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Solution Manual for Thomas' Calculus, SI Units, 15th edition Joel R. Hass Christopher E. Heil Maurice D. Weir R331,96
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Solution Manual for Thomas' Calculus, SI Units, 15th edition Joel R. Hass Christopher E. Heil Maurice D. Weir

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Solution Manual for Thomas' Calculus, SI Units, 15th edition Joel R. Hass Christopher E. Heil Maurice D. Weir

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  • September 15, 2024
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SOLUTION MANUAL
THOMAS' CALCULUS, SI UNITS, 15TH EDITION JOEL R. HASS
CHRISTOPHER E. HEIL MAURICE D. WEIR
CHAPTER 1-19



CHAPTER 1 FUNCTIONS

1.1 FUNCTIONS AND THEIR GRAPHS

1. domain  (, ); range  [1, ) 2. domain  [0, ); range  (, 1]

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


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

5. domain  (, 3)  (3, ); y in range and y  3 4 t , now if t  3  3  t  0  3 4 t  0, or if t  3 
3  t  0  3 4 t  0  y can be any nonzero real number  range  (, 0)  (0, ).


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 2  16 t 2  16
4  t  4  16  t 2  16  0   16
2  2 , or if t  4  t  16  0 
2 2  0  y can be any nonzero
t 2  16 t 2  16
real number  range  (,  18 ]  (0, ).

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.
(b) Not the graph of a function of x since it fails the vertical line test.



2
9. base  x; (height)2  2x  x 2  height  23 x; area is a( x)  12 (base)(height)  12 ( x) 23 x  43 x 2 ;  
perimeter is p( x)  x  x  x  3x.


10. s  side length  s 2  s 2  d 2  s  d ; and area is a  s 2  a  12 d 2 .
2


11. Let D  diagonal length of a face of the cube and  the length of an edge. Then 2  D2  d 2 and

 
2 2 3/2 3
D2  2 2  3 2  d 2   d . The surface area is 6 2
 6d3  2d 2 and the volume is 3
 d3  d .
3 3 3




© 2023 Pearson Education Ltd. All Rights Reserved.
1

,2 Chapter 1 Functions

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


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

20 x 2  20 x  25 20 x 2  20 x  25
 54 x 2  54 x  16
25 
16
 4


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


15. The domain is ( ,  ). 16. The domain is ( ,  ).




17. The domain is ( ,  ). 18. The domain is ( , 0].




19. The domain is (, 0)  (0, ). 20. The domain is (, 0)  (0, ).




21. The domain is 22. The range is [5, ) .
(, 5)  (5, 3]  [3, 5)  (5, ).




© 2023 Pearson Education Ltd. All Rights Reserved.

, Section 1.1 Functions and Their Graphs 3


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




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




 x  y 1   y  1 x 
   
x y 1  or  or 
 x  y  1  y  1  x 
   

25. x 0 1 2 26. x 0 1 2
y 0 1 0 y 1 0 0




 4  x , x  1  , x  0
2 1
27. F ( x)   28. G ( x)   x
 x  2 x, x  1
2
 x, 0  x




29. (a) Line through (0, 0) and (1, 1): y  x; Line through (1, 1) and (2, 0): y   x  2
 x, 0  x  1
f ( x)  
 x  2, 1  x  2




© 2023 Pearson Education Ltd. All Rights Reserved.

, 4 Chapter 1 Functions

 2, 0  x 1
 0, 1 x  2

(b) f ( x)  
 2, 2 x3
 0, 3 x 4

30. (a) Line through (0, 2) and (2, 0): y   x  2
0 1
Line through (2, 1) and (5, 0): m  5  2  31   13 , so y   1 ( x  2)  1   1 x  5
3 3 3
  x  2, 0  x  2
f ( x)   1
 3 x  3 , 2  x  5
5


3  0
(b) Line through (1, 0) and (0, 3): m  0  (1)  3, so y  3x  3
1  3
Line through (0, 3) and (2, 1) : m  2  0  24  2, so y  2 x  3
3x  3,  1  x  0
f ( x)  
2 x  3, 0  x  2

31. (a) Line through (1, 1) and (0, 0): y   x
Line through (0, 1) and (1, 1): y  1
0 1
Line through (1, 1) and (3, 0): m  3  1  21   12 , so y   12 ( x  1)  1   12 x  23
 x 1  x  0

f ( x)   1 0  x 1
 1 1 x  3
 2 x  2
3  1x 2  x  0
 2
(b) Line through ( 2, 1) and (0, 0): y  12 x f ( x)  2 x  2 0  x  1
Line through (0, 2) and (1, 0): y  2 x  2  1 1 x  3
Line through (1, 1) and (3, 1): y  1 


  1 0
 
32. (a) Line through T2 , 0 and (T, 1): m  T  (T /2)  T2 , so y  T2 x  T2  0  T2 x  1
 0, 0  x  T2

f ( x)  
 T x  1, 2  x  T
2 T

 A, 0  x  T
 2
  A, T  x  T
 2
(b) f ( x)  
 A, T  x  32T

  A, 32T  x  2T

33. (a)  x   0 for x  [0, 1) (b)  x   0 for x  (1, 0]

34.  x    x  only when x is an integer.

35. For any real number x, n  x  n  1, where n is an integer. Now: n  x  n  1   (n  1)   x   n.
By definition:   x   n and  x   n    x   n. So   x     x  for all real x.




© 2023 Pearson Education Ltd. All Rights Reserved.

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