Solution Manual For Calculus 5th Edition by James Stewart, Kokoska Chapter 1-13

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check



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



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website, in whole or in part.

,Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check


(a) f (2)  12 (b) f (2)  16 (c) f (a)  3a  a  2
2


(d) f (a)  3a 2  a  2 (e) f (a  1)  3a 2  5a  4 (f) 2 f ( x)  6a 2  2a  4
(g) f (2a)  12a 2  2a  2 (h) f (a 2 )  3a 4  a 2  2


(i)  f (a)  3a 2  a  2 
2
 9a 4  6a3  13a 2  4a  4
2



(j) f (a  h)  3  a  h    a  h   2  3a 2  3h 2  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  h 2
    (3  h)
h h h h


1.1.5

f (a  h)  f (a) a  3a h  3ah  h  a
3 2 2 3 3 h  3a 2  3ah  h 2 
   3a 2  3ah  h 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  x 1   x 1  
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

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



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website, in whole or in part.

,Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check




1.1.10

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


1.1.11

g  t   3  t  2  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,   .
4
x  5x
2



1.1.13

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

1.1.14

u 1
The domain of f (u )  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



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website, in whole or in part.

, Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check

(d)    
f g  f  2    f g  22  1  f 10  5   f  50   502  1  2501

1 1 1 1
(e)   
f  g  x   f 10 x  10 x   1 100 x 2  1
2



1.1.17

The domain of h( x)  4  x is 2  x  2 , and the range is
2


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)  1.6 x  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 2  1 is  x  | x  1,1 .




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website, in whole or in part.