SOLUTIONS MANUAL
Calculus of a Single Variable: Early Transcendental Functions, 8th Edition
by Ron Larson
LU
XE
LI
BR
AR
Y
, TABLE OF CONTENT
Chapter 1: Preparation for Calculus
Chapter 2: Limits and Their Properties
Chapter 3: Differentiation
Chapter 4: Applications of Differentiation
Chapter 5: Integration
LU
Chapter 6: Differential Equations
Chapter 7: Applications of Integration
XE
Chapter 8: Integration Techniques and Improper Integrals
Chapter 9: Infinite Series
Chapter 10: Conics, Parametric Equations, and Polar Coordinates
LI
BR
AR
Y
,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
LU
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)
XE
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).
LI
5. y = 3 − x 2
x-intercepts: ( )(
3, 0 , − 3, 0 )
y-intercept: (0, 3)
BR
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
AR
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
Y
2
, 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
LU
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
XE
x −6 −4 −3 −2 −1 0 2
y − 14 − 12 −1 Undef. 1 1
2
1
4
LI
13. y = x −6
x 0 1 4 9 16
BR
y −6 −5 −4 −3 −2 17. y = 5− x
AR
(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
Y
(a) (−0.5, y) = ( −0.5, 2.47)
(b) ( x, − 4) = ( −1.65, − 4) and ( x, − 4) = (1, − 4)
Calculus of a Single Variable: Early Transcendental Functions, 8th Edition
by Ron Larson
LU
XE
LI
BR
AR
Y
, TABLE OF CONTENT
Chapter 1: Preparation for Calculus
Chapter 2: Limits and Their Properties
Chapter 3: Differentiation
Chapter 4: Applications of Differentiation
Chapter 5: Integration
LU
Chapter 6: Differential Equations
Chapter 7: Applications of Integration
XE
Chapter 8: Integration Techniques and Improper Integrals
Chapter 9: Infinite Series
Chapter 10: Conics, Parametric Equations, and Polar Coordinates
LI
BR
AR
Y
,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
LU
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)
XE
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).
LI
5. y = 3 − x 2
x-intercepts: ( )(
3, 0 , − 3, 0 )
y-intercept: (0, 3)
BR
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
AR
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
Y
2
, 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
LU
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
XE
x −6 −4 −3 −2 −1 0 2
y − 14 − 12 −1 Undef. 1 1
2
1
4
LI
13. y = x −6
x 0 1 4 9 16
BR
y −6 −5 −4 −3 −2 17. y = 5− x
AR
(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
Y
(a) (−0.5, y) = ( −0.5, 2.47)
(b) ( x, − 4) = ( −1.65, − 4) and ( x, − 4) = (1, − 4)
