Solutions and Answer Guide for Calculus 5th Edition by James Stewart, Kokoska

SOLUTION AND ANSWER GUIDE
CALCULUS 5TH EDITION JAMES STEWART, KOKOSKA
Chapter 1-13



CHAPTER 1: SECTION 1.1

TABLE OF CONTENTS

End of Section Exercise Solutions .............................................................................................................. 1




END OF SECTION EXERCISE SOLUTIONS
1.1.1
(a) f (1)  3
(b) f (1)  0.2
(c) f (x)  1 when x = 0 and x = 3.
(d) f (x)  0 when x ≈ –0.8.
(e) The domain of f is 2  x  4. The range of f is 1  y  3.
(f) f is increasing on the interval2  x  1.

1.1.2
(a) f (4)  2; g(3)  4
(b) f (x)  g(x) when x = –2 and x = 2.

(c) f (x)  1 when x ≈ –3.4.

(d) f is decreasing on the interval 0  x  4.

(e) The domain of f is 4  x  4. The range of f is 2  y  3.

(f) The domain of g is 4  x  4. The range of g is 0.5  y  4.



1.1.3




1

, (a) f (2)  12 (b) f (2)  16 (c) f (a)  3a2  a  2
(d) f (a)  3a2  a  2 (e) f (a 1)  3a2  5a  4 (f) 2 f (x)  6a2  2a  4
(g) f (2a)  12a2  2a  2 (h) f (a2)  3a4  a2  2
 f (a)2  3a2  a  2
2
(i)  9a4  6a3 13a2  4a  4
(j) f (a  h)  3  a  h  a  h  2  3a2  3h2  6ah  a  h  2
2




1.1.4

f (3  h)  f (3) (4  3(3  h)  (3  h)2 )  4 9  3h  9  6h  h2) 3h  h2 
     (3  h)
h h h h


1.1.5 

f (a  h)  f (a)  a3  3a2h  3ah2  h3  a3  h 3a  3ah  h  3a2  3ah  h2
2 2


h h h


1.1.6

1 1 a x
 
f (x)  f (a)  x a ax ax  a  x   1

   




 
   




xa xa xa ax(x  a) ax



1.1.7

x  3 1 3 x  3 x  3  2x  2 x 1 x 1
 2
f (x)  f (1)  x 1 11  x 1 x 1   1
  x 1  x 1  
x 1 x 1 x 1 x 1 x 1 x 1 x 1


1.1.8

x4
The domain of f (x)  is x  | x  3,3.
x2  9

1.1.9 
 
2x3  5
The domain of f (x)  is x  | x  3, 2.
x2  x  6



2

,1.1.10 

The domain of f (t)  3 2t 1 is all real numbers.


1.1.11 

g t    is defined when 3  t  0  t  3 and 2  t  0  t  2. Thus, the domain is t  2,
or , 2.

1.1.12

1
The domain of h(x)  is , 0 5, .


1.1.13

The domain of F( p)  2  p is0  p  4.

1.1.14

The domain of f (u)  u 1 is u  | u  2, 1.
1
1
u 1
1.1.15
(a) This function shifts the graph of y = |x| down two units and to the left one unit.
(b) This function shifts the graph of y = |x| down two units
(c) This function reflects the graph of y = |x| about the x-axis, shifts it up 3 units and then to the left 2
units.
(d) This function reflects the graph of y = |x| about the x-axis and then shifts it up 4 units.
(e) This function reflects the graph of y = |x| about the x-axis, shifts it up 2 units then four units to the
left.
(f) This function is a parabola that opens up with vertex at (0, 5). It is not a transformation of y = |x|.


1.1.16

(a) g  f  x    g  x 2 1 10  x 2 1

(b) f g 4  f 104  402 1  1601

(c) g  g 1  g 101  1010  100



3

, (d)    
f g  f 2  f g 22 1   f 105  f 50  502 1  2501
(e) 1 1 1 1

    
f  g  x f 10x 10x
2
1 100x2  1

1.1.17

The domain of h(x)  4  x2 is 2  x  2, and the range is
0  y  2. The graph is the top half of a circle of radius 2 with center at
the origin.


1.1.18
The domain of f (x)  1.6x  2.4 is all real numbers.




1.1.19 
 
t 2 1
The domain of g(t)  ist  | t  1.
t 1




1.1.20 

x 1 
f (x) 
The domain of x 1 isx  | x  1,1.
2




4