SOLUTIONS MANUAL
SOLUTIONS MANUAL
Chapter 1. Limits And Their Properties. Chapter 2. Differentiation. Chapter 3. Applications Of Differentiation. Chapter 4. Integration. Chapter 5. Logarithmic, Exponential, And Other Transcendental Functions. Chapter 6. Differential Equations. Chapter 7. Applications Of Integration. Chapter 8. Integration Techniques And Improper Integrals. Chapter 9. Infinite Series. Chapter 10. Conics, Parametric Equations, And Polar Coordinates.
C H A P T E R 1 Preparation for Calculus Section 1.1
Graphs and Models................................................................................. 2
Section 1.2
Linear Models and Rates of Change ................................................... 11
Section 1.3
Functions and Their Graphs ................................................................. 22
Section 1.4
Review of Trigonometric Functions .................................................... 36
Section 1.5
Inverse Functions.................................................................................. 45
Section 1.6
Exponential and Logarithmic Functions ............................................. 62
Review Exercises .......................................................................................................... 72 Problem Solving ........................................................................................................... 87
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C H A P T E R 1 Preparation for Calculus Section 1.1 Graphs and Models 1. To find the x-intercepts of the graph of an equation, let y be zero and solve the equation for x. To find the y-intercepts of the graph of an equation, let x be zero and solve the equation for y.
8. y = 5 − 2 x x
−1
0
1
2
5 2
3
4
y
7
5
3
1
0
−1
−3
2. Symmetry helps in sketching a graph because you need only half as many points to plot. Answers will vary. 3. y = − 32 x + 3 x-intercept: ( 2, 0) y-intercept: (0, 3) Matches graph (b). 9. y = 4 − x 2
9 − x2
4. y =
x-intercepts: ( −3, 0), (3, 0)
x
−3
−2
0
2
3
y-intercept: (0, 3)
y
−5
0
4
0
−5
Matches graph (d). 5. y = 3 − x 2 x-intercepts:
( 3, 0), (− 3, 0)
y-intercept: (0, 3) Matches graph (a). 6. y = x3 − x x-intercepts: (0, 0), ( −1, 0), (1, 0)
10. y = ( x − 3)
2
y-intercept: (0, 0)
x
0
1
2
3
4
5
6
Matches graph (c).
y
9
4
1
0
1
4
9
7. y = 12 x + 2
2
x
−4
−2
0
2
4
y
0
1
2
3
4
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Section 1.1 11. y = x + 1
15. y =
x
−4
−3
−2
−1
0
1
2
y
3
2
1
0
1
2
3
Graphs and Models
3
3 x
x
−3
−2
−1
0
1
2
3
y
−1
− 32
−3
Undef.
3
3 2
1
12. y = x − 1 x
−3
−2
−1
0
1
2
3
y
2
1
0
−1
0
1
2
13. y =
16. y = x
−6
−4
−3
−2
−1
0
2
y
− 14
− 12
−1
Undef.
1
1 2
1 4
x −6
x
0
1
4
9
16
y
−6
−5
−4
−3
−2
17. y =
(a)
14. y =
1 x + 2
5− x
(2, y) = ( 2, 1.73)
(b) ( x, 3) = ( −4, 3)
x + 2
x
−2
−1
0
2
7
14
y
0
1
2
2
3
4
(y =
(3 =
5−2 = 5 − ( −4)
)
3 ≈ 1.73
)
18. y = x5 − 5 x
(a)
(−0.5, y) = (−0.5, 2.47)
(b) ( x, − 4) = ( −1.65, − 4) and ( x, − 4) = (1, − 4)
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4
Chapter 1
Preparation for Calculus
19. y = 2 x − 5 y-intercept: y = 2(0) − 5 = −5; (0, − 5) x-intercept: 0 = 2 x − 5
5 = 2x x = 52 ;
( 52 , 0)
25. y =
2− x 5x + 1
y -intercept: y =
2− 0 = 2; 5(0) + 1
x-intercept: 0 =
2− x 5x + 1
20. y = 4 x 2 + 3
y-intercept: y = 4(0) + 3 = 3; (0, 3) 2
x-intercept: 0 = 4 x 2 + 3
26. y =
−3 = 4 x 2 None. y cannot equal 0.
0 = 2−
x
x = 4;
(4, 0)
x 2 + 3x
(3x + 1)
2
y-intercept: y =
21. y = x 2 + x − 2
0 = ( x + 2)( x − 1) x = −2, 1; ( −2, 0), (1, 0)
y = 0; (0, 0) x-intercepts: 0 = x3 − 4 x 0 = x( x − 2)( x + 2)
2
x 2 + 3x
(3x + 1) x( x + 3) 0 = 2 (3x + 1) x = 0, − 3; (0, 0), ( −3, 0)
x-intercepts: 0 = x + x − 2
y-intercept: y 2 = 03 − 4(0)
3(0) + 1
x-intercepts: 0 =
2
22. y 2 = x3 − 4 x
02 + 3(0)
y = 0; (0, 0)
y-intercept: y = 02 + 0 − 2
y = −2; (0, − 2)
2
27. x 2 y − x 2 + 4 y = 0
y-intercept: 02 ( y ) − 02 + 4 y = 0 y = 0; (0, 0) x-intercept: x 2 (0) − x 2 + 4(0) = 0 x = 0; (0, 0)
x = 0, ± 2; (0, 0), ( ± 2, 0) 28. y = 2 x −
23. y = x 16 − x 2
y-intercept: y = 0 16 − 02 = 0; (0, 0) x-intercepts: 0 = x 16 − x 2
(4 − x)(4 + x) x = 0, 4, − 4; (0, 0), ( 4, 0), ( − 4, 0)
0 = x
x2 + 1
y-intercept: y = 2(0) −
x-intercept:
0 = 2x − 4x2 = x2 + 1
y-intercept: y = (0 − 1) 02 + 1
x2 =
x-intercept: 0 = ( x − 1)
x = 1; (1, 0)
x2 + 1
x2 + 1
2x = 3x 2 = 1
y = −1; (0, −1)
02 + 1
y = −1; (0, −1)
x2 + 1
24. y = ( x − 1)
(0, 2)
1 3
x = ±
3 3
x =
3 3 ; , 0 3 3
x2 + 1
Note: x = − 3 3 is an extraneous solution.
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Section 1.1 29. Symmetric with respect to the y-axis because y = ( − x) − 6 = x − 6. 2
2
Graphs and Models
5
41. y = 2 − 3 x y = 2 − 3(0) = 2, y -intercept 0 = 2 − 3( x) 3 x = 2 x = 23 , x-intercept
30. y = 9 x − x 2 No symmetry with respect to either axis or the origin. 31. Symmetric with respect to the x-axis because
Intercepts: (0, 2),
( 23 , 0)
Symmetry: none
(− y )2 = y 2 = x3 − 8 x. 32. Symmetric with respect to the origin because
( − y ) = ( − x )3 + ( − x ) − y = − x3 − x 42. y = 23 x + 1
y = x3 + x. 33. Symmetric with respect to the origin because ( − x)( − y ) = xy = 4. 34. Symmetric with respect to the x-axis because x( − y ) = xy = −10. 2
2
35. y = 4 −
y = 23 (0) + 1 = 1, y -intercept 0 = 23 x + 1 − 23 x = 1 x = − 32 , x-intercept
(
Intercepts: (0, 1), − 32 , 0
)
Symmetry: none
x +3
No symmetry with respect to either axis or the origin. 36. Symmetric with respect to the origin because
(− x)(− y ) −
4 − ( − x) = 0 2
xy −
4− x
2
= 0.
37. Symmetric with respect to the origin because −y =
−x
43. y = 9 − x 2 y = 9 − (0) = 9, y -intercept 2
0 = 9 − x 2 x 2 = 9 x = ± 3, x-intercepts Intercepts: (0, 9), (3, 0), ( −3, 0)
( − x) + 1 2
y = 9 − (− x) = 9 − x 2 2
x y = 2 . x +1
Symmetry: y-axis
38. Symmetric with respect to the origin because
−y =
( − x) 2 4 − ( − x)
−y =
− x5 4 − x2
5
44. y = 2 x 2 + x = x( 2 x + 1)
y = 0( 2(0) + 1) = 0, y -intercept
x5 . y = 4 − x2
0 = x( 2 x + 1) x = 0, − 12 , x-intercepts
39. y = x 3 + x is symmetric with respect to the y-axis because y = ( − x) + ( − x) = −( x3 + x) = x3 + x . 3
40.
(
Intercepts: (0, 0), − 12 , 0
)
Symmetry: none
y − x = 3 is symmetric with respect to the x-axis
because
−y − x = 3 y − x = 3.
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6
Chapter 1
Preparation for Calculus
45. y = x3 + 2 y = 03 + 2 = 2, y -intercept 0 = x + 2 x = −2 x = − 3
3
(
Intercepts: −
25 − x 2
48. y =
3
25 − 02 =
y = 3
2, x-intercept
25 − x 2 = 0
)
2, 0 , (0, 2)
Symmetry: none
25 = 5, y -intercept
25 − x 2 = 0
(5 + x)(5 − x) = 0 x = ± 5, x -intercept Intercepts: (0, 5), (5, 0), ( −5, 0)
y =
25 − ( − x) = 2
25 − x 2
Symmetry: y-axis
46. y = x 3 − 4 x
y = 03 − 4(0) = 0, y -intercept x3 − 4 x = 0 x ( x 2 − 4) = 0 x( x + 2)( x − 2) = 0 x = 0, ± 2, x -intercepts Intercepts: (0, 0), ( 2, 0), ( −2, 0) y = ( − x ) − 4( − x ) = − x 3 + 4 x = −( x 3 − 4 x) 3
Symmetry: origin
49. x = y 3
y 3 = 0 y = 0, y -intercept x = 0, x-intercept Intercept: (0, 0) − x = (− y ) − x = − y 3 3
Symmetry: origin
50. x = y 4 − 16 y 4 − 16 = 0
( y 2 − 4)( y 2 + 4) = 0 ( y − 2)( y + 2)( y 2 + 4) = 0 y = ± 2, y -intercepts
47. y = x
x +5
y = 0 0 + 5 = 0, y -intercept x
x + 5 = 0 x = 0, − 5, x-intercepts
Intercepts: (0, 0), ( −5, 0)
x = 0 4 − 16 = −16, x -intercept
Intercepts: (0, 2), (0, − 2), ( −16, 0) Symmetry: x-axis because x = ( − y ) − 16 = y 4 − 16 4
Symmetry: none
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Section 1.1
51. y =
8 x
7
54. y = 6 − x
8 y = Undefined no y -intercept 0 8 = 0 No solution no x-intercept x
Intercepts: none −y =
Graphs and Models
8 8 y = −x x
y = 6 − 0 = 6 = 6, y -intercept
6− x = 0 6− x = 0 6 = x, x-intercept Intercepts: (0, 6), (6, 0) Symmetry: none
Symmetry: origin
55. x 2 + y 2 = 9 y2 = 9 − x2
52. y = y =
10 x2 + 1 10 = 10, y -intercept 0 +1 2
10 = 0 No solution no x-intercepts 2 x +1
Intercept: (0, 10)
10
10 y = = 2 2 x +1 ( − x) + 1 Symmetry: y-axis
y = ± 9 − x2 y = ± 9 − 0 = ±3, y -intercepts ± 9− x
2
= 0
2
9− x = 0 9 = x2 ±3 = x, x-intercepts Intercepts: ( ±3, 0), (0, ± 3)
( − x)2 + y 2 = 9 x 2 + y 2 = 9 x2 + (− y) = 9 x 2 + y 2 = 9 2
(− x)2 + ( − y )2 = 9 x 2 + y 2 = 9 53. y = 6 − x
Symmetry: x-axis, y-axis, origin
y = 6 − 0 = 6, y -intercept
6− x = 0 6 = x x = ± 6, x-intercepts Intercepts: (0, 6), ( −6, 0), (6, 0)
y = 6 − −x = 6 − x Symmetry: y-axis
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8
Chapter 1
Preparation for Calculus 4 − x2 2
56. x 2 + 4 y 2 = 4 y = ±
58. 3x − 4 y 2 = 8
− 4 y 2 = − 3x + 8 y 2 = 34 x − 2
4 − 02 4 = ± = ±1, y -intercepts 2 2
y = ±
y = ±
x + 4(0) = 4 2
2
3 x − 2 4
y = ± 0− 2
x2 = 4
No real solution No y -intercepts
x = ± 2, x -intercepts Intercepts: ( −2, 0), ( 2, 0), (0, −1), (0, 1)
±
(− x)2 + 4( − y )2 = 4 x 2 + 4 y 2 = 4
3 x − 2 4
= 0
3 x − 2 4
= 0
x = 83 , x-intercept
Symmetry: origin and both axes
Intercept: ( 83 , 0) Symmetry: x-axis
59.
x + y = 8 y = 8− x 4x − y = 7 y = 4x − 7
8 − x = 4x − 7 15 = 5 x
57. 3 y 2 − x = 9
3 = x
3y2 = x + 9
The corresponding y-value is y = 5.
y 2 = 13 x + 3
±
y = ±
1x + 3 3
y = ±
0+3 = ±
Point of intersection: (3, 5)
3, y -intercepts
3x + 4 2 − 4 x − 10 4 x + 2 y = −10 y = 2
60. 3x − 2 y = − 4 y =
1 x +3 = 0 3 1 x +3 = 0 3
3x + 4 − 4 x − 10 = 2 2 3 x + 4 = − 4 x − 10
x = − 9, x-intercept
(
Intercepts: 0,
)(
3 , 0, −
)
3 , ( − 9, 0)
7 x = −14
3( − y ) − x = 3 y − x = 9 2
x = −2
2
The corresponding y-value is y = −1.
Symmetry: x-axis
Point of intersection: ( −2, −1)
61.
x 2 + y = 15 y = − x 2 + 15 − 3 x + y = 11 y = 3 x + 11 − x 2 + 15 = 3 x + 11 0 = x 2 + 3x − 4 0 = ( x + 4)( x − 1) x = − 4, 1 The corresponding y-values are y = −1 (for x = − 4) and y = 14 (for x = 1). Points of intersection: ( − 4, −1), (1, 14)
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Section 1.1
62.
x = 3 − y2 y2 = 3 − x
9
66. y = x 4 − 2 x 2 + 1
y = x −1 3 − x = ( x − 1)
Graphs and Models
y = 1 − x2 2
Points of intersection:
(−1, 0), (0, 1), (1, 0)
2
3 − x = x − 2x + 1 0 = x 2 − x − 2 = ( x + 1)( x − 2)
Analytically,
x = −1 or x = 2
1 − x2 = x4 − 2x2 + 1
The corresponding y-values are y = −2 (for x = −1)
0 = x4 − x2 0 = x 2 ( x + 1)( x − 1)
and y = 1 (for x = 2).
x = −1, 0, 1.
Points of intersection: ( −1, − 2), ( 2, 1)
67. y =
63. x 2 + y 2 = 5 y 2 = 5 − x 2
y =
x − y = 1 y = x −1 5 − x 2 = ( x − 1)
x+6 − x2 − 4x
2
5 − x2 = x2 − 2 x + 1 0 = 2 x 2 − 2 x − 4 = 2( x + 1)( x − 2) x = −1 or x = 2 The corresponding y-values are y = −2 (for x = −1) and y = 1 (for x = 2). Points of intersection: ( −1, − 2), ( 2, 1)
(
Points of intersection: ( −2, 2), −3,
− x2 − 4x
x + 6 = − x2 − 4x x2 + 5x + 6 = 0
64. x 2 + y 2 = 16 x + 2y = 4
x + 6 =
Analytically,
)
3 ≈ ( −3, 1.732)
( x + 3)( x + 2) = 0
x = 4 − 2y
x = −3, − 2.
( 4 − 2 y ) + y 2 = 16 2
5 y 2 − 16 y + 16 = 16 y(5 y − 16) = 0
68. y = − 2 x − 3 + 6
y = 0, 16 5
y = 6− x
x = 4 − 2(0) x = 4
( )
x = 4 − 2 16 x = − 12 5 5
(
Points of intersection: ( 4, 0), − 12 , 16 5 5
65. y = x3 − 2 x 2 + x − 1
y = − x 2 + 3x − 1
) Points of intersection: (3, 3), (1, 5) Analytically, − 2 x − 3 + 6 = 6 − x
Points of intersection:
(−1, − 5), (0, −1), (2, 1) Analytically,
2x − 3 = x 2 x − 3 = x or 2 x − 3 = − x x = 3 or
x = 1.
x3 − 2 x 2 + x − 1 = − x 2 + 3x − 1 x3 − x 2 − 2 x = 0 x( x − 2)( x + 1) = 0 x = −1, 0, 2.
© 2024 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.