Solutions Manual — Thomas' Calculus: Early Transcendentals, 15th Edition (Hass, Heil, Weir & Bogacki, 2022), Chapters 1-19 | All Chapter Solutions Covered

SOLUTIONS MANUAL

Thomas' Calculus: Early Transcendentals
Joel Hass, Christopher Heil, Maurice Weir, and Przemyslaw Bogacki
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15th Edition
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, TABLE OF CONTENTS
Thomas' Calculus: Early Transcendentals (15th Edition)
Joel Hass, Christopher Heil, Maurice D. Weir, and Przemyslaw Bogacki
M
Chapter 1 Functions

Chapter 2 Limits and Continuity

Chapter 3 Derivatives
ED
Chapter 4 Applications of Derivatives

Chapter 5 Integrals
GE
Chapter 6 Applications of Definite Integrals

Chapter 7 Integrals and Transcendental Functions

Chapter 8 Techniques of Integration
EK
Chapter 9 First-Order Differential Equations

Chapter 10 Infinite Sequences and Series

Chapter 11 Parametric Equations and Polar Coordinates
_S
Chapter 12 Vectors and the Geometry of Space

Chapter 13 Vector-Valued Functions and Motion in Space
TU
Chapter 14 Partial Derivatives

Chapter 15 Multiple Integrals

Chapter 16 Integrals and Vector Fields
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Chapter 17 Second-Order Differential Equations

Chapter 18 Complex Functions
A?
Chapter 19 Fourier Series and Wavelets
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, CHAPTER 1 FUNCTIONS

1.1 FUNCTIONS AND THEIR GRAPHS

1. domain  (, ); range  [1,) 2. domain  [0, ); range  (, 1]
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3. domain  [2, ); y in range and y  5x  10  0  y can be any positive real number  range  [0, ).


4. domain  (, 0] [3, ); y in range and y  x2  3x  0  y can be any positive real number 
range  [0, ).
ED
5. domain  (, 3)  (3, ); y in range and y  4 , now if t  3  3  t  0  4  0, or if t  3 
3 t 3t
3  t  0  3 4 t  0  y can be any nonzero real number  range  (, 0)  (0, ).
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6. domain  (, 4)  (4, 4)  (4,); y in range and y  2 , now if t  4  t 2 16  0  2  0, or if
t  16
2
t  16
2

4  t  4  16  t2  16  0   2  2 2
, or if t  4  t  16  0  2  0  y can be any nonzero
16 t  16
2
t  16
2


real number  range  (,  18]  (0, ).
EK
7. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Is the graph of a function of x since any vertical line intersects the graph at most once.

8. (a) Not the graph of a function of x since it fails the vertical line test.
_S
(b) Not the graph of a function of x since it fails the vertical line test.
2

9. base  x; (height)2  x  x2  height  3 x; area is a(x)  1 (base)(height)  1 (x)
2 2 2 2
 x 
3
2 4
3 2
x ;

perimeter is p(x)  x  x  x  3x.
TU
10.​ s  side length  s2  s2  d 2  s  d ; and area is a  s2  a  12 d 2 .
2

11. Let D  diagonal length of a face of the cube and ℓ  the length of an edge. Then ℓ2  D2  d 2 and
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 
2 3/2
d2 3
D2  2ℓ2  3ℓ2  d 2  ℓ  d . The surface area is 6ℓ2  6d  2d 2 and the volume is ℓ3   d .
3 3 3 3 3


 
12. The coordinates of P are x, x so the slope of the line joining P to the origin is m  x  1 (x  0).
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x x


Thus, x, x    1
m2
, 1
m .
2 2 2 2
13.​ 2x  4 y  5  y   12 x  54 ; L  (x  0)2  ( y  0)2  x  ( 12 x  54 )  x  14 x  54 x  25
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16
2 20x2  20x  25 20x2  20x  25
 54 x  54 x  16
25

16

4


14. y  x  3  y2  3  x; L  (x  4)2  ( y  0)2  ( y2  3  4)2  y2  ( y2 1)2  y2
 y4  2 y2 1 y2  y4  y2 1

, 2 Chapter 1 Functions

15. The domain is (, ). 16. The domain is (, ).
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17. The domain is (, ). 18. The domain is (, 0].
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19. The domain is (, 0)  (0, ). 20. The domain is (, 0)  (0, ).
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21. The domain is 22. The range is [5, ) .
(, 5)  (5, 3] [3, 5)  (5, ).

23. Neither graph passes the vertical line test.
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(a) (b)
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