Buy Official© Solutions Manual for Calculus Early Transcendental Functions,Larson,8e

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, 8. y = 5 − 2 x
let y be zero and solve the equation for x. To find the 5
y-intercepts of the graph of an equation, let x be zero x −1 0 1 2 2
3 4
and solve the equation for y. 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).

4. y = 9 − x2 9. y = 4 − x 2

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
10. y = ( x − 3)
2
x-intercepts: (0, 0), ( −1, 0), (1, 0)

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 = 1x +2
2

x −4 −2 0 2 4

y 0 1 2 3 4




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, Section 1.1 Graphs and Models 3


11. y = x + 1 3
15. y =
x
x −4 −3 −2 −1 0 1 2
x −3 −2 −1 0 1 2 3
y 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 1
16. y =
y 2 1 0 −1 0 1 2 x + 2

x −6 −4 −3 −2 −1 0 2

y − 14 − 12 −1 Undef. 1 1
2
1
4




13. y = x −6

x 0 1 4 9 16

y −6 −5 −4 −3 −2 17. y = 5− x




(a) (2, y) = ( 2, 1.73) (y = 5−2 = 3 ≈ 1.73 )
14. y = x + 2
(b) ( x, 3) = ( −4, 3) (3 = 5 − ( −4) )
x −2 −1 0 2 7 14 18. y = x5 − 5 x

y 0 1 2 2 3 4




(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 2− x
25. y =
y-intercept: y = 2(0) − 5 = −5; (0, − 5) 5x + 1

x-intercept: 0 = 2 x − 5 2− 0
y -intercept: y = = 2; (0, 2)
5 = 2x 5(0) + 1

x = 5;
2 ( 52 , 0) x-intercept: 0 =
2− x
5x + 1
20. y = 4 x 2 + 3 0 = 2− x
x = 4; (4, 0)
y-intercept: y = 4(0) + 3 = 3; (0, 3)
2



x-intercept: 0 = 4 x 2 + 3 x 2 + 3x
26. y =
(3 x + 1)
2
−3 = 4 x 2
None. y cannot equal 0. 02 + 3(0)
y-intercept: y = 2

21. y = x 2 + x − 2 3(0) + 1
y = 0; (0, 0)
y-intercept: y = 02 + 0 − 2
y = −2; (0, − 2) x 2 + 3x
x-intercepts: 0 =
(3x + 1)
2
2
x-intercepts: 0 = x + x − 2
x( x + 3)
0 = ( x + 2)( x − 1) 0 =
(3x + 1)
2

x = −2, 1; ( −2, 0), (1, 0)
x = 0, − 3; (0, 0), ( −3, 0)

22. y 2 = x3 − 4 x
27. x 2 y − x 2 + 4 y = 0
y-intercept: y 2 = 03 − 4(0)
y-intercept: 02 ( y ) − 02 + 4 y = 0
y = 0; (0, 0)
y = 0; (0, 0)
x-intercepts: 0 = x3 − 4 x
x-intercept: x 2 (0) − x 2 + 4(0) = 0
0 = x( x − 2)( x + 2)
x = 0; (0, 0)
x = 0, ± 2; (0, 0), ( ± 2, 0)

28. y = 2 x − x2 + 1
23. y = x 16 − x 2
y-intercept: y = 2(0) − 02 + 1
y-intercept: y = 0 16 − 02 = 0; (0, 0)
y = −1; (0, −1)
x-intercepts: 0 = x 16 − x 2
x-intercept: 0 = 2x − x2 + 1
0 = x (4 − x)(4 + x)
2x = x2 + 1
x = 0, 4, − 4; (0, 0), ( 4, 0), ( − 4, 0)
4x2 = x2 + 1

24. y = ( x − 1) x2 + 1 3x 2 = 1
1
x2 =
y-intercept: y = (0 − 1) 02 + 1 3
y = −1; (0, −1) 3
x = ±
3
x-intercept: 0 = ( x − 1) x2 + 1
3  3 
x = 1; (1, 0) x = ;  , 0 
3  3 

Note: x = − 3 3 is an extraneous solution.




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