Solutions Manual for Precalculus Mathematics for Calculus 7th Edition Stewart

SOLUTIONS MANUAL FOR
PRECALCULUS MATHEMATICS
FOR CALCULUS 7TH EDITION
STEWART

,Precalculus Mathematics for Calculus 7th Edition Stewart Solutions Manual




2 FUNCTIONS

2.1 FUNCTIONS
1. If f x  x 3  1, then
(a) the value of f at x  1 is f 1  13  1  0.
(b) the value of f at x  2 is f 2  23  1  9.
(c) the net change in the value of f between x  1 and x  2 is f 2  f 1  9  0  9.

2. For a function f , the set of all possible inputs is called the domain of f , and the set of all possible outputs is called the
range of f .

x 5
3. (a) f x  x 2  3x and g x  have 5 in their domain because they are defined when x  5. However,
x
  
h x  x  10 is undefined when x  5 because 5  10  5, so 5 is not in the domain of h.
55 0
(b) f 5  52  3 5  25  15  10 and g 5    0.
5 5

4. (a) Verbal: “Subtract 4, then square and add 3.”
(b) Numerical:
x f x
0 19
2 7
4 3
6 7

5. A function f is a rule that assigns to each element x in a set A exactly one element called f x in a set B. Table (i) defines
y as a function of x, but table (ii) does not, because f 1 is not uniquely defined.

6. (a) Yes, it is possible that f 1  f 2  5. [For instance, let f x  5 for all x.]
(b) No, it is not possible to have f 1  5 and f 1  6. A function assigns each value of x in its domain exactly one
value of f x.

7. Multiplying x by 3 gives 3x, then subtracting 5 gives f x  3x  5.

8. Squaring x gives x 2 , then adding two gives f x  x 2  2.

9. Subtracting 1 gives x  1, then squaring gives f x  x  12 .

 x 1
10. Adding 1 gives x  1, taking the square root gives x  1, then dividing by 6 gives f x  .
6
x 2
11. f x  2x  3: Multiply by 2, then add 3. 12. g x  : Add 2, then divide by 3.
3

x2  4
13. h x  5 x  1: Add 1, then multiply by 5. 14. k x  : Square, then subtract 4, then divide by 3.
3
141




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,142 CHAPTER 2 Functions

 3
15. Machine diagram for f x  x  1. 16. Machine diagram for f x  .
x 2
subtract 1,
subtract 2,
1 then take 0
square root 3 take reciprocal, 3
multiply by 3

subtract 1,
subtract 2,
2 then take 1
square root _1 take reciprocal, _1
multiply by 3

subtract 1,
subtract 2,
5 then take 2
square root 1 take reciprocal, _3
multiply by 3



17. f x  2 x  12 18. g x  2x  3

x f x x g x
1 2 1  12  8 3 2 3  3  3
0 2 12  2 2 2 2  3  1
1 2 1  12  0 0 2 0  3  3
2 2 2  12  2 1 2 1  3  5
3 2 3  12  8 3 2 3  3  9

19. f x  x 2  6; f 3  32  6  9  6  3; f 3  32  6  9  6  3; f 0  02  6  6;
   2
f 12  12  6  14  6   23 4.


20. f x  x 3  2x; f 2  23  2 2  8  4  12; f 1  13  2 1  1  2  3;
   3  
f 0  03  2 0  0; f 12  12  2 12  18  1  98 .

 
1  2x 1  2 2 1  2 2 5   1  2 12 1  2a
21. f x  ; f 2   1; f 2   ; f 1   0; f a  ;
3 3 3 3 2 3 3
1  2 a 1  2a 1  2 a  1 3  2a
f a   ; f a  1   .
3 3 3 3

x2  4 22  4 8 22  4 8 a2  4 x2  4 x2  4
22. h x  ; h 2   ; h 2   ; h a  ; h x   ;
5 5 5 5 5 5 5 5
 2
a  22  4 a 2  4a  8   x 4 x 4
h a  2   ;h x   .
5 5 5 5

23. f x  x 2  2x; f 0  02  2 0  0; f 3  32  2 3  9  6  15; f 3  32  2 3  9  6  3;
   2  
1 1 1 1 2
f a  a 2  2 a  a 2  2a; f x  x2  2 x  x 2  2x; f  2  2  .
a a a a a

1  
24. h x  x  1  1  1  2; h 2  2  1  5 ; h 1  1  1  1  2  5 ;
; h 1  1  1
x 2 2 2 2 1 2 2
2
 
1 1 1 1 1
h x  1  x  1  ;h     x.
x 1 x x 1 x
x

, SECTION 2.1 Functions 143
 
  1 1 1
1x 1  2 1 1 1  1 1 2 1
25. g x  ; g 2     ; g 1  , which is undefined; g     2  ;
1x 1  2 3 3 1  1 2 1 3 3
1 2 2
 
1  a 1a 1  a  1 1a1 2a  2  1  x2  1 2  x2
g a   ; g a  1    ;g x 1     .
1  a 1a 1  a  1 1a1 a 1  x2  1 x2
t 2 2  2 22 02 a2
26. g t  ; g 2   0; g 2  , which is undefined; g 0   1; g a  ;
t 2 2  2 22 02 a2
  a2  2  2 a2 a12 a3
g a2  2  2  2 ; g a  1   .
a 22 a 4 a12 a1
27. k x  x 2  2x  3; k 0  02  2 0  3  3; k 2  22  2 2  3  5; k 2   22  2 2  3  3;
   2   
k 2   2 2 2  3  1  2 2; k a  2   a  22  2 a  2  3  a 2  6a  5;
   2  
k x   x2  2 x  3  x 2  2x  3; k x 2   x 2  2 x 2  3  x 4  2x 2  3.

28. k x  2x 3  3x 2 ; k 0  2 03  3 02  0; k 3  2 33  3 32  27; k 3  2 33  3 32  81;
   3  2    3  2 a 3  3a 2
k 12  2 12  3 12   12 ; k a2  2 a2  3 a2  ; k x  2 x3  3 x2  2x 3  3x 2 ;
4
   3  2
k x 3  2 x 3  3 x 3  2x 9  3x 6 .
29. f x  2 x  1; f 2  2 2  1  2 3  6; f 0  2 0  1  2 1  2;
     
 
f 12  2  12  1  2 12  1; f 2  2 2  1  2 1  2; f x  1  2 x  1  1  2 x;
      
   
f x 2  2  2  x 2  2  1  2 x 2  1  2x 2  2 (since x 2  1  0 ).
x 2 2 1 1
30. f x  ; f 2    1; f 1    1; f x is not defined at x  0;
x 2 2 1 1
 
5 5   x 2  x2
 
1 1x x
f 5    1; f x 2  2  2  1 since x 2  0, x  0; f   .
5 5 x x x 1x x
31. Since 2  0, we have f 2  22  4. Since 1  0, we have f 1  12  1. Since 0  0, we have
f 0  0  1  1. Since 1  0, we have f 1  1  1  2. Since 2  0, we have f 2  2  1  3.
32. Since 3  2, we have f 3  5. Since 0  2, we have f 0  5. Since 2  2, we have f 2  5. Since 3  2, we
have f 3  2 3  3  3. Since 5  2, we have f 5  2 5  3  7.
33. Since 4  1, we have f 4  42  2 4  16  8  8. Since  32  1, we have
   2  
f  32   32  2  32  94  3   34 . Since 1  1, we have f 1  12  2 1  1  2  1. Since
1  0  1, we have f 0  0. Since 25  1, we have f 25  1.
34. Since 5  0, we have f 5  3 5  15. Since 0  0  2, we have f 0  0  1  1. Since 0  1  2, we have
f 1  1  1  2. Since 0  2  2, we have f 2  2  1  3. Since 5  2, we have f 5  5  22  9.
35. f x  2  x  22  1  x 2  4x  4  1  x 2  4x  5; f x  f 2  x 2  1  22  1  x 2  1  4  1  x 2  6.
36. f 2x  3 2x  1  6x  1; 2 f x  2 3x  1  6x  2.
   2
37. f x 2  x 2  4; f x  [x  4]2  x 2  8x  16.
x  x  f x 6x  18 3 2x  6
38. f 6  18  2x  18;    2x  6
3 3 3 3 3
39. f x  3x  2, so f 1  3 1  2  1 and f 5  3 5  2  13. Thus, the net change is f 5  f 1  13  1  12.
40. f x  4  5x, so f 3  4  5 3  11 and f 5  4  5 5  21. Thus, the net change is
f 5  f 3  21  11  10.