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

Solution nmand nmAnswer nmGuide: nmStewart nmKokoska, nmCalculus: nmConcepts nmand nmContexts, nm5e, nm2024, n m9780357632499, nmChapter nm2: nmSection
nm Concept nmCheck




SOLUTION AND ANSWER GUIDE nm nm nm




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




Chapter 1-13 nm




CHAPTER 1: SECTION 1.1 NM NM NM




NM TABLE OF CONTENTS NM NM




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




END OF SECTION EXERCISE SOLUTIONS
NM NM NM NM




1.1.1

(a) f (1)  3
nm nm nm



(b) f (1)  0.2
nm nm nm



(c) f (x)  1 when x = 0 and x = 3.
nm nm nm n m nm nm nm nm nm nm nm



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




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




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




1.1.2
(a) f (4)  2;
nm g(3)  4 nm nm n m nm nm



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




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




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




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




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




1.1.3



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nmpublicly nmaccessible

website, nmin nmwhole nmor nmin nmpart.

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


12 nm 16
nm


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



(g) 2 nm
(h)  4 nm nm



f (2a)  12a2  2anm nm nm nm nm f (a2)  3a4  a2  nm nm nm nm nm nm


2 nm nm 2 nm




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

(i) nm nm nm nm nm nm nm nm nm nm nm nm nm nm nm




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




1.1.4

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

h h h h


1.1.5 

f (a  h)  f (a) a3  3a2h  3ah2  h3  h  3a  3ah  
2
h2 n m nm nm nm nm

  3a2  3ah  h2
nm nm nm nm nm nm


nm nm nm nm nm nm nm nm

3 nm nm nm nm nm

a nm



h h h


1.1.6

1 1 a x
 
nm n m

f (x)  f (a)
nm nm nm
a nm
 
1 
 
nm  
n m
nm
 x a  axax 
n m
n m
nm nm n m



 x n m
 
nm



xa nm xa xa
nm ax(x  a) ax nm nm nm nm nm nm




1.1.7

x  3 1 3 x  3 x  3  2x  x 1 x 1
f (x)  f (1) x 1 11 x 1  2
nm nm nm nm nm nm nm nm nm nm nm nm nm nm



  x 1 1
nm nm
2 x

nm nm nm nm nm nm
nm nm

 
nm nm
nm nm

x 1 1
n m


x 1 nm x 1 x 1    x 1 nm
nm
nm
nm nm
nm
nm


x 1 x x 1 nm nm
nm

1 nm




1.1.8

nm x  nm



x4 nm nm nm
x 9 2
nm nm


The domain nm f (x) nm
is
of
nm 
nm




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nmpublicly nmaccessible

website, nmin nmwhole nmor nmin nmpart.

,Solution nmand nmAnswer nmGuide: nmStewart nmKokoska, nmCalculus: nmConcepts nmand nmContexts, nm5e, nm2024, n m9780357632499, nmChapter nm2: nmSection
nm Concept nmCheck
| nmx 3,3.
nm

nm 

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


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




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website, nmin nmwhole nmor nmin nmpart.

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nm Concept nmCheck




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


of
nm 
nm 1

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



or n m ,
2.
nm




1.1.12

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


nm  nm




1.1.13

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


p) 
nm nm




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

The domain of nm nm n m nm nm is nm nm nm nm nm nm

1
1
u 1 nm




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


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


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


then to the left 2 units.
nm nm nm nm nm nm


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


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


four units to the left.
nm nm nm nm nm


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




1.1.16

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




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nmpublicly nmaccessible

website, nmin nmwhole nmor nmin nmpart.