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

Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




SOLUTION AND ANSWER GUIDE c4 c4 c4




CALCULUS5THEDITION JAMESSTEWART,KOKOSKA 4
C C
4 C4 C4 4
C




Chapter 1-13 c4




CHAPTER1:SECTION 1.1 4
C 4
C C4




C4 TABLE OF CONTENTS C4 C4




End of Section Exercise Solutions ............................................................................................................... 1
c4 c4 c4 c4




END OF SECTION EXERCISE SOLUTIONS
C4 C4 C4 C4




1.1.1

(a) f(1)=3 c4 c4 c4




(b) f(−1)−0.2 c4 c4 c4




(c) f (x) =1 when x = 0 and x = 3. c4 c4 c4 c 4 c4 c4 c4 c4 c4 c4 c4




(d) f(x)=0when x ≈ –0.8. c4 c4 c4 c4 c4 c4 c4




(e) Thedomainoffis −2 x4.Therangeoff is −1 y3. c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




(f) f is increasing on the interval−2 x1.
c 4 c4 c4 c4 c4 c4 c4 c4 c4




1.1.2

(a) f(−4) =−2; g(3)=4 c4 c4 c4 c 4 c4 c4




(b) f (x) = g(x)when x = –2 and x = 2. c4 c4 c 4 c4 c4 c4 c4 c4 c4 c4 c4




(c) f(x)=−1 when x ≈ –3.4. c4 c4 c4 c4 c4 c4 c4




(d) fisdecreasing on theinterval 0 x4.
c4 c4 c4 c4 c4 c 4 c4 c4 c4 c4




(e) Thedomainof fis −4 x4.Therangeof fis −2 y3. c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




(f) Thedomainof gis −4 x4.Therangeofgis 0.5 y4. c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




1.1.3



© 2024 Cengage. All Rights Reserved. Maynot be scanned, copied or duplicated, or posted to apublicly accessible
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4

1
website, in whole or in part. c4 c4 c4 c4 c4

,Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




(a) f (2)=12 c4 c4 c4



(b) f (2)=16 c4 c4 c4



(c) f (a) =3a2 −a+2
c4 c4 c4 c4 c4 c4 c4




(d) f (−a) =3a2 +a+2
c4 c4 c4 c4 c4 c4 c4
(e) f (a+1)=3a2 +5a+4 c4 c4 c4 c4 c4 c4 c4 c4 (f) 2f (x) = 6a2 −2a+4 c4 c4 c4 c4 c4 c4 c4 c4




(g) f (2a) =12a2 −2a+2
c4 c4 c4 c4 c4 c4 c4
(h) f (a2) =3a4 −a2 +2
c4 c4 c4 c4 c4 c4 c4




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




c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4
c4




(j) f (a+h) = 3 ( a +h) −(a+h)+2 = 3a2 +3h2 +6ah−a−h+2
2 c 4




c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c 4 c4 c4 c4 c4 c4 c4 c4 c4 c4
c4




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)
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c 4




c4 c4 c4 c4




h h h h


1.1.5

f(a+h)− f(a) a3+3a2h+3ah2 +h3 −a3 h ( 3a2 +3ah+h2 ) =3a2+3ah+h2
=
c 4 c4 c4 c4 c4




=
c4 c4 c4 c4 c 4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4



c4 c4 c4 c4 c4




h h h


1.1.6

1 1 a x
− −
c4 c 4




f(x)− f(a) 1
a−x =−
  
c4 c 4




c4 c4 c 4 c4 c4




= x a = ax ax c 4



c 4
c4 c4
c4 c 4




c



= 4







x −a c4 c4 x −a x −a ax(x−a) ax
c4 c4 c4 c4 c4 c4




1.1.7

x+3 1+3 x +3 x+3−2x−2 −x +1 x−1
− −2
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




f(x)− f(1) x+1 1+1 x+1 x+1 x +1 = − x +1 = − 1
c4 c4




= = =
c4 c 4




=
c4 c4 c4 c4 c4 c4
c4 c4 c4

c4 c4 c 4 c4




x−1 x−1 x−1 x−1 x−1 x+1
c 4




c4


x−1 c4
c4
c4 c4 c4 c4
c4




1.1.8
x +4
is x | x  −3,3.
c4 c4 c 4




The domain of f (x) = c 4 c4




x2 −9
c4 c4 c4 c4 c4
c4 c4




c4 c4




1.1.9

2x3 −5
The domain of f (x) = isx |x−3,2.
c4 c4




x 2 +x −6
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




c4 c4 c4 c4




© 2024 Cengage. All Rights Reserved. Maynot be scanned, copied or duplicated, or posted to apublicly accessible
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4

2
website, in whole or in part. c4 c4 c4 c4 c4

,Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




1.1.10

The domain of f(t)= c4 c4
c4 c4
3
c4 2t−1 is all real numbers.
c4 c4 c4 c4




1.1.11

g(t)=
c4 c4
c4
− isdefinedwhen 3−t0t3and 2−t0t 2.Thus,thedomainis t2,
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c 4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c 4 c4 c4




or c 4
(−,
2.
c4




1.1.12

1
The domain of h(x) =− c4 c4 c 4 c4 c 4 is (−,0) (5,). c4 c4
c4
c4




1.1.13

The domain of F( p) = c4 c4 c 4 c4 c4 2− c4 p is0 p4. c4 c 4 c4 c4




1.1.14
u+1
The domain of f (u) = isu |u  −2,−1.
c4




c4 c4 c 4 c4 c4 c4 c4 c4 c4 c4 c4




1
1+
u+1 c4




1.1.15
(a) Thisfunction shifts the graph of y = |x| down two units and to the left one unit.
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




(b) This function shifts the graph of y = |x|down two units
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




(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.
c 4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




(d) This function reflects the graph of y = |x| about the x-axis and then shifts it up 4 units.
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




(e) This function reflects the graph of y = |x| about the x-axis, shifts it up 2 units then four units to the left.
c 4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




(f) This function is a parabola that opens up with vertex at (0, 5). It is not a transformation of y = |x|.
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c 4 c4 c4




1.1.16

(a) g(f( x ) ) =g( x 2 +1)=10 ( x 2 +1)
c4
c4
c4
c4
c4 c4 c4
c4
c4




(b) f(g(4))= f (10(4))=402 +1=1601
c4 c4
c4
c4 c4
c4
c4 c4 c4 c4




(c) g( g (−1))= g(10(−1))=10(−10)= −100
c4 c4
c4
c 4 c4
c4
c4
c4
c4




© 2024 Cengage. All Rights Reserved. Maynot be scanned, copied or duplicated, or posted to apublicly accessible
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4

3
website, in whole or in part. c4 c4 c4 c4 c4

, Solution and Answer Guide: Stewart Kokoska, Calculus: Concepts and Contexts, 5e, 2024, 9780357632499, Chapter 2: Section Concept Check
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




(d) (
f g (f(2)) = f g 22+1
c4 c4
c4
c4
) (( c4
c4 c4 c4 c4
))= f(10(5))= f (50)=502+1=2501
c4
c4 c4
c4
c 4 c4
c4
c4 c4 c4 c4




(e) 1 1 1 1
=  = =
f ( g (x)) f (10x) (10x) +1 100x +1
2 c 4
2
c 4 c4

c 4

c4 c4 c4




1.1.17

The domain of h(x)=
c4 c4 c 4 c4 4−x2 is−2x2,andtherangeis
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4




0 y2.Thegraphisthetophalfofacircleofradius2with centerat the origin.
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4
c4




1.1.18

The domain of f(x)=1.6x−2.4 is all real numbers.
c4 c4 c4 c4 c4 c4 c4 c 4 c4 c4 c4




1.1.19

t 2 −1
The domain of g(t) = ist  |t  −1.
c4




t+1
c4 c4 c4 c4 c4 c4
c 4




c4




1.1.20

x−1
f (x) =
c4




x −1 isx | x  −1,1.
c4 c4




2
The domain of
c 4




c 4
c4 c4 c4 c4 c4





c4




© 2024 Cengage. All Rights Reserved. Maynot be scanned, copied or duplicated, or posted to apublicly accessible
c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4 c4

4
website, in whole or in part. c4 c4 c4 c4 c4