 ¨ ¸ 2 1 sin 2 ¨ ¸ 16 ¬«
©8¹
©4¹
© 8 ¹
º
2 » | 1.910
¼»
S ª
§S ·
§S ·
§ 3S ·
«1 4 1 sin 2 ¨ ¸ 2 1 sin 2 ¨ ¸ 4 1 sin 2 ¨ ¸ 24 ¬«
8
4
© ¹
© ¹
© 8 ¹
º
2 » | 1.910
¼»
1 sin 2 x dx |
0
S 2
1 sin 2 x dx |
0
Sª
Graphing utility: 1.910
19. Trapezoidal: ³
Simpson’s: ³
x tan x dx |
S ª
§S · §S ·
§ 2S · § 2S ·
§ 3S ·
§ 3S · S º
0 2¨ ¸ tan ¨ ¸ 2¨ ¸ tan¨ ¸ 2¨ ¸ tan ¨ ¸ » | 0.194
32 «¬
16
16
16
16
16
© ¹ © ¹
© ¹ © ¹
© ¹
© 16 ¹ 4 ¼
x tan x dx |
S ª
§S ·
§S ·
§ 2S ·
§ 2S ·
§ 3S · § 3S · S º
0 4¨ ¸ tan ¨ ¸ 2¨ ¸ tan ¨ ¸ 4¨ ¸ tan ¨ ¸ » | 0.186
48 «¬
© 16 ¹
© 16 ¹
© 16 ¹
© 16 ¹
© 16 ¹ © 16 ¹ 4 ¼
S 4
0
S 4
0
Graphing utility: 0.186
20. Trapezoidal: ³
Simpson’s: ³
S sin x
x
0
S sin x
x
0
dx |
dx |
Sª
«1 8¬
2 sin S 4
2 sin S 2
2 sin 3S 4
º
0» | 1.836
S 4
S 2
3S 4
¼
4 sin S 4
2 sin S 2
4 sin 3S 4
º
Sª
0» | 1.852
«1 S 4
S 2
12 ¬
3S 4
¼
Graphing utility: 1.852
21. Trapezoidal: Linear polynomials
f x
2 x3
fc x
6x2
f cc x
12 x
f ccc x
12
4
0
23.
Simpson’s: Quadratic polynomials
22. For a linear function, the Trapezoidal Rule is exact. The
3
ba
ªmax f cc x º
¼
12n 2 ¬
0 for a linear function. Geometrically, a
error formula says that E d
and f cc x
linear function is approximated exactly by trapezoids:
f
x
(a) Trapezoidal: Error d
y
31
3
36
12 42
1.5 because
f cc x is maximum in [1, 3] when x
(b) Simpson’s: Error d
f 4 x
31
180 4
4
3.
5
0
0 because
0.
x
a
b
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
422
Chapter 4
24.
f x
NOT FOR SALE
Integration
egration
5x 2
27.
f x
x 1 ,
2
fc x
5
fc x
x
f cc x
0
f cc x
2 x 3
f ccc x
6 x 4
f 4 x
24 x 5
The error is 0 for both rules.
2
f x
x 1
fc x
2 x 1
f cc x
6 x 1
f ccc x
24 x 1
5
f 4 x
120 x 1
6
25.
1d x d 3
4
(a) Trapezoidal: Error d
Trapezoidal:
23
Error d
2 d 0.00001, n 2 t 133,333.33,
12n 2
n t 365.15 Let n
366.
4 2
3
(b) Simpson’s: Error d
42
Simpson’s: Error d
2.
120
1
because
12
f 4 x is a maximum of 120 at x
2.
28.
f x
1 x
1
fc x
1 x
f cc x
21 x
3
f ccc x
6 1 x
4
24 1 x
5
cos x
fc x
sin x
f cc x
cos x
f 4 x
f ccc x
sin x
(a) Maximum of f cc x
4
cos x
x
S 0
3
1
12 42
S
3
192
| 0.1615
Error d
21 x
3
is 2.
1
2 d 0.00001
12n 2
n 2 t 16,666.67
because f cc x is at most 1 on >0, S @.
n t 129.10. Let n
(b) Simpson’s:
S 0
180 44
because f
26.
Trapezoidal:
(a) Trapezoidal: Error d
Error d
25
24 d 0.00001,
180n 4
, 0 d x d1
2
f x
f
24 x 5 is 24.
n 4 t 426,666.67, n t 25.56 Let n
5
180 44
(b) Maximum of f 4 x
1
because
4
6
12 42
f cc x is a maximum of 6 at x
26.
2 x 3 is 2.
(a) Maximum of f cc x
3
4
5
1
S5
46,080
| 0.006641
x is at most 1 on >0, S @.
(b) Maximum of f
4
x
24 1 x
130.
5
is 24.
Simpson’s:
Error d
1
24 d 0.00001
180n 4
n 4 t 13,333.33
n t 10.75
Let n
12. (In Simpson’s Rule n must be even.)
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.6
29.
12
f x
x 2
fc x
1
1 2
x 2
2
1
3 2
x 2
4
3
5 2
x 2
8
15
7 2
x 2
16
f cc x
f ccc x
f 4 x
, 0 d x d 2
(a) Maximum of f cc x
Numerical
Numerica Integration
f x
sin x, 0 d x d
fc x
cos x
f cc x
sin x
f ccc x
cos x
4
sin x
30.
f
x
2
(a) Trapezoidal:
1
4x2
d
is
32
Error
S 2
3
1 d 0.00001
12n 2
n2 t
S3
105
96
n t 179.7. Let n
Trapezoidal:
20
12n 2
S
All derivatives are bounded by 1.
2
| 0.0884.
16
Error d
3
d
8 2 5
n2 t
10
12 16
2 5
10
24
n t 76.8. Let n
77.
15
16 x 2
72
is
15 2
| 0.0829.
256
Error
Simpson’s:
5
180n 4
1 d 0.00001
S5
105
5760
n t 8.5. Let n
1
41 x
32
Trapezoidal: Error d
32 15 2 5
10
180 256
10 even .
1 x
31. f x
in [0, 2].
f cc x is maximum when x
25 § 15 2 ·
Error d
¨
¸ d 0.00001
180n 4 ¨© 256 ¸¹
2 5
10
96
n t 6.2. Let n
S 2
n4 t
(a) f cc x
n4 t
180.
(b) Simpson’s:
§ 2·
¨¨
¸¸ d 0.00001
© 16 ¹
(b) Maximum of f 4 x
423
0 and f cc 0
8 §1·
¨ ¸ d 0.00001,
12n 2 © 4 ¹
n 2 t 16,666.67, n t 129.10; let n
(b) f 4 x
8 even .
15
16 1 x
72
130.
in [0, 2]
f 4 x is maximum when x
f 4 0
1
.
4
0 and
15
.
16
Simpson’s: Error d
32 § 15 ·
¨ ¸ d 0.00001,
180n 4 © 16 ¹
n 4 t 16,666.67, n t 11.36; let n
12.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
424
NOT FOR SALE
Chapter 4
Integration
egration
x 1
32. f x
(a) f cc x
23
2
9 x 1
43
in >0, 2@.
0 and f cc 0
f cc x is maximum when x
8 § 2·
¨ ¸ d 0.00001,
12n 4 © 9 ¹
Trapezoidal: Error d
n 2 t 14,814.81, n t 121.72; let n
(b) f 4 x
56
81 x 1
10 3
122.
in [0, 2].
f 4 x is maximum when x
Simpson’s: Error d
2
.
9
0 and f 4 0
56
.
81
32 § 56 ·
¨ ¸ d 0.00001,
180n 4 © 81 ¹
n 4 t 12,290.81, n t 10.53; let n
12. (In Simpson’s Rule n must be even.)
tan x 2
33. f x
(a) f cc x
2 sec2 x 2 ª¬1 4 x 2 tan x 2 º¼ in >0, 1@.
f cc x is maximum when x
Trapezoidal: Error d
10
12n 2
1 and f cc 1 | 49.5305.
3
49.5305 d 0.00001, n 2 t 412,754.17, n t 642.46; let n
643.
8 sec 2 x 2 ª¬12 x 2 3 32 x 4 tan x 2 36 x 2 tan 2 x 2 48 x 4 tan 3 x 2 º¼ in [0, 1]
(b) f 4 x
f 4 x is maximum when x
1 and f 4 1 | 9184.4734.
5
Simpson’s: Error d
10
9184.4734 d 0.00001, n 4 t 5,102,485.22, n t 47.53; let n
180n 4
48.
sin x 2
34. f x
(a) f cc x
2 ª¬2 x 2 sin x 2 cos x 2 º¼ in [0, 1]. f cc x is maximum when x
Trapezoidal: Error d
(b) f 4 x
10
12n 2
1 and f cc 1 | 2.2853.
3
2.2853 d 0.00001, n 2 t 19,044.17, n t 138.00; let n
139.
16 x 4 12 sin x 2 48 x 2 cos x 2 in [0, 1]
f 4 x is maximum when x | 0.852 and f 4 0.852 | 28.4285.
5
Simpson’s: Error d
35. n
4, b a
40
10
28.4285 d 0.00001, n 4 t 15,793.61, n t 11.21; Let n
180n 4
4
4
(a)
³ 0 f x dx | 84 ª¬3 2 7 2 9 2 7 0º¼
(b)
³ 0 f x dx | 124 ª¬3 4 7 2 9 4 7 0º¼
4
12.
1 49
2
77
3
49
2
24.5
| 25.67
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 4.6
8, b a
36. n
80
Numerical
Numerica Integration
8
8
(a)
³ 0 f x dx | 168 ª¬0 2 1.5 2 3 2 5.5 2 9 2 10 2 9 2 6 0º¼
(b)
³ 0 f x dx | 248 ª¬0 4 1.5 2 3 4 5.5 2 9 4 10 2 9 4 6 0º¼
8
S 2
³0
37. A
S 2
1 88
2
44
1 134
3
134
3
x cos x dx
Simpson’s Rule: n
³0
425
14
x cos x dx |
Sª
« 0 cos 0 4
84 ¬
S
28
cos
S
28
2
S
14
cos
S
14
4
3S
3S
cos
"
28
28
S
2
cos
Sº
» | 0.701
2¼
y
1
1
2
π
π
4
2
x
38. Simpson’s Rule: n
8 3³
S 2
0
1
8
3S ª
2
2
2
S
2
2 S
2 1 sin 2
"
« 1 sin 0 4 1 sin
6 ¬
3
3
16
3
8
2
sin 2 T dT |
3
1
2
S º
sin 2 »
3
2¼
| 17.476
39. Area |
1000
ª125 2 125 2 120 2 112 2 90 2 90 2 95 2 88 2 75 2 35 º¼
2 10 ¬
89,250 m 2
2
40. (a) The integral ³ f x dx would be overestimated because the trapezoids would be above the curve. Similarly, the integral
0
2
³ 0 g x dx would be underestimated.
(b) Simpson’s Rule would be more accurate because it takes into account the curvature of the graph.
41. W
5
³ 0 100 x 125 x dx
3
Simpson’s Rule: n
5
12
5 ª
§5·
§5·
3
§ 10 ·
§ 10 ·
³ 0 100 x 125 x dx | 3 12 ««0 400¨© 12 ¸¹ 125 ¨© 12 ¸¹ 200¨© 12 ¸¹ 125 ¨© 12 ¸¹
3
3
¬
3
º
§ 15 ·
§ 15 ·
400¨ ¸ 125 ¨ ¸ " 0» | 10,233.58 ft-lb
»
© 12 ¹
© 12 ¹
¼
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 4
426
Integration
egration
42. (a) Trapezoidal:
2
2
³ 0 f x dx | 2 8 ª¬4.32 2 4.36 2 4.58 2 5.79 2 6.14 2 7.25 2 7.64 2 8.08 8.14º¼ | 12.518
Simpson’s:
2
2
³ 0 f x dx | 3 8 ª¬4.32 4 4.36 2 4.58 4 5.79 2 6.14 4 7.25 2 7.64 4 8.08 8.14º¼ | 12.592
(b) Using a graphing utility,
1.37266 x3 4.0092 x 2 0.620 x 4.28. Integrating, ³
y
43. ³
12
6
0
1 x2
dx Simpson’s Rule, n
2
0
y dx | 12.521.
6
§1
·
¨ 0¸
2
©
¹ ª6 4 6.0209 2 6.0851 4 6.1968 2 6.3640 4 6.6002 6.9282º | 1 113.098 | 3.1416
S |
>
@
¼
36 ¬
36
44. Simpson’s Rule: n
4³
S
t
45. ³ sin
0
1
1
0 1 x2
x dx
6
dx |
4 ª
4
2
4
2
4
1º
«1 » | 3.14159
2
2
2
2
2
36 «
2»
1 16
1 26
1 36
1 46
1 56
¬
¼
2, n
10
By trial and error, you obtain t | 2.477.
Ax3 Bx 2 Cx D. Then f 4 x
46. Let f x
Simpson’s: Error d
ba
180n 4
0.
5
0
0
So, Simpson’s Rule is exact when approximating the integral of a cubic polynomial.
1
Example: ³ x3 dx
0
3
º
1ª
§1·
«0 4¨ ¸ 1»
6 ¬«
© 2¹
»¼
1
4
This is the exact value of the integral.
47. The quadratic polynomial
p x
x x2 x x3
x x1 x x3
x x1 x x2
y1 y2 y3
x1 x2 x1 x3
x2 x1 x2 x3
x3 x1 x3 x2
passes through the three points.
Review Exercises for Chapter 4
1. ³ x 6 dx
x2
6x C
2
4. ³ 3
2. ³ x 4 3 dx
x5
3x C
5
5. ³
3. ³ 4 x 2 x 3 dx
6
dx
x
x4 8
dx
x3
³ 6x
1 3
6˜
dx
³ x 8x
3
dx
x2 3
C
23
9 x2 3 C
1 2
4
x 2 C
2
x
4 x3 1 x 2 3x C
3
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 4
6. ³
x2 2 x 6
dx
x4
³ x
2
2 x 3 6 x 4 dx
1
2
13. (a) Answers will vary. Sample answer:
y
3
2
x
x
x
2˜
6˜
C
1
2
3
7. ³ 2 x 9 sin x dx
6 x, f 1
f x
3x 2 C
f 1
2
f x
3x 2 1
2
5 sin x 2 tan x C
C Ÿ C
1
dy
dx
2 x 4, 4, 2
y
³ 2 x 4 dx
2
16 16 C Ÿ C
y
x 4x 2
x2 4 x C
3 x3 x C
f 0
7
f x
3 x3 x 7
f cc x
24 x, f c 1
fc x
12 x 2 C1
f c 1
7
fc x
12 x 5
f x
4 x 3 5 x C2
7
−7
0C Ÿ C
12 1
8
7
14. (a) Answers will vary. Sample answer:
y
4
7, f 1
2
C1 Ÿ C1
6
5
(6, 2)
2
x
3
f 1
4
f x
4x 5x 3
41
7
−2
5 1 C2 Ÿ C2
3
12. f cc x
2cos x, f c 0
fc x
2sin x C1
fc 0
4
fc x
2sin x 4
f x
2cos x 4 x C2
f 0
5
4, f 0
2sin 0 C1 Ÿ C1
5
3
(b)
dy
dx
1 2
x 2 x, 6, 2
2
y
³ ¨© 2 x 2 x ¸¹dx
2
4
y
§1 2
·
1 3
x x2 C
6
1 3
6 62 C Ÿ C
6
1 3
x x2 2
6
2
4
2cos 0 4 0 C2
2 C2
Ÿ C2
2
2
1
f x
f x
(b)
−4
9 x 2 1, f 0
11.
−6
2
10. f c x
2
6
x 2 9 cos x C
3 1
30
x
−2
1
1
2
2 3 C
x
x
x
8. ³ 5 cos x 2 sec 2 x dx
9. f c x
427
−3
9
3
2cos x 4 x 3
−4
INSTRUCTOR USE ONLY
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428
Chapter 4
NOT FOR SALE
Integration
egration
32
15. a t
17. a t
a
vt
32t 96
vt
at C1
st
16t 96t
³ a dt
v0
0 C1
0 when C1
0.
vt
at
st
³ at dt
s0
0 C2
a 2
t C2
2
0 when C2
0.
st
a 2
t
2
s 30
a
2
30
2
2
32t 96
0 when t
144 288
144 ft
(b) v t
32t 96
96 when t
2
(c) s 32
16 94 96 32
(a) v t
s3
16. 45 mi h
66 ft sec
30 mi h
44 ft sec
3 sec.
3 sec.
2
108 ft
a
at
a
vt
at 66 because v 0
66 ft sec.
st
a
t 2 66t because s 0
2
0.
3600 or
2 3600
30
v 30
8 30
2
8 ft sec 2 .
240 ft sec
Solving the system
vt
at 66
st
a
t 2 66t
2
you obtain t
264
24 5 and a
55 12 t 66
§ 72 ·
s¨ ¸
© 5¹
44
55 12. Now solve
0 and get t
72 5. So,
2
55 12 § 72 ·
§ 72 ·
¨ ¸ 66¨ ¸ | 475.2 ft.
2 © 5¹
© 5¹
Stopping distance from 30 mi h to rest is
475.2 264
18.
211.2 ft.
1 mi h 5280 ft mi
3600 sec h
(a)
22
ft sec
15
T
0
5
10
15
20
25
30
V1 ft sec
0
3.67
10.27
23.47
42.53
66
95.33
V2 ft sec
0
30.8
55.73
74.8
88
93.87
95.33
(b) V1 t
0.1068t 2 0.0416t 0.3679
V2 t
0.1208t 2 6.7991t 0.0707
(c) S1 t
³ V1 t dt
0.1068 3 0.0416 2
t t 0.3679t
3
2
S2 t
³ V2 t dt
0.1208t 3
6.7991t 2
0.0707t
3
2
>In both cases, the constant of integration is 0 because S1 0
S2 0
0.º¼
S1 30 | 953.5 feet
S 2 30 | 1970.3 feet
INSTRUCTOR USE ONLY
The second car was going faster than the first until the end.
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 4
5
19. ¦ 5i 3
2 7 12 17 22
10
21. ¦
60
3
20. ¦ k 2 1
1 2 5 10
n
§ 3 ·§ i 1 ·
22. ¦ ¨ ¸¨
¸
i 1 © n ¹© n ¹
2
1
i 1 3i
i 1
429
1
1
1
"
31
32
3 10
18
k 0
2
2
3 § 1 1·
3 § 2 1·
3 § n 1·
¨
¸ ¨
¸ " ¨
¸
n© n ¹
n© n ¹
n© n ¹
24
30
23. ¦ 8
8 24
26. ¦ 3i 4
192
i 1
i 1
75
75
24. ¦ 5i
5 ¦i
i 1
20
5˜
75 76
§ 20 21 ·
2¨
¸
© 2 ¹
30
i 1
i 1
30 31
4 30
2
1395 120 1275
20
27. ¦ i 1
420
30
3¦ i 4 ¦ 1
3˜
14,250
2
i 1
25. ¦ 2i
i 1
2
2
i 1
20
¦ i 2 2i 1
i 1
20 21 41
20 21
2
20
6
2
2870 420 20 3310
12
28. ¦ i i 2 1
i 1
12
¦ i3 i
i 1
122 132
4
6084 78
29. y
10
, 'x
x2 1
1
,n
2
12 13
2
6006
4
S n
S 4
º
1 ª10
10
10
10
« 2
» | 13.0385
2
2
2« 1
12 1
1 1
3 2 1»¼
¬
sn
s4
1 ª 10
10
10
10 º
«
2
» | 9.0385
2
2
2« 1 2 1 1 1
3 2 1 2 1»¼
¬
9.0385 Area of Region 13.0385
30. y
9 14 x 2 , 'x
1, n
4
S 4
1ª 9 14 4
¬
9 14 9
9 14 16
9 14 25 º | 22.5
¼
s4
1ª 9 14 9
¬
9 14 16
9 14 25
9 9 º | 14.5
¼
14.5 Area of Region 22.5
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
430
Chapter 4
31. y
Integration
egration
3
, right endpoints
n
8 2 x, 'x
3
n
5 x 2 , 'x
33. y
y
n
n
Area
lim ¦ f ci 'x
nof
lim ¦ f ci 'x
Area
i 1
n of
8
§
§ 3i · · 3
lim ¦ ¨ 8 2¨ ¸ ¸
nof
© n ¹¹ n
i 1 ©
n
2
4
2
3 n §
6i ·
lim ¦ ¨ 8 ¸
nof n
n¹
©
i 1
−1
3 n ª
12i 9i 2 º
2»
1
¦
«
n of n
n
n ¼
i 1¬
x
1
−2
2
3
4
5
lim
3ª
6n n 1º
«8n »
2
n
¬
¼
lim
3ª
12 n n 1
9 n n 1 2n 1 º
2
lim «n »
2
6
n
n
¬
¼
nof n
n 1º
ª
lim 24 9
n »¼
nof«
¬
32. y
n of n
24 9
n of
y
y
n
Area lim ¦ f ci 'x
n of
ª
n 1 9 n 1 2n 1 º
lim «3 18
»
2
n
n2
¬
¼
3 18 9 12
15
2
, right endpoints
n
x 2 3, 'x
4
10
ª§ 2i ·
º§ 2 ·
lim ¦ «¨ ¸ 3»¨ ¸
n of
n
i 1¬
«© ¹
»¼© n ¹
2
lim
6
12
i 1
2
n
n of n
n
ª 4i
2
º
¦ « n2 3»
i 1¬
i 1
n ª
3i · º§ 3 ·
§
lim ¦ «5 ¨ 2 ¸ »¨ ¸
n of
n ¹ ¼»© n ¹
©
«
i 1¬
6
3
8
2
1
6
x
4
−4 −3
−1
2
1
2
3
4
−2
x
¼
1
2
8
6
3
26
3
º
2 ª 4 n n 1 2n 1
3n»
lim «
n of n n 2
6
¬
¼
ª 4 n 1 2n 1
º
6»
lim «
2
3
n
¬
¼
n of
34. y
1 3
x , 'x
4
2
n
y
n
Area
lim ¦ f ci 'x
n of
20
i 1
n
1§
2i · § 2 ·
lim ¦ ¨ 2 ¸ ¨ ¸
n of
4
n ¹ ©n¹
i 1 ©
15
1 n ª
24i
24i 2
8i 3 º
¦ «8 n n2 n3 »
n of 2 n
i 1¬
¼
5
3
10
lim
x
1
2
3
4
4 n ª
3i 3i 2
i3 º
2 3»
1
¦
«
n of n
n
n
n ¼
i 1¬
lim
2
4ª
3n n 1
3 n n 1 2n 1
1 n2 n 1 º
«n 2
3
»
n of n «
2
6
4
n
n
n
¬
¼»
lim
4 6 41
15
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 4
3
n
5 y y 2 , 2 d y d 5, 'y
35. x
431
y
2
ª §
3i · §
3i · º§ 3 ·
lim ¦ «5¨ 2 ¸ ¨ 2 ¸ »¨ ¸
n of
n¹ ©
n ¹ ¼»© n ¹
« ©
i 1¬
n
Area
6
4
3 n ª
15i
i
9i 2 º
4 12 2 »
lim ¦ «10 n of n
n
n
n ¼
i 1¬
3
2
1
3 n ª
3i 9i 2 º
¦ «6 n n2 »
n of n
i 1¬
¼
x
lim
lim
1
2
3
4
5
6
3ª
3n n 1
9 n n 1 2n 1 º
2
«6n »
2
6
n
n
¬
¼
n of n
9
ª
º
«18 2 9»
¬
¼
27
2
§ b ·§ b ·
§ 2b ·§ b ·
§ 3b ·§ b ·
§ 4b ·§ b ·
m¨ ¸¨ ¸ m¨ ¸¨ ¸ m¨ ¸¨ ¸ m¨ ¸¨ ¸
4
4
4
4
4
4
© ¹© ¹
© ¹© ¹
© ¹© ¹
© 4 ¹© 4 ¹
36. (a) S
§b·
§ b ·§ b ·
§ 2b ·§ b ·
§ 3b ·§ b ·
m 0 ¨ ¸ m¨ ¸¨ ¸ m¨ ¸¨ ¸ m¨ ¸¨ ¸
4
4
4
4
4
© ¹
© ¹© ¹
© ¹© ¹
© 4 ¹© 4 ¹
s
2
i 1
§b· n
m¨ ¸ ¦ i
©n¹ i 1
mb 2 § n n 1 ·
¨
¸
n2 ©
2
¹
mb 2 n 1
2n
n 1
n 1
§ bi ·§ b ·
¦ m¨© n ¸¨
¸
¹© n ¹
i 0
§ b · n 1
m¨ ¸ ¦ i
©n¹ i 0
2
mb 2 § n 1 n ·
¨
¸
2
n2 ©
¹
mb 2 n 1
2n
1 2
mb
2
1
b mb
2
(c) Area
mb 2 n 1
n of
2n
4
38. ³
10
3mb 2
8
§ mbi ·§ b ·
§ bi ·§ b ·
¦ f ¨© n ¸¨
¸
¹© n ¹
i 0
0
mb 2
1 23
16
¦ ¨© n ¸¨
¸
¹© n ¹
§ bi ·§ b ·
sn
37. ³
n
i 1
mb 2 n 1
n of
2n
lim
lim
y
y = mx
x=b
x
1
base height
2
y
40.
2 x 8 dx
10
5mb 2
8
¦ f ¨© n ¸¨
¸
¹© n ¹
n
(b) S n
mb 2
1 23 4
16
8
100 x 2 dx
4
2
x
39.
−6 −4 −2
y
−2
2
4
6
−4
12
9
6
6
³ 6
Triangle
3
3
6
1S 6 2
2
18S
semicircle
x
−3
36 x 2 dx
9
−3
5
³ 0 5 x 5 dx
1 5
2
25
2
5
(triangle)
8
³ 4 f x dx ³ 4 g x dx
8
³ 4 f x dx ³ 4 g x dx
41. (a)
³ 4 ¬ª f x g x º¼ dx
(b)
³ 4 ª¬ f x g x º¼ dx
(c)
³ 4 ª¬2 f x 3g x º¼ dx
(d)
³ 4 7 f x dx
8
8
7³
8
8
8
12 5
17
8
8
12 5
7
2³
f x dx
8
4
8
f x dx 3³ g x dx
4
7 12
2 12 3 5
9
84
INSTRUCTOR USE ONLY
4
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Chapter 4
432
Integration
egration
6
³ 0 f x dx ³ 3 f x dx
3
3
³
42. (a)
³ 0 f x dx
(b)
³ 6 f x dx
(c)
³ 4 f x dx
(d)
³ 3 10 f x dx
4
6
6
3
1
f x dx
8
0
6
10³
6
f x dx
3
3 x dx
ª
x2 º
«3x »
2 ¼0
¬
t 2 1 dt
ªt 3
º
« t»
3
¬
¼2
24 3
3
44. ³
2
45. ³
1
1
1
64
2
3
1
56
2
9
47. ³ x
x dx
4
9
³4
9
x3 2 dx
3S 4
49. ³
0
50. ³
S 4
S 4
4
³1 x
3
ª2 5 2º
«5 x »
¬
¼4
x dx
sin T dT
>cos T @30S 4
sec 2 t dt
>tan t@S 4
S 4
2
3
16
3
§ 243
· § 32
·
18 18 ¸ ¨
8 12 ¸
¨
5
5
©
¹ ©
¹
2ª
5 «¬
9
5
4
4 § 1
·
48. ³ ¨ 3 x ¸ dx
1
x
©
¹
6
0
ª x5
º
2
« 2 x 6 x»
5
¬
¼2
x 4 4 x 6 dx
10
§ 33
· § 23
·
2¸
¨ 3¸ ¨
3
3
©
¹ ©
¹
ª¬t 4 t 2 º¼
1
4t 3 2t dt
10 1
3
46. ³
3
0
8
43. ³
4 1
1 1
1ª 2
1º
x 2»
2 ¬«
x ¼1
1
2
2
231
5
1 ª§
1·
º
¨16 ¸ 1 1»
2 «¬©
16 ¹
¼
255
32
2
243 32
5
4
ª x2
x2 º
«
»
2 ¼1
¬ 2
§
2·
¨¨ ¸¸ 1
2
©
¹
5
4 ȼ
¼
211
4
5
2 2
2
53. A
2
422
5
6
³ 0 8 x dx
6
³ 0 sin x dx
ª
x2 º
«8 x »
2 ¼0
¬
>cos x@02
48 18 0
cos 2 1
30
2
51. Area
1 cos 2
54. y
| 1.416
A
52. Area
S 2
³0
S 2
x cos x dx
ª x2
º
« sin x»
2
¬
¼0
S2
8
x2 x 6
3
x 3 x 2
³ 2 x x 6 dx
2
3
1
ª x3
º
x2
6 x»
«
3
2
¬
¼2
9
§
· §8
·
¨ 9 18 ¸ ¨ 2 12 ¸
2
©
¹ ©3
¹
125
6
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 4
1
³ 0 x x dx
55. A
3
59. F c x
x 2 1 x3
60. F c x
1
x2
61. F c x
x 2 3x 2
62. F c x
csc 2 x
63. u
x3 3, du
3x 2 dx
x2
³ x 3
1
ª x2 x4 º
« »
4 ¼0
¬2
§1 1·
¨ ¸0
© 2 4¹
1
4
1
³0
56. A
1
³0
x 1 x dx
x1 2 x3 2 dx
³
1
ª2 3 2 2 5 2º
«3 x 5 x »
¬
¼0
x 3
3
3 x 4 2, du
64. u
ª1
«5 2
¬
9 1
1
dx
9 4³4
x
º
x»
¼4
³ 6x
3
12
1
3 x 4 2 12 x3 dx
2³
3x 4 2 dx
2
5
4
1 3x 2
˜
2
32
65. u
1 3 x 2 , du
³ x 1 3x
2
4
dx
y
2
) 254 , 25 )
66. u
4
6
58. Average value:
8
x 2 8 x 7, du
x 4
x
³ x 2 8 x 7 2 dx
10
2
1
x 3 dx
³
0
20
2
ª x4 º
« »
¬ 8 ¼0
x3
2
x
3
5
5
6 x dx
C
C
2 x 8 dx
2
1
x 2 8 x 7 2 x 8 dx
2³
1
1 2
x 8x 7 C
2
1
C
2 x2 8x 7
2
2
67. ³ sin 3 x cos x dx
y
4
16 ³ 1 3 x 2
1 3x 2 1
30
2
C
6 x dx
1 1 3x 2
30
1
32
32
1 4
3x 2
C
3
5
2
25
4
x
x 2 dx
12 x3 dx
1
x
x
1 2
9
2
32
5
2
5
3
1 2
1
3 x 2 dx
x3 3
³
3
12
2 3
C
x 3
3
§ 2 2·
¨ ¸ 0
© 3 5¹
4
15
57. Average value:
dx
433
1 sin 4 x C
4
8
6
4
2
( 3 2 , 2)
x
1
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 4
434
Integration
egration
68. ³ x sin 3 x 2 dx
1
6
³ sin 3x
cos T
dT
1 sin T
69. ³
2
16 cos 3 x 2 C
6 x dx
³ 1 sin T
1 2
2 1 sin T
12
70. ³
cos T dT
sin x
dx
cos x
C
2 1 sin T C
71. ³ 1 sec S x
2
1
S³
sec S x tan S x dx
72. ³ sec 2 x tan 2 x dx
1
2
³ cos x
1 2
2 cos x
12
sin x dx
C
2 cos x C
2
1 sec S x
³ sec 2 x tan 2 x 2 dx
1
3
1 sec S x C
3S
S sec S x tan S x dx
1 sec 2 x C
2
73. (a) Answers will vary. Sample answer:
y
2
x
−3
(b)
dy
dx
3
x 9 x 2 , 0, 4
3
y
³ 9x
4
y
2
12
1 9 x
2
32
2
x dx
1
32
90
C
3
32
1
9 x2
5
3
32
C
1
27 C Ÿ C
3
5
32
1
9 x2
C
3
−6
6
−5
74. Answers will vary. Sample answer:
(a)
(b)
y
dy
dx
1
x sin x 2 , 0, 0
2
y
³
3
x
1
sin x 2 2 x dx u
4³
1
cos x 2 C
4
1
cos x 2 C
4
−3
0
y
1
75. ³ 3x 1 dx
0
5
2
1
x sin x 2 dx
2
1 1
5
3 x 1 3dx
3 ³0
1
cos 0 C Ÿ C
4
1
1
cos x 2 4
4
1 ª 3x 1
«
3«
6
¬
6 1
º
»
»¼ 0
1 ª 46
1º
« »
3¬6
6¼
x2
−3
3
−2
1
4
455
2
INSTRUCTOR USE ONLY
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Review Exercises ffor Chapter 4
1
435
3
76. ³ x 2 x3 2 dx
0
x3 2, du
u
When x
77. ³
4 1
u º
»
12 ¼ 2
1
3 1
3
1
dx
1 x
78. ³
0
2. When x
0, u
³ 2 u 3 du
6
x
3
3 x2 8
3
When y
2S ³
1
1 u , dy
6
1, u
9
83. ³ x x 1
0
1, u
0. When x
x 1 dx
2S ³
13
1
8
8
0
u 1
2
u du
2S ³
0
0
u 3 2 2u1 2 du
1
1.
2S ³
u du
1
0
1
S
ª
§ x ·º
«2 sin ¨ 2 ¸»
© ¹¼ 0
¬
2
x 1, du
dx. Let u
85. ³
dx.
du
2
0
3
3
128 16
7
4
S 2
468
7
cos x sin 2 x dx
2
St
0
2
dt
2
(b)
³ 2 x dx ³ 0 x dx
(c)
³0 3x dx
(d)
³ 2 5 x dx ³ 0 5 x dx
0
2
0
4
2
S
4
4
0
32
5
4
2
3 ³ x 4 dx
0
2
64
5
4
§ 32 ·
3¨ ¸
©5¹
5 ³
96
5
2
0
x 4 dx
§ 32 ·
5¨ ¸
©5¹
32
2
1.75 1 1
2 ³ x 4 dx
2
4
S 2
2ª
St º 2
«1.75 cos »
2 ¼0
S¬
§ 32 ·
2¨ ¸
©5¹
2
³ 2 x dx
ª¬sin x 12 cos 2 x º¼
0
1 12 0 12
86. ³ 1.75 sin
32
5
x 4 dx
(a)
u1 3 du
ª 3u 7 3
3u º
«
»
4 ¼0
¬ 7
0
32S
105
4
2
ª2
º
2S « u 7 2 u 5 2 u 3 2 »
7
5
3
¬
¼0
u 5 2 2u 3 2 u1 2 du
43 8
84. ³
28S
15
4
ª2
º
2S « u 5 2 u 3 2 »
3
¬5
¼1
0 because sin 2x is an odd function.
13
43
1
0, u
S
§ x·1
2³ cos¨ ¸ dx
0
© 2¹ 2
³0 u 1 u
³0 u
1
2 7 1
3
0.
du
sin 2 x dx
2
du
1. When y
S
§ x·
81. ³ cos¨ ¸ dx
0
© 2¹
42
1 2º
ª1 2
«3 x 8 »
¬
¼3
u 1, dx
1
A
3
ª2 1 x 1 2 º
¬
¼0
dx
x 1, x
0
S 4
5
4
2S ³ ª¬ 1 u 1º¼
1
2S ³ x 2
S 4
1 y dy
When x
82. ³
15
12
1 2
1 6 2
x 8
2 x dx
6³3
dx
0, u
y 1
0
80. u
1 2
1
1, u
1
16
12 12
³0 1 x
1 y, y
79. u
1 du
3
3 x 2 dx, x 2 dx
7
S
| 2.2282 liters
INSTRUCTOR USE ONLY
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436
NOT FOR SALE
Chapter 4
Integration
egration
4: ³
87. Trapezoidal Rule n
|
dx
§
·
§
·
§
·
1ª 2
2
2
2
2 º
«
» | 0.285
¸ 2¨
¸ 2¨
¸ 2¨
2¸
2¸
2¸
2
¨
¨
¨
8 «1 2
1 32 »
©1 9 4 ¹
©1 5 2 ¹
© 1 11 4 ¹
¬
¼
4: ³
Simpson’s Rule n
|
2
3
2 1 x2
2
3
2 1 x2
dx
§
·
§
·
§
·
1ª 2
2
2
2
2 º
«
» | 0.284
¨
¸
¨
¸
¨
¸
4
2
4
2¸
2¸
2¸
¨
¨
¨
12 «1 22
1 32 »
©1 9 4 ¹
©1 5 2 ¹
© 1 11 4 ¹
¬
¼
Graphing utility: 0.284
88. Trapezoidal Rule n
4: ³
Simpson’s Rule n
4: ³
32
32
32
214
212
234
1ª
1º
x3 2
0
dx
|
«
» | 0.172
2
2
2
0 3 x2
8«
2»
3 14
3 12
3 34
¬
¼
1
32
32
32
414
212
434
x3 2
1ª
1º
dx |
«0 » | 0.166
2
2
2
2
0 3 x
12 «
2»
3 14
3 12
3 34
¬
¼
1
Graphing utility: 0.166
S 2
89. Trapezoidal Rule: n
4: ³
Simpson’s Rule n
4 : 0.685
0
x cos x dx | 0.637
Graphing Utility: 0.704
90. Trapezoidal Rule n
Simpson’s Rule: n
4: ³
S
0
1 sin 2 x dx | 3.820
4 : 3.820
Graphing utility: 3.820
Problem Solving for Chapter 4
11
³ 1 t dt
1. (a) L 1
x
(d) First show that ³ 1
0
1
1
by the Second Fundamental Theorem of
x
Calculus.
(b) Lc x
Lc 1
1
(c) L x
1
³ 1 t dt for x | 2.718
dt
0.999896
2.718 1
³1
t
x 1
1
1
³1 x1 t dt.
t
and du
x1
To see this, let u
Then
x1 1
³ 1 t dt
1
dt
t
1
1
1
dt.
x1
1
³1 x1 ux1 x1 du
1
³ 1 x1 t dt.
§
t ·
¸
x1 ¹
³ 1 x1 u du
1
1
Now,
L x1 x2
(Note: The exact value of x is e, the base of the
natural logarithm function.)
x1x2 1
x2
1
1
x2
1
x1 1
x2
1
³1
t
1
dt
³ 1 x1 u du ¨© using u
³ 1 x1 u du ³ 1 u du
³ 1 u du ³1 u du
L x1 L x2 .
INSTRUCTOR USE ONLY
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NOT FOR SALE
Problem Solving ffor Chapter 4
2. (a)
437
y
10
7
6
5
4
3
2
1
−4
x
−2 −1
1 2
4 5
3
Area
(b) Base
3
³ 3
2³
9 x 2 dx
6, height
3
0
2 bh
3
9, Area
2³
ba
0
2 6
3
9
2>27 9@
36
36
b 2 a 2 x 2 , a, b ! 0.
(c) Let the parabola be given by y
Area
ª
x3 º
2 «9 x »
3 ¼0
¬
9 x 2 dx
b 2 a 2 x 2 dx
ba
y
ª
x3 º
2 «b 2 x a 2 »
3 ¼0
¬
b2
ª § b · a 2 § b ·3 º
2 «b 2 ¨ ¸ ¨ ¸ »
3 © a ¹ »¼
«¬ © a ¹
ª b3 1 b3 º
2« »
3 a¼
¬a
Base
3. y
2b
, height
a
4 b3
3 a
−
2 § 2b · 2
¨ ¸b
3© a ¹
x 4 4 x3 4 x 2 , >0, 2@, ci
2i
n
2
, f x
n
lim ¦ f ci 'x
nof
x
4 b3
3 a
x 4 4 x3 4 x 2
n
A
b
a
b2
Archimedes’ Formula: Area
(a) 'x
b
a
i 1
4
3
2
n ª
§ 2i ·
§ 2i ·
§ 2i · º 2
lim ¦ «¨ ¸ 4¨ ¸ 4¨ ¸ »
nof
©n¹
© n ¹ »¼ n
i 1«
¬© n ¹
ª 32 n
64 n
32 n º
(b) « 5 ¦ i 4 4 ¦ i 3 3 ¦ i 2 »
n i 1
n i1 ¼
¬n i 1
lim
nof
2
ª 32 ˜ n n 1 2n 1 3n 2 3n 1
64 ˜ n 2 n 1
32 ˜ n n 1 2n 1 º
«
»
n5 ˜ 30
n4 ˜ 4
n3 ˜ 6
«¬
»¼
ª16 n 1 6n3 9n 2 n 1 16 n 1 2 16 n 1 2n 1 º
«
»
15n 4
n2
3n 2
«¬
»¼
ª16 n 1 n3 n 2 n 1 º
«
»
15n 4
«¬
»¼
(c) A
ª16n 4 16 º
lim «
»
4
nof
¬ 15n
¼
16n 4 16
15n 4
16
15
INSTRUCTOR USE ONLY
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438
NOT FOR SALE
Chapter 4
4. y
Integration
egration
2i
, 'x
n
1 5
x 2 x3 , >0, 2@, ci
2
n ª
1 § 2i ·
§ 2i · º 2
lim ¦ « ¨ ¸ 2¨ ¸ »
nof
© n ¹ ¼» n
«2© n ¹
i 1¬
5
n
lim ¦ f ci 'x
(a) A
nof
2
n
i 1
3
lim
nof
ª 32 n
32 n º
(b) « 6 ¦ i 5 4 ¦ i 3 »
n i 1 ¼
¬n i 1
2
ª 32 ˜ n 2 n 1 2 2n 2 2n 1
32 ˜ n 2 n 1 º
«
»
«
n6 ˜ 12
n4 ˜ 4
»
¬
¼
2
ª 8 n 1 2 2n 2 2 n 1
8n 1 º
«
»
«
n2
»
3n 4
¬
¼
ª 8 n 1 2 5n 2 2 n 1 º
«
»
«
»
3n 4
¬
¼
ª 40n 4 96n3 64n 2 8 º
lim «
»
nof
3n 4
¬
¼
(c) A
x
40
3
§St2 ·
³ 0 sin¨© 2 ¸¹ dt
5. S x
y
(a)
2
1
x
1
3
−1
−2
(b)
y
1.00
0.75
0.50
0.25
x
1
2
3
2
5
6
72 23
−0.25
The zeros of y
(c) S c x
sin
sin
S x2
S x2
2
0 Ÿ
2
Relative maxima at x
Relative minima at x
(d) S cc x
§S x ·
cos¨
¸ Sx
© 2 ¹
2
Points of inflection at x
correspond to the relative extrema of S(x).
S x2
nS Ÿ x 2
2
2n Ÿ x
2 | 1.4142 and x
1,
6 | 2.4495
2 2 | 2.8284
2 and x
0 Ÿ
2n , n integer
S x2
S
2
2
3,
5, and
nS Ÿ x 2
1 2n Ÿ x
1 2n , n integer
7.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Problem Solving ffor Chapter 4
6. (a)
§
1
1 ·
§ 1 ·
¸ cos¨
¸
3¹
© 3¹
³ 1 cos x dx | cos¨© 1
sin xº
»¼ 1
1
³ 1 cos x dx
1
1
§ 1 ·
2 cos¨
¸ | 1.6758
© 3¹
2 sin 1 | 1.6829
Error: 1.6829 1.6758
(b)
439
0.0071
1
1
3
2
³ 1 1 x 2 dx | 1 1 3 1 1 3
Note: exact answer is S 2 | 1.5708
ax3 bx 2 cx d .
(c) Let p x
1
ª ax 4
º
bx 3
cx 2
dx»
«
4
3
2
¬
¼ 1
1
³ 1 p x dx
§ 1 ·
§ 1 ·
p¨ ¸ p¨
¸
3¹
©
© 3¹
2b
2d
3
§b
· §b
·
¨ d¸ ¨ d¸
©3
¹ ©3
¹
2b
2d
3
y
7. (a)
5
4
3
2
1
(6, 2)
(8, 3)
f
(0, 0)
x
2
−1
−2
−3
−4
−5
(b)
4 5 6 7 8 9
(2, −2)
x
0
1
2
3
4
5
6
7
8
F x
0
12
2
72
4
72
2
1
4
3
(c) f x
­ x,
0 d x 2
°
®x 4, 2 d x 6
° 1 x 1, 6 d x d 8
¯2
x
­ x2 2 ,
0 d x 2
°
° 2
® x 2 4 x 4, 2 d x 6
°
2
°̄ 1 4 x x 5, 6 d x d 8
F x
³ 0 f t dt
Fc x
f x . F is decreasing on (0, 4) and increasing on (4, 8). Therefore, the minimum is 4 at x
maximum is 3 at x
(d) F cc x
x
fc x
4, and the
8.
­1, 0 x 2
°
2 x 6
®1,
°1 , 6 x 8
¯2
2 is a point of inflection, whereas x
6 is not.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Chapter 4
440
Integration
egration
8. Let d be the distance traversed and a be the uniform acceleration. You can assume that v 0
at
a
vt
at
st
1 2
at .
2
st
d when t
2d
.
a
The highest speed is v
a
The lowest speed is v
0.
The mean speed is
1
2
2d
a
2ad .
2ad 0
ad
.
2
d
ad 2
The time necessary to traverse the distance d at the mean speed is t
0 and s 0
0. Then
2d
a
which is the same as the time calculated above.
9. ³
x
0
So,
x
x
³ 0 xf t dt ³ 0 tf t dt
f t x t dt
d x
f t x t dt
dx ³ 0
xf x ³
x
0
x³
x
0
x
f t dt ³ tf t dt
0
f t dt x f x
x
³ 0 f t dt
Differentiating the other integral,
d x x
f v dv dt
dx ³ 0 ³ 0
x
³ 0 f v dv.
So, the two original integrals have equal derivatives,
x
x
³ 0 f t x t dt
t
³ 0 ³ 0 f v dv dt C.
0, you see that C
Letting x
0.
2
ª¬ f x º¼ Ÿ F c x
10. Consider F x
b
³ a f x f c x dx
2 f x f c x . So,
b
³ a 12 F c x dx
b
ª¬ 12 F x º¼
a
1 ªF b
2¬
1ª f
2¬
11. Consider ³
S n
1
0
1ª
«
n¬
b
F a º¼
2
f a º.
¼
2
1
x dx
1
n
2 x3 2 º
3
»¼
0
2
"
n
2 . The corresponding Riemann Sum using right-hand endpoints is
3
nº
»
n¼
1 ª
1
n3 2 ¬
2 "
n ¼º. So, lim
1
n of
2 "
n3 2
n
2
.
3
INSTRUCTOR USE ONLY
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NOT FOR SALE
Problem Solving ffor Chapter 4
1
x6 º
»
6 ¼0
1
12. Consider ³ x5 dx
0
441
1
.
6
The corresponding Riemann Sum using right endpoints is
5
5
5
1 ª§ 1 ·
§ 2·
§n· º
«¨ ¸ ¨ ¸ " ¨ ¸ »
n «¬© n ¹
©n¹
© n ¹ »¼
S n
1 5
ª1 25 " n5 º¼. So, lim S n
n of
n6 ¬
13. By Theorem 4.8, 0 f x d M Ÿ ³
1
So, 1 d ³
b
2.
Note: ³
M ba.
b
1
0
1 x4 d
2 and b a
1.
1 x 4 dx | 1.0894
f x
b
b x, du
dx.
f bu
0
³ b f b u f u du
f bu
b
f x
b
b
f b x
b
³ 0 f b u f u du
³ 0 f b x f x dx
f b x
³ 0 f x f b x dx ³ 0 f b x f x dx
Then, 2 A
b
³ 0 1 dx
b.
b
.
2
So, A
(b) b
1Ÿ ³
(c) b
3, f x
sin x
dx
sin x
1
0 sin 1 x
1
2
x
x
3
³0
1
.
6
³ 0 f x f b x dx.
Let u
A
a
f x dx d M b a . On the interval [0, 1], 1 d
1 x 4 dx d
0
14. (a) Let A
15. (a)
b
a
b
f x dx d ³ M dx
15 25 " n5
n6
³ a m dx d ³ a f x dx.
Similarly, m d f x Ÿ m b a
So, m b a d ³
b
a
lim
n of
x 3 x
0.2
0.6
3
2
dx
v
100
80
60
40
20
t
0.4
0.8
1.0
(b) v is increasing (positive acceleration) on (0, 0.4) and (0.7, 1.0).
(c) Average acceleration
v 0.4 v 0
0.4 0
60 0
0.4
150 mi h 2
(d) This integral is the total distance traveled in miles.
1
³ 0 v t dt | 101 ª¬0 2 20 2 60 2 40 2 40 65º¼
385
10
38.5 miles
(e) One approximation is
v 0.9 v 0.8
0.9 0.8
other answers possible
a 0.8 |
50 40
0.1
100 mi h
2
INSTRUCTOR USE ONLY
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442
NOT FOR SALE
Chapter 4
Integration
egration
16. Because f x d f x d f x , ³
17. (a) 1 i
3
f x dx d ³
1 3i 3i 2 i 3 Ÿ 1 i
3i 2 3i 1
(b)
b
a
i 1
n
3
i3
f x dx d ³
f x dx Ÿ ³
b
a
f x dx d ³
b
a
f x dx.
3i 2 3i 1
¦ ª¬ i 1 i3 º¼
i 1
3
i 1
23 13 33 23 " ª« n 1
¬
So, n 1
b
a
i3
n
¦ 3i 2 3i 1
3
b
a
3
n3 º»
¼
n1
3
1
n
¦ 3i 2 3i 1 1.
3
i 1
3
n 1
(c)
n
n
¦ 3i 2 3i 1
1
¦ 3i 2 i 1
i 1
3n n 1
n
2
3n n 1
n3 3n 2 3n n
2
n
Ÿ ¦ 3i 2
i 1
2n3 6n 2 6n 3n 2 3n 2n
2
3
2
2n 3n n
2
n n 1 2n 1
2
n n 1 2n 1
6
n
Ÿ ¦ i2
i 1
x sin t
³0
18. Si x
t
(a)
dt
2
− 12
12
−2
(b) Sic x
sin x
Sic x
x
0 for x
For positive x, x
2n 1 S
For negative x, x
2nS
2nS
Maxima at S , 3S , 5S , ! and 2S , 4S , 6S , !
(c) Sicc x
x cos x
x cos x sin x
x2
0
sin x for x | 4.4934
Si 4.4934 | 1.6556
(d) Horizontal asymptotes at y
lim Si x
x of
lim Si x
x of
r
S
2
S
2
S
2
INSTRUCTOR USE ONLY
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Problem Solving ffor Chapter 4
443
19. (a) R I T L
40
1ª
§ 1 · 1º
ª f 0 4 f 1 2 f 2 4 f 3 f 4 º¼ | «4 4 2 2 1 4¨ ¸ » | 5.417
34 ¬
3¬
© 2 ¹ 4¼
(b) S 4
x 2 16 is a parabola:
20. The graph of y
y
8
4
−8 −6
x
−2
2
6
8
−8
−20
−24
The integral ³
b
a
x 2 16 dx will be a minimum when a
4 and b
4, as indicated in the figure.
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NOT FOR SALE
C H A P T E R 5
Logarithmic, Exponential,
and Other Transcendental Functions
Section 5.1
The Natural Logarithmic Function: Differentiation..........................445
Section 5.2
The Natural Logarithmic Function: Integration................................457
Section 5.3
Inverse Functions................................................................................468
Section 5.4
Exponential Functions: Differentiation and Integration ...................481
Section 5.5
Bases Other than e and Applications.................................................495
Section 5.6
Inverse Trigonometric Functions: Differentiation ............................510
Section 5.7
Inverse Trigonometric Functions: Integration...................................524
Section 5.8
Hyperbolic Functions .........................................................................534
Review Exercises ........................................................................................................545
Problem Solving .........................................................................................................554
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C H A P T E R 5
Logarithmic, Exponential, and Other Transcendental Functions
Section 5.1 The Natural Logarithmic Function: Differentiation
1. (a) ln 45 | 3.8067
Domain: x ! 0
45 1
³ 1 t dt | 3.8067
(b)
y
2. (a) ln 8.3 | 2.1163
2
1
8.3 1
³ 1 t dt | 2.1163
(b)
2 ln x
10. f x
x
1
2
3
4
−1
3. (a) ln 0.8 | 0.2231
−2
0.8 1
³ 1 t dt | 0.2231
(b)
11. f x
4. (a) ln 0.6 | 0.5108
ln 2 x
Domain: x ! 0
y
0.6 1
³ 1 t dt | 0.5108
(b)
2
ln x 1
5. f x
1
x
Vertical shift 1 unit upward
1
Matches (b)
2
3
−1
ln x
6. f x
12. f x
Reflection in the x-axis
ln x
Domain: x z 0
Matches (d)
y
ln x 1
7. f x
3
Horizontal shift 1 unit to the right
2
Matches (a)
1
−3
ln x
8. f x
−2
x
−1
1
2
3
−2
Reflection in the y-axis and the x-axis
−3
Matches (c)
9. f x
ln x 3
13. f x
3 ln x
Domain: x ! 3
Domain: x ! 0
y
y
4
3
3
2
2
1
1
x
x
−1
−2
−3
1
2
3
4
5
−1
1
2
3
4
5
6
7
−2
−3
−4
INSTRUCTOR USE ONLY
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445
446
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Function
Functions
ln x 4
14. f x
xy
z
ln x ln y ln z
22. ln xyz
ln x ln y ln z
21. ln
Domain: x ! 0
y
x
1
−1
2
3
4
5
6
x2 5
23. ln x
−2
ln x ln x 2 5
−3
12
ln x 12 ln x 2 5
−4
−5
−6
ln x 2
15. h x
12
24. ln
a 1
ln a 1
25. ln
x 1
x
1 § x 1·
§ x 1·
ln ¨
ln ¨
¸
¸
x
2 © x ¹
©
¹
1
ªln x 1 ln xº¼
2¬
1
1
ln x 1 ln x
2
2
Domain: x ! 2
ln a 1
1
2
12
y
3
2
1
x
−3 −2
1
−1
2
3
26. ln 3e 2
ln 3 2 ln e
2 ln 3
−2
−3
27. ln z z 1
2
Domain: x ! 2
28. ln
y
1
e
ln 1 ln e
1
4
2
4
ln
30. 3 ln x 2 ln y 4 ln z
ln x3 ln y 2 ln z 4
6
−2
−4
ln
(b) ln 23
ln 2 ln 3 | 0.4055
(c) ln 81
ln 34
(d) ln
3
ln 31 2
18. (a) ln 0.25
ln 14
(c) ln 3 12
1
(d) ln 72
20. ln
4 ln 3 | 4.3944
1 ln 3 | 0.5493
2
ln 1 2 ln 2 | 1.3862
3 ln 2 ln 3 | 3.1779
(b) ln 24
x
4
x3 y 2
z4
ln 2 ln 3 | 1.7917
17. (a) ln 6
19. ln
x 2
x 2
29. ln x 2 ln x 2
x
−2
2
ln z 2 ln z 1
ln x 2 1
16. f x
ln z ln z 1
1 2 ln 2 ln 3
3
| 0.8283
ln 1 3 ln 2 2 ln 3 | 4.2765
ln x ln 4
x5
ln x5 2
5
ln x
2
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NOT FOR SALE
Section 5.1
31.
1 x x 3
ln 2
x 1
3
1ª
2 ln x 3 ln x ln x 2 1 º¼
3¬
ln 3
32. 2 ª¬ln x ln x 1 ln x 1 º¼
2 ln
34.
1
ln x 2 1
2
ln 9 ln
x2 1
3ª
ln x 2 1 ln x 1 ln x 1 º¼
2¬
2
x 1
2
x
x 1 x 1
ln
2
9
x 1
2
x 1
3
ln
x 1 x 1
2
ln
35. (a)
§ x2 1·
¨ 2
¸
© x 1¹
3
f x
ln x 1
fc x
1
x 1
43. g x
ln x 2
42.
3
f=g
9
0
−3
2
x
ln x 2 ln 4
4
because x ! 0.
(b) f x
36. (a)
2 ln x ln 4
ln
2 ln x
2
x
gc x
g x
44. h x
ln 2 x 2 1
hc x
1
4x
2x2 1
y
ln x
3
f=g
−1
5
45.
−1
(b) f x
ln
x x 1
2
37. lim ln x 3
f
38. lim ln 6 x
f
x o 3
x o 6
39. lim ln ª¬ x 2 3 x º¼
x o 5
x
x 4
f x
ln 3 x
fc x
1
3
3x
g x
46. y
yc
47. y
ln 4 | 1.3863
x o 2
40. lim ln
dy
dx
1 ª 2
ln x x 1 º¼
2 ¬
1ª
ln x ln x 2 1 º¼
2¬
41.
447
2
x x 3
§ x ·
ln ¨ 2
¸
© x 1¹
33. 2 ln 3 The Natural Logarithmic Function: D
Differentiation
Dif
ln 5 | 1.6094
48.
4x
2x2 1
4
3§ 1 ·
4 ln x ¨ ¸
© x¹
4 ln x
3
x
x 2 ln x
§1·
x 2 ¨ ¸ 2 x ln x
© x¹
ln t 1
1
t 1
2
x 2 x ln x
x 1 2 ln x
2 ln t 1
2
t 1
yc
2
y
ln
dy
dx
1§ 2x ·
¨
¸
2 © x2 4 ¹
x2 4
1
ln x 2 4
2
x
x2 4
1
x
INSTRUCTOR USE ONLY
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NOT FOR SALE
448
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
49.
y
ln ª x
¬
dy
dx
1
1 § 2x ·
¨ 2
¸
2 © x 1¹
x
50. y
3
ln ªt t 2 3 º
¬«
¼»
yc
1
3
2
2t
t
t 3
51.
52.
1
ln x 2 1
2
2x2 1
x x2 1
x 2 1º
¼
ln x x
x2 1
ln x ln x 2 1
fc x
1
2x
2
x
x 1
1 x2
x x2 1
gc t
54. h t
hc t
55.
56.
x cos x
sin x
62. y
ln csc x
csc x ˜ cot x
csc x
y
ln
3
x x 3
ln t
t2
t 1 t 2t ln t
t
4
t
1 2 ln t
t3
4 x2
cot x
sin x
sin x
cos x
cos x 1
dy
dx
1 ln t
t2
2
ln ln x 2
dy
dx
1 d
ln x 2
ln x 2 dx
y
ln ln x
dy
dx
1x
ln x
64.
2
x ln x 2
ln x 2
y
ln sec x tan x
dy
dx
sec x tan x sec 2 x
sec x tan x
1
x ln x
1
x ln x
x 1
x 1
1
ªln x 1 ln x 1 º¼
2¬
dy
dx
1ª 1
1 º
«
2 ¬ x 1 x 1»¼
x 1
x 1
1
1 x2
y
ln x 4
dy
dx
4
x
1ª 1
1 º
«
3 ¬ x 1 x 1¼»
1 2
3 x2 1
2
1,
4 ln x, 1, 0
dy
dx
4.
Tangent line: y 0
4 x 1
y
4x 4
(b)
1
ªln x 1 ln x 1 º¼
3¬
sec x
sec x tan x
65. (a)
ln
sin x
cos x 1
sec x sec x tan x
2 x x2
y
yc
1
cos x
cos x 1
tan x ln t
t
t 1 t ln t
ln 3
2
ln cos x ln cos x 1
2
y
58. y
·
¸
4 x ¹
x
cot x
yc
When x
57.
§
¨1 4 x ©
2
dy
dx
ln 2 x ln x 3
4 x2
1
ln sin x
63.
53. g t
ln x f x
y
61.
1
ln 4 x 2 ln x
2
4
2
x x 4
ln
fc x
ln
fc x
60.
1
6t
2
t
t 3
§ 2x ·
ln ¨
¸
© x 3¹
1
1
x
x 3
4 x2
x
1
x
2
x
4 x
f x
fc x
ln t 3 ln t 2 3
f x
f x
59.
5
−5
(1, 0)
5
−5
3x 1
2
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NOT FOR SALE
Section 5.1
y
ln x3 2
dy
dx
3
2x
66. (a)
When x
3
ln x, 1, 0
2
1,
dy
dx
f x
69. (a)
2 sin x cos x
fc x
3
x 1
2
3
3
x 2
2
y
2 1 sin 2 x
§S ·
f c¨ ¸
©4¹
2 2
(1, 0)
Tangent line: y ln
1§
S·
¨x ¸
3©
4¹
y
1
1 § 3·
S
x ln ¨ ¸ 3
2 © 2 ¹ 12
2
(
3
π
, ln
4
2
(
−2
y
3x ln x,
dy
dx
6x 2
2
1, 3
−2
1
x
dy
1,
dx
f x
sin 2 x ln x 2
2 sin 2 x ln x,
fc x
4 cos 2 x ln x 2 sin 2 x
2 sin 2
70. (a)
5.
Tangent line: y 3
5 x 1
y
5x 2
fc1
0
5x y 2
Tangent line: y 0
(b)
3
2
3
−2
When x
1
3
32
(b)
67. (a)
§S
3·
¨¨ , ln
¸
2 ¸¹
©4
sin x cos x
1 sin 2 x
2 2
2
−1
449
1 sin 2 x
ln
1
ln 1 sin 2 x ,
2
3
.
2
Tangent line: y 0
(b)
The Natural Logarithmic
Loga rithmic Function: D
Dif
Differentiation
4
1, 0
x
2 sin 2 x 1
2 sin 2 x 2 sin 2
y
(1, 3)
(b)
−1
1
2
(1, 0)
0
3
−3
y
§1
·
4 x 2 ln ¨ x 1¸,
©2
¹
dy
dx
2 x 68. (a)
When x
0,
1
§1·
¨ ¸
1 2 x 1© 2 ¹
dy
dx
2 x 1
x 2
71. (a)
1
.
2
1
x 0
2
1
x 4
2
Tangent line: y 4
y
8
(b)
−2
0, 4
f x
x3 ln x,
fc x
3 x ln x x 2
fc1
1
1, 0
2
Tangent line: y 0
1x 1
y
x 1
(b)
2
−1
(1, 0)
3
(0, 4)
−4
4
−2
−4
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450
Chapter 5
f x
72. (a)
fc x
f c 1
NOT FOR SALE
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
1
x ln x 2 , 1, 0
2
1
1 § 2x ·
ln x 2 x¨ 2 ¸
2
2 ©x ¹
1
ln x 2 1
2
7
4 xy 2 ln x ln y
7
2
1
yc
x
y
0
4 xyc 4 y 1
Tangent line: y 0
1x 1
y
x 1
(b)
4 xy ln x 2 y
76.
§
1·
¨ 4 x ¸ yc
y¹
©
yc
2
−3
(− 1, 0)
3
yc
−2
73.
77. y
x 2 3 ln y y 2
10
dy
3 dy
2y
2x y dx
dx
0
·
dy § 3
¨ 2y¸
dx © y
¹
2x
dy
dx
74.
2x
3 y 2y
ln xy 5 x
30
ln x ln y 5 x
30
1
1 dy
5
x
y dx
0
1 dy
y dx
dy
dx
2 xy
3 2 y2
y
5y
x
2
x
ycc
4 xy 2 2 y
4x2 y x
2
x2
xycc yc
§ 2· 2
x¨ 2 ¸ x
© x ¹
0
x ln x 4 x
§1·
x¨ ¸ ln x 4
3 ln x
© x¹
x y xyc
x x ln x 4 x x 3 ln x
79. y
§ y 5 xy ·
¨
¸
x
©
¹
0
x2
ln x
2
Domain: x ! 0
yc
1
x
x 1 x 1
x 75. 4 x3 ln y 2 2 y
2x
2
yc 2 yc
y
2
§2
·
¨ 2 ¸ yc
©y
¹
2 12 x 2
ycc
yc
2 12 x 2
2 y 2
§ 1·
Relative minimum: ¨1, ¸
© 2¹
yc
y 6 yx 2
1 y
12 x 2 2
x
1
4x y
4 y yc
1
5
x
2
x
2 ln x 3
yc
78. y
4 y x
0 when x
y 1 6x2
1
1.
1
! 0
x2
2
1 y
(1, 12 )
0
3
0
INSTRUCTOR USE ONLY
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Section 5.1
80. y
2 x ln 2 x
2 x ln 2 ln x
The Natural Logarithmic
Loga rithmic Function: D
Dif
Differentiation
83. y
Domain: x ! 0
2x 1
x
1
x
yc
2 ycc
1
! 0
x2
0 when x
x
ln x
Domain: 0 x 1, x ! 1
1
.
2
yc
ycc
§1 ·
Relative minimum: ¨ , 1¸
©2 ¹
ln x 1 x 1 x
ln x
ln x
2
0
x ln x
( e, e )
§1·
x¨ ¸ ln x
© x¹
1
! 0
x
0 when x
4
0 when x
e2 .
9
−4
84. y
yc
1
3
0
(e2, e2/2)
e 1.
2
(e−1 , −e−1 )
ycc
−1
ycc
ln x
x
x
x 2 ln , Domain: x ! 0
4
x
x·
§1·
§
x 2 ¨ ¸ 2 x ln
x¨1 2 ln ¸
x
4
4
© ¹
©
¹
0 when:
1
x
x
4e1 2
Ÿ ln
Ÿ x
4
4
2
x
x
§1·
1 2 ln 2 x¨ ¸
3 2 ln
4
4
© x¹
2 ln
0 when x
4e 3 2
Relative minimum: 4e 1 2 , 8e 1
Domain: x ! 0
x 1 x ln x
x2
1 ln x
x2
0 when x
e.
Point of inflection: 4e3 2 , 24e 3
4 (4e−3/2, −24e−3)
x 1 x 1 ln x 2 x
2
ycc
3
0
1 ln x
Relative minimum: e 1 , e 1
yc
e.
4
Domain: x ! 0
82. y
0 when x
§
e2 ·
Point of inflection: ¨ e 2 , ¸
2¹
©
6
0
ycc
2
Relative minimum: e, e
(12, 1)
yc
ln x
ln x
x ln x
ln x 1
2
1 x ln x 1 2 x ln x
2 ln x
4
81. y
451
x4
2 ln x 3
−4
0 when x
x3
8
32
e .
−4
Relative maximum: e, e
1
(4e−1/2, −8e−1)
3
§
·
Point of inflection: ¨ e3 2 , e 3 2 ¸
2
©
¹
2
−3
(e, e−1) ( e 2 , 23 e 2 )
3
−1
6
−4
INSTRUCTOR USE ONLY
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NOT FOR SALE
452
Chapter 5
85.
f x
ln x,
f 1
0
fc x
1
,
x
fc1
1
f cc x
1
,
x2
f 1 fc1 x 1
f cc 1
1
P1 1
0
P2 1
0
P1c 1
1
P2c 1
1
P2cc 1
1
P1 x
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
x 1,
1
2
f 1 f c 1 x 1 f cc 1 x 1
2
1
2
x 1 x 1 ,
2
P2 x
P1c x
1,
P2c x
1 x 1
P2cc x
1,
2 x,
The values of f , P1 , P2 , and their first derivatives agree at x
derivatives of f and P2 agree at x
1. The values of the second
1.
2
P1
f
−1
5
P2
−2
f x
x ln x,
f 1
0
fc x
1 ln x,
fc1
1
f cc x
f cc 1
1
P1 x
1
,
x
f 1 fc1 x 1
P1 1
0
P2 x
f 1 fc1 x 1 P2 1
0
P1c 1
1
P2c 1
1
P2cc 1
1
86.
x 1 P1c x
1,
P2c x
1 x 1
P2cc x
x,
1
2
x 1 ,
2
x,
x 1,
1
2
f cc 1 x 1
2
The values of f , P1 , P2 , and their first derivatives agree at x
x
1. The values of the second derivatives of f and P2 agree at
1.
3
f
P1
P2
−2
4
−1
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 5.1
x.
87. Find x such that ln x
f x
ln x x
fc x
1
1
x
xn 1
xn n
1
2
xn
0.5
f xn
–0.1931
The Natural Logarithmic
Loga rithmic Function: D
Dif
Differentiation
f x
x ln x 3
fc x
1
xn 1
xn 3
n
1
2
3
0.5644
0.5671
xn
2
2.2046
2.2079
–0.0076
–0.0001
f xn
–0.3069
–0.0049
0.0000
ª1 ln xn º
xn «
»
¬ 1 xn ¼
f xn
f c xn
Approximate root: x | 0.567
y
ln y
1 § dy ·
¨ ¸
y © dx ¹
dy
dx
1
x
ª 4 ln xn º
xn «
»
¬ 1 xn ¼
f xn
f c xn
x2 1
x
1
ln x 2 1
2
1
x
2
x
x 1
ln x ª 2 x2 1 º
»
y« 2
«¬ x x 1 »¼
2 x2 1
x2 1
x2 x 1 x 2 , x ! 0
y
x2 x 1 x 2
y2
2 ln y
2 ln x ln x 1 ln x 2
2 dy
y dx
2
1
1
x
x 1 x 2
y ª2
1
1 º
«
2 ¬x
x 1 x 2 »¼
dy
dx
x2 x 1 x 2 ª 2 x 1 x 2 x x 2 x x 1 º
«
»
2
x x 1 x 2
¬
¼
dy
dx
91.
0
Approximate root: x | 2.208
89.
90.
3 x.
88. Find x such that ln x
0
453
y
ln y
1 § dy ·
¨ ¸
y © dx ¹
dy
dx
x2
3x 2
x 1
92.
2
1
ln 3 x 2 2 ln x 1
2
2
3
2
x
x 1
2 3x 2
2 ln x ª 3x 2 15 x 8 º
y«
»
¬« 2 x 3 x 2 x 1 ¼»
4 x2 9 x 4
x 1 x 2
2
y
ln y
1 § dy ·
¨ ¸
y © dx ¹
dy
dx
x2 1
x2 1
1ª
ln x 2 1 ln x 2 1 º¼
2¬
1 ª 2x
2x º
2 ¬« x 2 1 x 2 1¼»
x2 1 ª 2x º
x 2 1 «¬ x 4 1»¼
x2 1
3 x 3 15 x 2 8 x
2 x 1
3
3x 2
x 1
12
x2 1
32
2
12
2x
x 1 x2 1
2
2x
x2 1
12
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
454
Chapter 5
93.
y
NOT FOR SALE
Logarithmic,
arithmic, Exponential,
Exponen
and Other Transcendental Functions
Function
x x 1
96. The base of the natural logarithmic function is e.
32
x 1
3
1
ln x ln x 1 ln x 1
2
2
1 3§ 1 · 1§ 1 ·
¨
¸ ¨
¸
x 2 © x 1¹ 2 © x 1¹
ln y
1 § dy ·
¨ ¸
y © dx ¹
y ª 4x2 4x 2 º
«
»
2 « x x2 1 »
¬
¼
1 § dy ·
¨ ¸
y © dx ¹
2x 2x 1
x 1
x 1
increasing.
32
98. (a) lim T
The temperature of the object seems to approach
20qC, which is the temperature of the surrounding
medium.
(b) The temperature changes most rapidly when it is first
removed from the furnace. The slope is steepest at
h
0.
99. False
ln x ln 25
2x 2
100. False. The property is
ln xy
ln x ln y (for x, y ! 0 ). As a counter
example, let x
y
e. Then
continuous, increasing, and one-to-one, and its
graph is concave downward. In addition, if a and b
are positive numbers and n is rational, then ln 1
0,
ln a ˜ b
ln a ln b, ln a
ln a b
ln a ln b.
n
ln xy
ln e 2
x
§
·
13.375 ln ¨
¸,
© x 1250 ¹
d
>ln S @
dx
n ln a, and
and
ln x ln y
1˜1
1.
0
ln e
1, then yc
0.
x ! 1250
(b) When x
50
1398.43: t | 30 years
Total amount paid
1000
2
101. False; S is a constant.
102. False. If y
(a)
ln 25 x z ln x 25
2
95. The domain of the natural logarithmic function
is 0, f and the range is f, f . The function is
103. t
20
hof
2
x2
ln x 2 1 is not concave up.
g x
ª
º
2 x2 4
»
y« 2
2
«¬ x 1 x 4 »¼
x 1 x 2
2x2 4
˜
x 1 x 2
x 1 x 1 x 2 x 2
2
x 2 1 (positive and concave up).
(b) No. Let f x
4 º
ª 2
2
y« 2
»
¬ x 1 x 4¼
x 1
gc x f x
and so, f c x ! 0. Therefore, the graph of f is
2
1
1
1
1
x 1 x 2 x 1 x 2
dy
dx
f x
Because f x ! 0, you know that f c x
ln x 1 ln x 2 ln x 1 ln x 2
ln y
fc x
(a) Yes. If the graph of g is increasing, then g c x ! 0.
x 1 x 2
x 1 x 2
y
f x ! 0
ln f x ,
gc x
3
1 º
y ª2
2 «¬ x x 1 x 1»¼
dy
dx
94.
97. g x
1398.43 30 12
$503,434.80
3000
0
(c) When x
1611.19: t | 20 years
Total amount paid
1611.19 20 12
$386,685.60
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.1
(d)
dt
dx
d
ª13.375 ln x ln x 1250 ¼º
dx ¬
When x
1398.43:
dt
| 0.0805
dx
When x
1611.19:
dt
| 0.0287
dx
The Natural Logarithmic
Loga rithmic Function: D
Dif
Differentiation
455
16718.75
x x 1250
1
ª1
º
13.375« x 1250 ¼»
¬x
(e) The benefits include a shorter term, and a lower total amount paid.
10
§ I ·
ln ¨
¸
ln 10 © 1016 ¹
104. (a) E
106. (a) You get an error message because ln h does not
exist for h
10
ªln I ln 1016 º¼
ln 10 ¬
0.
(b) Reversing the data, you obtain
0.8627 6.4474 ln p.
h
10
>ln I 16ln 10@
ln 10
[Note: Fit a line to the data x, y
(c)
10
ln I 160
ln 10
ln p, h . ]
350
10 log10 I 160
(b) E 10 5
10
ln 10 5 160
ln 10
50 160
105. (a)
0
100
0
110 decibels
(d) If p
0.75, h | 2.72 km.
(e) If h
13 km, p | 0.15 atmosphere.
350
h
0.8627 6.4474 ln p
1
6.4474
34.96
3.955
p
p
dp
dh
p
6.4474
For h
5, p
T c 10 | 4.75 deg/lb/in.2
dp dh
0.0816 atmos/km.
T c 70 | 0.97 deg/lb/in.2
For h
20, p
dp dh
0.0080 atmos/km.
(f)
0
100
0
(b) T c p
(c)
1 dp
implicit differentiation
p dh
0.5264 and
0.0514 and
30
As the altitude increases, the rate of change of
pressure decreases.
0
100
0
lim T c p
p of
0
Answers will vary. Sample answer: As the pounds
per square inch approach infinity, the temperature
will not change.
§ 10 10 ln ¨
¨
©
107. y
(a)
100 x 2 ·
¸ ¸
x
¹
100 x 2
10 ªln 10 ¬«
100 x 2 ln xº ¼»
100 x 2
20
INSTRUCTOR USE ONLY
0
10
10
0
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© Cengage Learning. All Rights Reserved.
456
NOT FOR SALE
Chapter 5
dy
dx
(b)
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
ª
x
10 «
« 100 x 2 10 «¬
100 x 2
1 º»
x»
»¼
ª
º 10
10
«
» 2
x
100 x ¬10 100 x ¼
x
100 x 2
x
2
x
100 x 2
ª
º 10
10
1» «
2
x
100 x ¬10 100 x
¼
x
2
ª
100 x 2 º 10
«
» x
100 x 2 ¬«10 100 x 2 ¼»
x
x
10 100 x 2
x 10 100 x 2
x2
(c)
10
x
10
x
When x
5, dy dx
3.
When x
9, dy dx
19 9.
lim
dy
100 x 2
x
0
x o10 dx
x
ln x
108. p x
§1·
ln x 1 x ¨ ¸
© x¹
2
ln x
pc x
ln x 1
ln x
ln 1000 1
(a) pc 1000
2
ln 1000
2
| 0.1238
About 12.4 primes per 100 integers
ln 1,000,000 1
(b) pc 1,000,000
ln 1,000,000
2
| 0.0671
About 6.7 primes per 100 integers
(c) pc 1,000,000,000
ln 1,000,000,000 1
ln 1,000,000,000
2
| 0.0459
About 4.6 primes per 100 integers
109. (a) f x
ln x, g x
x
25
g
f
0
500
0
fc x
1
, gc x
x
1
2
x
For x ! 4, g c x ! f c x . g is increasing at a faster rate than f for "large" values of x.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 5.2
(b) f x
4
ln x, g x
The Natural Logarithmic Function
Function: Integration
457
x
15
g
f
0
20,000
0
1
, gc x
x
fc x
1
4
4
x3
For x ! 256, g c x ! f c x . g is increasing at a faster rate than f for "large" values of x. f x
ln x increases very
slowly for "large" values of x.
Section 5.2 The Natural Logarithmic Function: Integration
1. ³
5
dx
x
5³
10
2. ³
dx
x
3. u
x 1, du
1
5. u
5 4 x, du
x
³ x 2 3 dx
x2 4
dx
x
3 x 2 6 x dx
3 x 2 2 x dx
1
1
3 x 2 6 x dx
³
3
3 x 3x 2
1
ln x3 3x 2 C
3
§
4·
³ ¨© x x ¸¹ dx
x2
4 ln x C
2
x2
ln x 4 C
2
9
1
³
4dx
4 5 4x
9
ln 5 4 x C
4
2 x dx
1
1
2 x dx
2 ³ x2 3
1
ln x 2 3 C
2
5 x3 , du
x2
11. ³
4 dx
x 2 3, du
³ 5 x3 dx
x3 3x 2 , du
x2 2x
³ x3 3x 2 dx
2 dx
3
ln x 4 3 x C
1
1
2 dx
2 ³ 2x 5
1
ln 2 x 5 C
2
9
³ 5 4 x dx
8. u
³ x 4 3x 4 x 3 dx
ln x 5 C
2 x 5, du
4 x3 3 dx
1
³ x 4 3x dx
10. u
dx
1
7. u
10 ln x C
ln x 1 C
³ 2 x 5 dx
6. u
x 4 3 x, du
4 x3 3
dx
x 5, du
³ x 5 dx
9. u
5 ln x C
1
10 ³ dx
x
1
³ x 1 dx
4. u
1
dx
x
12. ³
x3 8 x
dx
x2
§
8·
³ ¨© x x ¸¹ dx
x2
8 ln x C
2
13. u
x3 3 x 2 9 x, du
x2 2x 3
³ x3 3x 2 9 x dx
3 x 2 2 x 3 dx
2
1 3 x 2x 3
dx
3 ³ x3 3x 2 9 x
1
ln x3 3 x 2 9 x C
3
3 x 2 dx
1
1
3 x 2 dx
3 ³ 5 x3
1
ln 5 x3 C
3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
458
Chapter 5
NOT FOR SALE
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
x3 6 x 2 5, du
14. u
x2 4x
3 x 2 4 x dx
1
1
3 x 2 4 x dx
3 ³ x3 6 x 2 5
1
ln x3 6 x 2 5 C
3
³ x3 6 x 2 5 dx
15. ³
3 x 2 12 x dx
x 2 3x 2
dx
x 1
§
6 ·
³ ¨© x 4 x 1 ¸¹ dx
x2
4 x 6 ln x 1 C
2
16. ³
2 x2 7 x 3
dx
x 2
§
23. u
13
³
1
x13
x
19 ·
24. u
x3 3x 2 5
dx
x 3
§ 2
5
x 3 6 x 20
dx
x5
§ 2
1
³ x 2 3 1 x1 3 dx
x4 x 4
dx
x2 2
115 ·
³ ¨© x 5 x 19 x 5 ¸¹ dx
§ 2
x
20. ³
25. ³
2x
2
x 1
dx
ln x
x
dx
·
1 § 1 ·
¨
¸ dx
1 x1 3 © 3x 2 3 ¹
2x 2 2
³
x 1
2
§
x2 2 C
x
·
26. ³
³ ¨© x 4 x 2 5 ¸¹ dx
x x 2
x 1
3
dx
1
1
dx 2 ³
dx
2
x 1
x 1
2
C
x 1
x2 2x 1 1
x 1
dx
3
2
1
³ x 1 3 dx ³ x 1 3 dx
1
1
³ x 1 dx ³ x 1 3 dx
ln x 1 1
3
ln x C
3
1
ln ln x C
3
³
x 1
1
dx
x
1 1
1
˜ dx
3 ³ ln x x
dx
1
2 ln x 1 27. u
1
dx
22. ³
x ln x3
3³
2 x 1
1
x
4 x ln x 2 5 C
2
2
³
x C
1
x3
2 x ln x 2 2 C
3
2
x 4 x 4 x 20
dx
x2 5
2
2
ln 1 3
3
2³
2
ln x, du
³ x 1 2 dx 2³ x 1 2 dx
2
21. u
§ 3 ·
¨
¸ dx
x ©¨ 2 x ¹¸
³ ¨© x 2 x 2 2 ¸¹ dx
x3
2 x ln
3
3
2
1
³
3 13
3 ln 1 x1 3 C
x3 5 x 2
19 x 115 ln x 5 C
3
2
19. ³
1
dx
3x 2 3
1 x1 3 , du
·
³ ¨© x x 3 ¸¹ dx
x3
5 ln x 3 C
3
18. ³
dx
³ ¨© 2 x 11 x 2 ¸¹ dx
x 2 11x 19 ln x 2 C
17. ³
3
2 x
x , du
1
³1 2x
dx
2 x 1
2
C
1
dx Ÿ u 1 du
2x
2 x , du
1
1
³
u 1
u
du
§
dx
1·
³ ©¨1 u ¹¸ du
u ln u C1
1
2 x ln 1 2 x ln 1 2 x C1
2x C
C1 1.
INSTRUCTOR USE ONLY
where C
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 5.2
28. u
1
³1 3
dx Ÿ dx
2 3x
3x , du
1
3x
The Natural Logarithmic Function
Function: Integration
459
2
u 1 du
3
12
³ u 3 u 1 du
dx
2 §
1·
¨1 ¸ du
³
3 ©
u¹
2
ªu ln u ¼º C
3¬
2ª
1 3 x ln 1 3x º C
¼
3¬
2
2
3 x ln 1 3 x C1
3
3
³
1
x 3, du
29. u
x
dx
x 3
2
x
dx Ÿ 2 u 3 du
dx
2
2³
u 3
u
2³
u 2 6u 9
du
u
du
9·
§
2 ³ ¨ u 6 ¸ du
u
©
¹
ªu 2
º
2 « 6u 9 ln u » C1
¬2
¼
u 2 12u 18 ln u C1
x 3
x 6
where C
3
12
x 3 18 ln
x 18 ln
x 3 C
x 3 C1
C1 27.
1
dx Ÿ dx
3x 2 3
x1 3 1, du
30. u
2
x
³ 3 x 1 dx
³
2
3 u 1 du
u 1
2
3 u 1 du
u
u 1 2
u 2u 1 du
u
1·
§
3³ ¨ u 2 3u 3 ¸ du
u¹
©
3³
ªu3
º
3u 2
3« 3u ln u » C
2
¬3
¼
2
ª x1 3 1 3
º
3 x1 3 1
«
3
3 x1 3 1 ln x1 3 1» C
«
»
3
2
¬
¼
3 ln x1 3 1 §T ·
31. ³ cot ¨ ¸ dT
© 3¹
3x 2 3
3 x1 3 x C1
2
§ T ·§ 1 ·
3³ cot ¨ ¸¨ ¸ dT
© 3 ¹© 3 ¹
3 ln sin
T
C
3
32. ³ tan 5T dT
1 5 sin 5T
dT
5 ³ cos 5T
1
ln cos 5T C
5
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
460
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Function
Functions
x§ 1 ·
2³ sec ¨ ¸ dx
2© 2 ¹
x
34. ³ sec dx
2
35. ³ cos 3T 1 dT
1
3
³ cos 3T 3 dT ³ dT
T·
§
36. ³ ¨ 2 tan ¸ dT
4¹
©
T
4
sec x 1, du
³ sec 2 x tan 2 x 2 dx
ln sec x 1 C
1 ln sec 2 x tan 2 x ln cos 2 x C
2
y
43.
2x
³ x 2 9 dx
ln x 2 9 C
1
dx
x 2
3 ln x 2 C
3³
ln 0 9 C Ÿ C
0, 4 : 4
3 ln 1 2 C Ÿ C
y
0
3 ln x 2
y
sec x tan x dx
C
3
1, 0 : 0
ln cot t C
³ sec x 1 dx
³ 2 x dx
y
csc 2 t
sec x tan x
2T 4 ln cos
41.
csc 2 t dt
³ cot t dt
³ 2dT 4³ tan 4 ¨© 4 ¸¹ dT
1
2
cot t , du
38. u
T §1·
40. ³ sec 2 x tan 2 x dx
ln 1 sin t C
39. u
T C
cos t dt
cos t
³ 1 sin t dt
x
x
2 ln sec tan C
2
2
1 sin 3T
3
1 sin t , du
37. u
1
csc 2 x 2 dx
2³
1
ln csc 2 x cot 2 x C
2
33. ³ csc 2 x dx
4 ln 9
ln x 2 9 4 ln 9
8
(0, 4)
10
−9
(1, 0)
− 10
9
10
−4
− 10
y
42.
³
x2
dx
x
1, 0 : 0
y
§
2·
³ ¨©1 x ¸¹ dx
1 2 ln 1 C
sec 2 t
³ tan t 1 dt
44. r
x 2 ln x C
1 C Ÿ C
1
ln tan t 1 C
S, 4 : 4
x 2 ln x 1
r
8
ln 0 1 C Ÿ C
4
ln tan t 1 4
10
(− 1, 0)
−9
9
−8
−4
(π , 4)
8
−2
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 5.2
2
2 x 2 , x ! 0
x2
2
C
x
1 2 C Ÿ C
3
45. f cc x
fc x
fc1
f x
2
3
x
2 ln x 3 x C1
f 1
1
f x
2 ln x 3 x 2
fc x
47.
x 1
2
2
4
2x C
x 1
fc 2
0
fc x
f x
4
2x
x 1
4 ln x 1 x 2 C1
f 2
3
f x
4 ln x 1 x 7
44C Ÿ C
ln x
, 1, 2
x
y
1
x
4
−1
−2
2
2
2,
x !1
ln x
³ x dx
ln x
y1
2 Ÿ 2
ln 1
ln x
11
−4
7
x
49. ³
5
dx
0 3x 1
50. ³
1 2 x 3
4
ª5
º
« 3 ln 3 x 1»
¬
¼0
4
1
1
³1
1 ln x
x
52. u
ln x, du
e
−3
ln 2 C Ÿ C
So, y
ln x 2 1 ln 2
1 ln 2
§ x 2·
ln ¨
¸ 1.
© 2 ¹
e
e2
3º
ª1
« 3 1 ln x »
¬
¼1
dx
7
3
1
dx
x
e2 §
1
³ e x ln x dx
3
−3
1
dx
x
2
ln x 2 C
1Ÿ1
5
ln 13 | 4.275
3
1
1
ªln 2 x 3 ¼º 1
2¬
1
1
ln 5 | 0.805
>ln 5 ln 1@
2
2
dx
1 ln x, du
51. u
4
y0
2
2.
−1
y
1
C Ÿ C
2
0
3
³ x 2 dx
2
2
2
(0, 1)
y
C
2
4
1
, 0, 1
x 2
(b)
2
y
(b)
So, y
4 0 4 C1 Ÿ C1
−2
461
2
2
dy
dx
dy
dx
(a)
4 x 1
fc x
(a)
48.
2 0 3 C1 Ÿ C1
4
46. f cc x
The Natural Logarithmic Function
Function: Integration
1 ·1
e2
³ e ¨© ln x ¸¹ x dx
ªln ln x º
¬
¼e
ln 2
| 0.693
6
53. ³
2
dx
x 1
2§
2 x2
0
1 ·
³ 0 ¨© x 1 x 1 ¸¹ dx
−3
2
ª1 2
º
« 2 x x ln x 1 »
¬
¼0
ln 3
| 1.099
54. ³
1 x 1
0 x 1
dx
1
1
2
³ 0 1 dx ³ 0 x 1 dx
1
ª¬ x 2 ln x 1 º¼ 0
1 2 ln 2
| 0.386
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
462
Chapter 5
55. ³
NOT FOR SALE
Logarithmic,
arithmic, Exponential,
Exponen
and Other Transcendental Functions
Function
T
dT
T sin T
2 1 cos
1
2
ª¬ln T sin T º¼1
2 sin 2
| 1.929
1 sin 1
ln
2T , du
56. u
S
2 dT , T
8
S 4
S
Ÿ u
4
S
,T
S
Ÿ u
4
2
S 2
1
csc u cot u du
2 ³S 4
S 2
1
ª ln csc u cot u ln sin u º¼S 4
2¬
³ S 8 csc 2T cot 2T dT
1ª
« ln 1 0 ln 1 ln
2 «¬
1ª
«ln
2 «¬
2 1 ln
1 §
ln ¨1 2 ¨©
57. ³
1
1
x
1
58. ³
1
dx
x 2 ln 1 2º
»
2 »¼
2º
»
2 »¼
2·
¸
2 ¸¹
x C
x
³ 0 tan t dt
64. F x
Fc x
x
dx
x
4
§
ln ¨¨
©
x
dx
x 1
59. ³
2
2 1 ln
x x 4 ln 1 x 1·
¸ 2
x 1 ¸¹
x C
tan x
3x 1
³1 t dt
65. F x
1
3
3x
Fc x
x C
1
x
(by Second Fundamental Theorem of Calculus)
x2
dx
60. ³
x 1
S 2
x2
x C
ln x 1 2
61. ³
S 4
62. ³
sin 2 x cos 2 x
dx
S 4
cos x
csc x sin x dx
ln
S 4
Alternate Solution:
2 1 2
| 0.174
2
§ 2 1·
ln ¨¨
¸¸ 2 2
© 2 1¹
| 1.066
³ 1 t dt
Fc x
1
3
3x
ln 3 x
1
x
x2 1
2x
x2
Fc x
67. A
3x
ª¬ln t º¼1
³1 t dt
66. F x
Note: In Exercises 63–66, you can use the Second
Fundamental Theorem of Calculus or integrate the
function.
2
x
36
³ 1 x dx
3
ª¬6 ln x º¼1
6 ln 3
x1
³1 t dt
63. F x
1
x
Fc x
68. A
3x 1
F x
4
2
³ 2 x ln x dx
S 4
³0
2³
4
1 1
2 ln x x
dx
S 4
ln cos x º
»¼ 0
4
2 ln ln x ¼º 2
ln
2
0
2
2 ª¬ln ln 4 ln ln 2 º¼
§ 2 ln 2 ·
2 ln ¨
¸
© ln 2 ¹
2 ln 2
ln 2
2
INSTRUCTOR
R
R USE ONLY
69. A
tan x dx
ln
2
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NOT FOR SALE
Section
ction 5.2
3S 4
sin x
3S 4
³ S 4 1 cos x dx
70. A
4 x2 4
³1
71. A
x
ln 1 cos x º¼S 4
4§
4·
³1 ¨© x x ¸¹ dx
dx
The Natural Logarithmic Function
Function: Integration
§
ln ¨¨1 ©
§
2·
¸ ln ¨¨1 2 ¸¹
©
4
ª x2
º
« 4 ln x»
¬2
¼1
8 4 ln 4 1
2
2·
¸
2 ¸¹
§2 ln ¨¨
©2 2·
¸
2 ¸¹
463
ln 3 2 2
15
8 ln 2 | 13.045
2
10
0
6
0
5x
³ 1 x 2 2 dx
5 5 1
2 x dx
2 ³1 x2 2
5
72. A
5
ª5
º
2
« 2 ln x 2 »
¬
¼1
5
ln 27 ln 3
2
5
ln 9
2
5 ln 3 | 5.4931
4
0
6
0
10
2
Sx
0
6
73. ³ 2 sec
12
2
S ³0
dx
2
§ S x ·S
sec ¨ ¸ dx
© 6 ¹6
12 ª
Sx
Sx º
ln sec
tan
6
6 »¼ 0
S «¬
·
12 §
S
S
ln 1 0 ¸
¨ ln sec tan
3
3
S©
¹
74. ³
4
1
4
2 x tan 0.3x dx
ª 2 10
º
« x 3 ln cos 0.3x »
¬
¼1
12
S
ln 2 3 | 5.03041
0
4
0
10
10
ª
º ª
º
«16 3 ln cos 1.2 » «1 3 ln cos 0.3 » | 11.7686
¬
¼ ¬
¼
8
0
5
−2
75. f x
12
,b a
x
4, n
4
4
ª f 1 2 f 2 2 f 3 2 f 4 f 5 º¼
24 ¬
1
>12 12 8 6 2.4@
2
4
ª f 1 4 f 2 2 f 3 4 f 4 f 5 º¼
34 ¬
1
>12 24 8 12 2.4@ | 19.4667
3
Trapezoid:
Simpson:
51
Calculator: ³
5 12
1
x
20.2
dx | 19.3133
Exact: 12 ln 5
INSTRUCTOR USE ONLY
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464
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
8x
,b a
x 4
76. f x
40
2
Trapezoid:
Simpson:
4, n
4
4
ª f 0 2 f 1 2 f 2 2 f 3 f 4 º¼
24 ¬
1
>0 3.2 4 3.6923 1.6@ | 6.2462
2
4
ª f 0 4 f 1 2 f 2 4 f 3 f 4 º¼ | 6.4615
34 ¬
Calculator: ³
4
0 x2
8x
dx | 6.438
4
Exact: 4 ln 5
ln x, b a
77. f x
Trapezoid:
Simpson:
6 2
4, n
4
4
ª f 2 2 f 3 2 f 4 2 f 5 f 6 º¼
24 ¬
1
>0.6931 2.1972 2.7726 3.2189 1.7918@ | 5.3368
2
4
ª f 2 4 f 3 2 f 4 4 f 5 f 6 º¼ | 5.3632
34 ¬
6
Calculator: ³ ln x dx | 5.3643
2
S
sec x, b a
78. f x
Trapezoid:
Simpson:
3
§ S·
¨ ¸
© 3¹
2S
,n
3
4
2S 3 ª § S ·
S
§ S·
§S ·
§ S ·º
f ¨ ¸ 2 f ¨ ¸ 2 f 0 2 f ¨ ¸ f ¨ ¸» |
>2 2.3094 2 2.3094 2@ | 2.780
2 4 «¬ © 3 ¹
6
6
3
12
©
¹
© ¹
© ¹¼
2S 3 ª § S ·
§ S·
§S ·
§ S ·º
f ¨ ¸ 4 f ¨ ¸ 2 f 0 4 f ¨ ¸ f ¨ ¸» | 2.6595
3 4 «¬ © 3 ¹
6
6
©
¹
© ¹
© 3 ¹¼
Calculator: ³
S 3
S 3
sec x dx | 2.6339
79. Power Rule
84.
80. Substitution: u
x 2 4 and Power Rule
81. Substitution: u
x 2 4 and Log Rule
82. Substitution: u
tan x and Log Rule
y
2
1
x
1
2
3
4
−1
−2
y
83.
A | 3; Matches (a)
2
85.
x 3
x
ª¬ln t º¼1 4
x
−1
2
1
2
1
A | 1.25; Matches (d)
1
³ 1 4 t dt
ª¬3 ln t º¼1
−1
x
³ 1 t dt
3 ln x
2 ln x
ln x
x
x
§1·
ln x ln ¨ ¸
© 4¹
1
§ ·
ln ¨ ¸
ln 4
© 4¹
1
ln 4 ln 2
2
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
ction 5.2
86. ³
x1
1
t
dt
x
ª¬ln t º¼1
ln 5 Ÿ x
(b) ln x
1Ÿ x
88. ³ csc u du
87. ³ cot u du
assume x ! 0
ln x
(a) ln x
The Natural Logarithmic Function:
Function Integration
5
cos u
³ sin u du
465
ln sin u C
Alternate solution:
d
ªln sin u C º¼
du ¬
e
§ csc u cot u ·
³ csc u¨© csc u cot u ¸¹ du
³
1
csc u cot u csc 2 u du
csc u cot u
1
cos u C
sin u
cot u C
ln csc u cot u C
Alternate solution:
d
ªln csc u cot u C º¼
du ¬
89. ln cos x C
90. ln sin x C
1
C
csc x
91. ln sec x tan x C
92. ln csc x cot x C
93. Average value
1
csc u cot u csc2 u
csc u cot u
1
C
cos x
ln
ln
csc u cot u csc u
csc u cot u
ln sec x C
ln csc x C
ln
sec x tan x sec x tan x
C
sec x tan x
ln
sec 2 x tan 2 x
C
sec x tan x
ln
1
C
sec x tan x
ln sec x tan x C
csc x cot x csc x cot x
ln
csc x cot x
ln
csc2 x cot 2 x
C
csc x cot x
ln
1
C
csc x cot x
4 8
1
dx
³
2
4 2 x2
C
ln csc x cot x C
94. Average value
4
4 ³ x 2 dx
2
4
ª 1º
«4 x »
¬
¼2
§1 1·
4¨ ¸
©4 2¹
csc u
44 x 1
1
dx
4 2 ³ 2 x2
4§ 1
1·
2 ³ ¨ 2 ¸ dx
2 x
x
©
¹
4
1º
ª
2 «ln x »
x ¼2
¬
1
1
1º
ª
2 «ln 4 ln 2 »
4
2¼
¬
1º
ª
2 «ln 2 »
4¼
¬
ln 4 1
| 1.8863
2
INSTRUCTOR USE ONLY
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466
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
e 2 ln x
1
dx
e 1³ 1 x
95. Average value
dS
dt
k
t
S t
³ t dt
S 2
k ln 2 C
200
S 4
k ln 4 C
300
98.
e
2
2 ª ln x º
«
»
e 1« 2 »
¬
¼1
1
10
e 1
1
| 0.582
e 1
k ln t C
Solving this system yields k
C
2
1
Sx
dx
sec
³
0
20
6
96. Average value
k
S t
k ln t C because t ! 1.
100 ln 2 and
100. So,
100 ln t
100
ln 2
ª ln t
º
1».
100 «
ln
2
¬
¼
2
ª1 § 6 ·
Sx
Sx º
« 2 ¨ S ¸ ln sec 6 tan 6 »
¬ © ¹
¼0
3ª
ln 2 S¬
3
S
ln 2 3000
³ 1 0.25t dt
97. P t
99. t
3 ln 1 0 º
¼
300
10
ªln T 100 º¼ 250
ln 2 ¬
10
>ln 200 ln 150@
ln 2
10 ª § 4 ·º
ln ¨ ¸ | 4.1504 min
ln 2 «¬ © 3 ¹»¼
3
3000 4 ³
0.25
dt
1 0.25t
100.
12,000 ln 1 0.25t C
P0
10 300
1
dT
ln 2 ³ 250 T 100
50 90,000
1
dx
50 40 ³ 40 400 3 x
50
ª¬3000 ln 400 3 x º¼ 40
| $168.27
12,000 ln 1 0.25 0 C
1000
C
1000
12,000 ln 1 0.25t 1000
Pt
1000 ª¬12 ln 1 0.25t 1º¼
1000 ª¬12 ln 1.75 1º¼ | 7715
P3
x
1 x2
101. f x
y
1
0.5
x
5
10
1
x intersects f x
2
1
x
x
2
1 x2
1 x2
2
(a) y
x
A
x
:
1 x2
1
1§ª
x º 1 ·
³ 0 ¨© «¬1 x 2 »¼ 2 x ¸¹ dx
1
ª1
x2 º
2
« ln x 1 »
4 ¼0
¬2
1
1
ln 2 2
4
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
ction 5.2
1 x2 x 2x
(b) f c x
1 x2
fc 0
2
The Natural Logarithmic Function
Function: Integration
467
1 x2
1 x2
2
1
So, for 0 m 1, the graphs of f and y
mx enclose a finite region.
y
(c)
f(x) = 2x
x +1
0.5
y = mx
x
1−m
m
x
intersects y
x2 1
f x
x
1 x2
mx
1
m mx 2
x2
1 m
m
1 m
m
x
A
³0
mx :
1 m m §
x
·
mx ¸ dx,
¨
2
x
1
©
¹
ª1
mx 2 º
2
« ln 1 x »
2 ¼0
¬2
0 m 1
1 m m
1 §
1 m · 1 §1 m ·
ln ¨1 ¸ m¨
¸
m ¹ 2 © m ¹
2 ©
1 §1· 1
ln ¨ ¸ 1 m
2 ©m¹ 2
1
ªm ln m 1º¼
2¬
102. (a) At x
The slope of f at x
104. False
1
.
2
1, f c 1 |
1
1 is approximately .
2
(b) Since the slope is positive for x ! 2, f is
increasing on 2, f . Similarly, f is decreasing on
f, 2 .
1
x
105. True
1
³ x dx
ln x C1
ln x ln C
ln Cx , C z 0
106. False; the integrand has a nonremovable discontinuity at
x
0.
103. False
1 ln x
2
d
>ln x@
dx
ln x1 2 z ln x
12
INSTRUCTOR USE ONLY
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468
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
ln t on > x, y@,
107. Let f t
0 x y.
2x 1
³ x t dt ,
108. F x
By the Mean Value Theorem,
f y f x
fc c ,
y x
1
1
2 x
2x
Fc x
x c y,
x ! 0
0 Ÿ F is constant on 0, f .
Alternate Solution:
ln y ln x
y x
1
.
c
>ln t º¼ x
2x
F x
Because 0 x c y,
ln 2 x ln x
ln 2 ln x ln x
1
1
1
!
! . So,
x
c
y
ln 2
1
ln y ln x
1
.
y
y x
x
Section 5.3 Inverse Functions
1. (a)
5x 1
f x
x 1
5
§ x 1·
f¨
¸
© 5 ¹
g x
f g x
§ x 1·
5¨
¸ 1
© 5 ¹
5x 1 1
g 5x 1
g f x
f x
x3
g x
3
x
f g x
f
3
g f x
g x3
3. (a)
5
x
3
x
x
x
(b)
3
x3
x
y
3
y
(b)
3
x
f
2
3
g
1
f
2
1
x
−3 −2
g
x
−3
1
2
2
1
3
−2
3
−3
f x
1 x3
g x
3
1 x
f g x
f
3
4. (a)
2. (a)
3 4x
f x
3 x
4
§3 x·
f¨
¸
© 4 ¹
g x
f g x
3 3 4x
g 3 4x
g f x
(b)
§3 x·
3 4¨
¸
© 4 ¹
4
1
1 1 x
x
3
1 x
3
x
g 1 x3
g f x
3 1 x
(b)
y
1 x
1 x3
3
x3
x
y
f 3
8
f
2
g
4
2
−2
x
−2
2
−2
4
g
8
x
−1
2
3
−1
−2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 5.3
5. (a)
f x
x4
g x
x 2 4,
8. (a)
x t 0
g x
f x 4
f g x
x2 4 4
g f x
1
xt0
,
1 x
1 x
,
0 x d1
x
1
§1 x ·
f¨
¸
© x ¹ 1 1 x
x
1
1
§ 1 ·
1 x
g¨
¸
1
©1 x ¹
1 x
f x
2
x2
x
f g x
x4
g
x 4
2
4
x 4 4
x
g f x
y
(b)
12
g
(b)
10
Inver
Inverse Functions
1
1
x
469
x
x
1 x
˜
1 x
1
x
y
8
3
6
4
f
g
2
2
x
2
4
6
8
10
1
12
f
x
6. (a)
f x
16 x 2 ,
g x
16 x
f g x
x t 0
16 x
f
16 16 x
16 16 x
x
g 16 x 2
16 16 x 2
g f x
2
2
3
9. Matches (c)
10. Matches (b)
11. Matches (a)
12. Matches (d)
x2
x
3
x 6
4
13. f x
y
(b)
1
20
7
16
12
8
f
− 10
g
2
−1
x
8
7. (a)
12
16
1
x
1
x
1
1x
f x
g x
f g x
(b)
One-to-one; has an inverse
1
−4
−5
x
One-to-one; has an inverse
15. f T
y
x
1
2
sin T
1.5
f=g
1
−1
5
x
3
2
5x 3
14. f x
1
1x
g f x
20
−
2
5
2
3
− 1.5
Not one-to-one; does not have an inverse
INSTRUCTOR USE ONLY
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470
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
6x
x2 4
16. f x
22. h x
x 4 x 4
9
6
−9
−9
9
9
−9
−6
Not one-to-one; does not have an inverse
1
3
s 2
17. h s
Not one-to-one; does not have an inverse
23.
1
−4
f x
2 x x3
fc x
1 3 x 2 0 for all x
f is decreasing on f, f . Therefore, f is strictly
8
monotonic and has an inverse.
24.
−7
One-to-one; has an inverse
18. g t
f x
x3 6 x 2 12 x
fc x
3 x 2 12 x 12
2
t 0 for all x
1
f is increasing on f, f . Therefore, f is strictly
t2 1
monotonic and has an inverse.
3
25.
−3
f x
fc x
3
x4
2x2
4
0 when x
x3 4 x
0, 2, 2
f is not strictly monotonic on f, f . Therefore, f does
−1
Not one-to-one; does not have an inverse
19. f x
3x 2
ln x
not have an inverse.
26.
2
−1
f x
x5 2 x3
fc x
5 x 4 6 x 2 t 0 for all x
f is increasing on f, f . Therefore, f is strictly
5
monotonic and has an inverse.
−2
27.
f x
ln x 3 , x ! 3
fc x
1
! 0 for x ! 3
x 3
One-to-one; has an inverse
20. f x
5x
x 1
f is increasing on 3, f . Therefore, f is strictly
12
monotonic and has an inverse.
28.
0
f x
cos
fc x
6
0
One-to-one; has an inverse
21. g x
x 5
3x
2
3
3x
sin
2
2
0 when x
0,
2S 4S
,
,"
3 3
f is not strictly monotonic on f, f . Therefore, f does
3
not have an inverse.
200
− 10
2
−50
One-to-one; has an inverse
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 5.3
29.
2
f x
x 4
on [4, f)
fc x
2 x 4 ! 0 on [4, f )
Inver
Inverse Functions
2x 3
35. (a) f x
y
y 3
2
x 3
2
x 3
2
x
f is increasing on [4, f). Therefore, f is strictly
y
monotonic and has an inverse.
30.
f x
fc x
f 1 x
x 2 on [2, f)
x 2
x 2
1
1 ! 0 on [2, f)
y
(b)
4
f is increasing on [2, f). Therefore, f is strictly
2
f −1
monotonic and has an inverse.
x
−2
31.
f x
fc x
4
on 0, f
x2
8
3 0 on 0, f
x
monotonic and has an inverse.
f
(c) The graphs of f and f 1 are reflections of each other
across the line y
x.
(d) Domain of f :
all real numbers
Range of f :
all real numbers
f x
cot x on 0, S
Domain of f
1
: all real numbers
fc x
csc 2 x 0 on 0, S
Range of f 1 :
all real numbers
f is decreasing on 0, S . Therefore, f is strictly
7 4x
36. (a) f x
monotonic and has an inverse.
33.
4
2
−2
f is decreasing on 0, f . Therefore, f is strictly
32.
f x
cos x on >0, S @
fc x
sin x 0 on 0, S
y
7 y
4
7 x
4
7 x
4
x
y
f is decreasing on >0, S @. Therefore, f is strictly
f 1 x
monotonic and has an inverse.
34.
y
(b)
f x
ª S·
sec x on «0, ¸
¬ 2¹
fc x
§ S·
sec x tan x ! 0 on ¨ 0, ¸
© 2¹
5
4
f is increasing on [0, S 2). Therefore, f is strictly
monotonic and has an inverse.
471
f
3
f −1
1
x
−2 − 1
−1
1
3
4
5
−2
(c) The graphs of f and f 1 are reflections of each other
across the line y
x.
(d) Domain of f :
all real numbers
Range of f :
all real numbers
Domain of f 1 : all real numbers
Range of f 1 :
all real numbers
INSTRUCTOR USE ONLY
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472
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
x5
y
x
5
y
f 1 x
37. (a) f x
x
y
y
x
y2
5
x
y
x2
5
x
f 1 x
x2 ,
x1 5
x t 0
y
(b)
y
(b)
f x
39. (a)
f
3
2
f −1
f −1
1
2
f
x
−2
1
1
2
x
1
−2
(c) The graphs of f and f 1 are reflections of each other
across the line y
x.
(d) Domain of f :
all real numbers
Range of f :
all real numbers
38. (a) f x
(d) Domain of f :
x t 0
Range of f :
y t 0
Domain of f
Range of f
1
: x t 0
y t 0
:
x
3
y 1
y
3
x 1
f 1 x
3
x 1
x 1
13
(b)
y, x t 0
x2
40. (a) f x
y
x
y
y
x
f 1 x
x
y
y
f
4
5
4
3
2
1
all real numbers
x3 1
(b)
3
(c) The graphs of f and f 1 are reflections of each other
across the line y
x.
Domain of f 1 : all real numbers
Range of f 1 :
2
3
f −1
f
2
−1
x
−5 −4 −3
2 3 4 5
f
1
x
−4
−5
1
(c) The graphs of f and f 1 are reflections of each other
across the line y
x.
(d) Domain of f :
all real numbers
Range of f :
all real numbers
Domain of f 1 : all real numbers
Range of f 1 :
all real numbers
2
3
4
(c) The graphs of f and f 1 are reflections of each other
x.
across the line y
(d) Domain of f :
x t 0
Range of f :
y t 0
Domain of f 1 : x t 0
Range of f 1 :
y t 0
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 5.3
41. (a) f x
4 x2
y,
4 x
2
2
y
0 d x d 2
3
43. (a) f x
x
4 y2
y
4 x2
f 1 x
4 x2 ,
0 d x d 2
(b)
x 1
y3
x
y3 1
y
x3 1
f 1 x
x3 1
f −1
3
2
f
x
3
−3 −2
f = f −1
2
2
3
−2
−3
1
(c) The graphs of f and f 1 are reflections of each other
x
across the line y
x
1
2
3
(c) The graphs of f and f 1 are reflections of each other
across the line y
x. In fact, the graphs are
identical.
(d) Domain of f :
0 d x d 2
Range of f :
0 d y d 2
:
0 d y d 2
x2 4
y,
Range of f
42. (a) f x
1
all real numbers
Range of f :
all real numbers
Domain of f
x t 2
x
y 4
y
x 4
f 1 x
x 2 4,
1
1
all real numbers
x t 0
y,
32
x
y
y
x3 2
f 1 x
x3 2 , x t 0
y
(b)
2
: all real numbers
:
x2 3
44. (a) f x
y2 4
x2
(d) Domain of f :
Range of f
Domain of f 1 : 0 d x d 2
(b)
y
473
y
y
(b)
x 1
4 y2
x2
Inver
Inverse Functions
6
f −1
5
2
4
x t 0
y
3
f
2
1
x
5
1
f −1
4
3
4
5
6
(c) The graphs of f and f 1 are reflections of each other
x
across the line y
f
3
2
2
1
x
1
2
3
4
(d) Domain of f :
x t 0
Range of f :
y t 0
5
1
: x t 0
Range of f 1 :
y t 0
Domain of f
(c) The graphs of f and f 1 are reflections of each other
x
across the line y
(d) Domain of f :
x t 2
Range of f :
y t 0
1
: x t 0
Range of f 1 :
y t 2
Domain of f
INSTRUCTOR USE ONLY
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474
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
x
45. (a) f x
x 7
y
x2
y 2 x2 7
2
y 2 x2 7 y 2
7y
2
1 x2
,
1 x 1
f −1
6
4
f
2
f −1
3
x z 1
y
(b)
7x
y
(b)
2
x 1
2
,
x 1
f 1 x
1 x
x
2
y 1
y
7x
x z 0
y,
x
1 y2
y
f
yx
x1 y
7 y2
x
1
x2 7
x
x2 1 y2
x 2
x
x 2
46. (a) f x
y
2
−6
x
−4
2
4
6
2
f
1
−4
−6
x
−3
1
2
3
(c) The graphs of f and f 1 are reflections of each other
x
in the line y
(d) Domain of f :
(c) The graphs of f and f 1 are reflections of each other
x
in the line y
all x z 0
Range of f :
all y z 1
(d) Domain of f :
all real numbers
Range of f 1 :
Range of f :
1 y 1
Domain of f 1 : 1 x 1
Range of f 1 :
Domain of f 1 : all x z 1
all y z 0
y
47.
x
0
1
2
3
f x
1
2
3
4
(4, 4)
4
all real numbers
x
1
2
3
3
4
2
(3, 2)
(2, 1)
1
f 1 x
0
1
2
4
(1, 0)
1
48.
x
0
2
6
f x
4
2
0
x
f
1
x
0
2
4
6
2
0
x
2
3
4
y
8
6
(0, 6)
4
(2, 2)
2
(4, 0)
2
4
6
x
8
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 5.3
49. (a) Let x be the number of pounds of the commodity
costing 1.25 per pound. Because there are 50 pounds
total, the amount of the second commodity is 50 x.
The total cost is
0.35 x 80,
0 d x d 50.
(b) Find the inverse of the original function.
0.35 x 80
y
80 y
0.35 x
Inverse: y
x 2, x d 2
53. f x
x 2
2 x
2 x
y
2 y
x
f 1 x
2 x,
f is one-to-one; has an inverse
100
80 x
35
20
80 x
7
ax b
x represents cost and y represents pounds.
y
y b
a
x b
a
x b
, a z 0
a
x
(c) Domain of inverse is 62.5 d x d 80.
(d) If x
y
5
F 32 ,
9
50. C
(a)
y
73 in the inverse function,
100
80 73
35
9
C
5
F 32
F
32 95 C
100
5
20 pounds.
f 1 x
F t 459.6
x 3
5
F 32
9
273 19 .
32 95 22
71.6qF.
22q, then F
f
1
f x
x 2, Domain: x t 2
fc x
1
! 0 for x ! 2
x 2
2
f is one-to-one; has an inverse
x 2
y
x 3
x
x 3,
x 2
16 x 4 is one-to-one for x t 0.
f
1
52. f x
16 x 4
y
16 y
x4
4 16 x
y2 2
y
x 2
x
x 2 2,
y
x
16 x
y
1
4
f
y2
x t 0
(Answer is not unique.)
4
y
y
y 3
56. f x
51.
y
x
t 273.1 1.
Therefore, domain is C t 273. 1
2
x 3
(b) The inverse function gives the temperature F
corresponding to the Celsius temperature C.
(d) If C
2
x 3 is one-to-one for x t 3.
55. f x
(c) For F t 459.6, C
x t 0
ax b
54. f x
100
80 y
35
x
475
f is one-to-one; has an inverse
1.25 x 1.60 50 x
y
Inver
Inverse Functions
x
16 x ,
x d 16
(Answer is not unique.)
2
x 3 is one-to-one for x t 3.
57. f x
x t 0
x 3
3
Not one-to-one; does not have an inverse
y
x
y 3
y
x 3
f 1 x
x 3,
x t 0
(Answer is not unique.)
INSTRUCTOR USE ONLY
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476
NOT FOR SALE
Chapter 5
x 3 is one-to-one for x t 3.
58. f x
x 3
f
1
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
65.
1 5
x 2 x3 ,
27
1
5x4 6 x2
27
f x
y
x
y 3
y
x 3
x
x 3,
fc x
f is monotonic (increasing) on f, f therefore f has an
x t 0
inverse.
(Answer is not unique.)
1
243 54
27
1
1
f c f 11
f 3
59. Yes, the volume is an increasing function, and therefore
one-to-one. The inverse function gives the time t
corresponding to the volume V.
f 1 c 11
h t2 .
61. No, C t is not one-to-one because long distance costs
66. f x
are step functions. A call lasting 2.1 minutes costs the
same as one lasting 2.2 minutes.
fc x
63.
f x
5 2x ,
fc x
6 x 2
a
x 4,
2
64.
f 1 c 7
1
1
f c 1
f c f 1 7
f x
x3 2 x 1,
fc x
3x 2 ! 0
a
1
6 1
2
1
6
2
2
f is monotonic (increasing) on f, f therefore f has an
inverse.
f 1
f
1 c
2
2 Ÿ f 1 2
1
1
1
c
f 1
f c f 1 2
1
3 12 2
1
5
1
17
1
! 0 on 4, f
x 4
2 Ÿ f 1 2
fc 8
1
2 8 4
1
f c f 1 2
inverse.
1
1
1
459
27
x t 4
2,
f 8
f 1 c 2
f is monotonic (decreasing) on f, f therefore f has an
7 Ÿ f 1 7
1
f c 3
inverse.
7
f 1
a
3
f is monotonic (increasing) on [4, f) therefore f has an
62. Yes, the area function is increasing and therefore one-toone. The inverse function gives the radius r corresponding
to the area A.
3
11 Ÿ f 1 11
1
1
4
2
5 3 6 3
27
60. No, there could be two times t1 z t2 for which
h t1
11
a
67.
f x
fc x
sin x,
a
8
1
4
1
c
f 8
1 2, S
2
1
14
d x d
4
S
2
§ S S·
cos x ! 0 on ¨ , ¸
© 2 2¹
ª S Sº
f is monotonic (increasing) on « , » therefore f has
¬ 2 2¼
an inverse.
§S ·
f¨ ¸
©6¹
§1·
f 1 c ¨ ¸
© 2¹
sin
S
6
S
1
§1·
Ÿ f 1 ¨ ¸
2
© 2¹
6
1
§ 1 § 1 · ·
f c¨ f ¨ ¸ ¸
© 2 ¹¹
©
1
S·
§
f c¨ ¸
©6¹
1
§S ·
cos¨ ¸
©6¹
2
3
2 3
3
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 5.3
68.
1, 0 d x d S 2
f x
cos 2 x,
a
fc x
2 sin 2 x 0 on 0, S 2
f
1 c
477
f, f
f, f
y
(c)
3
1Ÿ f
f
1
1
2
0
f −1
1
1
1
Range f 1
(b) Range f
an inverse.
f 0
Domain f 1
71. (a) Domain f
f is monotonic (decreasing) on >0, S 2@ therefore f has
Inver
Inverse Functions
1
fc 0
f c f 1 1
1
2 sin 0
1
0
x
−3 −2
1
2
3
−2
So, f
1 c
x 6
,
x 2
69. f x
−3
1 is undefined.
x ! 0, a
(d)
x2
8
2
x 2
2
0 on 2, f
6
f 1 c 3
1
f c f 1 3
1
fc 6
x 3
,
x 1
70. f x
fc x
x ! 1, a
1
8 6 2
2
2
x 1
2
§1·
f c¨ ¸
© 2¹
3
4
2
3
2
x
−5 −4 −3 −2 −1
1
c
f 1
f, f
f
0 on 1, f
1
f, f
y
f −1
1
fc f
Domain f 1
Range f 1
(c)
2
2 Ÿ f 1 2
1
4
3
(b) Range f
2
(d)
1
2
11
2
2
2 3
−2
−3
−4
−5
inverse.
f 1 c 2
§1 1·
x, ¨ , ¸
©8 2¹
1
33 x
72. (a) Domain f
f is monotonic (decreasing) on 1, f therefore f has an
f 1
§ 1 1·
¨ , ¸
© 2 8¹
3
§1·
f 1 c ¨ ¸
©8¹
2
x 1 1 x 3 1
x 1
3x 2
f 1 c x
inverse.
3 Ÿ f 1 3
fc x
f 1 x
f is monotonic (decreasing) on 2, f therefore f has an
f 6
x3 ,
3
x 2 1 x6 1
fc x
f x
f x
3 4 x,
fc x
4
fc1
4
f 1 x
f 1 c x
f 1 c 1
1, 1
3 x
,
4
1
4
1
4
1, 1
INSTRUCTOR USE ONLY
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478
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
[4, f), Domain f 1
73. (a) Domain f
[0, f), Range f 1
(b) Range f
[0, f)
[4, f)
y
(c)
In Exercises 75–78, use the following.
f x = 18 x 3 and g x = x 3
f 1 x = 8 x + 3 and g 1 x = 3 x
12
f −1
10
8
75.
f 1 D g 1 1
f 1 g 1 1
f 1 1
76.
g 1 D f 1 3
g 1 f 1 3
g 1 0
0
77.
f 1 D f 1 6
f 1 f 1 6
f 1 72
600
78.
g 1 D g 1 4
g 1 g 1 4
g 1 3 4
32
6
4
f
2
x
2
(d)
4
6
8
10
12
f x
x 4,
fc x
1
x 4
2
1
2
fc 5
f
1
5, 1
3 3
x 4,
x
f 1 c x
2x
f 1 c 1
2
1, 5
f x = x + 4 and g x = 2 x 5
f 1 x = x 4 and g 1 x =
[0, f), Domain f
74. (a) Domain f
(b) Range f
(0, 4], Range f
1
1
g 1 D f 1 x
x + 5
2
g 1 f 1 x
(0, 4]
g 1 x 4
[0, f)
x 4 5
2
x 1
2
y
4
3
2
80.
f −1
1
f 1 D g 1 x
f 1 g 1 x
§ x 5·
f 1 ¨
¸
© 2 ¹
x 5
4
2
x 3
2
f
x
1
(d)
9 4
In Exercises 79–82, use the following.
2
79.
(c)
4
f x
fc x
fc1
2
3
4
1 x2
8 x
x2 1
2
2
81.
f D g x
f g x
f 2x 5
4 x
x
2
f 1 x
f 1 c x
f 1 c 2
4
x
2
1
2
4 x x
2x 5 4
2x 1
So, f D g
1
Note: f D g
x
1
x 1
.
2
g 1 D f 1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.3
82.
g D f x
89. False. Let f x
g x 4
90. True; if f has a y-intercept.
91. True
2x 3
1
1
Note: g D f
83. Let y
x
92. False. Let f x
x 3
.
2
f does not pass the horizontal line test.
90
f 1 x . Let the
of y. Interchange x and y to get y
−6
domain of f 1 be the range of f . Verify that
x and f 1 f x
Example:
f x
x3 ; y
f 1 x
3
x3 ; x
3
1 x.
2 x3 3x 2 36 x
93. (a) f x
f 1 D g 1
x or g x
f x be one-to-one. Solve for x as a function
f f 1 x
479
x2.
g f x
2x 4 5
So, g D f
Inver
Inverse Functions
5
x.
−45
3
y; y
(b) f c x
6 x 2 6 x 36
x;
6 x2 x 6
x
fc x
84. The graphs of f and f 1 are mirror images with respect
to the line y
x.
6x 3 x 2
2, 3
0 at x
On the interval 2, 2 , f is one-to-one, so, c
2.
94. Let f and g be one-to-one functions.
85. f is not one-to-one because many different x-values yield
the same y-value.
f S
Example: f 0
Not continuous at
(a) Let
0
2n 1 S
2
f D g x1
f D g x2
f g x1
f g x2
, where n is an integer.
g x1
x1
86. f is not one-to-one because different x-values yield the
same y-value.
§ 4·
f ¨ ¸
© 3¹
Example: f 3
f 2
3
k 2 2 2
3
f g x
y
1
12
m
f
1
at
1
.
4
1, 12
12 , 1
is
1
y . Also:
g 1 f 1 y
x
g 1 D f 1 y
1
1
y
g 1 D f 1 y and
g 1 D f 1.
95. If f has an inverse, then f and f 1 are both one-to-one.
2.
Let f 1
(b) Since the slope of the tangent line to f at 2, 1 is 2,
the slope of the tangent line to f
m
f D g
f D g
f 1 y
x
So, f D g
is
y, then x
y
g x
12k Ÿ k
88. (a) Since the slope of the tangent line to f at
1 , the slope of the tangent line to
2
Because g is one-to-one.
f D g x
k 2 x x is one-to-one. Because
2,
x2
(b) Let f D g x
3
5
3
f 1 3
Because f is one-to-one.
So, f D g is one-to-one.
Not continuous at r 2.
87. f x
g x2
1
f 1
1
1
x
y then x
f 1 y and f x
y. So,
f.
at 1, 2 is
1
.
2
INSTRUCTOR USE ONLY
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480
Chapter 5
NOT FOR SALE
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
96. Suppose g x and h x are both inverses of f x . Then
the graph of f x contains the point a, b if and only if
fc x
the graphs of g x and h x contain the point b, a .
Because the graphs of g x and h x are the same,
97. If f has an inverse and f x1
f
1
f x1
f
1
f x2
f x2 , then
Ÿ x1
101.
x2 . Therefore, f is
inverses, g
a. By the reflexive property of
g cc x
1
fc g x
fc g x
0 f cc g x g c x
ªfc g x º
¬
¼
1
fc 2
0
1
5
5
5
x 2
x 1
y 2
y 1
xy x
y 2
xy y
x 2
y
x 2
x 1
So, if f x
2
˜ ª1 f c g x
¬
2
ªfc g x º
¬
¼
f cc g x
y
x
99. From Theorem 5.9, you have:
gc x
f 1 c 0
102.
f is one-to-one, but not strictly monotonic.
2
1 x2
f .
0 d x d1
­ x,
.
®
¯1 x, 1 x d 2
Let f x
x
17
f, f Ÿ f is one-to-one.
1
98. Not true.
0
f c x ! 0 for all x Ÿ f increasing on
in the range, there corresponds exactly one value a in the
domain. Define g x such that the domain of g equals
the range of f and g b
, f 2
³ 2 1 t dt , f 2
f x
fc x
one-to-one. If f x is one-to-one, then for every value b
1 t4
1
1 x4
1
1
fc 2
1 17
f 1 c 0
h x . Therefore, the inverse of f x is unique.
g x
dt
x
³2
f x
100.
x 2
, then f 1 x
x 1
f x.
The graph of f is symmetric about the line y
º
¼
x.
f cc g x
ªfc g x º
¬
¼
3
If f is increasing and concave down, then f c ! 0 and
f cc 0 which implies that g is increasing and concave up.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section 5.4
Exponential Functions: Differentiation an
and Integration
481
ax b
cx d
103. f x
(a) Assume bc ad z 0 and f x1
ax1 b
cx1 d
f x2 . Then
ax2 b
cx2 d
acx1 x2 bcx2 adx1 bd
acx1 x2 adx2 bcx1 bd
ad bc x1
ad bc x2
x1
because ad bc z 0
x2
So, f is one-to-one.
0, then either b
0 or c
Now assume f is one-to-one. Suppose, on the contrary, that ad
bc. If d
0 and f is not one-to-one. So consider
f is not one-to-one. Similarly, if b
0, then a
0 or d
ax b
cx d
f x
adx bd b
˜
bcx bd d
bcx bd b
˜
bcx bd d
0. In both cases,
b
,
d
which is not one-to-one.
Alternate Solution:
ax b
Ÿ fc x
cx d
f x
ad bc
cx d
2
f is monotonic (and therefore one-to-one) if and only if ad bc z 0.
cyx dy
ax b
cx d
ax b
cy a x
b dy
x
b dy
cy a
y
(b)
(c)
b dx
,
cx a
f 1 x
y
ax b
cx d
b dx
cx a
acx 2 bcx a 2 x ab
bc ad z 0
bcx cdx 2 bd d 2 x
ac cd x 2 d 2 a 2 x bd ab
0
c a d x2 d a d a x b a d
So, f
f 1 if a
d , or if c
b
0 and a
0
d.
Section 5.4 Exponential Functions: Differentiation and Integration
1. eln x
4
3. e x
x
4
x
12
ln 12 | 2.485
2. eln 3 x
24
4. 5e x
36
3x
24
x
x
8
36
5
x
ln 36
| 1.974
5
e
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
482
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
5. 9 2e x
7
13. ln x 3
2
x
2
x 3
e2
ex
1
x
x
0
6. 8e 12
7
2e
x
8e
x
19
e
x
19
8
x
ln 19
8
14. ln 4 x
1
4x
e1
x
e
| 0.680
4
15. ln
| 0.865
7. 50e x
1
x 2
e1
x 2
2
3
5
x
ln 53
2
x 2
2
e12
x 2
e6
| 0.511
8. 100e 2 x
35
2 x
35
100
2 x
7
ln 20
12
2 e6 | 405.429
x
17. y
7
20
7
12 ln 20
e
e 2 2 | 5.389
16. ln x 2
ln 53
x
e
x
e x
e
e
x 2
30
x
3 e 2 | 10.389
e x
y
4
1 ln 20
2
7
3
| 0.525
9.
800
100 e x 2
800
50
50
100 e x 2
x
5000
1 e2 x
5000
2
2499
2
3
1
2
3
y
x
2
2 ln 84 | 8.862
ln 84
1
1 ex
2
18. y
ex 2
84
10.
x
−1
4
3
2
1
2
x
−1
1 e2 x
19. y
e
2x
ex 2
y
2x
ln 2499
6
1
ln 2499 | 3.912
2
x
5
4
3
11. ln x
2
2
e | 7.389
1
2
x
−3 −2 −1
12. ln x 2
10
2
e10
x
x
1
2
3
INSTRUCTOR USE ONLY
x
r e5 | r148.413
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.4
Exponential Functions: Differentiation an
and Integration
e x 1
20. y
24. (a)
483
10
y
3
2
−8
10
1
−2
x
−3 −2 −1
−1
1
2
3
Horizontal asymptotes: y
−2
(b)
−3
e x
21. y
0 and y
8
10
2
−8
Symmetric with respect to the y-axis
Horizontal asymptote: y
10
1
−2
0
Horizontal asymptote: y
y
4
Ce ax
25. y
2
Horizontal asymptote: y
0
Matches (c)
x
−1
Ce ax
26. y
1
Horizontal asymptote: y
22. y
0
Reflection in the y-axis
ex 2
Matches (d)
y
C 1 e ax
27. y
4
3
Vertical shift C units
2
Reflection in both the x- and y-axes
Matches (a)
x
−2
−1
1
23. (a)
C
1 e ax
C
lim
x o f1 e ax
C
lim
x o f 1 e ax
2
28. y
7
f
g
−5
7
−1
0
Horizontal asymptotes: y
Horizontal shift 2 units to the right
(b)
C
C and y
0
Matches (b)
3
f
−2
4
29. f x
e2 x
g x
ln
1
ln x
2
x
h
y
−3
6
A reflection in the x-axis and a vertical shrink
(c)
f
4
7
f
2
g
q
x
−2
−4
8
2
4
6
−2
−1
Vertical shift 3 units upward and a reflection in the
y-axis
axis
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
484
NOT FOR SALE
Chapter 5
30. f x
ex 3
g x
ln x3
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
38. y
3 ln x
yc
2
5e x 5
2
2
5e x 5 2 x
10 xe x 5
y
39. y
8
yc
6
4
e x ln x
§1·
e x ¨ ¸ e x ln x
© x¹
§1
·
e x ¨ ln x ¸
©x
¹
2
f
x
−2
g
−2
2
4
6
yc
ex 1
31. f x
40. y
8
41. y
ln x 1
g x
yc
xe4 x
4 xe 4 x e 4 x
e4 x 4 x 1
x3e x
x3e x 3 x 2 e x
x 2e x x 3
e x x3 3 x 2
y
42. y
6
f
x 2e x
yc
x 2 e x 2 xe x
43. g t
e t et
gc t
3 e t et
4
xe x 2 x
g
2
x
2
32. f x
e
4
6
44. g t
x 1
1 ln x
g x
3
e 3 t
2
et e t
2
6
2
gc t
e 3 t 6t 3
y
ln 1 e 2 x
dy
dx
2e 2 x
1 e2 x
y
§1 ex ·
ln ¨
x¸
©1 e ¹
dy
dx
ex
ex
x
1 e
1 ex
y
2
e x e x
dy
dx
2 e x e x
t 3e 3 t
2
y
4
45.
f
3
2
g
1
x
−1
1
3
2
−1
33.
f x
e2 x
fc x
2e 2 x
4
46.
47.
34. y
yc
35.
e
8 x
8e 8 x
y
e x
dy
dx
e x
2 x
36. y
yc
e
48.
y
dy
dx
2 x3
3
6 x 2e 2 x
37. y
ex 4
yc
x4
e
49. y
yc
ln 1 e x ln 1 e x
2e x
1 e2 x
2 e x e x
2
e x e x
1
2 e x e x
e x e x
2
e x e x
2
e x e x
2
ex 1
ex 1
ex 1 ex ex 1 ex
ex 1
2
2e x
ex 1
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.4
50. y
yc
51.
e2 x
e 1
59.
e 2 x 1 2e 2 x e 2 x 2 e 2 x
e2 x 1
2e 2 x
2
e2 x 1
2
dy
dx
e x cos x sin x sin x cos x e x
e x 2 cos x
2e x cos x
ln
ln x
e2 x
x
0, 0
0
0
ln e
f x
e3 x , 0, 1
1 2e
x 2e x 2 xe x 2e x ,
yc
x e 2 xe 2 xe 2e x 2e x
y
ln e
2x
fc x
2e 2 x , f c 0
Tangent line: y 1
63.
2
2x 1
f x
e1 x ,
1, 1
fc x
1 x
, fc1
1 x 1
y
x 2
yc
xe e e x
x
2
Tangent line: y 1
2x 2
y
2x 3
1, 0
x
xe x
e
Tangent line: y 0
e x 1
y
ex e
xe y 10 x 3 y
dy
dx
64.
2, 1
2 x 2 e 2 x x , yc 2
ex
xe x e x ,
1
Tangent line: y 1
e x 1
dy
dy
e y 10 3
xe y
dx
dx
dy y
xe 3
dx
2 x 0
2
x 2e x
e
y
yc 1
1, e
x
y
3x 1
e 2 x , 0, 1
x
1
3x 0
f x
y
2e
2x
3
Tangent line: y 1
2 x
Tangent line: y e
2x
62.
3e3 x , f c 0
,
y
yc 1
Fc x
e
61.
cos x
1
x
2x
e
1
1
x e
e
e x e x
,
2
1
Tangent line: y
³0
2 x x 2
e 1 x 1
t
cos eln x ˜
fc x
§1
·
e x ¨ ln x ¸
©x
¹
ªe x e x º¼
ª e x e x 2º ¬
¬
¼
yc 0
³S cos e dt
ln t 1 dt
yc
e 1
yc
e 2 x ª¬2 sec 2 2 xº¼ 2e 2 x tan 2 x
54. F x
58. y
fc1
60. y
e tan 2 x
Fc x
57.
§1·
e x ¨ ¸ e x ln x
© x¹
2x
53. F x
56.
fc x
485
1, 0
y
2e 2 x ª¬sec 2 2 x tan 2 xº¼
55.
e x ln x,
Tangent line: y 0
e x sin x cos x
yc
f x
2x
y
52. y
Exponential Functions:
Func
Fun tions:
ons: Differentiation an
and Integration
0
0
10 e y
10 e y
xe y 3
e xy x 2 y 2
dy
§ dy
·
y ¸e xy 2 x 2 y
¨x
dx
© dx
¹
dy xy
xe 2 y
dx
dy
dx
10
0
ye xy 2 x
ye xy 2 x
xe xy 2 y
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
486
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
xe y ye x
1,
xe y yc e y ye x yce x
0
65.
At (0, 1): e 1 yc
0, 1
fc x
0
yc
e x e x
2
e x e x
0 when x
2
e x e x
! 0
2
f x
71.
e 1
Tangent line: y 1
e 1 x 0
y
e 1 x 1
f cc x
0.
Relative minimum: 0, 1
6
66.
1 ln xy
ex y ,
1
> xyc y@
xy
e x y >1 yc@
At (1, 1): > yc 1@
1, 1
yc
e x e x
2
e x e x
! 0
2
e x e x
0 when x
2
f x
72.
0 x 1
y
3
0
0
Tangent line: y 1
67.
(0, 1)
−3
1 yc
1
fc x
3 x
f x
3 2x e
fc x
3 2 x 3e 3 x 2e 3 x
7 6 x e 3 x
f cc x
7 6 x 3e 3 x 6e 3 x
3 6 x 5 e 3 x
f cc x
0.
Point of inflection: 0, 0
2
68.
g x
x e x ln x
gc x
1
2
g cc x
69.
70.
1
4x
x
4e x
yc
4e x
ycc
4e x
ex 2x 1
4e x 4e x
y
e 3 x e 3 x
yc
3e3 x 3e 3 x
ycc
9e3 x 9e 3 x
ycc 9 y
x2
3
−2
1
xe x e x
ex
e x ln x
32
2
4x
x
x
y
ycc y
(0, 0)
−3
ex
e x ln x
x
x
e ln x
1 x2 2 2
e
2S
2
1
x 2 e x 2 2
2S
2
1
x 1 x 3 e x 2 2
2S
g x
73.
x
gc x
g cc x
§
Relative maximum: ¨ 2,
©
0
1 ·
¸ | 2, 0.399
2S ¹
Points of inflection:
1 1 2 · §
1 1 2 ·
§
e ¸, ¨ 3,
e ¸ | 1, 0.242 , 3, 0.242
¨1,
S
2
2S
©
¹ ©
¹
9e3 x 9e 3 x 9 e3 x e 3 x
(
2,
0
0.8
(
1, e
− 0.5
2π
1
2π
( (
(
3, e
−0.5
0
2π
(
4
0
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.4
1 x3 2 2
e
2S
2
1
x 3 e x 3 2
2S
2
1
x 2 x 4 e x 3 2
2S
g x
74.
Exponential Functions: Differentiation an
and Integration
gc x
g cc x
§
Relative maximum: ¨ 3,
©
(
1
2π
e−0.5
2π
( (
3,
(
2,
xe x e x
f cc x
e
1 ·
¸ | 3, 0.399
2S ¹
e
1 x
0 when x
(1, e −1)
2.
(2, 2e − 2)
−2
e−0.5
2π
1.
2
4
(
−2
g t
1 2 t et
gc t
1 t et
g cc t
te t
77.
6
0
f x
x 2e x
fc x
x 2e x 2 xe x
xe x 2 x
Relative maximum: 1, 1 e | 1, 3.718
0 when x
Point of inflection: 0, 3
0, 2.
5
e x 2 x x 2 e x 2 2 x
f cc x
0 when x
x
Relative maximum: 1, e 1
1 1 2 ·
e ¸ | 2, 0.242 , 4, 0.242
2S
¹
0
75.
x
e x x 2
(
4,
fc x
Point of inflection: 2, 2e2
1 1 2 · §
e ¸, ¨ 4,
2S
¹ ©
0.8
xe x
e x 1 x
Points of inflection:
§
¨ 2,
©
f x
76.
487
e x x 2 4 x 2
0 when x
(−1, 1 + e)
2r
(0, 3)
2.
−6
6
Relative minimum: 0, 0
−3
Relative maximum: 2, 4e2
x
2r
2
y
2 r
2 e
78.
2 2r
2
6r 4 2 e
2
2 e3 x 4 2 x
fc x
e3 x 2 3e3 x 4 2 x
e3 x 10 6 x
Points of inflection:
§2 r
¨
©
2r
f x
2, 6 r 4 2 e
2r
2 ·
f cc x
5.
3
e3 x 6 3e3 x 10 6 x
¸
¹
| 3.414, 0.384 , 0.586, 0.191
0 when x
e3 x 24 18 x
0 when x
Relative maximum:
5
, 96.942
3
Point of inflection:
4
, 70.798
3
4.
3
3
(0, 0)
−1
) 2, 4 e −2 )
( 53 , 96.942)
(
)
100
4
, 70.798
3
5
0
)2 ± 2, (6 ± 4 2)e− (2 ± 2))
− 0.5
2.5
0
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
488
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
79.
A
base height
dA
dx
4 x 2e x 2e x
2 xe x
2
2
81.
2
f x
e2 x
fc x
2e 2 x
x, e 2 x be the point on the graph where the
Let x, y
2e
x2
1 2x
2
2
.
2
0 when x
tangent line passes through the origin. Equating slopes,
2e 1 2
A
e2 x 0
x 0
1
x
1
e,
, y
2
2e 2 x
y
2
3
2
x
( ,e )
2
2
−1
2
1
2
−1
−1
80. (a)
10ce c
8
c x
ec x
cec x
f(x) = e2x
f(x) = (2e)x
c x ec
( (
1
,e
2
c x
ce x
ce c
x
0
2
0
x
82. (a) f is increasing on f, f . g is decreasing on
x
e 1
c
(b) A x
y
10 c x e c x
c
ec
1·
§
2e¨ x ¸
2¹
©
2ex
Tangent line: y e
f c x
f c
2e.
§1 ·
Point: ¨ , e ¸
©2 ¹
x
−2
yc
f, f .
x
xf c
ª § x ·
x «10¨ x
¸e
¬ © e 1¹
x e x 1
º
»
¼
(b) f and g are both concave upward on f, f .
15,000e 0.6286t , 0 d t d 10
83. V
10 x 2 x 1 e x
e
ex 1
(a)
20,000
10 x 2 x 1 e x
e
ex 1
(c) A x
6
0
(2.118, 4.591)
10
0
(b)
dV
dt
9429e 0.6286t
9
0
When t
1,
dV
| 5028.84.
dt
When t
5,
dV
| 406.89.
dt
0
The maximum area is 4.591 for x
f x
2.547.
2
x
ex 1
(d) c
lim c
1
lim c
0
x o 0
xof
2.118 and
(c)
0
20,000
4
0
0
Answers will vary. Sample answer:
10
0
As x approaches 0 from the right, the height of the
rectangle approaches 1.
As x approaches f, the height of the rectangle
approaches 0.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.4
Exponential Functions: Differentiation an
and Integration
84. 1.56e 0.22t cos 4.9t d 0.25 (3 inches equals one-fourth
1686.8t 32,561
86. (a) Linear model: V
foot.) Using a graphing utility or Newton's Method, you
have t t 7.79 seconds.
489
Quadratic model: V
109.52t 2 3658.2t 40,995
25,000
2
Quadratic
10
0
Linear
5
10,000
−2
85.
13
(b) The slope represents the average loss in value per year.
h
0
5
10
15
20
P
10,332
5583
2376
1240
517
ln P
9.243
8.627
7.773
7.123
6.248
(c) Exponential model:
V
(a)
40,955.46 0.90724
t
40,955.46 e 0.09735t
(d) As t o f, V o 0 for the exponential model. The
value of the car tends to zero.
40,955.46 0.09735 e 0.09735t
(e) V c
12
3987.01e 0.09735t
−2
22
When t
7, V c | 2017 dollars/year.
When t
11, V c | 1366 dollars/year.
0
0.1499h 9.3018 is the regression line for
y
data h, ln P .
ah b
(b) ln P
P
e ah b
ebe ah
P
Ce ah , C
eb
0.1499 and C
a
So, P
e9.3018
10,957.7.
87.
f x
ex
f 0
1
fc x
e
x
fc 0
1
f cc x
e
x
f cc 0
1
P1 x
11x 0
1 x
P2 x
11x 0
1
2
1 x 0
2
10,957.7e 0.1499 h
1 x x2
2
8
f
(c) 12,000
P2
P1
−6
4
−1
0
The values of f, P1 , and P2 and their first derivatives
agree at x
0.
22
0
(d)
dP
dh
10,957.71 0.1499 e 0.1499 h
1642.56e 0.1499 h
dP
dh
For h
5,
For h
18,
776.3.
dP
| 110.6.
dh
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
490
Chapter 5
88.
f x
ex 2 ,
f 0
1
93. Let u
fc x
1 x2
e ,
2
1 x2
e ,
4
1
x
1 x 0
1,
2
2
1
,
2
1
1
2
1 x 0 x 0
2
8
fc 0
1
2
1
4
³e
f cc x
P1 x
P1c x
P2 x
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
f cc 0
P2cc x
1
P1c 0
1
2
P2 0
1
2 x 1
P2cc 0
1
2
1
4
The values of f , P1 , P2 and their first derivatives agree
at x
0. The values of the second derivatives of f and
P2 agree at x
0.
7
³e
1 3 x
x3 , du
2 x3
³ x e dx
97. Let u
n
12
12!
12 ˜ 11 ˜ 10 " 3 ˜ 2 ˜ 1
479,001,600
2
e x
³ 1 e x dx
15!
15 ˜ 14 " 3 ˜ 2 ˜ 1
1,307,674,368,000
Stirlings Formula:
15
§ 15 ·
15! | ¨ ¸
©e¹
2S 15 | 1,300,430,722,200
| 1.3004 u 10
12
5 x, du
5 dx.
3
3
C
dx.
2e x C
2
dx.
x3
³
100. Let u
1 2
e1 x C
2
e x dx.
e x
dx
1 e x
1 e 2 x , du
e2 x
³ 1 e2 x dx
ln 1 e x C
101. Let u
³e
x
x 4 , du
4 x3dx.
x4
4 x 3 dx
e x C
4
102. Let u
1 e x , du
1 e x dx
e x e x
1
ln 1 e2 x C
2
e x dx.
³ 1 e x
12
e x dx
23 1 e x
32
C
e x e x , du
³ e x e x dx
103. Let u
2e 2 x dx.
1 2e 2 x
dx
2 ³ 1 e2 x
e5 x C
92. Let u
³e
x
ex 1
x ln e x 1 C
15
5x
³ e 5 dx
1
2
x
§ ex ·
ln ¨ x
¸C
© e 1¹
2S 12 | 475,687,487
n
91. Let u
³ e 1 e dx
1 e x , du
12
90.
2
1 1 x2 § 2 ·
e ¨ 3 ¸ dx
2³
©x ¹
Stirlings Formula:
§ 12 ·
12! | ¨ ¸
©e¹
e x dx.
x
1
, du
x2
99. Let u
1 e x3 C
3
3 x 2 dx
§ 1 ·
2³ e x ¨
¸ dx
©2 x ¹
e x
dx
x
−1
89.
x3
x , du
³ x3 dx
6
³e
2
x
e1 x
−6
1
3
13 e1 3 x C
3x 2 dx.
³ e e 1 dx
³
1 e 2 x 1 C
2
2 dx
3 dx.
e x 1, du
98. Let u
P1
2 x 1
13 ³ e1 3 x 3 dx
dx
95. Let u
f
P2
³e
2 dx.
1 3 x, du
94. Let u
x
P2c 0
1
2
dx
96. Let u
x2
x
1
8
2
1
1
x ,
4
2
1
,
4
P2c x
P1 0
2 x 1, du
e x e x dx.
ln e x e x C
e x e x , du
e x e x dx.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.4
e x e x , du
104. Let u
2e x 2e x
³ e x e x 2 dx
Exponential Functions: Differentiation an
and Integration
e x e x dx.
2³ e x e x
2
3
, du
x
113. Let u
e x e x dx
3 e3 x
³ 1 x 2 dx
5 ex
dx
e2 x
³ 5e
2 x
dx ³ e x dx
³ e 2e
x
³0
x
1
109. ³ e 2 x dx
0
115. Let u
³ ª¬tan e x º¼ e x dx
1
111. ³ xe
x2
dx
0
0
2
³
3
1 e 2 x , du
2e 2 x
³ 0 1 e2 x dx
2
0
2
e x 2 x dx
2
1 e 1
116. Let u
3
ªln 1 e 2 x º
¬
¼0
1
5 e x , du
ex
³ 0 5 e x dx
1
e 1
e
2e 2 x dx.
ln 1 e6 ln 2
1
csc e 2 x 2e 2 x dx
2³
1
ln csc e2 x cot e 2 x C
2
³
1
§ 1 e6 ·
ln ¨
¸
© 2 ¹
e x dx.
1
0 5 ex
e x dx
1
ª 1 2 x º
« 2 e »
¬
¼0
1 1 2 x
2 dx
e
2³0
x dx.
ªln 5 e x º
¬
¼0
ln 5 e ln 4
e2 1
2e 2
§ 4 ·
ln ¨
¸
©5 e¹
2
2
110. ³ e5 x 3 dx
112. ³
2
xe x 2 dx
dx
1
1 e 2
2
1
2
e 2
e 1
3
ªe x2 2 º
«¬
»¼ 0
ln cos e x C
108. ³ e2 x csc e 2 x dx
x2
, du
2
114. Let u
e x 2 x e x C
107. ³ e x tan e x dx
1 3 3 x§ 3 ·
e ¨ 2 ¸ dx
3 ³1
© x ¹
3
5
e 2 x e x C
2
e 2 x 2e x 1
106. ³
dx
ex
3
dx.
x2
ª 1 3 xº
« 3 e »
¬
¼1
2
C
x
e e x
105. ³
491
3
x 2e x 2 dx
ª1 5 x 3 º
«5 e
»
¬
¼1
117. Let u
1 7
e e 2
5
³0
sin S x, du
S 2 sin S x
e
1
S 2 sin S x
1
S 2
S ³0
cos S x dx
e
S cos S x dx
ªesin S x º¼
0
S¬
1 1 2
³ e x 2 x dx
2 0
1 ª x2 º1
«e »
¼0
2¬
1
ª¬e 1 1º¼
2
1 1e
e 1
2
2e
º
1 ª sin S 2 2
1»
e
S «¬
¼
118. Let u
sec 2 x, du
S 2 sec2 x
³S 3 e
2 sec 2 x tan 2 x dx.
sec 2 x tan 2 x dx
2 0 x3 2 § 3 2 ·
e ¨ x ¸ dx
3 ³ 2
©2 ¹
2 ª x3 2 º 0
e »
¼ 2
3 «¬
2
ª1 e4 º¼
3¬
2ª
1º
1 4»
«
3¬
e ¼
S cos S x dx.
1 S 2 sec2 x
e
2sec 2 x tan 2 x dx
2 ³S 3
1 sec 2 x S 2
ªe
º¼
S 3
2¬
1 1
ªe e2 º¼
2¬
1 ª1 1 º e 1
2 «¬ e e2 »¼
2e 2
2 e4 1
INSTRUCTOR USE ONLY
3e 4
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492
NOT FOR SALE
Chapter 5
119. (a)
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
y
2
³ e e dx
2 x
2x
³ e 2 e dx
x
x
122. y
5
1 e 2 x 2 x 1 e 2 x C
2
2
(0, 1)
x
−2
−2
dy
(b)
dx
y
2e x 2 ,
0, 1
³ 2e
dx
4e
x 2
x 2
5
4e x 2 5
y
6
−4
1 e x e x
2
C1
1 e x e x dx
2
1 e x e x
2
C2
f x
³
f 0
1 C2
f x
1 e x e x
2
124. f c x
³ sin x e
2x
dx
fc 0
1 12 C1
1
2
fc x
cos x 12 e 2 x 1
f x
³ cos x 12 e
0
1 Ÿ C2
2x
0
cos x 12 e 2 x C1
Ÿ C1
1
1 dx
sin x 14 e 2 x x C2
8
f 0
1 C
2
4
f x
x sin x 14 e 2 x
−2
120. (a)
x
C1
C
4 C Ÿ C
dx
x
fc 0
§ 1 ·
4 ³ e x 2 ¨ dx ¸
© 2 ¹
4e0 C
0, 1 : 1
³ 12 e e
123. f c x
5
y
Ÿ C2
1
4
0
4
5
5
125. ³ e x dx
ª¬e x º¼
0
0
e5 1 | 147.413
x
−4
150
4
−4
(b)
2
3·
§
xe0.2 x , ¨ 0, ¸
2¹
©
2
2
1
0.2 x
dx
e 0.2 x 0.4 x dx
³ xe
³
0.4
2
1 0.2 x2
e
C
2.5e0.2 x C
0.4
dy
dx
y
3· 3
§
¨ 0, ¸: 2¹ 2
©
2.5e0 C
2.5 C Ÿ C
3
3
³ 1 e
126. A
2 x
ª 1 2 x º
« 2 e »
¬
¼ 1
dx
1 6
e e 2 | 3.693
2
8
1
4
0
4
127. ³
−6
6
0
−2
2
2.5e 0.2 x 1
y
0
6
0
ª2e x2 4 º
¬«
¼» 0
2
xe x 4 dx
6
2e 3 2 2 | 1.554
6
3
−4
− 4.5
121. Let u
y
ax 2 , du
³ xe
ax 2
2ax dx.
4.5
Assume a z 0.
−3
dx
1 ax2
2ax dx
e
2a ³
1 ax2
e
C
2a
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.4
128. ³
2
0
134.
2
ª¬ 12 e 2 x 2 xº¼
0
e 2 x 2 dx
Exponential Functions: Differentiation an
and Integration
12 e 4 4 12 | 4.491
4
t
0
1
2
3
4
R
425
240
118
71
36
ln R
6.052
5.481
4.771
4.263
3.584
0.6155t 6.0609
(a) ln R
e 0.6155t 6.0609
R
−2
(b)
4
493
428.78e 0.6155t
450
0
129. ³
4
xe x dx, n
0
12
−1
Midpoint Rule:
92.1898
Trapezoidal Rule: 93.8371
Simpson's Rule: 92.7385
Graphing utility: 92.7437
2
130. ³ 2 xe x dx, n
48
e
lim e x
x o f
x
ª¬e 0.3t º¼
0
e 0.3 x 1
e
0.3 x
Let z
8
3
a
³ ae
Ce x , C a constant.
(b) Substitution: u
138. (a)
ex 1
x2
4
−4
5
−2
(b) When x increases without bound, 1 x approaches
ln
x
xof
137. (a) Log Rule: u
1
ln 2
2
ln 2
| 2.31 minutes
0.3
0.3 x
133. Area
47.72%
f.
0 and lim e x
136. Yes. f x
dt
1
2
1
2
1
2
1
2
x
0
dt
continuous, increasing, one-to-one, and concave upward
on its entire domain.
Graphing utility: 0.4772
132. ³ 0.3e 0.3t dt
0.6155t
e x . Domain is f, f and range is 0, f . f is
135. f x
60 0.0139 t 48 2
4
³ 0 428.78e
| 637.2 liters
Midpoint Rule: 1.1906
Trapezoidal Rule: 1.1827
Simpson's Rule: 1.1880
Graphing utility: 1.18799
131. 0.0665³
4
³ 0 R t dt
(c)
12
0
5
0
x
a
e x º¼
a
dx
zero, and e1 x approaches 1. Therefore, f x
approaches 2 1 1
asymptote at y
1. So, f x has a horizontal
1. As x approaches zero from the
right, 1 x approaches f, e1 x approaches f and
e a e a
f x approaches zero. As x approaches zero from
a
e :
the left, 1 x approaches f, e1 x approaches zero,
8
3
1
z
z
8
z
3
3z 2 8 z 3
1 z
3z 1 z 3
z
§
¨z
©
1
Ÿ ea
3
So, a
ln 3.
and f x approaches 2. The limit does not exist
because the left limit does not equal the right limit.
Therefore, x
0 is a nonremovable discontinuity.
2
0
3 Ÿ ea
x
x
0
0
139. ³ et dt t ³ 1 dt
0
ª¬et º¼ t >t@ 0
0
x
3 Ÿ a
1
·
impossible ¸
3
¹
ln 3
x
e x 1 t x Ÿ e x t 1 x for x t 0
INSTRUCTOR USE ONLY
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NOT FOR SALE
494
Chapter 5
140.
e x
x Ÿ f x
fc x
1 e x
xn 1
xn Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
x e x
Ae kt Be kt ; A, B, k ! 0
141. x t
Ake kt Bke kt
(a) xc t
f xn
f c xn
xn f x1
xn
xn e
1 e xn
Ae
e 2 kt
x1
1
x2
x1 | 0.5379
2kt
t
x3
f x2
x2 | 0.5670
f c x2
x4
f x3
x3 | 0.5671
f c x3
f c x1
0
Be kt
B
A
§B·
ln ¨ ¸
© A¹
kt
1 §B·
ln ¨ ¸
2k © A ¹
By the first Derivative Test, this is a minimum.
Ak 2e kt Bk 2e kt
(b) xcc t
k 2 Ae kt Be kt
Approximate the root of f to be x | 0.567.
k2x t
k 2 is the constant of proportionality.
ln x
x
142. f x
1 ln x
0 when x
e.
x2
On 0, e , f c x ! 0 Ÿ f is increasing.
(a) f c x
On e, f , f c x 0 Ÿ f is decreasing.
(b) For e d A B, you have
ln A
ln B
!
A
B
B ln A ! A ln B
ln AB ! ln B A
AB ! B A .
(c) Because e S , from part (b) you have eS ! S e .
y
1
2
2
e
x
4
6
8
− 12
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.5
143.
Bases Other than e and Applications
495
L
, a ! 0, b ! 0, L ! 0
1 ae x b
§ a
·
aL x b
L¨ e x b ¸
e
© b
¹
b
y
yc
2
1 ae x b
1 ae x b
2
2 § aL
· § aL
·
§ a
·
1 ae x b ¨ 2 e x b ¸ ¨ e x b ¸ 2 1 ae x b ¨ e x b ¸
© b
¹ © b
¹
© b
¹
ycc
1 ae x b
§ aL
·
§ aL
·§ a
·
1 ae x b ¨ 2 e x b ¸ 2¨ e x b ¸¨ e x b ¸
© b
¹
© b
¹© b
¹
Lae x b ª¬ae x b 1º¼
1 ae x b
1 ae x b b 2
0 if ae x b
ycc
4
x
b
1Ÿ
L
y b ln a
1 ae b ln a b
3
3
§1·
ln ¨ ¸ Ÿ x
©a¹
L
1 a1a
b ln a
L
2
Therefore, the y-coordinate of the inflection point is L 2.
Section 5.5 Bases Other than e and Applications
1. log 2 18
log 2 23
2. log 27 9
log 27 27 2 3
3
2
3
(b)
3. log 7 1
4. log a
5. (a)
(b)
0
1
a
log a 1 log a a
23
8
log 2 8
3
1
3
1
3
log 3 13
1
27 2 3
9
log 27 9
2
3
163 4
8
log16 8
3
4
2
8. (a) log 3 19
1
32
1
9
491 2
7
log 49 7
1
2
2x
9. y
x
–2
1
0
1
2
y
1
4
1
2
1
2
4
y
5
4
6. (a)
(b)
(b) log 0.5 8
0.5
1
2
3
3
2
x
−3 −2 −1
−1
1
2
3
2
7. (a) log10 0.01
10
3
2
0.01
3
8
8
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
496
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
4 x 1
10. y
5x 2
13. h x
x
–1
0
1
2
3
x
–1
0
1
2
3
y
1
16
1
4
1
4
16
y
1
125
1
25
1
5
1
5
y
y
16
14
4
12
10
3
8
2
6
4
1
2
x
−2
−1
1
x
y
3
x
3 x
–2
1
1
3
11. y
2
9
x
4
1
3
1
2
1
1
3
1
9
3
4
3 x
14. y
0
2
x
0
r1
r2
y
1
1
3
1
9
y
y
1
4
3
x
−1
1
2
−1
−2
x
−1
1
2
15. f x
2x
12. y
2
3x
Increasing function that passes through 0, 1 and
1, 3 . Matches (d).
x
–2
1
0
1
2
y
16
2
1
2
16
16. f x
3 x
Decreasing function that passes through 0, 1 and
y
1, 13 . Matches (c).
8
6
17. f x
4
3x 1
Increasing function that passes through 0, 0 and
2
1, 2 . Matches (b).
−2
−1
x
1
2
18. f x
3x 1
Increasing function that passes through 1, 1 and
2, 3 . Matches (a).
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.5
x
19. (a) log10 1000
x
1000
x
3
10
10 x
0.1
x
1
x
x
1
81
x
4
(b) log 6 36
x
6x
36
x
2
log 3 ¬ª x x 2 º¼
1
x x 2
31
x2 2 x 3
0
x 1 x 3
0
1 OR x
x
(b) log10 x 3 log10 x
1
1
x
3
9x
x
1
3
x
1
3
3
4
x
x
1
16
22. (a) log b 27
3
25.
27
b
3
(b) log b 125
3
26.
b3
125
b
5
log 5 25
x2 x
log 5 52
x2 x 2
0
x 1 x 2
0
x
(b) 3x 5
log 2 64
3x 5
log 2 26
1
x
1
3
75
2 x ln 3
ln 75
1
101
10 x
§ 1 · ln 75
| 1.965
¨ ¸
© 2 ¹ ln 3
56 x
8320
6 x ln 5
ln 8320
ln 8320
| 0.935
6 ln 5
x
x2 x
3x
32 x
x
b3
23. (a)
log10
4
(b) log 2 x
3
1
x 3
x
x 3
x
x 3
21. (a) log 3 x
2
1
497
3 is the only solution because the domain of the
x
logarithmic function is the set of all positive real
numbers.
1
log 3 81
3
24. (a) log 3 x log 3 x 2
x
(b) log10 0.1
20. (a)
Bases Other than e and Applications
27.
2
1 OR x
6
23 z
3 z ln 2
ln 625
3 z
ln 625
ln 2
z
2
28.
625
3
3 5 x 1
86
5 x 1
86
3
x 1 ln 5
x 1
x
ln 625
| 6.288
ln 2
§ 86 ·
ln ¨ ¸
©3¹
ln 86 3
ln 5
1
ln 86 3
ln 5
| 3.085
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
498
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
12 t
29.
0.09 ·
§
¨1 ¸
12 ¹
©
3
0.09 ·
§
12t ln ¨1 ¸
12 ¹
©
ln 3
30.
0.10 ·
§
¨1 ¸
365 ¹
©
5
x 1
25
x
33
32. log10 t 3
2.6
t 3
102.6
4.5
3
x
r
x
1
0
1
2
1
f x
1
16
1
4
1
2
4
x
1
16
1
4
1
2
4
g x
–2
–1
0
1
2
1
y
ln 2
f
3
ln 2
§ 1 ·
| 6.932
¨
¸
365
§
©
¹ ln 1 0.10 ·
¨
¸
365 ¹
©
2
g
x
−1
2
32
3
3x
36. f x
3 10
4.5
–2
−1
31. log 2 x 1
2
x
2
t
33. log 3 x 2
log 4 x
365t
0.10 ·
§
365t ln ¨1 ¸
365 ¹
©
t
g x
ln 3
§1·
| 12.253
¨ ¸
© 12 ¹ ln §1 0.09 ·
¨
¸
12 ¹
©
t
4x
35. f x
2.6
g x
log 3 x
x
–2
1
0
1
2
f x
1
9
1
3
1
3
9
x
1
9
1
3
1
3
9
g x
–2
–1
0
1
2
| 401.107
34.5 | r11.845
y
34. log 5
x 4
3.2
x 4
53.2
x 4
53.2
x
9
f
6
2
56.4
3
45
g
6.4
x
−3
| 29,748.593
3
6
9
−3
37.
38.
f x
4x
fc x
ln 4 4 x
f x
34 x
fc x
4 ln 3 34 x
39. y
yc
4 ln 3 81x
5 4x
4 ln 5 5 4 x
4 ln 5
625 x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.5
40. y
41.
6 3x 4
49. h t
3x 4
yc
3 ln 6 6
f x
x9 x
fc x
x ln 9 9 x 9 x
hc t
50. g t
9 x 1 x ln 9
42.
y
x 62 x
dy
dx
x ª¬2 ln 6 62 x º¼ 62 x
51.
62 x 1 2 x ln 6
gc t
52.
t 2 2t
t 2 ln 2 2t 2t 2t
2
2 log 5 4 t
1
ln 5 4 t
2
t 4 ln 5
log 5 4 t
2
log 2 t 2 7
3
y
log 5
f x
fc x
t 2t t ln 2 2
f t
fc t
3
t
t2
3
54.
2t ln 3 1
t2
45. h T
hc T
1
log 2 2 x 1
3
1
1
2
2
3 2 x 1 ln 2
3 2 x 1 ln 2
gc D
47. y
yc
y
48. y
yc
log 3 x 2 3 x
1
2x 3
x 2 3 x ln 3
x2 1
log10 x 2 1 log10 x
x
2x
1
x ln 10
x 2 1 ln 10
log10
1 ª x2 1 º
«
»
ln 10 « x x 2 1 »
¬
¼
55. h x
log 4 5 x 1
5
ln 4 5 x 1
x 2
ln 2 x x 1
1 ª 2x
1º
ln 10 «¬ x 2 1 x »¼
5D 2 sin 2D
1
5
5 x 1 ln 4
2 log 2 x log 2 x 1
2
1
x ln 2
x 1 ln 2
2T S sin ST ln 2 2T cos ST
5D 2 2 cos 2D 12 ln 5 5D 2 sin 2D
x2
x 1
fc x
2T ª¬ ln 2 cos ST S sin ST º¼
46. g D
x2 1
log 2
dy
dx
2T cos ST
6t
t 2 7 ln 2
f x
t 2 ln 3 32t 32t
2t
3 log 2 t 2 7
log 2 3 2 x 1
2t
44.
499
1
log 5 x 2 1
2
x
1
2x
˜
2 x 2 1 ln 5
x 2 1 ln 5
2t t 2 t ln 2
53.
3
2t
ln 2 t 2 7
gc t
dy
dx
62 x ª
¬ 2 x ln 6 1º¼
43. g t
Bases Other than e and Applications
hc x
x 1
2
1
log 3 x log 3 x 1 log 3 2
2
1
1
1
˜
0
2
x ln 3
x 1 ln 3
log 3
x
º
1 ª1
1
« »
ln 3 «¬ x
2 x 1 »¼
1 ª 3x 2 º
«
»
ln 3 «¬ 2 x x 1 »¼
2x 3
x x 3 ln 3
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
500
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
4
§
·
log 5 ¨ 2
¸
©x 1 x¹
56. g x
62. y
log 5 4 log 5 x log 5
yc
1 x
2
1
log 5 4 2 log 5 x log 5 1 x
2
1
1
1
2
1
x ln 5 2 1 x ln 5
gc x
log10 2 x ,
1
x ln 10
1
.
5 ln 10
At 5, 1 , yc
63.
1 § dy ·
¨ ¸
y © dx ¹
10
>1 ln t@
t 2 ln 4
5
1 ln t
t 2 ln 2
64.
58.
t 1
t 3 2 log 2
fc t
1 ª 32 1
3
º
t
t1 2 ln t 1 »
2 ln 2 «¬ t 1 2
¼
2 x ,
yc
x
t3 2
1 ln t 1
2 ln 2
f t
59. y
y
ln y
10 ª t 1 t ln t º
«
»
t2
ln 4 ¬
¼
gc t
y
x x 1
2 ln 2 .
y
yc
x2
5
65.
dy
dx
ªx 1
º
y«
ln x»
¬ x
¼
yc
1
x ln 3
At 27, 3 , yc
y
dy
dx
x ln 5 1 2 ln 5
66.
y
ln y
1 § dy ·
¨ ¸
y © dx ¹
dy
dx
1
.
27 ln 3
y
x 1
x 1 ln x 2
§ 1 ·
x 1¨
¸ ln x 2
© x 2¹
ªx 1
º
ln x 2 »
y«
¬x 2
¼
x 1 ª x 1
º
« x 2 ln x 2 »
¬
¼
ln 5 x 2
27, 3
Tangent line: y 3
x 2
x 2
ln 5.
y
log 3 x,
x 1 ln x
§1·
x 1 ¨ ¸ ln x
© x¹
1 § dy ·
¨ ¸
y © dx ¹
2 x ln 2 2 2 ln 2
2, 1
Tangent line: y 1
61. y
2 x 2/ x 2 1 ln x
1 § dy ·
¨ ¸
y © dx ¹
ln y
2 ln 2 x 1
ln 5
At 2, 1 , yc
2
1 ln x
x2
x x 2 x 1 x ln x
Tangent line: y 2
5x 2 ,
2
ln x
x
2§ 1 ·
§ 2·
¨ ¸ ln x¨ 2 ¸
x© x ¹
© x ¹
2y
1 ln x
x2
2 ln 2
60. y
x2 x
dy
dx
ln y
1, 2
At 1, 2 , yc
1
1
x 1
5 ln 10
ln 10
y
10 § ln t ·
¨
¸
ln 4 © t ¹
10 log 4 t
t
1
x 5
5 ln 10
Tangent line: y 1
2
1
x ln 5
2 1 x ln 5
57. g t
5, 1
1
x 27
27 ln 3
1 x
1x
1
ln 1 x
x
1§ 1 ·
§ 1·
¨
¸ ln 1 x ¨ 2 ¸
x ©1 x ¹
© x ¹
ln x 1 º
yª 1
«
»
x ¬x 1
x
¼
1 x
x
1x
ln x 1 º
ª 1
«
»
x
1
x
¬
¼
1
1
x 3
27 ln 3
ln 3
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
Section 5.5
67.
ln y
§S S ·
¨ , ¸
©2 2¹
sin x ln x
yc
y
sin x
cos x ln x
x
y
yc
1
S 2
S
y
sin x
ln y
2x
ln y
yc
y
ln x
x4 2
6 x4
C
2 ln 6
1
cos x
77. ³
e, 1
,
yc
y
cos e
x e
e
cos e
x 1 cos e
e
1, 1
At 1, 1 , yc
1 32 x , du
2 ln 3 32 x
1
dx
³
2 ln 3 1 32 x
2 ln 3 32 x dx
1
ln 1 32 x C
2 ln 3
sin x, du
cos x dx
sin x
2
C
ln 2
2
2
79. ³ 2 x dx
1
80. ³
1
ln x
x
1
ln x
2
2
x
x
32 x
dx, u
1 32 x
78. ³ 2sin x cos x dx, u
1
0.
e
y
ln y
C
2
cos e
x1 x ,
2
1
6 x 4 2 x 4 dx
³
2
§ 1 ·6
¨ ¸
© 2 ¹ ln 6
1
cos x ˜
sin x ˜ ln ln x
x ln x
y
2
§ 1 · 5 x
¨ ¸
C
© 2 ¹ ln 5
1 x2
C
5
2 ln 5
2
Tangent line: y 1
70.
1 x2
5
2 x dx
2³
dx
76. ³ x x 4 6 x 4 dx
cos x ˜ ln ln x
At e, 1 , yc
2
0.
Tangent line: y
y
75. ³ x 5 x
§S ·
, ¨ , 1¸
©2 ¹
2x
cos x 2 ln sin x
sin x
§S ·
At ¨ , 1¸, yc
©2 ¹
69.
0
2 x ln sin x
yc
y
x5
5x
C
5
ln 5
74. ³ x 4 5 x dx
S·
§
1¨ x ¸
2¹
©
x
2
y
68.
1 3
2 x
C
x 3
ln 2
1
Tangent line: y 501
x3
1 x
2 C
3
ln 2
73. ³ x 2 2 x dx
xsin x ,
yc
§S S ·
At ¨ , ¸ :
©2 2¹ S 2
Bases Other than e and Applications
4
4
3x 4 dx
ª 2x º
«
»
¬ ln 2 ¼ 1
4³
4
4
1 ª
1º
4 »
ln 2 «¬
2¼
7
2 ln 2
7
ln 4
§1 ·
3x 4 ¨ dx ¸
©4 ¹
4
ª 1 x 4º
3 »
«4
¬ ln 3
¼4
10
Tangent line: y 1
y
71. ³ 3x dx
3x
C
ln 3
72. ³ 8 x dx
1.
1x 1
x
4
3 31
ln 3
32
3 ln 3
8 x
C
ln 8
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
502
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
3
1
ª 5x
3x º
«
»
¬ ln 5 ln 3 ¼ 0
1
81. ³ 5 x 3x dx
0
82. ³
3
ª 7x
4x º
«
»
ln 4 ¼1
¬ ln 7
7 x 4 x dx
1
§ 5
3 · § 1
1 ·
¨
¸¨
¸
ln
5
ln
3
ln
5
ln
3¹
©
¹ ©
4
2
ln 5 ln 3
3
ª 1 xº
3 »
«
¬ ln 3 ¼ 0
3
x
³ 3 dx
83. Area
0
1
27 1
ln 3
§ 73
43 · § 7
4 ·
¨
¸ ¨
¸
ln
7
ln
4
ln
7
ln
4¹
©
¹ ©
336
60
ln 7
ln 4
26
| 23.6662
ln 3
y
30
20
10
x
1
84. Area
2
S
3
³0 3
cos x
sin x dx
S
ª 3cos x º
«
»
¬ ln 3 ¼ 0
1 1
ª3 3º¼
ln 3 ¬
8
| 2.4273
3 ln 3
y
2
1
π
2
85. f x
x
π
log10 x
86. f x
(a) Domain: x ! 0
(c)
x x Ÿ gc x
log10 x
g x
10 y
x
Note: Let y
f 1 x
10 x
ln y
ln x x
1
yc
y
x˜
y
(b)
log10 1000
log10 103
3
log10 10,000
4
4
log10 10
If 1000 d x d 10,000, then 3 d f x d 4.
(d) If f x 0, then 0 x 1.
(e) f x 1
log10 x log10 10
log10 10 x
x must have been increased by a factor of 10.
§x ·
(f) log10 ¨ 1 ¸
© x2 ¹
log10 x1 log10 x2
3n n
So, x1 x2
102 n
1
x ln 2
log 2 x Ÿ f c x
x x 1 ln x
g x . Then:
x ln x
1
ln x
x
yc
y 1 ln x
yc
x x 1 ln x
gc x
h x
x Ÿ hc x
2x
k x
2 Ÿ kc x
ln 2 2 x
2
x
From greatest to least rate of growth:
g x,k x,h x, f x
2n
100n.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.5
87. C t
P 1.05
(a) C 10
t
24.95 1.05
dC
dt
10
(a)
P ln 1.05 1.05
When t
1,
When t
dC
(c)
dt
§ 3·
25,000¨ ¸
© 4¹
88. V t
| $40.64
(b)
Bases Other than e and Applications
t
25,000
t
(2, 14,062.5)
dC
| 0.051P.
dt
8,
503
0
12
0
V 2
dC
| 0.072 P.
dt
ln 1.05 ªP 1.05 º
¬
¼
t
(b)
ln 1.05 C t
The constant of proportionality is ln 1.05.
(c)
§ 3·
25,000¨ ¸
© 4¹
2
$14,062.50
t
§ 3 ·§ 3 ·
25,000¨ ln ¸¨ ¸
© 4 ¹© 4 ¹
dV
When t 1,
| 5394.04.
dt
dV
When t
4,
| 2275.61.
dt
dV
dt
0
0
12
−8000
Horizontal asymptote: V c
0
As the car ages, it is worth less each year and
depreciates less each year, but the value of the car
will never reach $0.
89. P
$1000, r
3 12 %
0.035, t
10
10 n
A
0.035 ·
§
1000¨1 ¸
n ¹
©
A
1000e 0.035 10
n
1
2
4
12
365
Continuous
A
$1410.60
$1414.78
$1416.91
$1418.34
$1419.04
$1419.07
90. P
$2500, r
0.06, t
6%
A
0.06 ·
§
2500¨1 ¸
n ¹
©
A
2500e 0.06 20
20
20 n
n
1
2
4
12
365
Continuous
A
$8017.84
$8155.09
$8226.66
$8275.51
$8299.47
$8300.29
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
504
Chapter 5
91. P
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Function
Functions
$1000, r
0.05, t
5%
A
0.05 ·
§
1000¨1 ¸
n ¹
©
A
1000e 0.05 30
30
30 n
n
1
2
4
12
365
Continuous
A
$4321.94
$4399.79
$4440.21
$4467.74
$4481.23
$4481.69
92. P
$4000, r
0.04, t
4%
15
15 n
A
0.04 ·
§
4000¨1 ¸
n ¹
©
A
4000e0.04 15
n
1
2
4
12
365
Continuous
A
$7203.77
$7245.45
$7266.79
$7281.21
$7288.24
$7288.48
93. 100,000
100,000e 0.05t
Pe0.05t Ÿ P
t
1
10
20
30
40
50
P
$95,122.94
$60,653.07
$36,787.94
$22,313.02
$13,533.53
$8208.50
94. 100,000
100,000e 0.03t
Pe0.03t Ÿ P
t
1
10
20
30
40
50
P
$97,044.55
$74,081.82
$54.881.16
$40,656.97
$30,119.42
$22,313.02
95. 100,000
P 1 0.05
12
12t
Ÿ P
100,000 1 0.05
12
12 t
t
1
10
20
30
40
50
P
$95,132.82
$60,716.10
$36,864.45
$22,382.66
$13,589.88
$8251.24
96. 100,000
P 1 0.02
365
365t
Ÿ P
100,000 1 0.02
365
365t
t
1
10
20
30
40
50
P
$98,019.92
$81,873.52
$67,032.74
$54,882.07
$44,933.88
$36,788.95
97. (a) A
20,000 1 0.06
365
365 8
| $32,320.21
(b) A
$30,000
(c) A
8000 1 0.06
365
365 8
20,000 1 0.06
365
9000 ª« 1 0.06
365
¬
Take option (c).
365 8
1 0.06
365
(d) A
365 4
365 4
| $12,928.09 25,424.48
$38,352.57
1º» | $34,985.11
¼
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.5
98. Let P
$100, 0 d t d 20.
A
(a)
100e0.03t
(a)
A
505
10,000
1 19e t 5
101. p t
A 20 | 182.21
(b)
Bases Other than e and Applications
12,000
100e0.05t
A 20 | 271.83
0
A
(c)
40
0
100e0.06t
(b) Limiting size: 10,000 fish
A 20 | 332.01
10,000
1 19e t 5
et 5
pt
(c)
400
A = 100 e 0.06t
A = 100 e 0.05t
pc t
A = 100 e 0.03t
0
t of
6.7 million ft 3 /acre
(d) pcc t
ª
º
38,000 t 5 « 1 19e t 5 »
e
« 1 19e t 5 3 »
5
¬
¼
0
19e t 5
1
V c 60 | 0.040 million ft /acre/yr
t
5
t
3
0.86
1 e 0.25n
0.86
(a) lim
n o f 1 e 0.25 n
(b) In the long run,
2
pc 10 | 403.2 fish/month
6.7e0
322.27 48.1 t
(b)
Vc
e
t2
V c 20 | 0.073 million ft 3 /acre/yr
100. P
1 19e t 5
pc 1 | 113.5 fish/month
20
0
99. (a) lim 6.7e 48.1 t
1 19e t 5
38,000e t 5
§ 19 ·
¸ 10,000
©5¹
2¨
ln19
5 ln19
| 14.72
0.86
1
0.86, or 86%
102. (a) B
dP
o 0. The graph gets flatter.
dn
(b)
4.7539 6.7744
d
4.7539e1.9132 d
120
2
0
0
(c)
Bc d
9.0952e1.9132 d
Bc 0.8 | 42.03 tons/inch
Bc 1.5 | 160.38 tons/inch
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
506
Chapter 5
40.0 x 743
103. (a) y1
(b)
NOT FOR SALE
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Function
Functions
105.
y2
968 265.5 ln x
y3
836.817 0.917
y4
1344.888 x 0.569
x
t
0
1
2
3
4
y
1200
720
432
259.20
155.52
C kt
y
When t
700
720
1200
Let k
12
0
300
y
700
1200.
1200 k t
y
y1
1200 Ÿ C
0, y
0.6,
432
720
0.6,
259.20
432
0.6,
155.52
259.20
0.6
0.6.
t
1200 0.6
y2
106.
12
0
300
0
1
2
3
4
y
600
630
661.50
694.58
729.30
600 Ÿ C
600.
C kt
y
700
y3
700
y4
0, y
y
t
600 k
y
12
0
300
When t
30
661.50
1.05,
600
630
729.30
| 1.05
694.58
Let k 1.05.
12
0
300
Answers will vary.
600 1.05
51 ln 5
107.
(c) The slope 40.0 indicates that the number of
1
>ln 5@
ln 5
ln x
y1c | 40.0
1
ln x
y2c | 33.2
x
e
8 and
y3c | 36.3
108.
y4c | 29.3
x
1 x
1x
6ln 10 ln 6
ln ª¬6
y1 is decreasing at the greatest rate.
x
º¼
ln x
ln 10
ln 6
ln 6
ln x
ln 10 ln 6
1
101
102
104
106
ln 10
ln x
2
2.594
2.705
2.718
2.718
x
10
109.
694.58
| 1.05,
661.50
x
ln x
(d) For 2008, x
1.05,
t
ln ª¬51 ln 5 º¼
transplants decreases by 40 each year.
104.
t
91 ln 3
ln ª¬9
º¼
1 ln 3
x
ln x
1
ln 32
ln 3
ln x
1
2 ln 3
ln 3
ln x
ln x
2
INSTRUCTOR USE ONLY
x
e2
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.5
321 ln 2
110.
ln ª¬32
x
º¼
ln x
1
ªln 25 º¼
ln 2 ¬
ln x
5
ln x
x
5
1 ln 2
117. (a)
Bases Other than e and Applications
23
2
2
82
64
29
512
32
e
g x
x
ln e n 1 ln e n
n 1 n
x
g f x
ln 2 e x 2
2 x2
ln e x
2, f 2
y for n
g 2
x
ln y
x 2 ln x
x
ex
d
e and ª¬e x º¼
dx
x
e x when x
e0 e 0
e
x
116. True
0 Ÿ
f x
g x ex
g x
0 because e x ! 0 for all x.
x
§1·
2 x ln x x 2 ¨ ¸
© x¹
yc
x
x2
x
x2
x 2 ln x 1
2
x x 1 2 ln x 1
g x
ln y
x x ln x
(Note:
d x
ªx º
dx ¬ ¼
yc
y
x x 1 ln x ln x x x
yc
x
x
xx
x
xx
y
0.
1
16.
yc
y
1, 2, 3, !
115. True
19683, whereas
xx
f x
(ii)
d x
ªe º
dx ¬ ¼
39
y
114. True
Ce x
and
are not the same.
113. True
2 eln x 2
x2
x
1
(c) (i)
f g x
x
327. Note that when
g 3
112. True
f e n 1 f e n
xx
For example, f 3
111. False. e is an irrational number.
dny
dx n
xx
(b) In general, f x
507
x x 1 ln x from Example 5)
1
x
1º
ª
˜ x x « 1 ln x ln x »
x¼
¬
x x x 1
ª x ln x 2 x ln x 1º
¬
¼
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
508
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
118.
y
ax 1
ax 1
yc
f x
a x 1 a x ln a a x 1 a x ln a
ax 1
2
2a x ln a
2
ax 1
For 0 a 1, yc 0 Ÿ one-to-one and has an inverse
For a ! 1, yc ! 0 Ÿ one-to-one and has an inverse
y ax 1
ax 1
ax y 1
1 y
ax
y 1
1 y
x ln a
§ y 1·
ln ¨
¸
©1 y ¹
x
§ y 1·
1
ln ¨
¸
ln a © 1 y ¹
f 1 x
1
§ x 1·
ln ¨
¸
ln a © 1 x ¹
8 §5
·
y¨ y ¸, y 0
25 © 4
¹
dy
dt
119.
dy
y ª¬ 5 4 yº¼
8
dt
25
·
4 §1
1
¨ ¸ dy
5 ³ ¨© y
5 4 y ¸¹
³ 25 dt
§5
·
ln y ln ¨ y ¸
©4
¹
2
t C
5
§
·
y
ln ¨
¨ 5 4 y ¸¸
©
¹
y
54 y
2
t C
5
When t
0, y
8
e 2 5 t C
1 Ÿ C1
C1e 2 5 t
y
54 y
4 Ÿ 4e 2 5 t
§5
·
4e 2 5 t ¨ y ¸
©4
¹
5e 2 5 t
y
120. f x
1
y
4e 2 5 t y y
5e 2 5 t
4e
25t
1
4e 2 5 t 1 y
5
4 e 0.4t
1.25
1 0.25e 0.4t
ax
au v
(a) f u v
(b) f 2 x
a2x
au av
ax
2
f u f v
ª¬ f x º¼
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 5.5
121. (a)
yx
y ln x
yc
ln y
y
y
yc ln x
x
ªx
º
yc« ln x»
y
¬
¼
y
ln y
x
y x ln y
yc
x y ln x
y 2 xy ln y
x 2 xy ln x
yc
At c, c : yc
c 2 c 2 ln c
c 2 c 2 ln c
1, c z 0, e
(ii) At 2, 4 : yc
16 8 ln 4
4 8 ln 2
4 4 ln 2
| 3.1774
1 2 ln 2
(iii) At 4, 2 : yc
4 8 ln 2
16 8 ln 4
1 2 ln 2
| 0.3147
4 4 ln 2
(b) (i)
509
xy
x ln y
x
Bases Other than e and Applications
(c) yc is undefined for
x2
xy ln x
x
y ln x
ex
x y.
ln x y
At e, e , yc is undefined.
ln x
, x ! 0.
x
122. Let f x
1 ln x
0 for x ! e Ÿ f is decreasing for
x2
x t e. So, for e d x y :
fc x
f x ! f y
ln x
ln y
!
x
y
xy
ln x
ln y
! xy
x
y
ln x y ! ln y x
xy ! yx
For n t 8, e letting x
n
Note:
n n, y
n 1
8
n 1,
8 | 2.828 and so
n 1, you have
!
n 1
9
| 22.6 and
n
.
9
8
| 22.4.
Note: This same argument shows eS ! S e .
INSTRUCTOR USE ONLY
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510
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponential,
Exponen
and Other Transcendental Functions
Function
1·
§
123. log e ¨1 ¸
x¹
©
1·
§
ln ¨1 ¸
x¹
©
1 x dt
³x
! ³
t
1 x
x
dt
1 x
1 x
ª t º
«1 x »
¬
¼x
§ because 1 x t t ·
¨
¸
© on x d t d 1 x ¹
1 x
x
1 x 1 x
Note: You can confirm this result by graphing y1
1
1 x
1·
§
ln ¨1 ¸ and y2
x¹
©
1
.
1 x
Section 5.6 Inverse Trigonometric Functions: Differentiation
1. y
12. arcsin 0.39 | 0.40
arccos x
§
2 3S ·
§ 3S ·
,
¨¨ ¸¸ because cos¨ ¸
2
4
© 4 ¹
©
¹
§1 S ·
§S ·
¨ , ¸ because cos¨ ¸
©2 3¹
©3¹
14. arctan 5 | 1.37
3
2
In Exercises 15–20, use the triangle.
1
S·
§ S·
3, ¸ because tan ¨ ¸
3¹
© 3¹
1
2
S
4. arcsin 0
0
6
1 − x2
y
1
§
3 S·
§ S·
, ¸¸ because tan ¨ ¸
¨¨ 3
6
© 6¹
©
¹
3. arcsin
x
3
3
3
15.
y
cos y
arccos x
x
16. sin y
1 x2
17. tan y
1 x2
x
x
18. cot y
5. arccos
3
19. sec y
6. arccos 1
1 x2
S
1
2
0
S
3
7. arctan
3
§ 1 ·
arccos¨
¸ | 0.66
© 1.269 ¹
13. arcsec 1.269
arctan x
§ S·
§S ·
¨1, ¸ because tan¨ ¸
© 4¹
©4¹
§
¨
©
2
2
1
2
§ 3 S·
§S ·
, ¸¸ because cos¨ ¸
¨¨
2
6
6¹
©
©
¹
2. y
20. csc y
6
8. arccot 3
5S
6
9. arccsc 2
10. arcsec 2
3S
4
1
x
1
1 x2
S
4
11. arccos 0.8 | 2.50
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
on 5.6
21. (a) sin arctan 34
Inverse Trigonometric Functions: D
Differentiation
Dif
ª
§ 3 ·º
24. (a) sec «arctan ¨ ¸»
© 5 ¹¼
¬
3
5
511
34
5
5
θ
5
−3
3
34
θ
ª
§ 5 ·º
(b) tan «arcsin ¨ ¸»
© 6 ¹¼
¬
4
5
3
(b) sec arcsin 54
5 11
11
11
θ
6
5
−5
4
θ
3
25. y
§
2·
22. (a) tan ¨¨ arccos
¸
2 ¸¹
©
§S ·
tan ¨ ¸
©4¹
1
cos arcsin 2 x
T
arcsin 2 x
y
cos T
1
2
1 4x2
2x
θ
2
1− 4x 2
θ
26. y
2
5·
§
(b) cos¨ arcsin ¸
13 ¹
©
12
13
sec arctan 4 x
T
arctan 4 x
y
sec T
13
1 16 x 2
1 + 16x 2
5
4x
θ
θ
12
1
ª
§ 1 ·º
23. (a) cot «arcsin ¨ ¸»
© 2 ¹¼
¬
§ S·
cot ¨ ¸
© 6¹
3
θ
3
27. y
sin arcsec x
T
arcsec x, 0 d T d S , T z
y
x2 1
x
−1
2
ª
§ 5 ·º
(b) csc «arctan ¨ ¸»
© 12 ¹¼
¬
13
5
sin T
S
2
The absolute value bars on x are necessary because of the
restriction 0 d T d S , T z S 2, and sin T for this
domain must always be nonnegative.
12
θ
13
−5
x
x2 − 1
θ
1
INSTRUCTOR USE ONLY
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512
Chapter 5
28. y
NOT FOR SALE
Logarithmic,
arithmic, Exponential,
Exponen
and Other Transcendental Functions
Function
cos arccot x
T
arccot x
y
cos T
x h·
§
cos¨ arcsin
¸
r ¹
©
x h
arcsin
r
32. y
x
x 1
T
2
x2 + 1
r2 x h
cos T
y
2
r
1
θ
x
r
x·
§
tan ¨ arcsec ¸
3¹
©
x
arcsec
3
29. y
T
r 2 − (x − h)2
33. arcsin 3 x S
x 9
3
2
tan T
y
x−h
θ
1
2
3x S
sin 12
1ª
sin 12
3¬
x
x
x2 − 9
34. arctan 2 x 5
θ
3
T
arcsin x 1
y
sec T
1
1
35. arcsin
2 x x2
x−1
2x − x 2
x ·
§
csc¨ arctan
¸
2¹
©
x
arctan
2
T
x2 2
x
csc T
y
2x
arccos
2x
sin arccos
2x
1 x,
2x
1 x
3x
1
x
1
3
5 | 1.721
x
x
0 d x d1
1−x
θ
x
36. arccos x
arcsec x
x
x2 + 2
x2 − 1
cos arcsec x
x
1
x
x2
1
x
r1
θ
2
1 tan 1
2
1
x
x
tan 1
x
θ
31. y
1
2x 5
sec ª¬arcsin x 1 º¼
30. y
S º | 1.207
¼
θ
1
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
on 5.6
1
arcsin ,
x
37. (a) arccsc x
Let y
S
2
Inverse Trigonometric
Trigo nometric Functions: D
Dif
Differentiation
42.
x t1
arccsc x. Then for
d y 0 and 0 y d
x Ÿ sin y
csc y
2
,
43.
1 x. So, y
Therefore, arccsc x
(b) arctan x arctan
S
arcsin 1 x .
f x
arcsec 2 x
fc x
2
S
2
x ! 0
,
44.
Let y
arctan x arctan 1 x . Then,
tan y
tan arctan x tan ª¬arctan 1 x º¼
1 tan arctan x tan ª¬arctan 1 x º¼
arctan e x
fc x
1
arctan
fc x
§ 1 ·§ 1 ·
¨
¸¨
¸
© 1 x ¹© 2 x ¹
x 1x
gc x
x 1x
which is undefined .
0
S 2.
arcsin x,
sin y Ÿ x
So, y
x
sin y Ÿ x
cos y Ÿ x
fc x
2
41. g x
gc x
47. h t
hc t
x2
1 9 x 2 arcsin 3x
x 2 arctan 5 x
2 x arctan 5 x x 2
S arccos x.
1 x 1
2x x2
1 5x
5x2
1 25 x 2
sin arccos t
1 t2
2
5
1 2
1
1 t2
2t
2
t
1 t2
48.
f x
arcsin x arccos x
fc x
0
49. y
2
2
1
2 x arctan 5 x cos S y .
S arccos x.
arccos x Ÿ y
2 arcsin x 1
fc t
arcsin x. Therefore,
cos y Ÿ x
f x
f t
sin y .
arccos x . Then,
Therefore, arccos x
40.
hc x
S arccos x, x d 1
So, S y
39.
46. h x
arcsin x.
(b) arccos x
2
1 9 x 2 arcsin 3 x
3x x d1
arcsin x Ÿ y
arcsin x
Let y
x3
arcsin x . Then,
Let y
1
x1 x
x2 1 9 x2
arctan x arctan 1 x
38. (a) arcsin x
x
arcsin 3x
x
S 2. Therefore,
So, y
x
1 ex
ex
1 e2 x
ex
2
4x2 1
x
f x
45. g x
1 x1x
4x 1
f x
arcsin 1 x .
1
x
1
2
2x
513
yc
2
2 x arccos x 2 1 x 2
1 2
§1·
2¨ ¸ 1 x 2
2 x
2
©
¹
1 x
2x
2x
2 arccos x
2
1 x
1 x2
2 arccos x 2 x
2 arccos x arcsin t 2
S
1
2
2t
1 t4
50. y
x
2
3 1 2
3
1 x 4
4 x2
yc
3 arccos
2
t
1
arctan
2
2
2t
1
1
§1·
˜
¨ ¸
t2 4
2 1 t 2 2 © 2¹
ln t 2 4 2t
1
t2 4 t2 4
2t 1
t2 4
INSTRUCTOR USE ONLY
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514
Chapter 5
51.
y
1§ 1 x 1
·
arctan x ¸
¨ ln
2© 2 x 1
¹
dy
dx
1§ 1
1 ·
12
¨
¸
4 © x 1 x 1¹ 1 x2
52. y
1ª
§ x ·º
x 4 x 2 4 arcsin ¨ ¸»
2 «¬
© 2 ¹¼
53.
54.
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
yc
ª
1 2
1« 1
x 4 x2
2 x 2« 2
¬
y
x arcsin x dy
dx
§
x¨
©
y
dy
dx
55. y
yc
1ª
«
2¬
x2
4 x
2
4 x2 º
»
4 x ¼
4
2
4 x2
x
1 x2
1
ln 1 4 x 2
4
2x
1 § 8x ·
arctan 2 x ¨
¸
1 4x2
4 © 1 4 x2 ¹
arcsin x
x arctan 2 x 8 arcsin
2
1
1 x 4
25 arcsin
5
2
2
1 2
16 x 2
x
16 x 2
2 x
2
4
16 x 2
x2
2
2 16 x 2
16 16 x 2 x 2
x2
2 16 x
16 x 2
2
x
x 25 x 2
5
1
1 x 2
arctan x 2
25 x 2 x
25 x 2
25 x 2
x
1 x2
1 x2 x 2 x
1
2
2
1 x
1 x2
1 x2 1 x2
1 x2
2
1 2
1
2 x
25 x 2
2
x2
2x2
25 x 2
25 x 2
58. y
yc
arctan
x
1
2
2 x2 4
2
1
1
1
x2 4 2 x
2
21 x 2
2
x
2
2
2
x2 4
x 4
2x2 8 x
2
1 x2
arctan 2 x
x
x 16 x 2
4
2
25 x 2
yc
º
»
2»
1 x 2 ¼
1
1 x2
·
¸ arcsin x 1 x ¹
25
57. y
4 x2 2
2
16 x
yc
1
1 x4
1
8
56. y
1
1
ªln x 1 ln x 1 º¼ arctan x
4¬
2
2
x2 4
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
on 5.6
59. y
yc
§1 S ·
¨ , ¸
©2 3¹
2 arcsin x,
1 x2
2
S
4 §
1·
¨x ¸
2¹
3©
S
4
2
x 3
3
3
3
y
Tangent line: y yc
2S 4 x 4
§1 S ·
¨ , ¸
©2 4¹
3 x arcsin x,
3x
1
3 arcsin x
1 x2
1
§S ·
3¨ ¸
34
©6¹
3
2
S
Tangent line: y 4
y
1
2 12
3S
8
y
61. y
y
2 1 x2
§
2 3S ·
At ¨¨ ,
¸, yc
2
8 ¸¹
©
2
.
2
2§
¨x 2 ¨©
2·
¸
2 ¸¹
2
3S
1
x 2
8
2
§ S·
¨ 2, ¸
© 4¹
§ x·
arctan ¨ ¸,
© 2¹
1
§1·
¨ ¸
1 x2 4 © 2 ¹
2
4 x2
65.
4 arccos x 1
4 2S .
§1 S ·
At ¨ , ¸, yc
©2 4¹
1
2
2S 4 x 1
yc
§
2 3S ·
,
¨¨ ¸¸
© 2 8 ¹
1
arccos x,
2
1 x 1
Tangent line: y 2S
64. y
S
4 3
2 3
x 3
3
3
y
4x
At 1, 2S , yc
4
.
3
1 14
Tangent line: y yc
yc
1
f x
arctan x,
f 0
0
fc x
1
,
1 x2
2 x
f cc x
515
1, 2S
4 x arccos x 1 ,
63. y
2
§1 S ·
At ¨ , ¸, yc
©2 3¹
60. y
Inverse Trigonometric Functions: D
Differentiation
Dif
1 x2
§
¨
©
3 §
¨
©
3 a
2
3 S
2
.
S ·§
1·
¸¨ x ¸
2 ¹©
2¹
S·
3
¸x 2¹
2
0
fc 0
1
, f cc 0
0
P1 x
f 0 fc 0 x
x
P2 x
f 0 fc 0 x 1
f cc 0 x 2
2
x
y
§ S·
At ¨ 2, ¸, yc
© 4¹
2
4 4
Tangent line: y S
yc
4
4
1
16 x 1
2
§ 2 S·
, ¸¸, yc
At ¨¨
© 4 4¹
Tangent line: y P1 = P 2
1.0
f
0.5
x
− 1.0
0.5 1.0 1.5
−1.0
−1.5
§ 2 S·
, ¸¸
¨¨
© 4 4¹
arcsec 4 x ,
4x
1.5
1
x 2
4
1
1
S
x 4
4
2
y
62. y
1
.
4
x 16 x 2 1
1
2 4
2 1
4
§
2 2 ¨¨ x ©
y
2 2x S
for x ! 0
2 2.
2·
¸
4 ¸¹
S
1
INSTRUCTOR USE ONLY
4
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
516
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
66.
f x
arccos x,
f 0
S
a
68.
0
2
1
fc x
1 x2
x
f cc x
,
32
1 x2
,
fc x
1
1 x2
2 x
1
f cc x
f cc 0
0
P1 x
S
x
2
1
f 0 f c 0 x f cc 0 x 2
2
P2 x
arctan x,
fc 0
f 0 fc 0 x
P1 x
f x
1 x2
2
1
2
S
1
x 1
4
2
1
2
f 1 f c 1 x 1 f cc 1 x 1
2
1
1
S
2
x 1 x 1
4
2
4
f 1 fc1 x 1
P2 x
S
a
x
y
y
P1(x)
π
2
f 3
f
π
4
P1 = P2
1
−4
−2
x
−2
2
P2(x)
x
−1
1
−1
67.
f x
arcsin x,
fc x
1
1
2
1 x
2
S
6
arcsec x x
1
x2 1
x
1
x2 x2 1
1
x4 x2 1
0 when x 2
S
2 3§
1·
¨x ¸
6
3 ©
2¹
2 3§
1· 2 3§
1·
¨x ¸ ¨x ¸
3 ©
2¹
9 ©
2¹
1
r
x
2
1
5
2
5
2
x2 1
0 when x
32
1 · 1 § 1 ·§
1·
§1·
§ 1 ·§
f ¨ ¸ f c¨ ¸¨ x ¸ f cc¨ ¸¨ x ¸
2 ¹ 2 © 2 ¹©
2¹
© 2¹
© 2 ¹©
P2 x
f x
2
1·
§1·
§ 1 ·§
f ¨ ¸ f c¨ ¸¨ x ¸
2¹
© 2¹
© 2 ¹©
P1 x
69.
fc x
1 x
x
f cc x
a
1
or
r1.272
Relative maximum: 1.272, 0.606
Relative minimum: 1.272, 3.747
2
70.
f x
arcsin x 2 x
y
1.5
1.0
fc x
P1
1
2
1 x2
0.5
0 when
x
0.5 1.0 1.5
P2
− 1.0
f
− 1.5
f cc x
1 x2
1
or x
2
r
3
2
x
1 x2
32
§ 3·
f cc¨¨
¸¸ ! 0
© 2 ¹
§ 3
·
Relative minimum: ¨¨
, 0.68 ¸¸
2
©
¹
§
3·
f cc¨¨ ¸¸ 0
© 2 ¹
§
·
3
Relative maximum: ¨¨ , 0.68 ¸¸
© 2
¹
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
on 5.6
71. f x
arctan x arctan x 4
fc x
1
1
1 x2 1 x 4 2
1 x
Inverse Trigonometric Functions: D
Differentiation
Dif
74.
0
2
1 x 4
0
8 x 16
2
x
f x
arctan x fc x
1
1 x2
2 x
2
f cc x
1 x2
2
2
Increasing on f, f
maximum.
No relative extrema
§ S·
Point of inflection: ¨ 0, ¸
© 2¹
arcsin x 2 arctan x
1
fc x
2
1 x2
0
1 x2
2 1 x2
Domain: f, f
1 2x2 x4
4 1 x2
Range: 0, S
x 6x 3
x
0
1 x2
4
2
Horizontal asymptotes: y
f is arctan x shifted
r0.681
By the First Derivative Test, 0.681, 0.447 is a relative
f x
arcsin x 1
fc x
1
S
2
0 and y
units upward.
π
)0, π2 )
π
2
1
1 x 1
2
2 x x2
x
−6 −4 −2
2
4
6
x 1
f cc x
2 x x2
32
75.
f x
arcsec 2 x
1
§ S·
Maximum: ¨ 2, ¸
© 2¹
fc x
S·
§
Minimum: ¨ 0, ¸
2¹
©
1º
§
ª1 ·
Domain: ¨ f, » ‰ « , f ¸
2¼
©
¬2 ¹
Point of inflection: 1, 0
ª S·
§S º
Range: «0, ¸ ‰ ¨ , S »
2
¬
¹
©2 ¼
Domain: >0, 2@
x
The graph of f is y
arcsin x shifted 1 unit to the right.
S
2
y
π
π
2
(− 12 , π (
π
2
) )
2,
(1, 0)
1
−π
§1 ·
Minimum: ¨ , 0 ¸
©2 ¹
Horizontal asymptote: y
y
π
−
2
4x2 1
§ 1 ·
Maximum: ¨ , S ¸
© 2 ¹
ª S Sº
Range: « , »
¬ 2 2¼
−1
S
y
maximum and 0.681, 0.447 is a relative minimum.
73.
S
By the First Derivative Test, 2, 2.214 is a relative
72. f x
517
π
2
x
2
3
)0, − π2 )
( 12 , 0(
−2
−1
x
1
2
INSTRUCTOR USE ONLY
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518
Chapter 5
76.
f x
NOT FOR SALE
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
x
4
arccos
79.
1
fc x
0
16 x 2
x
f cc x
1
1 x
2
1
1
§ 2
2·
,
At ¨¨
¸ : yc
2 ¸¹
© 2
Range: >0, S @
Maximum: 4, S
0
1
1 x2
1
Tangent line: y Minimum: 4, 0
2
2
§
1¨¨ x ©
2·
¸
2 ¸¹
y
x 2
Point of Inflection: 0, S 2
y
(−4, π )
80.
y2 1
2 yyc
(4, 0)
−2
2
2
1 x y
) )
−6 −4
arctan x y
π
π
0,
2
4
At 1, 0 :
x
6
>1 yc@
1
>1 yc@
2
Tangent line: y 0
x 2 x arctan y
77.
2 x arctan y x
yc
1 y2
§
x ·
¨1 ¸ yc
y2 ¹
1
©
yc
§ S ·
At ¨ , 1¸: yc
© 4 ¹
S
2
y
1 xy
2 x arctan y
x
1
1 y2
S
S
2
2
1
1 x y
1 yc Ÿ y c
Tangent line: y
2S
8S
4
82. arctan 0
0 Ÿ yc
,
1, 0
1
1 x 1
x 1
0. S is not in the range of y
83. (a) arcsin arcsin 0.5
arctan x.
| 0.551
arcsin arcsin 1.0 does not exist
(b) In order for f x
arcsin arcsin x to be real, you
must have 1 d arcsin x d 1.
Because arcsin x
arcsin x
2S
S2
x 1
8S
16 2S
> y xyc@
At 0, 0 : 0
S
2S §
S·
¨x ¸
8 S©
4¹
arcsin x y ,
2
4
on which they are one-to-one.
2 x arctan y
arctan xy
1
y
S
81. The trigonometric functions are not one-to-one on
f, f , so their domains must be restricted to intervals
yc
4
S 4
1
2
Tangent line: y 1
78.
§ S ·
¨ , 1¸
© 4 ¹
y 1,
§ 2
2·
,
¨¨
¸¸
2
2
©
¹
,
yc
1 y2
Domain: >4, 4@
2
yc
1 y2
32
16 x 2
S
arcsin x arcsin y
1 Ÿ sin 1
1 Ÿ sin 1
x and
sin 1
x, you have
sin 1 d x d sin 1
0.84147 d x d 0.84147
0, 0
2
>1 yc@
1
x
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
on 5.6
§
2 S·
84. (a) ¨¨ , ¸¸ and 0, 0 lie on the graph of
2
4¹
©
§ S·
arcsin x because sin ¨ ¸
© 4¹
y
0, and 0 and sin 0
Inverse Trigonometric Functions: Dif
Differentiation
D
ª S Sº
lie in the internal « , ».
4
¬ 2 2¼
S
§ 1 2S ·
§ S·
(b) ¨ ,
¸ and ¨ 0, ¸ lie on the graph of
© 2 3 ¹
© 2¹
2S
S
and lie in the
y
arccos x because both
3
2
§1 S ·
interval >0, S @. ¨ , ¸ does not lie on the graph
©2 3¹
of y
arccos x because S
3
(b)
dT
dt
1 5
1 x5
5 dx
x 2 25 dt
dx
dt
dx
dt
400 and x
10,
If
dx
dt
400 and x
3,
dT
dt
16 rad/h.
dT
| 58.824 rad/h.
dt
x
3
§ x·
arccot ¨ ¸
© 3¹
T
(b)
2
If
92. (a) cot T
dT
dt
3 dx
x 2 9 dt
If x
10,
If x
3,
is not in the interval
>0, S @.
dT
| 11.001 rad/h.
dt
dT
| 66.667 rad/h.
dt
A lower altitude results in a greater rate of change of T .
85. False
arccos
§ x·
arccot ¨ ¸
©5¹
T
§ 3 2S ·
,
¨¨
¸ does not lie on the graph of
3 ¸¹
© 2
2S
y
arcsin x because
is not in the interval
3
ª S Sº
« 2 , 2 ».
¬
¼
x
5
91. (a) cot T
2
and
2
519
1
2
S
16t 2 256
because the range is >0, S @.
16t 2 256
0 when t
4 sec
h
86. False
§S ·
sin ¨ ¸
©4¹
ht
93. (a)
3
θ
§ 2·
2
, so arcsin ¨¨
¸¸
2
© 2 ¹
S
4
500
.
(b) tan T
87. True
d
>arctan x@
dx
1
! 0 for all x.
1 x2
T
dT
dt
88. False
The range of y
ª S Sº
arcsin x is « , ».
¬ 2 2¼
1 ª¬ 4 125 t 2 16 º¼
15,625 16 16 t 2
sec 2 x
1 tan 2 x
sec2 x
sec2 x
2
1000t
89. True
d
ªarctan tan x º¼
dx ¬
16t 2 256
500
ª 16
º
arctan «
t 2 16 »
500
¬
¼
8t 125
h
500
1
2
When t
1, dT dt | 0.0520 rad/sec.
When t
2, dT dt | 0.1116 rad/sec.
90. False
2
arcsin 2 0 arccos 2 0
§S ·
0¨ ¸ z 1
©2¹
INSTRUCTOR USE ONLY
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520
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
800
s
94. cos T
s
T
§ 800 ·
arccos¨
¸
© s ¹
dT
dt
dT ds
˜
ds dt
800
1
§ 800 · ds
¸
s 2 ¹ dt
©
1 800 s
40
x
40 85
x
tan D
95.
θ
tan D T
x
tan D tan T
1 tan D tan T
tan D T
40 x tan T
40
1
tan T
x
x 40 x tan T
125
x
125 x 40 tan T
dT
dt
30 2S
tan T
x
50
dT
dt
85 x
x 2 5000 tan T
T
85 x
§
·
arctan ¨ 2
¸
© x 5000 ¹
dT
dx
85 5000 x 2
x 2 1600 x 2 15625
0 Ÿ x
dx
dt
50 2 | 70.71 ft.
θ
50
dx
50
x 2 2500 dt
dx
dt
x 2 2500 dT
dt
50
50
45q
2
S
4
,x
2500
50
x
50:
60S
6000S ft/min
97. (a) tan arctan x arctan y
tan arctan x tan arctan y
1 tan arctan x tan arctan y
Therefore, arctan x arctan y
(b) Let x
50 2
60S rad/min
dT dx
dx dt
When T
5000
85
40 x tan T
x 40 tan T
§ x·
arctan ¨ ¸
© 50 ¹
T
s ! 800
α θ
By the First Derivative Test, this is a maximum x
96.
,
s 2 8002 dt
s
40
125
x
ds
800
2¨
1
and y
2
x y
,
1 xy
xy z 1
§x y·
arctan¨
¸, xy z 1.
© 1 xy ¹
1
.
3
§1·
§1·
arctan ¨ ¸ arctan ¨ ¸
© 2¹
© 3¹
arctan
12 13
1 ª¬ 1 2 ˜ 1 3 º¼
arctan
56
1 16
arctan
56
56
arctan 1
S
INSTRUCTOR USE
E ONLY
4
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
on 5.6
98. (a) Let y
Inverse Trigonometric Functions: D
Differentiation
Dif
521
arctan u. Then
tan y
u
dy
dx
dy
dx
uc
sec 2 y
uc
sec 2 y
uc
.
1 u2
1 + u2
u
y
1
(b) Let y
arccot u. Then
cot y
u
dy
dx
dy
dx
uc
csc 2 y
uc
csc 2 y
uc
.
1 u2
1 + u2
1
y
u
(c) Let y
arcsec u. Then
sec y
u
dy
sec y tan y
dx
dy
dx
uc
uc
sec y tan y
uc
u2 1
u
.
u
u2 − 1
y
1
Note: The absolute value notation in the formula for the derivative of arcsec u is necessary
because the inverse secant function has a positive slope at every value in its domain.
(d) Let y
arccsc u. Then
csc y
u
dy
csc y cot y
dx
dy
dx
uc
u
uc
csc y cot y
uc
u
u2 1
1
y
u2 − 1
Note: The absolute value notation in the formula for the derivative of arccsc u is necessary
because the inverse cosecant function has a negative slope at every value in its domain.
INSTRUCTOR USE ONLY
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522
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
arccos x arcsin x
99. (a) f x
y
101.
y
(4, 2)
(0, 2)
2
1
θ
θ1
θ2
x
(c, 0)
(4, 0)
x
−1
1
2
, tan T 2
c
tan T1
S 2.
(b) The graph of f is the constant function y
u
(c) Let
cos u
arccos x
and
v
x
and
sin v
1
arcsin x
x.
1
1 − x2
x
u
To maximize T , minimize f c
fc c
1 − x2
sin u v
sin u cos v sin v cos u
1 x2
1
c2 4
4c
c 4
c 8c 16 4
2
1 x2 x ˜ x
1 x x
1
So, u v S 2. Therefore,
2
arccos x arcsin x
2
S 2.
By the First Derivative Test, c
c, f c
100. f x
sin x
g x
T1 T 2 .
§ 2·
§ 2 ·
arctan ¨ ¸ arctan ¨
¸
c
© ¹
©4 c¹
2
2
0
2
c2 4
4c 4
f c
v
x
2
, 0 c 4
4c
1
2
4
2
8c
16
c
2
2 is a minimum. So,
2, S 2 is a relative maximum for the angle
T . Checking the endpoints:
arcsin sin x
(a) The range of y
arcsin x is S 2 d y d S 2.
(b) Maximum: S 2
c
0: tan T
c
4: tan T
c
2: T
4
2
4
2
Minimum: S 2
3
f
− 2
2
2 Ÿ T | 1.107
2 Ÿ T | 1.107
S T1 T 2
S
2
| 1.5708
So, 2, S 2 is the absolute maximum.
g
−3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
on 5.6
102.
Inverse Trigonometric Functions: D
Differentiation
Dif
523
2
Q
3−c
θ1
θ
R
θ2
3
c
P
5
2
, tan T 2
3c
tan T1
5
, 0 c 3
c
To maximize T , minimize f c
T1 T 2 .
§ 2 ·
§ 5·
arctan ¨
¸ arctan ¨ ¸
3
c
©
¹
©c¹
f c
2
fc c
3c
2
4
5
c 2 25
0
2 c 2 25
5 c 2 6c 9 4
3c 2 30c 15
0
c 10c 5
0
2
5 2 5 | 0.5279 because c  >0, 3@
c
T1 T 2 | 2.1458 and
T | S T1 T 2 | 0.9958
Checking the endpoints:
c
3: tan T
c
0: tan T
3
Ÿ T | 0.5404
5
3
Ÿ T | 0.9828
2
5 2 5 yields the absolute maximum.
So, c
T
103. Let
tan T
sin T
arcsin x
So, arcsin x
§
arctan ¨
©
x
·
x
¸,
1 x2 ¹
1 x 1
1 x2
x
1
T.
x
§
arctan ¨
©
·
¸ for 1 x 1.
1 x ¹
x
2
1
x
θ
1 − x2
INSTRUCTOR USE ONLY
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524
NOT FOR SALE
Chapter 5
104. f x
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Function
Functions
S
0 d x sec x,
2
,S d x 3S
2
y
arcsec x
x
sec y
1
sec y tan y ˜ yc
yc
1
sec y tan y
(b)
y
4
2
−π
2
−2
1
x
π
2
x
tan 2 y 1
−4
(a) y
x d 1
arcsec x,
0 d y S
S d y or
2
sec 2 y
r
tan y
x t1
or
x2 1
sec 2 y 1
On 0 d y S 2 and S d y 3S 2, tan y t 0.
3S
2
y
3π
2
π
2
x
−6 −4 −2
2
4
6
Section 5.7 Inverse Trigonometric Functions: Integration
1. ³
9 x
2
dx
2. ³
1 4x2
3. ³
x 4 x2 1
4. ³
12
dx
1 9x2
5. ³
6. ³
1
1
2³
4³
1 x 1
1
4 x 3
2
t 2 , du
t
1 t4
dt
2
2
3
dx
1 9 x2
dx
1
arcsin 2 x C
2
dx
2
2x
1
dx
arcsec 2 x C
1
2³
³ x x 4 4 dx
t
9. ³ 4
dt
t 25
x
x
2
2
2 x dx
22
1
1
2 dt
³
2
2
2 t
52
§ t2 ·
1 1
arctan ¨ ¸ C
2 5
©5¹
§ t2 ·
1
arctan ¨ ¸ C
10
©5¹
arcsin x 1 C
10. ³
1
x 1 ln x
2
dx
³
1
1 ln x
2
˜
1
dx
x
arcsin ln x C
2t dt.
1
2³
1
2
1
x2
arcsec
C
4
2
4 arctan 3 x C
1
§ x 3·
arctan ¨
¸ C
2
© 2 ¹
dx
2 x dx.
1
2
2x
x 2 , du
8. Let u
1 4 x2
³
dx
1
7. Let u
³
§ x·
arcsin ¨ ¸ C
© 3¹
dx
1
1 t2
2
2t dt
1
arcsin t 2 C
2
11. Let u
e 2 x , du
e2 x
³ 4 e4 x dx
2e 2 x dx.
1
2e 2 x
dx
³
2 4 e2 x 2
1
e2 x
C
arctan
4
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ection 5.7
12. u
3 x, du
3 dx, a
2
³ x 9 x 2 25 dx
Inverse Trigonometric Functions
Functions: Integration
5
14. ³
1
2³
3x
3x
2
sin x
dx
7 cos 2 x
sec 2 x
25 tan 2 x
sec2 x
³
dx
7
3 dx
52
3x
2
arcsec
C
5
5
13. ³
1
³
cos 2 x
sin x dx
1
§ cos x ·
arctan ¨
¸ C
7
© 7 ¹
§
7
arctan ¨¨
7
©
7 cos x ·
¸¸ C
7
¹
dx
2
52 tan x
2
525
§ tan x ·
arcsin ¨
¸ C
© 5 ¹
15.
1
dx, u
x 1 x
1
³ u 1 u 2 2u du
³
16. ³
du
2³
3
dx, u
x1 x
2
u 2 , dx
x, x
2 arcsin u C
1 u2
1
x , du
x
2
dx, dx
3
2u du
2 ³ u 1 u2
3³
x 3
dx
x2 1
1
2x
1
dx 3³ 2
dx
2 ³ x2 1
x 1
17. ³
18. ³
x2 3
x 4
2
x
du
1 u2
2u du
3 arctan u C
x2
1
ln x 2 1 3 arctan x C
2
3
x2 4 x 5
2
³
dx
x 2
20. ³
x 1
2
4
x C
3 arctan
³ x x 2 4 dx ³ x x 2 4 dx
dx
9 x 3
x C
2u du
1 2
1
x2 4
2 x dx 3 ³
³
2
x
19. ³
2 arcsin
dx
1
x2 4
dx
x
3
arcsec
C
2
2
x 3
9 x 3
2
9 x 3
2
dx ³
8
9 x 3
2
§ x 3·
8 arcsin ¨
¸C
© 3 ¹
dx
§x
·
6 x x 2 8 arcsin ¨ 1¸ C
©3
¹
1
2x 2
3
dx ³
dx
³
2
2
2 x 1 4
x 1 4
1
3
§ x 1·
ln x 2 2 x 5 arctan ¨
¸ C
2
2
© 2 ¹
21. Let u
16
³0
3 x, du
3
1 9x
2
3 dx.
dx
16
³0
1
1 3x
16
ª¬arcsin 3x º¼ 0
2
3 dx
22. ³
2
0
1
4 x2
dx
S
6
xº
ª
«arcsin 2 »
¬
¼0
arcsin
2
2
arcsin 0
2
S
4
INSTRUCTOR USE ONLY
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526
NOT FOR SALE
Chapter 5
23. Let u
³0
3 2
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Function
Functions
2 x, du
2 dx.
1
dx
1 4x2
24. ³
1 32
2
dx
2
2³0
1 2x
ª1
º
« 2 arctan 2 x »
¬
¼0
3 2
1
3
3
x 4 x2 9
ª1
2x º
« arcsec
»
3 ¼
¬3
dx
3
3
1
1
2 3
arcsec 2 arcsec
3
3
3
S
1§ S · 1§ S ·
¨ ¸ ¨ ¸
3© 3 ¹ 3© 6 ¹
18
S
6
25. ³
6
1
3
25 x 3
6
ª1
§ x 3 ·º
« 5 arctan ¨ 5 ¸»
©
¹¼ 3
¬
dx
2
1
arctan 3 5
5
| 0.108
26. ³
4
1
1
x 16 x 5
4 dx
4
³1
dx
2
4x
2
4x
5
2
4
ª§ 1 ·
4x º
Ǭ
»
¸ arcsec
5
5 ¼1
©
¹
¬
27. Let u
ln 5
e x , du
e x dx
ex
ªarctan e x º
¬
¼0
ln 5
³ 0 1 e2 x dx
28. Let u
e x , du
e x
ln 4
³ ln 2
1e
29. Let u
S
1
³0
2
³0
33. ³
2
0
2
S
arcsin x
1 x
2
1
1 x2
6
§1·
arcsin¨ ¸ | 0.271
© 4¹
ª1
2 º
« 2 arcsin x»
¬
¼0
³
2
1
1
1 x2
S
4
2
| 0.308
dx.
1 x
1
S2
32
2 arccos x
0
4
dx.
1
dx
S
S
S
S 2
dx
dx
x2 2x 2
| 0.588
ª
¬ arctan cos x º¼S 2
dx
ª¬arctan sin x º¼ 0
arccos x, du
arccos x
sin x
S 2 1 cos 2 x
arcsin x, du
1 x2
32. Let u
1
³
cos x
dx
1 sin 2 x
31. Let u
4
sin x dx.
cos x, du
sin x
S 2
0
S
§1·
§1·
arcsin ¨ ¸ arcsin¨ ¸
© 4¹
© 2¹
ln 4
x º
ª
¬arcsin e ¼ ln 2
³ S 2 1 cos2 x dx
30. ³
arctan 5 e x dx
dx
2 x
1
16
1
§ 4 ·
arcsec
arcsec¨
¸ | 0.091
5
5
5
© 5¹
2
³ 0 1 x 1 2 dx
1
dx
2
ª 1
2 º
« 2 arccos x»
¬
¼0
2
ª¬arctan x 1 º¼ 0
3S 2
| 0.925
32
S
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
ection 5.7
Inverse Trigonometric Functions
Functions: Integration
2
dx
³ 2 x 2 2 9
ª1
§ x 2 ·º
« 3 arctan ¨ 3 ¸»
©
¹¼ 2
¬
2x 6
1
2x 6
1
§ x 3·
ln x 2 6 x 13 3 arctan ¨
¸C
© 2 ¹
2x 2
1
ln x 2 2 x 2 7 arctan x 1 C
1
§ 4·
arctan¨ ¸
3
© 3¹
34. ³
2
dx
2 x 2 4 x 13
35. ³
2x
dx
x 2 6 x 13
³ x 2 6 x 13 dx 6³ x 2 6 x 13 dx
2
³ x 2 6 x 13 dx 6³ 4 x 3 2 dx
36. ³
2x 5
dx
x2 2 x 2
37. ³
x2 4 x
38. ³
x 4x
39. ³
1
2
2
3
2x 3
2
4x x
2
³ x 2 2 x 2 dx 7 ³ 1 x 1 2 dx
1
dx
³
dx
³
dx
³2
4 x 2
§ x 2·
arcsin ¨
¸C
© 2 ¹
dx
2
2
4 x 4x 4
2x 4
4x x
³
3
2
4x x2
2
dx ³
1/2
2
³
dx
2
3
4 x 2
3
1
2
4x x2
4 2 x dx ³
2
3
1
2
4 x 2
x 1
41. Let u
x 2x
2
³
1
x 1
x 1
x 2 1, du
2 x dx.
x
1
2x
dx
2 ³ x2 1 2 1
x 2 4, du
2 x dx.
³ x 4 2 x 2 2 dx
42. Let u
³
dx
x
9 8x2 x4
dx
1
2³
25 x 2 4
et 3. Then u 2 3
³
³ u 2 3 du
2u 2
1
et , 2u du
x 2
x 2, u 2 2
³ x 1 dx
2u 2
³ u 2 3 du
S
6
| 1.059
arcsec x 1 C
dx
2
dx
§ x2 4 ·
1
arcsin ¨
¸ C
2
© 5 ¹
et dt , and
2u du
u2 3
dt.
1
³ 2 du ³ 6 u 2 3 du
2u 2 3 arctan
44. Let u
42 3 1
arctan x 2 1 C
2
2x
43. Let u
et 3 dt
2
dx
2
3
1
§ x 2·
2 arcsin ¨
¸C
© 2 ¹
dx
dx
ª
§ x 2 ·º
2
«2 4 x x arcsin ¨ 2 ¸»
©
¹¼ 2
¬
40. ³
527
x, 2u du
³
u
C
3
2 et 3 2 3 arctan
et 3
C
3
dx.
2u 2 6 6
du
u2 3
2³ du 6³
2u 1
du
u2 3
6
u
arctan
C
3
3
x 2 2 3 arctan
x 2
C
3
INSTRUCTOR
RU
USE
E ONLY
ON
N
2
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528
NOT FOR SALE
Chapter 5
45. ³
dx
x1 x
3
1
49. (a)
x , u2
Let u
3
S·
§S
2¨ ¸
4¹
©3
S
x 1, u 2
3 x
³1
³1
2 4u u
2
§ u5
u3 ·
2¨
¸C
3¹
©5
2 3 2
u 3u 5 C
15
2
32
x 1 ª¬3 x 1 5º¼ C
15
2
32
x 1
3x 2 C
15
dx,
du
2
(c) Let u
x 1. Then x
dx
2u du.
4 u2
§ u ·º
arcsin ¨ ¸»
© 2 ¹¼1
2
³
x
dx
x 1
³
§ 2·
§1·
arcsin ¨¨
¸¸ arcsin ¨ ¸
2
© 2¹
©
¹
S
4
1
S
S
6
12
47. (a)
³
(b)
³
(c)
³ x 1 x 2 dx cannot be evaluated using the basic
1 x2
x
1 x2
dx
arcsin x C ,
dx
1 x2 C,
u
u2 1
2u du
u
§ u3
·
u¸ C
2¨
3
©
¹
2
u u2 3 C
3
2
x 1 x 2 C
3
1 x2
Note: In (b) and (c), substitution was necessary
before the basic integration rules could be used.
1
50. (a)
integration rules.
48. (a)
u 2 1 and
2 ³ u 2 1 du
x
u
u 2 1 and
2
6
x 1, 2u du
x 1
³ u 1 u 2u du
2 ³ u 4 u 2 du
4 u2 .
2u du
2
2
32
x 1
C, u
3
x 1 dx
³ x x 1 dx
ª¬2 arctan u º¼ 1
Let u
³
(b) Let u
x 1. Then x
dx
2u du.
2
3
dx
3 x x 1
1
1 u 2.
³ 1 1 u 2 du
u 1 u2
0 2
46. ³
dx, 1 x
x, 2u du
2u du
3
³1
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Function
Functions
integration rules.
x2
³ e dx cannot be evaluated using the basic
(b)
integration rules.
1 x2
e C,
2
x2
(b)
³ xe dx
(c)
³ x2 e
1
1/ x
dx
1
³ 1 x 4 dx cannot be evaluated using the basic
u
x2
e1 x C , u
1
x
x
³ 1 x 4 dx
1
2x
dx
2 ³ 1 x2 2
1
arctan x 2 C , u
2
(c)
x3
³ 1 x 4 dx
1 4 x3
dx
4 ³ 1 x4
1
ln 1 x 4 C , u
4
x2
1 x4
51. No. This integral does not correspond to any of the basic
differentiation rules.
52. The area is approximately the area of a square of side 1.
So, c best approximates the area.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
ection 5.7
1
53. yc
³
y
4 x
1
y
0, y
x
−4
S
4
−4
When x
2, y
S:
1
§ 2·
arctan ¨ ¸ C
2
© 2¹
S
S
55. (a)
25 x 2
5, S
,
2
§ x·
2 arcsin ¨ ¸ C
©5¹
y
³
S
2 arcsin 1 C Ÿ C
y
§ x·
2 arcsin ¨ ¸
©5¹
25 x
dx
2
0
5
7S
8
C Ÿ C
8
1
§ x · 7S
arctan ¨ ¸ 2
8
© 2¹
y
2
(b) yc
1
, 2, S
4 x2
1
1
x
³ 4 x 2 dx 2 arctan 2 C
y
S
S Ÿ C
§ x·
arcsin ¨ ¸ S
© 2¹
54. yc
(5, π )
§ x·
arcsin ¨ ¸ C
©2¹
dx
529
y
4
4 x2
When x
56. (a)
0, S
,
2
Inverse Trigonometric Functions
Functions: Integration
−5
5
−5
y
4
dy
dx
57.
10
x
x2 1
,
3, 0
4
x
4
−6
12
−4
(b) yc
y
2
, 0, 2
9 x2
2
2
§ x·
³ 9 x 2 dx 3 arctan¨© 3 ¸¹ C
2
C
y
2
§ x·
arctan ¨ ¸ 2
3
© 3¹
−8
58.
dy
dx
1
,
12 x 2
4, 2
4
−6
6
5
−4
−4
4
59.
dy
dx
−1
2y
16 x 2
,
0, 2
3
−3
3
−1
INSTRUCTOR USE ONLY
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530
Chapter 5
60.
dy
dx
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Function
Functions
y
,
1 x2
63. Area
0, 4
1
3
1
3
³1 x 2 2 x 5 dx ³1 x 1 2 4 dx
3
ª1
§ x 1 ·º
« 2 arctan ¨ 2 ¸»
©
¹¼1
¬
1
1
arctan 1 arctan 0
2
2
7
−1
6
S
0
8
2
1
³0
61. Area
4 x
dx
2
64. Area
1
ª
§ x ·º
«2 arcsin ¨ 2 ¸»
© ¹¼ 0
¬
2
0
³ 2 x 2 2 4 dx
0
ª
§ x 2 ·º
«arctan ¨ 2 ¸»
©
¹¼ 2
¬
arctan 1 arctan 0
§1·
2 arcsin ¨ ¸ 2 arcsin 0
©2¹
§S ·
2¨ ¸
©6¹
S
S
3
4
2
1
³ 2 2 x x 2 1 dx
62. Area
65. Area
S 2
3 cos x
³ S 2 1 sin 2 x dx
3
S
S
4
12
S 2
1
S 2 1 sin 2 x
cos x dx
S 2
3 arctan 1 3 arctan 1
§ 2 ·
arcsec 2 arcsec¨
¸
© 2¹
3³
ª¬3 arctan sin x º¼ S 2
>arcsec x@22 2
S
2
0
³ 2 x 2 4 x 8 dx
3S 3S
4
4
66. Area
ln
³0
3
4 ex
dx,
1 e2 x
4 ª¬arctan e x º¼
ln
u
3S
2
ex
3
0
4 ªarctan 3 arctan 1 º
¬
¼
S· S
§S
4¨ ¸
4¹
3
©3
67. (a)
y
2
1
x
1
2
1
Shaded area is given by ³ arcsin x dx.
0
(b)
1
³ 0 arcsin x dx | 0.5708
(c) Divide the rectangle into two regions.
Area rectangle
§S ·
1¨ ¸
©2¹
base height
S
2
S 2
1
³ 0 arcsin x dx ³ 0 sin y dy
Area rectangle
S
S 2
1
³ 0 arcsin x dx cos y º¼ 0
2
1
S
0
2
So, ³ arcsin x dx
1
³ 0 arcsin x dx 1
INSTRUCTOR USE ONLY
1,
1
| 0.5708 .
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ection 5.7
4
1
³ 0 1 x 2 dx >4 arctan x@0
1
68. (a)
(b) Let n
4³
§S ·
4¨ ¸ 4 0
©4¹
531
S
6.
1
1
4 arctan 1 4 arctan 0
Inverse Trigonometric Functions
Functions: Integration
0 1 x2
4
2
4
2
4
1º
§ 1 ·ª
dx | 4¨ ¸ «1 » | 3.1415918
18
1
1
36
1
1
9
1
1
4
1
4
9
1
25
36
2 »¼
© ¹ ¬«
(c) 3.1415927
1 x2 2
dt
2³ x t2 1
69. F x
(a) F x represents the average value of f x over the interval > x, x 2@. Maximum at x
1, because the graph is greatest
on >1, 1@.
>arctan t@xx 2
(b) F x
1
Fc x
70. ³
1 x 2
1
6x x2
(a) 6 x x 2
2
1
1 x2
1 x2 x2 4x 5
4 x 1
x 1 x 4x 5
x 1 x2 4 x 5
2
2
2
0 when x
1.
dx
9 x2 6x 9
1
³
6 x x2
(b) u
x , u2
³
6u u
9 x 3
x, 2u du
4
9 x 3
2
dx
³
2u du
2
§ x 3·
arcsin¨
¸ C
© 3 ¹
dx
³
dx
1
2
arctan x 2 arctan x
2
6u
2
§ u ·
2 arcsin ¨
¸C
© 6¹
du
§
2 arcsin ¨¨
©
x·
¸C
6 ¹¸
4
(c)
y2
y1
−1
7
−2
The antiderivatives differ by a constant, S 2.
Domain: >0, 6@
71. False, ³
72. False, ³
3 x 9 x 2 16
dx
dx
25 x 2
75.
3x
1
C
arcsec
12
4
dx
º
d ª
§u·
arcsin ¨ ¸ C »
dx «¬
©a¹
¼
x
1
arctan C
5
5
So, ³
73. True
d ª
x
º
arccos C »
2
dx «¬
¼
74. False. Use substitution: u
12
du
a2 u 2
§ uc ·
¨ ¸
1 u2 a2 © a ¹
1
uc
a u2
§u·
arcsin ¨ ¸ C.
©a¹
1
1 x 2
2
9 e 2 x , du
4 x2
2e 2 x dx
INSTRUCTOR USE ONLY
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NOT FOR SALE
532
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
76.
d ª1
u
º
arctan C »
«
dx ¬ a
a
¼
1 ª uc a º
«
»
a «1 u a 2 »
¬
¼
º
1ª
uc
«
»
2
a « a2 u 2 a2 »
¬
¼
So, ³
uc
du
a u2
³ a 2 u 2 dx
2
uc
a2 u2
1
u
arctan C.
a
a
77. Assume u ! 0.
ª
1«
a« u a
¬«
d ª1
u
º
arcsec C »
«
dx ¬ a
a
¼
ª
1«
a «u
¬«
º
»
2
»
u a 1 ¼»
uc a
uc
u a
2
2
º
»
a 2 »»
¼
uc
u u a2
2
.
The case u 0 is handled in a similar manner.
So, ³
uc
du
u u a
2
arctan x 78. Let f x
y
1
1 x
2
1 x2
1 x2
2
2x
! 0 for x ! 0.
1 x2
0 and f is increasing for x ! 0,
Because f 0
arctan x x
1 x2
2
fc x
u
1
C.
arcsec
a
a
³ u u 2 a 2 dx
2
x
x
.
! 0 for x ! 0. So, arctan x !
1 x2
1 x2
y3
4
3
2
y2
1
y1
2
4
6
x
8
10
x arctan x
Let g x
gc x
5
1
1
1 x2
x2
! 0 for x ! 0.
1 x2
0 and g is increasing for x ! 0, x arctan x ! 0 for x ! 0. So, x ! arctan x. Therefore,
Because g 0
x
arctan x x.
1 x2
79. (a) Area
1
1
³ 0 1 x 2 dx
(b) Trapezoidal Rule: n
8, b a
10
1
Area | 0.7847
(c) Because
1
1
³ 0 1 x 2 dx >arctan x@0
1
S
4
,
you can use the Trapezoidal Rule to approximate S 4, and therefore, S . For example, using n
S | 4 0.785397
200, you obtain
3.141588.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ection 5.7
Inverse Trigonometric Functions
Functions: Integration
533
32t 500
80. (a) v t
550
0
20
0
³ v t dt
(b) s t
³ 32t 500 dt
16t 2 500t C
s0
16 0 500 0 C
st
16t 2 500t
0 Ÿ C
0
When the object reaches its maximum height, v t
32t 500
vt
32t
500
t
15.625
s 15.625
16 15.625
0.
0
2
500 15.625
3906.25 ft Maximum height
1
³ 32 kv2 dv
(c)
³ dt
§
1
arctan ¨¨
32k
©
k ·
v¸
32 ¸¹
t C1
§
arctan ¨¨
©
k ·
v¸
32 ¸¹
32kt C
k
v
32
tan C v
When t
0, v
vt
§
32 ª
tan «arctan ¨¨ 500
k
©
¬«
(d) When k
500, C
32kt
32
tan C k
32kt
arctan 500 k 32 , and you have
k ·
¸ 32 ¸¹
º
32kt ».
¼»
0.001:
32,000 tan ªarctan 500 0.00003125 ¬
vt
0.032t º
¼
500
0
7
0
0 when t0 | 6.86 sec.
vt
(e) h
6.86
³0
32,000 tan ªarctan 500 0.00003125 ¬
Simpson’s Rule: n
0.032 t º dt
¼
10; h | 1088 feet
(f) Air resistance lowers the maximum height.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
534
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
Section 5.8 Hyperbolic Functions
e3 e 3
| 10.018
2
1. (a) sinh 3
sinh 2
(b) tanh 2
e 2 e2
| 0.964
e 2 e 2
cosh 2
e0 e0
2
2. (a) cosh 0
3. (a) csch ln 2
(b) coth ln 5
1 § 4·
ln ¨ ¸ | 0.347
2 © 2¹
2
1
7. tanh 2 x sech 2 x
cosh ln 5
eln 5 e ln 5
eln 5 e ln 5
1
§1 ln ¨
¨
©
§ e x e x ·
¨
¸
2
©
¹
2
cosh 2 x 1
sinh 2 x
3 | 1.317
9.
1 cosh 2 x
2
1
1 e 2 x e 2 x 2
e
1 49 ·
¸ | 0.962
¸
23
¹
2x
2
2 e 2 x
4
§ e x e x ·
¨
¸
2
©
¹
2
cosh 2 x
e2 x 2 e 2 x
4
§ e 2 x e 2 x ·
1 ¨
¸
2
©
¹
2
1 cosh 2 x
2
1
cosh 2 x
1
sinh 2 x
sinh 2 x
sinh 2 x
sinh 2 x
ln 2 2
3
So, sinh 2 x
8. coth 2 x csch 2 x
0
5. (a) cosh 1 2
2
2
e2 x 2 e 2 x
e2 x 2 e 2 x
0
0
(b) sech 1
4
3
13
12
5 15
4. (a) sinh 1 0
e x e x
2
2 12
sinh ln 5
§ e x e x ·
2
§
·
¨ x
¨ x
x ¸
x ¸
©e e ¹
©e e ¹
e2 x 2 e 2 x 4
2
eln 2 e ln 2
5 15
10. sinh 2 x
(b) coth 1 3
§1 5 ·
ln ¨¨
¸ | 0.481
2 ¸¹
©
2
| 0.648
e e 1
(b) sech 1
(b) tanh
6. (a) csch 1 2
2 e 2 x e 2 x
4
1 cosh 2 x
.
2
11. 2 sinh x cosh x
12. sinh 2 x cosh 2 x
§ e x e x ·§ e x e x ·
2¨
¸¨
¸
2
2
©
¹©
¹
e 2 x e 2 x
2
e 2 x e 2 x
e 2 x e 2 x
2
2
13. sinh x cosh y cosh x sinh y
sinh 2 x
e2 x
§ e x e x ·§ e y e y · § e x e x ·§ e y e y ·
¨
¸¨
¸ ¨
¸¨
¸
2
2
2
2
©
¹©
¹ ©
¹©
¹
1ª x y
x
y
e x y e x y e
e x y e x y e x y e x y º
e
¼
4¬
1 ª x y
2e
e x y º
¼
4¬
e x y e x y
2
sinh x y
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 5.8
14. 2 cosh
x y
x y
cosh
2
2
§ 3·
cosh 2 x ¨ ¸
© 2¹
2
csch x
sech x
coth x
1
13 2
2 13
13
1
13
3
3
13
1
2
sinh x
x o0
x
2
csch x
1
3 2
1
12
3
Ÿ sech x
4
1 Ÿ sech 2 x
1
3 3
tanh x
§ 1 ·§ 2 3 ·
¸
¨ ¸¨¨
© 2 ¹© 3 ¸¹
coth x
f
18. lim tanh x
e x e x
x of e x e x
19. lim sech x
0
20. lim csch x
0
x of
x of
24.
fc x
3 cosh 3 x
f x
cosh 8 x 1
fc x
8 sinh 8 x 1
3
2
26.
2
1
sech 5 x 2
sech 5 x 2 tanh 5 x 2 10 x
10 x sech 5 x 2 tanh 5 x 2
27.
lim
sinh 3 x
yc
3
sech x
17. lim sinh x
x of
3
3
f x
25. y
csch x
2 3
3
1
2
x of
xo0
3
3
3
1
22. lim coth x does not exist.
23.
tanh x cosh x
e x e x
xo0
2x
lim
coth x o f for x o 0 , coth x o f for x o 0
2
sinh x
3
2
2 3
3
Putting these in order:
cosh x
13
2
21. lim
§1·
2
¨ ¸ sech x
©2¹
sinh x
13
Ÿ cosh x
4
1 Ÿ cosh 2 x
32
3 13
13
13 2
1
2
32
3
tanh x
coth x
e x e x
e y e y
2
2
3
2
15. sinh x
cosh x
535
ª e x y 2 e x y 2 º ª e x y 2 e x y 2 º
2«
»«
»
2
2
¬«
¼» ¬«
¼»
ª e x e y e y e x º
2«
»
4
¬
¼
cosh x cosh y
16. tanh x
Hyperbolic Functions
Hyperbol
Hyperbo
f x
tanh 4 x 2 3 x
fc x
8 x 3 sech 2 4 x 2 3 x
f x
ln sinh x
fc x
1
cosh x
sinh x
coth x
INSTRUCTOR USE ONLY
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536
NOT FOR SALE
Chapter 5
28. y
yc
Logarithmic,
arithmic, Exponential,
Exponen
and Other Transcendental Functions
Function
x·
§
ln ¨ tanh ¸
2¹
©
33.
12
§ x·
sech 2 ¨ ¸
tanh x 2
© 2¹
sinh 1 x 2 , 1, 0
yc
cosh 1 x 2 2 x
yc 1
1
2 sinh x 2 cosh x 2
1
sinh x
y
csch x
Tangent line: y 0
2 x 1
y
2 x 2
y
34.
ln y
29. h x
1
x
sinh 2 x 4
2
hc x
1
1
cosh 2 x 2
2
30. y
yc
cosh 2 x 1
2
yc
y
sinh 2 x
cosh x ln x
cosh x
sinh x ln x
x
cosh 1 .
cosh 1 x 1
cosh 1 x cosh 1 1
y
x sinh x cosh x cosh x
Note: cosh 1 | 1.5431
arctan sinh t
fc t
1
cosh t
1 sinh 2 t
gc x
1, 1
Tangent line: y 1
x cosh x sinh x
f t
32. g x
x cosh x ,
At 1, 1 , yc
x sinh x
31.
2
2
cosh x sinh x ,
35. y
cosh t
cosh 2 t
yc
sech t
2 cosh x sinh x sinh x cosh x
At 0, 1 , yc
2
sech 3 x
2 sech 3 x sech 3 x tanh 3 x 3
0, 1
2 1 1
2.
Tangent line: y 1
2 x 0
y
2 x 1
6 sech 2 3x tanh 3 x
36.
y
esinh x ,
0, 1
yc
sinh x
cosh x
yc 0
e
0
e 1
1
Tangent line: y 1
y
37.
f x
sin x sinh x cos x cosh x,
fc x
sin x cosh x cos x sinh x cos x sinh x sin x cosh x
2 sin x cosh x
0 when x
4 d x d 4
(−π , cosh π )
1x 0
x 1
12 (π , cosh π )
0, r S .
− 2
Relative maxima: rS , cosh S
(0, − 1)
2
−2
Relative minimum: 0, 1
38.
f x
x sinh x 1 cosh x 1
fc x
x cosh x 1 sinh x 1 sinh x 1
fc x
0 for x
x cosh x 1
0.
By the First Derivative Test, 0, cosh 1
| 0, 1.543 is a relative minimum.
6
−6
6
(0, −1.543)
−2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.8
39.
g x
x sech x
gc x
sech x x sech x tanh x
x tanh x
x
,
25
18 25 cosh
42. (a) y
sech x 1 x tanh x
Hyperbol
Hyperbo
Hyperbolic Functions
537
25 d x d 25
y
0
80
1
60
Using a graphing utility, x | r1.1997.
By the First Derivative Test, 1.1997, 0.6627 is a
20
relative maximum and 1.1997, 0.6627 is a relative
x
−20
−10
10
20
minimum.
r25, y
(b) At x
1
(1.20, 0.66)
−
At x
18 25 cosh 1 | 56.577.
18 25
0, y
(c) yc
sinh
x
. At x
25
43.
25, yc
sinh 1 | 1.175.
(− 1.20, − 0.66) − 1
40.
h x
2 tanh x x
hc x
2 sech 2 x 1
43. ³ cosh 2 x dx
³ cosh 2 x 2 dx
1 sinh 2 x C
2
0
44. ³ sech 2 3 x dx
1
2
sech 2 x
1
2
1
3
³ sech 3x 3 dx
2
1 tanh
3
Using a graphing utility, x | 0.8814.
From the First Derivative Test, 0.8814, 0.5328 is a
relative maximum and 0.8814, 0.5328 is a relative
minimum.
1 2 x, du
45. Let u
³ sinh 1 2 x dx
3x C
2 dx.
12 ³ sinh 1 2 x 2 dx
12 cosh 1 2 x C
2
(0.88, 0.53)
−3
3
46. Let u
(− 0.88, − 0.53)
−2
³
x
cosh
x
x
, 15 d x d 15
15
10 15 cosh
41. (a) y
y
At x
48. Let u
(c) yc
§ 1 ·
x¨
¸ dx
©2 x ¹
2 sinh
x C
r15, y
cosh x, du
sinh
x
10
0, y
2³ cosh
sinh x 1 dx.
1 cosh 3 x 1
3
2
10
(b) At x
dx.
³ cosh x 1 sinh x 1 dx
20
−10
dx
x
2
cosh x 1 , du
47. Let u
30
1
x , du
³ 1 sinh 2 x dx
20
x
sinh . At x
15
15, yc
25.
sinh 1 | 1.175.
sinh x dx.
sinh x
³ cosh 2 x dx
1
C
cosh x
sech x C
10 15 cosh 1 | 33.146.
10 15 cosh 0
C
49. Let u
sinh x, du
cosh x
³ sinh x dx
ln sinh x C
2 x 1, du
50. Let u
cosh x dx.
³ sech 2 x 1 dx
2
2 dx.
1
2
³ sech 2 x 1 2 dx
2
1 tanh 2 x 1
2
C
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
538
Chapter 5
NOT FOR SALE
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
x2
, du
2
51. Let u
§
2x ·
³ ¨© csch 2 ¸¹ x dx
2
2
2x
³ x csch 2 dx
52. Let u
59. Let u
x dx.
³0
2
x
coth
C
2
³ sech x tanh x dx
csch 1 x coth 1 x
x2
sinh x, du
cosh x
³
9 sinh 2 x
ln 2
0
x
1
1§ 1 ·
coth ¨ 2 ¸ dx
x
x© x ¹
2 4
ª e x e x º
2e x «
»
2
¬
¼
ln 2
³0
2
2 dx
S
4
1 e 2 x
1 e 2 x dx
ª
1 § 1 ·º ª
1º
«ln 2 2 ¨ 4 ¸» «0 2 »
© ¹¼ ¬
¼
¬
3
ln 2
8
1
C
x
cosh x dx.
ln 2 sinh x
³ 0 cosh x dx, u
tanh x dx
1 2x
ln 2
³ csch
dx
1
2 4
1 2 x º
ª
«x 2 e »
¬
¼0
61. Answers will vary.
62. f x
§ e x e x ·
arcsin ¨
¸ C
6
©
¹
55. ³
³0
dx
2e x cosh x
ln 2
§ sinh x ·
arcsin ¨
¸ C
© 3 ¹
dx
2
³ 0 2e cosh x dx
1
dx.
x2
csch
54. Let u
1 4x
60.
13 sech 3 x C
³
2
2 4
ª¬arcsin 2 x º¼ 0
³ sech 2 x sech x tanh x dx
3
1
, du
x
2 dx.
sech x tanh x dx.
sech x, du
53. Let u
2 x, du
cosh x and f x
values. f x
sech x take on only positive
sinh x and f x
tanh x are
increasing functions.
63. The derivatives of f x
cosh x
f x
cosh x and
sech x differ by a minus sign.
ln 2
ª¬ln cosh x º¼ 0
§5·
ln ¨ ¸ 0
© 4¹
1 1 cosh 2 x
1
56. ³ cosh 2 x dx
³0
0
increasing on 0, f .
§5·
ln ¨ ¸
© 4¹
2
2
concave downward on 0, f .
65. y
4
5 xº
ª1
«10 ln 5 x »
¬
¼0
58. ³
4
1
0
25 x 2
yc
1
1
1
1
dx ³
dx
10 ³ 5 x
10 5 x
dx
4
dx
xº
ª
«arcsin 5 »
¬
¼0
3
1
ln 9
10
4
arcsin
5
9x2 1
tanh 1
66. y
1 1
sinh 1 cosh 1
2 2
1
cosh 1 3 x
yc
1ª
1
º
1 sinh 2 »
2 «¬
2
¼
4
tanh x is concave upward on f, 0 and
g x
dx
1
0 25 x 2
cosh x is concave upward on f, f .
(b) f x
5
4
1ª
1
º
x sinh 2 x »
2 «¬
2
¼0
57. ³
tanh x is increasing on f, f .
g x
2 1/2
eln 2 e ln 2
2
Note: cosh ln 2
cosh x is decreasing on f, 0 and
64. (a) f x
ln cosh ln 2 ln cosh 0
1
ln 3
5
1
§1·
2¨ ¸
1 x 2 © 2¹
tanh 1
y
67.
yc
x
2
1
1
2
2
4 x2
x
§ 1 1 2 ·
x ¸
2¨
¹
x ©2
1
x1 x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.8
68.
coth 1 x 2
f x
1
fc x
69. y
yc
70. y
yc
71. y
yc
72. y
yc
77.
2x
2
1 x2
2x
1 x4
74. y
1
sec 2 x
tan 2 x 1
sec x
1
2 cos 2 x
1 sin 2 2 x
csch
1
x
75. ³
2 sec 2 x
76. ³
·
¸
1 x 2 ¸¹
sech 1 cos 2 x ,
2 csch 1 x
cos 2 x 1 cos 2 2 x
1 x2
x
0 x 1
x
dx
9 x4
1 e2 x ·
¸C
¸
ex
¹
§
ln ¨
©1 ex
·
¸C
2x
1 e ¹
1 e2 x 1 x C
2 x
1
dx
2 ³ 9 x2 2
1§ 1 ·
3 x2
¨ ¸ ln
C
2© 6 ¹
3 x2
S
4
2 sec 2 x,
79. Let u
³
1 4 x2
§
2 x¨
©
·
1
¸ 2 sinh 2 x 2
1 4x ¹
2 sinh
1
1
3 x2
ln
C
12
3 x2
1
x , du
1
dx
x 1 x
dx.
x
2
2³
1 4x2
§ 1 ·
¸ dx
x¹
©
1
1
x C
x 2 ln
2 ¨2
x
2 sinh 1
4x
2
1 x C
2x
1
x tanh x ln 1 x
1
ln 1 x 2
2
³
x
1 x3
dx
tan 1 x
1 1
ln
32 3
3 3x
C
3 3x
3
1
ln
18
1
3x
C
3x
81. ³
1
dx
4x x2
1
1
2 dx
2 ³ 2x 1 2x 2
1 ª1 ln «
2 ¬«
2
3³
x dx.
§3
1
1 x
32
2¨
©2
·
x ¸ dx
¹
2
sinh 1 x 3 2 C
3
2
ln x3 2 1 x3 C
3
1
1
3 dx
3 ³ 3 3x 2
dx
3
2
x3 2 , du
80. Let u
2
x
§ 1 ·
tanh 1 x x¨
2¸
1 x2
©1 x ¹
2x 1 4x2
78. ³
2 sin 2 x
2
cos 2 x
2 x sinh 1 2 x 1
§1 ln ¨
¨
©
ln
1
dx
2
§ e x e x 1 e x ·
¸C
ln ¨
¨
¸
e2 x
©
¹
2
§
2 csch 1 x¨
¨x
©
1
dx
3 9 x2
1 ex
e
tanh 1 sin 2 x
x tanh 1 x yc
1e
ex
³ x
dx
2x
539
csch 1 e x C
since sin 2 x t 0 for 0 x S 4.
yc
1
sinh 1 tan x
2 sin 2 x
cos 2 x sin 2 x
73. y
³
Hyperbol
Hyperbo
Hyperbolic Functions
1
³ x 2 2 4 dx
x 2 2
1
ln
4
x2 2
1
x4
C
ln
4
x
1 4x2 º
» C
2x
¼»
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
540
NOT FOR SALE
Chapter 5
dx
82. ³
83. ³
84. ³
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
x 2
7
³
x2 4 x 8
1
dx
ª 1 §2 « ln ¨
2 ©¨
¬«
3
1
1
x 4 x2
2
x 2
7
ªln x ¬«
x 4
2
x 2
dx
3
x2 4 º
¼» 3
4
ln 7 1
1
1 16 9 x 2
dx
45 ln 3 x 2
x 2
y
1
ª1 1 1
4 3x º
« 3 4 2 ln 4 3 x »
¬
¼ 1
1
0
1
25 x 1
2
dx
1 1
5³0
1
5x
ª1
« 5 ln 5 x ¬
1
ln 5 5
89. y
x 3 21x
³ 5 4 x x 2 dx
2
1
§
ln ¨¨
©
4 x 1, du
4 dx.
80 8 x 16 x 2
4
81 4 x 1
2 x 1 , du
88. Let u
y
1ª
§ 1 ·º
ln 7 ln ¨ ¸»
«
24 ¬
© 7 ¹¼
86. ³
45 ·
¸
5 ¸¹
2
5 3·
¸
2 ¸¹
dx
dx
1
§ 4x 1·
arcsin ¨
¸ C
4
© 9 ¹
1 1
1
3 dx
3 ³ 1 42 3 x 2
1
>ln 7 ln 1 ln 7@
24
4 ·¸
C
¸¸
¹
1
³
1
4³
5
2
§7 ln ¨¨
© 3
5
87. Let u
3
4 x 2 ·º
¸»
¸
x
¹¼»1
1 § 2 13 · 1
ln ¨¨
¸¸ ln 2 2 ©
3
¹ 2
85. ³
§2 1
ln ¨
2 ¨¨
©
dx
³ x 1
³
1
ln 7
12
5 dx
2 dx.
1
4 x 2 8 x 1
dx
2
2 x 1
1
ln
3
3
3 2
dx
ª¬2 x 1 º¼
2
4 x 2 8 x 1
C
2 x 1
1
º
25 x 2 1 »
¼0
26
§
20
·
³ ©¨ x 4 5 4 x x 2 ¸¹ dx
1
³ x 4 dx 20³ 32 x 2 2 dx
3 x 2
x2
20
4x C
ln
2
6
3 x 2
x2
10
1 x
4x C
ln
2
3
5 x
x2
10
5 x
4x C
ln
x 1
2
3
90. y
1 2x
³ 4 x x 2 dx
4 2x
1
ln 4 x x 2 x 2 2
3
ln
C
x 2 2
4
ln 4 x x 2 x 4
3
ln
C
x
4
³ 4 x x 2 dx 3³ x 2 2 4 dx
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section 5.8
91. A
x
dx
2
4
2
dx
2³ x 2
0 e
ex 2
4
2 ³ sech
93. A
0
4³
4
e
0
ex 2
x2
2
dx
1
4
ª8 arctan e x 2 º
¬
¼0
8 arctan e 2 2S | 5.207
92. A
2
³0
94. A
e 2 x
³ 0 e2 x e2 x dx
2 e2 x
tanh 2 x dx
1 2
1
2 e 2 x e 2 x dx
2 ³ 0 e 2 x e 2 x
2
ª1
2 x º
2x
« 2 ln e e »
¬
¼0
e4 e 4
| 1.654
2
ln
95. ³
1
1
ln e 4 e 4 ln 2
2
2
3k
dt
16
³ x 2 12 x 32 dx
3kt
16
³ x 6 2 4 dx
0
C
541
5x
dx
x4 1
5 2
2x
dx
2
2³0
x2 1
2
ª5
2
« 2 ln x ¬
º
x4 1 »
¼0
5
ln 4 2
17 | 5.237
5
³3
6
x2 4
dx
5
ª6 ln x ¬«
x2 4 º
¼» 3
6 ln 5 21 6 ln 3 §5 6 ln ¨¨
©3
21 ·
¸ | 3.626
5 ¸¹
5
1
x 6 2
1
ln
C
22
x 6 2
1
When x
t
2
³0
Hyperbolic Functions
Hyperbol
Hyperbo
0:
When x
t
1
ln 2
4
30k
16
k
1
x 8
ln
C
4
x 4
When t
1:
10
1
7
1
ln
ln 2
4
4
3
1 §7·
ln ¨ ¸
4 ©6¹
20:
§ 3 ·§ 2 · § 7 ·
¨ ¸¨ ¸ ln ¨ ¸ 20
© 16 ¹© 15 ¹ © 6 ¹
2 §7·
ln ¨ ¸
15 © 6 ¹
x 8
1
ln
4 2x 8
2
§7·
ln ¨ ¸
©6¹
49
36
62 x
x
ln
x 8
2x 8
x 8
2x 8
104
104
62
52
| 1.677 kg
31
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
542
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
32t
96. (a) v t
³ v t dt
(b) s t
2
³ 32t dt
s0
16 0
st
16t 2 400
dv
dt
(c)
C
dv
³ dt
dv
³ dt
³ 32 kv 2
k v, then du
Because v 0
ln
0, C
k v
k v
32 32 kv
kv
2 32k t
32 32 kv
kv
e 2 32 k t
32 kv
e 2 32 k t
32
tanh
k
32 e 2 32 k t 1
k e 2 32 k t 1
º
32k t »
»¼
The velocity is bounded by (e) Because ³ tanh ct dt
st
³
When t
s0
When k
32 kv
32 e 2 32 k t 1
v
ª
(d) lim «
t of«
¬
t C
0.
k e2 32 k t
k 400
k dv.
32 32 1
1
ln
˜
k 2 32
v
400 Ÿ C
32 kv 2
³ kv 2 32
Let u
16t 2 C
˜
e 32 k t
e 32 k t
32 k t
ª
e 32 k t º
32 « e
»
k « e 32 k t e 32 k t »
«¬
»¼
32
tanh
k
32k t
32
k
k.
32
1 c ln cosh ct (which can be verified by differentiation), then
32
tanh
k
32k t dt
32
k
1
ln ªcosh
32k ¬
32k t º C
¼
1
ln ªcosh
k ¬
32k t º C.
¼
0,
C
400 Ÿ 400 1 k ln ªcosh
¬
32k t º.
¼
0.01:
s2 t
400 100 ln cosh
s1 t
16t 2 400
s1 t
0 when t
s2 t
0 when t | 8.3 seconds
0.32 t
5 seconds
When air resistance is not neglected, it takes approximately 3.3 more seconds to reach the ground.
(f ) As k increases, the time required for the object to reach the ground increases.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 5.8
x
a
a sech 1
97. (a) y
a2 x2 ,
1
dy
dx
1 x2 a
x a
2
a sech 1
0, y
543
a ! 0
a 2
x
a x
2
x0
a
x a x
2
2
2
x0 , y0 : y a sech 1
(b) Equation of tangent line through P
When x
Hyperbolic Functions
Hyperbol
Hyperbo
a 2 x02 x2 a2
x
a x
2
x0
a
a 2 x02
x a x
2
2
a 2 x02
a sech 1
2
a2 x2
x
a 2 x02
x x0
x0
x0
.
a
So, Q is the point ª¬0, a sech 1 x0 a º¼.
x0 0
Distance from P to Q: d
2
y0 asech 1 x0 a
x02 a 2 x02
2
a2
a
y
Q
a
P
(a, 0)
x
L
98. In Example 5, a
20. From Exercise 97(a), yc
202 x 2
.
x
y
(0, y1)
(x, y)
x
10
20
The slope of the line connecting x, y and 0, y1 can be determined by analyzing the shaded triangle. From Exercise 97(b),
the hypotenuse is a.
20
20 2 − x 2
x
m
202 x 2
x
yc
Hence, the boat is always pointing toward the person.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
544
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
tanh 1 x, 1 x 1
u
99. Let
tanh u
x.
sinh u
cosh u
eu e u
eu e u
x
eu e u
xeu xe u
e 2u 1
xe 2u x
e 2u 1 x
1 x
1 x
§1 x ·
ln ¨
¸
©1 x ¹
2u
sinh
2t
e 2u 2teu 1
0
2t r
4t 4
2
t r
t2 1
t t 2 1 because eu ! 0
tanh x
arctan sinh x
x
e +e
sinh x
cosh x
sinh x
coth x
sinh 2 x cosh 2 x
sinh 2 x
1
sinh 2 x
2 e x e x
2
e x e x
sech x tanh x
cosh 1 x
y
cosh y
x
sinh y yc
1
yc
1
sinh y
y
sinh 1 x
sinh y
x
cosh y yc
1
yc
csch 2 x
2
e x e x
sech x
1
cosh y
y
108.
x
e x e x
sinh x.
2
arctan sinh x . Therefore,
So, y
107.
t 1
e e
and
e x e x
x
e x e x
2
§ 2 ·§ e x e x ·
¨ x
x ¸¨ x
x ¸
© e e ¹© e e ¹
2
arcsin tanh x . Then,
tan y
yc
e x e x
2
2
ln t u
sin y
105. y
cosh x
106.
t
eu
101. Let y
yc
t. Then
eu e u
2
eu e u
sinh u
104. y
1 §1 x ·
ln ¨
¸, 1 x 1
2 ©1 x ¹
u
100. Let u
yc
1 x
e 2u
1
103. y
1
1
sinh y 1
x 1
2
sech 1 x
sech y
x
1
arcsin tanh x .
1
x2 1
2
sech y tanh y yc
yc
1
cosh 2 y 1
1
sech y tanh y
1
sech y
1 sech 2 y
1
−x
x
e −e
−x
x 1 x2
y
2
b
102. ³
b
b
e xt dt
ª e xt º
« »
¬ x ¼ b
e xb
e xb
x
x
2 ª e xb e xb º
«
»
x¬
2
¼
2
sinh xb
x
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 5
109. y
110. There is no such common normal. To see this, assume
there is a common normal.
x
c
c cosh
y cosh a
The slope at P is sinh x1 c . The equation of line L is
When y
y
§x ·
c sinh ¨ 1 ¸. The
©c¹
length of L is
§x ·
c 2 sinh 2 ¨ 1 ¸ c 2
©c¹
(c, sinh c)
y = cosh x
x
y = sinh x
y sinh c
x
Ÿ x
sinh x1 c
0, c
1
x a.
sinh a
(a, cosh a)
Similarly,
1
x 0.
sinh x1 c
y c
sinh x
Normal line at a, cosh a is
x
c
sinh
cosh x Ÿ yc
y
Let P x1 , y1 be a point on the catenary.
yc
545
x
c ˜ cosh 1
c
1
x c
cosh c
is normal at c, sinh c . Also,
1
sinh a
1
Ÿ cosh c
cosh c
sinh a.
y1 ,
The slope between the points is
the ordinate y1 of the point P.
Therefore, y
a c
cosh a sinh c
sinh c cosh a
.
c a
cosh c
sinh a.
cosh c ! 0 Ÿ a ! 0
P(x1, y1)
sinh x cosh x for all
(0, c)
x Ÿ sinh c cosh c
c a. But,
L
x
sinh a cosh a. So,
a c
0, a contradiction.
cosh a sinh c
Review Exercises for Chapter 5
ln x 3
1. f x
3. ln 5
Vertical shift 3 units downward
Vertical asymptote: x
4x2 1
4x2 1
0
y
−1
4. ln ª¬ x 2 1 x 1 º¼
x
−1
−2
1
2
3
4
2x 1 2x 1
1
ln
5
4x2 1
1ª
ln 2 x 1 ln 2 x 1 ln 4 x 2 1 º¼
5¬
5
ln x 2 1 ln x 1
x=0
−3
5. ln 3 −4
1
ln 4 x 2 ln x
3
−5
ln 3 ln 3 4 x 2 ln x
§ 3 3 4 x2 ·
¸
ln ¨
¨
¸
x
©
¹
−6
ln x 3
2. f x
Horizontal shift 3 units to the left.
3
Vertical asymptote: x
y
x = −3 3
2
−1
x
−1
1
2
3
−2
−3
INSTRUCTOR USE ONLY
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546
Chapter 5
NOT FOR SALE
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
6. 3ª¬ln x 2 ln x 2 1 º¼ 2 ln 5
3 ln x 6 ln x 2 1 ln 52
ln x3 ln x 2 1
7. g x
ln
ln 2 x
1 1
2
2 2x
gc x
8.
2x
1
ln 2 x
2
12
6
ln 25
ª 25 x3 º
»
ln «
6
« x2 1 »
¬
¼
1
2x
f x
ln 3x 2 2 x
fc x
1
6x 2
3x 2 2 x
f x
x ln x
fc x
1 2 § 1 ·
§ x·
¨ ¸ ln x
¨ ¸ © 2¹
© x¹
ln x
1
2 ln x
1 2 ln x
2 ln x
y
2 x 2 ln x 2
yc
4x yc 1
4 2
14.
10.
ln x
ª¬ln 2 x º¼
fc x
3ª¬ln 2 x º¼
2
1
2
2x
y
6x 4
7 x 2, du
ln
1
1
1
7 dx
7 ³ 7x 2
1
ln 7 x 2 C
7
16. ³
x2
dx
x 1
1
1
3 x 2 dx
3 ³ x3 1
1
ln x3 1 C
3
17. ³
sin x
dx
1 cos x
³
sin x
dx
1 cos x
ln 1 cos x C
18. u
ln x, du
1
dx
x
³
x2 4
x2 4
1ª
ln x 2 4 ln x 2 4 º¼
2¬
1ª 1
1
º
˜ 2x 2
˜ 2 x»
x 4
2 «¬ x 2 4
¼
yc
7 dx
³ 7 x 2 dx
3
2
ln 2 x
x
11. y
6
6 x 1
6x 2
3x 2 2 x
3
f x
2
x
Tangent line: y 2
15. u
9.
19. ³
3
x
ln
dx
x
4 2x 1
2x
1
1
§1·
ln x ¨ ¸ dx
2³
© x¹
4§
§ 4x ·
ln ¨
¸
© x 6¹
1
1
x
x 6
12. y
4
yc
1
21. ³
0
22.
13.
y
yc
yc 1
2
ln 2 x ,
2 x
1
2
2
2 x
2 x
1 2
1
Tangent line: y 2
y
1 x 1
1, 2
e ln x
20. ³
ln 4 ln x ln x 6
6
x6 x
1 ·
1
ª
º
« x 2 ln x »
¬
¼1
4
8x
x 4 16
1
2
ln x C
4
³ 1 ¨©1 2 x ¸¹ dx
dx
1 ª 2 x3 8 x 2 x3 8 x º
«
»
x 4 16
2¬
¼
2 x 2 2 ln x, 1, 2
x
S 3
S
dx
1
ln 4 1
2
1§ 1 ·
e
³ 1 ln x ¨© x ¸¹ dx
sec T dT
T
³ 0 tan 3 dT
3 ln 2
e
2º
ª1
« ln x »
¬2
¼1
S 3
ª¬ln sec T tan T º¼ 0
S
3 ³ tan
0
1
2
ln 2 3
T §1·
¨ ¸ dT
3 © 3¹
S
ª
T º
« 3 ln cos 3 »
¬
¼0
§1·
3 ln ¨ ¸ 3 ln 1
©2¹
3 ln 2
x 1
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 5
23. (a)
f x
1x 3
2
y
1x 3
2
2 y 3
2x 3
f 1 x
25. (a)
x 1
y
x 1
x
y 1
x
y
x 1
y
2
2
f
2x 6
(b)
f x
1
y
f −1
4
f −1
8
x 2 1, x t 0
x
(b)
y
547
3
6
f
2
f
2
1
x
−8 −6
−2
6
x
8
−2 −1
−6
1
−1
2
3
4
−2
−8
(c) f 1 f x
(c) f 1 f x
f f
1
f 1 12 x 3
f 2x 6
x
2 12 x 3 6
1 2x 6
2
3
x
f f 1 x
y
5x 7
y 7
5
x 7
5
x
y
f 1 x
(b)
6
x3 2
y
x3 2
y 2
x
3
x 2
y
1
3
x
x
x2 1 1
x 2
y
(b)
5
4
f
3
f −1
f
1
4
f −1
2
2
4
x
−4 −3 −2 −1
x
−8 −6 −4 −2
1
x for x t 0.
f x
3
f
y
2
Domain f 1 : x t 0; Range f 1 : y t 1
26. (a)
x 7
5
x2 1
(d) Domain f : x t 1; Range f : y t 0
numbers
5x 7
f x2 1
x2
Domain f 1 : all real numbers; Range f 1 : all real
f x
x 1
x
(d) Domain f: all real numbers; Range f: all real numbers
24. (a)
f 1
3 4 5
−2
−3
−4
6
−4
−6
(c) f 1 f x
−8
(c) f 1 f x
f 1 5 x 7
f f 1 x
§ x 7·
f¨
¸
© 5 ¹
5x 7 7
5
§ x 7·
5¨
¸ 7
© 5 ¹
x
f f 1 x
f 1 x3 2
3
x3 2 2
x 2
3
x 2
f
3
3
2
x
x
(d) Domain f: all real numbers; Range f: all real numbers
x
(d) Domain f: all real numbers; Range f: all real numbers
Domain f 1 : all real numbers; Range f 1 : all real
numbers
Domain f 1 : all real numbers; Range f 1 : all real
numbers
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
548
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
f x
x3 2, a
fc x
3x ! 0
1
f x
3
x 1
y
3
x 1
y 1
x
f is monotonic (increasing) on f, f therefore f has
x 1
y
an inverse.
f 1 x
x3 1
f 31 3
1 Ÿ f 1 1
f −1
f c 31 3
32 3
27. (a)
3
3
29.
y
(b)
4
f 1 c 1
3
f
2
2
31 3
1
1
1
f c f 1 1
f c 31 3
3 32 3
1
35 3
x
−2
2
3
4
30.
−2
(c) f
1
f x
f
1 3
x 1
x 1
3
3
1
x 3,
f x
x
fc x
1
x
2
a
4
1
x 3
x 3 ! 0
f is monotonic (increasing) on [3, f) therefore f has an
x
inverse.
f 4
4 Ÿ f 1 4
(d) Domain f: all real numbers; Range f: all real numbers
fc 4
21
Domain f 1 : all real numbers; Range f 1 : all real
f 1 c 4
f f
1
f x 1
3
x
3
x 1 1
3
x
28. (a)
f x
x 2 5,
y
x 5
x
x 5
y
f
31.
y
6
tan x,
fc x
§ S S·
sec 2 x ! 0 on ¨ , ¸
© 4 4¹
f
f −1
−2
x
4
−2
6
−4
f f
x
§ 3·
3
Ÿ f 1 ¨¨
¸¸
3
© 3 ¹
§S ·
f c¨ ¸
©6¹
4
3
1
§
§ 3 ··
f c¨ f 1 ¨¨
¸¸ ¸¸
¨
© 3 ¹¹
©
f 1 x 2 5
x2 5 5
1
a
§S ·
f¨ ¸
©6¹
§ 3·
f 1 ¨¨
¸¸
© 3 ¹
−6
(c) f 1 f x
3 S
S
, d x d
3
4
4
f x
4
−4
f
1
3
ª S Sº
f is monotonic (increasing) on « , » therefore f has
¬ 4 4¼
an inverse.
x 5
x
(b)
−6
1
fc 4
2
y 5
1
x t 0
3
1
f c f 1 4
numbers
4
x 5
x for x t 0.
x 5
2
5
(d) Domain f : x t 0; Range f : y t 5
Domain f 1 : x t 5; Range f 1 : y t 0
32.
x
S
6
1
§S ·
f c¨ ¸
©6¹
1
§ 4·
¨ ¸
© 3¹
3
4
0, 0 d x d S
f x
cos x, a
fc x
sin x on 0, S
f is monotonic (decreasing) on >0, S @ therefore f has an
inverse.
§S ·
f¨ ¸
©2¹
0 Ÿ f 1 0
§S ·
f c¨ ¸
©2¹
1
f 1 0
1
f c f 1 0
S
2
1
S·
§
f c¨ ¸
©2¹
1
1
1
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 5
e3 x
33.
30
3x
ln 30
3x
ln 30
x
1
ln 30 | 1.134
3
ln e
34. 4 3e 2 x
6
3e
2x
10
e
2x
10
3
ln e 2 x
ln 10
3
2x
ln 10
3
x
35. ln
x 1
2
x 1
e2
x 1
e4
36. ln x ln x 3
0
ln x x 3
0
x x 3
e0
x 2 3x 1
0
x
x
3r
13
2
3
13
2
38. g x
§ ex ·
ln ¨
x¸
©1 e ¹
40. h z
hc z
41. g x
gc x
42. y
yc
2
1
0.
tet t 2
ln e x ln 1 e x
yc
13
43.
t 2et 2tet
39. y
3
only because
t 2 et
gc t
gc x
12 ln 10
| 0.602
3
e 4 1 | 53.598
x
37. g t
549
ex
1 ex
f x
e6 x , 0, 1
fc x
6e 6 x
fc 0
6e 0
6
Tangent line: y 1
6x 0
y
6x 1
x ln 1 e x
1
1 ex
44.
e 2 x e 2 x
1 2
1 2x
e e 2 x
2e 2 x 2e 2 x
2
2
e z 2
f x
e x 4 , 4, 1
fc x
ex 4
fc 4
e0
1
e 2 x e 2 x
Tangent line: y 1
1x 4
e 2 x e 2 x
y
x 3
y ln x y 2
45.
§1·
§ dy ·
§ dy ·
y¨ ¸ ln x ¨ ¸ 2 y¨ ¸
© x¹
© dx ¹
© dx ¹
dy
2 y ln x
dx
dy
dx
2
ze z 2
x2
ex
e x 2 x x 2e x
x2 x
2x
x
e
e
46.
3e 3 t
3e 3 t 3t 2
9e 3 t
t2
cos x 2
xe y
2 x sin x 2
xe y
dy
dx
0
0
y
x
y
x 2 y ln x
dy
ey
dx
2 x sin x 2 e y
xxe y
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
550
Chapter 5
2
2
47. ³ xe1 x dx
2 x3 1
³x e
2
12 ³ e1 x 2 x dx
x3 1, du
48. Let u
49. ³
NOT FOR SALE
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
dx
1
3
³e
12 e1 x C
3 x 2 dx.
x3 1
e4 x e2 x 1
dx
ex
1 e x3 1 C
3
3x 2 dx
e2 x e 2 x
e 2 x 1, du
2e 2 x dx.
2
ª1
º
2x
« 2 ln e 1 »
¬
¼0
1 2 2e 2 x
dx
2 ³ 0 e2 x 1
e x e x dx
1
1
ln e 4 1 ln 2
2
2
1 3x
e e x e x C
3
1 § e4 1·
ln ¨
¸ | 1.663
2 © 2 ¹
³e
3x
e 2 x e 2 x , du
2
³ 0 2e
55. A
x
2
ª¬ 2e x º¼
0
dx
2e2 x e 2 x dx.
2e 2 2 | 1.729
1 2e 2 x 2e 2 x
dx
2 ³ e 2 x e 2 x
1
ln e 2 x e 2 x C
2
³ e2 x e2 x dx
e2 x
dx
1
2
0 e2 x
Let u
e 4 x 3e 2 x 3
C
3e x
50. Let u
54. ³
56. (a), (c) 10,000
5
0
1
51. ³ xe
3 x 2
0
dx
2
1 1
³ e 3 x 6 x dx
0
6
0
V
9000e 0.6t ,
ª 1 3 x2 º
« 6 e
»
¬
¼0
Vc t
5400e
0.6t
1
ª¬e 3 1º¼
6
Vc 1
2963.58 dollars year
Vc 4
489.88 dollars year
(b)
1
1§
1·
¨1 3 ¸ | 0.158
6©
e ¹
e1 x
dx
1 2 x2
1
1
Let u
, du
dx.
x
x2
1
x
Ÿ u 2, x 2 Ÿ u
2
52. ³
53. ³
3x 2
57. y
y
6
2
³
12
2
eu du
12
ª¬eu º¼
2
0 d t d 5
5
4
3
2
x
−4 −3 − 2 −1
1
2
e1 2 e 2
1
2
3 4
1
2
−2
e2 e | 5.740
1
4
58. y
x
y
6
ex
dx
1 ex 1
3
4
e 1, du
x
Let u
3
5
ex
³ 1 e x 1 dx
3
x
e dx.
3
1
ªln e x 1 º
¬
¼1
−3
ln e 1 ln e 1
−2
x
−1
3
3
§ e3 1 ·
ln ¨
¸
© e 1¹
59. f x
fc x
3x 1
3x 1 ln 3
ln e 2 e 1 | 2.408
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 5
60. f x
53 x
fc x
67. t
3x
3 ln 5 5
3 ln 5 125
x
551
§ 18,000 ·
50 log10 ¨
¸
© 18,000 h ¹
(a) Domain: 0 d h 18,000
y
61.
62.
x 2 x 1
(b)
ln y
2 x 1 ln x
yc
y
2x 1
2 ln x
x
yc
§ 2x 1
·
2 ln x ¸
y¨
© x
¹
f x
x 43 x
fc x
x 3 ln 4 43 x 4 3 x
4
gc x
§ 2x 1
·
2 ln x ¸
x 2 x 1 ¨
© x
¹
Vertical asymptote: h
(c)
10t 50
18,000 h
1
log 3 1 x
2
1
1
§1·
¨ ¸
x
x
2
1
ln
3
2
1 ln 3
© ¹
dh
dt
x
x 1
18,000 1 10t 50
§1·
360 ln 10¨ ¸
© 10 ¹
P
log 5 x log 5 x 1
1 ª1
1 º
x 1»¼
ln 5 «¬ x
2
18,000 10t 50
h
68. (a) 10,000
1 ª 1 º
«
»
ln 5 «¬ x x 1 »¼
(b)
§ 1 · 1 x 1 2
5
C
¨ ¸
© 2 ¹ ln 5
1 1 t
2
C
ln 2
18,000
§ 18,000 ·
50 log10 ¨
¸
© 18,000 h ¹
18,000
18,000 h
t
3x ln 4 1
65. ³ x 1 5 x 1 dx
21 t
dt
t2
20,000
− 20
1 3 x ln 4
64 x
log 5
hc x
66. ³
− 2,000
log 3 1 x
63. g x
64. h x
3x
100
t 50
is greatest when t
0.
Pe 0.05 15
10,000
| 4723.67
e0.75
2P
Pe10 r
2
e10 r
ln 2
10r
ln 2
| 6.93%
10
r
69. (a)
Let T
arcsin
sin T
1
2
1·
§
sin ¨ arcsin ¸
2¹
©
1
2
1
.
2
sin T
2
1
θ
3
(b)
Let T
arcsin
sin T
1
2
1·
§
cos¨ arcsin ¸
2¹
©
cos T
1
2
3
.
2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
552
Chapter 5
NOT FOR SALE
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Function
Functions
Let T
70. (a)
arccot 2
cot T
71. y
1
.
2
tan T
1 x2
yc
5
72. y
1
θ
Let T
arcsec
sec T
5
5
73. y
1
74. y
θ
yc
1
2
yc
2 x arcsin x
1 x
x 4
2
x 2
1
x
1 x
4
dx
1
x 9 x 49
2
4x
x arcsec x
x
x2 1
x
arcsec x
1
arctan e2 x
2
1§ 1 · 2x
¨
¸ 2e
2 © 1 e4 x ¹
e2 x
1 e4 x
1 x
2
2x
1 x2
arcsin x
arcsin x
4
x2 4
2
x2
x 4
2
1
x
x 4
2
x
x 4
2
x2 4
x
1
1
2e 2 x dx
2 ³ 1 e2 x 2
1
arctan e 2 x C
2
5 dx.
1
5³
x 2 , du
79. Let u
2 1 x2
x
2
³ 1 e4 x dx
5 x, du
³ 3 25x 2 dx
1
2 x 4
e2 x
1
78. Let u
3 2
2e 2 x dx.
³ e2 x e2 x dx
80. ³
2
1
e 2 x , du
77. Let u
³
2
arcsin x
2
x
yc
1 x2
2 x 2 1 x 2 arcsin x
x
x 2 4 2 arcsec ,
2
76. y
2
2x 3
2
2
x arcsin x
1 2
arctan 2 x 2 3
yc
5
75. y
x2 1 x2
4x
4 x 4 12 x 2 10
2x
4
2x 6 x2 5
1
.
5
cos T
5
12
1 x2
yc
2
cos arcsec
1 x2
2
tan arccot 2
(b)
x
tan arcsin x
1
2
3
5x
2
1
5x
C
arctan
5 3
3
5 dx
2 x dx.
1
2³
dx
1
1 x2
2
³
1
3x
3x
2 x dx
2
1
arcsin x 2 C
2
3 dx
72
3x
1
arcsec
C
7
7
INSTRUCTOR USE ONLY
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Review Exercises ffor Chapter 5
§ x·
arctan ¨ ¸, du
© 2¹
81. Let u
82. Let u
1 4x2
4³
4 x
arcsin 2 x
4
2
C
dx
dx 2
1 2
1 1
4 x2
2 x dx
2³0
1
ª
§ x·
«4 arcsin ¨ 2 ¸ © ¹
¬
º
4 x2 »
¼0
84. A
6
³ 0 16 x 2 dx
ª6
§ x ·º
« 4 arctan ¨ 4 ¸»
© ¹¼ 0
¬
§
§1·
¨ 4 arcsin¨ ¸ © 2¹
©
85. y
sech 4 x 1
4
4
yc
2
x·
1§
¨ arctan ¸ C
4©
2¹
dx
2
4 x2
1
1
0
1 4x2
1 ¬ªarcsin 2 x ¼º
C
2
2
dx
4 x
1
³0
83. A
2
arcsin 2 x , du
arcsin 2 x
³
2
dx.
4 x2
x ·§ 2 ·
1 §
¨ arctan ¸¨
¸ dx
2³ ©
2 ¹© 4 x 2 ¹
arctan x 2
³ 4 x 2 dx
·
3¸ 2
¹
yc
87. y
yc
2
1
2
x
yc
3§ S ·
¨ ¸
2© 4 ¹
sinh 1 4 x
2
x
sinh x
2 x
16 x 2 1
1
§ 2 ·
tanh 1 2 x
x¨
2¸
©1 4x ¹
yc
91. Let u
16 x
4
2
x tanh 1 2 x
90. y
coth 8 x 2
csch 2 8 x 2
4
4x
x
sinh
3S
8
yc
x3 , du
³ x sech x
2
16 x csch 2 8 x 2
88. y
3 2 | 1.8264
89. y
sech 4 x 1 tanh 4 x 1 4
2 x cosh
2S
3
3
arctan 1 arctan 0
2
4 sech 4 x 1 tanh 4 x 1
86. y
553
3
2
2x
tanh 1 2 x
1 4x2
3x 2 dx.
dx
2
1
sech x3 3x 2 dx
3³
1
tanh x3 C
3
ln cosh x
1
sinh x
cosh x
92. ³ sinh 6 x dx
tanh x
93. Let u
1 cosh 6 x C
6
tanh x, du
sech 2 x dx.
sech 2 x
³ tanh x dx
1
³ tanh x sech x dx
94. Let u
csch 3 x , du
2
ln tanh x C
3 csch 3x coth 3x dx.
³ csch 3x coth 3x dx
4
1
csch 3 3x 3csch 3x coth 3 x dx
3³
csch 4 3 x
12
C
INSTRUCTOR USE ONLY
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554
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Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
2
x, du
3
95. Let u
1
³ 9 4 x 2 dx
2
dx.
3
19
dx
§4 ·
1 ¨ x2 ¸
©9 ¹
³
1
§2 ·
tanh 1 ¨ x ¸ C
6
©3 ¹
Alternate solution: ³
x 2 , du
96. Let u
³
x
x 1
4
1
32 2 x
2
1
3 2x
ln
C
12
3 2x
dx
2 x dx.
1
2³
dx
1
x2
1
ln x 2 2
2
2 x dx
1
x4 1 C
Problem Solving for Chapter 5
1. f x
f 0
f x
fc x
fc 0
f cc x
f cc 0
a bx
1 cx
1Ÿ a
e0
1 bx
1 cx
b c
1 cx
2
b c
1
2c c b
1 cx
1
Since b c
So, f x
1
3
2c c b
2c 1 Ÿ c
1, 1
1
and therefore, b
2
1
.
2
§1 ·
1 ¨ x¸
© 2 ¹.
§1 ·
1 ¨ x¸
©2 ¹
6
f
ex
−5
2
−2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Problem Solving ffor Chapter 5
2. (a)
(b)
y
555
y
4
2
3
1
π
2
−1
π
3π
2
2
x
2π
1
−2
π
2
S
³ 0 sin x dx
(c)
³
2S
sin x dx Ÿ ³
S
2S
0
sin x dx
2S
³0
0
1
π
2
1
2
x
1
³ 1 arccos x dx
§S ·
2¨ ¸
©2¹
1
π
4
S
x
2π
2 2S
π
2
1
y
1 tan x
S 2
2
ln x 1
x
x
§S 1 ·
symmetric with respect to point ¨ , ¸.
© 4 2¹
1
³ 0 1 tan x 2 dx
3. (a) f x
4S
y
π
−1
3π
2
sin x 2 dx
(d)
y
π
S §1·
¨ ¸
2© 2¹
S
4
1 d x d 1
,
2
−1
1
0
(b) lim f x
1
(c) Let g x
ln x, g c x
xo0
gc 1
lim
xo0
4. f x
g1 x g1
xo0
So, lim f x
1. From the definition of derivative
ln 1 x
x
.
1.
sin ln x
1
1
0, f
sin ln x Ÿ ln x
Two values are x
(c) f x
lim
xo0
x
(a) Domain: x ! 0 or
(b) f x
1 x, and g c 1
2
2kS
eS 2 , e S 2 2S .
sin ln x Ÿ ln x
Two values are x
S
3S
2kS
2
e S 2 , e3S 2 .
(d) Because the range of the sine function is >1, 1@, parts (b) and (c) show that the range of f is >1, 1@.
INSTRUCTOR USE ONLY
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556
NOT FOR SALE
Chapter 5
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
1
cos ln x
x
0 Ÿ cos ln x
0
fc x
(e)
fc x
S
Ÿ ln x
2
kS
eS 2 on >1, 10@
Ÿ x
½
°°
f 1 0
¾ Maximum is 1 at x
f 10 | 0.7440°°
¿
f eS 2
(f )
1
eS 2 | 4.8105
2
5
0
−2
1
lim f x seems to be . (This is incorrect.)
2
xo0
(g) For the points x
eS 2 , e 3S 2 , e 7S 2 , ! you have f x
1.
For the points x
e S 2 , e 5S 2 , e 9S 2 , ! you have f x
1.
That is, as x o 0 , there is an infinite number of points where f x
1,
1. So, lim sin ln x does not exist.
and an infinite number where f x
xo0
You can verify this by graphing f x on small intervals close to the origin.
0.5 x and y
5. y
ax
x Ÿ a
yc
x
1.2 x intersect y
2 x does not intersect y
x. y
x. Suppose y
x is tangent to y
a x at x, y .
x1 x .
1 Ÿ x ln x1 x
a ln a
1 Ÿ ln x
For 0 a d e1 e | 1.445, the curve y
1Ÿ x
a x intersects y
e1 e
e, a
x.
y
6
5
y=x
a=2
4
a = 0.5
3
2
a = 1.2
x
−4 −3 −2
1
2
3
4
−2
6. (a)
t
Ÿ Area sector
2S
Area sector
Area circle
(b) Area AOP
At
Ac t
At
But, A 0
t
S
2S
t
2
cosh t
1
base height ³
x 2 1 dx
1
2
cosh t
1
cosh t ˜ sinh t ³
x 2 1 dx
1
2
1
ªcosh 2 t sinh 2 t º¼ cosh 2 t 1 sinh t
2¬
1
ªcosh 2 t sinh 2 t º¼ sinh 2 t
2¬
1
ªcosh 2 t sinh 2 t º¼
2¬
1
2
1
t C
2
C
0 Ÿ C
0 So, A t
1
t or t
2
INSTRUCTOR USE ONLY
2A t .
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NOT FOR SALE
Problem Solving ffor Chapter 5
557
ln x is continuous on >1, e@ and differentiable on 1, e .
7. f x
By the Mean Value Theorem, there exists c  1, e such that
f e f 1
fc c
e 1
1
e 1
e 1.
1
c
c
8.
10
e 1
f x
ln x n
x
fc x
ª §1·
º
ln x »
x
n « ¨© x ¸¹
«
»
x2
¬
¼
n ln x
,
x
x ! e, n ! 0
n
1 ln x
x2
For x ! e, ln x ! 1 and f c x 0.
So, f is decreasing for x ! e.
9. (a) y
f x
sin y
(b)
arcsin x
x
S 4
Area A
S 4
³ S 6 sin y ˜ dy >cos y@ S 6
Area B
§ 1 ·§ S ·
¨ ¸¨ ¸
© 2 ¹© 6 ¹
2 2
³1 2
arcsin x dx
S
12
2
8
(c) Area A
ln 3
(d)
tan y
Area A
2
| 0.1589
2
1·
¸
12 ¹
© 8
2 2
3
§ S ·§ 2 ·
¸ A B
¨ ¸¨
© 4 ¹© 2 ¹
3 2
2
S
12
§
S¨
³ 0 e dy
y
ª¬e y º¼ ln 3
0
Area B
3 2
2
3
2
2
| 0.2618
Area C
S
3
³1 ln x dx
| 0.1346
y
31
y = ln x
2
ln 3
3 ln 3 A
3 ln 3 2
ln 27 2 | 1.2958
ey = x
A
B
x
x
1
S 3
2
3
³ S 4 tan y dy
S 3
ª¬ln cos y º¼ S / 4
ln
Area C
1
2
ln
2
2
3
³1 arctan x dx
S
12
4 3 3 y
ln
§S ·
¨ ¸
©3¹
2
1
ln 2
2
1
§S ·
3 ln 2 ¨ ¸ 1
2
©4¹
y = arctan x
π
3
π
4
1
ln 2 | 0.6818
2
A
C
B
x
1
3
INSTRUCTOR USE ONLY
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558
NOT FOR SALE
Chapter 5
10. y
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Functions
Function
ln x
11. y
ex
yc
ex
1
x
yc
y
If x
y
If y
b b 1
arctan sinh x
(a)
4
−6
e a x ae a b
aea b
bx
ab b
x
a 1
c
a 1
1.
So, a c
12. gd x
ea x a
0: e a x
b 1.
0, c
So, b c
Tangent line: y b
1
x a
a
1
x b 1
a
Tangent line: y b
a a 1
b
ea
1.
6
−4
(b) gd x
arctan sinh x
arctan sinh x
because sinh is odd
arctan sinh x
because arctan is odd
gd x
So, gd x is an odd function
(c)
d
gd x
dx
1
cosh x
1 sinh 2 x
cosh x
cosh 2 x
1
cosh x
sech x ! 0
So, gd x is monotonic (increasing) on f, f
(d)
d2
gd x
dx 2
d
sech x
dx
sech x tanh x
For x 0, sech x tanh x ! 0, and for
x ! 0, sech x tanh x 0. So, 0, 0 is the point of inflection.
(e) Let y
arcsin tanh x . Then
sin y
e x e x
, as indicated in the figure.
e x e x
tanh x
e x + e −x
e x − e −x
y
2
The third side of the triangle is
e x e x
2
e x e x
e e
2
x
Finally, tan y
y
2
arcsin tanh x
4
2
x
sinh x and
arctan sinh x
gd x
(f) From part (c),
d
gd x
dx
x
1
1
Ÿ ³
dt
0 cosh t
cosh x
>gd t @0
x
gd x gd 0
gd x
INSTRUCTOR USE ONLY
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NOT FOR SALE
Problem Solving ffor Chapter 5
13. Let u
1
Area
³1
x,
u 1, x
x
1
dx
x x
4
u 2 2u 1, dx
559
2u 2 du.
2u 2
3
³ 2 u 1 u 2 2u 1 du
32 u 1
³ 2 u 2 u du
32
3
³ 2 u du
ª¬2 ln u º¼ 2
§ 3·
2 ln ¨ ¸
© 2¹
2 ln 3 2 ln 2
| 0.8109
14. Let u
Area
sec 2 x dx.
tan x, du
S 4
³0
1
dx
sin x 4 cos 2 x
2
S 4
sec 2 x
dx
tan 2 x 4
1
du
³0
³ 0 u2 4
1
ª1
§ u ·º
« 2 arctan ¨© 2 ¸¹»
¬
¼0
1
§1·
arctan ¨ ¸
© 2¹
2
15. (a) (i) y
y1
ex
1 x
4
y
y1
−2
2
−1
(ii) y
y2
ex
§ x2 ·
1 x ¨ ¸
© 2¹
4
y
y2
−2
2
−1
(iii) y
y3
ex
1 x x2
x3
2
6
4
y
−2
y3
2
−1
(b) nth term is x n n! in polynomial: y4
(c) Conjecture: e x
1 x 1 x x2
x3
x4
2!
3!
4!
x2
x3
"
2!
3!
INSTRUCTOR USE ONLY
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560
Chapter 5
NOT FOR SALE
Logarithmic,
arithmic, Exponen
Exponential, and Other Transcendental Function
Functions
2 t
§
120,000 0.095 ·§
0.095 ·
985.93 ¨ 985.93 ¸¨1 ¸
12
12 ¹
©
¹©
16. (a) u
2 t
§
120,000 0.095 ·§
0.095 ·
¨ 985.93 ¸¨1 ¸
12
12 ¹
©
¹©
v
1000
u
v
0
35
0
(b) The larger part goes for interest. The curves intersect when t | 27.7 years.
(c) The slopes are negatives of each other. Analytically,
u
uc 15
(d) t
du
dt
14.06.
985.93 v Ÿ
vc 15
dv
dt
12.7 years
Again, the larger part goes for interest.
INSTRUCTOR USE ONLY
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C H A P T E R 6
Differential Equations
Section 6.1
Slope Fields and Euler’s Method.......................................................562
Section 6.2
Differential Equations: Growth and Decay.......................................574
Section 6.3
Separation of Variables and Logistic Equation.................................583
Section 6.4
First-Order Linear Differential Equations .........................................596
Review Exercises ........................................................................................................606
Problem Solving .........................................................................................................613
INSTRUCTOR USE ONLY
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NOT FOR SALE
C H A P T E R 6
Differential Equations
Section 6.1 Slope Fields and Euler’s Method
1. Differential equation: yc
4y
4. Differential Equation:
4x
Solution: y
Ce
Check: yc
4Ce 4 x
Solution: y 2 2 ln y
4y
yc
Check: 2 yyc e 2 x
2. Differential Equation: 3 yc 5 y
Solution: y
e 2 x
2e
Check: 3 2e 2 x 5 e 2 x
e 2 x
2xy
x2 y2
Solution: x 2 y 2
Cy
Check: 2 x 2 yyc
Cyc
yc
2 x
2y C
yc
2 xy
2 y 2 Cy
2
yc
y
xy
y2 1
x2
2x
§
1·
x
¨ y ¸ yc
y¹
©
x
yc
1
y y
2 x
3. Differential equation: yc
dy
dx
yc
xy
y2 1
2 xy
2 y x2 y2
2
2 xy
y 2 x2
2 xy
x2 y 2
5. Differential Equation: ycc y
0
Solution: y
C1 sin x C2 cos x
yc
C1 cos x C2 sin x
ycc
C1 sin x C2 cos x
Check: ycc y
C1 sin x C2 cos x C1 sin x C2 cos x
6. Differential equation: ycc 2 yc 2 y
Solution: y
Check:
0
0
C1e x cos x C2e x sin x
yc
C1 C2 e x sin x C1 C2 e x cos x
ycc
2C1e x sin x 2C2e x cos x
ycc 2 yc 2 y
2C1e x sin x 2C2e x cos x 2 C1 C2 e x sin x C1 C2 e x cos x 2 C1e x cos x C2e x sin x
2C1 2C1 2C2 2C2 e x sin x 2C2 2C1 2C2 2C1 e x cos x
0
INSTRUCTOR USE ONLY
562
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NOT FOR SALE
Section 6.1
7. Differential Equation: ycc y
Slope Fields and Eu
Euler's Method
563
tan x
cos x ln sec x tan x
Solution: y
yc
cos x
1
sec x ˜ tan x sec 2 x sin x ln sec x tan x
sec x tan x
cos x
sec x tan x
sec x tan x sec x sin x ln sec x tan x
1 sin x ln sec x tan x
ycc
sin x
1
sec x ˜ tan x sec 2 x cos x ln sec x tan x
sec x tan x
sin x sec x cos x ln sec x tan x
Check: ycc y
sin x sec x cos x ln sec x tan x cos x ln sec x tan x
8. Differential Equation: ycc 4 yc
Solution: y
yc
ycc
2e x
2 4 x
e
ex
5
2
4e 4 x e x
5
32 4 x
2
e
ex
5
5
Check: ycc 4 yc
tan x.
8
2
e 4 x e x
5
5
2 ·
2 ·
§ 32 4 x
§ 8
e x ¸ 4¨ e4 x e x ¸
¨ e
5 ¹
5 ¹
©5
© 5
§ 2 8· x
¨ ¸e
5¹
©5
2e x
sin x cos x cos 2 x
9. y
yc
sin 2 x cos 2 x 2 cos x sin x
1 2 cos 2 x sin 2 x
Differential Equation:
2 y yc
2 sin x cos x cos 2 x 1 2 cos 2 x sin 2 x
2 sin x cos x 1 sin 2 x
2 sin 2 x 1
§S ·
Initial condition ¨ , 0 ¸ :
©4 ¹
sin
10. y
yc
S
4
cos
S
4
cos 2
2
2 § 2·
˜
¨¨
¸¸
2
2
© 2 ¹
S
4
6 x 4 sin x 1
yc
Differential equation: yc
6 4 cos x
Initial condition 0, 1 : 0 0 1
yc
6 x 2
4e
12 x
48 xe
e cos x
e cos x sin x
sin x ˜ e cos x
Differential Equation:
yc
1
sin x ˜ e cos x
sin x y
§S ·
Initial condition ¨ , 1¸: e cos S 2
©2 ¹
2
4e 6 x
0
12. y
6 4 cos x
11. y
2
6 x2
y sin x
e0
1
Differential equation:
yc
12 xy
12 x 4e 6 x
2
48 xe 6 x
2
INSTRUCTOR USE ONLY
4e 0
Initial condition 0, 4 : 4e
4
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564
NOT FOR SALE
Chapter 6
Differential
ferential Equations
Equation
In Exercises 13–20, the differential equation
is y 4 16 y = 0.
13.
y
y
4
16.
y
4
y
3 cos x
y
4
3 cos x
Yes
16 y
45 cos x z 0,
14.
y
y
4
3 sin 2 x
48 sin 2 x
16 y
e 2 x
y4
16e 2 x
y
2 sin x
y 4 16 y
4
2 sin x
Yes
16 y
2 sin x 16 2 sin x z 0
48 sin 2 x 16 3 sin 2 x
y
17.
No
y
4
16e 2 x 16e 2 x
y
18.
No
y
y
3 cos 2 x
4
48 cos 2 x
y 4 16 y
30
x4
30
4 80 ln x z 0,
x
y 4 16 y
48 cos 2 x 48 cos 2 x
0,
5 ln x
y4
15.
0
No
0,
Yes
C1e 2 x C2e 2 x C3 sin 2 x C4 cos 2 x
y
19.
16C1e 2 x 16C2e 2 x 16C3 sin 2 x 16C4 cos 2 x
y4
y 4 16 y
0,
Yes
y
3e 2 x 4 sin 2 x
y4
48e 2 x 64 sin 2 x
20.
y 4 16 y
48e 2 x 64 sin 2 x 16 3e 2 x 4 sin 2 x
0,
Yes
In Exercises 21–28, the differential equation is
xyc 2 y = x 3e x .
21. y
x 2 , yc
x 2x 2 x2
0 z x 3e x ,
xyc 2 y
26. y
x 3 , yc
x 3x 2 2 x3
x3 z x3e x
27. y
ln x, yc
No
23. y
x 2 e x , yc
xyc 2 y
x 2e x 2 xe x
x ex x2 2x
e x x2 2 x
2 x 2e x
x 3e x ,
xyc 2 y
x 2 2 e x , yc
xyc 2 y
x2 ex 2 x 2 ex
x ª¬ x 2e x 2 xe x 4 xº¼ 2 ª¬ x 2e x 2 x 2 º¼
x3e x ,
x sin x 2 cos x z x3e x
1
x
§1·
x¨ ¸ 2 ln x z x3e x ,
© x¹
No
28. y
Yes
24. y
sin x
No
3x 2
xyc 2 y
cos x
x cos x 2 sin x z x3e x ,
cos x, yc
xyc 2 y
No
22. y
sin x, yc
No
2x
xyc 2 y
25. y
x 2e x 5 x 2 , yc
xyc 2 y
x 2e x 2 xe x 10 x
x ª¬ x 2e x 2 xe x 10 xº¼ 2 ª¬ x 2e x 5 x 2 º¼
x3e x ,
Yes
INSTRUCTOR USE ONLY
Yes
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 6.1
29. y
3
Ce x 2 passes through 0, 3 .
C Ÿ C
Ce0
Particular solution: y
30. y x 2 y
31. y 2
C Ÿ C
3e x 2
Particular solution: y 2
32. 2x 2 y 2
4
33. Differential equation: 4 yyc x
0
General solution: 4 y x
2
Particular solutions: C
2
1
4
1 x 3 or 4 y 2
4
x3
C passes through 3, 4 .
2 9 16
Particular solution: y x 2 y
C Ÿ C
2
Particular solution: 2 x 2 y 2
2
C
0, Two intersecting lines
r1, C
C
r4, Hyperbolas
2
2
2
C = −1
C=1
C=0
−3
C 64 Ÿ C
16
4
565
Cx3 passes through 4, 4 .
3
C passes through 0, 2 .
20 2
Slope Fields and Eu
Euler's Method
−3
3
3
−2
−2
2
2
−3
3
−2
C = −4
C=4
−3
−3
3
3
−2
−2
34. Differential equation: yyc x
General solution: x y
2
Particular solutions: C
C 1, C
4, Circles
2
0
C
0, Point
35. Differential equation: yc 2 y
General solution: y
yc 2 y
Ce
C 2 e 2 x 2 Ce 2 x
Initial condition 0, 3 : 3
y
2
0
2x
Particular solution: y
0
Ce0
C
3e2 x
1
1
2
x
36. Differential equation: 3x 2 yyc
General solution: 3x 2 2 y 2
6 x 4 yyc
0
C
0
2 3x 2 yyc
0
3 x 2 yyc
0
Initial condition 1, 3 :
2
23
2
3 18
21
C
Particular solution: 3x 2 2 y 2
21
31
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
566
Chapter 6
NOT FOR SALE
Differential
ferential Equation
Equations
37. Differential equation: ycc 9 y
0
C1 sin 3 x C2 cos 3x
General solution: y
yc
3C1 cos 3x 3C2 sin 3x,
ycc
9C1 sin 3 x 9C2 cos 3 x
ycc 9 y
9C1 sin 3 x 9C2 cos 3x 9 C1 sin 3 x C2 cos 3x
§S ·
Initial conditions ¨ , 2 ¸ and yc
©6 ¹
2
yc
1
1 when x
§S ·
§S ·
C1 sin ¨ ¸ C2 cos¨ ¸ Ÿ C1
©2¹
©2¹
3C1 cos 3 x 3C2 sin 3 x
§S ·
§S ·
3C1 cos¨ ¸ 3C2 sin ¨ ¸
©2¹
©2¹
38. Differential equation: xycc yc
§1·
C2 ¨ ¸, ycc
© x¹
xycc yc
yc
C2
x
C2
Ÿ C2
2
1
2
1, C1
0
1
when x
2
General solution: y
C1 x C2 x
C1 3C2 x 2 , ycc
x ycc 3xyc 3 y
2
ln
yc
C1 3C2 x 2
4
C1 12C2
C1 4C2
C1 12C2
0½
¾ C2
4¿
Particular solution: y
x
2
0
3
6C2 x
x 6C2 x 3x C1 3C2 x 2 3 C1 x C2 x3
2
Initial conditions 2, 0 and yc
2C1 8C2
2:
ln 2
39. Differential equation: x 2 ycc 3xyc 3 y
0
1
3
1
cos 3x
3
ln 2 ln x
Particular solution: y
yc
§1·
C2 ¨ 2 ¸
©x ¹
Initial conditions 2, 0 and yc
C1 C2 ln 2
2
0
1·
1
§
x¨ C2 2 ¸ C2
x ¹
x
©
0
:
C1 C2 ln x
General solution: y
yc
6
3C2 Ÿ C2
2 sin 3 x Particular solution: y
S
0
1, C
1
2
4 when x
0
2:
2
2 x 12 x3
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 6.1
40. Differential equation: 9 ycc 12 yc 4 y
General solution: y
e 2 x 3 C1 C2 x
yc
2 e2 x 3 C C x
1
2
3
C2 e 2 x 3
ycc
2 e2 x 3 2 C C 2 C x
2
3
3 1
3 2
9 ycc 12 yc 4 y
9 23 e2 x 3
Slope Fields and Eu
Euler's Method
567
0
e 2 x 3 23 C1 C2 23 C2 x
e2 x 3 23 C2
2 e 2 x 3 2 C 2C 2 C x
2
3
3 1
3 2
2 C 2C 2 C x
2
3 1
3 2
12 e 2 x 3
2C C 2C x
2
3 1
3 2
4 e 2 x 3 C1 C2 x
0
Initial conditions 0, 4 and 3, 0 :
0
e 2 C1 3C2
4
1 C1 0 Ÿ C1
0
e 4 3C2 Ÿ C2
42.
43
e 2 x 3 4 43 x
Particular solution: y
41.
4
2
dy
dx
6x2
y
³ 6 x dx
dy
dx
10 x 4 2 x3
y
³ 10 x 2 x dx
dy
43.
dx
y
u
47.
2 x3 C
2
4
3
2 x5 x4
C
2
x
1 x2
x
³ 1 x 2 dx
1
ln 1 x 2 C
2
1 x 2 , du
2 x dx
48.
49.
dy
dx
sin 2 x
y
³ sin 2 x dx
u
2 x, du
2 dx
dy
dx
tan 2 x
sec 2 x 1
y
³ sec x 1 dx
dy
dx
x
44.
45.
dy
dx
ex
4 ex
y
ex
³ 4 e x dx
ln 4 e x C
dy
dx
x 2
x
2
x
y
³ ¨©1 x ¸¹ dx
§
1
46.
dy
dx
x cos x 2
y
³ x cos x
u
x 2 , du
2
dx
tan x x C
x 6
x 6, then x
³ x x 6 dx
2·
x 2 ln x C
1
cos 2 x C
2
2
Let u
y
x ln x 2 C
u 2 6 and dx
2u du.
³ u 6 u 2u du
2³ u 4 6u 2 du
2
§ u5
·
2u 3 ¸ C
2¨
©5
¹
2
52
32
x6
4x 6
C
5
2
32
x6
x 6 10 C
5
2
32
x6
x 4 C
5
1
sin x 2 C
2
2 x dx
INSTRUCTOR USE ONLY
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NOT FOR SALE
568
Chapter 6
Differential
ferential Equation
Equations
50.
dy
dx
2 x 4 x2 1
y
³ 2 x 4 x 1 dx
58.
2
1
4³
4 x 2 1 8 x dx
1 4x 1
4
32
2
59.
32
C
60.
dy
dx
xe x
y
³ xe dx
u
x 2 , du
1
cos x
2
For x
0,
dy
dx
e 2 x
2
dy
dx
1 x2
e C
2
1
. Matches (c).
2
dy
o 0. Matches (d).
dx
1
x
For x
x2
dy
dx
As x o f,
32
1 2
4x 1
C
6
51.
dy
dx
0,
dy
is undefined (vertical tangent). Matches (a).
dx
61. (a), (b)
2 x dx
y
(4, 2)
5
52.
53.
dy
dx
5e x 2
y
x 2
³ 5e dx
§ 1·
5 2 ³ e x 2 ¨ ¸ dx
© 2¹
10e x 2 C
x
–4
–2
0
2
4
8
y
2
0
4
4
6
8
1
4
3
2
dy dx
–4
Undef.
0
x
−2
8
(c) As x o f, y o f
As x o f, y o f
62. (a), (b)
y
54.
4
x
–4
–2
0
2
4
8
y
2
0
4
4
6
8
dy dx
6
2
4
2
2
0
(1, 1)
x
4
−4
55.
x
–4
–2
0
2
4
8
y
2
0
4
4
6
8
dy dx
2 2
–2
0
0
2 2
–8
(c) As x o f, y o f
As x o f, y o f
63. (a), (b)
y
(2, 2)
5
56.
x
–4
–2
0
2
4
8
y
2
0
4
4
6
8
0
3
dy dx
3
3
0
3
x
−4
4
−3
57.
dy
dx
For x
(c) As x o f, y o f
sin 2 x
As x o f, y o f
dy
0,
dx
0. Matches (b).
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 6.1
64. (a), (b)
569
1
, 0, 1
y
66. (a) yc
y
2
Slope Fields and Eu
Euler's Method
(0, − 4)
y
(0, 1)
−4
3
x
−2
2
−2
x
−4
3
−6
−3
(c) As x o f, y o f
As x o f, y o f
As x o f, y o f
65. (a) yc
1
, 1, 1
y
(b) yc
1
, 1, 0
x
y
(1, 1)
y
3
(1, 0)
3
2
1
x
3
x
6
−1
−2
−3
−3
As x o f, y o f
As x o f, y o f
[Note: The solution is y
ln x. ]
67.
1
, 2, 1
x
(b) yc
dy
dx
0.25 y, y 0
4
(a), (b)
12
y
(2, −1)
3
−6
2
6
1
−4
x
−1
6
−2
−3
68.
As x o f, y o f
dy
dx
4 y, y 0
6
(a), (b)
10
−5
5
0
69.
dy
dx
0.02 y 10 y , y 0
2
(a), (b)
12
−12
48
−2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
570
Chapter 6
70.
dy
dx
Differential
ferential Equation
Equations
0.2 x 2 y , y 0
9
71.
(a), (b)
dy
dx
0.4 y 3 x , y 0
1
(a), (b)
8
10
−2
−5
8
5
−2
0
72.
1 x 8
Sy
e
sin
,y0
2
4
dy
dx
2
(a), (b)
5
−3
3
−3
73. yc
x y,
y0
n
2,
10, h
0.1
y1
y0 hF x0 , y0
2 0.1 0 2
y2
y1 hF x1 , y1
2.2 0.1 0.1 2.2
2.2
2.43, etc.
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
2
2.2
2.43
2.693
2.992
3.332
3.715
4.146
4.631
5.174
5.781
74. yc
x y,
y0
2, n
20,
h
0.05
y1
y0 hF x0 , y0
2 0.05 0 2
y2
y1 hF x1 , y1
2.1 0.05 0.05 2.1
2.1
2.2075, etc.
0, 2, 4, ! , 20.
The table shows the values for n
n
0
2
4
6
8
10
12
14
16
18
20
xn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
2
2.208
2.447
2.720
3.032
3.387
3.788
4.240
4.749
5.320
5.960
n
h
75. yc
3 x 2 y,
y0
3,
10,
0.05
y1
y0 hF x0 , y0
3 0.05 3 0 2 3
y2
y1 hF x1 , y1
2.7 0.05 3 0.05 2 2.7
2.7
2.4375, etc.
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
yn
3
2.7
2.438
2.209
2.010
1.839
1.693
1.569
1.464
1.378
1.308
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 6.1
76. yc
0.5 x 3 y ,
h
5,
1 0.4 0.5 0 3 1
y2
y1 hF x1 , y1
1 0.4 0.5 0.4 3 1
1
n
0
1
2
3
4
5
xn
0
0.4
0.8
1.2
1.6
2.0
yn
1
1
1.16
1.454
1.825
2.201
e xy ,
y0
1,
n
10, h
1.16, etc.
0.1
y1
y0 hF x0 , y0
1 0.1 e
y2
y1 hF x1 , y1
1.1 0.1 e 0.1 1.1 | 1.2116, etc.
01
1.1
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
1
1.1
1.212
1.339
1.488
1.670
1.900
2.213
2.684
3.540
5.958
cos x sin y, y 0
5, n
10, h
0.1
y1
y0 hF x0 , y0
5 0.1 cos 0 sin 5 | 5.0041
y2
y1 hF x1 , y1
5.0041 0.1 cos 0.1 sin 5.0041
| 5.0078, etc.
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
5
5.004
5.008
5.010
5.010
5.007
4.999
4.985
4.965
4.938
4.903
dy
dx
y, y
3e x , 0, 3
x
0
0.2
0.4
0.6
0.8
1
y x (exact)
3
3.6642
4.4755
5.4664
6.6766
8.1548
y x h
0.2
3
3.6000
4.3200
5.1840
6.2208
7.4650
y x h
0.1
3
3.6300
4.3923
5.3147
6.4308
7.7812
dy
dx
571
0.4
y0 hF x0 , y0
78. yc
80.
n
1,
y1
77. yc
79.
y0
Slope Fields and Eu
Euler's Method
2x
,y
y
2 x 2 4,
0, 2
x
0
0.2
0.4
0.6
0.8
1
y x (exact)
2
2.0199
2.0785
2.1726
2.2978
2.4495
y x h
0.2
2
2.000
2.0400
2.1184
2.2317
2.3751
y x h
0.1
2
2.0100
2.0595
2.1460
2.2655
2.4131
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
572
Chapter 6
81.
dy
dx
NOT FOR SALE
Differential
ferential Equation
Equations
1
sin x cos x e x ,
2
y cos x, y
0, 0
x
0
0.2
0.4
0.6
0.8
1
y x (exact)
0
0.2200
0.4801
0.7807
1.1231
1.5097
y x h
0.2
0
0.2000
0.4360
0.7074
1.0140
1.3561
y x h
0.1
0
0.2095
0.4568
0.7418
1.0649
1.4273
82. As h increases (from 0.1 to 0.2), the error increases.
83.
dy
dt
(a)
1
y 72 ,
2
0, 140 , h
87. Consider yc
0.1
0
1
2
3
Euler
140
112.7
96.4
86.6
72 68e t 2
generating the sequence of points
xn 1 , yn 1
xn h, yn hF xn , yn .
exact
t
0
1
2
3
Exact
140
113.24
97.016
87.173
y
Ce kx
dy
dx
Cke kx
Because dy dx
dy
dt
1
y 72 ,
2
0, 140 , h
0.07 y, you have Cke kx
0.07Ce kx .
0.07.
C cannot be determined.
0
1
2
3
Euler
140
112.98
96.7
86.9
The approximations are better using h
0, yc
So, k
0.05
t
84. When x
y0 .
Then, using a step size of h, find the point
x1 , y1
x0 h, y0 hF x0 , y0 . Continue
88.
(c)
y0 . Begin with a point
x0 , y0 that satisfies the initial condition, y x0
t
(b) y
F x, y , y x0
89. False. Consider Example 2. y
xyc 3 y
0.05.
90. True
0, therefore (d) is not possible.
91. True
When x, y ! 0, yc 0 (decreasing function) therefore
(c) is the equation.
0, but y
x3 is a solution to
x3 1 is not a solution.
92. False. The slope field could represent many different
differential equations, such as yc
2 x 4 y.
85. The general solution is a family of curves that satisfies
the differential equation. A particular solution is one
member of the family that satisfies given conditions.
86. A slope field for the differential equation
yc
F x, y consists of small line segments at various
points x, y in the plane. The line segment equals the
slope yc
F x, y of the solution y at the point x, y .
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 6.1
93.
dy
dx
2 y, y 0
(a)
573
4e 2 x
4, y
x
0
0.2
0.4
0.6
0.8
1
y
4
2.6813
1.7973
1.2048
0.8076
0.5413
y1
4
2.5600
1.6384
1.0486
0.6711
0.4295
y2
4
2.4000
1.4400
0.8640
0.5184
0.3110
e1
0
0.1213
0.1589
0.1562
0.1365
0.1118
e2
0
0.2813
0.3573
0.3408
0.2892
0.2303
0.4312
0.4447
0.4583
0.4720
0.4855
r
Slope Fields and Eu
Euler's Method
(b) If h is halved, then the error is approximately halved r | 0.5 .
(c) When h
94.
dy
dx
0.05, the errors will again be approximately halved.
x y, y 0
(a)
x 1 2e x
1, y
x
0
0.2
0.4
0.6
0.8
1
y
1
0.8375
0.7406
0.6976
0.6987
0.7358
y1
1
0.8200
0.7122
0.6629
0.6609
0.6974
y2
1
0.8000
0.6800
0.6240
0.6192
0.6554
e1
0
0.0175
0.0284
0.0347
0.0378
0.0384
e2
0
0.0375
0.0606
0.0736
0.0795
0.0804
0.47
0.47
0.47
0.48
0.48
r
(b) If h is halved, then the error is halved r | 0.5 .
(c) When h
0.05, the error will again be approximately halved.
dI
RI
dt
dI
12 I
4
dt
dI
dt
95. (a) L
96.
Et
24
1
24 12 I
4
6 3I
y
e kt
yc
ke kt
ycc
k 2e kt
ycc 16 y
0
k 2e kt 16e kt
0
k 2 16
0
k
r4
I
because e kt z 0
3
97. y
t
−3
3
−3
A sin Z t
yc
AZ cos Z t
ycc
AZ 2 sin Z t
ycc 16 y
(b) As t o f, I o 2. That is, lim I t
t of
I
2. In fact,
2 is a solution to the differential equation.
0
AZ 2 sin Z t 16 A sin Z t
A sin Z t ª¬16 Z º¼
2
If A z 0, then Z
0
0
r4
INSTRUCTOR USE ONLY
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574
NOT FOR SALE
Chapter 6
Differential
ferential Equation
Equations
f x f cc x
98.
xg x f c x ,
g x t 0
2
2 f x f c x 2 f c x f cc x
2 xg x ª¬ f c x º¼
d ª
2
2
f x fc x º
¼
dx ¬
2 x g x ª¬ f c x º¼
2
2
For x 0, 2 x g x ª¬ f c x º¼ t 0
2
For x ! 0, 2 x g x ª¬ f c x º¼ d 0
So, f x
f x
2
2
2
f c x is increasing for x 0 and decreasing for x ! 0.
2
f c x has a maximum at x
99. Let the vertical line x
0. So, it is bounded by its value at x
k cut the graph of the solution y
0, f 0
2
2
f c 0 . So, f (and f c ) is bounded.
f x at k , t . The tangent line at k , t is
fc k x k
y t
Because yc p x y
y t
q x , you have
ª¬q k p k t º¼ x k
§
1 qk ·
For any value of t, this line passes through the point ¨ k ,
¸.
¨
p
k p k ¸¹
©
To see this, note that
qk
t
pk
?
?
§
·
1
ª¬q k p k t º¼ ¨¨ k k ¸¸
p
k
©
¹
q k k p k tk qk
t kq k p k kt
pk
qk
t.
pk
Section 6.2 Differential Equations: Growth and Decay
1.
2.
3.
dy
dx
x 3
y
³ x 3 dx
dy
dx
5 8x
y
³ 5 8 x dx
4.
x2
3x C
2
dy
dx
dy
6 y
6 y
dx
1
dy
dx
dy
y 3
1
ln 6 y dy
x C1
ln y 3
x C1
e x C1
Ce x 3
e x C1
6 y
5.
Ce x
Ce x
6 Ce x
y
dx
³ dx
y
³ dx
y 3
³ y 3 dy
y 3
5x 4 x2 C
³ 6 y dy
yc
5x
y
yyc
5x
³ yyc dx
³ y dy
³ 5 x dx
³ 5 x dx
1 2
y
2
y 2 5x2
5 2
x C1
2
C
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 6.2
yc
x
4y
4 y yc
x
³ 4 y dy
³
6.
6 y 2 2 x3 2
xy
yc
y
x
³
x dx
³y
³
x dx
ln y
2 32
x C1
3
dy
y
e
Ce
yc
8.
yc
1 y
yc
³ 1 y dx
yc
³ 1 x 2 dx
³y
dy
³ 1 x 2 dx
ln y
ln 1 x 2 C1
ln y
ln 1 x 2 ln C
ln y
ln ª¬C 1 x 2 º¼
xy yc
10.
³ 100 y dx
³ x dx
1
³ x dx
x1 y
³ 100 y dy
x
ln 100 y
³1 y
³ x dx
100 y
ln 1 y
x2
C1
2
y
e
x 2 2 C1
2
eC1 e x 2 1
11.
2
Ce x 2 1
dQ
dt
dQ
³ dt dt
³ dQ
Q
dP
dt
12.
x2
C1
2
x2
C1
2
e
x 2 2 C1
x2 2
y
e C1 e
y
100 Ce x 2
k
t2
k
³ t 2 dt
k
C
t
k
C
t
k 25 t
³ k 25 t dt
³ dP
P
100
2
dP
³ dt dt
x 100 y
x
yc
3
ln 100 y
1 y
C 1 x2
100 x xy
yc
100 y
³ x dx
dy
2x
100 x
yc
2 3 x3 2
2 x3 2
2x
³ y dx
y
2 3 x3 2 C1
eC1 e
2 xy
1 x2
2x
1 x2
yc
y
x dx
575
0
yc
C
yc
yc
³ y dx
9. 1 x 2 yc 2 xy
2
x3 2 C1
3
2 y2
7.
Differential Equations: Growt
Grow
Growth and Decay
k
2
25 t C
2
k
2
25 t C
2
INSTRUCTOR USE ONLY
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576
NOT FOR SALE
Chapter 6
Differential
ferential Equations
Equation
13. (a)
dy
dt
y
15.
9
1
t , 0, 10
2
1
³ 2 t dt
1 2
t C
4
1 2
0 C Ÿ C
4
1 2
t 10
4
³ dy
y
x
−5
5
−1
dy
dx
dy
y 6
(b)
10
(0, 0)
x6 y,
0, 0
y
16
x dx
ln y 6
x2
C
2
y 6
ex 2 C
(0, 10)
2
−4
2
C1e x 2
4
−1
2
y
6 C1e x 2
(0, 0): 0
6 C1 Ÿ C1
6
16.
2
6 6e x 2
y
7
−6
dy
dt
9 t ,
³ dy
³ 9 t dt
y
6t 3 2 C
10
0C Ÿ C
y
6t 3 2 10
6
0, 10
10
12
−1
(0, 10)
y
14. (a)
4
−1
3
−2
(0, 12 )
x
−4
17.
4
−4
(b)
10
dy
dx
xy,
dy
y
x dx
§ 1·
¨ 0, ¸
© 2¹
dy
dt
dy
³y
1
y,
0, 10
2
1
³ 2 dt
ln y
1
t C1
2
y
e t 2 C1
eC1 et 2
10
Ce0 Ÿ C
10
y
t 2
2
ln y
x
C
2
y
ex 2 C
2
10e
Ce t 2
16
2
C1e x 2
§ 1· 1
C1e0 Ÿ C1
¨ 0, ¸:
© 2¹ 2
1 x2 2
y
e
2
(0, 10)
1
2
−1
10
−1
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 6.2
18.
dy
dt
dy
³y
3
y, 0, 10
4
3
³ 4 dt
ln y
3
t C1
4
y
e 3 4 t C1
1
2
1 kt
e
2
1 5k
e
2
ln 10
5
1 ¬ª ln 10 5¼ºt
e
2
C
y
5
Ce3t 4
10
Ce0 Ÿ C
10
y
3t 4
k
y
40
(0, 10)
−5
5
−5
19.
dN
dt
kN
N
Ce kt
Theorem 6.1
0, 250 : C
250
1, 400 : 400
250e Ÿ k
N
250e
ln 8 5 t
When t
kP
P
Ce kt
| 250e
400
ln
250
8
ln
5
0.4700t
250e4 ln 8 5
1
2
4e5k
250e
4
ln 8 5 4
8192
.
5
§ 19 ·
ln ¨ ¸
© 20 ¹
5000e k Ÿ k
0.0513t
ln 19 20 t
| 5000e
5, P
5000eln 19 20 5
5000e
ln 1 8
5
| 0.4159
4e 0.4159t
Ce kt ,
23. y
1, 5 , 5, 2
5
Ce Ÿ 10
2Ce k
2
Ce5 k Ÿ 10
5Ce k
2Ce k
5Ce5k
2e k
5e5k
2
5
e4k
k
1 § 2·
ln ¨ ¸
4 ©5¹
k
14
§ 2·
ln ¨ ¸
©5¹
5000
1, 4750 : 4750
When t
4e kt
Theorem 6.1
0, 5000 : C
P
y
y
§8·
250¨ ¸
©5¹
dP
20.
dt
4
k
k
4, N
C
1 t5
1
10 or y | e0.4605t
2
2
§ 1·
0, 4 , ¨ 5, ¸
© 2¹
Ce kt ,
22. y
577
§ 1·
¨ 0, ¸, 5, 5
© 2¹
Ce kt ,
21. y
eC1 e 3 4 t
10e
Differential Equations: Growth
Growt
Grow and Decay
5e 1 4 ln 2 5
§ 2·
5¨ ¸
©5¹
1 4
14
§5·
5¨ ¸
© 2¹
C
5e k
y
§ 5 · ª1 4 ln 2 5 º¼t
5¨ ¸ e ¬
| 6.2872 e 0.2291t
© 2¹
14
5
§ 19 ·
5000¨ ¸ | 3868.905.
© 20 ¹
INSTRUCTOR USE ONLY
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578
NOT FOR SALE
Chapter 6
24. y
Differential
ferential Equation
Equations
§ 1·
¨ 3, ¸, 4, 5
© 2¹
Ce kt ,
30. Because the half-life is 1599 years,
1
2
1e k 1599
1 ln 1 .
1599
2
1
2
Ce3k Ÿ 1
2Ce3k
k
5
Ce 4 k Ÿ 1
1 4k
Ce
5
Because there are 1.5 g after 1000 years,
2Ce3k
1 4k
Ce
5
10e3k
e4 k
10
ek
k
ln 10 | 2.3026
C | 2.314.
So, the initial quantity is approximately 2.314 g.
5
2.3026 4
| 0.03 g.
C | 0.0005
0.0005e 2.3026t
25. In the model y
y (when t
ªln 1 2 1599º¼ 10,000
2.314e ¬
10,000, y
31. Because the half-life is 1599 years,
Ce .3026t
y
ªln 1 2 1599¼º 1000
Ce ¬
When t
y
Ce
1.5
1
2
1e k 1599
k
1 ln 1 .
1599
2
Because there are 0.1 gram after 10,000 years,
Ce kt , C represents the initial value of
0 ). k is the proportionality constant.
0.1
ªln 1 2 1599º¼ 10,000
Ce ¬
C | 7.63.
So, the initial quantity is approximately 7.63 g.
26. yc
dy
dt
dy
27.
dx
1
xy
2
ky
When t
dy
dx
ln 1 2 1599¼º 1000
32. Because the half-life is 5715 years,
1 2
x y
2
1
2
1e k 5715
k
1 ln 1 .
5715
2
Because there are 3 grams after 10,000 years,
dy
! 0 when y ! 0. Quadrants I and II.
dx
29. Because the initial quantity is 20 grams,
y
7.63e »
| 4.95 g.
dy
! 0 when xy ! 0. Quadrants I and III.
dx
28.
1000, y
3
ªln 1 2 5715¼º 10,000
Ce ¬
C | 10.089.
So, the initial quantity is approximately 10.09 g.
When t
1000, y
kt
20e .
10.09e ª¬
ln 1 2 5715º¼ 1000
| 8.94 g.
Because the half-life is 1599 years,
10
20e k 1599
k
1 ln 1 .
1599
2
So, y
20e »
33. Because the initial quantity is 5 grams, C
5.
Because the half-life is 5715 years,
2.5
ln 1 2 1599¼ºt
.
k
ªln 1 2 1599¼º 1000
20e ¬
When t
1000, y
When t
10,000, y | 0.26g.
| 12.96 g .
5e k 5715
1 ln 1 .
5715
2
When t
1000 years, y
When t
10,000 years, y
ªln 1 2 5715¼º 1000
5e ¬
| 4.43 g.
ªln 1 2 5715¼º 10,000
5e ¬
| 1.49 g.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 6.2
34. Because the half-life is 5715 years,
1e
k
1 ln 1 .
5715
2
Because there are 1.6 grams when t
ªln 1 2 5715º¼ 1000
C | 1.806.
So, the initial quantity is approximately 1.806 g.
1.806e »
ln 1 2 5715¼º 10,000
10,000, y
| 0.54 g.
35. Because the half-life is 24,100 years,
1
2
1e k 24,100
k
1
ln 12 .
24,100
ªln 1 2 24,100¼º 1000
Ce ¬
C | 2.161.
Ce k 5715
k
1
§1·
ln ¨ ¸
5715 © 2 ¹
0.15C
Ce[ln 1 2 5715]t
1e k 24,100
k
1
ln 12 .
24,100
e0.06t
ln 2
0.06t
| 1.62 g.
36,000
C | 0.533.
So, the initial quantity is approximately 0.533 g.
1000, y
0.533e »
ln 1 2 24,100¼º 1000
| 0.52 g.
37.
18,000e0.055t
e0.055t
ln 2
0.055t
ln 2
| 12.6 years.
0.055
Amount after 10 years:
A 18,000e 0.055 10 | $31,198.55
41. Because A 750ert and A 1500 when
t
7.75, you have the following.
750e7.75r
1500
2
e7.75r
Cekt
ln 2
7.75r
1C
2
Cek 1599
r
k
1 ln 1
1599
2
y
When t
100, y
ln 2
| 0.0894
7.75
Amount after 10 years: A
Ce »
ln 1 2 1599¼º 100
| 0.9576 C
Therefore, 95.76% remains after 100 years.
4000e 0.06 10 | $7288.48
18,000e0.055t , the time to double is given
2
t
ªln 1 2 24,100º¼ 10,000
Ce ¬
When t
ln 2
| 11.55 years.
0.06
t
40. Because A
by
ªln 1 2 24,100º¼ 10,000
Because there are 0.4 grams after 10,000 years,
0.4
4000e0.06t
2
36. Because the half-life is 24,100 years,
1
2
4000e0.06t , the time to double is given by
39. Because A
2.161e ¬
10,000, y
579
§1·
ln ¨ ¸t
©2¹
ln 0.15
5715
t | 15,641.8 years
Amount after 10 years: A
So, the initial quantity is approximately 2.161 g.
When t
1
C
2
8000
Because there are 2.1 grams after 1000 years,
2.1
Ce kt
1000 years,
Ce ¬
When t
y
38.
k 5715
1
2
1.6
Differential Equations: Growth
Growt
Grow and Decay
8.94%
750e0.0894 10 | $1833.67
42. Because A 12,500e rt and A
25,000 when
t
20, you have the following.
25,000
12,500e 20 r
2
e 20 r
ln 2
20r
r
ln 2
| 0.03466 | 3.47%
20
Amount after 10 years:
A 12,500
, e0.03466 10 | $17,678.14
,
INSTRUCTOR USE ONLY
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580
Chapter 6
Differential
ferential Equation
Equations
43. Because A 500e rt and A 1292.85 when
t 10, you have the following.
1292.85
500e10 r
2.5857
e10 r
ln 2.5857
10r
P
ln 2.5857
r
10
| 0.0950
9.50%
ln 2
500e0.0950t
2
e0.0950t
ln 2
0.0950t
t
0.09 ·
§
P¨1 ¸
12 ¹
©
12 25
0.09 ·
§
1,000,000¨1 ¸
12 ¹
©
| $106,287.83
1000 1 0.07
300
t
1.07t
t ln 1.07
ln 2
| 10.24 years
ln 1.07
12t
ln 2
| 7.30 years.
0.095
t
49. (a) 2000
2
The time to double is given by
1000
48. 1,000,000
(b) 2000
0.07 ·
§
1000¨1 ¸
12 ¹
©
12 t
rt
44. Because A 6000e and A 8950.95 when
t 10, you have the following.
ln 2
10 r
6000e
8950.95
8950.95
6000
§ 8950.95 ·
ln ¨
¸
© 6000 ¹
2
e10 r
t
10r
(c) 2000
1
8950.95
ln
10
6000
r
0.04
4%
2
The time to double is given by
6000e0.04t
12,000
2
e0.04t
ln 2
0.04t
ln 2
t
ln 2
| 17.33 years.
0.04
t
45. 1,000,000
0.075 ·
§
P¨1 ¸
12 ¹
©
(d) 2000
12 20
0.075 ·
§
P 1,000,000¨1 ¸
12 ¹
©
| $224,174.18
46. 1,000,000
P
47. 1,000,000
P
0.06 ·
§
P¨1 ¸
12 ¹
©
0.07 ·
§
12t ln ¨1 ¸
12 ¹
©
ln 2
12 ln 1 0.07 12
0.07 ·
§
1000¨1 ¸
365 ¹
©
0.07 ·
§
¨1 ¸
365 ¹
©
365t
365t
0.07 ·
§
365t ln ¨1 ¸
365 ¹
©
ln 2
365 ln 1 0.07 365
| 9.90 years
1000e 0.07 t
2
e 0.07t
ln 2
0.07t
t
| 9.93 years
ln 2
| 9.90 years
0.07
12 40
1,000,000 1.005
0.08 ·
§
P¨1 ¸
12 ¹
©
240
0.007 ·
§
¨1 ¸
12 ¹
©
480
| $91,262.08
12 35
0.08 ·
§
1,000,000¨1 ¸
12 ¹
©
$61,377.75
420
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 6.2
1000 1 0.055
50. (a) 2000
2
ln 2
t
34.6
t ln 1.055
P
33.38e0.036t
0.055 ·
§
12t ln ¨1 ¸
12 ¹
©
1
ln 2
| 12.63 years
0.055 ·
12 §
ln ¨1 ¸
12 ¹
©
t
0.055 ·
§
1000¨1 ¸
365 ¹
©
(c) 2000
0.055 ·
§
¨1 ¸
365 ¹
©
2
365t
t
Ce 0.002 1 Ÿ C | 10.02
P1
10.0
P
10.02e 0.002t
(b) For 2020, t
10 and
10.02e 0.002 10 | 9.82 million.
P
(c) Because k 0, the population is decreasing.
100.1596 1.2455
(b) N
365t
Ce 0.002t
Cekt
55. (a) N
400 when t
6.3 hours (graphing utility)
400
100.1596 1.2455
400
100.1596
ln 3.9936
1.2455t
t ln 1.2455
Ce kt .
56. (a) Let y
e0.055t
ln 2
0.055t
At time 2: 125
ln 2
| 12.60 years
0.055
At time 4:
350 Cek 4 Ÿ 350
t
Ce 0.006 1 Ÿ C | 2.21
2.2
P
2.21e0.006t
(b) For 2020, t
10 and
2.21e0.006 10 | 2.08 million.
C
P1
P
kt
Ce
82.1
80.47e
(b) For 2020, t
0.020 t
125e 2 k e 4 k
14
5
e2k
2k
ln 14
5
k
1 ln 14
2
5
| 0.5148
125e 2 k
5
125 14
625
14
| 44.64
Approximately 45 bacteria at time 0.
Ce0.020 1 Ÿ C | 80.47
(b) y
0.020 t
625 1 2 ln 14 5 t
e
14
(c) When t
10 and
y
(c) Because k ! 0, the population is increasing.
(d) 25,000
| 44.64e0.5148t
8,
625 e 1 2 ln 14 5 8
14
80.47e0.020 10 | 98.29 million.
P
125e 2 k
125e 2 1 2 ln 14 5
(c) Because k 0, the population is decreasing.
Ce
Cek 2 Ÿ C
Ce 0.006t
Ce kt
P1
52. (a) P
3.9936
ln 3.9936
| 6.3 hours
ln 1.2455
t
2
P
t
1000e0.055t
(d) 2000
51. (a) P
t
Analytically,
0.055 ·
§
365t ln ¨1 ¸
365 ¹
©
1
ln 2
| 12.60 years
0.055 ·
365 §
ln ¨1 ¸
365 ¹
©
ln 2
10 and
33.38e0.036 10 | 47.84 million.
P
54. (a) P
12t
ln 2
(b) For 2020, t
(c) Because k ! 0, the population is increasing.
0.055 ·
§
¨1 ¸
12 ¹
©
2
Ce0.036 1 Ÿ C | 33.38
P1
12 t
581
Ce0.036t
1.055t
0.055 ·
§
1000¨1 ¸
12 ¹
©
(b) 2000
Ce kt
53. (a) P
ln 2
| 12.95 years
ln 1.055
t
Differential Equations: Growth
Growt
Grow and Decay
625 14
14 5
625 1 2 ln 14 5 t
e
14
4
2744.
Ÿ t | 12.29 hours
INSTRUCTOR USE ONLY
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582
NOT FOR SALE
Chapter 6
57. (a)
Differential
ferential Equations
Equation
Ce kt
P1
181e kt
1
205
ln 181
| 0.01245
10
181e10 k Ÿ k
205
P1 | 181e0.01245t | 181 1.01253
Ce kt
P1
61. (a)
t
(b) Using a graphing utility, P2 | 182.3248 1.01091
e10 k
1 § 123 ·
ln ¨
¸ | 0.01487
10 © 106 ¹
Ÿ k
t
1
§ 123 ·
ln¨
¸t
106e10 ©106 ¹
106e0.01487t
P1
(c)
123
106
106e k 10 Ÿ
123
0 l 1920
106e kt t
106 1.01499
300
t
t
(b) Using a graphing utility, P2 | 107.2727 1.01215 .
P1
P2
(c)
350
P1
0
150
50
P2
The model P2 fits the data better.
(d) Using the model P2 ,
320
75
182.3248 1.01091
320
182.3248
1.01091
t
The model P2 fits the data better.
P2
(d)
t
30e30 k
ln 1 3
30
N | 30 1 e
25
(b)
62. A t
10
k
e0.0366t
t
ln 3
| 0.0366
30
0.0366 t
t
ln 400 107.2727
ln 1.01215
1
6
ln 6
| 49 days
0.0366
(b) Although the percentage increase is constant each
month, the rate of growth is not constant. The rate of
change of y is given by
ry
which is an exponential model.
60. (a) Both functions represent exponential growth because
the graphs are increasing.
V t e 0.10t
100,000e0.8 t e 0.10t
dA
dt
dA
dt
30 1 e 0.0366t
59. (a) Because the population increases by a constant each
month, the rate of change from month to month will
always be the same. So, the slope is constant, and the
model is linear.
dy
dt
t
| 109, or 2029.
30 1 e30 k
20
107.2727 1.01215
1.01215
t
| 51.8 years, or 2011.
58. (a)
400
400
107.2727
ln 320 182.3248
ln 1.01091
t
100
0
100,000e0.8 t 0.10t
§ 0.4
·
100,000¨
0.10 ¸e0.8 t 0.10t
t
©
¹
0.4
0 when
0.10 Ÿ t 16.
t
The timber should be harvested in the year 2026
2010 16 .
Note: You could also use a graphing utility to graph
A t and find the maximum value. Use a viewing
window of 0 d x d 30, 0 d y d 600,000.
I
, I0
I0
1016
(a) E 1014
10 log10
1014
1016
20 decibels
(b) E 109
10 log10
109
1016
70 decibels
(c) E 106.5
10 log10
106.5
1016
95 decibels
(d) E 104
10 log10
104
1016
120 decibels
63. E I
10 log10
(b) g has a greater k value because its graph is increasing
at a greater rate than the graph of f.
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 6.3
64.
I
1016
Separation of Variables and the Logis
Logist
Logistic Equations
dy
dt
k y 20
y
20 Ce kt
160
20 Ce
60
20 140e
2
7
e5 k
k
1 § 2·
ln ¨ ¸ | 0.25055
5 ©7¹
k y 80
30
20 140e 1 5 ln 2 7 t
1
³ y 80 dy
³ k dt
1
14
eln 2 7
ln y 80
kt C.
1
14
2
t
ln
5 7
1
5 ln
14
2
ln
7
93
10 log10
6.7
log10 I Ÿ I
80
I
1016
10 log10
8
log10 I Ÿ I
10 log10 I 16
66.
106.7
10 log10 I 16
108
§ 106.7 108 ·
Percentage decrease: ¨
¸ 100 | 95%
106.7
©
¹
dy
dt
65. Because
ln
When t
0, y
1500. So, C
When t
1, y
1120. So,
k 1 ln 1420
ln 1040 ln 1420
ªln 104 142 º¼t
1420e ¬
When t
t
ln 1120 80
k
So, y
ln 1420.
80.
5, y | 379.2qF.
ln
104
.
142
See Example 6.
Ÿ C
k 0
t 5
140
k5
§ 2·
¨ ¸
©7¹
t 5
5 ln 14
| 10.53 minutes
7
ln
2
It will take 10.53 5
67. False. If y
583
5.53 minutes longer.
Ce kt , yc
Ckekt z constant.
68. True
69. False. The prices are rising at a rate of 6.2% per year.
70. True
Section 6.3 Separation of Variables and the Logistic Equation
1.
dy
dx
x
y
³ y dy
³ x dx
y2
2
y2 x2
x2
C1
2
C
dy
dx
2.
³ 3x dx
y3
3
x3 C1
y 3 3x3
2
C
0
x2
³ 5 y dy
³ x dx
5 y2
2
x3
C1
3
2
15 y 2 2 x3
3x 2
y2
³ y dy
2
dy
dx
dy
5y
dx
3. x 2 5 y
C
dy
dx
6 x2
2 y3
³ 2 y dy
³ 6x
4.
3
y4
2
3 y 4 2 x3 36 x
6x 2
dx
x3
C1
3
C
INSTRUCTOR USE ONLY
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584
5.
6.
Chapter 6
dr
ds
dr
³r
ln r
Differential
ferential Equation
Equations
11.
0.75 r
1 4 x 2 yc
³ dy
0.75 s C1
0.75 s C1
e
r
Ce0.75 s
dr
ds
0.75 s
³ dr
³ 0.75 s ds
12.
x 2 16 yc
3 ln 2 x ln C
8.
xyc
y
dy
³y
dx
³x
ln y
ln x ln C
y
9.
y
0
ln x
ln y
1
2
ln x C1
2
e 1 2 ln x
14. 12 yyc 7e x
2 C
1
7e x
³ 12 y dy
³ 7e dx
dy
dx
4 sin x
6 y2
7e x C
15. yyc 2e x
4 cos x C1
C 8 cos x
³ y dy
³ 2e dx
yyc
8 cos S x
y2
2
2e x C
dy
dx
8 cos S x
Initial condition 0, 3 :
³ 8 cos S x dx
Particular solution:
y
2
y2
8 sin S x
S
16
S
2 2
0
y
2
y2
2
Ce ln x
dx ·
¸
x¹
x
dy
y
dx
³ y dy
ln x, du
0
dy
12 y
dx
4 sin x
³ 4 sin x dx
§
¨u
©
³ x dx
yyc
2
y
ln Cx
dx
dy
³y
y
Cx
³ y dy
10.
3
C x 2
y
ln C 2 x
x 2 16
11 x 2 16 C
13. y ln x xyc
3
x 2 16
11x
³
y
3
³ 2 x dx
ln y
11x
³ dy
3y
dy
³y
11x
dy
dx
s
C
2
0.375 s 2 C
7. 2 x yc
dx
1 4x2
1 2
1
1 4 x2
8 x dx
³
8
1
1 4 x2 C
4
y
0.75
r
³
dx
1 4x2
x
2
r
x
dy
³ 0.75 ds
r
x
2e x
x
C
y2
2
y2
9
2
2 C Ÿ C
5
2
5
2
4e x 5
2e x sin S x C
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
ction 6.3
x 16.
y yc
³ y dy
y
32
0
³ x
2 32
y
3
2
x3 2 C1
3
x
C
12
dx
dy
dx
x 1 y2
y dy
³ 1 x
y 1 x2
20.
12
32
Separation of Variables and Logis
Logistic Equation
³ 1 y
2
1 2
1 y2
12
1 x2
Initial condition 0, 1 :
1 2
2
12
x dx
C
1 C Ÿ C
0
585
1
Initial condition 1, 9 :
9
32
1
32
Particular solution:
27 1
21.
28
0
dy
³y
³ x 1 dx
ln y
Ce 1 2 , C
Initial condition 2, 1 : 1
18. 2 xyc ln x
ª1 x 1 2 º 2
»¼
e «¬
e1 2
e
2 ln x
³ dy
³ x dx
2
Particular solution: u
e
dr
ds
e1 2
2
³e
e r
1
e 2 s C
2
2 s
ds
1
2
ln x 2
2
y
dy
1 y2
x
dx
1 x2
0: 1
1
ln 1 y 2
2
1
ln 1 x 2 C1
2
ln 1 y 2
ln 1 x 2 ln C
ln ª¬C 1 x 2 º¼
C 1 x2
3: 1 3
C Ÿ C
Particular solution: 1 y
2
41 x
y
2
3 4x
2
1
C Ÿ C
2
1
2
Particular solution:
1
1
e r
e 2 s 2
2
1 2 s
1
r
e
e
2
2
C
x 1 y2
Initial condition 0,
1 cos v2
1
e 1 2
er 2 s
r
r 0
C
y 1 x 2 yc
1 y2
2
Initial condition:
2
Particular solution: y
cos v 2
1: C
ln x
ln x
Ce
³ e dr
Initial condition 1, 2 : 2
19.
1
cos v 2 C1
2
x2 2 x 2
0
dy
2x
dx
y
2
Initial condition: u 0
22.
2
³ v sin v dv
u
2 2
Particular solution: y
uv sin v 2
ln u
C1
2
Ce x 1
y
du
dv
du
³u
2
x 1
1 x2 1
C
28
Particular solution: y 3 2 x3 2
17. y x 1 yc
1 y2
4
r
1·
§1
ln ¨ e 2 s ¸
2¹
©2
r
2 ·
§
ln ¨
2 s ¸
©1 e ¹
23. dP kP dt
§ 1 e 2 s ·
ln ¨
¸
2 ¹
©
0
dP
³P
ln P
k ³ dt
P
Ce kt
kt C1
Initial condition: P 0
P0 , P0
Particular solution: P
P0e kt
Ce0
C
2
INSTRUCTOR USE ONLY
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586
NOT FOR SALE
Chapter 6
Differential
ferential Equation
Equations
24. dT k T 70 dt
0
dT
³ T 70
k ³ dt
ln T 70
kt C1
Ce 0
70
³ x dx
2 y2
x2
C
2
70 1 e kt
0C Ÿ C
2y
4 y 2 x2
³ 16 y dy
8 y2
ln y
y
9 2
x C
2
9
C, C
2
8
25
9 2
x 2
2
25
8 y2
2
³ y dy
dy
y
dx
2x
1
³ x dx
2 ln y
ln x C1
25
2
31.
y 0
x 0
dy
dx
³ 9 x dx
dy
dx
y
2
Ce x 2
dy
16 y 2 9 x 2
y2
1
x C1
2
³y
Particular solution:
yc
ln y
9 x
16 y
Initial condition 1, 1 :
27.
³ 2 dx
30. m
1 23
x
2
x2 , y
1
8
1
³y
y
x2
8
2
16
2
8
C 82 , C
0 y
x 2 x
dy
dx
m
dy
Particular solution:
dy
dx
23
Particular solution: 8 y 3
29.
Initial condition 0, 2 : 2 22
26.
Cx 2
C
x
4y
³ 4 y dy
ln x 2 ln C
Initial condition 8, 2 :
Particular solution:
T 70
70e kt , T
dy
dx
ln y 3
3
y3
Initial condition:
T 0
140: 140 70
yc
³ y dy
2y
3x
2
³ x dx
Ce kt
T 70
25.
dy
dx
28.
y
x
dx
³x
ln x C1
ln x ln C
ln Cx
Cx
x
y
2
x
−2
2
−2
y
ln x ln C
³ x dx
1 2
x C
2
Cx
9C Ÿ C
Initial condition 9, 1 : 1
Particular solution:
y2
1
x
9
9 y2 x
0
y
1
3
1
9
x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section
ction 6.3
32.
dy
dx
x
y
Separation of Variables and Logis
Logistic Equation
36. (a)
y
ky 2
(b) The direction field satisfies dy dx
0 along
y
0, and grows more positive as y increases.
Matches (d).
4
2
−4
dy
dx
587
x
−2
2
4
37.
−2
dy
dt
ky ,
Ce kt
y
−4
Initial amount: y 0
y dy
2
y
2
y2 x2
dy
33. (a)
dx
x dx
Half-life:
x
C1
2
C
2
y
k y 4
34. (a)
y
4; but not along y
dy
dx
k x 4
0 along
4. Matches (b).
dy
dx
ky y 4
0 and y
dw
dt
39. (a)
ln 1 2 1599¼ºt
50, y
ky , y
0.9786C or 97.86%.
Ce kt
dw
ln 1200 w
kt C1
1200 w
e kt C1
1200 Ce kt
60
1200 C Ÿ C
1200 1140e
10
1200 60
k
ln
7
8
40et ln 7 8
10
40et ln 7 8
1
4
et ln 7 8
ln 1 4
ln 7 8
| 10.38 hours
1140
1400
10
0
0
0.8
40e
C
k
kt
1400
0
35
35
Ce kt
w
1400
0
Ce
0
k 1200 w
³ k dt
w0
40, y 1
When 75% has been changed:
0 along
4. Matches (c).
³ 1200 w
k
1
§1·
ln ¨ ¸
1599 © 2 ¹
Particular solution: y
t
w
k
0 along
(b) The direction field satisfies dy dx
(b)
y0ek 1599
40
x
y
dy
dt
y0
2
Initial conditions: y 0
(b) The direction field satisfies dy dx
35. (a)
38.
0. Matches (a).
C
Ce »
When t
(b) The direction field satisfies dy dx
y0
10
0
0
k
0.9
k
1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
588
Chapter 6
Differential
ferential Equation
Equations
(c) k
0.8:
t
1.31 years
k
0.9:
t
1.16 years
k
1.0:
t
1.05 years
(d) Maximum weight: 1200 pounds
lim w
1200
xof
40. From Exercise 39
w
1200 Ce kt , k
w
1200 Ce t
w0
w
Cy
2x
Cyc
yc
2x
C
1
1200 C Ÿ C
w0
43. Given family (parabolas): x 2
1200 1200 w0 e
41. Given family (circles):
1200 w0
t
C
2 x 2 yyc
0
yc
4
2³ y dy
−6
x
y
yc
y
x
dy
³y
dx
³x
ln y
ln x ln K
y
44. Given family (parabolas):
y2
2 yyc
yc
Kx
4
Orthogonal trajectory (ellipse):
−6
42. Given family (hyperbolas): x 2 2 y 2
C
2 x 4 yyc
0
yc
x
2y
³y
2 y
x
2
³ dx
x
ln y
2 ln x ln k
yc
Orthogonal trajectory:
dy
y
kx
2
6
−4
y2
Cx 3
2 yyc
3Cx 2
yc
3Cx 2
2y
45. Given family:
Orthogonal trajectory (ellipses):
K
2Cx
y2 § 1 ·
¨ ¸
2x © y ¹
C
y
yc
−3
2x
y
y
2
2
2x y2
x 2 K1
3x 2 § y 2 ·
¨ ¸
2 y © x3 ¹
3y
2x
K
yc
2x
3y
2 ³ x dx
x 2 K1
K
4
3
−2
y
2x
³ 2 x dx
3y2
2
3 y 2 2x2
2
x2
K1
2
2C
3³ y dy
k
x2
x
2y
³ x dx
x2 2 y 2
2
−4
³ y dy
4
6
2y
x
y2
6
−4
−6
yc
Orthogonal trajectory (ellipses):
x2 y 2
Orthogonal trajectory (lines):
2x
x2 y
−6
6
−4
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section
ction 6.3
46. Given family (exponential functions): y
Ce x
yc
Ce x
yc
Orthogonal trajectory (parabolas):
³ y dy
y2
2
y2
4
−6
6
Separation of Variables and Logis
Logistic Equation
2100
1 29e 0.75t
51. P t
y
1
y
³ dx
x K1
(a) k
0.75
(b) L
2100
(c) P 0
2100
1 29
(d)
1050
2 x K
e 0.75t
1
29
12
1 e x
t
6, it matches (c) or (d).
(e)
Because (d) approaches its horizontal asymptote slower
than (c), it matches (d).
48. y
Because y 0
49. y
3, it matches (a).
12
1
1 e x
2
Because y 0
50. y
12
4
12
§3·
¨ ¸
©2¹
0.2
(b) L
5000
(c) P 0
5000
1 39
(d)
2500
53.
1
39
t
12 faster for (c), it matches (c).
P ·
§
3P¨1 ¸ (a) k
100
©
¹
dP
dt
(b) L
(c)
e 0.2t
6, it matches (c) or (d).
(e)
5000
1 39e 0.2t
2
0.2t
Because y approaches L
70
125
1 39e 0.2t
12
1 e 2 x
Because y 0
P0
5000
1 39e 0.2t
(a) k
8, it matches (b).
§ 1·
ln 29
ln ¨ ¸
© 29 ¹
ln 29
| 4.4897 yr
0.75
P ·
§
0.75 P¨1 ¸,
2100 ¹
©
dP
dt
52. P t
12
1 3e x
2100
1 29e 0.75t
2
0.75t
Because y 0
70
1 29e 0.75t
−4
47. y
589
dP
dt
§1·
ln 39
ln¨ ¸
© 39 ¹
ln 39
| 18.3178
0.2
P ·
§
0.2 P¨1 ¸,
5000
©
¹
P0
125
3
100
120
5
0
0
INSTRUCTOR USE ONLY
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590
NOT FOR SALE
Chapter 6
(d)
Differential
ferential Equation
Equations
P ·
§
§ Pc ·
3Pc¨1 ¸ 3P ¨
¸
100 ¹
©
© 100 ¹
d 2P
dt 2
ª §
P ἤ
P ·
P ·º
3P ª §
3«3P¨1 3P¨1 ¸ ¨1 ¸
¸
100 ¹»¼©
100 ¹ 100 «¬ ©
100 ¹»¼
¬ ©
d 2P
dt 2
54.
dP
dt
§
50, and by the first Derivative Test, this is a maximum. ¨ Note: P
©
0 for P
0.1P 0.0004 P 2
56.
dy
dt
0.1P 1 0.004 P
k
P ·
§
0.1P¨1 ¸
250 ¹
©
y
(a) k
0.1
(b) L
250
(c)
P ·§
P
P ·
§
9 P¨1 ¸¨1 ¸
100 ¹©
100 100 ¹
©
y·
§
2.8 y¨1 ¸,
10 ¹
©
2.8, L 10
Solution: y
300
57.
100
0
k
250
2
(d) P
y0
125. (Same argument as in Exercise 77)
4
5
7
10
Ÿ b
7
3
7
10
§ 3 · 2.8t
1 ¨ ¸e
©7¹
4y
y2
5
150
dy
dt
100 ·
¸
2 ¹
L
2
10
L
1 be kt
1 be 2.8t
10
0, 7 : 7
Ÿ 1b
1b
1
10
0
50
P ·§
2P ·
§
9 P¨1 ¸¨1 ¸
100 ¹©
100 ¹
©
0.8, L
4 §
y ·
y ¨1 ¸,
5 ©
120 ¹
y0
8
120
L
120
1 be kt
1 be 0.8t
120
Ÿ b 14
0, 8 : 8
1b
y
55.
dy
dt
k
y
y·
§
y¨1 ¸,
36 ¹
©
1, L
36
L
1 be kt
y0
4
Solution: y
36
1 be t
0, 4 : 4
36
Ÿ b
1b
Solution: y
36
1 8e t
8
58.
dy
dt
120
1 14e 0.8t
3y
y2
20 1600
3 §
y ·
y¨1 ¸,
20 ©
240 ¹
y0
15
3
,L
240
20
240
L
y
1 be kt
1 be 3 20 t
240
0, 15 : 15
Ÿ b 15
1b
k
Solution: y
240
1 15e
3 20 t
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 6.3
L
,L
1 be kt
59. (a) P
200, P 0
200
Ÿ b
1b
200
25
39
1 7e k 2
200
39
23
39
1 § 23 ·
ln ¨ ¸
2 © 39 ¹
1 7e 2 k
e 2 k
k
Separation of Variables and Logistic
Logis
Equation
591
25
7
1 § 39 ·
ln ¨ ¸ | 0.2640
2 © 23 ¹
200
1 7e 0.2640t
P
(b) For t
5, P | 70 panthers.
(c)
100
200
1 7e 0.264t
1 7e 0.264t
2
§1·
ln ¨ ¸
©7¹
t | 7.37 years
0.264t
(d)
P·
§
kP¨1 ¸
L¹
©
dP
dt
P ·
§
0.264 P¨1 ¸, P 0
200 ¹
©
25
Using Euler's Method, P | 65.6 when t
(e) P is increasing most rapidly where P
60. (a)
y
1
4
L
,L
1 be kt
20
Ÿ b
1b
20
1 19e2 k
1 19e 2 k
5
2 k
4
k
y
20
1 19e0.7791t
19e
(b) For t
1 §4·
ln ¨ ¸
2 © 19 ¹
20, y 0
5.
200 2
1, y 2
100, corresponds to t | 7.37 years.
4
19
1 § 19 ·
ln ¨ ¸ | 0.7791
2 ©4¹
5, y | 14.43 grams
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
592
NOT FOR SALE
Chapter 6
Differential
ferential Equation
Equations
(c)
18
20
1 19e 0.7791t
20
10
18
9
1
9
1
171
1
§ 1 ·
ln ¨
¸ | 6.60 hours
0.7791 © 171 ¹
1 19e 0.7791t
19e 0.7791t
e 0.7791t
t
(d)
dy
dt
y·
§
ky¨1 ¸
L¹
©
y·
1 § 19 · §
ln ¨ ¸ y¨1 ¸
2 ©4¹ ©
20 ¹
t
0
1
2
3
4
5
Exact
1
2.06
4.00
7.05
10.86
14.43
Euler
1
1.74
2.98
4.95
7.86
11.57
(e) The weight is increasing most rapidly when y
L 2
61. A differential equation can be solved by separation of
variables if it can be written in the form
M x N y
dy
dx
0.
20 2
64.
10, corresponding to t | 3.78 hours.
dy
dt
y·
§
ky ¨1 ¸, y 0 L
L
©
¹
d2y
dt 2
y·
§
§ yc ·
kyc ¨1 ¸ ky¨ ¸
L
©
¹
© L¹
To solve a separable equation, rewrite as,
M x dx
ª
y ·º
§
2
« ky ¨1 L ¸ »
y
§
·
©
¹»
k 2 y¨1 ¸ ky «
L¹
L
«
»
©
«¬
»¼
N y dy
and integrate both sides.
62. Two families of curves are mutually orthogonal if each
curve in the first family intersects each curve in the
second family at right angles.
63. y
yc
2y ·
y· §
§
k 2 ¨1 ¸ y ¨1 ¸
L¹ ©
L¹
©
1
1 be kt
1
1 be kt
y · ª§
y·
yº
§
k 2 ¨1 ¸ y «¨1 ¸ »
L ¹ ¬©
L ¹ L¼
©
2
bke
So,
kt
d2y
dt 2
0 when 1 2y
L
0 Ÿ y
L
.
2
By the First Derivative Test, this is a maximum.
k
be kt
˜
kt
1 be
1 be kt
1 be kt 1
k
˜
1 be kt
1 be kt
k
1
§
·
˜ ¨1 ¸
1 be kt ¹
1 be kt ©
ky 1 y
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 6.3
65. (a)
dv
dt
dv
³W v
ln W v
Separation of Variables and Logistic
Logis
Equation
71.
kW v
kt C1
2 ln>txty@
20, v
72.
0 and v
0 when t
0.5 so, C
20, k
10
f x, y
tan x y
f tx, ty
tan tx ty
73.
f x, y
2 ln
x
y
f tx, ty
2 ln
tx
ty
or
v
(b) s
1.386 t
dt | 20 t 0.7215e 1.386t C
74.
0, C | 14.43 and you have
Because s 0
2 ln
x
y
Homogeneous of degree 0
20 1 e 1.386t
³ 20 1 e
tan ª¬t x y º¼
Not homogeneous
ln 4.
t
§
§1· ·
20¨1 ¨ ¸ ¸
¨
© 4 ¹ ¸¹
©
20 1 e ln 4 t
2 ln t 2 ln xy
Not homogeneous
W Ce kt
Particular solution:
v
f tx, ty
2 ln ª¬t 2 xyº¼
Initial conditions:
when t
2 ln xy
³ k dt
v
W
f x, y
593
f x, y
f tx, ty
s | 20t 14.43 e 1.386t 1 .
y
x
ty
tan
tx
tan
tan
y
x
Homogeneous of degree 0
66. Answers will vary. Sample answer: There might be
limits on available food or space.
67.
f x, y
x3 4 xy 2 y 3
f tx, ty
t 3 x3 4txt 2 y 2 t 3 y 3
x y dx 2 x dy
f x, y
x3 3x 2 y 2 2 y 2
f tx, ty
t 3 x3 3t 4 x 2 y 2 2t 2 y 2
0
1 u dx 2 x du 2u dx
0
Not homogeneous
ln Cx
x2 y 2
f x, y
Cx
x2 y2
t 4 x2 y 2
f tx, ty
t
t 2 x2 t 2 y 2
x2 y2
3
x2 y2
x
Homogeneous of degree 3
70.
Cx
x du u dx
ux, dy
x ux dx 2 x x du u dx
1
dx
x
1
³ x dx
ln x ln C
Homogeneous of degree 3
69.
0, y
1 u dx
t 3 x3 4 xy 2 y 3
68.
75.
2 x du
2
du
1u
1
du
2³
1u
2 ln 1 u
ln 1 u
2
1
1u
1
2
ª¬1 y x º¼
2
x2
x y
2
C x y
2
xy
f x, y
x y2
2
tx ty
f tx, ty
t 2 x2 t 2 y 2
t 2 xy
t
x y
2
2
t
xy
x y2
2
Homogeneous of degree 1
INSTRUCTOR USE ONLY
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594
Chapter 6
NOT FOR SALE
Differential
ferential Equation
Equations
76. x3 y 3 dx xy 2 dy
0, y
ª x3 ux 3 º dx x ux 2 x du u dx
¬
¼
0
1 u dx u x du u dx
0
3
2
2
0, y
0
1 u dx 1 u x du u dx
0
1 2u u 2 dx
dx
³x
³ u du
dx
x
dx
³
x
2
1§ y ·
¨ ¸
3© x ¹
u3
3
3
3
3x3 ln x Cx3
0
1 u 2 dx 2u x du u dx
0
1 u 2 dx
x 1 u du
1u
du
u 2 2u 1
u 1
³ u 2 2u 1 du
1
ln u 2 2u 1
2
C
x
ln u 2 2u 1
C2
x2
u 2 2u 1
C
x2
§ y·
§ y·
¨ ¸ 2¨ ¸ 1
x
© ¹
© x¹
C
y 2 2 yx x 2
12
2
2ux du
dx
x
dx
³
x
2u
du
1 u2
2u du
³ 1 u2
ln x ln C
ln 1 u 2
C
x
ln u 2 1
C
x
u2 1
Cx
y2 x2
ln
x du u dx
x du u dx
dx 2 x ux x du u dx
ln
ln x ln C
3 ln x C
ux, dy
ux, dy
x ux dx x ux x du u dx
xu du
y3
x 2 ux
x y dx x y dy
dx
§ y·
¨ ¸
© x¹
0, y
77.
2
ln x C1
78. x 2 y 2 dx 2 x dy
x du u dx
ux, dy
ln ª¬u 2 1º¼
ln u 2 1
2
§ y·
¨ ¸ 1
© x¹
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
Section
ction 6.3
79. xydx y 2 x 2 dy
x ux dx ª ux
¬
0, y
0
u dx u 2 1 x du u dx
0
3
u dx
u 2 1 x du
dx
x
dx
³x
1 u2
du
u3
1·
§ 3
³ ¨© u u ¸¹ du
ln x ln C1
ln C1 xu
80. 2 x 3 y dx x dy
y
Ce
2 3u dx x du u dx
0
2 2u dx
2 y x
x2
2 y2
2 y2
x du
2dx
x
1
2 ³ dx
x
2ln x ln C
du
1u
1
³ u 1 du
ln u 1
ln x 2 C
ln u 1
1u
x2 C
y
x
y
x
x2 C
1
x2
2
x du u dx
ux, dy
0
Cx 2 1
Cx3 x
y
dy
dx
1
2 x 3ux dx x x du u dx
81. False.
1
ln u
2u 2
1
2
2u
ln C1 y
0, y
595
x du u dx
ux, dy
x 2 º x du u dx
¼
2
Separation of Variables and Logistic
Logis
Equation
x
is separable, but y
y
0 is not a solution.
82. True
dy
dx
x 2 y 1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
596
Chapter 6
Differential
ferential Equations
Equation
83. True
fg c gf c
84.
x2 y2
x2 y 2
2Cy
dy
dx
x
C y
2 Kx
K x
y
dy
dx
x
K x
˜
C y
y
f f c g c gf c
0
fc
g
f fc
0
gc 2
gc
g
x y 2x
x2 y 2 2 y 2
2
2
2
fc
fc f
1 2x ex
2 xe x
2
2x 1 e
2
z 0, so
x2
1
1
2x 1
12
Ce x 2 x 1
g x
1
2
1
ln 2 x 1 C1
2
x ln g x
y 2 x2
x2 y 2
e x 2 xe x
1
.
2
avoid x
2 Kx 2 x
2Cy 2 y 2
Product Rule
2
Need f f c
Kx x 2
Cy y 2
f cg c
So there exists g and interval a, b , as long as
1
 a, b .
2
Section 6.4 First-Order Linear Differential Equations
1. x3 yc xy
ex 1
1
yc 2 y
x
6.
1 x
e 1
x3
dy
2
y
dx
x
Integrating factor: e ³
Linear
C
3 2
5
x x 2
4
3
x
ln x yc 1 2 x y
0
y
0
7. yc y
1 2x
ln x
y
3. yc y sin x
xy 2
5x
2 yc
5 xy
yc 5 xy
2
ye x
³ 16e
8. yc 2 xy
ye x
dy § 1 ·
5.
¨ ¸y
dx © x ¹
6x 2
Integrating factor: e ³
xy
³ x 6 x 2 dx
y
2x2 x 1 x dx
eln x
x2
x
e x
16e x C
dx
16 Ce x
10 x
Integrating factor: e ³
Linear
1 dx
16e x
y
2 yc
y
2
16
e x yc e x y
Not linear, because of the xy 2 -term.
4.
2
Integrating factor: e ³
Linear
eln x
3 4
5 x3
x C
4
3
³ x 3x 5 dx
y
yc 2 x dx
x2 y
2 xy yc ln x
2.
3x 5
2
³ 10 xe
x2
y
5 Ce x
dx
2 x dx
ex
2
2
5e x C
2
x
2x x C
3
2
C
x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 6.4
9.
y 1 cos x dx
yc
yc cos x y
y
y cos x cos x
5
cos x
Integrating factor: e ³
cos x dx
e sin x
yce sin x cos x e sin x y
ye
cos x e
³ cos x e
sin x
sin x
10. ª¬ y 1 sin xº¼ dx dy
yc sin x y
−3
0
³ sin xe
e
C
cos x
x 1 yc y
x2 1
§ 1 ·
yc ¨
¸y
© x 1¹
x 1
ª1 x 1 º¼ dx
y x 1
³ x 1 dx
y
x3 3x C
3x 1
x 1
ex
³ e dx
ye x
2x
e C
2
2x
1
C Ÿ C
2
1Ÿ1
ye x
1 2x
1
e 2
2
1 x
1 x
e e
2
2
1
2
1 x
e e x
2
6
(c)
−6
6
−2
e3 x
Integrating factor: e ³
3 dx
³ e e dx
ye3 x
eln x 1
dx
1 3
x x C1
3
2
12. yc 3 y
ye x
y0
y
Integrating factor: e ³ ¬
Integrating factor: e ³
e2 x
cos x
1 Ce cos x
y
11.
e
dx
dy
ex y
dx
dy
y
ex
dx
e x yc e x y
sin x dx
Integrating factor: e ³
ye
(b)
sin x
cos x
4
dx
1 Cesin x
y
x
−4
sin x
e sin x C
cos x
597
15. (a) Answers will vary.
dy
y 1 cos x
First-Order Linear Differenti
Differential Equations
e3 x
³ e dx
3x 3x
6x
1 6x
e C
6
1 3x
e Ce3 x
6
y
3
13. yc 3 x 2 y
ex
3x 2 dx
Integrating factor: e ³
3
³e e
x3 x3
y
x C ex
ye x
14. yc y tan x
y
3
x C
3
sec x
Integrating factor: e ³
y sec x
³ dx
dx
e x
tan x dx
³ sec x dx
2
e ln cos x
sec x
tan x C
sin x C ˜ cos x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
598
NOT FOR SALE
Chapter 6
Differential
ferential Equation
Equations
16. (a)
19. yc y tan x
y
4
sec x cos x
Integrating factor: e ³
x
−4
(b) yc 1
y
x
u x
e³
1
,Q x
x
sin x 2 , P x
1 x dx
ycx y
e ln x
sin x 2
x sin x 2
1
cos x 2 C
2
yx
2
³ x sin x dx
y
1ª 1
º
cos x 2 C »
x «¬ 2
¼
0
1 ª 1
º
cos S C » Ÿ C
«
2
S¬
¼
y
1ª 1
1º
cos x 2 »
2¼
x «¬ 2
(c)
y
sin x x cos x C cos x
1, 1
Particular solution: y
sin x x 1 cos x
³ sec x tan x sec x dx
y sec x tan x
sec x tan x C
y
1
2
1
Initial condition: y 0
1
§1·
21. yc ¨ ¸ y
© x¹
1
1
3 cos x
1 sin x
3
0
Integrating factor: e ³
yc sec 2 x y
sec 2 x
Separation of variables:
Integrating factor: e ³
³ sec xe
2
sec 2 x dx
tan x
e tan x C
dx
1 Ce tan x
5, C
Particular solution: y
1 4e tan x
18. x3 yc 2 y
e1 x
§2·
yc ¨ 3 ¸ y
©x ¹
4
2
1 1 x2
e
x3
Integrating factor: e ³
1
³ x3 dx
2 x3 dx
dy
dx
e tan x
Initial condition: y 0
y
C
,C
1 0
4, 4
3
sec x tan x
0
2
C
sec x tan x
Particular solution:
17. yc cos 2 x y 1
ye 1 x
sec x tan x
4
−4
y
C
sec x
Integrating factor:
sec x dx
e³
eln sec x tan x
y
ye tan x
tan x x C
Initial condition: y 0
4
−4
sec x
³ sec x sec x cos x dx
20. yc y sec x
x
eln sec x
y sec x
4
−4
tan x dx
e
1
³ y dy
1 x dx
eln x
x
y
x
1
³ x dx
ln y
ln x ln C
ln xy
ln C
xy
C
Initial condition: y 2
2, C
Particular solution: xy
4
4
1 x2
1
C1
2x2
2
2 § Cx 1 ·
e1 x ¨
¸
2
© 2x ¹
Initial condition: y 1
e, C
Particular solution: y
2
2 § 3x 1 ·
e1 x ¨
¸
2
© 2x ¹
3
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 6.4
22. yc 2 x 1 y
0
Integrating factor: e
2
ye x x
First-Order Linear Differenti
Differential Equations
25.
³ 2 x 1 dx
e
x2 x
C
Ce x x
y
2
Separation of variables:
1
³ y dy
³ 1 2 x dx
ln y ln C1
x x
kP N
2
2
yC1
ex x
y
Ce x x
P
2
Initial condition: y 1
2, 2
Particular solution: y
2e x x
x y 2
x
dy
1
y
dx
x
1
e³
1 x dx
u x
y
P
C
When t
2
26.
1
x
2·
§1
x ³ ¨ 2 ¸ dx
x ¹
©x
2
ª
º
C»
x «ln x x
¬
¼
2 x ln x Cx
y1
y
10
2 C Ÿ C
y
§ x2
1· 1
x1 2 ³ ¨
¸ 1 2 dx
2¹x
©2
y
2
P0
N
N
Ÿ C
P0 k
k
N · kt
N
§
¨ P0 ¸e k¹
k
©
C t C1
rA P
e rt C2
³ dt
rt C2
C3e rt P
r
P
Ce rt r
A
When t
Linear
§ x3 2
x 1 2 ·
x1 2 ³ ¨
¸ dx
2 ¹
© 2
27. (a) A
A
x
64
4 2C Ÿ C
5
17
x3
x 5
5
C3e kt N
k
N
Ce kt k
1
ln rA P
r
ln rA P
12
ª x5 2
º
x1 2 «
x1 2 C »
5
¬
¼
y4
e kt C2
dt
A
x2
1
2
2
1
x
dx
1
2
e³
x1 2
x3
x C
5
kt C2
rA P
x3 x
dy
1
y
dx
2x
u x
t C1
dA
dt
dA
rA P
dA
³ rA P
2 x ln x 12 x
24. 2 xyc y
³ dt
P
Linear
2·1
§
x ³ ¨1 ¸ dx
x¹x
©
dt
P0
y
2
1
x
x
2
x
kP N , N constant
0: P
x y 2 dx
23. x dy
dy
dx
dP
dt
dP
kP N
1
³ kP N dP
1
ln kP N
k
ln kP N
17
5
(b) A
599
0: A
0
0
C A
P rt
e 1
r
P
Ÿ C
r
P
r
P rt
e 1
r
275,000 0.08 10
e
1 | $4,212,796.94
0.06
550,000 0.059 25
e
1 | $31,424,909.75
0.05
x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
600
NOT FOR SALE
Chapter 6
Differential
ferential Equations
Equation
31. From Example 3,
125,000 0.08t
e
1
0.08
28. 1,000,000
dv
kv
g
dt
m
mg
v
1 e kt m ,
k
e0.08t
1.64
ln 1.64
t
| 6.18 years
0.08
8, v 5
k 75 N
(b) N c kN
k dt
N ce kt kNe kt
75 ke kt
Ne kt c
75 ke kt
Ne kt
³ 75 ke
(c) For t
1, N
For t
As t o f, v o 159.47 ft/sec. The graph of v is
shown below.
75 e kt C
kt
50
kt
0
75 Ce k Ÿ 55
40
20, N
35:
Ce k
−200
75 Ce 20 k Ÿ 40
Ce 20 k
³ v t dt
0.2007 t
dt
³ 159.47 1 e
32. s t
1 § 11 ·
ln ¨ ¸
19 © 8 ¹
11
Ÿ k
8
| 0.0168
55
159.47t 794.57e 0.2007t C
s0
794.57 C Ÿ C
5000
159.47t 794.57e
st
55e k | 55.9296
C
159.47 1 e 0.2007t .
v
20
Ce k
Ÿ e19 k
Ce 20 k
Ce k
1
4
20:
35
55
40
8
g
Using a graphing utility, k | 0.050165, and
e kt
75 Ce
N
101, m
8
1 e 5k 1 4 .
k
implies that 101
75k
Integrating factor: e ³
N
32, mg
g
dN
29. (a)
dt
Solution
0.2007 t
5794.57
5794.57
The graph of s t is shown below.
75 55.9296 e 0.0168t
6000
dQ
30. (a)
dt
q kQ, q constant
(b) Qc kQ
q
0
Let P t
k, Q t
is u t
kt
e .
Q
e kt ³ qe kt dt
When t
0: Q
Q
t of
0 when t | 36.33 sec.
st
Q0
(c) lim Q
q, then the integrating factor
40
−500
§q
·
e kt ¨ e kt C ¸
©k
¹
q
Ce kt
k
33. L
dI
RI
dt
E0 , I c Integrating factor: e ³
Q0
q
q
C Ÿ C
Q0 k
k
q §
q · kt
¨ Q0 ¸e
k
k¹
©
I e Rt L
I
E0 Rt L
³ Le
R
I
L
R L dt
dt
E0
L
e Rt L
E0 Rt L
e
C
R
E0
Ce Rt L
R
q
k
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 6.4
34. I 0
0, E0
L
4 henrys
600 ohms,
1
5
E0
Ce Rt L
R
120
1
C Ÿ C
600
5
1 1 150t
e
5 5
1
amp
5
1
0.18
1 e 150t
5
0.9
1 e 150t
150 t
0.1
I
0
I
lim I
t of
0.90
e
120 volts, R
150t
t
First-Order Linear Differenti
Differential Equations
37. (a) From Exercise 35,
r2 Q
dQ
dt
u0 r1 r2 t
You have Q 0 q0
r1
r2
150
q1r1
25, q1
0, u0
200, and
10. Hence, the linear differential
equation is
dQ
1
Q
dt
20
0.
By separating variables,
³ Q
dQ
³
ln Q
ln 0.1
ln 0.1
601
1
dt
20
1
t ln C1
20
1t
Q
Ce 20 .
The initial condition Q 0
| 0.0154 sec
C
35. Let Q be the number of pounds of concentrate in the
solution at any time t. Because the number of gallons of
solution in the tank at any time t is v0 r1 r2 t and
because the tank loses r2 gallons of solution per minute,
it must lose concentrate at the rate
ª
º
Q
«
» r2 .
«¬ v0 r1 r2 t »¼
(b) 15
25. Hence, Q
Ÿ t
(c) lim Q
t of
1t
25e 20 Ÿ
3
5
25 implies that
1t
25e 20 .
1t
§ 3·
e 20 Ÿ ln ¨ ¸
©5¹
1
t
20
§ 3·
20 ln ¨ ¸ | 10.2 minutes
©5¹
1t
lim 25e 20
t of
0
The solution gains concentrate at the rate r1q1. Therefore,
the net rate of change is
dQ
dt
ª
º
Q
q1r1 «
» r2
«¬ v0 r1 r2 t »¼
or
dQ
r2Q
dt
v0 r1 r2 t
q1r1.
36. From Exercise 35, and using r1
dQ
rQ
dt
v0
r2
r,
q1r.
INSTRUCTOR USE ONLY
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602
NOT FOR SALE
Chapter 6
Differential
ferential Equations
Equation
38. (a) The volume of the solution in the tank is given by
v0 r1 r2 t. Therefore, 100 5 3 t
200 or
t
39. yc P x y
r2Q
v0 r1 r2 t
Q0
q0 , q0
0.5,v0
3
3, Qc Q
100 2t
r2
100, r1
³ 2.5 50 t
32
50 t
32
50 t
52
dt
50 t C 50 t
Q
50 C 50
0, 0
3 2
52
100
(c) At t
50
41.
52
3 2
0. It takes 25 minutes to empty the
e³
Q x
Standard form
P x dx
Integrating factor
50
0
³ dy
³ 2 x dx
y
x2 C
Matches (c).
Q0
q0
Qc 3Q
100 2t
q1r1
44. yc 2 y
1, v0
dy
³y
0, q1
³ 2 dx
ln y
2 x C1
Q 50 t
32
Q
100, r1
5, r2
3
5
0
Ce 2 x
y
Integrating factor is 50 t
³ 5 50 t
32
32
.
dt
Matches (d).
2 50 t
2 50 t C 50 t
52
C
45. yc 2 xy
3 2
0:
3 2
0
100 C 50
Q
2 50 t 2 50
Q
dy
P x y
dx
43. yc 2 x
25
| 82.32 lb
2
r2Q
v0 r1 r2 t
When t
25, Q
42. The term “first-order” means that the derivative in the
equation is first order.
3 2
minutes.
Q0
20 pounds.
tank.
(c) The volume of the solution is given by
v0 r1 r2 t 100 5 3 t
200 Ÿ t
Qc 0, Q
u x
100 50
100 P xu
(b) The rate of solution withdrawn is greater.
C
3 2
,C
50 t 505 2 50 t
Q 50
Q xu
40. (a) At t
Particular solution:
Q
uy c
so uc x
Initial condition:
Q0
Q xu
P x dx
Answer (a)
ª3 100 2 t ¼º dt
32
5,
2.5
Integrating factor: e ³ ¬
Q 50 t
ycu P x yu
q1r1
0, q1
e³
Integrating factor: u
50 minutes.
(b) Qc Q x
ŸC
52
100 50
50 t
32
2 50
50
2
| 164.64 lb (double the answer to part (b))
200 2 50
52
100
3 2
200 ³ 2 x dx
ln y
x 2 C1
52
Ce x
y
3 2
50,
0
dy
³y
2
Matches (a).
46.
yc 2 xy
dy
³ 2y 1
1
ln 2 y 1
2
2y 1
y
x
³ x dx
1 2
x C1
2
C2 e x
2
2
1
Ce x
2
Matches (b).
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 6.4
47. (a)
First-Order Linear Differenti
Differential Equations
603
10
−4
4
−6
dy
1
(b)
y
dx
x
x2
Integrating factor: e 1 x dx
1
1
yc 2 y
x
x
§1 ·
¨ y¸
©x ¹
x
x2
C
2
³ x dx
x3
Cx
2
y
2, 4 : 4
8
2C Ÿ C
2
2, 8 : 8
8
2C Ÿ C
2
(c)
1
x
e ln x
x3
1
4x
x x2 8
2
2
x3
1
2x
x x2 4
2
2
4 Ÿ y
2 Ÿ y
10
−4
4
−6
48. (a)
5
−1
3
−1
(b) yc 4 x3 y
x3
Integrating factor: e ³
4
4 x3 dx
yce x 4 x3 ye x
4
x 3e x
ye x
4
³x e
y
0, 72 : 72
0, 12 : 12
(c)
ex
4
4
3 x4
1 Ce
4
dx
1 e x4
4
C
x4
1 C Ÿ C
4
1 C Ÿ C
4
13 Ÿ
4
y
1 13 e x4
4
4
43 Ÿ y
1 3 e x4
4
4
5
−1
3
−1
INSTRUCTOR USE ONLY
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604
NOT FOR SALE
Chapter 6
Differential
ferential Equations
Equation
3
49. (a)
−2
6
−3
(b) yc cot x y
2
Integrating factor: e ³
cot x dx
ycsin x cos x y
sin x
2 sin x
³ 2 sin x dx
2 cos x C
2 cot 1 C csc 1 Ÿ C
1 2 cot 1
csc 1
y sin x
2 cot x C csc x
y
1, 1 : 1
eln sin x
y
2 cot x sin 1 2 cos 1 csc x
3, 1 : 1
2 cot 3 C csc 3 Ÿ C
sin 1 2 cos 1
2 cot 3 1
csc 3
2 cos 3 sin 3
2 cot x 2 cos 3 sin 3 csc x
y
3
(c)
−2
6
−3
50. (a)
7
−5
5
−3
(b) yc 2 xy
xy 2
Bernoulli equation, n
2
1 x2
e
C
2
1
y
2
1
Ce x
2
2
y 1e x
y
0, 3 : 3
y
0, 1 : 1
y
(c)
y1 2
2 letting z
1 2Ce x
2
y 1 , you obtain e 2 x dx
2
2
e x and ³ 1 xe x dx
1 x2
e . The solution is:
2
2
2
1 2Ce x
2
Ÿ 1 2C
1 2C
2
6
2
1 ex 3
3 ex
2
Ÿ 1 2C
1 2C
2
1 ex
2
Ÿ C
3
2 Ÿ C
1
2
1
6
2
2
7
INSTRUCTOR USE ONLY
−5
5
5
−3
−3
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NOT FOR SALE
Section 6.4
51. e2 x y dx e x y dy
56.
0
Separation of variables:
x y
2x y
e e dx
ee
dy
³ e dx
³e
dy
x
2e x e 2 y
y
C
1
1
x
³ x dx
1
y
x ln x C
y3
2 y2
3
x2
3 x C1
2
dy
dx
3
yc y
x
2 y 3 12 y 2
3x 2 18 x C
Integrating factor: e ³
y cos x cos x dx dy
ycx3 0
ln y 1
sin x ln C
y
54. yc
Ce
sin x
y
12 2
C
x 3
5
x
1
³ 2 x dx
arcsin y
x C
y
³ y e
1
ln x 2 1
2
1
y 2 e y C1
2
ln x 2 1 y 2 2e y
59. yc 3 x 2 y
sin x 2 C
0
§ 2·
Linear: yc ¨ ¸ y
© x¹
1 x
e
x
Integrating factor: e ³
2 x dx
y
³ x2 1 dx
y
dy
C
2
55. 2 y e x dx x dy
21 x
0
Separation of variables:
n
yx 2
12 5
x C
5
4
x
dy
1 y2
x3
12 x 4
58. x dx y e y x 2 1 dy
Separation of variables:
³
e3 ln x
³ 12 x dx
2x 1 y2
1
3 x dx
yx3
ln y 1 ln C
sin x
1
x
12 x
12 x x3
1
³ y 1 dy
ln x 1
3 y 12 x 2
3 3
x y
x
Separation of variables:
³ cos x dx
e
x dy
x
³ x 3 dx
2
1 x dx
ln x C
57. 3 y 4 x 2 dx
³ y 4 y dy
605
0
Integrating factor: e ³
Separation of variables:
53.
1
y
x
Linear: yc x 3
y y 4
dy
dx
x y dx x dy
12 e 2 y C1
ex
52.
2 y
First-Order Linear Differenti
Differential Equations
³ x x e dx
x2 , P
3, Q
y 2e ³
2 3 x 2 dx
2
y e
eln x
2
x2
y 2e 2 x
3x 2
2 3 x 2 dx
³ 2 x e ³
3
³ 2 x 2e 2 x dx
2 2 x3
ex x 1 C
ex
C
x 1 2
x2
x
x2 y3
dx
1 2 x3
e
C
3
3
1
Ce 2 x
3
3
y 2
3
1
1
Ce 2 x 2
y
3
INSTRUCTOR USE ONLY
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606
NOT FOR SALE
Chapter 6
Differential
ferential Equations
Equation
xy 1
60. yc xy
n
1, Q
x, P
y 2e x
2
x2
y
³ 2 xe
1 Ce
2
§1·
61. yc ¨ ¸ y
© x¹
n
2, Q
e³
1 x dx
y 1 x 1
x, e ³
2 x dx
ex
63. xyc y
xy 3
yc 1
y
x
y3
n
3, Q
1, P
1
,
x
³ 2x
dx C
2
2
ex C
dx
x2
y 2 x 2
xy 2
x, P
x 1
e ln x
x 1
³ x x
1
1
y
x Cx
y
1
Cx x 2
x C
dx
y 2
2 x Cx 2
y2
1
2 x Cx 2
64. yc y
n
e 1 2 1 x dx
12 12
y x
y
x
1,
³ 2 e
dx
x
e 1 2 ln x
1 Ce 2 x
y2
1
1 Ce 2 x
x5 2 C
65. yc y
ex 3 y , n
e³
e 2 3 x
x
1 12
³ 2 x x dx
x5 2 C
1 52
x C1
5
5
2
2x
y 2
y
x, P
2 3 dx
y 2 3e 2 3 x
³ 23 e e
23 23x
13x
y
2e
e³
x 2
2 x Cx 2
2 1 dx
e2 x
e 2 x C
1
,Q
3
dx
1
ex , P
³ 23 e
13x
dx
C
2e Ce 2 x 3
x
23
66. yyc 2 y 2
n
1
y2
e³
x 23x
y e
25 x
e 2 ln x
2 x 1 C
or
1, Q
3, P
y 2e 2 x
1
³ x dx
y3
2
§1·
62. yc ¨ ¸ y
© x¹
1
n
,Q
2
2
2
e
ex
yc 2 y
e x y 1
1, Q
ex , P
2 2 dx
e 4 x
y 2e 4 x
³ 2e
2
4 x x
e dx
23 e3 x C
23 e x Ce 4 x
y2
67. False. The equation contains
68. True. yc x e x y
y.
0 is linear.
Review Exercises for Chapter 6
1. y
x 3 , yc
2 xyc 4 y
2.
3x 2
2 x 3x 2 4 x3
Yes, it is a solution.
10 x3 .
y
2 sin 2 x
yc
4 cos 2 x
ycc
8 sin 2 x
yccc
16 cos 2 x
yccc 8 y
16 cos 2 x 8 2 sin 2 x z 0
Not a solution
INSTRUCTOR USE ONLY
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NOT FOR SALE
Review Exercises ffor Chapter 6
3.
dy
dx
4x2 7
y
³ 4 x 7 dx
dy
4.
dx
y
dy
5.
dx
6.
7.
8.
dy
dx
2x y
x
–4
2
0
2
4
8
y
2
0
4
4
6
8
dy dx
–10
–4
–4
0
2
8
4 x3
7x C
3
2
3x 8 x
3
3 4
x 4 x2 C
4
³ 3x 8 x dx
3
10.
cos 2 x
y
³ cos 2 x dx
dy
dx
2 sin x
y
³ 2 sin x dx
dy
dx
e2 x
dy
dx
§S y ·
x sin ¨ ¸
© 4 ¹
x
–4
2
0
2
4
8
y
2
0
4
4
6
8
dy dx
–4
0
0
0
–4
0
1
sin 2 x C
2
2 cos x C
11. yc
2 x 2 x,
0, 2
(a) and (b)
y
(0, 2)
y
³e
dy
dx
2 e3 x
y
9.
607
2 x
dx
e2 x C
3x
³ 2e dx
2 3x
e C
3
5
x
−3
12. yc
3
−1
y 4 x,
1, 1
(a) and (b)
y
(−1, 1)
2
x
−3
3
−4
13. yc
x y, y 0
4, n
10, h
0.05
y1
y0 hf x0 , y0
4 0.05 0 4
y2
y1 hf x1 , y1
3.8 0.05 0.05 3.8
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
yn
4
3.8
3.6125
3.437
3.273
3.119
2.975
2.842
2.717
2.601
2.494
3.8
3.6125, etc.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
608
NOT FOR SALE
Chapter 6
Differential
ferential Equations
Equation
14. yc
5 x 2 y, y 0
y1
y0 hf x0 , y0
2 0.1 5 0 2 2
y2
y1 hf x1 , y1
1.6 0.1 5 0.1 2 1.6
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
2
1.6
1.33
1.164
1.081
1.065
1.102
1.182
1.295
1.436
1.599
15.
16.
dy
dx
2x 5x2
y
³ 2 x 5 x dx
dy
dx
dy
³y8
³ dx
ln y 8
x C1
y 8
e x C1
3 y
dy
dx
3 y
dy
³ dx
1
3 10
y
y
0
dy
dx
xy
2 x
1
dy
y
x
dx
2 x
1
dy
y
2 ·
§
¨1 ¸ dx
x¹
2
©
ln y
x 2 ln 2 x C1
2
21.
³ 10 dx
10 x C1
5x C
dy
dx
dy
³y
dy
dt
§
¨C
©
2
C1 ·
¸
2¹
22.
³
2 x
2
ln y
x ln x C1
x 1
dx
x
Cxe x
k
t3
3
dt
k
C
2t 2
y
dy
dt
k 50 t
³ dy
³ k 50 t dt
y
Ce x
x 1y
y
1
x C
2
0
x
³ dy ³ kt
5x C
Ce x 2 x
y
x C
y
y1 2
19. 2 x yc xy
20. xyc x 1 y
3 y
2 y1 2
1.33, etc.
Ce x
1
x C
1 2
³ y dy
1.6
2
8 Ce x
2
dy
dx
0.1
y 8
17.
³ 3 y
10, h
5
x x3 C
3
2
y
18.
2, n
50kt ³ 50k kt dt
k 2
t C
2
(Alternate form: y
k
2
50 t C1 )
2
INSTRUCTOR USE ONLY
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NOT FOR SALE
Review Exercises ffor Chapter 6
Cekt
23. y
27.
0, 34 : 34
C
5, 5 : 5
3 k5
e
4
e
k
1 ln 20
5
3
5e
1
30
e
k
1 ln 1
5
30
y
5e ¬
15ekt
7.5
15e k 1599
| 0.000433
1
ln 12
1599
So, C
3 2 k
e
2
3 2 k 4 k
e
e
2
2C
Ce0.0185t
2
e0.0185t
ln 2
0.0185t
e2k Ÿ k
1 ln 10
2
3
3 2 1 2 ln 10 3
e
2
3 3
2 10
9 1 2 ln 10 3 t
e
20
ln 2
| 37.5 years
0.0185
t
3 2k
e
2
9
.
20
9 0.602t
| 20
e
30. A
1000e 0.04 8 | $1377.13
31. S
Ce k t
(a)
Cekt
lim Ce
1, 4 : 4
Ce k 1
Ce k Ÿ C
4, 1 : 1
Ce k 4
Ce4 k
1
4e
k
3k
Ÿ k
1 ln 1
3
4
k
0.4621
4e
e
4e
4k
4e
S
5 when t
k
5
Ce
C
5
30e k
k
ln 16 | 1.7918
S
30e 1.7918 t
3k
(b) When t
| 6.3496.
(c)
1
k t
t of
4e k
13 ln 4 | 0.4621
15e0.000433 750 | 10.84 g .
750, y
Ce0.0185t
kt
10
3
y
30e 35,000 ln 2 18,000 | 7.79 inches
P
29.
| 5e 0.6802t
Ce 4 k
So, C
ln 2
18,000
Ce kt
When t
15 ln 30
4, 5 : 5
e
ln 1 2
5k
Ce 2 k Ÿ C
1
4
15
k5
2, 32 : 32
26. y
30e18,000 k
30e h ln 2 18,000
P 35,000
k
30
18,000
Ph
| 34 e0.379t
ª ln 30 5¼º t
y
30e kh
k
28. y
5, 16 : 16
Ce
Ph
C
0, 5 : 5
25. y
P0
5k
Cekt
24. y
kp,
P 18,000
20
3
3 ª¬ln 20 3 5º¼t
e
4
y
dP
dh
609
30
5, S | 20.9646 which is 20,965 units.
30
6.3496e 0.4621t
0
40
0
32. S
25 1 e kt
(a) 4
S
25 1 e k 1
Ÿ 1 ek
4 Ÿ ek
25
21 Ÿ k
25
21 | 0.1744
ln 25
25 1 e 0.1744t
(b) 25,000 units lim S
t of
25
INSTRUCTOR USE ONLY
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610
NOT FOR SALE
Chapter 6
Differential
ferential Equations
Equation
(c) When t
5, S | 14.545 which is 14,545 units.
(d)
25
0
8
0
33.
dy
dx
37. y 3 yc 3 x
5x
y
y3
³ y dy
³ 5 x dx
y2
2
y2
5x2
C1
2
5x2 C
³ y dy
3
4
y
4
y4
x3
2 y2
dy
dx
34.
2
3x
³ 3x dx
3x 2
C1
2
6x2 C
Initial condition: y 2
2
³ x dx
2 y3
3
x4
C1
4
8 y3
3x 2 C
3
35. yc 16 xy
dy
dx
16 xy
³ y dy
³ 5e
8 x 2 C1
y2
2
y2
3
0, y 0
5e 2 x
ln y
2x
dx
5 2x
e C1
2
5e 2 x C
Ce8 x
36. yc e y sin x
5 C
C
4
2
Particular solution: y 2
0
5e2 x 4
39. y 3 x 4 1 yc x3 y 4 1
dy
dx
e sin x
dy
³ sin x dx
y
cos x C1
ey
1
cos x C
y
ln
e
2
3 : 3
Initial condition: y 0
y
y
y
38. yyc 5e 2 x
0
³ 16 x dx
2
8
6x2 8
Particular solution: y 4
dy
y
dx
1
³ y dy
e8 x C1
24 C
2 : 16
C
³ 2 y dy
³e
dy
dx
0, y 2
y
1
cos x C
y3 x4 1
C
C1
ln cos x C
dy
dx
0, y 0
1
x3 y 4 1
y3
³ y 4 1 dy
x3
³ x4 1 dx
1
ln y 4 1
4
1
1
ln x 4 1 ln C1
4
4
ln y 4 1
y4 1
ln ª¬C x 4 1 º¼
C x4 1
1 :1 1
Initial condition: y 0
C
C 01
2
Particular solution: y 1
2x 1
y4
2x4 1
4
4
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Review Exercises ffor Chapter 6
40. yyc x cos x 2
y
dy
dx
2
0, y 0
0.55
5250
³ x cos x dx
(b) L
y2
2
y2
1
sin x 2 C1
2
sin x 2 C
(c) P 0
5250
1 34
(d)
2625
2
2 : 4
C
Particular solution: y
dy
dx
(a) k
³ y dy
Initial condition: y 0
41.
5250
1 34e 0.55t
43. P t
x cos x 2
sin 0 C
1
34
1
§1·
ln ¨ ¸ | 6.41 yr
0.55 © 34 ¹
t
4 x
y
³ y dy
(e)
³ 4 x dx
y2
2
4 x2 y 2
2 x 2 C1
C
ellipses
P ·
§
0.55P¨1 ¸
5250
©
¹
dP
dt
4800
1 14e 0.15t
44. P t
y
(a) k
0.15
(b) L
4800
4
4800
1 14
(c) P 0
x
−4
(d)
4
−4
dy
42.
dx
2
e 0.55t
2
1
ln 2 y 3
2
ln 2 y 3
2y 3
(e)
³ dx
x C1
2 x 2C1
C2 e
2 x
2y
3 C2e 2 x
y
3
Ce 2 x
2
45.
dy
dt
14e 0.15t
1
t
1
§1·
ln ¨ ¸ | 17.59 yr
0.15 © 14 ¹
P ·
§
0.15 P¨1 ¸
4800 ¹
©
dP
dt
y·
§
y¨1 ¸,
80 ¹
©
k
1, L
y
L
1 be kt
y0
8: 8
Solution: y
y
320
4800
1 14e0.15t
2400
3 2y
dy
³ 2y 3
150
5250
1 34e 0.55t
1 34e 0.55t
4
sin x 4
2
611
0, 8
80
80
1 be t
80
Ÿ b
1b
9
80
1 9e t
x
4
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
612
Chapter 6
Differential
ferential Equations
Equation
46.
dy
dt
y·
§
1.76 y ¨1 ¸,
8¹
©
k
1.76, L
y
L
1 be kt
y0
3:
8
1 be 1.76t
8
Ÿ b
1b
3
y
20,400
1 be kt
y0
1200
y1
5
3
16e
k
k
1200, y 1
48.
dy
dt
53.
ex y y
1
y
4
1 x4
e
4
1 x4
e
4
u x
e³
e 1 4 x
1 4 dx
1
e
1 x 4 14 x
14 x
³ 4e e
dy
5y
2
dx
x
P x
u x
e³
5 x 2 dx
e5 x
1 1 5x
e dx
e5 x ³ x 2
1
1
x 2
6
8
Exact
1200
3241
7414
12,915
17,117
dy
1
y
dx
x 2
Euler
1200
2743
5853
10,869
16,170
P x
1
,Q x
x 2
u x
e³
1, Q( x)
u x
e³
dx
y
10
1 § 1 5x
·
¨ e C ¸
e5 x © 5
¹
x 2 yc y
4
P x
dx
x2
5
,Q x
x2
2
10
1 x
e Ce 4 x
3
1
x2
0
49. yc y
y
4 yc
t
Euler's method gives y 8 | 16,170 trout.
§1
·
e 4 x ¨ e3 x C ¸
©3
¹
1
e x e 4 x dx
e4 x ³
1
,Q x
4
y
1200
1.
e4 x
§1
·
e 1 4 x¨ x C¸
4
©
¹
1 x4
x4
xe Ce
4
20,400
Ÿ t | 4.94 yr
1 16e 0.553t
y ·
§
0.553 y ¨1 ¸, y 0
20,400 ¹
©
4 dx
e x
P x
y
52.
Use Euler's method with h
e³
2000
(b) y 8 | 17,118 trout
(c) 10,000
u x
yc 20,400
1 16e 0.553t
y
4, Q( x)
51.
20,400
Ÿ b 16
1b
20,400
1 16e k
46
5
23
40
ln
ln
| 0.553
40
23
2000
e x
P x
y
8
§ 5 · 1.76t
1 ¨ ¸e
© 3¹
20,400, y 0
1
yc 4 y
8
Solution: y
47. (a) L
50. e x yc 4e x y
0, 3
1 x 2 dx
1
Ce5 x
5
x 2
eln x 2
1
§ 1 ·
¨
¸ x 2 dx
³
x 2 © x 2¹
x 2
1
xC
x 2
e x
1
10e x dx
e x ³
e x 10e x C
10 Ce x
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffor Chapter 6
54.
x 3 yc 2 y
dy
2
y
dx
x 3
2x 3
2 x 3
P x
2
,Q x
x 3
u x
e³
y
55. yc 5 y
2
3
P x
5, Q x
e
5x
u x
5 dx
e³
5x
2 x 3
2 x 3 dx
1
e5 x , y 0
e 2 ln x 3
1
e5 x e5 x dx
e5 x ³
1
e10 x dx
e5 x ³
1
§ 1 10 x
·
¨ e C¸
e5 x ³ © 10
¹
1 5x
e Ce 5 x
10
y
x 3
2
³ 2 x 3 x 3 dx
2
x 3
2
ª x 34
º
«
C»
2
2
»¼
x 3 «¬
1
x 3
2
2
Initial condition:
1 0
29
e Ce0 Ÿ C
10
10
1 5x
29 5 x
Particular solution: y
e e
10
10
C
x 3
e
613
y0
2
3 :3
§ 3·
56. yc ¨ ¸ y
© x¹
2 x3 , y 1
1
P x
3
,Q x
x
2 x3
u x
e³
3 dx
x
e 3 ln x
x 3
1
2 x3 x 3 dx
x 3 ³
y
x3 ³ 2 dx
x3 2 x C
2 x 4 Cx3
2C Ÿ C
Initial condition: y 1
1 :1
Particular solution: y
2x x
4
1
3
Problem Solving for Chapter 6
1. (a)
³y
1.01
y 0.01
y
So, y
³ dt
y H
H
kt C1
t C1
y H
H kt C
dy
y 0.01
0.01
1
y 0.01
y0
³ k dt
y1.01
(b)
³y
1 H
dy
dy
dt
y
0.01t C
1
C 0.01t
1
y0
C 0.01t
100
1
Ÿ C
C100
1
.
100
1 0.01t
1: 1
y0
So, y
1
For t o
1
C H kt
1H
1
Ÿ C1 H
C1 H
1
1H
§ 1
·
¨ H H kt ¸
y
© 0
¹
1
Ÿ C
y0
H
§1·
¨ ¸
© y0 ¹
.
1
, y o f.
y0H H k
f.
INSTRUCTOR USE ONLY
For T
100, lim y
t oT © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
614
Chapter 6
2. (a)
S
dS
dt
NOT FOR SALE
Differential
ferential Equation
Equations
k1S L S
L
is a solution because
1 Ce kt
2
dS
L 1 Ce kt
Cke kt
dt
LC ke kt
1 Ce
kt
ln 1 9
k dt
ln >ln L ln y@
kt C1
ln
(b)
y
Le Ce
kt
kt
(c) As t o f, y o L, the carrying capacity.
(d) y0
5000e C Ÿ eC
500
10 Ÿ C
ln 10
7000
0
500
0
dy
dt
§L·
k ln ¨ ¸ y
© y¹
d2y
dt 2
§ L · dy
1 § L · dy
k ln ¨ ¸
ky
¨ ¸
L y © y 2 ¹ dt
© y ¹ dt
k
d2y
dt 2
L
e
º
dy ª § L ·
«ln ¨ ¸ 1»
dt ¬ © y ¹
¼
º
§ L· ª § L·
k 2 ln ¨ ¸ y «ln ¨ ¸ 1»
y
y
© ¹ ¬ © ¹
¼
0 when
1Ÿ
L
y
e Ÿ y
L
.
e
5000
| 1839.4 and t | 41.7.
e
The graph is concave upward on 0, 41.7 and
10
0
(d)
eCe
500
y
0
L
y
0
§ L·
ln ¨ ¸
© y¹
125
Ce kt
0
So,
t | 2.7 months
(This is the point of inflection.)
L
y
2000
t
ln 4 9
(c)
dy
y>ln L ln y@
2
L
C Le kt
§k·
˜
¨ ¸
kt
1 Ce kt
© L ¹1 Ce
L
L
§k·
§
·
˜ ¨L ¨ ¸
¸
kt
1 Ce kt ¹
© L ¹1 Ce
©
k
.
k1S L S , where k1
L
L 100. Also, S
10 when t
0 Ÿ C
9.
4
And, S
20 when t 1 Ÿ k
ln .
9
100
100
Particular Solution: S
ln 4 9 t
1
9
e 0.8109t
1 9e
dS
(b)
k1S 100 S
dt
ª § dS ·
d 2S
dS º
k1 «S ¨ ¸ 100 S
dt 2
dt
dt »¼
¹
¬ ©
dS
k1 100 2 S
dt
dS
0 when S
50 or
0.
dt
Choosing S
50, you have:
100
50
1 9eln 4 9 t
2 1 9eln 4 9 t
§ L·
k ln ¨ ¸ y
© y¹
dy
dt
3. (a)
downward on 41.7, f .
S
140
120
100
80
60
40
20
t
1
2
3
4
(e) Sales will decrease toward the line S
L.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffor Chapter 6
4. ª¬ f x g x º¼c
?
2
f c x gc x
5. k
x, g c x
(a) Let g x
1, then
ª¬ f x xº¼c
fc x
fc x x f x
df
x 1
dx
df
³ f
x2 y 6
2
fc x
x
2
f x
Area of cross section: A h
dx
1
1 x
fg c
f cg c
f cg fg c
f cg c
f c g gc
fg c
fc
f
³ 18h
216h1 2 dh
³ dt
k
2 gh
1
S 64h
144
1
h1 2
18
t C
t C
63 2
504 | 1481.45.
5
The tank is completely drained when
gc
h
e , then g c x g x
e e
x
(c) If g x
12 y y 2
12h h 2 S
32
2
36 y 6
dh
dt
dh
2
12h h S
dt
2 dh
12h h
dt
When h
dx
³
e gc g
f
Equation of tank
36 5 2
h 144h3 2
5
h3 2
36h 720
5
gc
gc g
gc
³ g c g dx
ln f
36
Ah
ln 1 x
f x
(b)
g
§1·
¨ ¸ S
© 12 ¹
32
³1 x
ln f x
615
x
6, t
0 Ÿ t
0 and C
1481.45 sec | 24 min, 41 sec
y
x
0
6 ft
Therefore, no f can exist.
h
x
x 2 + (y − 6)2 = 36
dh
dt
2 dh
Sr
dt
k
2 gh
k
64h
6. (a) A h
h 1 2 dh
8k
Sr2
dt
2 h
Ct C1
2 18
C1
30 60
2 12
3600 sec Ÿ 2 h
0.000865 3600 6 2
Ÿ h | 7.21 ft
C dt ,
C
0, h
18
8k
r
Sr2
18 ft
h
1800, h
12:
1800 C 6 2
6 2 4 3
1800
So, 2 h
h
t
Ct 6 2.
So, 2 h
At t
at t
(b)
0 Ÿ t
C | 0.000865
0.000865t 6 2.
6 2
0.000865
| 9809.1 seconds 2 h, 43 min, 29 sec
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
616
7.
NOT FOR SALE
Chapter 6
dh
dt
dh
S 64
dt
k
Ah
³h
1 2
2 gh
dh
0 Ÿ t
e 2u
k t ln b k
3.5 0.019s
(a)
³ 3.5 0.019s
ds
eln b e kt
§1 §
1 ª
ln b · ·º
L «1 tanh ¨ k ¨ t ¸ ¸»
2 ¬
2
k ¹ ¹¼
©
©
4 5
2
1 e 2u
be kt
L
>1 tanh u@
2
2
L§
·
¨
¸
2 © 1 be kt ¹
L
.
1 be kt
4 5 288
ds
dt
Notice the graph of the logistics function is just a shift of
the graph of the hyperbolic tangent. (See Section 5.8.)
³ dt
1
ln 3.5 0.019 s
0.019
ln 3.5 0.019s
t C1
3.5 0.019 s
C3e 0.019t
0.019 s
s
(b)
eu e u
eu e u
Finally,
| 2575.95 sec | 42 min, 56 sec
9.
e
1
t
4 5
288
2 h
h
1 tanh u
C
20: 2 20
1 §
ln b ·
k¨t ¸.
2 ©
k ¹
8. Let u
S
8 h
36
1
³ 288 dt
t
C
288
2 h
h
Differential
ferential Equation
Equations
0.019t C2
3.5 C3e 0.019t
184.21 Ce 0.019t
400
0
200
0
(c) As t o f, Ce
10. (a)
dC
³C
ln C
C
0.019 t
o 0, and s o 184.21.
R
³ V dt
R
t K1
V
Ke Rt V
C0 when t
0, it follows that K
Since C
(b) Finally, as t o f, we have
lim C
t of
lim C0e Rt V
t of
C0 and the function is C
C0e Rt V .
0.
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Problem Solving ffor Chapter 6
C0e Rt V .
11. From Exercises 10, you have C
(a) For V
2, R
617
0.5, and C0
0.6, you have C
0.6e 0.25t
0.6, you have C
0.6e 0.75t .
0.8
0
4
0
(b) For V
2, R
1.5, and C0
0.8
0
4
0
12. (a)
³ Q RC dC
1
³ V dt
1
ln Q RC
R
t
K1
V
Q RC
C
Because C
1
e
R ª¬ t V K1º¼
1
R ª t V K1º¼
Q e ¬
R
0 when t
1
Q Ke Rt V
R
0, it follows that K
Q and you have C
Q
1 e Rt V .
R
(b) As t o f, the limit of C is Q R.
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
C H A P T E R 7
Applications of Integration
Section 7.1
Area of a Region Between Two Curves............................................619
Section 7.2
Volume: The Disk Method ................................................................634
Section 7.3
Volume: The Shell Method................................................................650
Section 7.4
Arc Length and Surfaces of Revolution ............................................662
Section 7.5
Work....................................................................................................675
Section 7.6
Moments, Centers of Mass, and Centroids .......................................681
Section 7.7
Fluid Pressure and Fluid Force ..........................................................694
Review Exercises ........................................................................................................699
Problem Solving .........................................................................................................707
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
C H A P T E R 7
Applications of Integration
Section 7.1 Area of a Region Between Two Curves
6
6
1. A
³ 0 ¬ª0 x 6 x ¼º dx
2. A
³ 2 ª¬ 2 x 5 x 2 x 1 º¼ dx
³
2
2
0
x 2 6 x dx
3 ª§ x 3
· xº
9. ³ «¨
x ¸ » dx
2
¹ 3 ¼»
¬«© 3
y
2
2
7
6
³ 2 x 4 dx
2
5
4
3
3. A
3
2
2
³ 0 ª¬ x 2 x 3 x 4 x 3 º¼ dx
3
³0
x
10. ³
4. A
³ 0 x x dx
5. A
2 ³ 3 x3 x dx
2
1
2
−1
2 x 2 6 x dx
1
2
3
S 4
S 4
4
5
6
7
sec 2 x cos x dx
y
0
1
or 6 ³
6. A
1
0
6³
0
1
x3 x dx
3
x3 x dx
1
2³ ª x 1
0¬
2
x 1 º dx
¼
3
4ª
xº
7. ³ « x 1 » dx
0
2
¬
¼
−π
4
11. ³
y
x
π
4
1
ª 2 y y 2 º¼ dy
2 ¬
y
5
4
3
3
2
2
x
−1
1
1
−1
2
3
4
5
x
1
2
3
4
5
−3
1
8. ³ ª¬ 2 x 2 x 2 º¼ dx
1
y
12. ³
4
0
y y dy
2
y
3
4
1
−2
3
x
−1
1
−1
2
2
1
x
1
2
3
4
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
619
620
NOT FOR SALE
Chapter 7
Applications
lications of Integration
x 1
13. f x
x 1
g x
y
15. (a)
(3, 4)
3
2
4 y
2
A 4
(0, 1)
x
y 2
2
y 2
y y 6
0
y 3 y 2
0
Intersection points: 0, 2 and 5, 3
x
3
2
1
2
0
4
³ 5 ª¬ x 2 A
2 12 x
g x
4 y2
2
Matches (d)
14. f x
x
61 32
6
3
x
4 x dx ³ 2 4 x dx
0
125
6
y
A |1
6
Matches (a)
4
(0, 2)
y
x
4
− 6 −4
6
(−5, −3)
3
−6
(0, 2)
1
(b) A
(4, 0)
x
2
1
y 2 º¼ dy
125
6
6 x
6 x Ÿ x2 x 6
2
2
(c) The second method is simpler. Explanations will
vary.
x 2 and y
16. (a) y
x
3
2
³ 3 ª¬ 4 y
0 Ÿ x 3 x 2
0
Intersection points: 2, 4 and 3, 9
y
10
(−3, 9)
8
6
(2, 4)
4
x
−6 −4 −2
A
(b) A
2
−2
4
6
2
³ 3 ª¬ 6 x x º¼ dx
2
4
125
6
9
y dy ³ ª¬ 6 y 4
³0 2
y dy
32 61
3
6
125
6
(c) The first method is simpler. Explanations will vary.
y
17.
A
6
1
³ 0 ª¬ x 2 x 1 º¼ dx
1
³ 0 x x 3 dx
4
x
−2
2
−2
2
1
2
−4
2
4
ª x3
º
x2
3 x»
«
3
2
¬
¼0
§ 1 1
·
¨ 3¸ 0
© 3 2
¹
13
6
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 7.1
y
18.
Area of a Region Between Two Curves
20. The points of intersection are given by:
x 2 3x 1
4
(−1, 3)
x 1
x 4x
0
x4 x
0 when x
2
(1, 1)
x
−4
−2
4
−2
0, 4
y
(1, −2)
(− 1, −4)
4
1
³ 1 ª¬ x 2 x 3 º¼ dx
A
621
3
1
(0, 1)
x
−2
4
³ 1 x x 5 dx
6
3
(4, − 3)
1
ª x4
º
x2
5 x»
«
2
¬ 4
¼ 1
4
³ 0 ª¬ x 3x 1 1 x º¼ dx
A
1
1
§ 1
· § 1
·
¨ 5 ¸ ¨ 5¸
2
2
© 4
¹ © 4
¹
2
4
³ 0 x 4 x dx
10
2
4
ª x3
º
2x2 »
«
¬ 3
¼0
19. The points of intersection are given by:
x2 2 x
x 2
x x 2
2
x 2 x 1
0
0 when x
2, 1
64
32
3
32
3
21. The points of intersection are given by:
y
x
6
x
2 x and
x
0 and 2 x
0
1
x
0
x
2
4
y
(1, 3)
2
(−2, 0)
3
x
−4
2
4
2
−2
A
(1, 1)
1
1
³ 2 ª¬g x f x º¼ dx
(2, 0)
(0, 0) 1
1
2
³ 2 ª¬ x 2 x 2 x º¼ dx
9
2
x
3
1
³ 0 ¬ª 2 y y º¼ dy
A
1
ª x3
º
x2
2 x»
«
3
2
¬
¼ 2
§ 1 1
· §8
·
¨ 2¸ ¨ 2 4¸
3
2
3
©
¹ ©
¹
2
1
ª¬2 y y 2 º¼
0
1
Note that if you integrate with respect to x, you need two
integrals. Also, note that the region is a triangle.
22.
y
(1, 4)
4
3
2
(4, 161 (
1
x
1
A
2
4
3
4
³ 1 x3 dx
4
4
³1 4x
3
dx
4
ª¬ 2 x 2 º¼
1
4
ª 2º
« x2 »
¬ ¼1
2
2
16
15
8
INSTRUCTOR USE ONLY
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622
NOT FOR SALE
Chapter 7
Applications
lications of Integration
23. The points of intersection are given by:
1
x 3
2
1
x
2
x2
when x
4
x 3
x
x
25. The points of intersection are given by:
y 2
y2
y 2 y 1
0 when y
1, 2
y
3
0, 4
(4, 2)
2
1
y
x
6
1
(4, 5)
2
3
4
5
−1
5
(1, −1)
4
−3
(0, 3)
2
1
1
2
3
4
2
³ 1 ª¬g y f y º¼ dy
A
x
−2 −1
−1
5
2
4ª
³ 0 «¬
A
³ 1 ª¬ y 2 y º¼ dy
§1
·º
x 3 ¨ x 3¸» dx
©2
¹¼
2 4
ª2 3 2
x º
« x »
4 ¼0
¬3
16
4
3
4
3
2
ª
y2
y3 º
«2 y »
2
3 ¼ 1
¬
x 1
x 1
x 1
x 1
x3 3 x 2 2 x
0
x x 2 3x 2
0
x x 2 x 1
0
3
2 y y2
y
y y 3
0 when y
x3 3x 2 3x 1
0, 3
y
(−3, 3)
when x
9
2
26. The points of intersection are given by:
24. The points of intersection are given by:
3
2
3
1
0, 1, 2
(0, 0)
y
−3
x
−2
1
−1
1
(2, 1)
(1, 0)
3
³ 0 ª¬ f y g y º¼ dy
A
x
2
3
³ 0 ª¬ 2 y y
(0, −1)
3
A
1
2³ ª¬ x 1 0
3
³ 0 3y y
x 1º¼ dx
2
dy
3
ª 3 y 2 1 y3 º
3 ¼0
¬2
1
ª x2
3
4 3º
2«
x x 1 »
2
4
¬
¼0
ª§ 1
· § 3 ·º
2 «¨ 1 0 ¸ ¨ ¸»
¹ © 4 ¹¼
© 2
y º¼ dy
2
9
2
y
27.
3
1
2
(0, 2)
(5, 2)
1
x
2
−2
A
3
4
(0, − 1)
5
6
(2, −1)
2
³ 1 ¬ª f y g y º¼ dy
2
³ 1 ª¬ y 1 0º¼ dy
2
2
ª y3
º
y»
«
3
¬
¼ 1
6
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 7.1
28.
Area of a Region Between Two Curves
10
Ÿ x
x
y
29. y
4
10
y
10 10
y
( 37 , 3 )
3
³ 2 y dy
A
12
2
(0, 10)
1
8
(1, 10)
>10 ln y@102
10 ln 10 ln 2
6
x
−1
A
1
4
3
2
(0, 2)
10 ln 5 | 16.0944
(5, 2)
3
³ 0 ª¬ f y g y º¼ dy
3ª
¼
1
³ 16 y 2
2 0
3
1 2
3
ª 16 y 2 º
¬
¼0
2 y dy
4
2
4
6
8
30. The point of intersection is given by:
³ 0 «« 16 y 2 0»» dy
¬
x
−4 −2
º
y
7 | 1.354
4
2 x
4
4
4
2 x
0
when x
1
y
A
(1, 4)
1§
4
·
³ 0 ¨© 4 2 x ¸¹ dx
1
ª¬4 x 4 ln 2 x º¼ 0
3
4 4 ln 2
(0, 2)
| 1.227
1
x
−1
31. (a)
623
3
1
11
(3, 9)
−6
(0, 0)
(1, 1)
12
−1
(b) The points of intersection are given by:
x3 3 x 2 3 x
x2
x x 1 x 3
0
A
when x
1
0, 1, 3
3
³ 0 ª¬ f x g x º¼ dx ³ 1 ª¬g x f x º¼ dx
1
3
³ 0 ª¬ x 3x 3x x º¼ dx ³ 1 ª¬ x x 3x 3x º¼ dx
3
2
2
2
3
2
1
1
³0
x3 4 x 2 3 x dx ³
3
1
x3 4 x 2 3 x dx
3
ª x4 4 3 3 2 º
ª x4
4
3 º
x3 x 2 »
« x x » «
4
3
2
4
3
2 ¼1
¬
¼0 ¬
5
8
12 3
37
12
(c) Numerical approximation: 0.417 2.667 | 3.083
32. (a)
(b) The points of intersection are given by:
10
(−2, 8)
(2, 8)
−4
x4 2x2
2x2
x2 x2 4
0
when x
0, r 2
4
−2
2
(0, 0)
A
2
2 ³ ª¬2 x 2 x 4 2 x 2 º¼ dx
0
2³
2
0
4 x 2 x 4 dx
ª 4 x3
x5 º
»
2«
5 ¼0
¬ 3
128
15
(c) Numerical approximation: 8.533
INSTRUCTOR USE ONLY
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624
NOT FOR SALE
Chapter 7
Applications
lications of Integration
x4 4 x2 , g x
33. (a) f x
x2 4
2
−4
(−2, 0)
4
(2, 0)
(−1, −3)
(1, − 3)
−5
(b) The points of intersection are given by:
x4 4x2
x4 5x2 4
x 4 x 1
2
2
x2 4
0
r 2, r1
0 when x
By symmetry:
A
1
2
2 ³ ª¬ x 4 4 x 2 x 2 4 º¼ dx 2 ³ ª¬ x 2 4 x 4 4 x 2 º¼ dx
0
1
2³
1
0
x 4 5 x 2 4 dx 2 ³
2
1
x 4 5 x 2 4 dx
1
2
ª x5
º
ª x5
º
5 x3
5 x3
2« 4 x» 2 «
4 x»
3
3
¬5
¼0
¬ 5
¼1
ª§ 32
40
ª1 5
º
· § 1 5
·º
8 ¸ ¨ 4 ¸»
2 « 4» 2 «¨ 5
3
5
3
5
3
¬
¼
¹ ©
¹¼
©
8
(c) Numerical approximation:
5.067 2.933
34. (a)
(−3, 0)
8.0
15
(0, 0)
−5
(3, 0)
5
(1, − 8)
− 25
(b) The points of intersection are given by:
x4 9x2
x3 9 x
x 4 x3 9 x 2 9 x
0
x x 3 x 1 x 3
0
A
0
when x
3, 0, 1, 3
1
³ 3 ¬ª x 9 x x 9 x ¼º dx ³ 0 ¬ª x 9 x
3
4
2
0
4
2
3
x3 9 x ¼º dx ³ ¬ª x3 9 x x 4 9 x 2 ¼º dx
1
1
3
ª x4
º
ª x5
ª x4
º
9 x2
x5
x4
9 x2 º
9 x2
x5
3x3 » « 3x3 3x3 »
«
» «
2
5
4
2 ¼0 ¬ 4
2
5
¬4
¼ 3 ¬ 5
¼1
1053 29
68
20
20
5
677
10
(c) Numerical approximation: 67.7
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 7.1
35. (a)
3
(
− 1, 1
2
Area of a Region Between Two Curves
S 6
³ S 2 cos 2 x sin x dx
38. A
(( (
1, 1
2
S 6
−3
ª1
º
« 2 sin 2 x cos x»
¬
¼ S 2
3
§ 3
3·
¨¨
¸ 0
2 ¸¹
© 4
−1
(b) The points of intersection are given by:
1
1 x2
x2
2
x x 2
0
4
2
x 2 x 1
2
2
( π6 , 12 )
0 when x
r1
−π
2
39. A
3 1
ª
x º
2 «arctan x »
6 ¼0
¬
S
2
−1
(− π2 , −1)
1ª
x2 º
1
2³ «
» dx
2
0 1 x
2¼
¬
x
π
6
1
0
1·
§S
2¨ ¸
6¹
©4
3 3
| 1.299
4
y
2³ ª¬ f x g x º¼ dx
A
625
S 3
2³
0
2³
0
S 3
ª¬ f x g x º¼ dx
2 sin x tan x dx
S 3
1
| 1.237
3
2 ª¬2 cos x ln cos x º¼ 0
2 1 ln 2 | 0.614
y
(c) Numerical approximation: 1.237
g
4
36. (a)
3
( π3 , 3 (
3
( 3, 95 )
2
f
1
−1
5
(0, 0)
−
π
2
(0, 0)
−1
6x
º
³ 0 «¬ x 2 1 0»¼ dx
(b) A
(− π3 , − 3 (
−3
3ª
x
π
2
−4
3
ª3 ln x 2 1 º
¬
¼0
1ª
³ 0 «¬
40. A
3 ln 10
| 6.908
(c) Numerical approximation: 6.908
2S
³ 0 ª¬ 2 cos x cos xº¼ dx
37. A
2³
2S
1 cos x dx
0
2S
2> x sin x@0
2 4 x 4 sec
Sx
4
tan
4 »¼
dx
1
ª
«
¬«
S xº
2 4 2
4
x 4 x sec
»
S
2
4 ¼» 0
§
¨¨
©
2 4
4
4
S
2
· § 4·
2 ¸¸ ¨ ¸
¹ © S¹
2
4
2 1
S
2
2 | 2.1797
4S | 12.566
S xº
y
y
4
3
3
(0, 1)
2
g
(2π, 1)
f
−1
π
2
π
2
(1,
2)
1
2π
x
x
1
INSTRUCTOR USE ONLY
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626
NOT FOR SALE
Chapter 7
³ 0 ª¬«xe
1
41. A
Applications
lications of Integration
x2
45. (a)
0º» dx
¼
1
ª 1 x2 º
« 2 e »
¬
¼0
4
(1, e)
1§
1·
¨1 ¸ | 0.316
2©
e¹
(3, 0.155)
0
6
y
0
(b) A
1
3
1
³1 x2 e
1x
dx
3
ª¬e 1 x º¼
1
)1, 1e )
(0, 0)
e e1 3
x
1
(c) Numerical approximation: 1.323
46. (a)
2
(5, 1.29)
2 ª§ 3
·
xº
³ 0 «¬¨© 2 x 1¸¹ 2 »¼ dx
42. A
0
6
(1, 0)
2
ª3x 2
2x º
x «
»
ln 2 ¼ 0
¬ 4
−2
§
4 ·
1
¨3 2 ¸ ln
2
ln
2
©
¹
3
| 0.672
5
ln 2
(b) A
y
43. (a)
dx
ª2 ln x 2 º
¬
¼1
2
From the graph, f and g
intersect at x
0 and
x
2.
x
5
(2, 4)
4
5 4 ln x
³1
2 ln 5
(0, 1)
x
2
(c) Numerical approximation: 5.181
2
47. (a)
6
3
−1
4
−1
0
(b) The integral
0
(b) A
S
³ 0 2 sin x sin 2 x dx
A
S
ª2 cos x 1 cos 2 xº
2
¬
¼0
2 12
2 12
x3
dx
4 x
3
³0
does not have an elementary antiderivative.
(c) A | 4.7721
4
48. (a)
(c) Numerical approximation: 4.0
4
(1, e)
44. (a)
2
(0, 1)
(π , 1)
−1
5
4
−
4
−1
(b) The integral
−2
(b) A
2
S
³0
A
2 sin x cos 2 x dx
S
ª2 cos x 1 sin 2 xº
2
¬
¼0
1
³0
xe x dx
does not have an elementary antiderivative.
4
(c) 1.2556
(c) Numerical approximation: 4
INSTRUCTOR USE ONLY
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NOT FOR SALE
Section 7.1
49. (a)
Area of a Region Between Two Curves
627
5
−3
3
−1
(b) The intersection points are difficult to determine by hand.
c
³ c ª¬4 cos x x º¼ dx | 6.3043 where c | 1.201538.
(c) Area
2
4
50. (a)
−4
3
−1
(b) The intersection points are difficult to determine.
(c) Intersection points: 1.164035, 1.3549778 and 1.4526269, 2.1101248
1.4526269
³ 1.164035 ª¬ 3 x x º¼ dx | 3.0578
A
2
x
ªt 2
º
« t»
¬4
¼0
x§1
·
³ 0 ¨© 2 t 1¸¹ dt
51. F x
(a) F 0
x2
x
4
·
2
³ 0 ¨© 2 t 2 ¸¹ dt
(a) F 0
0
x
x §1
52. F x
y
ª1 3
º
« 6 t 2t »
¬
¼0
x3
2x
6
0
y
6
20
5
4
16
3
12
2
8
t
−1
−1
1
2
3
4
5
4
6
x
1
22
2
4
(b) F 2
3
2
3
5
4
6
43
24
6
(b) F 4
y
56
3
y
6
5
4
20
3
16
2
12
8
t
−1
−1
1
2
3
4
5
6
4
x
1
62
6
4
(c) F 6
15
2
5
4
6
36 12
(c) F 6
y
3
48
y
6
5
20
4
16
3
12
2
8
−1
−1
t
1
2
3
4
5
4
6
x
1
2
3
4
5
6
INSTRUCTOR USE ONLY
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628
NOT FOR SALE
Chapter 7
Applications
lications of Integration
D
ST
D
ST º
ª2
«S sin 2 »
¬
¼ 1
³ 1 cos 2 dT
53. F D
(a) F 1
2
S
sin
SD
2
2
y
³ 1 4e
54. F y
S
(a) F 1
0
x 2
y
ª¬8e x 2 º¼
1
dx
8e y 2 8e 1 2
0
y
y
30
3
2
25
20
15
1
2
10
−1
2
1
2
−1
5
θ
1
x
−1
1
2
3
4
2
2
(b) F 0
S
8 8e 1 2 | 3.1478
(b) F 0
| 0.6366
y
y
30
25
3
2
20
15
10
1
2
−1
2
5
1
2
−1
θ
1
x
−1
1
2
2
2
| 1.0868
S
3
4
8e 2 8e 1 2 | 54.2602
(c) F 4
§1·
(c) F ¨ ¸
© 2¹
2
y
30
y
25
20
3
2
15
10
5
1
2
x
−1
1
6
77
2
3
4
θ
−1
2
1
2
−1
2
1
4
6
³ 2 ª¬ 92 x 12 x 5 º¼ dx ³ 4 ª¬ 52 x 16 x 5 º¼ dx
55. A
4
6
³ 2 72 x 7 dx ³ 4 72 x 21 dx
y
4
ª 7 x 2 7 xº ª 7 x 2 21xº
¬4
¼2 ¬ 4
¼4
14
y = 29 x − 12
(4, 6)
6
y = − 25 x + 16
4
2
(6, 1)
x
6
2
8
10
−2
−4
56. A
(2, −3)
y=x−5
6§
3 ·
³ 0 4 x dx ³ 4 ¨© 9 2 x ¸¹ dx
y
4 3
2 4
4
2 6
ª3x º
ª
3x º
«
» «9 x »
8
4 ¼4
¬
¼0 ¬
(4, 3)
2
(6, 0)
6 54 27 36 12
x
INSTRUCTOR
T
USE ONLY
63
9
(0, 0) 2
4
6
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NOT FOR SALE
Section 7.1
57.
y
4
3
(0, 2)
Left boundary line: y
x 2 œ x
y 2
Right boundary line: y
x 2 œ x
y 2
(4, 2)
2
A
1
x
−4
−2 −1
(−4, − 2)
2
3
4
629
2
³ 2 ª¬ y 2 y 2 º¼ dy
2
³ 2 4 dy >4 y@ 2
2
(0, −2)
−3
Area of a Region Between Two Curves
8 8
16
−4
3ª
1
7 ·º
§1
³ 0 ª¬2 x 3x º¼ dx ³ 1 «¬ 2 x 4 ¨© 2 x 2 ¸¹»¼ dx
58. A
3§
1
15 ·
5
³ 0 5 x dx ³1 ¨© 2 x 2 ¸¹ dx
y
3
ª5x2 º 1
ª 5x2
15 º
x»
«
» «
2
4
2 ¼1
¬
¼0
¬
2
(1, 2)
1
(0, 0)
−2 −1
5 § 45
45
5 15 ·
¨
¸
2 © 4
2
4
2¹
x
2
−1
15
2
−4
59. Answers will vary. Sample answer: If you let 'x
4
(3, −2)
−2
−3
3
(1, −3)
6 and n
10, b a
10 6
60.
(a) Area | 260
ª0 2 14 2 14 2 12 2 12 2 15 2 20 2 23 2 25 2 26 0º¼
10 ¬
3>322@
966 ft 2
ª0 4 14 2 14 4 12 2 12 4 15 2 20 4 23 2 25 4 26 0º¼
(b) Area | 360
10 ¬
2>502@
1004 ft 2
60. Answers will vary. Sample answer: 'x
(a) Area |
32
ª0 2 11
28 ¬
2>190.8@
4, n
8, b a
8 4
32
2 13.5 2 14.2 2 14 2 14.2 2 15 2 13.5 0º¼
381.6 mi 2
(b) Area | 32 ª¬0 4 11 2 13.5 4 14.2 2 14 4 14.2 2 15 4 13.5 0º¼
38
>
@
4 296.6
3
61.
f x
x3
fc x
3x 2
At 1, 1 , f c 1
395.5 mi 2
3.
Tangent line: y 1
3 x 1 or y
The tangent line intersects f x
3x 2
x 3 at x
2.
1
A
1
3
³ 2 ª¬x 3x 2 º¼ dx
ª x4
º
3x 2
2 x»
«
2
¬4
¼ 2
27
4
y
8
6
4
y = 3x − 2
2
(1, 1)
x
−4 −3 −2
1
2
3
4
f (x) = x3
−6
INSTRUCTOR
ST
USE ONLY
(− 2,
2 − 8)
−8
−8
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NOT FOR SALE
630
Chapter 7
Applications
lications of Integration
62.
y
x 3 2 x,
yc
3x 2
1, 1
2
yc 1
3 2
1
Tangent line: y 1
1x 1 Ÿ y
y
x 2
5
4
Intersection points: 1, 1 and 2, 4
(2, 4)
y=x+2 3
2
2
2
(−1, 1)
³ 1 x 3x 2 dx
3
ª x4
º
3x 2
2 x»
«
2
¬ 4
¼ 1
63.
2
³ 1 ª¬ x 2 x 2 x º¼ dx
A
3
ª
3
§ 1
·º
« 4 6 4 ¨ 4 2 2 ¸»
©
¹¼
¬
1
x2 1
2x
2
2
x 1
f x
fc x
x
−4
2
−1
−2
27
4
−3
3
4
y = x 3 − 2x
y
(0, 1)
§ 1·
At ¨1, ¸, f c 1
© 2¹
1
.
2
Tangent line: y 1
2
1
f ( x) = 2
x +1
3
4
(1, 12 )
1
2
1
x 1 or y
2
1
x 1
2
1
4
y=− 1x+1
2
x
1
at x
x2 1
The tangent line intersects f x
A
1ª
1
§ 1
·º
³ 0 «¬ x 2 1 ¨© 2 x 1¸¹»¼ dx
y
64.
yc
§1·
yc¨ ¸
© 2¹
1
2
0.
1
1
S 3
ª
º
x2
x»
«arctan x 4
¬
¼0
4
3
2
2
| 0.0354
§1 ·
¨ , 1¸
©2 ¹
2
,
1 4 x2
16 x
1 4x2
8
22
2
y
2
(0, 2) y =
y = −2x + 2
Tangent line: y 1
y
1·
§
2¨ x ¸
2¹
©
2 x 2
1
2
1 + 4x 2
( 12 , 1(
x
−1
§1 ·
Intersection points: ¨ , 1¸, 0, 2
©2 ¹
A
12 ª
2
º
³ 0 «¬1 4 x 2 2 x 2 »¼ dx
1
12
ª¬arctan 2 x x 2 2 xº¼
0
arctan 1 1
1
4
S
4
3
| 0.0354
4
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 7.1
65. x 4 2 x 2 1 d 1 x 2 on >1, 1@
A
1
³ 1 ª¬ 1 x
1
2
³ 1 x x
2
4
2
dx
1
631
y
x 4 2 x 2 1 º¼ dx
ª x3
x5 º
« »
5 ¼ 1
¬3
Area of a Region Between Two Curves
(0, 1)
4
15
x
( 1, 0)
(1, 0)
You can use a single integral because x 4 2 x 2 1 d 1 x 2 on >1, 1@.
66. x3 t x on >1, 0@, x3 d x on >0, 1@
y
Both functions symmetric to origin.
0
³ 1 x x dx
³
3
Thus, ³
1
1
1
0
x x dx
(0, 0)
0.
x x dx
ª x2
x4 º
2« »
4 ¼0
¬2
A
67. (a)
2³
0
3
5
³ 0 ¬ªv1 t v2 t º¼ dt
³ 0 ¬ªv1 t v2 t º¼ dt
1
−1
1
2
10 means that Car 1 traveled
10 more meters than Car 2 on the interval
0 d t d 5.
10
x
−1
x x dx
3
1
1
(1, 1)
1
3
30 means that Car 1
(− 1, − 1)
30
36
bº¼ dx
18
ª¬ 9 b x 2 º¼ dx
9
9b
³0
9b
2
9b
ª
x3 º
«9 b x »
3 ¼0
¬
5 means that Car 2
9
2
32
9b
3
traveled 5 more meters than Car 1 on the interval
20 d t d 30.
9b
(b) No, it is not possible because you do not know the
initial distance between the cars.
(c) At t
2
dx
³ 9 b ª¬ 9 x
traveled 30 more meters than Car 2 on the interval
0 d t d 10.
³ 20 ª¬v1 t v2 t º¼ dt
3
³ 3 9 x
A
69.
9 b
10, Car 1 is ahead by 30 meters.
9 3
b
(d) At t
20, Car 1 is ahead of Car 2 by 13 meters.
From part (a), at t
30, Car 1 is ahead by
13 5 8 meters.
27
2
9
3
4
9
| 3.330
4
y
10
68. (a) The area between the two curves represents the
difference between the accumulated deficit under
the two plans.
(b) Proposal 2 is better because the cumulative deficit
(the area under the curve) is less.
32
9
6
4
2
x
−6
−2
2
6
(−
9 − b, b)
(
9 − b, b)
INSTRUCTOR USE ONLY
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632
NOT FOR SALE
Chapter 7
Applications
lications of Integration
9
2³
70. A
9
0
ª
x2 º
2 «9 x »
2 ¼0
¬
9 x dx
9b
2³
0
2³
0
9b
ª¬ 9 x bº¼ dx
ª¬ 9 b xº¼ dx
9b
9b 9b
81
2
9
2
9
2³
4
4a
42 3
4a
4 y 2 dy
8º
ª
2 «8 »
3¼
¬
4
4 x dx
a
32
3
4
3 2º
4x »
3
¼a
4
32
4a
3
32
y
3
4 42 3 | 1.48
1
x
−1
−1
1
2
3
4
5
a
9
| 2.636
2
−3
y
n
73.
lim ¦ xi xi2 'x
' o0
9
−6
2
0
4
16
3
a
12
6
(− (9 − b), b)
2³
dy
2
81
2
b
2
ª
y3 º
2 «4 y »
3 ¼0
¬
81
2
81
2
ª
x2 º
2« 9 b x »
2 ¼0
¬
9b
2
³2 4 y
72. Total area
81
i 1
i
and 'x
n
where xi
(9 − b, b)
x
−3
3
1
is the same as
n
1
6
1
³0
−3
−6
x x 2 dx
ª x2
x3 º
»
«
3 ¼0
¬2
1
.
6
y
71. Area of triangle OAB is
a
4
³0
a 2 8a 8
0
1
4 4
2
4 x dx
0.6
8.
0.2
2 a
ª
x º
«4 x »
2 ¼0
¬
(1, 0)
a2
4a 2
0.4
0.2
0.6
n
Because 0 a 4, select a
¦ 4 xi2 'x
' o0
lim
i 1
4 2 2 | 1.172.
2 where xi
y
B
4i
and 'x
n
4
is the same as
n
2
A
2
³2
3
4 x 2 dx
ª
x3 º
«4 x »
3 ¼ 2
¬
32
.
3
y
2
1
x
0.8 1.0
(0, 0)
74.
4r2 2
a
f(x) = x − x 2
0.4
5
a
O
f ( x) = 4 − x 2
x
1
2
3
3
4
2
1
(− 2, 0)
−3
(2, 0)
x
−1
1
3
−1
75. R1 projects the greater revenue because the area under
the curve is greater.
20
³15 ¬ª 7.21 0.58 t 7.21 0.45 t º¼ dt
20
20
³15
0.13 t dt
ª 0.13 t 2 º
«
»
¬ 2 ¼155
$11.375 billion
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section 7.1
Area of a Region Between Two Curves
633
76. R2 projects the greater revenue because the area under the curve is greater.
20
³15 ª¬ 7.21 0.26 t 0.02 t
20
7.21 0.1t 0.01t 2 º¼ dt
2
³15 0.01t 0.16 t dt
2
20
ª 0.01t 3
0.16 t 2 º
«
» | $29.417 billion
2 ¼15
¬ 3
0.0124 x 2 0.385 x 7.85
77. (a) y1
y
y
(c)
Percents of total income
Percents of total income
(b)
100
80
60
40
20
x
20
40
60
(d) Income inequality
100
60
40
20
x
20
80 100
15.9e0.05t
3.5%: P2
100
80
40
60
80 100
Percents of families
Percents of families
78. 5%: P1
³ 0 > x y1@ dx | 2006.7
(in millions)
15.9e0.035t
(in millions)
Difference in profits over 5 years:
5
5
³0
5
³0
P1 P2 dt
15.9 e0.05t e0.035t dt
ª e0.05t
e0.035t º
15.9 «
» | $3.44 million
0.035 ¼ 0
¬ 0.05
5.5
ª 5§
º
1
·
2 «³ ¨1 5 x ¸ dx ³ 1 0 dx»
0
5
3
¹
¬ ©
¼
79. (a) A
5
§ª
2
5.5 ·
3 2º
2¨ « x 5 x » > x@5 ¸
¨¬
¸
9
¼0
©
¹
§
·
10 5
2¨¨ 5 5.5 5 ¸¸ | 6.031 m 2
9
©
¹
2 A | 2 6.031 | 12.062 m3
(b) V
(c) 5000 V | 5000 12.062
60,310 pounds
80. The curves intersect at the point where the slope of y2 equals that of y1 , 1.
y2
0.08 x 2 k Ÿ y2c
0.16 x
1Ÿ x
1
0.16
6.25
(a) The value of k is given by
y1
6.25
k
(b) Area
y2
0.08 6.25
2
k
3.125.
6.25
2³
0
2³
0
6.25
y2 y1 dx
0.08 x 2 3.125 x dx
6.25
ª 0.08 x3
x2 º
2«
3.125 x »
2 ¼0
¬ 3
INSTRUCTOR USE ONLY
2 6.510417 | 13.02083
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634
NOT FOR SALE
Chapter 7
Applications
lications of Integration
3
x
7S
81. Line: y
2 x x 2 , f and g
intersect at 1, 1 , the midpoint of >0, 2@, but
7S 6 ª
3x º
«sin x 7S » dx
¬
¼
³0
A
x and g x
85. False. Let f x
b
7S 6
2
³ a ª¬ f x g x º¼ dx ³ 0 ª¬x 2 x x º¼ dx
y
2
2 z 0.
3
1
ª
3x 2 º
«cos x »
14S ¼ 0
¬
86. True. The area under f x between 0 and 1 is 16 . The
1
2
(0, 0)
3
7S
1
2
24
| 2.7823
π
6
x
4π
3
( 76π , − 12 (
−1
1 1 3 , and the area between
2
curves intersect at x
1 12
y
13
13
1.
x and f on the interval ª0, 12 º is 12
«¬
»¼
87. You want to find c such that:
x2
4 ³ b 1 2 dx
0
a
4b a
a ³0
a
82. A
a
³0
a 2 x 2 dx is the area of
4b § S a 2 ·
¨
¸
a © 4 ¹
So, A
a x dx
2
b
2
1
of a circle
4
³ 0 ª¬ 2 x 3x
S a2
4
cº¼ dx
0
b
.
S ab.
3 4
ª 2
º
¬ x 4 x cx¼ 0
0
b 2 34 b 4 cb
0
2b 3b3 because b, c is on the graph.
But, c
y
2
x
1− 2
a
y=b
3
b 2 34 b 4 2b 3b3 b
0
4 3b 2 8 12b 2
0
2
4
b
2
3
c
4
9
9b
b
x
a
y
y = 2x − 3x 3
83. True. The region has been shifted C units upward (if
C ! 0 ), or C units downward (if C 0 ).
(b, c)
c
84. True. This is a property of integrals.
x
Section 7.2 Volume: The Disk Method
1
1
1. V
S ³ x 1 dx
2. V
S³
3. V
S³
4. V
S³
S³
2
0
1
0
ª x3
º
x 2 x»
¬3
¼0
S«
x 2 2 x 1 dx
S
3
2
2
0
4
1
3
0
4 x2
x
2
2
S³
dx
2
x 4 8 x 2 16 dx
0
4
4
S ³ x dx
dx
9 x2
1
2
dx
S³
3
0
S«
15S
2
9 x 2 dx
S «9 x ª x2 º
»
¬ 2 ¼1
ª
¬
ª x5
º
8 x3
16 x»
3
¬5
¼0
S«
256S
15
3
x3 º
»
3 ¼0
18S
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 7.2
S ³ ª x2
1
5. V
2
x 5 º dx
¼»
2
«
0¬
7. y
V
1
S ³ x 4 x10 dx
0
2
x
4
16 x 2
2
8
x
r2 2
x
2S ³
S³
4
0
0
16 y 2
4
ª
¬
9. y
2
S³
dy
4
0
16 y 2 dy
128S
3
y3 º
»
3 ¼0
x2 3 Ÿ x
16 y 2
y3 2
1
2 ª x4
º
2 x 12» dx
«
¬16
¼
2
4
S ³ y dy
dy
8S
S «16 y V
S³
10. V
S³
2
º
2 2 ª§
x2 ·
2
«¨ 4 S³
2 » dx
¸
2 2 «
4¹
»¼
©
V
y
2
16 x 2 Ÿ x
8. y
V
8
y
0
0
2
2
ªx
º
2x
12 x»
2S « 80
3
¬
¼0
5
635
4
6S
55
1·
§1
¸
© 5 11 ¹
4
S³
4
S«
S¨
2
x2 Ÿ x
ª y2 º
»
¬ 2 ¼0
1
ª x5
x11 º
S« »
11 ¼ 0
¬5
6.
Volume: The Disk Method
3
1
0
4
1
y3 2
2
y2 4 y
S«
0
2
S³
dy
4
ª y5
16 y 3 º
2 y4 »
3 ¼1
¬5
S«
2
ª y4 º
»
¬ 4 ¼0
1
S ³ y 3 dy
dy
4
1
S
4
y 4 8 y 3 16 y 2 dy
459S
15
153S
5
ª128 2
º
32 2
24 2 »
2S «
3
«¬ 80
»¼
448 2
S | 132.69
15
11. y
x, y
0, x
(a) R x
3
x, r x
V
S³
1
2
3
0
2
x
0
3
dx
ª x2 º
»
¬ 2 ¼0
3
S ³ x dx
S«
0
9S
2
y
2
1
x
3
−1
(b) R y
S³
V
y2
3, r y
3
0
ª32 y 2 2 º dy
«¬
»¼
S³
3
0
9 y 4 dy
ª
S «9 y ¬
y5 º
»
5 ¼0
3
ª
¬
S «9 3 9
5
º
3»
¼
36 3S
5
y
2
1
x
1
2
3
INSTRUCTOR
S
USE ONLY
−1
−1
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© Cengage Learning. All Rights Reserved.
636
NOT FOR SALE
Chapter 7
(c) R y
Applications
lications of Integration
3 y2, r y
0
3
2
S³
V
0
3 y2
y
S³
dy
3
0
9 6 y 2 y 4 dy
2
1
ª
y5 º
S «9 y 2 y 3 »
5 ¼0
¬
3
ª
9 3º
S «9 3 6 3 »
5 ¼»
¬«
24 3S
5
3 3 y2
(d) R y
S³
V
3
0
6 y2, r y
ª 6 y 2 2 32 º dy
¬«
¼»
ªy
º
4 y 3 27 y»
¬5
¼0
5
S«
3
S³
2x2 , y
(a) R y
4
y 4 12 y 2 27 dy
3
2
ª9 3
º
S«
12 3 27 3 »
«¬ 5
»¼
1
x
1
−1
2
3
4
5
6
−2
0, x
2
2, r y
(c) R x
y 2
8§
y·
S ³ ¨ 4 ¸ dy
0
2¹
©
V
3
y
3
0
2
−1
3
84 3S
5
12. y
x
1
2
S ³ ª¬64 64 32 x 2 4 x 4 º¼ dx
V
8
ª
y2 º
S «4 y »
4 ¼0
¬
16S
8 2x2
8, r x
0
S³
y
2
0
4S ³
32 x 2 4 x 4 dx
2
0
8 x 2 x 4 dx
2
1 º
ª8
4S « x3 x5 »
5 ¼0
¬3
8
6
y
896S
15
4
6
2
4
−4
x
−2
2
4
(b) R x
2x2 , r x
0
V
S ³ 4 x 4 dx
2
S«
2
−4
2
0
ª 4 x5 º
»
¬ 5 ¼0
128S
5
2
(d) R y
y
8§
8§
©
−2
8
ª
2
S «4 y x
2
¬
4
0
y
y·
¸¸ dy
2
2¹
S ³ ¨¨ 4 4
0
4
4
y·
¸ dy
2 ¸¹
©
6
2
2
S ³ ¨¨ 2 0
V
−4
y 2, r y
x
−2
4 2 32
y2 º
y »
3
4 ¼0
16S
3
y
8
6
4
2
x
−4
−2
4
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 7.2
13. y
4 x x 2 intersect at 0, 0 and 2, 4 .
x2 , y
4x x2 , r x
(a) R x
x2
¬«
0
4 2x x2 , y
14. y
3
2
«¬
0
0
ª16
º
S « x3 2 x 4 »
3
¬
¼0
4 x
3
2
2
S ³ ª 4 2 x x 2 4 x º dx
V
S ³ 16 x 8 x dx
2
4 x intersect at
4 2 x x2 , r x
(a) R x
¼»
2
637
0, 4 and 3, 1 .
2
2
S ³ ª 4 x x 2 x 4 º dx
V
Volume: The Disk Method
S³
32S
3
3
0
»¼
x 4 4 x3 5 x 2 24 x dx
3
153S
5
ª x5
º
5 x3
x4 12 x 2 »
3
¬5
¼0
S«
y
4
y
3
6
5
2
4
3
1
2
1
x
−1
1
2
3
x
−2 −1
−1
6 x2 , r x
(b) R x
6 4x x2
S ³ ª 6 x2
0 «
¬
2
V
8S ³
2
2
6 4x x
1
2
ª x4
º
5
x3 3x 2 »
8S «
3
¬4
¼0
3
2
º dx
»¼
4
5
4 2 x x 2 1, r x
(b) R x
2
S ³ ª 3 2 x x2
3
V
2
0«
¬
x3 5 x 2 6 x dx
0
2
4 x 1
2
3 x º dx
»¼
3
S ³ x 4 4 x3 3 x 2 18 x dx
0
64S
3
3
ª x5
º
x 4 x3 9 x 2 »
¬5
¼0
S«
y
108S
5
y
6
5
5
4
4
3
3
2
2
1
1
−2
−1
x
1
2
3
4
x
−2 −1
−1
1
2
3
4
5
4 x, r x
15. R x
1
3
2
2
S ³ ª 4 x 1 º dx
V
0 ¬
S³
3
0
¼
x 2 8 x 15 dx
3
ª x3
º
4 x 2 15 x»
3
¬
¼0
S«
18S
y
5
3
2
1
−1
x
INSTRUCTOR USE ONLY
1
2
3
4
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© Cengage Learning. All Rights Reserved.
638
Chapter 7
Applications
lications of Integration
x3
,r x
2
4
16. R x
18. R x
0
V
3 2
1§
x ·
S ³ ¨ 4 ¸ dx
0
2¹
©
V
2ª
x6 º
» dx
4¼
S ³ «16 4 x3 0
¬
¬
S 3
S³
0
S³
0
ª 4 2 4 sec x 2 º dx
¬
¼
S 3
8 sec x sec 2 x dx
S 3
S ¬ª8 ln sec x tan x tan x¼º 0
2
ª
S «16 x x 4 4 sec x
4, r x
x7 º
»
28 ¼ 0
S ª« 8 ln 2 3 3 8 ln 1 0 0 º»
¼
S ª8 ln 2 3 3 º | 27.66
¼
π
9
5π
9
¬
¬
128 ·
§
S ¨ 32 16 ¸
28 ¹
©
y
5
144
S
7
3
y
2
(2, 4)
1
3
2
x
π
3
2π
9
4π
9
1
x
−1
1
2
3
V
17. R x
V
3
4
1 x
4, r x
3ª
2
3 · º
§
S ³ «4 ¨ 4 » dx
¸
0
1 x ¹ ¼»
©
¬«
3ª
24
9 º
» dx
2
0 1 x
1 x ¼»
¬«
ª§
©
27 ·
§
¨ 48 ln 2 ¸S | 83.318
4¹
©
0
4
dy
25 10 y y 2 dy
y
ª
¬
124S
3
9 º
1 x »¼ 0
º
9·
¸ 9»
4¹
¼
S³
5 y
S «100 80 3
S Ǭ 24 ln 4 0
4
S³ «
ª
¬
4
S³
0
ª
y3 º
S «25 y 5 y 2 »
3 ¼0
¬
2
S «24 ln 1 x 5 y, r y
19. R y
4
−1
R y
3
2
64 º
3 »¼
1
x
1
2
3
4
5
−1
3 x, x
20. y
4
3 y
5 3 y
5, r y
2 y
2
S ³ ª52 2 y º dy
V
0 ¬
S³
y
4
2
0
¼
y 2 4 y 21 dy
2
ª y3
º
2 y 2 21 y»
¬ 3
¼0
S«
3
2
94S
3
y
1
6
−1
x
1
2
3
5
4
3
2
1
x
−1
1
2
3
4
5
6
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 7.2
5 y2 , r y
21. R y
S³
V
ª 5 y2
2
1
«
2 ¬
1
,r x
x 1
23. R x
1º dy
¼»
2
2
2S ³ ª¬ y 4 10 y 2 24º¼ dy
0
S³
2
ª y 5 10 y 3
º
24 y»
2S «
5
3
¬
¼0
1
dx
x 1
0
4
S ª¬ln x 1 º¼ 0
S ln 5
y
832S
15
ª 32 80
º
48»
2S «
3
¬5
¼
0
1 ·
¸ dx
x 1¹
©
4
639
2
4§
S³ ¨
0
V
Volume: The Disk Method
y
4
3
2
3
1
2
x
1
−1
1
2
3
4
x
1
−1
2
3
5
−2
24. R x
−3
V
22. xy
3
y
3, x
4§
©
S ³ ¨ 25 1
©
9
30 ·
¸ dy
y2
y¹
¬
º
9
30 ln y»
y
¼1
ª§
©
º
9
·
30 ln 4 ¸ 25 9 »
4
¹
¼
S Ǭ100 0
3
5 2
25. R x
1
,r x
x
V
S ³ ¨ ¸ dx
1 x
3§1·
1
x
−1
1
2
3
−2
128S
15
−3
0
y
2
2
© ¹
ª 1º
2
−3
1
3
S « »
¬ x¼
x
1
1
ª 327
º
S«
30 ln 4» | 126.17
¬ 4
¼
y
dx
4 x 2 x 4 dx
ª 32 32 º
2S «
5 »¼
¬3
4
ª
S «25 y 2
2
x 4 x2
0
0
ª4x
x º
2S «
»
3
5 ¼0
¬
0
3·
¸ dy
y¹
4§
2
3
2
S ³ ¨5 1
V
2S ³
2S ³
3
5 ,r y
y
R y
x 4 x2 , r x
2
3
−1
ª 1
º
S « 1»
3
¬
¼
2
S
3
y
5
26. R x
2
,r x
x 1
V
dx
S³ ¨
0 x 1¸
©
¹
4
3
2
1
x
1
2
3
4
5
6§
4S ³
6
0
2 ·
2
x 1
2
0
dx
4
3
6
ª 1 º
4S «
»
¬ x 1¼ 0
ª 1
º
4S « 1»
¬ 7
¼
y
2
1
x
24S
7
−1
1
2
3
4
5
6
−2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
640
NOT FOR SALE
Chapter 7
Applications
lications of Integration
e x , r x
27. R x
y
0
ex 4 , r x
28. R x
0
y
2
1
2
S ³ e x
V
0
V
dx
1
S ³ e 2 x dx
1
ª S 2 x º
« 2 e »
¬
¼0
2
1e
S
x
1
2
2
ex 4
5
dx
4
6
3
0
2
1
6
ª¬2e x 2 º¼
0
x
1
0
x x 2
0
x 2 x 1
0
2
0 ¬
«
0
2
0
x 4 x 16
4 ª§
S ³ «¨ 4 0
«¬©
8
4
2
2
3
−1
x
1
−2
2
16 is an extraneous root.)
2º
8ª
x » dx S ³ «
4
»¼
«¬
2
1 ·
x¸ 2 ¹
x
2
y
2
1 · º
§
¨ 4 x ¸ » dx
2 ¹ »¼
©
4
8 § x2
·
·
S³ ¨
5 x 16 ¸ dx S ³ ¨ 5 x 16 ¸ dx
0
4
©4
¹
© 4
¹
4
8
ª x3
º
ª x3
º
5x2
5x2
16 x» S «
16 x»
2
2
¬12
¼0
¬ 12
¼4
S«
(4, 2)
3
4 § x2
88
56
S S
3
3
(2, 5)
6
4 x 8 x 20 x 24 dx
3
277S
3
152
125
S
3
3
The curves intersect at 4, 2 . (Note x
V
3
2
0
x
3
6
10
¼»
8 3
8
º
ª
º
x 10 x 2 24 x» S « x x3 10 x 2 24 x»
3
3
¼0
¬
¼2
1
x 4
2
1 2
x 4 x 16
4
x 2 20 x 64
x
30.
2 ¬
«
4 x 8 x 20 x 24 dx S ³
ª
¬
S
¼»
3
S « x 5
y
2
3
2
2
2
2
S ³ ª 5 2 x x x 2 1 º dx S ³ ª x 1 5 2 x x 2 º dx
S³
4
S 2e3 2 | 119.92
2
The curves intersect at 1, 2 and 2, 5 .
2
3
x2 2 x 5
2 x2 2 x 4
V
2
| 1.358
x2 1
29.
6
0
S ³ e x 2 dx
1
0
S
S³
2
1
x
−2
2
4
6
8
10
−1
48S
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 7.2
1
6 y
3
6 3x Ÿ x
31. y
6 ª1
º
34. V
2
S ³ « 6 y » dy
0 3
¬
¼
S 6
2
V
0
S³
0
ª36 12 y y º¼ dy
9 ³0 ¬
6
Sª
y3 º
2
«36 y 6 y »
9¬
3 ¼0
S 4
S³
Volume: The Disk Method
641
cos 2 2 x dx
S 4 1 cos 4 x
2
dx
S 4
Sª
x 2 «¬
sin 4 x º
4 »¼ 0
S ªS º
S2
2 «¬ 4 »¼
8
Numerical approximation: 1.2337
Sª
216 º
216 216 «
9¬
3 »¼
1 2
8S
S r h, Volume of cone
3
y
1
y
x
π
4
6
−1
5
4
3
2
35. V
1
x
1
3
4
5
32. y
9 x ,y
x
9 y
6
2
S³
V
5
2, x
3
º
e2 x 2 »
2
¼1
2
2º» dy
¼
2
S
2
e2 1
Numerical approximation: 10.0359
5
y2 º
»
2 ¼0
§
©
S ¨ 25 25S
2
25 ·
¸
2¹
36. V
2
2
S ³ ª¬e x 2 e x 2 º¼ dx
1
2
S ³ ª¬e x e x 2º¼ dx
y
9
8
7
6
5
4
3
2
1
dx
2
S
5
¬
2
1
0
ª
e x 1
S ³ e 2 x 2 dx
S ³ 5 y dy
S «5 y 2
1
2
0, x
9 y
0
S³
1
S ª¬e x e x 2 xº¼
2
1
S ª¬ e 2 e2 4 e 1 e 2 º¼
(2, 5)
S e 2 e 6 e 2 e 1
Numerical approximation: 49.0218
x
1 2 3 4 5 6 7 8 9
33. V
S
S³
0
S³
0
sin x
2
S 1 cos 2 x
2
2
dx
37. V
2
2
S ³ ª«e x º» dx | 1.9686
38. V
S ³ >ln x@ dx | 3.2332
39. V
S ³ ª¬2 arctan 0.2 x º¼ dx
dx
S
Sª
1
º
x sin 2 x»
2 «¬
2
¼0
S
2
>S @
Numerical approximation: 4.9348
y
S2
2
0 ¬
3
¼
2
1
5
2
0
| 15.4115
3
2
1
INSTRUCTOR
NST
NS
N
ST
S
TR
T
USE ONLY
x
1
2
3
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
642
NOT FOR SALE
Chapter 7
40. x 2
Applications
lications of Integration
2x
x4
2x
3
2
x
21 3 | 1.2599
V
S³
x
1
S ³ 1 y dy
45. V
0
1
ª
S «y 21 3 ª
2
x 2 º» dx | 2.9922
¼
2
2x
¬«
0
¬
y2 º
»
2 ¼0
1
S³ 1 46. V
1
S³ 1 2
S «y y=
1
¬
2x
2
dy
4 32
y2 º
y »
3
2 ¼0
4
1·
§
S ¨1 ¸
3
2¹
©
x
1
S
1
ª
2
1·
¸
2¹
y y dy
0
y = x2
2
y
0
y
§
©
S ¨1 2
S
6
41. V
42. V
1
S ³ y 2 dy
S
0
3 1
y º
»
3 ¼0
1
S
S³
47. V
3
1
ª y2
y3 º
»
3 ¼0
¬2
S ³ ª12 1 y º dy
¼
1
S ³ ª¬2 y y 2 º¼ dy
1
¬
y3 º
»
3 ¼0
1·
§
S ¨1 ¸
3¹
©
2
y º» dy
¼
1
1
S ³ ª¬2
0
y 3 y y 2 º¼ dy
ª4
¬3
3 y2
y3 º
»
2
3 ¼0
1
S « y3 2 1
0
3 1·
§4
¸
2 3¹
©3
S¨
1
ª x3
x5 º
»
5 ¼0
¬3
S
§1 1·
S¨ ¸
©3 5¹
2S
15
1
6
1
S ³ x 2 x 4 dx
«
0¬
S
0
2
S
3
S ³ ª 1 x2
§ 1 1·
¸
© 2 3¹
S¨
S ³ ª¬1 2 y y 2 1 2 y yº¼ dy
S«
44. V
2
0¬
1
ª
43. V
S ³ ª« 1 y
48. V
0
S «y2 y y 2 dy
S«
2
0¬
1
0
6
49. S ³
2
2
1 x º dx
¼»
S 2
0
sin 2 x dx represents the volume of the solid
generated by revolving the region bounded by
y
sin x, y
0, x
0, x S 2 about the x-axis.
y
1
S ³ ª¬1 2 x 2 x 4 1 2 x x 2 º¼ dx
0
1
1
S ³ ª¬2 x 3 x 2 x 4 º¼ dx
0
1
ª
S « x 2 x3 ¬
§1·
S¨ ¸
©5¹
x5 º
»
5 ¼0
π
4
π
2
x
S
5
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 7.2
Volume: The Disk Method
4
53. (a) True. Answers will vary.
2
(b) False. Answers will vary.
50. S ³ y 4 dy represents the volume of the solid generated
by revolving the region bounded by
x
y2 , x
0, y
2, y
4 about the y-axis.
643
54. (a) Matches (ii) because the axis of rotation is vertical,
and this is the washer method.
y
(b) Matches (iv) because the axis of rotation is
horizontal, and this is the washer method.
4
3
(c) Matches (i) because the axis of rotation is horizontal.
2
(d) Matches (iii) because the axis of rotation is vertical.
1
x
4
8
12
16
4
2
x
0
4
0
c
ªS x 2 º
«
»
¬ 2 ¼0
c
S ³ x dx
0
3
c2
2
1
−2
ªS x 2 º
«
»
¬ 2 ¼0
4
S ³ x dx
dx
8S
Let 0 c 4 and set
y
51.
S³
55. V
S c2
4S .
2
8
c
8
2 2
x
−1
2
1
So, when x
2 2, the solid is divided into two parts
of equal volume.
y
8S
(one third of the volume).
3
c
56. Set S ³ x dx
4
0
3
Then
2
S c2
8S 2
,c
3
2
16
,c
3
4
3
4 3
.
3
1
16S
3
d
To find the other value, set S ³ x dx
x
2
1
3
4
0
(two thirds of the volume).
The volumes are the same because the solid has been
translated horizontally. 4 x x
4 x 2
2
2
Then
Sd
16S 2
,d
3
2
32
,d
3
32
3
4 6
.
3
The x-values that divide the solid into three parts of
y
52.
equal volume are x
4 3 3 and x
4 6 3.
(3, 9)
9
y = x2
6
57. V
x= y
3
6
9
(a) Around x-axis:
3
2
V
S ³ ª92 x 2 º dx
«
0 ¬
¼»
972
S
5
194.4S
(b) Around y-axis:
V
S³
9
0
(c) Around x
V
y
2
dy
R2 r 2
81
S
2
40.5S
ª
R2 x2
R2 r 2 «
2S ³
x
3
S³
¬
R2 r 2
0
2
º
r 2 » dx
¼
R 2 r 2 x 2 dx
ª
x3 º
2S « R 2 r 2 x »
3 ¼0
¬
R2 r 2
32
ª
R2 r 2 º
32
»
2S « R 2 r 2
«
»
3
¬
¼
32
4
S R2 r 2
3
3:
9
S 32 9 ³ S
0
2
y 3 dy
81S 27
S
2
135S
| 67.5S
2
R
r
INSTRUCTOR USE
SE ONLY
SE
So, b c a.
R2 − r 2
R
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
644
NOT FOR SALE
Chapter 7
58. Let R
Applications
lications of Integration
6 in the previous Exercise.
4
S 36 r 2
3
36 r
2
32
3
1 4
S 6
2 3
32
36 r
108
h r2
0 h
2
0
r 2 x 2 dx
r
r 2 x 2 dx
0
r
0
2S r 3 13 r 3
4S r 3
3
y
r
x, r x
h
S³
V
r
r
2S ª¬r 2 x 13 x3 º¼
36 1082 3 | 3.65
r
S³
2S ³
23
36 1082 3
r2
59. R x
V
108
2
r 2 x2 , r x
60. R x
0
y = r2 − x2
h
ª r 2S 3 º
« 2x »
¬ 3h
¼0
x 2 dx
r 2S 3
h
3h 2
1 2
Sr h
3
x
y
(−r, 0)
y= r x
h
r
(r, 0)
(h, r)
x
h
61. x
V
r
y
H
y·
§
r ¨1 ¸, R y
H¹
©
hª §
y ·º
¬ ©
¹¼
r y·
§
r ¨1 ¸, r y
H¹
©
2
S ³ «r ¨1 ¸» dy
0
H
h§
©
r
S³
h
S³
h
r
r 2 y2
ª
¬
S Ǭ r 3 ©
h
h
§
h2
h3 ·
¸
3H 2 ¹
H
r 2 y2 , r y
§
S r 2 h¨1 ©
h
h2 ·
¸
3H 2 ¹
H
−r
r
x
0
2
dy
r 2 y 2 dy
S «r 2 y ª§
H
1 2
1 3º
y y
3H 2 »¼ 0
H
©
V
·
¹
ª
¬
S r2¨h r 2 y2 , R y
1
y
S r 2 ³ ¨1 y 2 y 2 ¸ dy
0
H
H
S r2 «y 62. x
2
0
y
3 r
y º
»
3 ¼h
r
h
r3 · § 2
h3 ·º
¸ ¨r h ¸»
3¹ ©
3 ¹¼
§ 2r 3
h3 ·
r h S¨
¸
3¹
© 3
S
3
x
2r 3 3r 2 h h3
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 7.2
63.
Volume: The Disk Method
645
0.5
0
2
− 0.25
2
2 §1
·
2 x ¸ dx
¹
S ³ ¨ x2
0 8
©
V
S
2
64 ³ 0
°­ 0.1x3 2.2 x 2 10.9 x 22.2,
®
°̄2.95,
64. y
S³
V
11.5
«
64 ¬ 5
2
S
x6 º
»
6 ¼0
30
m3
0 d x d 11.5
11.5 x d 15
0.1x3 2.2 x 2 10.9 x 22.2
0
S ª 2 x5
x 4 2 x dx
2
dx S ³
15
11.5
2.952 dx
11.5
ª 0.1x 4
º
2.2 x3 10.9 x 2
22.2 x»
3
2
¬ 4
¼0
S«
15
S ª¬2.952 xº¼
11.5
| 1031.9016 cubic centimeters
y
8
6
4
2
x
4
8
12
3
5
65. (a) R x
16
25 x 2 , r x
9S 5
25 x 2 dx
25 ³ 5
V
0
18S 5
25 x 2 dx
25 ³ 0
5
18S ª
x3 º
«25 x »
25 ¬
3 ¼0
60S
y
8
6
4
2
−6
−4
x
−2
2
4
6
−2
−4
5
3
(b) R y
9 y2 , r y
25S 3
9 y 2 dy
9 ³0
V
0, x t 0
3
25S ª
y3 º
«9 y »
9 ¬
3 ¼0
50S
y
6
4
2
INSTRUCTOR
STR
USE ONLY
x
2
4
6
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
646
NOT FOR SALE
Chapter 7
Applications
lications of Integration
3
4S 50
66. Total volume: V
500,000S 3
ft
3
3
Volume of water in the tank:
y
S³ 0
2500 y 2
50
2
y
ª
y
S ³ 0 2500 y 2 dy
dy
S «2500 y 50
¬
y3 º 0
»
3 ¼ 50
§
S ¨ 2500 y0 ©
y0 3
250,000 ·
¸
3
3
¹
When the tank is one-fourth of its capacity:
1 § 500,000S ·
¨
¸
4©
3
¹
§
S ¨ 2500 y0 ©
y03
250,000 ·
¸
3
3
¹
y
60
40
7500 y0 y03 250,000
125,000
y03 7500 y0 125,000
20
0
−60
y0 | 17.36
32.64 feet
67. (a) First find where y
x2
x
4
16 4b
x
2 4b
³0
60
67.36 feet.
b intersects the parabola:
2
4
2
40
−60
When the tank is three-fourths of its capacity the depth is 100 32.64
V
x
20
−40
Depth: 17.36 50
b
−20
4b
44b
2
ª
S «4 ¬
2
4
º
ª
x2
x2 º
b» dx ³
S «b 4 » dx
2 4b
4
4¼
¼
¬
z
5
2
ª
º
x2
³ 0 S «¬4 4 b»¼ dx
4
4 ª x4
S³ «
0
¬16
2x2 º
bx 2
b 2 8b 16» dx
2
¼
2
2
4
y
x
4
ª x5
º
2 x3
bx3
b 2 x 8bx 16 x»
3
6
¬ 80
¼0
S«
§ 64 128 32
·
b 4b 2 32b 64 ¸
3
3
© 5
¹
S¨
(b) Graph of V b
§
©
S ¨ 4b 2 §
©
S ¨ 4b 2 64
512 ·
b ¸
3
15 ¹
64
512 ·
b ¸
3
15 ¹
120
0
4
0
Minimum volume is 17.87 for b
(c) V c b
V cc b
§
©
S ¨ 8b 64 ·
¸
3¹
8S ! 0 Ÿ b
0 Ÿ b
2.67.
64 3
8
8
3
2
2
3
8
is a relative minimum.
3
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 7.2
10
Volume: The Disk Method
647
2
³ 0 S ª¬ f x º¼ dx
68. (a) V
Simpson’s Rule: b a
V |
|
Sª
2.1
3¬
S
2
4 1.9
10 0
2
2 2.1
2
10, n
4 2.35
10
2
2 2.6
2
4 2.85
2
2 2.9
2
4 2.7
2
2 2.45
2
4 2.2
2
2
2.3 º
¼
178.405 | 186.83 cm3
3
0.00249 x 4 0.0529 x3 0.3314 x 2 0.4999 x 2.112
(b) f x
6
0
10
0
10
(c) V | ³ S f x
2
0
h
69. (a) S ³ r 2 dx
dx | 186.35 cm3
2
ii
0
is the volume of a right circular cylinder with radius
r and height h.
h § rx ·
(d) S ³ ¨ ¸ dx
0
©h¹
i
is the volume of a right circular cone with the radius
of the base as r and height h.
y
y
y=r
(h, r)
(h, r)
y= r x
h
x
x
2
§
x2 ·
(b) S ³ ¨ a 1 2 ¸ dx
b ¨
b ¸¹
©
b
(e) S ³
iv
ª
¬
R
r 2 x2
2
y
2
y=a 1− x
b2
y
r 2 − x2
R+
R
x
(−b, 0)
(b, 0)
R−
(c) S ³
r
r
2
º
r 2 x 2 » dx v
¼
R
is the volume of a torus with the radius of its circular
cross section as r and the distance from the axis of
the torus to the center of its cross section as R.
is the volume of an ellipsoid with axes 2a and 2b.
(0, a)
r
r «
r x
2
2
−r
2
dx
r
x
iii
is the volume of a sphere with radius r.
70. Let A1 x and A2 x equal the areas of the cross sections
of the two solids for a d x d b. Because
A1 x
A2 x , you have
y
y=
r2− x2
r2 − x2
V1
x
b
³ a A1 x dx
b
³ a A2 x dx
V2 .
So, the volumes are the same.
INSTRUCTOR
T
USE ONLY
(−r, 0)
(r, 0)
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© Cengage Learning. All Rights Reserved.
648
Chapter 7
Applications
lications of Integration
y
71.
4
3
2
x
2
3
4
x 1 x2 1
Base of cross section
4 4 x 3x 2 2 x3 x 4
2
2
³ 1
V
2
2 x x2
b2
(a) A x
2 x x2
1 4 1 5º
ª
2
3
«4 x 2 x x 2 x 5 x »
¬
¼ 1
4 4 x 3 x 2 2 x3 x 4 dx
81
10
2 + x − x2
2 + x − x2
(b) A x
2 x x2 1
bh
2
V
2
³ 1
ª
x 2 x3 º
»
«2 x 2
3 ¼ 1
¬
2 x x 2 dx
1
9
2
2 + x − x2
y
72.
3
1
x
−3
3
1
−1
−3
2 4 x2
Base of cross section
2 4 x2
b2
(a) A x
2
2 4 − x2
V
2
³2 4 4 x
2
dx
2
ª
x3 º
4 «4 x »
3 ¼ 2
¬
2 4 − x2
128
3
(b) A x
V
1
bh
2
3³
2
2
1
2 4 x2
2
3
4 x2
3 4 x2
4 x 2 dx
2
ª
x3 º
3 «4 x »
3 ¼ 2
¬
32 3
3
2 4 − x2
2 4 − x2
2 4 − x2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
ction 7.2
S
1 2
Sr
2
(c) A x
S
2
2 ³ 2
V
2
2
2
4 x2
16S
3
x3 º
«4 x »
2¬
3 ¼ 2
1
bh
2
2 4 − x2
1
2 4 x2
2
2
4 x2
4 x2
³ 2 4 x dx
V
2
2
ª
x3 º
«4 x »
3 ¼ 2
¬
32
3
4 − x2
2 4 − x2
73. The cross sections are squares. By symmetry, you can set
up an integral for an eighth of the volume and multiply
by 8.
1: x y
74. (a) When a
When a
2: x
2
r 2 y2
b2
A y
8³
r
0
r y
2
2
8ª¬r 2 y 13 y 3 º¼
2
1 represents a square.
y
2
1 represents a circle.
y
a=2
1
V
649
4 x 2 dx
Sª
(d) A x
S
2
4 x2
Volume: The Disk Method
a=1
dy
r
−1
1
x
0
16 r 3
3
−1
y
1 x
A
2³
(b)
y
x
1
1
a 1a
1 x
a 1a
dx
1
4³ 1 x a
0
1a
dx
To approximate the volume of the solid, from n
slices, each of whose area is approximated by the
integral above. Then sum the volumes of these n
slices.
75. (a) Because the cross sections are isosceles right triangles:
Ax
V
1
bh
2
1
2
r 2 y2
1 r 2
r y 2 dy
2³r
1 2
r y2
2
r 2 y2
r
r 2 y 2 dy
ª 2
y3 º
«r y »
3 ¼0
¬
r 2 y 2 tan T
tan T 2
r y2
2
r
³0
2 3
r
3
x
y
(b) A x
V
1
bh
2
1
2
r 2 y2
tan T r 2
r y 2 dy
2 ³ r
r
tan T ³
r
0
r 2 y 2 dy
ª
y3 º
tan T «r 2 y »
3 ¼0
¬
2 3
r tan T
3
INSTRUCTOR USE ONLY
As T o 90q, V o f.
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© Cengage Learning. All Rights Reserved.
650
NOT FOR SALE
Chapter 7
76. (a)
x R
Applications
lications of Integration
2
y2
r2
R r
x
r§
2S ³ ¨ ªR 0 ¬
©
V
r 2 y2
2
·
r 2 y 2 º ¸ dy
¼ ¹
2
r 2 y 2 º ªR ¼
¬
r
2S ³ 4 R
8S R ³
r 2 y 2 dy
0
r
r 2 y 2 dy
0
y
x
R
(b)
r
³0
r 2 y 2 dy is one-quarter of the area of a circle of radius r , 14 S r 2 .
V
8S R 14 S r 2
2S 2 r 2 R
Section 7.3 Volume: The Shell Method
1. p x
x, h x
x
V
2S ³ x x dx
5. p x
2
2. p x
V
ª 2S x3 º
«
»
¬ 3 ¼0
2
0
16S
3
V
1
0
32S
1
3. p x
V
4. p x
V
0
ª x2
x3 º
2S « »
3 ¼0
¬2
x x 2 dx
x, h x
4
2S ³ x
0
x, h x
S
3
x
4
2S ³ x3 2 dx
x dx
0
§1
·
3 ¨ x 2 1¸
©2
¹
4
ª 4S 5 2 º
«5 x »
¬
¼0
2
2 §
1 ·
2S ³ x¨ 2 x 2 ¸ dx
0
2 ¹
©
3
2
2S 4 2
4S
1
x
1
1 3
x
2
6. p x
x, h x
V
§ x3 ·
2S ³ x¨ ¸ dx
0
©2¹
15
3
10
128S
5
4
5
243S
5
V
3
y
ª x5 º
»
¬ 5 ¼0
1 2
x
2
2
3
S«
7. p x
2
ª
x4 º
2S « x 2 »
8 ¼0
¬
4
§1 ·
2S ³ x¨ x 2 ¸ dx
0
©4 ¹
4
« »
2 ¬ 4 ¼0
1
y
S ªx º
2S ³ x 1 x dx
2S ³
1 2
x
4
4 4
1 x
x, h x
x, h x
x
1
x, h x
2
4 x x2 x2
3
4 x 2 x2
2
2S ³ x 4 x 2 x 2 dx
0
4S ³
2
0
2 x 2 x3 dx
2
1 º
ª2
4S « x3 x 4 »
4 ¼0
¬3
16S
3
y
4
3
2
1
INSTRUCTOR USE ONLY
N
−
1
−1
x
1
2
3
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
tion 7.3
8. p x
9 x2
x, h x
y
3
2S ³ x 9 x 2 dx
V
11. p x
10
4
6
3
ª9 x2
x4 º
2S «
»
4 ¼0
¬ 2
2
81S
2
ª81 81º
2S « »
4¼
¬2
x
−3
−1
1
3
5
Let u
x 2, x
When x
2, u
0.
When x
4, u
2.
2S ³
V
9. p x
x
2
0
u 2, du
dx.
u 2 u1 2 du
2
4
ª2
º
2S « u 5 2 u 3 2 »
5
3
¬
¼0
4 4x x2
h x
x 2 dx
2
4
651
x 2
x, h x
2S ³ x
V
0
Volume: The SShell Method
x2 4 x 4
2
V
2S ³ x x 4 x 4 dx
V
2S ³
4 3 2º
ª2 5 2
2S « 2
2 »
5
3
¬
¼
2
0
2
0
8S
3
128 2S
15
y
2
ª x4
º
4
2S «
x3 2 x 2 »
4
3
¬
¼0
4 º
ª2
2« 4 2 »
3 ¼
¬5
2S
x3 4 x 2 4 x dx
4
3
2
1
y
x
4
1
2
3
4
3
12. p x
2
1 x2
x, h x
y
1
1
2S ³ x 1 x
V
1
10. p x
2
3
ªx
x º
2S «
»
4 ¼0
¬2
S
1·
§1
2S ¨ ¸
4¹
©2
4
2S ³ x 8 x3 2 dx
V
2
4 1
2
8 x3 2
x, h x
dx
0
x
−1
2
0
x
−1
2
1
4
2
ª
º
2S «4 x 2 x 7 2 »
7
¬
¼0
2
ª
º
2S «64 128 »
7
¬
¼
13. p x
384S
7
V
1 x2 2
e
2S
x, h x
1 §
1 x2 2 ·
e
2S ³ x¨
¸ dx
0
© 2S
¹
y
1
0
8
ª
«¬
6
4
2
x
2
4
y
2
2S ³ e x 2 x dx
1
2S e x 2 º»
¼0
2
1 ·
§
2S ¨1 ¸
e¹
©
| 0.986
1
3
4
1
2
1
4
x
1
4
1
2
3
4
1
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
652
NOT FOR SALE
Chapter 7
14. p x
Applications
lications of Integration
y
sin x
x
x, h x
18. p y
ª sin x º
2S ³ x «
» dx
0
¬ x ¼
S
V
V
2
S
π
4
0
>2S cos x@0
15. p y
x
3π
4
0
4
0
16 y y
3
dy
π
ª
y º
2S «8 y 2 »
4 ¼0
¬
x
4
−1
8
12
−2
−3
−4
128S
y
3
1
2S 128 64
2 y
y
4
2
4 4
−1
4S
y, h y
π
2
4
2S ³ y 16 y 2 dy
2S ³
1
2S ³ sin x dx
S
16 y 2
y, h y
3
2
2S ³ y 2 y dy
V
2
0
2S ³
2
0
2 y y dy
2
1
8S
3
ª
y3 º
2S « y 2 »
3 ¼0
¬
y, h y
V
2S ³ y 3 y dy
y
0
2S ³ y
2
y
8
8
x
1
3
19. p y
43
0
8
dy
6
4
8
3 1 y
h y
V
So, p y t 0 on >2, 0@
y
16. p y
0
2S ³
2
2S ³
2 ¬
0
ª § 3 · 7 3º
«2S ¨ 7 ¸ y »
¬ © ¹
¼0
2 y
y 2 y dy
ª2 y y º¼ dy
20. y
2
6S 7
2
7
768S
7
4x2 , x
y
2
x
−2
2
4
6
0
ª
y3 º
2S « y 2 »
3 ¼ 2
¬
2
ª
8 ·º
§
2S «0 ¨ 4 ¸»
3 ¹¼
©
¬
2S
p y
y
V
1
x
4
3
1
2
ª2
y and h y
1 if 0 d y p y
y and h y
1
1
1 if
d y d 1.
y
2
0
y dy 2S ³
12
12
1
y
21. p y
V
1 y dy
ª y2 º
ª
y2 º
2S « » 2S « y »
2 ¼1 2
¬ 2 ¼0
¬
S
4
3
4
1
4
º
x
−1
0
S
S
4
2
y, h y
1
4 y y
2
4 2y
2
2S ³ y 4 2 y dy
0
2S ³
2
0
4y 2y
2
y
dy
4
3
2
2 º
ª
2S «2 y 2 y 3 »
3 ¼0
¬
16 ·
§
2S ¨ 8 ¸
3¹
©
1
2
64S
5
1
.
2
17. p y
2S ³
3
0
S « y5 2 »
¬5
¼
1
4
4
−2
12
y
§ y·
dy
2S ³ y ¨
¨ 2 ¸¸
0
©
¹
4
S ³ y 3 2 dy
−1
8S
3
V
4
y
2
y , h( y )
16S
3
(2, 2)
2
1
x
−1
1
2
3
4
−1
1
2
1
4
x
1
2
1
3
2
2
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
tion 7.3
22. p y
y y2 2
y, h y
2 y y2
2S ³
2
0
y
2 y y 2 y 3 dy
(2, 2)
4x x2 x2
653
4 x 2 x2
4 x 4 x 2 x 2 dx
2
0
x3 6 x 2 8 x dx
2
2
2
2
0
2S 2 ³
3
ª
y3
y4 º
2S « y 2 »
3
4 ¼0
¬
ª x4
º
2 x3 4 x 2 »
4S «
4
¬
¼0
1
16S
x
8
§
·
2S ¨ 4 4 ¸
3
©
¹
16S
3
4 x, h x
2 x x2
23. p x
2S ³
V
0
4 x, h x
25. p x
2
2S ³ y 2 y y 2 dy
V
Volume: The SShell Method
−2
−1
1
y
2
−1
4
3
2
V
2
2S ³
0
2S ³
0
2
4 x 2 x x 2 dx
1
8 x 6 x 2 x3 dx
x
1
2
3
2
ª
x4 º
2S «4 x 2 2 x3 »
4 ¼0
¬
2S >16 16 4@
3 x, h x
26. p x
8S
V
y
2S ³
3
0
6 x x2 1 3
x
3
§
x3 ·
3 x ¨ 6x x2 ¸ dx
3¹
©
3 § x4
·
9 x 2 18 x ¸ dx
2S ³ ¨
0
3
©
¹
3
2
3
ª x5
º
2S « 3x3 9 x 2 »
15
¬
¼0
1
162S
5
x
1
2
3
y
6 x, h x
24. p x
x
8
2S ³
V
2S ³
4
0
4
0
6 x
x dx
6
4
6 x1 2 x3 2 dx
4
2
192S
5
2
ª
º
2S «4 x 3 2 x5 2 »
5
¬
¼0
x
1
y
2
3
27. The shell method would be easier:
4
V
3
4
2
2S ³ ª4 y 2 º y dy
0 ¬
¼
2
Using the disk method:
1
x
1
2
3
4
V
5
−1
Using the disk method, x
0 ¬
ª
«Note: V
¬
−2
28. The shell method is easier: V
4
S ³ ª« 2 2S ³
ln 4
0
4 x
2
2
4 x
2
dx
128S º
3 »¼
x 4 e x dx
ln 4 y and V
S³
3
0
ln 4 y
2
dy.
ªNote: V
¬«
S ª8 ln 2
¬
8 ln 2 3ºº»
¼¼
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
654
NOT FOR SALE
Chapter 7
Applications
lications of Integration
29. (a) Disk
30. (a) Disk
3
R x
x ,r x
0
128S
7
ªx º
»
¬ 7 ¼0
2
S ³ x 6 dx
V
7 2
S«
0
R x
10
,r x
x2
V
S ³ ¨ 2 ¸ dx
5 § 10 ·
1
y
0
2
©x ¹
5
100S ³ x 4 dx
1
8
5
ª x 3 º
100S « »
¬ 3 ¼1
6
4
2
x
−1
1
2
3
100S § 1
·
1¸
¨
3 © 125
¹
496
S
15
y
10
(b) Shell
8
p x
x3
x, h x
6
4
2
2
2S ³ x dx
V
4
0
ª x5 º
2S « »
¬ 5 ¼0
64S
5
2
−1
x
−2
1
2
3
4
5
y
(b) Shell
8
R x
6
4
V
2
x
−1
1
2
3
x, r x
0
5 § 10 ·
2S ³ x¨ 2 ¸ dx
1
©x ¹
5 1
20S ³
dx
1 x
5
20S ª¬ln x º¼1
(c) Shell
4 x, h x
p x
V
2
2S ³
0
2S ³
0
2
x3
(c) Disk
4 x x3 dx
R x
4 x3 x 4 dx
2
1 º
ª
2S « x 4 x5 »
5 ¼0
¬
96S
5
y
V
20S ln 5
10 10, r x
5ª
§
©
S ³ «102 ¨10 1
«¬
5
200 º
ª100
3
x »¼1
¬ 3x
S«
10
x2
2
10 · º
¸ » dx
x 2 ¹ »¼
1904
S
15
8
6
4
2
x
1
2
3
4
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
tion 7.3
31. (a) Shell
655
32. (a) Disk
p y
a1 2 y1 2
y, h y
2
a
a
0
S³
V
0
2S ³
a2 3 x2 3
R x
2S ³ y a 2a1 2 y1 2 y dy
V
Volume: The SShell Method
a2 3 x2 3
a
2S ³
ay 2a1 2 y 3 2 y 2 dy
a
,r x
3
0
dx
a 2 3a 4 3 x 2 3 3a 2 3 x 4 3 x 2 dx
0
a
9
9
1 º
ª
2S «a 2 x a 4 3 x5 3 a 2 3 x 7 3 x3 »
5
7
3 ¼0
¬
a
ªa
4a1 2 5 2
y3 º
2S « y 2 y »
5
3 ¼0
¬2
9
9
1 ·
§
2S ¨ a 3 a 3 a 3 a 3 ¸
5
7
3 ¹
©
S a3
§ a3
4a 3
a3 ·
2S ¨
¸
5
3¹
©2
a
32
15
32S a 3
105
y
y
(0, a)
(0, a)
(−a, 0)
(a, 0)
x
(a, 0)
x
(b) Same as part (a) by symmetry
(b) Same as part (a) by symmetry
y
(c) Shell
(0, a)
a x, h x
p x
12
a
x
12
2
(a, 0)
V
a
2S ³
0
2S ³
0
a
a x a1 2 x1 2
2
x
dx
a 2 2a 3 2 x1 2 2a1 2 x3 2 x 2 dx
(0, −a)
a
4
4
1 º
ª
2S «a 2 x a 3 2 x3 2 a1 2 x5 2 x3 »
3
5
3 ¼0
¬
4S a 3
15
33. (a)
1.5
y = (1 − x 4/3 ) 3/4
y
− 0.25
(0, a)
1.5
−0.25
(b) x 4 3 y 4 3
(a, 0)
0, y
0
34
y
1 x4 3
V
2S ³ x 1 x 4 3
x
34. (a)
1, x
1
0
34
dx | 1.5056
1.5
y=
−1
1 − x3
2
−0.5
(b) V
1
2S ³ x 1 x3 dx | 2.3222
0
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
656
NOT FOR SALE
Chapter 7
35. (a)
Applications
lications of Integration
4 § x·
40. 2S ³ x¨ ¸ dx
0
© 2¹
7
y=
3
(x − 2)2 (x − 6)2
−1
This integral represents the volume of the solid generated
by revolving the region bounded by y
x 2, y
0,
7
−1
6
2S ³ x 3 x 2
(b) V
4 about the y-axis by using the shell method.
and x
2
2
x 6
2
2
2
2
2
S ³ ª 4 2 y º dy
2
S ³ ª16 2 y º dy
dx | 187.249
0 ¬
¼
0 ¬
¼
represents this same volume by using the disk method.
36. (a)
3
y
y=
2
1 + e1/x
4
3
−1
5
2
−1
1
2S ³
(b) V
3
x
2x
1 1 e1 x
dx | 19.0162
1
(a) The rectangles would be vertical.
height
b
2
1
(b, k)
k
x
1
x
b
39. S ³
5
S³
x 1 dx
1
2
3
4
−1
k
height
5
3
y
b
(b) radius
4
y
41.
(b) The rectangles would be horizontal.
k
3
Disk method
37. Answers will vary.
38. (a) radius
2
−1
5
1
x 1
2
S³
(a) Around x-axis: V
2
y
4
dx
ª 5 9 5º
«S 9 x »
¬
¼0
4
0
4
ª 5 12 5 º
«2S 12 x » | 23.2147S
¬
¼0
(c) Around x
represents this same volume by using the shell method.
0
2
2S ³ x x 2 5 dx
(b) Around y-axis: V
using the disk method.
2S ³ y ª¬5 y 2 1 º¼ dy
0
x2 5
5
95
S 4
| 6.7365S
9
dx
This integral represents the volume of the solid generated
by revolving the region bounded by
y
x 1, y
0, and x
5 about the x-axis by
4
V
2S ³
4:
4
0
4 x x 2 5 dx | 16.5819S
So, a c b .
4
3
42. (a) The figure will be a circle of radius AB and center A.
2
(b) The figure will be a circular cylinder of radius AB.
1
x
1
2
3
4
5
(c) Disk method: V
−1
Disk method
Shell method: V
3
2
S ³ ª¬ g y º¼ dy
0
2S ³
2.45
0
x f x dx
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
tion 7.3
2
2
43. 2S ³ x 3 dx
2S ³ x x dx
0
46. 2S ³
0
(a) Plane region bounded by
2
y
x ,y
0, x
0, x
1
Volume: The SShell Method
657
4 x e x dx
0
(a) Plane region bounded by
y e x , y 0, x 0, x 1
2
(b) Revolved about the line x
(b) Revolved about the y-axis
4
y
y
y = ex
4
3
3
2
2
1
1
x
1
x
−1
1
2
3
Other answers possible.
44. 2S ³
1
0
47. p x
1
2S ³ y 1 y y 3 2 dy
0
(a) Plane region bounded by x
y dy
y, x
1, y
2 12 x 2
x, h x
2
2S ³
0
2
0
2 x 12 x3 dx
2S ª¬ x 2 18 x 4 º¼
0
x
1
2 ª¬ x 2 18 x 4 º¼
1
2 x0 2 14 x0 4
x0
6 y dy
0
6 y, x
x0
0
0
4r 2 3
2
Quadratic Formula
Take x0
4 2 3 | 0.73205, because the other
root is too large.
(a) Plane region bounded by
x
total volume
2S ³ 0 2 x 12 x3 dx
x0 4 8 x0 2 4
Other answers possible.
y 2
4S
S
1
6
2
0
Now find x0 such that:
(1, 1)
x
45. 2S ³
4
0
y
x= y
3
2S ³ x 2 12 x 2 dx
V
(b) Revolved about the x-axis
1
2
4
0, y
0
(b) Revolved around line y
2
Diameter: 2 4 2 3 | 1.464
y
y
6
2
x= 6−y
5
4
1
3
2
1
−3 −2 −1
x
1
2
3
4
5
x
1
2
−2
Other answers possible
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
658
NOT FOR SALE
Chapter 7
Applications
lications of Integration
48. Total volume of the hemisphere is
2S 3 3
3
1 4 S r3
2 3
p x
18S . By the Shell Method,
So, ³ x sin x dx
x
2S ³ 0 x 9 x 2 dx
6S
0
³
6
x0
0
9x
ª 2 9 x
¬« 3
32
2
2
12
(b) (i) p x
2 x dx
3 2 x0
º
¼» 0
d
>sin x x cos x C@
dx
18 23 9 x0
2
32
sin x x cos x C.
x, h x
2S ³
V
S 2
0
sin x
x sin x dx
S 2
2S >sin x x cos x@0
2S ª¬ 1 0 0º¼
18
2S
y
9 182 3 | 1.460
x0
cos x x sin x cos x
x sin x
9 x 2 . Find x0 such that:
x, h x
9 x0 2
51. (a)
1.0
Diameter: 2 9 182 3 | 2.920
0.5
y
−π
4
π
4
π
2
3π
4
π
x
2
−1.0
1
x
−3 −2 −1
−1
1
2
3
(ii) p x
2 sin x sin x
x, h x
−2
−3
49. V
4S ³
8S ³
S
2S ³ x 3 sin x dx
V
0
S
1
2 x
1
1
1 x 2 dx 4S ³
1
6S ³ x sin x dx
1 x 2 dx
0
1
1
S
6S >sin x x cos x@0
x 1 x 2 dx
6S S
1
12
§S ·
8S ¨ ¸ 2S ³ x 1 x 2
2 dx
1
2
© ¹
4S ³
r
R x
r
4S R ³
6S 2
y
2
1
3 2º
ª § 2·
4S 2 «2S ¨ ¸ 1 x 2 »
¬ © 3¹
¼ 1
50. V
3 sin x
4S 2
1
−π
4
r 2 x 2 dx
−1
π
2
π
5π
4
x
−2
r
r
r x dx 4S ³
2
2
r
r
x r x dx
2
2
r
3 2º
§Sr2 · ª § 2 · 2
2
4S R¨
¸ «2S ¨ ¸ r x
»
2
3
¼r
©
¹ ¬ © ¹
2S 2 r 2 R
INSTRUCTOR USE ONLY
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
tion 7.3
d
>cos x x sin x C@
dx
52. (a)
sin x sin x x cos x
54.
x2
y2
2
2
a
b
x cos x
Hence, ³ x cos x dx
cos x x sin x C.
V | 2 2S ³
0.8241
0
1
y
rb 1 x ª¬cos x x 2 º¼ dx
659
1
y2
b2
cos x Ÿ x | r 0.8241
(b) (i) x 2
Volume: The SShell Method
x2
a2
x2
a2
y
0.8241
ª
x4 º
4S «cos x x sin x »
4 ¼0
¬
| 2.1205
b
x
a
y
2
p x
x
−1
b 1
x, h x
1
−1
V
x 2
(ii) 4 cos x
V | 2S ³
2S ³
1.511
0
2
Ÿ x
0
a
2
x ª4 cos x x 2 º dx
¬
¼
6.2993
y
x2
dx
a2
a 2 x 2 x dx
a
1.511
«4 cos x 4 x sin x «¬
0
4S b
a ³0
0, 1.5110
1.511 ª
a
2 2S ³ xb 1 x2
a2
3
x2 º
»
3 »
¼0
32 º
2
2
§
·
4S b ¨ a x
¸»
¸¸»
a ¨¨
3
©
¹¼» 0
4S b 3
a
3a
4 2
Sa b
3
3
If the region is revolved about the x-axis, then by
4
symmetry the volume would be V
S ab 2 .
3
2
Note: If a
4
b, then volume is that of a sphere.
1
x
−2
−1
1
2
53. Disk Method
r 2 y2
R y
r y
V
0
S³
r
r h
r 2 y 2 dy
r
ª
S «r 2 y ¬
y3 º
»
3 ¼r h
1 2
S h 3r h
3
y
r
−r
x
r
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
660
NOT FOR SALE
Chapter 7
Applications
lications of Integration
b
r
x·
§
56. (a) 2S ³ hx¨1 ¸ dx
0
r¹
©
³ 0 ª¬ab ax º¼ dx
55. (a) Area of region
n
n
b
ª n
x n 1 º
«ab x a
»
n 1¼ 0
¬
is the volume of a right
circular cone with the
radius of the base as r
and height h.
b n 1
n 1
1 ·
§
ab n 1 ¨1 ¸
n 1¹
©
ab n 1 a
(b) 2S ³
(b)
lim R1 n
ab b
n
lim
lim ab n b
f
nof
(
(r, 0)
x
r
R x 2 r x
2
r
n
n 1
n
nof n 1
(
2
dx
v
y
ab n 1 ª¬n n 1 º¼
nof
y=h 1− x
r
(0, h)
is the volume of a torus with the radius of its circular
cross section as r and the distance from the axis of
the torus to the center of its cross section as R.
§ n ·
ab n 1 ¨
¸
© n 1¹
R1 n
y
ii
x=R
r2 − x2
y=
1
x
(r, 0)
(−r, 0)
y=−
r2 − x2
(c) Disk Method:
V
r
(c) 2S ³ 2 x r 2 x 2 dx
b
2S ³ x ab n ax n dx
0
0
2S a ³
b
0
is the volume of a sphere with radius r.
xb n x n 1 dx
n2
y
ªbn 2
bn 2 º
2S a «
»
n 2¼
¬ 2
R2 n
(d) lim R2 n
nof
lim S b 2 ab n
nof
S ab
S ab n 2 ª¬n n 2 º¼
(r, 0)
n2§
n ·
¨
¸
n
2¹
©
x
y=−
§ n ·
¨
¸
© n 2¹
S b 2 ab n
§ n ·
lim ¨
¸
n o f© n 2 ¹
r2 − x2
y=
b
ªb
º
x
2S a « x 2 »
n 2¼0
¬2
n
iii
r2 − x2
r
(d) 2S ³ hx dx
i
0
is the volume of a right circular cylinder with a
radius of r and a height of h.
1
f
y
(e) As n o f, the graph approaches the line x
(r, h)
b.
x
b
(e) 2S ³ 2ax 1 x 2 b 2 dx
0
iv
is the volume of an ellipsoid with axes 2a and 2b.
y
y =a
(0, a)
2
1 − x2
b
(b, 0)
(0, −a)
y = −a
x
2
1− x
b2
INSTRUCTOR USE ONLY
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© Cengage Learning. All Rights Reserved.
NOT FOR SALE
Section
tion 7.3
2S 40
ª0 4 10 45 2 20 40 4 30 20 0º¼
34 ¬
4
2S