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

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

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Institution
Calculus
Course
Calculus

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Solutionand Answer Guide: StewartKokoska,Calculus:Conceptsand Contexts,5e,2024, 9780357632499,Chapter2:Section Concept Check
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




SOLUTION AND ANSWER GUIDE ,l , l ,l




CALCULUS5THEDITIONJAMESSTEWART,KOKOSKA ,L ,L ,L ,L ,L




Chapter1-13 ,l




CHAPTER1:SECTION1.1 ,L ,L ,L




TABLE OF CONTENTS
,L ,L ,L




Endof SectionExercise Solutions................................................................................................................ 1
,l ,l ,l ,l




ENDOF SECTIONEXERCISE SOLUTIONS,L ,L ,L ,L




1.1.1

(a) f(1)=3 ,l ,l ,l




(b) f(−1)−0.2 ,l ,l ,l




(c) f (x) =1 when x = 0 and x = 3. ,l ,l ,l , l ,l ,l ,l ,l ,l ,l ,l




(d) f(x)=0whenx≈–0.8. ,l ,l ,l ,l ,l ,l ,l




(e) Thedomainoffis −2x4.Therangeoffis −1 y3. ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




(f) f isincreasingontheinterval−2 x1.
, l ,l ,l ,l ,l ,l ,l ,l ,l




1.1.2

(a) f(−4)=−2; g(3)=4 ,l ,l ,l , l ,l ,l




(b) f (x) = g(x)when x = –2 and x = 2. ,l ,l , l ,l ,l , l ,l ,l ,l , l ,l




(c) f(x)=−1 whenx≈ –3.4. ,l ,l ,l ,l ,l ,l ,l




(d) fisdecreasingontheinterval 0x4.
,l ,l ,l ,l ,l , l ,l ,l ,l ,l




(e) Thedomainoffis −4 x4.Therangeoffis −2 y3. ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




(f) Thedomainofgis −4x4.Therangeofgis 0.5 y4. ,l ,l ,l ,l , l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




1.1.3



©2024Cengage. AllRightsReserved. Maynotbescanned,copied orduplicated,orposted toapubliclyaccessible
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l

1
website,inwholeorin part. ,l ,l ,l ,l ,l

,Solutionand Answer Guide: StewartKokoska,Calculus:Conceptsand Contexts,5e,2024, 9780357632499,Chapter2:Section Concept Check
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




(a) f(2)=12 ,l ,l ,l



(b) f(2)=16 ,l ,l ,l



(c) f (a) =3a2 −a+2
,l ,l ,l ,l ,l ,l ,l




(d) f (−a)=3a2 +a+2
,l ,l ,l ,l ,l ,l ,l
(e) f(a+1)=3a2 +5a+4 ,l ,l ,l ,l ,l ,l ,l ,l (f) 2f (x)=6a2 −2a+4 ,l ,l ,l ,l ,l ,l ,l ,l




(g) f (2a) =12a2 −2a+2
,l ,l ,l ,l ,l ,l ,l
(h) f (a2) =3a4 −a2 +2
,l ,l ,l ,l ,l ,l ,l




(i) f(a) = 3a2−a+2 ( )
2 2
=9a4−6a3+13a2−4a+4
, l
, l




,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l
,l




(j) f (a+h) =3 ( a +h) −(a+h)+2 =3a2 +3h2 +6ah−a−h+2
2 , l




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,l




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)
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l , l




,l ,l ,l ,l




h h h h


1.1.5

f(a+h)− f(a) a3+3a2h+3ah2+h3−a3 h ( 3a2+3ah+h2 ) =3a2+3ah+h2
=
, l ,l ,l ,l ,l




=
,l ,l ,l ,l , l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l



,l ,l ,l ,l ,l




h h h


1.1.6

1 1 a x
− −
,l , l




f(x)− f(a) 1
a−x =−
  
,l , l




,l ,l , l ,l , l




= x a = ax ax , l



, l
,l ,l
,l , l




,



= l







x −a ,l ,l x −a x −a ax(x−a) ax ,l ,l ,l ,l ,l ,l




1.1.7

x+3 1+3 x +3 x+3−2x−2 −x +1 x−1
− −2
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




f(x)− f(1) x+1 1+1 = x+1 x+1 x + 1 = − x +1 = − 1
,l ,l




= = =
,l , l
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,l , l ,l ,l ,l

,l ,l , l ,l




x−1 ,l
x−1 x−1 ,l ,l
x−1 x −1 x−1 x+1 ,l ,l ,l ,l
,l




1.1.8
x +4
is x |x  −3,3.
,l ,l , l




Thedomain of f(x)= , l ,l




x −9
,l ,l ,l ,l ,l
,l ,l




2
,l ,l




1.1.9

2x3 −5
f (x) = isx |x−3,2.
,l ,l




Thedomain of
x2 +x−6
,l ,l ,l ,l ,l ,l ,l ,l
,l ,l




,l ,l ,l ,l




©2024Cengage. AllRightsReserved. Maynotbescanned,copied orduplicated,orposted toapubliclyaccessible
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l

2
website,inwholeorin part. ,l ,l ,l ,l ,l

,Solutionand Answer Guide: StewartKokoska,Calculus:Conceptsand Contexts,5e,2024, 9780357632499,Chapter2:Section Concept Check
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




1.1.10

Thedomain of ,l ,l
f(t)= ,l ,l
3
2t−1 isall realnumbers.
,l ,l ,l ,l ,l




1.1.11

g(t)=
,l ,l
,l
− isdefinedwhen3−t0t3and 2−t0t2.Thus,thedomainis t2,
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l , l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l , l ,l ,l




or (−,2.
, l ,l




1.1.12

1
The domain of h(x) = − ,l ,l , l ,l , l is (−,0)(5,).,l ,l
,l
,l




1.1.13

The domain of F( p)= ,l ,l , l ,l ,l 2− ,l p is0 p4. ,l , l ,l ,l




1.1.14
u+1
The domain of f (u)= isu |u  −2,−1.
,l




,l ,l , l ,l ,l ,l ,l ,l ,l ,l ,l




1
1+
u+1 ,l




1.1.15
(a) Thisfunction shifts thegraph of y= |x|down two units andtotheleft one unit.
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(b) Thisfunction shiftsthegraph of y= |x|down two units ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




(c) This function reflects the graph of y = |x|about the x-axis, shifts it up 3units and then totheleft2 units.
, l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




(d) Thisfunction reflectsthe graph of y= |x|about thex-axis andthen shifts it up4 units.
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




(e) This function reflects the graph of y = |x|about the x-axis, shifts it up 2unitsthen four units tothe left.
, l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




(f) Thisfunctionis aparabolathat opens upwith vertex at(0,5). It is not atransformation of y= |x|.
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l , l ,l ,l




1.1.16

(a) g(f( x ) ) =g( x 2 +1)=10 ( x 2 +1)
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(b) f(g(4))= f(10(4))=402+1=1601
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(c) g( g (−1))= g(10(−1))=10(−10)= −100
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,l




©2024Cengage. AllRightsReserved. Maynotbescanned,copied orduplicated,orposted toapubliclyaccessible
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l

3
website,inwholeorin part. ,l ,l ,l ,l ,l

, Solutionand Answer Guide: StewartKokoska,Calculus:Conceptsand Contexts,5e,2024, 9780357632499,Chapter2:Section Concept Check
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(d) (
f g (f(2)) = f g 22+1
,l ,l
,l
,l
) (( ,l
,l ,l ,l ,l
))= f(10(5))= f (50)=502+1=2501
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(e) 1 1 1 1
=  = =
f( g (x)) ,l ,l ,l
f (10x)
, l

(10x)
2 , l




+1 100x2 +1 , l ,l




1.1.17

The domain of h(x)=
,l ,l , l ,l 4−x2 is−2x2,andtherangeis
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l




0 y2.Thegraphisthetophalfofacircleofradius2withcenterat the origin.
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1.1.18

Thedomain of ,l ,l f(x)=1.6x−2.4 isall realnumbers.
,l ,l ,l ,l ,l , l ,l ,l ,l




1.1.19

t 2 −1
Thedomain of g(t) = ist |t  −1.
,l




t+1
,l ,l ,l ,l ,l ,l
, l




,l




1.1.20

x−1
f (x) =
,l




x −1 isx |x −1,1.
,l ,l




2
Thedomain of
, l




, l
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,l




©2024Cengage. AllRightsReserved. Maynotbescanned,copied orduplicated,orposted toapubliclyaccessible
,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l ,l

4
website,inwholeorin part. ,l ,l ,l ,l ,l

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Institution
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Course
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Uploaded on
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Number of pages
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Written in
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Type
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