Solution Manual for Calculus 5th Edition by Stewart Kokoska – Chapter 1 Exercises & Answers

SOLUTION AND ANSWER GUIDE . . .



CALCULUS5TH EDITION JAMES STEWART,KOKOSKA
. . . . .


Chapter.1-13



CHAPTER1:SECTION1.1 TAB. . . .




LE 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

, (c). . f.(a).=.3a −.a.+.2
2
(a). f.(2).=.12 (b). . f.(2).=.16 .


(d) f.(−a).=.3a2.+.a.+.2 (e) f.(a.+1).=.3a2.+.5a.+.4 (f) 2.f.(x).=.6a2. −.2a.+.4
(g) f.(2a).=.12a2. −.2a.+.2 (h) f.(a2).=.3a4. −.a2. +.2

(
(i) .f.(a) =. 3a2.−.a.+.2 )
2. 2.
=.9a4.−.6a3.+13a2.−.4a.+.4
(j) f. (a.+.h).=.3(a.+.h) −.(a.+.h)+
2.
. .2.=.3a . +.3h . +.6ah.−.a.−.h.+.2
2 2




1.1.4

f.(3.+.h).−.f.(3). (4.+.3(3.+.h).−.(3.+.h)2.).−.4. 9.+.3h.−.9.−.6h.−.h2). −3h.−.h2. h h
=. =. =. =.−.(3.+.h).
h h


1.1.5
f (a.+.h).−. f.(a) a3.+.3a2h.+.3ah2.+.h3.−.a3 h(3a2.+.3ah.+.h2.).
=. = . 3a 2 + 3ah + h2
. .

=
h h h


1.1.6
1 1 a x 1
f.(x).−. f.(a) x −a
.
. . .
  




ax − ax a.−.x =.−.
=  . = =
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

2x3.−.5
The.domain.of. f.(x).= is.x. |.x..−3,.2.
x2.+.x.−.6

,1.1.10

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


1.1.11

g.(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,.).


1.1.13

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

1.1.14

The.domain.of. f.(u).= u.+1 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.(x2. +1)=. 10(x2. +1)

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

(c) g.(g.(−1)).=. g.(10(−1)).=.10(−10).=.−100

, (d) (
f. g.(.f.(2)) = ) ((
. .f. g. 2 .+1
2
))= f (10(5))= f (50)=502 +1= 2501
. . . . . . . . . . .




(e) 1 1 1 1
= 
= =
f.(g.(.x)) f. (10x) (10x) 100x2. +.1
2.
+1

1.1.17

The.domain.of. h(x).= 4.−.x2 is.−2..x..2,.and.the.range.is
0  y  2.The graph is the top half of a circle of radius 2 with center at the orig
. . . . . . . . . . . . . . . . . . . . .


in.


1.1.18

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




1.1.19

t2. −1
The.domain.of. g(t). = ist.. |.t. .−1.
t.+1




1.1.20
x.−1
f. (x).=.
The.domain.of x2. −1. isx. |.x..−1,1.