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

Solution bnand bnAnswer bnGuide: bnStewart bnKokoska, bnCalculus: bnConcepts bnand bnContexts, bn5e, bn2024, bn 9780357632499, bnChapter bn2: bnSection
bn Concept bnCheck




SOLUTION AND ANSWER GUIDE bn bn bn




CALCULUS 5TH EDITION JAMES STEWART, KOKOSKA BN BN BN BN BN




Chapter 1-13 bn




CHAPTER 1: SECTION 1.1 BN BN BN




BN TABLE OF CONTENTS BN BN




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




END OF SECTION EXERCISE SOLUTIONS
BN BN BN BN




1.1.1

(a) f (1)  3
bn bn bn



(b) f (1)  0.2
bn bn bn



(c) f (x)  1 when x = 0 and x = 3.
bn bn bn bn bn bn bn bn bn bn bn



(d) f (x)  0 when x ≈ –0.8.
bn bn bn bn bn bn bn




(e) The domain of f isbn bn bn bn bn  2  bn bn x  4. The range of f is
bn bn bn bn bn bn bn bn 1  bn bn y  3.
bn bn




(f) f b n is increasing on the interval 2  x  1.
bn bn bn bn bn bn bn bn




1.1.2
(a) f (4)  2;
bn g(3)  4 bn bn b n bn bn



(b) f (x)  g(x) when x = –2 and x = 2.
bn bn bn bn bn bn bn bn bn bn bn




(c) f (x)  1 when x ≈ –3.4.
bn bn bn bn bn bn bn




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




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




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




1.1.3



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bnpublicly bnaccessible

website, bnin bnwhole bnor bnin bnpart.

,Solution bnand bnAnswer bnGuide: bnStewart bnKokoska, bnCalculus: bnConcepts bnand bnContexts, bn5e, bn2024, bn 9780357632499, bnChapter bn2: bnSection
bn Concept bnCheck



(a) f (2)  12 bn bn bn
(b) f (2)  bn bn
(c) f (a)  3a2  a  2
bn bn bn bn bn bn bn


16
bn


(d) f (a)  3a  a  2
bn bn bn
2
bn bn bn bn (e) f (a 1)  3a2  5a  bn bn bn bn bn bn bn (f) 2 f (x)  6a  2a  4
bn bn bn bn
2
bn bn bn bn



(g) f (2a)  12a  2a 
bn bn bn
2
bn bn bn
(h) 4 bn



2
bn f (a2)  3a4  a2  2 bn bn bn bn bn bn bn




(i)  f (a)  3a2  a  2
2 2
 9a4  6a3 13a2  4a  4
b n b n

bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn




(j) f (a  h)  3  a  h  a  h 2  3a2  3h2  6ah  a  h  2
2 b n
bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn




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)
bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn
bn bn bn bn

h h h h


1.1.5 

f (a  h)  f (a) a3  3a2h  3ah2  h3  h  3a2  3ah  h2
b n bn bn bn bn

  3a2  3ah  h2
bn bn bn bn bn bn


bn bn bn bn bn bn bn bn
bn bn bn bn bn

a3 bn



h h h


1.1.6

1 1 a x
 
bn b n

f (x)  f (a) 



a
bn
1 b n


   


bn bn bn bn


bn

x a  ax ax b n
bn
bn bn b n




b n x  
bn



xa bn bn xa xa bn ax(x  a)
bn ax bn bn bn bn




1.1.7

x  3 1 3 x 3 x  3  2x  2 x 1 x 1
 2
bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn


f (x)  f (1) x 1 11 x 1   x 1  1
bn bn


 x 1 x
bn bn bn bn


bn bn bn bn


 
bn
bn bn bn bn


x 1 1
bn


x 1 bn x 1 bn
x 1 bn
 x 1 bn bn
bn
bn


x x 1 bn
bn

1 bn




1.1.8

x 4
x | x  3,3.
bn bn bn

The domain bn f (x)bn
is bn
bn bn bn


of
bn 
bn
 bn


x2  9 bn bn




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bnpublicly bnaccessible

website, bnin bnwhole bnor bnin bnpart.

,Solution bnand bnAnswer bnGuide: bnStewart bnKokoska, bnCalculus: bnConcepts bnand bnContexts, bn5e, bn2024, bn 9780357632499, bnChapter bn2: bnSection
bn Concept bnCheck


1.1.9 
 
2x3  5
is  x | x  3, 2.
bn bn
The domain
bn f (x) 2
bn bn bn bn bn bn


 x  x
of 
bn bn bn
bn bn
bn
6 bn




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bnpublicly bnaccessible

website, bnin bnwhole bnor bnin bnpart.

, Solution bnand bnAnswer bnGuide: bnStewart bnKokoska, bnCalculus: bnConcepts bnand bnContexts, bn5e, bn2024, bn 9780357632499, bnChapter bn2: bnSection
bn Concept bnCheck




1.1.10 

3
The domain bn f (t)bn 2t
bn is all real numbers.
bn bn bn


of
bn 
bn 1

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



or b n ,
2.
bn




1.1.12

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


bn  bn




1.1.13

The domain of bn bn b n F( 2  p is 0  p  4.
bn bn bn bn bn


p) 
bn bn




1.1.14
u 1
f (u)  u  | u  2,1.
bn

The domain of bn bn b n bn bn is bn bn bn bn bn bn

1
1
u 1 bn




1.1.15
(a) This function shifts the graph of y = |x| down two units and to the left one unit.
bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn


(b) This function shifts the graph of y = |x| down two units
bn bn bn bn bn bn bn bn bn bn bn


(c) This function reflects the graph of y = |x| about the x-axis, shifts it up 3 units and then to
bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn


the left 2 units.
bn bn bn bn


(d) This function reflects the graph of y = |x| about the x-axis and then shifts it up 4 units.
bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn


(e) This function reflects the graph of y = |x| about the x-axis, shifts it up 2 units then four
bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn


units to the left.
bn bn bn bn


(f) This function is a parabola that opens up with vertex at (0, 5). It is not a transformation of
bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn bn b n y = |x|.
bn bn




1.1.16

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




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bnpublicly bnaccessible

website, bnin bnwhole bnor bnin bnpart.