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Solution Manual For Calculus of a Single Variable 8th Edition by Ron Larson, Bruce H. Edwards Chapte

Page 1

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

© 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.


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

© 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.


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)

© 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.


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.

© 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.


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.

© 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.


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

© 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.


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

© 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.


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