Chapter4 2.pdf微積分。。。。。。。。。。。。。。。。。。。。。。。。。

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118 SECTION 4.1
CHAPTER 4
SECTION 4.1
1.fis diﬀerentiable on (0,1),continuous on [0, 1]; andf(0) =f(1) = 0.
f
ﬀ
(c)=3c
2
−1; 3c
2
−1=0=⇒c=
√
3
3
ﬀ
−
√
3
3
/∈(0,1)
ﬃ
2.fis diﬀerentiable on (−2,2),continuous on [−2,2]; andf(−2) =f(2) = 0.
f
ﬀ
(c)=4c
3
−4c;4c(c
2
−1)=0=⇒c=0,±1
3.fis diﬀerentiable on (0,2π),continuous on [0,2π]; andf(0) =f(2π)=0.
f
ﬀ
(c) = 2 cos 2c; 2 cos 2c=0=⇒2c=
π
2
+nπ,andc=
π
4
+
nπ
2
,n=0,±1,±2...
Thus,c=
π
4
,
3π
4
,
5π
4
,
7π
4
4.fis diﬀerentiable on (0,8),continuous on [0,8]; andf(0) =f(8) = 0.
f
ﬀ
(c)=
2
3
c
−1/3
−
2
3
c
−2/3
=
2
3
c
1/3
−1
c
2/3
f
ﬀ
(c)=0 =⇒c=1.
5.f
ﬀ
(c)=2c,
f(b)−f(a)
b−a
=
4−1
2−1
=3; 2 c=3 =⇒c=3/2
6.f
ﬀ
(c)=
3
2
√
c
−4,
f(b)−f(a)
b−a
=
−10−(−1)
4−1
=−3;
3
2
√
c
−4=−3=⇒c=9/4
7.f
ﬀ
(c)=3c
2
,
f(b)−f(a)
b−a
=
27−1
3−1
= 13; 3c
2
=13 =⇒c=
1
3
√
39
Δ
−
1
3
√
39 is not in [a, b]
θ
8.f
ﬀ
(c)=
2
3
c
−1/3
,
f(b)−f(a)
b−a
=
4−1
8−1
=
3
7
;
2
3
c
−1/3
=
3
7
=⇒c=
(14)
3
9
3
9.f
ﬀ
(c)=
−c
√
1−c
2
,
f(b)−f(a)
b−a
=
0−1
1−0
=−1;
−c
√
1−c
2
=−1=⇒c=
1
2
√
2
(−
1
2
√
2 is not in [a, b])
10.f
ﬀ
(c)=3c
2
−3,
f(b)−f(a)
b−a
=
−2−2
1−(−1)
=−2; 3c
2
−3=−2=⇒c=±
√
3
3
11.fis continuous on [−1,1], diﬀerentiable on (−1,1) andf(−1) =f(1) = 0.
f
ﬀ
(x)=
−x(5−x
2
)
(3 +x
2
)
2
√
1−x
2
,f
ﬀ
(c) = 0 forcin (−1,1) impliesc=0.
12.(a)f
ﬀ
(x)=
2
3
x
−1/3
=
2
3x
1/3
δ= 0 for allx∈(−1,1).
(b)f
ﬀ
(0) does not exist. Therefore,fis not diﬀerentiable on (−1,1).

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SECTION 4.1 119
13.No. By the mean-value theorem there exists at least one numberc∈(0,2) such that
f
ﬀ
(c)=
f(2)−f(0)
2−0
=
3
2
>1.
14.No, by Rolle’s theorem:f(2) =f(3) = 1 but there is no valuec∈(2,3) such that
f
ﬀ
(c)=0.
15.By the mean-value theorem there is a numberc∈(2,6) such that
f(6)−f(2) =f
ﬀ
(c)(6−2) =f
ﬀ
(c)4.
Since 1≤f
ﬀ
(x)≤3 for allx∈(2,6), it follows that
4≤f(6)−f(2)≤12.
16.f(x)=x
2
+x+3,f
ﬀ
(x)=2x+1.
The slope of the line through (−1,3) and (2,9) is 2. Setting 2x+ 1 = 2, we getx=
1
2
. The point on
the graph offwhere the tangent line is parallel to the line through (−1,3) and (2,9) is: (1/2,15/4).
17.fis everywhere continuous and everywhere diﬀerentiable except possibly atx=−1.
fis continuous atx=−1: as you can check,
lim
x→−1
−
f(x)=0,lim
x→−1
+
f(x)=0,andf(−1) = 0.
fis diﬀerentiable atx=−1 andf
ﬀ
(−1) = 2: as you can check,
lim
h→0
−
f(−1+h)−f(−1)
h
= 2 and lim
h→0
+
f(−1+h)−f(−1)
h
=2.
Thusfsatisﬁes the conditions of the mean-value theorem on every closed interval [a, b].
f
ﬀ
(x)=
δ
2,x≤−1
3x
2
−1,x>−1
f(2)−f(−3)
2−(−3)
=
6−(−4)
2−(−3)
=2.
f
ﬀ
(c) = 2 withc∈(−3,2) iﬀc=1 or−3<c≤−1.
18.fis continuous and diﬀerentiable everywhere;
f
ﬀ
(x)=
δ
3x
2
,x≤1
3,x>1
f(2)−f(−1)
2−(−1)
=
6−1
3
=
5
3
Forc≤1,f
ﬀ
(c)=3c
2
=
5
3
=⇒c=±
√
5
3
Forc>1,f
ﬀ
(c)=3δ=
5
3

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120 SECTION 4.1
19.Letf(x)=Ax
2
+Bx+C.Thenf
ﬀ
(x)=2Ax+B.By the mean-value theorem
f
ﬀ
(c)=
f(b)−f(a)
b−a
=
(Ab
2
+Bb+C)−(Aa
2
+Ba+C)
b−a
=
A(b
2
−a
2
)+B(b−a)
b−a
=A(b+a)+B
Therefore, we have
2Ac+B=A(b+a)+B=⇒c=
a+b
2
20.
f(1)−f(−1)
1−(−1)
= 1 andf
ﬀ
(x)=−1/x
2
<0;fis not continuous at 0.
21.
f(1)−f(−1)
1−(−1)
= 0 and f
ﬀ
(x) is never zero. This result does not violate the mean-value theorem
sincefis not diﬀerentiable at 0; the theorem does not apply.
22.f(x)=
δ
2x−4,x≥1/2
−2x−2,x<1/2
f
ﬀ
(x)=
δ
2,x>1/2
−2,x<1/2
f
ﬀ
(x)δ= 0 for allxδ=
1
2
.
Rolle’s theorem is not violated becausefis
not diﬀerentiable atx=
1
2
.
23.SetP(x)=6x
4
−7x+ 1. If there existed three numbersa<b<cat whichP(x) = 0, then by Rolle’s
theoremP
ﬀ
(x) would have to be zero for somexin (a, b) and also for somexin (b, c).This is not the
case:P
ﬀ
(x)=24x
3
−7 is zero only atx=(7/24)
1/3
.
24.SetP(x)=6x
5
+13x+1.Note thatP(−1)<0 andP(0)>0.By the intermediate-value theorem,
the equationP(x) = 0 has at least one real rootc. If this equation had another real rootd,then
by Rolle’s theoremP
ﬀ
(x) would have to be zero for somexbetweencandd.This is not the case:
P
ﬀ
(x)=30x
4
+ 13 is never zero.
25.SetP(x)=x
3
+9x
2
+33x−8. Note thatP(0)<0 andP(1)>0. Thus, by the intermediate-value
theorem, there exists some numbercbetween 0 and 1 at whichP(x) = 0. If the equationP(x)=0
had an additional real root, then by Rolle’s theorem there would have to be some real number at
whichP
ﬀ
(x) = 0. This is not the case:P
ﬀ
(x)=3x
2
+18x+ 33 is never zero since the discriminant
b
2
−4ac= (18)
2
−12(33)<0.
26.(a) Suppose thatfhas two zeros,x1,x2∈(a, b).Then,fis diﬀerentiable on (x1,x2) and continuous
on [x1,x2].By Rolle’s theorem,f
ﬀ
has a zero in (x1,x2) which contradicts the hypothesis.
(b) Iffhad three zeros in (a, b),then, by Rolle’s’ theorem,f
ﬀ
would have at least two zeros
in (a, b) andf
ﬀﬀ
would have at least one zero in (a, b) which contradicts the hypothesis.

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SECTION 4.1 121
27.Letcanddbe two consecutive roots of the equationP
ﬀ
(x) = 0. The equationP(x) = 0 cannot have
two or more roots betweencanddfor then, by Rolle’s theorem,P
ﬀ
(x) would have to be zero somewhere
between these two roots and thus betweencandd. In this casecanddwould no longer be consecutive
roots ofP
ﬀ
(x)=0.
28.Iff(x)=0ata1,a2, ...,anthen by Rolle’s theorem,f
ﬀ
(x) is zero at some numberb1∈(a1,a2),at
some numberb2∈(a2,a3), ...,at some numberbn−1∈(an−1,an);f
ﬀﬀ
(x),in turn, must be zero at
some numberc1∈(b1,b2),at some numberb2∈(b2,b3),...,at some numbercn−2∈(bn−2,bn−1).
29.Suppose thatfhas two ﬁxed pointsa, b∈I,witha<b.Letg(x)=f(x)−x.Theng(a)=f(a)−a=
0 andg(b)=f(b)−b=0.Sincefis diﬀerentiable onI,we can conclude thatgis diﬀerentiable on
(a, b) and continuous on [a, b].By Rolle’s theorem, there exists a numberc∈(a, b) such thatg
ﬀ
(c)=
f
ﬀ
(c)−1=0 orf
ﬀ
(c)=1.This contradicts the assumption thatf
ﬀ
(x)<1onI.
30.SetP(x)=x
3
+ax+b.It is obvious that forxsuﬃciently large,P(x)>0 and forxsuﬃciently large
negative,P(x)<0.Thus, by the intermediate-value theorem, the equationP(x) = 0 has at least one
real root.
Ifa≥0,thenP
ﬀ
(x)=3x
2
+ais positive, except possibly at 0,where it remains nonnegative. It follows
thatPis everywhere increasing and therefore it cannot take on the value 0 more than once.
Suppose now thata<0.Then−
1
3
√
3|a|and
1
3
√
3|a|are consecutive roots of the equationP
ﬀ
(x)=0
and thus, by Exercise 27,Pcannot take on the value zero more than once between these two numbers.
31.(a)f
ﬀ
(x)=3x
2
−3<0 for all x in (−1,1).Also,fis diﬀerentiable on (−1,1) and continuous on
[−1,1].Thus there cannot beaandbin (−1,1) such thatf(a)=f(b)=0,or they would contradict
Rolle’s theorem.
(b) Whenf(x)=0,b=3x−x
3
=x(3−x
2
).Whenxis in (−1,1),then|x(3−x
2
)|<2.
Thus|b|<2.
32.f
ﬀ
(x)=3x
2
−3a
2
>0 for all x in(−a, a).Also,fis diﬀerentiable on (−a, a) and continuous on [−a, a].
Thus there cannot bebandcin (−a, a) such thatf(b)=f(c)=0,or they would contradict Rolle’s
theorem.
33.Forp(x)=x
n
+ax+b, p
ﬀ
(x)=nx
n−1
+a,which has at most one real zero forneven
ν
x=−
a
n
1
n−1
ω
.
If there were more than two distinct real roots ofp(x),then by Rolle’s theorem there would be more
than one zero ofp
ﬀ
(x).Thus there are at most two distinct real roots ofp(x).
34.Forp(x)=x
n
+ax+b, p
ﬀ
(x)=nx
n−1
+a,which has at most two real zeros fornodd. If there were
more than three distinct real roots ofp(x),then by Rolle’s theorem there would be more than two
zeros ofp
ﬀ
(x).Thus there are at most three distinct real roots ofp(x).

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122 SECTION 4.1
35.Ifx1=x2, then|f(x1)−f(x2)|and|x1−x2|are both 0 and the inequality holds. Ifx1δ=x2, then by
the mean-value theorem
f(x1)−f(x2)
x1−x2
=f
ﬀ
(c)
for some numbercbetweenx1andx2. Since|f
ﬀ
(c)|≤1:

f(x1)−f(x2)
x1−x2

≤1 and thus|f(x1)−f(x2)|≤|x1−x2|.
36.See the proof of Theorem 4.2.2.
37.Set, for instance,f(x)=
δ
1, a<x<b
0,x=a, b
38.(a) Letf(x) = cosx.Choose any numbersxandy,(assumex<y).By the mean-value theorem,
there is a numbercbetweenxandysuch that
f(y)−f(x)
y−x
=f
ﬀ
(c)⇒
|cosy−cosx|
|y−x|
=|−sinc|≤1⇒|cosx−cosy|≤|x−y|
(b) Repeat the in part (a) withf(x) = sinx.
39.(a) By the mean-value theorem, there exists a numberc∈(a, b) such thatf(b)−f(a)=f
ﬀ
(c)(b−a).
Iff
ﬀ
(x)≤Mfor allx∈(a, b),then it follows that
f(b)≤f(a)+M(b−a)
(b) Iff
ﬀ
(x)≥mfor allx∈(a, b),then it follows that
f(b)≥f(a)+m(b−a)
(c) If|f
ﬀ
(x)|≤Kon (a, b),then−K≤f
ﬀ
(x)≤Kon (a, b) and the result follows from parts (a)
and (b).
40.Assume thatg(x)δ= 0 for allx∈[a, b] and leth(x)=
f(x)
g(x)
.Thenhis deﬁned on [a, b] andh(a)=
h(b)=0.Therefore, by Rolle’s theorem, there exists a numberc∈(a, b) such that
h
ﬀ
(c)=
g(c)f
ﬀ
(c)−f(c)g
ﬀ
(c)
g
2
(c)
=0
Thusg(c)f
ﬀ
(c)−f(c)g
ﬀ
(c) = 0 which contradicts the given conditionf(x)g
ﬀ
(x)−g(x)f
ﬀ
(x)δ= 0 for all
x∈I.Thus,ghas at least one zero in (a, b).
By reversing the roles offandg,the same argument can be used to show thatgcannot have
two (or more) zeros on (a, b).
41.We show ﬁrst that the conditions onfandgimply thatfandgcannot be simultaneously 0.
Seth(x)=f
2
(x)+g(x).Then
h
ﬀ
(x)=2f(x)f
ﬀ
(x)+2g(x)g
ﬀ
(x)=2f(x)[g(x)] + 2g(x)[−f(x)]=0=⇒f
2
(x)+g
2
(x)=Cconstant.

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SECTION 4.1 123
IfC=0,thenf
2
(x)=−g
2
(x)=⇒f(x)≡0,contradicting the assumptions onf. Therefore,
f
2
(x)+g
2
(x)=C, C >0. Iff(α)=0,theng(αδ=0.
Assume thatf(a)=f(b) = 0 andf(x)δ= 0 on (a,b). Suppose thatg(x)δ= 0 on (a,b). We know also
thatg(a)δ= 0 andg(b)δ= 0. Seth(x)=f(x)/g(x).Thenhis continuous on the closed interval [a,b]
and diﬀerentiable on the open interval (a, b). Therefore, by Rolle’s theorem, there exists at least one
pointcΔ(a, b) such thath
ﬀ
(c)=0.But,
h
ﬀ
(x)=
g(x)f
ﬀ
(x)−f(x)g
ﬀ
(x)
g
2
(x)
=
g
2
(x)+f
2
(x)
g
2
(x)
=
C
g
2
(x)
>0
for allxΔ(a, b),and we have a contradiction. Thusgmust have at least one zero in (a, b).Since
g
ﬀ
(x)=−f(x)δ= 0 in (a, b),ghas exactly one zero in (a, b).
Simply reverse the roles offandgto show thatfhas exactly one zero between two consecutive
zeros ofg.
42.We prove the result forh>0.The proof forh<0 is similar. Iffis diﬀerentiable on (x, x+h),it is
continuous there and thus, by the hypothesis atxandx+hcontinuous on [x, x+h]. By the mean-value
theorem, there existscin (x, x+h) for which
f(x+h)−f(x)
x+h−x
=f
ﬀ
(c).
Multiplying through by (x+h)−x=h,we have
f(x+h)−f(x)=f
ﬀ
(c)h.
Sincecis betweenxandx+h, ccan be written
c=x+θhwith 0<θ<1.
43. f
ﬀ
(x0) = lim
h→0
f(x0+h)−f(x0)
h
= lim
h→0
f
ﬀ
(x0+θh)h
h
= lim
h→0
f
ﬀ
(x0+θh)
∧
(by the hint)
= lim
x→x0
f
ﬀ
(x)=L
∧
(by 2.2.6)
44.Suppose thatf(a)=f(b)=k,and letg(x)=f(x)−k.Thengis diﬀerentiable on (a, b),continuous
on [a, b],andg(a)=g(b)=0.Therefore, by Rolle’s theorem, there exists at least one numberc∈(a, b)
such thatg
ﬀ
(c)=0.Sinceg
ﬀ
(x)=f
ﬀ
(x),it follows thatf
ﬀ
(c)=0.
45.Using the hint,Fis continuous on [a, b],diﬀerentiable on (a, b),andF(a)=F(b).By Exercise 44,
there is a numbercin (a, b) such thatF
ﬀ
(c)=0.
Therefore [f(b)−f(a)]g
ﬀ
(c)−[g(b)−g(a)]f
ﬀ
(c) = 0 and
f(b)−f(a)
g(b)−g(a)
=
f
ﬀ
(c)
g
ﬀ
(c)
.

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124 SECTION 4.1
46.f(x)=2x
3
+3x
2
−3x−2 is diﬀerentiable on (−2,1),continuous on [−2,1],andf(−2) =f(1) = 0.
f
ﬀ
(x)=6x
2
+6x−3
f
ﬀ
(c)=0atc1
∼
=−1.366,c2
∼
=0.366
47.f(x)=1−x
3
−cos(πx/2) is diﬀerentiable on (0,1),continuous on [0,1],andf(0) =f(1) = 0.
f
ﬀ
(x)=−3x
2
+
π
2
sin(πx/2)
f
ﬀ
(c)=0atc
∼
=0.676
48.f(1) = 0;fsatisﬁes Rolle’s theorem on [0,1];f
ﬀ
(c)=0 atc
∼
=0.6058.
49.b
∼
=0.5437,c
∼
=0.3045,
f(0.3045)
∼
=−0.1749
0.2 0.4 0.6 0.8
x
y
50.x-intercepts:x=0,x=1;f
ﬀ
(x)=0atx
∼
=0.3874
51.0c=0
52.x-intercepts:x=−2,x=
4
5
x=2;f
ﬀ
(x)=0atx
∼
=−1.1005 andx
∼
=1.5577
53.f(x)=x
4
−7x
2
+2;f
ﬀ
(x)=4x
3
−14x
g(x)=4x
3
−14x−
f(3)−f(1)
3−1
=4x
3
−14x−12
g(c)=0 atc
∼
=2.205

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SECTION 4.2 125
54.f(x)=xcosx+ 4 sinx;f
ﬀ
(x) = cosx−xsinx+ 4 cosx= 5 cosx−xsinxand
g(x) = 5 cosx−xsinx−
f(π/2)−f(−π/2)
π
= 5 cosx−xsinx−
8
π
g(c)=0 atc1
∼
=−0.872,c2
∼
=0.872
55.f(x)=x
3
−x
2
+x−1;f
ﬀ
(x)=3x
2
−2x+1;c=8/3.
1 2 3 4
x
20
40
60
y
56.f(x)=x
4
−2x
3
−x
2
−x+1;f
ﬀ
(x)=4x
3
−6x
2
−2x−1;c=1/2,−0.6180,1.6180.
−2 −1 1 2 3
x
5
5
10
y
SECTION 4.2
1.f
ﬀ
(x)=3x
2
−3=3

x
2
−1

=3(x+ 1)(x−1)
fincreases on (−∞,−1] and [1,∞),decreases on [−1,1]
2.f
ﬀ
(x)=3x
2
−6x=3x(x−2)
fincreases on (−∞,0] and [2,∞),decreases on [0,2]
3.f
ﬀ
(x)=1−
1
x
2
=
x
2
−1
x
2
=
(x+ 1)(x−1)
x
2
fincreases on (−∞,−1] and [1,∞),decreases on [−1,0) and (0,1] (fis not deﬁned at 0)
4.f
ﬀ
(x)=3(x−3)
2
;fincreases on (−∞,∞)

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126 SECTION 4.2
5.f
ﬀ
(x)=3x
2
+4x
3
=x
2
(3+4x)
fincreases on

−
3
4
,∞

, decreases on

−∞,−
3
4

6.f
ﬀ
(x)=3x
2
+6x+2
fincreases on

−∞,−1−
1
3
√
3

and

−1+
1
3
√
3,∞

,decreases on

−1−
1
3
√
3,−1+
1
3
√
3

7.f
ﬀ
(x)=4(x+1)
3
fincreases on [−1,∞), decreases on (−∞,−1]
8.f
ﬀ
(x)=
2(x
3
+1)
x
3
fincreases on [−∞,−1] and (0,∞), decreases on [−1,0)
9.f(x)=
⎧
⎪
⎨
⎪
⎩
1
2−x
,x<2
1
x−2
,x>2
f
ﬀ
(x)=
⎧
⎪
⎨
⎪
⎩
1
(2−x)
2
,x<2
−1
(x−2)
2
,x>2
fincreases on (−∞,2), decreases on (2,∞)(fis not deﬁned at 2)
10.f
ﬀ
(x)=
(1 +x
2
)−x(2x)
(1 +x
2
)
2
; fincreases on [−1,1],decreases on (−∞,−1] and [1,∞)
11.f
ﬀ
(x)=−
4x
(x
2
−1)
2
fincreases on (−∞,−1) and (−1,0],decreases on [0,1) and (1,∞)(fis not deﬁned at±1)
12.f
ﬀ
(x)=
(x
2
+ 1)(2x)−x
2
(2x)
(x
2
+1)
2
=
2x
(x
2
+1)
2
fincreases on [0,∞),decreases on (−∞,0]
13.f(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
x
2
−5,x< −
√
5
−

x
2
−5

,−
√
5≤x≤
√
5
x
2
−5,
√
5<x
f
ﬀ
(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
2x, x < −
√
5
−2x,−
√
5<x<
√
5
2x,
√
5<x
fincreases on [−
√
5,0 ] and [
√
5,∞),decreases on (−∞,−
√
5] and [0,
√
5]
14.f
ﬀ
(x)=x
2
(2)(1 +x)+(x+1)
2
(2x)=2x(x+ 1)(2x+1)
fincreases on [−1,−1/2] and [0,∞),decreases on (−∞,−1] and [−1/2,0]
15.f
ﬀ
(x)=
2
(x+1)
2
;fincreases on (−∞,−1) and (−1,∞)(fis not deﬁned at−1)
16.f
ﬀ
(x)=2x−
32
x
3
=
2(x
4
−16)
x
3
=
2(x−2)(x+ 2)(x
2
+4)
x
3
fincreases on [−2,0) and [2,∞),decreases on (−∞,−2] and (0,2]
17.f
ﬀ
(x)=
x
(2 +x
2
)
2
⇐
2+x
2
1+x
2
fincreases on [0,∞),decreases on (−∞,0]

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SECTION 4.2 127
18.f(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
x
2
−x−2,x ≤−1
−x
2
+x+2,−1<x<2
x
2
−x−2,x ≥2
f
ﬀ
(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
2x−1,x ≤−1
−2x+1,−1<x<2
2x−1,x ≥2
fincreases on

−1,
1
2

and [2,∞),decreases on (−∞,−1] and

1
2
,2

19.f
ﬀ
(x) = 1 + sinx≥0;fincreases on [0,2π]
20.f
ﬀ
(x) = 1 + cosx≥0;fincreases on [0,2π]
25.f
ﬀ
(x)=−2 sin 2x−2 sinx=−2 sinx(2 cosx+ 1);fincreases on

2
3
π, π

, decreases on

0,
2
3
π

22.f
ﬀ
(x)=−2 cosxsinx=−2 sin 2x;fincreases on [π/2,π],decreases on [0,π/2]
23.f
ﬀ
(x)=
√
3 + 2 sin 2x;fincreases on

0,
2
3
π

and

5
6
π, π

, decreases on

2
3
π,
5
6
π

24.f
ﬀ
(x) = 2 sinxcosx−
√
3 cosx= cosx

2 sinx−
√
3

fincreases on [
1
3
π,
1
2
π] and [
2
3
π, π],decreases on [0,
1
3
π] and [
1
2
π,
2
3
π]
25.
d
dx
Δ
x
3
3
−x
θ
=f
ﬀ
(x)=⇒f(x)=
x
3
3
−x+C
f(1) = 2 =⇒2=
1
3
−1+C,soC=
8
3
.Thus,f(x)=
1
3
x
3
−x+
8
3
.
26.
d
dx

x
2
−5x

=f
ﬀ
(x)=⇒f(x)=x
2
−5x+C
f(2) = 4 =⇒4=4−10 +C,soC=10.Thus,f(x)=x
2
−5x+10.
27.
d
dx

x
5
+x
4
+x
3
+x
2
+x

=f
ﬀ
(x)=⇒f(x)=x
5
+x
4
+x
3
+x
2
+x+C
f(0) = 5 =⇒5=0+C,soC=5.Thus,f(x)=x
5
+x
4
+x
3
+x
2
+x+5.
28.
d
dx

−2x
−2

=f
ﬀ
(x)=⇒f(x)=−2x
−2
+C
f(1) = 0 =⇒0=−2+C,soC=2.Thus,f(x)=−2x
−2
+2,x>0.
29.
d
dx
Δ
3
4
x
4/3
−
2
3
x
3/2
θ
=f
ﬀ
(x)=⇒f(x)=
3
4
x
4/3
−
2
3
x
3/2
+C
f(0) = 1 =⇒1=0+C,soC=1.Thus,f(x)=
3
4
x
4/3
−
2
3
x
3/2
+1,x≥0.
30.
d
dx

−
1
4
x
−4
−
25
4
x
4/5

=f
ﬀ
(x)=⇒f(x)=−
1
4
x
−4
−
25
4
x
4/5
+C
f(1) = 0 =⇒0=−
1
4
−
25
2
+C,soC=
13
2
.Thus,f(x)=−
1
4
x
−4
−
25
4
x
4/5
+
13
2
,x>0.
31.
d
dx
(2x−cosx)=f
ﬀ
(x)=⇒f(x)=2x−cosx+C
f(0) = 3 =⇒3=0−1+C,soC=4.Thus,f(x)=2x−cosx+4.
32.
d
dx
(2x
2
+ sinx)=f
ﬀ
(x)=⇒f(x)=2x
2
+ sinx+C
f(0) = 1 =⇒1=0+C,soC=1.Thus,f(x)=2x
2
+ sinx+1.

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128 SECTION 4.2
33.f
ﬀ
(x)=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
1,x< −3
−1,−3<x<−1
1,−1<x<1
−2, 1<x
fincreases on (−∞,−3) and [−1,1];
decreases on [−3,−1] and [1,∞)
34.f
ﬀ
(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
2(x−1),x< 1
−1,1<x<3
−2,x> 3
fdecreases on (−∞,1),[1,3) and [3,∞).
35.f
ﬀ
(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
−2x, x < 1
−2,1<x<3
3, 3<x
fincreases on (−∞,0] and [3,∞);
decreases on [0,1) and [1,3]
36.f
ﬀ
(x)=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
1,x< 0
2(x−1),0<x<3
−1,3<x<7
2,x> 7
fincreases on (−∞,0],[1,3],(7,∞);
decreases on [0,1] and [3,7)
37. 38.

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SECTION 4.2 129
39. 40.
41. 42.
43. 44.
45.Not possible;fis increasing, sof(2) must be greater thanf(−1).
46.Not possible; by the intermediate-value theorem,fmust have a zero in (3,5).
47.(a) True. Letx1,x2∈[a, c],x1<x2.Ifx1,x2∈[a, b], or ifx1,x2∈[b, c], thenf(x1)<f(x2).
Ifx1∈[a, b) andx2∈[b, c], thenf(x1)<f(b)≤f(x2). Thereforefincreases on [a, c]
(b) False. A slight modiﬁcation of Example 6 is a counterexample. Letg(x)=
⎧
⎨
⎩
1
2
x+2,x≤1
x
3
,x> 1.
48.(a) True. Use the obvious modiﬁcation of the argument in Exercise 47(a).
(b) False. An example like 47(b):f(x)=
⎧
⎨
⎩
−x+2,x≤1
−x+3,x>1.
.
49.(a) True. Iff
ﬀ
(c)<0 at some numberc∈(a, b), then there exists a numberδsuch that
f(x)>f(c)>f(z) forx∈(c−δ, c) andz∈(c, c+δ) (Theorem 4.1.2).
(b) False.f(x)=x
3
increases on (−1,1) andf
ﬀ
(0) = 0.
50.False.f(x)=x+
1
2π
sin (2x−1)πis increasing on [0,4] andf
ﬀ
(x) = 1 + cos (2x−1)πx=0 at
x=1,x=2.x=3.

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130 SECTION 4.2
51.Letf(x)=x−sinx.Thenf
ﬀ
(x)=1−cosx.
(a)f
ﬀ
(x)≥0 for allx∈(−∞,∞) andf
ﬀ
(x) = 0 only atx=
π
2
+nπ, n=0,±1,±2, ...
It follows from Theorem 4.2.3 thatfis increasing on (−∞,∞).
(b) Sincefis increasing on (−∞,∞) andf(0) = 0−sin 0 = 0,we have:
f(x)>0 for allx>0⇒x>sinxon (0,∞);
f(x)<0 for allx<0⇒x<sinxon (−∞,0).
52.A proof is outlined just below the statement of the theorem.
53.f
ﬀ
(x) = 2 secx(secxtanx) = 2 sec
2
xtanxandg
ﬀ
(x) = 2 tanxsec
2
x.
Therefore,f
ﬀ
(x)=g
ﬀ
(x) for allx∈I.
54.Evaluating sec
2
x−tan
2
x=Catx= 0 givesC=1.
55.Letfandgbe functions such thatf
ﬀ
(x)=−g(x) andg
ﬀ
(x)=f(x).Then:
(a) Diﬀerentiatingf
2
(x)+g
2
(x) with respect tox,we have
2f(x)f
ﬀ
(x)+2g(x)g
ﬀ
(x)=−2f(x)g(x)+2g(x)f(x)=0.
Thus,f
2
(x)+g
2
(x)=C(constant).
(b)f(0) = 0 andg(0) = 1 impliesC=1.
(c) The functionsf(x) = sinx, g(x) = cosxhave these properties.
56.(a) Leth(x)=f(x)−g(x).Thenh
ﬀ
(x)=f
ﬀ
(x)−g
ﬀ
(x)>0on(0,c),andhis increasing on (0,c).
Sinceh(0) =f(0)−g(0) = 0, it follows thath(x)>0on(0,c).Thus,f(x)>g(x)on(0,c).
(b) Again leth(x)=f(x)−g(x).Thenhis increasing on (−c,0) which implies thath(x)<0 on this
interval sinceh(0) = 0.Therefore,f(x)<g(x)on(−c,0).
57.Letf(x) = tanxandg(x)=xforx∈[0,π/2).Thenf(0) =g(0) = 0 andf
ﬀ
(x) = sec
2
x>g
ﬀ
(x)=1
forx∈(0,π/2).Thus, tanx>xforx∈(0,π/2) by Exercise 56(a).
58.Letf(x) = cosx−

1−
1
2
x
2

forx∈[0,∞).Thenf(0) = 0 andf
ﬀ
(x)=−sinx+x=x−sinx>0
forx∈(0,∞) by Exercise 51 (b). Thus,f(x)>0 forx∈(0,∞) which implies cosx>1−
1
2
x
2
on
(0,∞).
59.Choose an integern>1.Letf(x) = (1 +x)
n
andg(x)=1+nx, x >0.Then,f(0) =g(0) = 1
andf
ﬀ
(x)=n(1 +x)
n−1
>g
ﬀ
(x)=nsince (1 +x)
n−1
>1 forx>0.The result follows from Exercise
56(a).
60.Letf(x) = sinx−

x−
1
6
x
3

.Thenf(0) = 0 andf
ﬀ
(x) = cosx−

1−
1
2
x
2

>0 by Exercise 58.
Therefore,f(x)>f(0) = 0 for allx∈(0,∞) which implies sinx>x−
1
6
x
3
on (0,∞).
61.4
◦∼
=0.06981 radians. By Exercises 51 and 60,
0.6981−
(0.6981)
3
6
=0.06975<sin 4
◦
<0.6981

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SECTION 4.2 131
62.(a) Letf(x) = cosx−(1−
1
2
x
2
+
1
24
x
4
).Thenf(0) = 0 andf
ﬀ
(x)=−sinx+x−
x
3
6
<0by
Exercise 60. Therefore,f(x)<f(0) = 0 on allx∈(0,∞),which implies cosx<1−
1
2
x
2
+
1
24
x
4
on (0,∞).
(b) 6
◦
=
π
30
.Using this forxin 1−
1
2
x
2
<cosx<1−
1
2
x
2
+
1
24
x
4
,
=⇒0.994517<cos 6
◦
<0.994522.
63.Letf(x)=3x
4
−10x
3
−4x
2
+10x+9,x∈[−2,5].Thenf
ﬀ
(x)=12x
3
−30x
2
−8x+10.
f
ﬀ
(x)=0atx
∼
=−0.633,0.5,2.633
fis decreasing on [−2,−0.633]
and [0.5,2.633]
fis increasing on [−0.633,0.5]
and [2.633,5]
64.Letf(x)=2x
3
−x
2
−13x−6,x∈[−3,4].Thenf
ﬀ
(x)=6x
2
−2x−13.
f
ﬀ
(x)=0atx
∼
=−1.315,1.648
fis decreasing on [−1.315,1.648]
fis increasing on [−3,−1.315] and [1.648,4]
65.Letf(x)=xcosx−3 sin 2x, x∈[0,6].Thenf
ﬀ
(x) = cosx−xsinx−6 cos 2x.
f
ﬀ
(x)=0atx
∼
=0.770,2.155,3.798,5.812
fis decreasing on [0,0.770],[2.155,3.798]
and [5.812,6]
fis increasing on [0.770,2.155]
and [3.798,5.812]
66.Letf(x)=x
4
+3x
3
−2x
2
+4x+4,x∈[−5,3].Thenf
ﬀ
(x)=4x
3
+9x
2
−4x+4.
f
ﬀ
(x)=0atx
∼
=−2.747
fis decreasing on [−5,−2.747]
fis increasing on [−2.747,3]

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132 SECTION 4.3
67.(a)f
ﬀ
(x)=0atx=0,
π
2
,π,
3π
2
,2π (b)f
ﬀ
(x)>0on

π,
3π
2

∪

3π
2
,2π

(c)f
ﬀ
(x)<0on

0,
π
2

∪

π
2
,π

68.(b)f
ﬀ
(x)>0on(−∞,∞)
69.(a)f
ﬀ
(x)=0atx= 0 (b) f
ﬀ
(x)>0on(0,∞)
(c)f
ﬀ
(x)<0on(−∞,0)
70.(a)f
ﬀ
(x)=0atx=−
1
2
,
8
5
,3 (b) f
ﬀ
(x)>0on

−∞,−
1
2

∪

−
1
2
,
8
5

∪(3,∞)
(c)f
ﬀ
(x)<0on

8
5
,3

71.f=C, constant;f
ﬀ
(x)≡0
SECTION 4.3
1.f
ﬀ
(x)=3x
2
+3>0; no critical pts, no local extreme values
2.f
ﬀ
(x)=8x
3
−8x=8x(x
2
−1); critical pts−1,0,1
f
ﬀﬀ
(x)=24x
2
−8;f
ﬀﬀ
(−1) =f
ﬀﬀ
(1) = 16>0,f
ﬀﬀ
(0) =−8<0;
f(0) = 6 local max,f(−1) = 4 local min,f(1) = 4 local min
3.f
ﬀ
(x)=1−
1
x
2
; critical pts−1,1
f
ﬀﬀ
(x)=
2
x
3
,f
ﬀﬀ
(−1) =−2,f
ﬀﬀ
(1) = 2f(−1) =−2 local max,f(1) = 2 local min
4.f
ﬀ
(x)=2x+
6
x
3
=
2x
4
+6
x
3
; no critical pts (note: 0 is not in the domain off),
no local extreme values
5.f
ﬀ
(x)=2x−3x
2
=x(2−3x); critical pts 0,
2
3
f
ﬀﬀ
(x)=2−6x;f
ﬀﬀ
(0) = 2,f
ﬀﬀ
(
2
3
)=−2
f(0) = 0 local min,f(
2
3
)=
4
27
local max
6.f
ﬀ
(x)=−2(1−x)(1 +x)+(1−x)
2
=(x−1)(3x+ 1); critical pts−
1
3
,1
f
ﬀﬀ
(x)=(1+3x)+3(x−1) = 2(3x−1);f
ﬀﬀ

−
1
3

=−4,f
ﬀﬀ
(1) = 4
f

−
1
3

=
32
27
local max,f(1) = 0 local min
7.f
ﬀ
(x)=
2
(1−x)
2
; no critical pts, no local extreme values
8.f
ﬀ
(x)=
(2 +x)(−3)−(2−3x)(1)
(2 +x)
2
=−
8
(2 +x)
2
; no critical pts (note:−2 is not
in the domain off), no local extreme values

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SECTION 4.3 133
9.f
ﬀ
(x)=−
2(2x+1)
x
2
(x+1)
2
; critical pt−
1
2
f

−
1
2

=−8 local max
10.f(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
x
2
−16,x< −4
16−x
2
,−4≤x<4
x
2
−16,x ≥4
f
ﬀ
(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
2x, x < −4
−2x,−4<x<4
2x, x > 4
critical pts−4,0,4;f(−4) =f(4) = 0 local minima,f(0) = 16 local max
11.f
ﬀ
(x)=x
2
(5x−3)(x−1); critical pts 0,
3
5
,1
f
Δ
3
5
θ
=
2
2
3
3
5
5
local max
f(1) = 0 local min
no local extreme at 0
12.f
ﬀ
(x)=3
Δ
x−2
x+2
θ
2
4
(x+2)
2
≥0; critical pt 2,no local extreme values
13.f
ﬀ
(x)=(5−8x)(x−1)
2
; critical pts
5
8
,1
f

5
8

=
27
2048
local max
no local extreme at 1
14.f
ﬀ
(x)=−(1 +x)
3
+(1−x)(3)(1 +x)
2
= 2(1 +x)
2
(1−2x); critical pts−1,
1
2
f

1
2

=
27
16
local max
no local extreme at−1
15.f
ﬀ
(x)=
x(2 +x)
(1 +x)
2
; critical pts−2,0
f(−2) =−4 local max
f(0) = 0 local min
16.f
ﬀ
(x)=(1−x)
1/3
−
1
3
x(1−x)
−2/3
=
3−4x
3(1−x)
2/3
; critical pts
3
4
,1
f(3/4) =
3
4
4/3local max
no local extreme at 1
17.f
ﬀ
(x)=
1
3
x(7x+ 12)(x+2)
−2/3
; critical pts−2,−
12
7
,0
f

−
12
7

=
144
49

2
7

1/3
local max
f(0) = 0 local min

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134 SECTION 4.3
18.f
ﬀ
(x)=
−1
(x+1)
2
+
1
(x−2)
2
=
3(2x−1)
(x+1)
2
(x−2)
2
; critical pt
1
2
f

1
2

=
4
3
local min
19.f(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
2−3x, x ≤−
1
2
x+4,−
1
2
<x<3
3x−2, 3≤x
f
ﬀ
(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
−3,x< −
1
2
1,−
1
2
<x<3
3, 3<x
critical pts−
1
2
,3
f

−
1
2

=
7
2
local min
no local extreme at 3
20.f
ﬀ
(x)=
7
3
x
4/3
−
7
3
x
−2/3
=
7
3
x
2
−1
x
2/3
; critical pts−1,0,1
f(−1) = 6 local max,
f(1) =−6 local min
no local extreme at 0
21.f
ﬀ
(x)=
2
3
x
−4/3
(x−1); critical pt 1
f(1) = 3 local min
no local extreme at 0
22.f
ﬀ
(x)=
(x+ 1)3x
2
−x
3
(x+1)
2
=
x
2
(2x+3)
(x+1)
2
; critical pts−
3
2
,0
f

−
3
2

=
27
4
local min
no local extreme at 0
23.f
ﬀ
(x) = cosx−sinx; critical pts
1
4
π,
5
4
π
f
ﬀﬀ
(x)=−sinx−cosx, f
ﬀﬀ

1
4
π

=−
√
2,f
ﬀﬀ

5
4
π

=
√
2
f(
1
4
π)=
√
2 local max,f(
5
4
π)=−
√
2 local min
24.f
ﬀ
(x)=1−2 sin 2x; critical pts
π
12
,
5π
12
f(
1
12
π)=
π
12
+
√
3
2
local max
f(
5
12
π)=
5π
12
−
√
3
2
local min

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.3 135
25.f
ﬀ
(x) = cosx(2 sinx−
√
3 ); critical pts
1
3
π,
1
2
π,
2
3
π
f(
1
3
π)=f(
2
3
π)=−
3
4
local mins
f(
1
2
π)=1−
√
3 local max
26.f
ﬀ
(x) = 2 sinxcosx; critical pts
1
2
π, π,
3
2
π
f

1
2
π

=1=f

3
2
π

local max
f(π) = 0 local min
27.f
ﬀ
(x) = cos
2
x−sin
2
x−3 cosx+ 2 = (2 cosx−1)(cosx−1) critical pts
1
3
π,
5
3
π
f

1
3
π

=
2
3
π−
5
4
√
3 local min
f

5
3
π

=
10
3
π+
5
4
√
3 local max
28.f
ﬀ
(x) = 6 sin
2
xcosx−3 cosx= 3 cosx(2 sin
2
x−1); critical pts
1
4
π,
1
2
π,
3
4
π
f(
1
4
π)=f(
3
4
π)=−
√
2 local mins
f(
1
2
π)=−1 local max
29.(a)fincreases on [−2,0] and [3,∞);fdecreases on (−∞,−2] and [0,3].
(b)f(−2) andf(3) are local minima;f(0) = 1 is a local maximum.
30.(a)fincreases on (−∞,−1] and [0,∞);fdecreases on [−1,0].
(b)f(−1) is a local maximum;f(0) = 1 is a local minimum.
31.Leth(x)=f(x)−g(x). Thenh(x) gives the vertical separation between graphs offandgatx.If
hhas a maximum atc, thenh
ﬀ
(c) = 0. Sinceh
ﬀ
(x)=f
ﬀ
(x)−g
ﬀ
(x),h
ﬀ
(c) = 0 impliesf
ﬀ
(c)=g
ﬀ
(c).
Thus the lines tangent to the graphs offandgare parallel atx=c.
32.Setg(x)=−f(−x) and apply the proof of the second derivative test already given.
33.Solvingf
ﬀ
(x)=2ax+b= 0 gives a critical point atx=−
b
2a
.Sincef
ﬀﬀ
(x)=2a,
fhas a local maximum at−
b
2a
ifa<0 and a local minimum at−
b
2a
ifa>0.
34.Settingf
ﬀ
(x)=3ax
2
+2bx+c= 0 and checking the discriminant, we get
(1) 2 local extrema ifb
2
>3ac
(2) 1 local extrema ifb
2
=3ac
(3) 0 local extrema ifb
2
<3ac

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
136 SECTION 4.3
35. P(x)=x
4
−8x
3
+22x
2
−24x+4
P
ﬀ
(x)=4x
3
−24x
2
+44x−24
P
ﬀﬀ
(x)=12x
2
−48x+44
SinceP
ﬀ
(1) = 0,P
ﬀ
(x) is divisible byx−1.Division byx−1 gives
P
ﬀ
(x)=(x−1)

4x
2
−20x+24

=4(x−1)(x−2)(x−3).
The critical pts are 1,2,3.Since
P
ﬀﬀ
(1)>0,P
ﬀﬀ
(2)<0,P
ﬀﬀ
(3)>0,
P(1) =−5 is a local min,P(2) =−4 is a local max, andP(3) =−5 is a local min.
SinceP
ﬀ
(x)<0 forx<0,Pdecreases on (−∞,0]. SinceP(0)>0,Pdoes not take on the value 0 on
(−∞,0].
SinceP(0)>0 andP(1)<0,Ptakes on the value 0 at least once on (0,1).SinceP
ﬀ
(x)<0on(0,1),
Pdecreases on [0,1].It follows thatPtakes on the value zero only once on [0,1].
SinceP
ﬀ
(x)>0on(1,2) andP
ﬀ
(x)<0on(2,3),Pincreases on [1,2] and decreases on [2,3].Since
P(1),P(2),P(3) are all negative,Pcannot take on the value 0 between 1 and 3.
SinceP(3)<0 andP(100)>0,Ptakes on the value 0 at least once on (3,100).SinceP
ﬀ
(x)>0on
(3,100),Pincreases on [3,100].It follows thatPtakes on the value zero only once on [3,100].
SinceP
ﬀ
(x)>0 on (100,∞),Pincreases on [100,∞). SinceP(100)>0,Pdoes not take on the value
0 on [100,∞).
36.fhas a local maximum atx=0;fhas a local minimum atx=−1 andx=2.
37.
38.Letf(x)=Ax
2
+Bx+C.Thenf
ﬀ
(x)=2Ax+B.
f(−1) = 3 =⇒A−B+C=3;f(3) =−1=⇒9A+3B+C=−1
Sincefhas a minimum atx=2,f
ﬀ
(2) = 4A+B=0
Solving forA, B, C,we getA=
1
2
,B=−2,C=
1
2
.

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.3 137
39.Letf(x)=
ax
x
2
+b
2
.Thenf
ﬀ
(x)=
a

b
2
−x
2

(b
2
+x
2
)
2
.Now
f
ﬀ
(0) =
a
b
2
=1⇒a=b
2
andf
ﬀ
(x)=
b
2

b
2
−x
2

(b
2
+x
2
)
2
f
ﬀ
(−2) =
b
2

b
2
−4

(b
2
+4)
2
=0⇒b=±2
Thus,a= 4 andb=±2.
40.(a)f(x)=x
p
(1−x)
q
,p,q≥2;f
ﬀ
(x)=x
p−1
(1−x)
q−1
[p−(p+q)x]
f
ﬀ
(x)=0 =⇒x=0,x=1,x=
p
p+q
(b)peven,p−1 odd:
fhas a local min atx=0
(c)qeven,q−1 odd:
fhas a local min atx=1
(d)f
ﬀﬀ
Δ
p
p+q
θ
=−(p+q)
Δ
p
p+q
θ
p−1Δ
q
p+q
θ
q−1
<0⇒fhas a local max atx=
p
p+q
.
41.Ifpis a polynomial of degreen,thenp
ﬀ
has degreen−1.This implies thatp
ﬀ
has at most
n−1 zeros, and it follows thatphas at mostn−1 local extreme values.
42.The functionD(x)=

x
2
+[f(x)]
2
gives the distance from the origin to the point (x, f(x)) on the
graph off.Since the graph offdoes not pass through the origin,
D
ﬀ
(x)=
x+f(x)f
ﬀ
(x)

x
2
+[f(x)]
2
is deﬁned for allx∈dom (f).Suppose thatDhas a local extreme value atc.Then
D
ﬀ
(c)=
c+f(c)f
ﬀ
(c)

c
2
+[f(c)]
2
=0⇒c+f(c)f
ﬀ
(c) = 0 andf
ﬀ
(c)=−
c
f(c)
Suppose thatcδ=0.The slope of the line through (0,0) and (c, f(c)) is given bym1=
f(c)
c
and the
slope of the tangent line to the graph offatx=cis given bym2=f
ﬀ
(c)=−
c
f(c)
.Sincem1m2=−1,
these two lines are perpendicular. Ifc=0,then the tangent line to the graph offis horizontal and
the line through (0,0) and (0,f(0)) is vertical.
43.Iff(x)=x
4
−7x
2
−8x−3, thenf
ﬀ
(x)=4x
3
−14x−8 and f
ﬀﬀ
(x)=12x
2
−14.Sincef
ﬀ
(2) =
−4<0 andf
ﬀ
(3) = 58>0,f
ﬀ
has at least one zero in (2,3).Sincef
ﬀﬀ
(x)>0 forx∈(2,3),f
ﬀ
is
increasing on this interval and so it has exactly one zero. Thus,fhas exactly one critical pointcin
(2,3).

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138 SECTION 4.3
44.Iff(x) = sinx+
x
2
2
−2x, thenf
ﬀ
(x) = cosx+x−2 andf
ﬀﬀ
(x)=−sinx+1.Sincef
ﬀ
(2) =
−0.4161<0 andf
ﬀ
(3) = 0.01>0,f
ﬀ
has at least one zero in (2,3).Sincef
ﬀﬀ
(x)>0 forx∈(2,3),
f
ﬀ
is increasing on this interval and so it has exactly one zero. Thus,fhas exactly one critical point
cin (2,3).
45.f(x)=
ax
2
+b
cx
2
+d
andf
ﬀ
(x)=
2(ad−bc)x
(cx
2
+d)
2
;x= 0 is a critical number.
f
ﬀﬀ
(x)=
2(ad−bc)(cx
2
−4cx+d)
(cx
2
+d)
3
;f
ﬀﬀ
(0) =
2(ad−bc)
d
2
Therefore,ad−bc >0 implies thatf(0) is a local minimum;ad−bc <0 implies thatf(0) is a
local maximum.
46.Letδbe any positive number and considerfon the interval (−δ, δ).Letnbe a positive integer
such that
0<
1
π
2
+2nπ
<δand 0<
1
−π
2
+2nπ
<δ.
Then
f
Δ
1
π
2
+2nπ
θ
>0 andf
Δ
1
−π
2
+2nπ
θ
<0.
Thusftakes on both positive and negative values in every interval centered at 0 and it follows that
fcannot have a local maximum or minimum at 0.
47.(a)
critical points:x1
∼
=−0.692,x2
∼
=2.248
local extreme values:f(−0.692)
∼
=29.342,f(2.248)
∼
=−8.766
(b)fis increasing on [−3,−0.692],and [2.248,4];fis decreasing on [−0.692,2.248]
48.(a)

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
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SECTION 4.3 139
critical points:x1
∼
=−2.085,x2
∼
=−1,x3
∼
=0.207,x4
∼
=1.096,x5=1.544
local extreme values:f(−2.085)
∼
=−6.255,f(−1) = 7,f(0.207)
∼
=0.621,f(1.096)
∼
=7.097,
f(1.544)
∼
=4.635
(b)fis increasing on [−2.085,−1],[0.207,1.096],and [1.544,4]
fis decreasing on [−4,−2.085],[−1,0.207],and [1.096,1.544]
49.(a)
critical points:x1
∼
=−2.201,x2
∼
=−0.654,x3
∼
=0.654,x4
∼
=2.201
local extreme values:f(−2.204)
∼
=2.226,f(−0.654)
∼
=−6.634,f(0.654)
∼
=6.634,
f(2.204)
∼
=−2.226
(b)fis increasing on [−3.−2.204],[−0.654,0.654],and [2.204,3]
fis decreasing on [−2.204,−0.654],and [0.654,2.204]
50.f
ﬀ
(x)=0 atx=2,3;f(2) = 0 is a local minimum.
51.f
ﬀ
(x)>0on

2
3
,∞

;fhas no local extrema.
52.f
ﬀ
(x) = 0 at the multiples of
1
4
π;f(0) = 1 is a local maximum,f(π/4) = 0 is a local minimum,
f(π/2) = 1 is a local maximum, and so on.
53.
critical points off:x1
∼
=−1.326,x2=0,x3
∼
=1.816
f
ﬀﬀ
(−1.326)
∼
=−4<0⇒fhas a local maximum atx=−1.326
f
ﬀﬀ
(0) = 4>0⇒fhas a local minimum atx=0
f
ﬀﬀ
(1.816)
∼
=−4⇒fhas a local maximum atx=1.816

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
140 SECTION 4.4
54.
critical number off:x1
∼
=−1.935
f
ﬀﬀ
(−1.935)
∼
=14.60>0⇒fhas a local minimum atx=−1.935
SECTION 4.4
1.f
ﬀ
(x)=
1
2
(x+2)
−1/2
,x>−2;
f(−2) = 0 endpt and abs min; asx→∞,f(x)→∞; so no abs max
2.f
ﬀ
(x)=2x−3; critical pt.
3
2
;f

3
2

=−
1
4
local and abs min
3.f
ﬀ
(x)=2x−4,x∈(0,3);
critical pt. 2;
f(0) = 1 endpt and abs max,f(2) =−3 local and abs min,f(3) =−2 endpt max
4.f
ﬀ
(x)=4x+5,x∈(−2,0);
critical pt.−
5
4
;
f(−2) =−3 endpt max,f

−
5
4

=−
33
8
local and abs min,f(0) =−1 endpt and abs max
5.f
ﬀ
(x)=2x−
1
x
2
=
2x
3
−1
x
2
,xδ=0;f
ﬀ
(x)=0 atx=2
−1/3
critical pt. 2
−1/3
;f
ﬀﬀ
(x)=2+
2
x
3
,f
ﬀﬀ
ν
2
−1/3
ω
=6
f

2
−1/3

=2
−2/3
+2
1/3
=2
−2/3
+2·2
−2/3
=3·2
−2/3
local min
6.f
ﬀ
(x)=1−
2
x
3
critical pt. 2
1/3
;f

2
1/3

= 3(2)
−2/3
local min
7.f
ﬀ
(x)=
2x
3
−1
x
2
,x∈
Δ
1
10
,2
θ
;
critical pt. 2
−1/3
;
f

1
10

=10
1
100
endpt and abs max,f

2
−1/3

=3·2
−2/3
local and abs min,
f(2) = 4
1
2
endpt max

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
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SECTION 4.4 141
8.f
ﬀ
(x)=1−
2
x
3,x∈(1,
√
2)
critical pt. 2
1/3
;f(1) = 2 endpt and abs max
f

2
1/3

= 3(2)
−2/3
local and abs min,f
√
2

=
√
2+
1
2
endpt max
9.f
ﬀ
(x)=2x−3,x∈(0,2);
critical pt.
3
2
;
f(0) = 2 endpt and abs max,f

3
2

=−
1
4
local and abs min,
f(2) = 0 endpt max
10.f
ﬀ
(x)=2(x−1)(x−2)(2x−3);x∈(0,4)
critical pts. 1,
3
2
,2,;f(0) = 4 endpt max,f(1) = 0 local and abs min,
f

3
2

=
1
16
local max,f(2) = 0 local and abs min,f(4) = 36 endpt and abs max
11.f
ﬀ
(x)=
(2−x)(2 +x)
(4 +x
2
)
2
,x∈(−3,1);
critical pt.−2;
f(−3) =−
3
13
endpt max,f(−2) =−
1
4
local and abs min,
f(1) =
1
5
endpt and abs max
12.f
ﬀ
(x)=
2x
(1 +x
2
)
2
,x∈(−1,2)
critical pt. 0;f(−1) =
1
2
endpt max,f(0) = 0 local and abs min,
f(2) =
4
5
endpt and abs max
13.f
ﬀ
(x)=2(x−
√
x)
Δ
1−
1
2
√
x
θ
,x>0;
critical pts.
1
4
,1;
f(0) = 0 endpt and abs min,f

1
4

=
1
16
local max,f(1) = 0 local and abs min;
asx→∞,f(x)→∞; so no abs max
14.f
ﬀ
(x)=
2(2−x
2
)
(4−x
2
)
1/2
,x∈(−2,2)
critical pts.−
√
2,
√
2;f(−2) = 0 endpt max,f

−
√
2

=−2 local and abs min,
f
√
2

= 2 local and abs max,f(2) = 0 endpt min
15.f
ﬀ
(x)=
3(2−x)
2
√
3−x
,x<3
critical pt. 2;
f(2) = 2 local and abs max,f(3) = 0 endpt min;
asx→−∞,f(x)→−∞; so no abs min

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142 SECTION 4.4
16.f
ﬀ
(x)=
1
2

x
−1/2
+x
−3/2

,x>0
no critical pts; no extreme values.
17.f
ﬀ
(x)=−
1
3
(x−1)
−2/3
,xδ=1;
critical pt. 1;
no local extremes;
asx→∞,f(x)→−∞
asx→−∞,f(x)→∞

no abs extremes
18.f
ﬀ
(x)=
8
3
3x−1
(4x−1)
2/3
(2x−1)
1/3
;
critical pts.
1
4
,
1
3
,
1
2
; no extreme value at
1
4
f

1
3

=
1
3
local max,f

1
2

= 0 local min
19.f
ﬀ
(x) = sinx

2 cosx+
√
3

,x∈(0,π);
critical pt.
5
6
π;
f(0) =−
√
3 endpt and abs min,f

5
6
π

=
7
4
local and abs max,f(π)=
√
3 endpt min
20.f
ﬀ
(x)=−csc
2
x+1,x∈(0,2π/3) ;
critical pt.
1
2
π(note: 0 is not in the domain)
No extreme value at
1
2
π, f

2
3
π

=
1
3

2π−
√
3

endpt and abs max
21.f
ﬀ
(x)=−3 sinx

2 cos
2
x+1

<0,x∈(0,π); no critical pts.
f(0) = 5 endpt and abs max,f(π)=−5 endpt and abs min
22.f
ﬀ
(x) = 2 cos 2x−1,x∈(0,π):
critical pts.
1
6
π,
5
6
π;f(0) = 0 endpt min,f

1
6
π

=
1
2
√
3−
1
6
πlocal and abs max,
f

5
6
π

=−
1
2
√
3−
5
6
πlocal and abs min,f(π)=−πendpt max
23.f
ﬀ
(x) = sec
2
x−1≥0,x∈

−
1
3
π,
1
2
π

; critical pt. 0 ;
f

−
1
3
π

=
1
3
π−
√
3 endpt and abs min, no abs max
24.f
ﬀ
(x) = 2 sinxcosx(2 sin
2
x−1),x∈

0,
2
3
π

;
critical pts.
1
4
π,
1
2
π;f(0) = 0 endpt and abs max,f

1
4
π

=−
1
4
local and abs min,
f

1
2
π

= 0 local and abs max,f

2
3
π

=−
3
16
endpt min

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SECTION 4.4 143
25.
f
ﬀ
(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
−2,0<x<1
1,1<x<4
−1,4<x<7
critical pts. 1,4;
f(0) = 0 endpt max,f(1) =−2 local and abs min,
f(4) = 1 local and absolute max,f(7) =−2 endpt and abs min
26.f
ﬀ
(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
1,−8<x<−3
2x+1,−3<x≤2
5, 2<x<5
critical pts.−3,−
1
2
;
f(−8) = 1 endpt min,f(−3) = 6 local max
f

−
1
2

=−
1
4
local and abs min
27.
f
ﬀ
(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
2x,−2<x<−1
2−2x,−1<x<3
1, 3<x<6
critical pts.−1,1,3
f(−2) = 5 endpt max,f(−1) = 2 local and abs min,
f(1) = 6 local and abs max,f(3) = 2 local and abs min
28.f
ﬀ
(x)=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
−2x−2,−2<x<0
−1,0<x<2
1,2<x<3
(x−2)
2
,3<x<4
critical pts.−1,0,2,3;
f(−2) = 2 endpt min,f(−1) = 3 local and abs max,
f(0) not an extreme value,f(2) = 0 local and abs min,
f(3) =
1
3
local min,f(4) =
8
3
endpt max

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144 SECTION 4.4
29.
f
ﬀ
(x)=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
−1,−3<x<−1
1,−1<x<0
2x−4, 0<x<3
2, 3≤x<4
critical pts.−1,0,2
f(−3) = 2 endpt and abs max,f(−1) = 0 local min,
f(0) = 2 local and abs max,f(2) =−2 local and abs min
30.f
ﬀ
(x)=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
−2x,0<x<1
−2,1<x<2
−x,2<x<3
Note: 1 is not in the domain off.
critical pts. 2;
f(0) = 0 endpt and abs max,f(2) =−2 local max,
f(3) =−
9
2
endpt and abs min
31. 32.
33.Not possible:f(1) =f(3) = 0 impliesf
ﬀ
(c) = 0 for somec∈(1,3) (Rolle’s theorem).
34.f(x)=x−
1
2π
sin 2πx
35.Letp(x)=x
3
+ax
2
+bx+c.Thenp
ﬀ
(x)=3x
2
+2ax+bis a quadratic with discriminant Δ = 4a
2
−
12b=4(a
2
−3b).Ifa
2
≤3b,then Δ≤0.This implies thatp
ﬀ
(x) does not change sign on (−∞,∞).
On the other hand, ifa
2
−3b>0,then Δ>0 andp
ﬀ
has two real zeros,c1andc2,from which it
follows thatphas extreme values atc1andc2.Therefore, ifphas no extreme values, then we must
havea
2
−3b≤0.

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SECTION 4.4 145
36.f(x) = (1 +x)
r
−(1 +rx),x≥−1.
f
ﬀ
(x)=r

(1 +x)
r−1
−1

;f
ﬀ
(x)=0 =⇒x=0
f
ﬀﬀ
(x)=r(r−1)(1 +x)
r−2
;f
ﬀﬀ
(0) =r(r−1)>0=⇒fhas a local minimum atx=0
Since 0 is the only critical number off, the local minimum must be an absolute minimum.
37.By contradiction. Iffis continuous atc, then, by the ﬁrst-derivative test (4.3.4),f(c)isnotalocal
maximum.
38.Sincef(c)≥f(x) for allxin some open interval aroundc,and likewisef(c)≤f(x) for allxin some
open interval aroundc, it follows thatfmust be constant on some open interval containingc.
39.Iffis not diﬀerentiable on (a, b), thenfhas a critical point at each pointcin (a, b) wheref
ﬀ
(c)does
not exist. Iffis diﬀerentiable on (a, b), then by the mean-value theorem there existscin (a, b) where
f
ﬀ
(c)=[f(b)−f(a)]/(b−a)=0.This meanscis a critical point off.
40.We give a proof by contradiction. Suppose for nocin (c1,c2)isf(c) a local minimum. By Theorem
2.6.2,fhas a minimum on [c1,c2] and therefore, this minimum must occur atc1orc2.Suppose that
f(c1) is an endpoint minimum. Then for someδ1>0,
(*) f(x)≥f(c),x∈[c1,c1+δ1).
Sincef(c1) is a local maximum, there existsδ2>0 such that
(**) f(x)≤f(c1),x∈(c1−δ2,c1+δ2).
Setδ= min [δ1,δ2].From (*) and (**), it follows that
f(x)=f(c1),x∈(c1,c1+δ).
This means thatfhas a local minimum on (c1,c2).The argument atc2is similar.
41.Letf(x)=
⎧
⎨
⎩
1,ifxis a rational number
0,ifxis an irrational number
42.f(x) = sinxandf(x) = cosxeach have an inﬁnite number of local maxima and local minima occur-
ring at distinct points. However, the local maximum values are all the same, 1, and the local minimum
values are all the same,−1. The functionf(x)=
1
2
x+ sinxhas an inﬁnite number of local maxima
and local minima, and the local maximum values and the local minimum values are all diﬀerent.
43.LetMbe a positive number. Then
P(x)−M≥anx
n
−

|an−1|x
n−1
+···+|a1|x+|a0|+M

forx>0
≥anx
n
−x
n−1
(|an−1|+···+|a1|+|a0|+M) forx>1
≥x
n−1
[anx−(|an−1|+···+|a1|+|a0|+M)]
It now follows that
P(x)−M≥0 for x≥K=
|an−1|+···+|a1|+|a0|+M
an
1/n
.

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146 SECTION 4.4
44.LetRbe a rectangle with its diagonals having lengthc,and letxbe the length of one of its sides.
Then the length of the other side isy=
√
c
2
−x
2
and the area ofRis given by
A(x)=x

c
2
−x
2
Now
A
ﬀ
(x)=

c
2
−x
2
−
x
2
√
c
2
−x
2
=
c
2
−2x
2
√
c
2
−x
2
,
and
A
ﬀ
(x)=0=⇒x=
√
2
2
c
It is easy to verify thatAhas a maximum atx=
√
2
2
c.Sincey=
√
2
2
cwhenx=
√
2
2
c,it follows that
the rectangle of maximum area is a square.
45.f(0) =f(1) = 0 andf(x)>0on(0,1).
f
ﬀ
(x)=−qx
p
(1−x)
q−1
+px
p−1
(1−x)
q
;f
ﬀ
(x) = 0 impliesx=
p
p+q
. The absolute maximum value
offisf(p/(p+q)) =
Δ
p
p+q
θ
p
·
Δ
q
p+q
θ
q
46.LetS=x
3
+y
3
wherex+y=16.ThenS(x)=x
3
+ (16−x)
3
and
S
ﬀ
(x)=3x
2
−3(16−x)
2
= 96(x−8);S
ﬀ
(x)=0 =⇒x=8 =⇒y=8;
S
ﬀﬀ
(x) = 96;S
ﬀﬀ
(8) = 96>0=⇒Shas a local minimum atx=8.
It now follows thatS(8) is the absolute minimum ofS.
47.SettingR
ﬀ
(θ)=
v
2
cos 2θ
16
=0,givesθ=
π
4
.SinceR
ﬀﬀ
ν
π
4
ω
=−
v
2
8
<0,θ=
π
4
is a maximum.
48.Cut the wire into two pieces, one of lengthxand the other of lengthL−x.Suppose that the wire of
lengthxis used to form the equilateral triangle, and the other piece is used to form the square. Then
the area of the triangle is
√
3x
2
/36,and the area of the square is (L−x)
2
/16.Now, let
S(x)=
√
3
36
x
2
+
1
16
(L−x)
2
Then
S
ﬀ
(x)=
√
3
18
x−
1
8
(L−x)
=
4
√
3+9
72
x−
1
8
L
SettingS
ﬀ
(x) = 0 we ﬁnd that
x=
9
4
√
3+9
L.
∼
=0.5650L
Now,
S(0) =
1
16
L
2
=0.0625L
2
(absolute maximum)

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SECTION 4.4 147
S
Δ
9
4
√
3+9
L
θ
∼
=0.0390L
2
(absolute minimum)
S(L)=
√
3
36
L
2
=0.0481L
2
To maximize the sum of the areas, use the wire to form a square; to minimize the sum, usex
∼
=0.5650L
to form the triangle and the remainder to form the square.
49. critical pts:x1=−1.452,x2=0.760
f(−1.452) local maximum
f(0.727) local minimum
f(3) absolute maximum
f(−2.5) absolute minimum
50. critical pts:x1=−2.179,x2=1,x3=1.158
f(−3) endpoint maximum
f(1) local maximum
f(−2.158),f(1.158) local minima
f(3) absolute maximum
f(−2.179) absolute minimum
51. critical points:x1=−1.683,x2=−0.284,
x3=0.645,x4=1.760
f(−1.683),f(0.645) local maxima
f(−0.284),f(1.760) local minima
f(π) absolute maximum
f(−π) absolute minimum
52. critical pts:x1=−0.667,x2=0
f(−0.667) local maximum
f(0) local minimum
f(1) absolute maximum
f(−3) absolute minimum

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148 SECTION 4.5
53.Yes;M=f(2) = 1;m=f(1) =f(3) = 0
1 2 3
x
0.5
1
y
54.No;fis discontinuous atx=3;M=f(3) =
7
2
;m=f(0) =−
19
4
55.Yes;M=f(6) = 2 +
√
3;m=f(1) =
3
2
1 4 6
x
1
2
3
y
SECTION 4.5
1.SetP=xyandy=40−x.We want to maximize
P(x)=x(40−x),0≤x≤40.
P
ﬀ
(x)=40−2x, P
ﬀ
(x)=0 =⇒x=20.
SincePincreases on (0,20] and decreases on [20,40),the abs max ofPoccurs whenx=20.Then,
y= 20 andxy= 400.
The maximal value ofxyis 400.
2.SetA=xyand 2x+2y=24ory=12−x.We want to maximize
A(x)=x(12−x),0≤x≤12.
A
ﬀ
(x)=12−2x, P
ﬀ
(x)=0 =⇒x=6.
SinceAincreases on [0,6] and decreases on [6,12],the abs max ofAoccurs whenx=6.Then,y=6.
The dimensions of the rectangle having perimeter 24 and maximum area are: 6×6.
3.
MinimizeP
P=x+2y,200 =xy, y= 200/x
P(x)=x+
400
x
,x>0.

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SECTION 4.5 149
P
ﬀ
(x)=1−
400
x
2
,P
ﬀ
(x)=0 =⇒x=20.
SincePdecreases on (0,20 ] and increases on [20,∞),the abs min ofPoccurs whenx=20.
To minimize the fencing, make the garden 20 ft (parallel to barn) by 10 ft.
4. MaximizeA
A=2xy, y=4−x
2
A(x)=2x(4−x
2
)=8x−2x
3
,0≤x≤2.
A
ﬀ
(x)=8−6x
2
,P
ﬀ
(x)=0 =⇒x=
2
√
3
.
SinceAincreases on [0,2/
√
3] and decreases on [2/
√
3,2],the abs max ofAoccurs whenx=2/
√
3.
The maximal area is
32
9
√
3.
5. MaximizeA
A=xy, x
2
+y
2
=8
2
,y=
√
64−x
2
A(x)=x
√
64−x
2
,0≤x≤8.
A
ﬀ
(x)=
√
64−x
2
+x
Δ
−x
√
64−x
2
θ
=
64−2x
2
√
64−x
2
,A
ﬀ
(x)=0 =⇒x=4
√
2.
SinceAincreases on (0,4
√
2 ] and decreases on [4
√
2,8),the abs max ofAoccurs whenx=4
√
2.
Then,y=4
√
2 andxy=32.
The maximal area is 32.
6.MaximizeP=2x+2y;xy=A(constant) =⇒y=
A
x
P(x)=2x+
2A
x
,x>0;P
ﬀ
(x)=2−
2A
x
2
;
P
ﬀ
(x)=0 =⇒x=
√
A=⇒y=
√
A
The rectangle having a given areaAand minimum perimeter is a square of side length
√
A.
7.
MinimizeP=3x+2y
A=xy=15,000,y=
15,000
x

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150 SECTION 4.5
P(x)=3x+
30,000
x
,0<x<∞.
P
ﬀ
(x)=3−
30,000
x
2
=
3(x
2
−10,000)
x
2
;P
ﬀ
(x)=0 =⇒x= 100.
SincePdecreases on (0,100] and increases on [100,∞),the abs min ofPoccurs whenx= 100.When
x= 100,y= 150; at least 600 feet of fencing is needed.
8.MinimizeC= 300y+ 400x
A= 5000 =xy⇒y=
5000
x
;p
C(x)=
1,500,000
x
+ 400x, x >0;
C
ﬀ
(x)=−
1,500,000
x
2
+ 400;
C
ﬀ
(x)=0 =⇒x
∼
=61.24.
C
ﬀﬀ
(x)=
3,000,000
x
3
;C
ﬀﬀ
(61.24)>0.
Chas an abs min atx=61.24.The dimensions that will minimize the cost are:x=61.24,y=81.65.
9.
MaximizeL
To account for the semi-circular portion admitting less
light per square foot, we multiply its area by 1/3.
L=2xy+
1
3
Δ
πx
2
2
θ
,
2x+2y+πx=p, y=
1
2
(p−2x−πx)
L=2x
Δ
p−2x−πx
2
θ
+
1
6
πx
2
L(x)=px−
Δ
2+
5
6
π
θ
x
2
,0≤x≤
p
2+π
.
L
ﬀ
(x)=p−
Δ
4+
5
3
π
θ
x;L
ﬀ
(x)=0 =⇒x=
3p
12 + 5π
.
SinceL
ﬀﬀ
(x)<0 for allxin the domain ofL, the local max atx=3p/(12+5π) is the abs max.
For the window that admits the most light, take the radius of the semicircle as
3p
12+5π
ft.
10. MaximizeA
A=xy, x+2y= 800
A(y) = (800−2y)y= 800y−2y
2
,0≤y≤400.
A
ﬀ
(y) = 800−4y, A
ﬀ
(y)=0 =⇒y= 200.
SinceAincreases on [0,200] and decreases on [200,400],the abs max ofAoccurs wheny= 200.
The dimensions of the ﬁeld of maximum area are: 200×400.

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SECTION 4.5 151
11. MaximizeA
A=xy,
3
4
=
y
4−x
(similar triangles)
y=
3
4
(4−x)
A(x)=
3x
4
(4−x),0≤x≤4.
A
ﬀ
(x)=3−
3x
2
,A
ﬀ
(x)=0 =⇒x=2.
SinceAincreases on (0,2] and decreases on [2,4),the abs max ofAoccurs whenx=2.
To maximize the area of the rectangle, takePas the point

2,
3
2

.
12.The equation of the third side is:y=mx+(1−m).
The base of the triangle is:b=
m−1
m
.
The two lines intersect when 3x=mx+(1−m);
=⇒x=
1−m
3−m
=⇒h=
3(1−m)
3−m
.
We want to minimize
A(m)=
1
2
m−1
m
3(1−m)
3−m
=−
3
2
(1−m)
2
3m−m
2
,m< 0.
A
ﬀ
(m)=−
3
2
(m+ 3)(m−1)
(3m−m
2
)
2
,A
ﬀ
(m)=0 =⇒m=−3.
The area of the triangle is a minimum when the slope of the line is−3.
13. MinimizeA
A=
1
2
(x-intercept) (y-intercept)
Equation of line:y−5=m(x−2)
x-intercept: 2−
5
m
y-intercept: 5−2m
A=
1
2
Δ
2−
5
m
θ
(5−2m)=10−2m−
25
2m
A(m)=10−2m−
25
2m
,m<0.
A
ﬀ
(m)=−2+
25
2m
2
,A
ﬀ
(m)=0 =⇒m=−
5
2
.
SinceA
ﬀﬀ
(m)=−25/m
3
>0 form<0,the local min atm=−5/2 is the abs min.
The triangle of minimal area is formed by the line of slope−5/2.

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152 SECTION 4.5
14.Since lim
m→0
−
A(m)=+∞,no minimum exists.
15. MaximizeV
V=2x
2
h,2

2x
2
+xh+2xh

= 100,h=
50−2x
2
3x
V=2x
2
Δ
50−2x
2
3x
θ
V(x)=
100
3
x−
4
3
x
3
,0≤x≤5.
V
ﬀ
(x)=
100
3
−4x
2
,V
ﬀ
(x)=0 =⇒x=
5
3
√
3.
SinceV
ﬀﬀ
(x)=−8x<0on(0,5),the local max atx=
5
3
√
3 is the abs max.
The base of the box of greatest volume measures
5
3
√
3 in. by
10
3
√
3 in.
16.With no top, we have 2x
2
+2xh+4xh= 100,orh=
50−x
2
3x
.
MaximizeV(x)=2x
2
Δ
50−x
2
3x
θ
=
2x
3
(50−x
2
),0≤x≤5
√
2.
V
ﬀ
(x)=
100
3
−2x
2
,V
ﬀ
(x)=0 =⇒x=
5
3
√
6.
SinceV
ﬀﬀ
(x)=−4x<0on(0,5
√
2),the local max atx=
5
3
√
6 is the abs max.
The base of the box of greatest volume measures
5
3
√
6 in. by
10
3
√
6 in.
17. MaximizeA
A=
1
2
hy
2x+y=12 =⇒y=12−2x
Pythagorean Theorem:
h
2
+
ν
y
2
ω
2
=x
2
=⇒h=

x
2
−
ν
y
2
ω
2
Thus,h=

x
2
−(6−x)
2
=
√
12x−36.
A(x)=(6−x)
√
12x−36,3≤x≤6.
A
ﬀ
(x)=−
√
12x−36+(6−x)
Δ
6
√
12x−36
θ
=
72−18x
√
12x−36
,
A
ﬀ
(x)=0 =⇒x=4.
SinceAincreases on (3,4] and decreases on [4,6),the abs max ofAoccurs atx=4.
The triangle of maximal area is equilateral with side of length 4.
18.It is suﬃcient to minimize the square of the distance:
S=(x−0)
2
+(y−6)
2
=8y+(y−6)
2
sincex
2
=8ywherey≥0.
S
ﬀ
(y)=2y−4,S
ﬀ
(y)=0 =⇒y=2.
The points on the parabola that are closest to (0,6) are: (4,2) and (−4,2).

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SECTION 4.5 153
19. Minimized
d=

(y
2
−0)
2
+(y−3)
2
The square-root function is increasing;
dis minimal whenD=d
2
is minimal.
D(y)=y
4
+(y−3)
2
,yreal.
D
ﬀ
(y)=4y
3
+2(y−3) = (y−1)

4y
2
+4y+6

,D
ﬀ
(y)=0 aty=1.
SinceD
ﬀﬀ
(y)=12y
2
+2>0,the local min aty= 1 is the abs min.
The point (1,1) is the point on the parabola closest to (0,3).
20.f(x)=Ax
−1/2
+Bx
1/2
,f(9) = 6 =⇒
1
3
A+3B=6.
f
ﬀ
(x)=
−A
2x
3/2
+
B
2x
1/2
;f
ﬀ
(9) = 0 =⇒
−A
54
+
B
6
=0.
Solving the two equations gives:A=9,B=1.
21.The ﬁgure shows a rectangle inscribed in the ellipse 16x
2
+9y
2
= 144. The area of the rectangle is:
A=(2x)(2y)=4xy. Solving the equation of the ellipse fory, we gety=
4
3
√
9−x
2
.
MaximizeA=
16
3
x
√
9−x
2
,0≤x≤3.
A
ﬀ
(x)=
16
3

9−x
2
−
x
2
√
9−x
2

=
16
3
Δ
9−2x
2
√
9−x
2
θ
A
ﬀ
(x)=0 =⇒x=3/
√
2.
SinceA(0) =A(3) = 0, we can conclude thatA(3/
√
2)
is the absolute maximum value ofA.
Atx=3/
√
2,y=4/
√
2 andA= 24; the maximum possible
area for a rectangle inscribed in the ellipse is 24 sq. units.
x
y
22.Simply repeat the solution of Exercise 21, replacing 9 bya
2
and 16 by b
2
. The result is:
x=a/
√
2,y=b/
√
2; the maximum possible area of a rectangle inscribed in the ellipse isA=
4(a/
√
2)(b/
√
2) = 2absquare units.
23. MaximizeA
A=xy+
√
3
4
x
2
,30 = 3x+2y, y=
30−3x
2
A(x)=15x−
3
2
x
2
+
√
3
4
x
2
,0≤x≤10.
A
ﬀ
(x)=15−3x+
√
3
2
x, A
ﬀ
(x)=0 =⇒x=
30
6−
√
3
=
10
11
(6 +
√
3).

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154 SECTION 4.5
SinceA
ﬀﬀ
(x)=−3+
√
3
2
<0on(0,10),the local max atx=
10
11
ν
6+
√
3
ω
is the abs max.
The pentagon of greatest area is composed of an equilateral triangle with side
10
11
ν
6+
√
3
ω
∼
=7.03
in. and rectangle with height
15
11
ν
5−
√
3
ω
∼
=4.46 in.
24.To maximize the area, use the cross-section that is wider at the top.
A(h)=4h+h
√
16−h
2
,0≤h≤4;
A
ﬀ
(h)=4+
16−2h
2
√
16−h
2
;
A
ﬀ
(h)=0 =⇒h=2
√
3.
The depth of the gutter that has maximum carrying capacity is: 2
√
3 inches.
25. MaximizeV
V=x(8−2x)(15−2x)
x≥0
8−2x≥0
15−2x≥0
⎫
⎪
⎪
⎬
⎪
⎪
⎭
=⇒0≤x≤4
V(x) = 120x−46x
2
+4x
3
,0≤x≤4.
V
ﬀ
(x) = 120−92x+12x
2
= 4(3x−5)(x−6),V
ﬀ
(x)=0 atx=
5
3
.
SinceVincreases on

0,
5
3

and decreases on [
5
3
,4),the abs max ofVoccurs whenx=
5
3
.
The box of maximal volume is made by cutting out squares 5/3 inches on a side.
26. MinimizeP
P=2x+2y;
(x−4)(y−6) = 81;y=
81
x−4
+6.
P(x)=2x+2
Δ
81
x−4
+6
θ
,x>4.
P
ﬀ
(x)=2−
162
(x−4)
2
,P
ﬀ
(x)=0 =⇒x=13.
SinceP
ﬀﬀ
(x)=
324
(x−4)
3
>0 whenx>4,x= 13 is the abs min.
The most economical page has dimensions: width 13 cm, length 15 cm.

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SECTION 4.5 155
27. MinimizeAP+BP+CP=S
lengthAP=

9+y
2
lengthBP=6−y
lengthCP=

9+y
2
S(y)=6−y+2

9+y
2
,0≤y≤6.
S
ﬀ
(y)=−1+
2y

9+y
2
,S
ﬀ
(y)=0 =⇒y=
√
3.
Since
S(0) = 12,S
ν√
3
ω
=6+3
√
3
∼
=11.2,andS(6) = 6
√
5
∼
=13.4,
the abs min ofSoccurs wheny=
√
3.
To minimize the sum of the distances, takePas the point

0,
√
3

.
28.Refer to Exercise 27. Here we want to minimize
S(y)=3−y+2

36 +y
2
,0≤y≤3.
S
ﬀ
(y)=−1+
2y

36 +y
2
,S
ﬀ
(y)=0 =⇒y=
√
12>3.
Thus, the minimum must occur at one of the endpoints:S(0) = 15,S(3) = 2
√
45<S(0).
To minimize the sum of the distances, takeP=(0,3).
29. MinimizeL
L
2
=y
2
+(x+1)
2
.
By similar triangles
y
x+1
=
8
x
,y=
8
x
(x+1).
L
2
=

8
x
θ
(x+1)

2
+(x+1)
2
=(x+1)
2
Δ
64
x
2
+1
θ
SinceLis minimal whenL
2
is minimal, we consider the function
f(x)=(x+1)
2
Δ
64
x
2
+1
θ
,x>0.
f
ﬀ
(x)=2(x+1)
Δ
64
x
2
+1
θ
+(x+1)
2
Δ
−128
x
3
θ
=
2(x+1)
x
3

x
3
−64

,f
ﬀ
(x)=0 =⇒x=4.
Sincefdecreases on (0,4] and increases on [4,∞),the abs min offoccurs whenx=4.
The shortest ladder is 5
√
5 ft long.

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156 SECTION 4.5
30.MaximizeL=x+y.
By similar triangles,
y
6
=
x
√
x
2
−64
L(x)=x+
6x
√
x
2
−64
,x>8
L
ﬀ
(x)=1−
384
(x
2
−64)
3/2
L
ﬀ
(x)=0 =⇒x=

64 + (384)
2/3∼
=10.81
L(10.81)
∼
=19.73; the longest ladder is approximately 19.7 ft.
31.
MaximizeA
(We use feet rather than inches to reduce arithmetic.)
A=(L−1)

W−
4
3

LW=27 =⇒W=
27
L
A=(L−1)
Δ
27
L
−
4
3
θ
=
85
3
−
27
L
−
4
3
L
A(L)=
85
3
−
27
L
−
4
3
L,1≤L≤
81
4
.
A
ﬀ
(L)=
27
L
2
−
4
3
,A
ﬀ
(L)=0 =⇒L=
9
2
.
SinceA
ﬀ
(L)=−54/L
3
<0 for 1<L<
81
4
,the max atL=
9
2
is the abs max.
The banner has length 9/2 ft = 54 in. and height 6 ft = 72 in.
32.AssumeV=
1
3
πr
2
h=1,orh=
3
πr
2
.
Minimize surface areaS=πr
√
r
2
+h
2
=πr

r
2
+
9
π
2
r
4
=
√
π
2
r
6
+9
r
.
dS
dr
=
2π
2
r
6
−9
r
2
√
π
2
r
6
+9
=0 =⇒r=
Δ
9
2π
2
θ
1/6
=
3
1/3
2
1/6
π
1/3
=⇒h=
3
π
Δ
9
2π
2
θ
1/3
=
3
1/3
2
1/3
π
1/3
=r
√
2.
33.
Find the extreme values ofA
A=πr
2
+x
2
2πr+4x=28 =⇒x=7−
1
2
πr.
A(r)=πr
2
+
Δ
7−
1
2
πr
θ
2
,0≤r≤
14
π
.

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SECTION 4.5 157
Note: the endpoints of the domain correspond to the instances when the string is not cut:r=0
when no circle is formed,r=14/πwhen no square is formed.
A
ﬀ
(r)=2πr−π
Δ
7−
1
2
πr
θ
,A
ﬀ
(r)=0 =⇒r=
14
4+π
.
SinceA
ﬀﬀ
(r)=2π+π
2
/2>0on(0,14/π),the abs min ofAoccurs whenr=14/(4 +π) and
the abs max ofAoccurs at one of the endpts:A(0) = 49,A(14/π) = 196/π >49.
(a) To maximize the sum of the two areas, use all of the string to form the circle.
(b) To minimize the sum of the two areas, use 2πr=28π/(4 +π)
∼
=12.32 inches of string for the
circle.
34.MaximizeV=x
2
hgiven thatx
2
+4xh=12 =⇒h=
12−x
2
4x
.
V(x)=x
2
Δ
12−x
2
4x
θ
=3x−
1
4
x
3
,0<x≤
√
12.
V
ﬀ
(x)=3−
3
4
x
2
,V
ﬀ
(x)=0 =⇒x=2.SinceVincreases on (0,2] and decreases on [2,
√
12),
Vhas an abs max atx= 2; the maximum volume isV(2) = 4 cu ft.
35.
MaximizeV
V=πr
2
h
By similar triangles
8
5
=
h
5−r
orh=
8
5
(5−r).
V(r)=
8π
5
r
2
(5−r),0≤r≤5.
V
ﬀ
(r)=
8π
5

10r−3r
2

,V
ﬀ
(r)=0 =⇒r=10/3.
SinceVincreases on (0,10/3] and decreases on [10/3,5),the abs max ofVoccurs whenr=10/3.
The cylinder with maximal volume has radius 10/3 and height 8/3.
36.MaximizeA=2πrh=
16π
5
r(5−r),0≤r≤5,

h=
8
5
(5−r) from Exercise 33

.
A
ﬀ
(r)=
16π
5
(5−2r),A
ﬀ
(r)=0 =⇒r=
5
2
The curved surface is a maximum whenr=
5
2
,h=4.
37. MinimizeC
In dollars,
C= cost base + cost top + cost sides
=.35

x
2

+.15

x
2

+.20(4xy)
=
1
2
x
2
+
4
5
xy
Volume =x
2
y= 1250y=
1250
x
2

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158 SECTION 4.5
C(x)=
1
2
x
2
+
1000
x
,x>0.
C
ﬀ
(x)=x−
1000
x
2
,C
ﬀ
(x)=0 =⇒x=10.
SinceC
ﬀﬀ
(x) = 1 + 2000/x
3
>0 forx>0,the local min ofCatx= 10 is the abs min.
The least expensive box is 12.5 ft tall with a square base 10 ft on a side.
38.MaximizeA=xy.By similar triangles
y
b−x
=
h
b
,so
A(x)=
h
b
x(b−x),0≤x≤b.
A
ﬀ
(x)=
h
b
(b−2x),A
ﬀ
(x)=0 ⇒x=
b
2
.
SinceAis increasing on [0,b/2] and decreasing on [b/2,b],Ahas an abs max atx=b/2
A(b/2) =
1
4
hb=
1
2
area of triangleABC.
39.
MinimizeA
A=
1
2
(h)(2x)=hx
TrianglesADCandABEare similar:
AD
DC
=
AB
BE
or
h
x
=
AB
r
.
Pythagorean Theorem:
r
2
+(AB)
2
=(h−r)
2
.
Thus
r
2
+
Δ
hr
x
θ
2
=(h−r)
2
.
Solving this equation forhwe ﬁnd that
h=
2x
2
r
x
2
−r
2
.
A(x)=
2x
3
r
x
2
−r
2
,x>r.
A
ﬀ
(x)=

x
2
−r
2

6x
2
r

−2x
3
r(2x)
(x
2
−r
2
)
2
=
2x
2
r

x
2
−3r
2

(x
2
−r
2
)
2
,
A
ﬀ
(x)=0 =⇒x=r
√
3.

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SECTION 4.5 159
SinceAdecreases on

r, r
√
3

and increases on

r
√
3,∞

,the local min atx=r
√
3 is the abs
min ofA. Whenx=r
√
3,we geth=3rso thatFC=2r
√
3 andAF=FC=
√
h
2
+x
2
=
2r
√
3.The triangle of least area is equilateral with side of length 2r
√
3.
40.MaximizeA(x)=
1
2
(r+x)2
√
r
2
−x
2
=(r+x)
√
r
2
−x
2
,0≤x≤r.
A
ﬀ
(x)=
r
2
−rx−2x
2
√
r
2
−x
2
;
A
ﬀ
(x)=0 =⇒x=
r
2
.
SinceAincreases on [0,r/2] and decreases on [r/2,r],Ahas an abs max atx=r/2;
A(r/2) =
3
√
3
4
r
2
.
41. MaximizeV
V=πr
2
h
By the Pythagorean Theorem,
(2r)
2
+h
2
=(2R)
2
so
h=2
√
R
2
−r
2
.
V(r)=2πr
2
√
R
2
−r
2
,0≤r≤R.
V
ﬀ
(r)=2π

2r
√
R
2
−r
2
−
r
3
√
R
2
−r
2

=
2πr

2R
2
−3r
2

√
R
2
−r
2
V
ﬀ
(r)=0 =⇒r=
1
3
R
√
6.
SinceVincreases on

0,
1
3
R
√
6

and decreases on

1
3
R
√
6,R

,the local max atr=
1
3
R
√
6is
the abs max.
The cylinder of maximal volume has base radius
1
3
R
√
6 and height
2
3
R
√
3.
42.MaximizeA=2πrh=4πr
√
R
2
−r
2
,0≤r≤R,

h=2
√
R
2
−r
2
from Exercise 39

.
A
ﬀ
(r)=
4π(R
2
−2r
2
)
√
R
2
−r
2
,A
ﬀ
(r)=0 =⇒r=
R
√
2
.
The curved surface is a maximum whenr=
R
√
2
,h=R
√
2.

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160 SECTION 4.5
43.
MaximizeV
V=
1
3
πr
2
h
Pythagorean Theorem
Case 1 : (h−R)
2
+r
2
=R
2
Case 2 : (R−h)
2
+r
2
=R
2
Case 1 :h≥RCase 2 :h≤R In both cases
r
2
=R
2
−(R−h)
2
=2hR−h
2
.
V(h)=
1
3
π

2h
2
R−h
3

,0≤h≤2R.
V
ﬀ
(h)=
1
3
π

4hR−3h
2

,V
ﬀ
(h)=0 at h=
4R
3
.
SinceVincreases on

0,
4
3
R

and decreases on

4
3
R,2R

,the local max ath=
4
3
Ris the abs
max.
The cone of maximal volume has height
4
3
Rand radius
2
3
R
√
2.
44.MaximizeV=
1
3
πr
2
h,wherer
2
+h
2
=a
2
.
V(h)=
1
3
π(a
2
−h
2
)h,0≤h≤a,
V
ﬀ
(h)=
1
3
π(a
2
−3h
2
),V
ﬀ
(h)=0 ⇒h=
a
√
3
.
Maximum volumeV

a/
√
3

=
2
27
πa
3
√
3.
45. MinimizeC
In units of $10,000,
C=
cost of cable
underground
+
cost of cable
under water
= 3(4−x)+5
√
x
2
+1.
Clearly, the cost is unnecessarily high if
x>4orx<0.
C(x)=12−3x+5
√
x
2
+1,0≤x≤4.
C
ﬀ
(x)=−3+
5x
√
x
2
+1
,C
ﬀ
(x)=0 =⇒x=3/4.

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SECTION 4.5 161
Since the domain ofCis closed, the abs min can be identiﬁed by evaluatingCat each critical point:
C(0) = 17,C

3
4

=16,C(4) = 5
√
17
∼
=20.6.
The minimum cost is $160,000.
46.Maximizeα−θ.
Since the tangent function is an increasing function
on [0,π/2), it suﬃces to maximize tan(α−θ).
tan(α−θ)=
tanα−tanθ
1 + tanαtanθ
; tanα=
16
x
,tanθ=
9
x
.
Thus, we maximizeT(x)=
16
x
−
9
x
1+
16
x
·
9
x
=
7x
x
2
+ 144
,x>0.
T
ﬀ
(x)=
7(144−x
2
)
(x
2
+ 144)
2
,T
ﬀ
(x)=0 =⇒x=12.
Stand 12 ft from the wall for the most favorable view.
47.P
ﬀ
(θ)=
−mW(mcosθ−sinθ)
(msinθ+ cosθ)
2
;Pis minimized when tanθ=m.
48.R(θ)=
2v
2
gcos
2
α
cosθsin(θ−α),0<θ<
1
2
π
R
ﬀ
(θ)=
2v
2
gcos
2
α
[−sinθsin(θ−α) + cosθcos(θ−α)]
=
2v
2
gcos
2
α
cos(2θ−α)
R
ﬀ
(θ)=0 =⇒2θ−α=
1
2
π=⇒θ=
1
4
π+
1
2
α.
49.MinimizeI=
a
x
2
+
b
(s−x)
2
.
I
ﬀ
(x)=−
2a
x
3
+
2b
(s−x)
3
,I
ﬀ
(x)=0 =⇒x=
a
1
3s
a
1
3+b
1
3
.
50.MinimizeD=(y−y1)
2
+(x−x1)
2
,wherey=−
1
B
(Ax+C).
D
ﬀ
=2
Δ
−
1
B
(Ax+C)−y1
∈√
−
A
B
θ
+2(x−x1)=0
=⇒x=
B
2
x1−AC−ABy1
A
2
+B
2
and thusy=
A
2
y1−BC−ABx1
A
2
+B
2
.
Thusd=
√
D=
⇐
Δ
A
2
y1−BC−ABx1
A
2
+B
2
−y1
θ
2
+
Δ
B
2
x1−AC−ABy1
A
2
+B
2
−x1
θ
2
=
|Ax1+By1+C|
√
A
2
+B
2
.

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
162 SECTION 4.5
51.The slope of the line through (a, b) and (x, f(x)) is
f(x)−b
x−a
.
LetD(x)=[x−a]
2
+[b−f(x)]
2
.ThenD
ﬀ
(x)=0
=⇒2[x−a]−2[b−f(x)]f
ﬀ
(x)=0
=⇒f
ﬀ
(x)=
x−a
b−f(x)
.
52.LetP=(x1,x
2
1) andQ=(x2,x
2
2).The slope of the linePQis
x
2
1−x
2
2
x1−x2
=x1+x2.This is
perpendicular to the slope of the tangent to the parabola through pointP,which is 2x1.Thus
x1+x2=−
1
2x1
=⇒x2=−x1−
1
2x1
.Now minimize:
D=(x
2
1−x
2
2)
2
+(x1−x2)
2
=(x
2
1−x
2
2)
2
(1+(x1+x2)
2
)
=
Δ
2x1+
1
2x1
θ
2Δ
1+
1
4x
2
1
θ
ThusD
ﬀ
=
Δ
2x1+
1
2x1
θ
2Δ
−
1
2x
3
1
θ
+2
Δ
2x1+
1
2x1
∈√
2−
1
2x
2
1
∈√
1+
1
4x
2
1
θ
=0
=⇒
Δ
2x1+
1
2x1
∈√
1
2x
3
1
θ
=2
Δ
2−
1
2x
2
1
∈√
1+
1
4x
2
1
θ
=⇒x1=±
√
2
2
,andy1=
1
2
.
53.SetF(x)=6x
4
−16x
3
+9x
2
. For integral values ofx,F(x)=f(x).
F
ﬀ
(x)=24x
3
−48x
2
+18x=6x(4x
2
−8x+3)=6x(2x−1)(2x−3)
F
ﬀ
(x)=0atx=0,1/2,3/2.
F
ﬀ
(x)<0on(−∞,0)∪(1/2,3/2);F
ﬀ
(x)>0on(0,1/2)∪(3/2,∞).Fhas local minima atx=0
andx=3/2;Fhas a local maximum atx=1/2.F(0) =f(0) = 0,F(1) =f(1) =−1,F(2) =
f(2) = 4;n= 1 minimizesf(n).
54.Letxbe the number of passengers andRthe revenue in dollars.
R(x)=
δ
37x,16≤x≤35

37−
1
2
(x−35)

x,35<x≤48;
R
ﬀ
(x)=
δ
37,16<x<35
109
2
−x,35<x<48.
The critical number is 35.FromR(16) = 592,R(35) = 1295,andR(48) = 1464 we conclude that
the revenue is maximized by taking a full load of 48 passengers.

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.5 163
55.Letxbe the number of customers andPthe net proﬁt in dollars. Then 0≤x≤250 and
P(x)=
δ
12x,0≤x≤50
[12−0.06(x−50)]x,50<x≤250;
P
ﬀ
(x)=
δ
12,0≤x≤50
15−0.12x,50<x≤250.
The critical points are:x=50,x= 125. FromP(0) = 0,P(50) = 600,P(125) = 937.50,and
P(250) = 0,we conclude that the net proﬁt is maximized by servicing 125 customers.
56.MaximizeP(x)=2cy+cx,wherecis the price of low-grade steel andy=
Δ
40−5x
10−x
θ
.
P(x)=2c(
40−5x
10−x
)+cx
P
ﬀ
(x)=2c

(10−x)(−5)−(40−5x)(−1)
(10−x)
2

+c,
P
ﬀ
(x)=0 =⇒x=10−
√
20 or about 5
1
2
tons.
57.y=mx−
1
400
(m
2
+1)x
2
.Wheny=0,x=
800m
m
2
+1
.
Diﬀerentiatingxwith respect tom, x
ﬀ
=
800−800m
2
(m
2
+1)
2
=0
=⇒m=1.
58.Whenx= 300,y= 300m−
225
2
(m
2
+1).
Diﬀerentiatingywith respect tom, y
ﬀ
= 300−225m=0
=⇒m=
4
3
.
59.Driving atνmph, the trip takes
300
ν
hours and uses
Δ
1+
1
400
ν
2
θ
300
ν
gallons of fuel.
Thus, the expenses are:
E(ν)=2.60
Δ
1+
1
400
ν
2
θ
300
ν
+20
Δ
300
ν
θ
=1.95ν+
6780
ν
,35≤x≤70.
Diﬀerentiating, we get
E
ﬀ
(ν)=1.95−
6780
ν
2
,andE
ﬀ
(ν)=0atν
∼
=59
Eis decreasing on [35,59] and increasing on [59,70]; the minimal expenses occur
when the truck is driven at 59 mph.
60.Letνbe the speed of the boat measured in kilometers per hour. A 100 kilometer trip will take
100/νhours.

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
164 SECTION 4.5
Fixed costs:F(ν) = 2500
Δ
100
ν
θ
dollars.
Fuel costs:G(ν)=kν
2
; 400 =k(10)
2
impliesk= 4. Therefore,G(ν)=4ν
2
Δ
100
ν
θ
dollars.
Total cost:C(ν) = 2500
Δ
100
ν
θ
+4ν
2
Δ
100
ν
θ
= 100
Δ
2500
ν
+4ν
θ
,0<ν<∞. Diﬀerentiating, we
get
C
ﬀ
(ν) = 100
Δ
−
2500
ν
2
+4
θ
= 100
Δ
4ν
2
−2500
ν
2
θ
andC
ﬀ
(ν)=0atν=25.
Cis decreasing on (0,25] and increasing on [25,∞); the speed that minimizes the expenses is 25
kilometers per hour. If we replace the 100 kilometers byMkilometers, the total cost will be
C(ν)=M
Δ
2500
ν
+4ν
θ
and we will get exactly the same result. The minimizing speed is independent of the length of the trip.
61.MinimizeSA=2πrh+2πr
2
,where 2r≤h<6 andπr
2
h=16π(henceh=
16
r
2
).
ThusSA=
32π
r
+2πr
2
.Diﬀerentiating,SA
ﬀ
=−
32π
r
2
+4πr=0
=⇒r
3
=8,sor= 2 feet andh= 4 feet. Thus no minimum exists.
62.We want to maximize the ratio
income
cost
=
200,000n
1,000,000n+ 100,000(1+2+···+n−1)+5,000,000
=
2n
10n+
1
2
(n−1)n+50
=
4n
n
2
+19n+ 100
.
Letf(x)=
4x
x
2
+19x+ 100
,x>0
Thenf
ﬀ
(x)=
(x
2
+19x+ 100)4−4x(2x+ 19)
(x
2
+19x+ 100)
2
=
4(100−x
2
)
(x
2
+19x+ 100)
2
=0
=⇒x=10.
Sincef
ﬀ
(x)>0 forx<0,f
ﬀ
(x)<0 forx>10,fhas an absolute maximum atx=10.
A ten story building provides the greatest return on investment.
63.Swimming at 2 miles per hour, Maggie will reach pointBin 1 hour; walking along the shore (a distance
ofπmiles), she will reach pointBinπ/5
∼
=0.63 hours. Suppose that she swims to a pointCand then
walks toB. Letθbe the central angle (measured in radians) determined by the pointsBandC. See
the ﬁgure. By the law of cosines, the square of the distance fromAtoCis given by
d
2
=1
2
+1
2
−2(1)(1) cos (π−θ) = 2 + 2 cosθ.
Therefore the distancedfromAtoCisd=
√
2 + 2 cosθ. The distance fromCtoBisθ.

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.5 165
Now, the total length of time to swim toCand then walk toBis
T(θ)=2
√
2+2cosθ+5θ
Maggie wants to minimizeT.
T
ﬀ
(θ)=
−2 sinθ
2
√
2 + 2 cosθ
+5.
SettingT
ﬀ
(θ) = 0, we get
−2 sinθ
2
√
2 + 2 cosθ
+5=0 A θ B
C
which reduces to 2 cos
2
θ+ 25 cosθ+23=0
and factors into (cosθ+ 1)(2 cosθ+ 23) = 0.
This impliesθ=π. Maggie should walk the entire distance!
64.We modify the solution of Exercise 63, replacing the walking rate of 2 miles per hour by the rowing
rate of 3 miles per hour.
The total length of time to row toCand then walk toBis
T(θ)=3
√
2+2cosθ+5θ
The derivative ofTis
T
ﬀ
(θ)=
−3 sinθ
2
√
2+2cosθ
+5.
SettingT
ﬀ
(θ) = 0, we get
−3 sinθ
2
√
2+2cosθ
+5=0,
which reduces to
9 cos
2
θ+ 50 cosθ+ 41 = (cosθ+ 1)(9 cosθ+ 41) = 0.
This impliesθ=π. Again, Maggie should walk the entire distance!
65.(b) The point on the graph offthat is closest toPis:

1+
√
2,2+
√
2

(c)lPQ:y−

2+
√
2

=
1−
√
2
3−
√
2

x−

1+
√
2

(d) and (e)lPQ=lN.
66.The point on the graph off(x)=x−x
3
that is closest to (1,8) is (approximately) (−2.1474,7.7548).
67.D(x)=

x
2
+(7−3x)
2
; the point on the line that is closest to the origin is:

21
10
,
7
10

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
166 SECTION 4.6
68.The point on the graph offthat minimizes the distance to the point (4,3) is (approximately)
(1.3918,2.0629).
PROJECT 4.5
1.Distance over water:
√
36 +x
2
.
Distance over land: 12−x.
Total energy:E(x)=W
√
36 +x
2
+L(12−x),0≤x≤12.
2.W=1.5L,soE(x)=1.5L
√
36 +x
2
+L(12−x),for 0≤x≤12.
E
ﬀ
(x)=
1.5Lx
√
36 +x
2
−L=0 =⇒x=
12
√
5
5.36.
E
ﬀ
(x)<0on(0,
12
√
5
) andE
ﬀ
(x)>0on(
12
√
5
,12),soEhas an absolute minimum at
12
√
5
.
3.(a)W=kL, k >1,soE(x)=kL
√
36 +x
2
+L(12−x),for 0≤x≤12.
E
ﬀ
(x)=
kLx
√
36 +x
2
−L=0 =⇒x=
6
√
k
2
−1
.
E
ﬀ
(x)<0on(0,
6
√
k
2
−1
) andE
ﬀ
(x)>0on(
6
√
k
2
−1
,12),soEhas an absolute minimum
at
6
√
k
2
−1
.
(b) Askincreases,xdecreases. Ask→1
+
,xincreases.
(c)x=12 =⇒k=
√
5
2
1.12.
(d) No
SECTION 4.6
1.(a)fis increasing on [a, b],[d, n];fis decreasing on [b, d],[n, p].
(b) The graph offis concave up on (c, k),(l, m);
The graph offis concave down on (a, c),(k, l),(m, p).
The x-coordinates of the points of inﬂection are:x=c, x=k, x=l, x=m.
2.(a)gis increasing on [a, b],[c, e],[m, n];gis decreasing on [b, c],[e, m].
(b) The graph ofgis concave up on (a, b),(b, d);
The graph ofgis concave down on (d, m),(m, n).
The x-coordinate of the point of inﬂection is:x=d.
3.(i)f
ﬀ
,(ii)f,(iii)f
ﬀﬀ
.
4.(a)x=−2 local max, no local minima.
(b) points of inﬂection atx=0,2.

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.6 167
(c)
−2 2
x
y
5.f
ﬀ
(x)=−x
−2
,f
ﬀﬀ
(x)=2x
−3
;
concave down on (−∞,0), concave up on (0,∞); no pts of inﬂection
6.f
ﬀ
(x)=1−x
−2
,f
ﬀﬀ
(x)=2x
−3
;
concave down on (−∞,0), concave up on (0,∞); no pts of inﬂection
7.f
ﬀ
(x)=3x
2
−3,f
ﬀﬀ
(x)=6x;
concave down on (−∞,0), concave up on (0,∞); pt of inﬂection (0,2)
8.f
ﬀ
(x)=4x−5,f
ﬀﬀ
(x) = 4; concave up on (−∞,∞)
9.f
ﬀ
(x)=x
3
−x, f
ﬀﬀ
(x)=3x
2
−1;
concave up on

−∞,−
1
3
√
3

and

1
3
√
3,∞

, concave down on

−
1
3
√
3,
1
3
√
3

;
pts of inﬂection

−
1
3
√
3,−
5
36

and

1
3
√
3,−
5
36

10.f
ﬀ
(x)=3x
2
−4x
3
,f
ﬀﬀ
(x)=6x−12x
2
=6x(1−2x);
concave down on (−∞,0), and

1
2
,∞

, concave up on

0,
1
2

;
pts of inﬂection (0,0),

1
2
,
1
16

11.f
ﬀ
(x)=−
x
2
+1
(x
2
−1)
2
,f
ﬀﬀ
(x)=
2x

x
2
+3

(x
2
−1)
3
;
concave down on (−∞,−1) and (0,1), concave up on (−1,0) and on (1,∞);
pts of inﬂection (0,0)
12.f
ﬀ
(x)=
−4
(x−2)
2
,f
ﬀﬀ
(x)=
8
(x−2)
3
;
concave down on (−∞,2), concave up on (2,∞); no pts of inﬂection
13.f
ﬀ
(x)=4x
3
−4x, f
ﬀﬀ
(x)=12x
2
−4;
concave up on

−∞,−
1
3
√
3

and

1
3
√
3,∞

, concave down on

−
1
3
√
3,
1
3
√
3

;
pts of inﬂection

−
1
3
√
3,
4
9

and

1
3
√
3,
4
9

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
168 SECTION 4.6
14.f
ﬀ
(x)=
6(1−x
2
)
(x
2
+1)
2
,f
ﬀﬀ
(x)=
12x(x
2
−3)
(x
2
+1)
3
;
concave down on

−∞,−
√
3

and

0,
√
3

,concave up on

−
√
3,0

and
√
3,∞

;
pts of inﬂection

−
√
3,−
3
2
√
3

,(0,0),
√
3,
3
2
√
3

15.f
ﬀ
(x)=
−1
√
x(1 +
√
x)
2
,f
ﬀﬀ
(x)=
1+3
√
x
2x
√
x(1 +
√
x)
3
;
concave up on (0,∞); no pts of inﬂection
16.f
ﬀ
(x)=
1
5
(x−3)
−4/5
,f
ﬀﬀ
(x)=−
4
25
(x−3)
−9/5
;
concave up on (−∞,3), concave down on (3,∞); pt of inﬂection (3,0)
17.f
ﬀ
(x)=
5
3
(x+2)
2/3
,f
ﬀﬀ
(x)=
10
9
(x+2)
−1/3
;
concave down on (−∞,−2), concave up on (−2,∞); pt of inﬂection (−2,0)
18.f
ﬀ
(x)=
4−2x
2
(4−x
2
)
1/2
,f
ﬀﬀ
(x)=
2x(x
2
−6)
(4−x
2
)
3/2
Note: dom (f)=[−2,2]
concave up on (−2,0),concave down on (0,2); pt of inﬂection (0,0)
19.f
ﬀ
(x) = 2 sinxcosx= sin 2x, f
ﬀﬀ
(x) = 2 cos 2x;
concave up on

0,
1
4
π

and

3
4
π, π

, concave down on

1
4
π,
3
4
π

;
pts of inﬂection

1
4
π,
1
2

and

3
4
π,
1
2

20.f
ﬀ
(x)=−4 cosxsinx−2x, f
ﬀﬀ
(x)=−4(cos
2
x−sin
2
x)−2=−4 cos 2x−2;
concave down on

0,
1
3
π

and

2
3
π, π

, concave up on

1
3
π,
2
3
π

;
pts of inﬂection
Δ
1
3
π,
9−2π
2
18
θ
and
Δ
2
3
π,
9−8π
2
18
θ
21.f
ﬀ
(x)=2x+ 2 cos 2x, f
ﬀﬀ
(x)=2−4 sin 2x;
concave up on

0,
1
12
π

and on

5
12
π, π

, concave down on

1
12
π,
5
12
π

;
pts of inﬂection
Δ
1
12
π,
72 +π
2
144
θ
and
Δ
5
12
π,
72+25π
2
144
θ
22.f
ﬀ
(x) = 4 sin
3
xcosx, f
ﬀﬀ
(x) = 4 sin
2
x[3 cos
2
x−sin
2
x];
concave up on

0,
1
3
π

and

2
3
π, π

, concave down on

1
3
π,
2
3
π

;
pts of inﬂection

1
3
π,
9
16

and

2
3
π,
9
16

23.points of inﬂection: (±3.94822,10.39228)
24.f
ﬀﬀ
(x)=0atx
∼
=±0.94,±2.57,±3.71,±5.35
25.points of inﬂection: (−3,0),(−2.11652,2,39953),(−0.28349,−18.43523)
26.f
ﬀﬀ
(x)>0 for allx∈domf

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.6 169
27.f(x)=x
3
−9x
(a)f
ﬀ
(x)=3x
2
−9=3(x
2
−3)
f
ﬀ
(x)≥0⇒x≤−
√
3orx≥
√
3;
f
ﬀ
(x)≤0⇒−
√
3≤x≤
√
3.
Thus,fis increasing on (−∞,−
√
3]∪[
√
3,∞)
and decreasing on [−
√
3,
√
3].
(b)f(−
√
3)
∼
=10.39 is a local maximum;
f(
√
3)
∼
=−10.39 is a local minimum.
(c)f
ﬀﬀ
(x)=6x;
The graph offis concave up on (0,∞) and concave down on (−∞,0).
(d) point of inﬂection: (0,0)
28.f(x)=3x
4
+4x
3
+1
(a)f
ﬀ
(x)=12x
3
+12x
2
=12x
2
(x+1)
f
ﬀ
(x)≥0⇒x≥−1;
f
ﬀ
(x)≤0⇒x≤−1.
Thus,fis increasing on [−1,∞)
and decreasing on (−∞,−1].
(b)f(−1) = 0 is a local minimum;
no local maxima.
(c)f
ﬀﬀ
(x)=36x
2
+24x=36x

x+
2
3

;
The graph offis concave up on

−∞,−
2
3

and (0,∞); and concave down on

−
2
3
,0

.
(d) points of inﬂection:

−
2
3
,
11
27

,(0,1)
29.f(x)=
2x
x
2
+1
(a)f
ﬀ
(x)=−
2(x+ 1)(x−1)
(x
2
+1)
2
f
ﬀ
(x)≥0⇒−1≤x≤1;
f
ﬀ
(x)≤0⇒x≤−1orx≥1.
Thus,fis increasing on [−1,1];
and decreasing on (−∞,−1]∪[1,∞).
(b)f(−1) =−1 is a local minimum;
f(1) = 1 is a local maximum.
(c)f
ﬀﬀ
(x)=
4x(x+
√
3)(x−
√
3)
(x
2
+1)
3
f
ﬀﬀ
(x)>0⇒x≤−
√
3orx≥
√
3;
f
ﬀﬀ
(x)<0⇒−
√
3<x<
√
3.
The graph offis concave up on (−
√
3,0)∪(
√
3,∞) and concave down
on (−∞,−
√
3)∪(0,
√
3).
(d) points of inﬂection: (−
√
3,−
√
3/2),(0,0),(
√
3,
√
3/2)

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
170 SECTION 4.6
30.f(x)=x
1/3
(x−6)
2/3
(a)f
ﬀ
(x)=
x−2
x
2/3
(x−6)
1/3
f
ﬀ
(x)≥0⇒x≤2,xδ=0,orx>6;
f
ﬀ
(x)≤0⇒2≤x<6.
Thus,fis increasing on (−∞,2]∪[6,∞)
and decreasing on [2,6].
(b)f(2) = 2(4)
1/3
is a local maximum;
f(6) = 0 is a local minimum.
(c)f
ﬀﬀ
(x)=
−8
x
5/3
(x−6)
4/3
;
The graph offis concave down on (0,∞) and concave down up (−∞,0).
(d) point of inﬂection: (0,0)
31.f(x)=x+ sinx, x∈[−π, π]
(a)f
ﬀ
(x) = 1 + cosx
f
ﬀ
(x)>0on(−π, π)
Thus,fis increasing on [−π, π].
(b) No local extrema
(c)f
ﬀﬀ
(x)=−sinx
f
ﬀﬀ
(x)>0 forx∈(−π,0);
f
ﬀﬀ
(x)<0 forx∈(0,π).
The graph offis concave up on (−π,0) and concave down on (0,π).
(d) point of inﬂection: (0,0)
32.f(x) = sinx+ cosx, x∈[0,2π]
(a)f
ﬀ
(x) = cosx−sinx
f
ﬀ
(x)≥0⇒0≤x≤
1
4
πor
5
4
π≤x≤2π
f
ﬀ
(x)≤0⇒
1
4
π≤x≤
5
4
π
Thus,fis increasing on

0,
1
4
π

∪

5
4
π,2π

;
fis decreasing on

1
4
π,
5
4
π

.
(b)f(π/4) =
√
2 is a local maximum;
f(5π/4) =−
√
2 is a local minimum.
(c)f
ﬀﬀ
(x)=−sinx−cosx
f
ﬀﬀ
(x)>0⇒
3
4
π<x<
7
4
π;
f
ﬀﬀ
(x)<0⇒0<x<
3
4
πor
7
4
π<x<2π.
The graph offis concave up on

3
4
π,
7
4
π

and concave down on

0,
3
4
π

∪

7
4
π,2π

.
(d) points of inﬂection:

3
4
π,0

,

7
4
π,0

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.6 171
33.f(x)=
δ
x
3
,x< 1
3x−2,x ≥1.
(a)f
ﬀ
(x)=
δ
3x
2
,x< 1
3,x ≥1;
f
ﬀ
(x)>0on(−∞,0)∪(0,∞)
Thus,fis increasing on (−∞,∞).
(b) No local extrema
(c)f
ﬀﬀ
(x)=
δ
6x, x <1
0,x ≥1;
f
ﬀﬀ
(x)>0 forx∈(0,1);f
ﬀﬀ
(x)<0 forx∈(−∞,0).
Thus, the graph offis concave up on (0,1) and concave down on (−∞,0).
The graph offis a straight line forx≥1.
(d) point of inﬂection: (0,0)
34.f(x)=
δ
2x+4,x ≤−1
3−x
2
,x> −1.
(a)f
ﬀ
(x)=
δ
2,x ≤−1
−2x, x >−1;
f
ﬀ
(x)≥0on(−∞,0];
f
ﬀ
(x)≤0on[0,∞).
Thus,fis increasing on (−∞,0] and decreasing on [0,∞).
(b)f(0) = 3 is a local maximum.
(c)f
ﬀﬀ
(x)=
δ
0,x< −1
−2,x> −1;
f
ﬀﬀ
(x)<0 forx∈(−1,∞).
Thus, the graph offis concave down on (−1,∞).
The graph offis a straight line forx≤−1.
(d) There are no points of inﬂection.
35. 36.

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172 SECTION 4.6
37. 38.
39.Sincef
ﬀﬀ
(x)=6x−2(a+b+c),setd=
1
3
(a+b+c).Note thatf
ﬀﬀ
(d) = 0 and thatfis concave
down on (−∞,d) and concave up on (d,∞); (d, f(d)) is a point of inﬂection.
40.f
ﬀ
(x)=2cx−2x
−3
,f
ﬀﬀ
(x)=2c+6x
−4
.To have a point of inﬂection at 1 we need
f
ﬀﬀ
(1) = 0 =⇒2c+6=0 = ⇒c=−3
41.Since (−1,1) lies on the graph, 1 =−a+b.
Sincef
ﬀﬀ
(x) exists for allxand there is a pt of inﬂection atx=
1
3
,we must havef
ﬀﬀ

1
3

=0.
Therefore
0=2a+2b.
Solving these two equations, we ﬁnda=−
1
2
andb=
1
2
.
Veriﬁcation: the function
f(x)=−
1
2
x
3
+
1
2
x
2
has second derivativef
ﬀﬀ
(x)=−3x+1.This does change sign atx=
1
3
.
42.f(x)=Ax
1/2
+Bx
−1/2
;f(1) = 4 =⇒A+B=4.
f
ﬀ
(x)=
1
2
Ax
−1/2
−
1
2
Bx
−3/2
,f
ﬀﬀ
(x)=−
1
4
Ax
−3/2
+
3
4
Bx
−5/2
.
To have a point of inﬂection at (1,4),we needf
ﬀﬀ
(1) = 0 =⇒−
1
4
A+
3
4
B=0.
Solving the two equations givesA=3,B=1.
43.First, we require that

1
6
π,5

lie on the curve:
5=
1
2
A+B.
Next we require that
d
2
y
dx
2
=−4Acos 2x−9Bsin 3xbe zero (and change sign) atx=
1
6
π:
0=−2A−9B.
Solving these two equations, we ﬁndA=18,B=−4.
Veriﬁcation: the function
f(x) = 18 cos 2x−4 sin 3x
has second derivativef
ﬀﬀ
(x)=−72 cos 2x+ 36 sin 3x.This does change sign atx=
1
6
π.

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.6 173
44.f(x)=Ax
2
+Bx+C;f
ﬀ
(x)=2Ax+B;f
ﬀﬀ
(x)=2A.
(a) Concave up =⇒f
ﬀﬀ
(x)>0=⇒A>0; to decrease betweenAandBwe need
f
ﬀ
(x)<0,forxbetweenAandB=⇒B≤−2A
2
.
(b) Concave down =⇒f
ﬀﬀ
(x)<0=⇒A<0; to havef
ﬀ
(x)>0 forxbetweenAandBwe
need 2A
2
+B≥0 and 2AB+B≥0,that is,B≥−2A
2
andB(2A+1)≥0.IfA>−
1
2
,then
we needB≥0 (and automaticallyB≥−2A
2
). IfA≤−
1
2
,then we needB≤0 andB≥−2A
2
.
The conditions are:−
1
2
<A<0,B≥0,orA≤−
1
2
,−2A
2
≤B≤0.
45.Letf
ﬀ
(x)=3x
2
−6x+3.Then we must havef(x)=x
3
−3x
2
+3x+cfor some constantc.Note
thatf
ﬀﬀ
(x)=6x−6 andf
ﬀﬀ
(1) = 0.Since (1,−2) is a point of inﬂection of the graph off,(1,−2)
must lie on the graph. Therefore,
1
3
−3(1)
2
+ 3(1) +c=−2 which impliesc=−3
and sof(x)=x
3
−3x
2
+3x−3.
46.f
ﬀ
(x) = cosxandf
ﬀﬀ
(x)=−sinx=−f(x).
Thusfis concave down whenf
ﬀﬀ
(x)<0=⇒f(x)>0.
Similarly,fis concave up whenf
ﬀﬀ
(x)>0=⇒f(x)<0.
g
ﬀ
(x)=−sinxandg
ﬀﬀ
(x)=−cosx=−g(x),sog(x) has the same property.
47.(a)p
ﬀﬀ
(x)=6x+2ais negative forx<−a/3,and positive forx>−a/3.Therefore, the graph ofp
has a point of inﬂection atx=−a/3.
(b)p
ﬀ
(x)=3x
2
+2ax+b. The discriminant of this quadratic is 4a
2
−12b=4(a
2
−3b). Thus,p
ﬀ
has
two real zeros iﬀa
2
>3b.
(c) Ifa
2
≤3b, thenb≥a
2
/3 andp
ﬀ
(x)=3x
2
+2ax+b≥3x
2
+2ax+a
2
/3=3(x+
1
3
a)
2
≥0 for
allx;pis increasing in this case.
48.It is suﬃcient to show that the x-coordinate of the point of inﬂection is the x-coordinate of the mid-
point of the line segment connecting the local extrema. It is easy to show that the x-coordinate of the
point of inﬂection isx0=−
1
3
a.Now suppose thatphas local extrema atx1andx2,x1δ=x2.Then
p
ﬀ
(x1)=p
ﬀ
(x2)=0⇒3x
2
1+2ax1+b−(3x
2
2+2ax2+b)=0⇒x1+x2=−
2
3
a.
Thus,
x1+x2
2
=−
1
3
a=x0.
49.(a)
(b) No. Iff
ﬀﬀ
(x)<0 andf
ﬀ
(x)<0 for allx,then
f(x)<f
ﬀ
(0)x+f(0) on (0,∞),which implies that
f(x)→−∞ asx→∞.

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
174 SECTION 4.6
50.Letf(x)=anx
n
+an−1x
n−1
+···+a2x
2
+a1x+a0.
Thenf
ﬀﬀ
(x)=n(n−1)anx
n−2
+···+2a2is a degreen−2 polynomial which can have at most
n−2 roots. Hencefhas at mostn−2 points of inﬂection.
51.
(a) concave up on (−4,−0.913)∪(0.913,4)
concave down on (−0.913,0.913)
(b) pts of inﬂection atx=−0.913,0.913
52.
(a) up: (−2π,−3.64),(−1.077,1.077),(3.64,2π)
down: (−3.64,−1.077)∪(1.077,3.64)
(b) pts of inﬂection atx=±3.64,±1.077
53.
(a) concave up on:
(−π,−1.996)∪(−.0345,2.550)
concave down on:
(−1.996,−0.345)∪(2.550,π)
(b) pts of inﬂection at:
x=−1.996,−0.345,2,550
54.
(a) concave up on (−5,5)
(b) no pts of inﬂection
55.(a)f
ﬀﬀ
(x)=0atx=0.68824,2.27492,4.00827,5.59494
(b)f
ﬀﬀ
(x)>0on(0.68824,2.27492)∪(4.00827,5.59494)
(c)f
ﬀﬀ
(x)<0on(0,0.68824)∪(2.27492,4.00827)∪(5.59494,2π)
56.(a)f
ﬀﬀ
(x)δ= 0 for allx (b)f
ﬀﬀ
(x)>0on(−∞,−1)∪(1,∞)
(c)f
ﬀﬀ
(x)<0on(−1,1)
57.(a)f
ﬀﬀ
(x)=0atx=±1,0,±0.32654,±0.71523
(b)f
ﬀﬀ
(x)>0on(−0.71523,−0.32654)∪(0,0.32654)∪(0.71523,1)∪(1,∞)
(c)f
ﬀﬀ
(x)<0on(−∞,−1)∪(−1,−0.71523)∪(−0.32654,0)∪(0,32654,0.71523)

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.7 175
58.(a)f
ﬀﬀ
(x)=0atx= 0 (b)f
ﬀﬀ
(x)>0on(−4,0)
(c)f
ﬀﬀ
(x)<0on(0,4)
SECTION 4.7
1.(a)∞ (b)−∞ (c)∞ (d) 1
(e) 0 (f) x=−1,x= 1 (g) y=0,y=1
2.(a)d (b)c (c)x=a, x=b
(d)y=d (e)p (f)q
3.vertical:x=
1
3
; horizontal:y=
1
3
4.vertical:x=−2; horizontal: none
5.vertical:x= 2 ; horizontal: none 6.vertical: none; horizontal:y=0
7.vertical:x=±3 ; horizontal:y=0 8.vertical:x= 16; horizontal:y=0
9.vertical:x=−
4
3
; horizontal:y=
4
9
10.vertical:x=
1
3
; horizontal:y=
4
9
11.vertical:x=
5
2
; horizontal:y=0 12.vertical:x=
1
2
; horizontal:y=−
1
8
13.vertical: none ; horizontal:y=±
3
2
14.vertical:x= 8; horizontal:y=0
15.vertical:x= 1 ; horizontal:y=0 16.vertical:x=±1; horizontal:y=±2
17.vertical: none ; horizontal:y=0 18.vertical: none; horizontal:y=0
19.vertical:x=

2n+
1
2

π; horizontal: none20.vertical:x=2nπ; horizontal: none
21.f
ﬀ
(x)=
4
3
(x+3)
1/3
; neither 22.f
ﬀ
(x)=
2
5
x
−3/5
; cusp
23.f
ﬀ
(x)=−
4
5
(2−x)
−1/5
; cusp 24.neither;f(−1) is not deﬁned
25.f
ﬀ
(x)=
6
5
x
−2/5

1−x
3/5

; tangent 26.f
ﬀ
(x)=
7
5
(x−5)
2/5
; neither
27.f(−2) undeﬁned; neither 28.f
ﬀ
(x)=
3
7
(2−x)
−4/7
; tangent
29.f
ﬀ
(x)=
⎧
⎨
⎩
1
2
(x−1)
−1/2
,x>1
−
1
2
(1−x)
−1/2
,x<1;
cusp
30.f
ﬀ
(x)=(4x−3)(x−1)
−2/3
; tangent

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
176 SECTION 4.7
31.f
ﬀ
(x)=
⎧
⎨
⎩
1
3
(x+8)
−23
,x>−8
−
1
3
(x+8)
−2/3
,x<−8;
cusp
32.fis not deﬁned forx>2; neither 33.fnot continuous at 0; neither
34.fis not continuous at 0; neither
35. 36.
37. 38.
39.f(x)=x−3x
1/3
(a)f
ﬀ
(x)=1−
1
x
2/3
fis increasing on (−∞,−1]∪[1,∞)
fis decreasing on [−1,1]
(b)f
ﬀﬀ
(x)=
2
3
x
−5/3
concave up on (0,∞); concave down on (−∞,0)
vertical tangent at (0,0)
40.f(x)=x
2/3
−x
1/3
(a)f
ﬀ
(x)=
2
3
x
−1/3
−
1
3
x
−2/3
=
2x
1/3
−1
3x
2/3
fis increasing on

1
8
,∞

fis decreasing on

−∞,
1
8

(b)f
ﬀﬀ
(x)=−
2
9
x
−4/3
+
2
9
x
−5/3
=
2(1−x
1/3
)
9x
5/3
concave up on (0,1)
concave down on (−∞,0)∪(1,∞)
vertical tangent at (0,0)

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.7 177
41.f(x)=
3
5
x
5/3
−3x
2/3
(a)f
ﬀ
(x)=x
2/3
−2x
−1/3
=
x−2
x
1/3
fis increasing on (−∞,0]∪[2,∞)
fis decreasing on [0,2]
(b)f
ﬀﬀ
(x)=
2
3
x
−1/3
+
2
3
x
−4/3
=
2x+2
3x
4/3
concave up on (−1,0)∪(0,∞); concave down on
(−∞,−1)
vertical cusp at (0,0)
42.f(x)=

|x|
=
δ
x
1/2
,x≥0
(−x)
1/2
,x<0.
(a)f
ﬀ
(x)=
δ
1
2
x
−1/2
,x>0
−
1
2
(−x)
−1/2
,x<0;
fis increasing on [0,∞)
fis decreasing on (−∞,0]
(b)f
ﬀﬀ
(x)=
δ
−
1
4
x
−3/2
,x>0
−
1
4
(−x)
−3/2
,x<0;
concave down on (−∞,0)∪(0,∞)
vertical cusp at (0,0)
43.
no asymptotes
vertical cusp at (0,1)
44.
vertical asymptotes:x=1,x=−1
horizontal asymptote:y=1
no vertical tangents or cusps

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178 SECTION 4.7
45.
horizontal asymptotes:y=−1,y=1
vertical tangent at (0,0)
46.
horizontal asymptote:y=4
vertical tangent at (0,1)
47.(a)podd; (b)peven.
48.[r(x)−(ax+b)] =
Q(x)
q(x)
→0asx→±∞,since deg [Q(x)]<deg [q(x)].
49.
vertical asymptote:x=0
oblique asymptote:y=x
50.
vertical asymptote:x=−1
oblique asymptote:y=2x+1
51.
vertical asymptote:x=1
oblique asymptote:y=x
52.
vertical asymptote:x=0
oblique asymptote:y=−3x+1
53.oblique asymptote:y=3x−4 54.oblique asymptote:y=5x−3

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.8 179
55.y= 1 is a horizontal asymptote.
f(x)=
√
x
2
+2x−x=
√
x
2
+2x−x

·
√
x
2
+2x+x
√
x
2
+2x+x
=
2x
√
x
2
+2x+x
→1.
56.y=−
1
2
is a horizontal asymptote.
f(x)=
√
x
4
−x
2
−x
2
=
√
x
4
−x
2
−x
2

·
√
x
4
−x
2
+x
2
√
x
4
−x
2
+x
2
=
−x
2
√
x
4
−x
2
+x
2
→−
1
2
.
SECTION 4.8
[Rough sketches; not scale drawings]
1.f(x)=(x−2)
2
f
ﬀ
(x)=2(x−2)
f
ﬀﬀ
(x)=2
2.f(x)=1−(x−2)
2
f
ﬀ
(x)=−2(x−2)
f
ﬀﬀ
(x)=−2
3.f(x)=x
3
−2x
2
+x+1
f
ﬀ
(x)=(3x−1)(x−1)
f
ﬀﬀ
(x)=6x−4

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180 SECTION 4.8
4.f(x)=x
3
−9x
2
+24x−7
f
ﬀ
(x)=3x
2
−18x+24
f
ﬀﬀ
(x)=6x−18
5.f(x)=x
3
+6x
2
,x∈[−4,4]
f
ﬀ
(x)=3x(x+4)
f
ﬀﬀ
(x)=6x+12
6.f(x)=x
4
−8x
2
,x∈[0,∞)
f
ﬀ
(x)=4x
3
−16x
f
ﬀﬀ
(x)=12x
2
−16
7.f(x)=
2
3
x
3
−
1
2
x
2
−10x−1
f
ﬀ
(x)=(2x−5)(x+2)
f
ﬀﬀ
(x)=4x−1

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SECTION 4.8 181
8.f(x)=x(x
2
+4)
2
=x
5
+8x
3
+16x
f
ﬀ
(x)=5x
4
+24x
2
+16
f
ﬀﬀ
(x)=20x
3
+48x=4x(5x
2
+ 12)
9.f(x)=x
2
+2x
−1
f
ﬀ
(x)=2x−2x
−2
=2

x
3
−1

/x
2
f
ﬀﬀ
(x)=2+4x
−3
asymptote:x=0
10.f(x)=x−x
−1
,
f
ﬀ
(x)=1+x
−2
f
ﬀﬀ
(x)=−x
−3
asymptotes:x=0,y=x

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182 SECTION 4.8
11.f(x)=(x−4)/x
2
f
ﬀ
(x)=(8−x)/x
3
f
ﬀﬀ
(x)=(2x−24)/x
4
asymptotes:x=0,y=0
12.f(x)=
x+2
x
3
=
1
x
2
+
2
x
3
f
ﬀ
(x)=−
2
x
3
−
6
x
4
=
−2x−6
x
4
f
ﬀﬀ
(x)=
6
x
4
+
24
x
5
=
6(x+4)
x
5
asymptotes:x=0,y=0
13.f(x)=2x
1/2
−x, x∈[0,4]
f
ﬀ
(x)=x
−1/2

1−x
1/2

f
ﬀﬀ
(x)=−
1
2
x
−3/2
14.f(x)=
1
4
x−
√
x, x∈[0,9]
f
ﬀ
(x)=
1
4
−
1
2
x
−1/2
f
ﬀﬀ
(x)=
1
4
x
−3/2

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SECTION 4.8 183
15.f(x)=2+(x+1)
6/5
f
ﬀ
(x)=
6
5
(x+1)
1/5
f
ﬀﬀ
(x)=
6
25
(x+1)
−4/5
16.f(x)=2+(x+1)
7/5
f
ﬀ
(x)=
7
5
(x+1)
2/5
f
ﬀﬀ
(x)=
14
25
(x+1)
−3/5
17.f(x)=3x
5
+5x
3
f
ﬀ
(x)=15x
2

x
2
+1

f
ﬀﬀ
(x)=30x

2x
2
+1

18.f(x)=3x
4
+4x
3
f
ﬀ
(x)=12x
3
+12x
2
=12x
2
(x+1)
f
ﬀﬀ
(x)=36x
2
+24x=12x(3x+2)

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184 SECTION 4.8
19.f(x)=1+(x−2)
5/3
f
ﬀ
(x)=
5
3
(x−2)
2/3
f
ﬀﬀ
(x)=
10
9
(x−2)
−1/3
20.f(x)=1+(x−2)
4/3
f
ﬀ
(x)=
4
3
(x−2)
1/3
f
ﬀﬀ
(x)=
4
9
(x−2)
−2/3
21.f
ﬀ
(x)=
8x
(x
2
+4)
2
f
ﬀﬀ
(x)=48
8(4−3x
2
)
(x
2
+4)
3
f
ﬀ
(x)<0on(−∞,0)
f
ﬀ
(x)>0on(0,∞)
f
ﬀﬀ
(x)<0on

−∞,−2/
√
3

∪

2/
√
3,∞

;
f
ﬀﬀ
(x)>0on

−2/
√
3,2/
√
3

;
asymptote:y=1
22.f(x)=
2x
2
x+1
f
ﬀ
(x)=
2x(x+2)
(x+1)
2
f
ﬀﬀ
(x)=
4
(x+1)
3
asymptote:y=2x−2

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SECTION 4.8 185
23.f(x)=
x
(x+3)
2
f
ﬀ
(x)=
3−x
(x+3)
3
f
ﬀﬀ
(x)=
2x−12
(x+3)
4
asymptotes:x=−3,y=0
24.f(x)=
x
x
2
+1
f
ﬀ
(x)=
1−x
2
(x
2
+1)
2
f
ﬀﬀ
(x)=
2x(x
2
−3)
(x
2
+1)
3
asymptote:y=0
25.f(x)=
x
2
x
2
−4
f
ﬀ
(x)=
−8x
(x
2
−4)
2
f
ﬀﬀ
(x)=
8(3x
2
+4)
(x
2
−4)
3
asymptotes:x=−2,x=2,y=1

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186 SECTION 4.8
26.f(x)=
1
x
3
−x
f
ﬀ
(x)=
1−3x
2
(x
3
−x)
2
increasing on

−1/
√
3,0

∪

0,1/
√
3

;
decreasing otherwise
f
ﬀﬀ
(x)=
2

6x
4
−3x
2
+1

(x
3
−x)
3
concave up on (−1,0)∪(1,∞)
concave down on (−∞,0)∪(0,1)
asymptotes:x=−1,x=0,x=1,y=0
27.f(x)=x(1−x)
1/2
f
ﬀ
(x)=
1
2
(1−x)
−1/2
(2−3x)
f
ﬀﬀ
(x)=
1
4
(1−x)
−3/2
(3x−4)
28.f(x)=(x−1)
4
−2(x−1)
2
f
ﬀ
(x)=4(x−1)
3
−4(x−1)
f
ﬀﬀ
(x) = 12(x−1)
2
−4
29.f(x)=x+ sin 2x, x∈[0,π]
f
ﬀ
(x) = 1 + 2 cos 2x
f
ﬀﬀ
(x)=−4 sin 2x

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SECTION 4.8 187
30.f(x) = cos
3
x+ 6 cosx, x∈[0,π]
f
ﬀ
(x)=−3 sinx(cos
2
x+2)
f
ﬀﬀ
(x)=−9 cos
3
x
31.f(x) = cos
4
x, x∈[0,π]
f
ﬀ
(x)=−4 cos
3
xsinx
f
ﬀﬀ
(x) = 4 cos
2
x

3 sin
2
x−cos
2
x

32.f(x)=
√
3x−cos 2x, x∈[0,π]
f
ﬀ
(x)=
√
3 + 2 sin 2x
f
ﬀﬀ
(x) = 4 cos 2x
33.f(x) = 2 sin
3
x+ 3 sinx, x∈[0,π]
f
ﬀ
(x) = 3 cosx

2 sin
2
x+1

f
ﬀﬀ
(x) = 9 sinx

1−2 sin
2
x

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188 SECTION 4.8
34.f(x) = sin
4
x, x∈[0,π]
f
ﬀ
(x) = 4 sin
3
xcosx
f
ﬀﬀ
(x) = 4 sin
2
x

3 cos
2
x−sin
2
x

35.f(x)=[(x+1)−1]
3
+1
f
ﬀ
(x)=3x
2
f
ﬀﬀ
(x)=6x
36.f(x)=x
3
(x+5)
2
f
ﬀ
(x)=5x
2
(x+ 3)(x+5)
f
ﬀﬀ
(x)=10x(2x
2
+12x+ 15)
37.f(x)=x
2
(5−x)
3
f
ﬀ
(x)=5x(2−x)(5−x)
2
f
ﬀﬀ
(x) = 10(5−x)

2x
2
−8x+5

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SECTION 4.8 189
38.f(x)=
δ
4−2x+x
2
,0<x<2
4+2x−x
2
,x≤0,x≥2
f
ﬀ
(x)=
δ
−2+2x,0<x<2
2−2x, x <0,x>2
f
ﬀﬀ
(x)=
δ
2,0<x<2
−2,x<0,x>2
39.f(x)=
δ
4−x
2
,|x|>1
x
2
+2,−1≤x≤1
f
ﬀ
(x)=
δ
−2x,|x|>1
2x,−1<x<1
f
ﬀﬀ
(x)=
δ
−2,|x|>1
2,−1<x<1
40.f(x)=x−x
1/3
f
ﬀ
(x)=1−
1
3
x
−2/3
f
ﬀﬀ
(x)=
2
9
x
−5/3
vertical tangent at (0,0)

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190 SECTION 4.8
41.f(x)=x(x−1)
1/5
f
ﬀ
(x)=
1
5
(x−1)
−4/5
(6x−5)
f
ﬀﬀ
(x)=
2
25
(x−1)
−9/5
(3x−5)
vertical tangent at (1,0)
42.f(x)=x
2
(x−7)
1/3
f
ﬀ
(x)=
7x(x−6)
3(x−7)
2/3
f
ﬀﬀ
(x)=
14(2x
2
−24x+ 63)
9(x−7)
5/3
vertical tangent at (7,0)
43.f(x)=x
2
−6x
1/3
f
ﬀ
(x)=2x
−2/3

x
5/3
−1

f
ﬀﬀ
(x)=
2
3
x
−5/3
ν
3x
5/3
+2
ω
vertical tangent at (0,0)

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SECTION 4.8 191
44.f(x)=
2x
√
x
2
+1
f
ﬀ
(x)=
2
(x
2
+1)
3/2
f
ﬀﬀ
(x)=
−6x
(x
2
+1)
5/2
asymptotes:y=2,y=−2
45.f(x)=
Δ
x
x−2
θ
1/2
;x≤0,x>2
f
ﬀ
(x)=−
Δ
x
x−2
θ
−1/2
(x−2)
−2
f
ﬀﬀ
(x)=(2x−1)
Δ
x
x−2
θ
−3/2
(x−2)
−4
asymptotes:x=2,y=1
46.f(x)=
Δ
x
x+4
θ
1/2
;x<−4,x≥0
f
ﬀ
(x)=2
Δ
x
x+4
θ
−1/2
(x+4)
−2
f
ﬀﬀ
(x)=
−4(x+1)
(x+4)
5/2
x
3/2
asymptotes:x=−4,y=1

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192 SECTION 4.8
47.f(x)=x
2

x
2
−2

−1/2
,|x|>
√
2
f
ﬀ
(x)=x

x
2
−4

x
2
−2

−3/2
f
ﬀﬀ
(x)=2

x
2
+4

x
2
−2

−5/2
asymptotes:x=−
√
2,x=
√
2
48.f(x) = 3 cos 4x, x∈[0,π]
f
ﬀ
(x)=−12 sin 4x
f
ﬀﬀ
(x)=−48 cos 4x
49.f(x) = 2 sin 3x, x∈[0,π]
f
ﬀ
(x) = 6 cos 3x
f
ﬀﬀ
(x)=−18 sin 3x
50.f(x)=3+2cotx+ csc
2
x, x∈(0,π/2)
f
ﬀ
(x)=−2 csc
2
x(1 + cotx)
f
ﬀﬀ
(x) = 2 csc
2
x

3 cot
2
x+ 2 cotx+1

asymptote:x=0

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SECTION 4.8 193
51.f(x) = 2 tanx−sec
2
x, x∈(0,π/2)
=−(1−tanx)
2
f
ﬀ
(x) = 2 sec
2
x(1−tanx)
f
ﬀﬀ
(x)=−2 sec
2
x

3 tan
2
x−2 tanx+1

asymptote:x=
1
2
π
52.f(x) = 2 cosx+ sin
2
x
f
ﬀ
(x) = 2 sinx(cosx−1)
f
ﬀﬀ
(x) = 2(2 cos
2
x−cosx−1)
53.f(x)=
sinx
1−sinx
,x∈(−π, π)
f
ﬀ
(x)=
cosx
(1−sinx)
2
f
ﬀﬀ
(x)=
1−sinx+ cos
2
x
(1−sinx)
3
asymptote:x=
1
2
π

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194 SECTION 4.8
54.f(x)=
1
1−cosx
,x∈(−π, π)
f
ﬀ
(x)=
−sinx
(1−cosx)
2
f
ﬀﬀ
(x)=
2−cosx−cos
2
x
(1−cosx)
3
asymptote:x=0
55.(a)fincreases on (−∞,−1]∪(0,1]∪[3,∞);
fdecreases on [−1,0)∪[1,3]; critical points:x=−1,0,1,3.
(b) concave up on ( −∞,−3)∪(2,∞)
concave down on (−3,0)∪(0,2).
(c) The graph does not necessarily have
a horizontal asymptote.
56.(a)

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SECTION 4.8 195
(b) (c)
(d)Fis discontinuous at 0; lim
x→0
sin(1/x) does not exist
Gis continuous at 0:|xsin(1/x)|=|x||sin(1/x)|≤|x|→0,
so that lim
x→0
G(x)=0=G(0).
His continuous at 0:

x
2
sin(1/x)

=|x|
2
|sin(1/x)|≤|x|
2
→0,
so that lim
x→0
H(x)=0=H(0).
(e)Fis not diﬀerentiable at 0 since it is not continuous at 0.
Gis not diﬀerentiable at 0: lim
h→0
G(h)−G(0)
h
= lim
h→0
sin(1/h) does not exist.
His diﬀerentiable at 0:H
ﬀ
(0) = lim
h→0
H(h)−H(0)
h
= lim
h→0
hsin(1/h)=0.
57.f(x)−x
2
=
x
3
−x
1/3
x
−x
2
=
x
3
−x
1/3
−x
3
x
=−
1
x
2/3
→0asx→±∞.
The ﬁgure shows the graphs offandg(x)=x
2
forx>0. Sincefandgare even functions,
the graphs are symmetric about they-axis.
x
y
58.(a)
−a a
x
b
y
(b)
b
a
Ł
x
2
−a
2
=
b
a
x
−
1−
a
2
x
2
→
b
a
xasx→∞. Therefore,
b
a
Ł
x
2
−a
2
−
b
a
x→0asx→∞.
(c) Ifx<0, then the second quadrant arc is given byy=−
b
a
x
−
1−
a
2
x
2
→−
b
a
xasx→−∞.

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196 SECTION 4.9
SECTION 4.9
1.x(5) =−6;v(t)=3−2tsov(5) =−7 and speed = 7;a(t)=−2soa(5) =−2.
2.x(3) =−12;v(t)=5−3t
2
sov(3) =−22 and speed = 22;a(t)=−6tsoa(3) =−18.
3.x(1) = 6;v(t)=−18/(t+2)
2
sov(1) =−2 and speed = 2,
a(t)=36/(t+2)
3
soa(1) = 4/3.
4.x(3) = 1;v(t)=
6
(t+3)
2
sov(3) = 1/6 = speed;a(t)=−12/(t+3)
3
soa(3) =−1/18.
5.x(1) = 0,v(t)=4t
3
+18t
2
+6t−10 sov(1) = 18 and speed = 18,
a(t)=12t
2
+36t+6soa(1) = 54.
6.x(2) =−20;v(t)=4t
3
−18tsov(2) =−4 and speed = 4,
a(t)=12t
2
−18 soa(2) = 30.
7.x(t)=5t+1,x
ﬀ
(t)=v(t)=5,x
ﬀﬀ
(t)=a(t)=0.
v(t)δ=0,a(t) = 0 for allt.
8.x(t)=4t
2
−t+3,x
ﬀ
(t)=v(t)=8t−1,x
ﬀﬀ
(t)=a(t)=8.
v(t)=0 att=1/8;a(t)δ= 0 for allt.
9.x(t)=t
3
−6t
2
+9t−1,x
ﬀ
(t)=v(t)=3t
2
−12t+9,x
ﬀﬀ
(t)=a(t)=6t−12.
v(t)=3(t−1)(t−3),v(t)=0 att=1,3;a(t)=0 att=2.
10.x(t)=t
4
−4t
3
+4t
2
+2,x
ﬀ
(t)=v(t)=4t
3
−12t
2
+8t, x
ﬀﬀ
(t)=a(t)=12t
2
−24t+8.
v(t)=4t(t−1)(t−2),v(t)=0 att=0,1,2;a(t)=0 att=1+
√
3/3.
11.A 12.C 13.A 14.C 15.AandB
16.B 17.A 18.A 19.AandC 20.B
21.The object is moving right whenv(t)>0. Here,
v(t)=4t
3
−36t
2
+56t=4t(t−2)(t−7);v(t)>0 when 0<t<2 and 7<t.
22.The object is moving left whenv(t)<0. Here,
v(t)=3t
2
−24t+21=3(t−7)(t−1);v(t)<0 when 1<t<7.
23.The object is speeding up whenv(t) anda(t) have the same sign.
v(t)=5t
3
(4−t) sign of v(t):
a(t)=20t
2
(3−t) sign of a(t):
Thus, 0<t<3 and 4<t.

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SECTION 4.9 197
24.The object is slowing down whenv(t) anda(t) have opposite sign.
v(t)=12t−4t
3
=4t(3−t
2
) sign of v(t):
a(t) = 12(1−t) sign of a(t):
Thus, 1<t<
√
3.
25.The object is moving left and slowing down whenv(t)<0 anda(t)>0.
v(t)=3(t−5)(t+ 1) sign of v(t):
a(t)=6(t−2) sign of a(t):
Thus, 2<t<5.
26.The object is moving right and slowing down whenv(t)>0 anda(t)<0.
v(t)=3(t−5)(t+ 1) sign of v(t):
a(t)=6(t−2) sign of a(t):
This never happens.
27.The object is moving right and speeding up whenv(t)>0 anda(t)>0.
v(t)=4t(t−2)(t−4) sign of v(t):
a(t) = 4(3t
2
−12t+ 8) sign of a(t):
Thus, 0<t<2−
2
3
√
3 and 4<t.
28.The object is moving left and speeding up whenv(t)<0 anda(t)<0.
v(t)=4t(t−2)(t−4) sign of v(t):
a(t) = 4(3t
2
−12t+ 8) sign of a(t):
Thus, 2<t<2+
2
3
√
3.
29.x(t)=(t+1)
2
(t−9)
3
,x
ﬀ
(t)=v(t)=3(t+1)
2
(t−9)
2
+2(t+ 1)(t−9)
3
=5(t+ 1)(t−9)
2
(t−3).
The object changes direction at timet=3.
30.x(t)=t(t−8)
3
,x
ﬀ
(t)=v(t)=(t−8)
3
+3t(t−8)
2
=4(t−8)
2
(t−2).
The object changes direction at timet=2.
31.x(t)=(t
3
−12t)
4
,x/(t)=v(t)=4(t
3
−12t)
3
(3t
2
−12) = 12t
3
(t+2
√
3)
3
(t−2
√
3)
3
(t+ 2)(t−2).
The object changes direction at timest=2,2
√
3.
32.x(t)=(t
2
−8t+ 15)
3
,x
ﬀ
(t)=v(t)=3(t
2
−8t+ 15)
2
(2t−8) = 6(t
2
−8t+ 15)
2
(t−4).
The object changes direction at timet=4.
Exercises 33–38.The object is moving right with increasing speed when bothv(t) anda(t) are positive.
33.x(t) = sin 3t, x
ﬀ
(t)=v(t) = 3 cos 3t, x
ﬀﬀ
(t)=a(t)=−9 sin 3t.
v(t)>0on(0,π/6)∪(π/2,5π/6)∪(7π/6,3π/2)∪(11π/6,2π).
a(t)>0on(π/3,2π/3)∪(π,4π/3)∪(5π/3,2π).
v(t) anda(t) are positive on (π/2,2π/3)∪(7π/6,4π/3)∪(11π/6,2π).

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198 SECTION 4.9
34.x(t) = cos 2t, x
ﬀ
(t)=v(t)=−2 sin 2t, x
ﬀﬀ
(t)=a(t)=−4 cos 2t.
v(t)>0on(π/2,π)∪(3π/2,2π);a(t)>0on(π/4,3π/4)∪(5π/4,7π/4).
v(t) anda(t) are positive on (π/2,3π/4)∪(3π/2,7π/4).
35.x(t) = sint−cost, x
ﬀ
(t)=v(t) = cost+ sint, x
ﬀﬀ
(t)=a(t)=−sint+ cost.
v(t)>0on(0,3π/4)∪(7π/4,2π);a(t)>0on(0,π/4)∪(5π/4,2π).
v(t) anda(t) are positive on (0,π/4)∪(7π/4,2π).
36.x(t) = sint+ cost, x
ﬀ
(t)=v(t) = cost−sint, x
ﬀﬀ
(t)=a(t)=−sint−cost.
v(t)>0on(0,π/4)∪(5π/4,2π);a(t)>0on(3π/4,7π/4).
v(t) anda(t) are positive on (5π/4,7π/4).
37.x(t)=t+ 2 cost, x
ﬀ
(t)=v(t)=1−2 sint, x
ﬀﬀ
(t)=a(t)=−2 cost.
v(t)>0on(0,π/6)∪(5π/6,2π);a(t)>0on(π/2,3π/2).
v(t) anda(t) are positive on (5π/6,3π/2).
38.x(t)=t−
√
2 sint, x
ﬀ
(t)=v(t)=1−
√
2 cost, x
ﬀﬀ
(t)=a(t)=
√
2 sint.
v(t)>0on(π/4,7π/4);a(t)>0on(0,π).
v(t) anda(t) are positive on (π/4,π).
39.Sincev0= 0 the equation of motion is
y(t)=−16t
2
+y0.
We want to ﬁndy0so thaty(6) = 0. From
0=−16(6)
2
+y0
we gety0= 576 feet.
40.The equation of motion is:y(t)=−4.9t
2
+y0.Therefore, the velocity is given byv(t)=−9.8t.
Since the object hits the ground at 98 m/sec., we have−9.8t=−98,andt=10.
Therefore,y(10) = 0 =−4.9(10)
2
+y0andy0= 490 meters.
41.The object’s height and velocity at timetare given by
y(t)=−
1
2
gt
2
+v0tand v(t)=−gt+v0
Since the object’s velocity at its maximum height is 0,it takesv0/gseconds to reach
maximum height, and
y(v0/g)=−
1
2
g(v0/g)
2
+v0(v0/g)=v
2
0/2gorv
2
0/19.6 (meters)

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SECTION 4.9 199
42.Sincey0=0,we havey(t)=−16t
2
+v0t=t(−16t+v0).Now,
y(8) = 0 =⇒v0= (16)8 = 128 =⇒the initial velocity was 128 ft/sec.
43.At timet,the object’s height isy(t)=−
1
2
gt
2
+v0t+y0,and its velocity isv(t)=−gt+v0.Suppose
thaty(t1)=y(t2),t1δ=t2.Then
−
1
2
gt
2
1+v0t1+y0=−
1
2
gt
2
2+v0t2+y0
1
2
g(t
2
2−t
2
1)=v0(t2−t1)
gt2+gt1=2v0
From this equation, we get−(−gt1+v0)=−gt2+v0and so|v(t1)|=|v(t2)|.
44.Sincey0=0,we havey(t)=−4.9t
2
+v0t=t(v0−4.9t) The object hits the ground att=v0/4.9 sec.,
that is, the object is in the air forv0/4.9 sec. At its maximum height, the velocity of the object is 0.
Sincev(t)=−9.8t+v0,we have−9.8t+v0= 0 andt=v0/9.8=
1
2
(v0/4.9).The result follows from
this.
45.In the equation
y(t)=−16t
2
+v0t+y0
we takev0=−80 andy0= 224. The ball ﬁrst strikes the ground when
−16t
2
−80t+ 224 = 0;
that is, att= 2. Since
v(t)=y
ﬀ
(t)=−32t−80,
we havev(2) =−144 so that the speed of the ball the ﬁrst time it strikes the
ground is 144 ft/sec. Thus, the speed of the ball the third time it strikes the
ground is
1
4

1
4
(144)

= 9 ft/sec.
46.Sincey0=0,we havey(t)=−16t
2
+v0t.
y(2) = 64 =⇒− 16(2)
2
+2v0=64 =⇒v0= 64 andy(t)=−16t
2
+64t
Now, at the maximum height,v(t)=−32t+64=0 = ⇒t=2.We already know
the height att=2,namely 64 ft.
47.The equation isy(t)=−16t
2
+32t. (Herey0= 0 andv0= 32.)
(a) We solvey(t) = 0 to ﬁnd that the stone strikes the ground att= 2 seconds.
(b) The stone attains its maximum height whenv(t) = 0. Solving
v(t)=−32t+ 32 = 0, we gett= 1 and, thus, the maximum height isy(1) = 16 feet.

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200 SECTION 4.9
(c) We want to choosev0in
y(t)=−16t
2
+v0t
so thaty(t0) = 36 when v(t0) = 0 for some timet0.
Fromv(t)=−32t+v0= 0 we get t0=v0/32 so that
−16
ν
v0
32
ω
2
+v0
ν
v0
32
ω
=36,or
v0
2
64
=36.
Thus,v0= 48 ft/sec.
48.(a) Measuring height from the water surface, we havey(t)=−16t
2
+y0,sincev0(0) = 0.
If the stone hits the water 3 seconds later, theny(3) =−16(3)
2
+y0=0.soy0= 144.
(b) It takesy0/1080 seconds for the sound of the splash to reach the man so the stone hits the
at timet=3−y0/1080.Thus,
y(t)=−16
ν
3−
y0
1080
ω
2
+y0=0 =⇒y0
∼
=132.47 ft.
49.For all three parts of the problem the basic equation is
y(t)=−16t
2
+v0t+y0
with
(∗) y(t0) = 100 andy(t0+2)=16
for some timet0>0.
We are asked to ﬁndy0for a given value ofv0.
From (∗)weget
16−100 =y(t0+2)−y(t0)
=[−16(t0+2)
2
+v0(t0+2)+y0]−[−16t0
2
+v0t0+y0]
=−64t0−64+2v0
so that
t0=
1
32
(v0+ 10).
Substituting this result in the basic equation and noting thaty(t0) = 100, we have
−16
Δ
v0+10
32
θ
2
+v0
Δ
v0+10
32
θ
+y0= 100
and therefore
(∗∗) y0= 100−
v0
2
64
+
25
16
.
We use (∗∗) to ﬁnd the answer to each part of the problem.
(a)v0=0soy0=
1625
16
ft (b) v0=−5soy0=
6475
64
ft (c) v0=10soy0= 100 ft

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SECTION 4.9 201
50.Letv0>0 be the initial velocity. The equation of motion prior to the impact is:y(t)=−16t
2
−v0t+4.
The ball hits the ground at timet=

v
2
0
+ 256−v0
32
with velocityv=

v
2
0
+ 256.The equation of
motion following the impact is:y(t)=−16t
2
+

v
2
0
+ 256
2
t.It reaches its maximum height at time
T=

v
2
0
+ 256
64
.Now,y(T)=4 =⇒v0=16
√
3.
51.Lety0>0 be the initial height. The equation of motion becomes:
0=−16(8)
2
+ 5(8) +y0,soy0= 984 ft.
52.Using 0 =−16t
2
−5t+ 984,yieldst=
123
16
or about 7.7 sec.
53.Letf1(t) andf2(t) be the positions of the horses at timet.Considerf(t)=f1(t)−f2(t).LetTbe
the time the horses ﬁnish the race. Thenf(0) =f(T) = 0. By the mean-value theorem there is acin
(0,T) such thatf
ﬀ
(c)=0.Hencef
ﬀ
1(t)=f
ﬀ
2(t),so the horses had the same speed at timec.
54.Apply the argument given in Exercise 53 to the velocity functionsv1(t),v2(t).
55.The driver must have exceeded the speed limit at some time during the trip. Lets(t) denote the car’s
position at timet, withs(0) = 0 ands(5/3) = 120. Then, by the mean-value theorem, there exists at
least one number (time)csuch that
v(c)=s
ﬀ
(c)=
s(5/3)−s(0)
5
3
−0
=
120
5
3
=
360
5
= 72 mi./hr.
56.Letf(t) be the function that gives the car’s velocity afterthours. Thenf(0) = 30 andf(
1
4
)=60.f
is diﬀerentiable on (1,
1
4
) and continuous on [1,
1
4
],so by the mean-value theorem there is acin (1,
1
4
)
such thatf
ﬀ
(c)=
60−30
0−t1
= 120.i.e. The acceleration at timecwas 120 mph.
57.If the speeds(t) of the car is less than 60 mi/hr= 1 mi/min, then the distance traveled in 20 minutes
is less than 20 miles. Therefore, the car must have gone at least 1 mi/min at some timet<20. Let
t1be the ﬁrst instant the car’s speed is 1 mi/min. Then the speeds(t) is less than 1 mi/min on the
interval [0,t1) and the distancertraveled int1minutes is less thant1miles. Now, by the mean-value
theorem, there is a timec∈[t1,20] such that
s
ﬀ
(c)=
20−r
20−t1
>
20−t1
20−t1
= 1 (= 60 mi/hr).
58.Lets(t) denote the distance that the car has traveled intseconds since applying the brakes, 0≤t≤6.
Thens(0) = 0 ands(6) = 280.Assume thatsis diﬀerentiable on (0,6) and continuous on [0,6].Then,
by the mean-value theorem, there exists a timec∈(0,6) such that
s
ﬀ
(c)=v(c)=
s(6)−s(0)
6−0
=
280
6
∼
=46.67 ft/sec
Nowv(0)≥v(c)=46.7 ft/sec. Thus, the driver must have been exceeding the speed limit (44 ft/sec)
at the instant he applied his brakes.

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202 SECTION 4.9
59.y(t)=Asin(ωt+φ0),y
ﬀ
(t)=v(t)=ωAcos(ωt+φ0),y
ﬀﬀ
(t)=a(t)=−ω
2
Asin(ωt+φ0)
(a)y
ﬀﬀ
(t)+ω
2
y(t)=−ω
2
Asin(ωt+φ0)+ω
2
Asin(ωt+φ0)=0.
(b)v
ﬀ
(t)=a(t)=−ω
2
Asin(ωt+φ0) = 0 whenωt+φ0=nπ;t=
nπ−φ0
ω
.
The extreme values ofvoccur at these times. Now
y
Δ
nπ−φ0
ω
θ
=Asin
Δ
ω
nπ−φ0
ω
+φ0
θ
=Asin(nπ)=0.
The bob attains maximum speed at the equilibrium position.
(c)a
ﬀ
(t)−ω
3
Acos(ωt+φ0) = 0 whenωt+φ0=(2n−1)π/2;t=
(2n−1)π/2−φ0
ω
.
The extreme values ofaoccur at these times. Now
a
Δ
(2n−1)π/2−φ0
ω
θ
=−ω
2
Asin
Δ
(2n−1)π/2−φ0
ω
+φ0
θ
=±ω
2
A.
The bob attains these values at
y
Δ
(2n−1)π/2−φ0
ω
θ
=Asin
Δ
ω
(2n−1)π/2−φ0
ω
+φ0
θ
=Asin (2n−1)π/2=±A.
60.(a)x(t)=t
3
−7t
2
+10t+5,v(t)=3t
2
−14t+10,0≤t≤5
(b) The object is moving to the right when 0<t<0.88 and when 3.79<t<5.
The object is moving to the left when 0.88<t<3.79
(c) The object stops at timest
∼
=0.88 andt
∼
=3.79.
The maximum speed isv
∼
=6.33 att
∼
=2.33.
(d)a(t)=6t−14
The object is speeding up whenv(t) anda(t) have
the same sign: 0.88<t<2.33 and 3.79<t<5.
The object is slowing down whenv(t) anda(t) have
opposite sign: 0<t<0.88 and 2.33<t<3.79.
PROJECT 4.9a
1.length of arc =rθ, speed =
d
dt
[rθ]=r
dθ
dt
=rω
2.v=rωsoKE=
1
2
mr
2
ω
2
.
3.We know thatdθ/dt=ωand, at timet=0,θ=θ0.Thereforeθ=ωt+θ0.It follows that
x(t)=rcos (ωt+θ0)and y(t)=rsin (ωt+θ0).
x(t)=rcos(ωt+θ0),y(t)=rsin(ωt+θ0)
v(t)=x
ﬀ
(t)=−rωsin(ωt+θ0)=−ωy(t)
a(t)=−rω
2
cos(ωt+θ0)=−ω
2
x(t)
v(t)=y
ﬀ
(t)=rωcos(ωt+θ0)=ωx(t)
a(t)=−rω
2
sin(ωt+θ0)=−ω
2
y(t)

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SECTION 4.10 203
4.For the sectorA=
1
2
r
2
θ,
dA
dt
=
1
2
r
2
dθ
dt
=
1
2
r
2
ωis constant.
For triangleT
A=
1
2
(2rsin
1
2
θ)(rcos
1
2
θ)
=
1
2
r
2
(2 sin
1
2
θcos
1
2
θ)=
1
2
r
2
sinθ,
dA
dt
=
1
2
r
2
cosθ
dθ
dt
=
1
2
r
2
ωcosθvaries withθ.
For segmentS
A=
1
2
r
2
θ−
1
2
r
2
sinθ=
1
2
r
2
(θ−sinθ),
dA
dt
=
1
2
r
2
Δ
dθ
dt
−cos
dθ
dt
θ
=
1
2
r
2
ω(1−cosθ) varies withθ.
5.From Exercise 4,
dAT
dt
=
1
2
r
2
ωcosθand
dAS
dt
=
1
2
r
2
ω−
1
2
r
2
ωcosθ
Now,
1
2
r
2
ωcosθ=
1
2
r
2
ω−
1
2
r
2
ωcosθ=⇒cosθ=
1
2
=⇒θ=
π
3
.
PROJECT 4.9B
1.
d
dt

mgy+
1
2
mv
2

=mg
dy
dt
+
1
2
m
d
dt
(v
2
)
=mgv+
1
2
m

2v
dv
dt

=mgv+mv
dv
dt
=mgv+mv(−g) (sincedv/dt=a=−g)
=mgv−mgv=0
2.By Problem 1,mgy+
1
2
mv
2
=C(constant). Sincev= 0 at heighty=y0,we haveC=mgy0.
Thus,
mgy0=mgy+
1
2
mv
2
and|v|=

2g(y0−y)
3.y(t)=−
1
2
gt
2
+y0=⇒gt=

2g(y0−y)
v(t)=y
ﬀ
(t)=−gtTherefore,|v(t)|=

2g(y0−y).
SECTION 4.10
1.x+2y=2,
dx
dt
+2
dy
dt
=0
(a) If
dx
dt
= 4, then
dy
dt
=−2 units/sec. (b) If
dy
dt
=−2, then
dx
dt
= 4 units/sec.

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204 SECTION 4.10
2.x
2
+y
2
=25,2x
dx
dt
+2y
dy
dt
= 0 and
dx
dt
=−
y
x
dy
dt
.
At the point (3,4),
dy
dt
=−2.Therefore,
dx
dt
=
8
3
; the x-coordinate is
increasing at the rate of 8/3 units per second.
3.y
2
=4(x+2),2y
dy
dt
=4
dx
dt
and
dx
dt
=
1
2
y
dy
dt
At the point (7,6),
dy
dt
=3.Therefore
dx
dt
=
1
2
·6·3 = 9 units/sec.
4.We are given
dx
dt
=2.Also 4y=(x+2)
2
so 4
dy
dt
=2(x+2)
dx
dt
or
dy
dt
=
1
2
(x+2)
dx
dt
.
Atx=2,
dy
dt
=4.The distance from a point on the parabola to the point (−2,0) is given by
S=

(x+2)
2
+y
2
=

4y+y
2
since (x+2)
2
=4y.Now
dS
dt
=
1
2
(4y+y
2
)
−1/2
(4+2y)
dy
dt
=
2+y

4y+y
2
dy
dt
.
Therefore, at the point (2,4),
dS
dt
=
6
√
32
4=3
√
2.
5.Lets=

x
2
+y
2
denote the distance to the origin at timet.Sincex= 4 costandy= 2 sint,
we have
s(t)=

16 cos
2
t+ 4 sin
2
t=

12 cos
2
t+4
ds
dt
=
1
2
(12 cos
2
t+4)
−1/2
(−24 costsint)
=
−12 costsint
√
12 cos
2
t+4
Att=π/4,
ds
dt
=
−12 cos(π/4) sin(π/4)

12 cos
2
(π/4)+4
=−
3
5
√
10.
6.y=x
√
x=x
3/2
;
dy
dt
=
3
2
x
1/2
dx
dt
.
Now
dx
dt
=
dy
dt
=zδ=0,=⇒
3
2
x
1/2
=1 =⇒x=
4
9
andy=
8
27
.
Both coordinates are changing at the same rate at the point (4/9,8/27).
7.Find
dx
dt
and
dS
dt
whenV= 27m
3
given that
dV
dt
=−2m
3
/min.
(∗)V=x
3
,S=6x
2
Diﬀerentiation of equations (∗) gives
dV
dt
=3x
2
dx
dt
and
dS
dt
=12x
dx
dt
.

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.10 205
WhenV=27,x= 3. Substitutingx= 3 anddV/dt=−2, we get
−2=27
dx
dt
so that
dx
dt
=−2/27 and
dS
dt
= 12(3)
Δ
−2
27
θ
=−8/3.
The rate of change of an edge is−2/27 m/min; the rate of change of the surface area is
−8/3m
2
/min.
8. Find
dr
dt
and
dS
dt
whenr=10ft
given that
dV
dt
=8ft
3
/min.
(∗)V=
4
3
πr
3
,S=4πr
2
Diﬀerentiation of equations (∗) with respect totgives
dV
dt
=4πr
2
dr
dt
and
dS
dt
=8πr
dr
dt
.
Substitutingr= 10 anddV/dt= 8, we get
8=4π(10)
2
dr
dt
so that
dr
dt
=
1
50π
and
dS
dt
=8π(10)
1
50π
=
8
5
.
The radius is increasing
1
50π
ft/min; the surface area is increasing
8
5
ft
2
/min.
9.The area of an equilateral triangle of sidexis given by
A=
1
2
x
ﬀ
x
√
3
2
ﬃ
=
√
3
4
x
2
.Thus
dA
dt
=
√
3
2
x
dx
dt
.
Whenx=α,
dx
dt
=k,and
dA
dt
=
√
3
2
αkcm
2
/min.
10.We have 2x+2y=24,orx+y=12.Thus,A=xy=x(12−x)=12x−x
2
.
WhenA=32,x=4orx=8,and it follows that
dA
dt
= (12−2x)
dx
dt
=±4
dx
dt
.
11. Find
dA
dt
whenl= 6 in.
given that
dl
dt
=−2 in./sec.
By the Pythagorean theorem
l
2
+w
2
= 100.
Also,A=lw. Thus,A=l
√
100−l
2
. Diﬀerentiation with respect totgives
dA
dt
=l
Δ
−l
√
100−l
2
θ
dl
dt
+

100−l
2
dl
dt
.
Substitutingl= 6 anddl/dt=−2, we get
dA
dt
=6
Δ
−6
8
θ
(−2) + (8)(−2) =−7.
The area is decreasing at the rate of 7 in.
2
/sec.

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206 SECTION 4.10
12. x
2
+6
2
=y
2
;2x
dx
dt
=2y
dy
dt
Ify= 30 ft and
dx
dt
= 8 ft/min,then
dy
dt
=
x
y
dy
dt
=

(30)
2
−36
30
8=
16
5
√
6 ft/min
13. Compare
dy
dt
to
dx
dt
=−13 mph
given thatz= 16 and
dz
dt
=−17
whenx=y.
By the Pythagorean theoremx
2
+y
2
=z
2
.Thus,
2x
dx
dt
+2y
dy
dt
=2z
dz
dt
.
Sincex=ywhenz= 16, we havex=y=8
√
2 and
2(8
√
2)(−13) + 2(8
√
2)
dy
dt
= 2(16)(−17).
Solving fordy/dt,weget
−13
√
2+
√
2
dy
dt
=−34 or
dy
dt
=
1
√
2
(13
√
2−34)
∼
=−11.
Thus, boatAwins the race.
14.V=
4
3
πr
3
,so
dV
dt
=4πr
2
dr
dt
=4πr
2
(−
1
5
).Thus atr= 12 we have
dV
dt
=−
576
5
πcm
3
/min.
15.We want to ﬁnd
dA
dt
when
dx
dt
= 2 andx= 12.
A=
1
2
x

169−x
2
,so
dA
dt
=

1
2

169−x
2
−
x
2
2
√
169−x
2

dx
dt
When
dx
dt
= 2 andx= 12,
dA
dt
=−
119
5
ft
2
/sec.
16. x
2
+y
2
= (13)
2
;2x
dx
dt
+2y
dy
dt
= 0 and
dy
dt
=−
x
2y
since
dx
dt
=0.5.Whenx=5,y= 12 and
dy
dt
=−
5
24
; the top of the ladder is
dropping 5/24 ft/sec.
17.We want to ﬁnddV/dtwhenV= 1000 ft
3
andP= 5 lb/in.
2
given thatdP/dt=−0.05 lb/in.
2
/hr.
DiﬀerentiatingPV=Cwith respect tot, we get
P
dV
dt
+V
dP
dt
= 0 so that 5
dV
dt
+ 1000(−0.05) = 0.Thus,
dV
dt
=10.
The volume increases at the rate of 10 ft
3
/hr.

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.10 207
18.PV
1.4
=C;V
1.4
dP
dt
+(1.4)PV
0.4
dV
dt
= 0 and
dP
dt
=−
1.4P
V
dV
dt
.
WithV=10,P= 50 and
dV
dt
=−1,we have
dP
dt
=−
(1.4)50
10
(−1) = 7
The pressure is increasing 7 lb/in
2
/sec.
19. Find
ds
dt
whenx= 3 ft (ands= 4 ft)
given that
dx
dt
= 400 ft/min.
By similar triangles
L
x+s
=
6
s
.
Substitution ofx= 3 ands= 4 gives us
L
7
=
6
4
so that the lamp post is
L=10.5 ft tall. Rewriting
10.5
x+s
=
6
s
ass=
4
3
x
and diﬀerentiating with respect tot, we ﬁnd that
ds
dt
=
4
3
dx
dt
=
1600
3
.
The shadow lengthens at the rate of 1600/3 ft/min.
If the tip of his shadow is at the pointz, then
z=x+sand
dz
dt
=
dx
dt
+
ds
dt
= 400 +
1600
3
=
2800
3
.
The tip of his shadow is moving at the rate of 2800/3 ft./min.
20. y(t)=64−16t
2
;
dy
dt
=−32t
By similar triangles,
y
x
=
64
20 +x
.
Thus,y=
64x
20 +x
.
Now,
dy
dt
=
(20 +x)(64)−64x
(20 +x)
2
dx
dt
=
1280
(20 +x)
2
=⇒
dx
dt
=
(20 +x)
2
1280
dy
dt
.
Att=1,y=48,x=60,and
dy
dt
=−32.Therefore,
dx
dt
=
(80)
2
1280
(−32) =−160 ft/sec.
21.LetW(t) = 150

1+
1
4000
r

−2
.We want to ﬁnddW/dtwhenr= 400 given that
dr/dt= 10 mi/sec. Diﬀerentiating with respect tot,we get
dW
dt
=−300
Δ
1+
1
4000
r
θ
−3Δ
1
4000
θ
dr
dt
Now setr= 400 anddr/dt=10.Then
dW
dt
=−300
Δ
1+
400
4000
θ
−3
10
4000
∼
=−0.5634 lbs/sec

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
208 SECTION 4.10
22.M=
m

1−v
2
/c
2
;
dM
dt
=m

−
1
2

1−v
2
/c
2

−3/2
−2v/c
2
dv
dt
=
mv
c
2
(1−v
2
/c
2
)
3/2
dv
dt
.
Ifv=
c
2
and
dv
dt
=
c
100
,then
dM
dt
=
m(c/2)
c
2
(1−c
2
/4c
2
)
3/2
c
100
=
√
3
225
m.
23. Find
dh
dt
whenh= 3 in.
given that
dV
dt
=−
1
2
cu in./min.
By similar triangles
r=
1
3
h.
ThusV=
1
3
πr
2
h=
1
27
πh
3
. Diﬀerentiating with respect tot,weget
dV
dt
=
1
9
πh
2
dh
dt
.
Whenh=3,
−
1
2
=
1
9
π(9)
dh
dt
and
dh
dt
=−
1
2π
.
The water level is dropping at the rate of 1/2πinches per minute.
24. V=
1
3
πr
2
hand
r
h
=
4
6
(similar triangles)
soV=
4
27
πh
3
.Thus
dV
dt
=
4
9
πh
2
dh
dt
,
so at
dh
dt
=0.5 andh=2,
dV
dt
=
8π
9
cubic ft per sec.
25.
dV
dt
=4πr
2
dr
dt
and
dSA
dt
=8πr
dr
dt
.Thus when
dSA
dt
= 4 and
dr
dt
=0.1
we getr=
5
π
and
dV
dt
=
10
π
cm
3
/min.
26.V=πrh
2
−
1
3
πh
3
;
dV
dt
=2πrh
dh
dt
−πh
2
dh
dt
,and
dh
dt
=
2
π(14h−h
2
)
sincer= 7 and
dV
dt
=2.
(a) Whenh=7/2,
dh
dt
=
8
147π
in./sec. (b) Whenh=7,
dh
dt
=
2
49π
in./sec.
27. Find
dθ
dt
whenx=4ft
given that
dx
dt
= 2 in./min.
(∗) tan
θ
2
=
3
x
Diﬀerentiation of (∗) with respect totgives
1
2
sec
2
θ
2
dθ
dt
=−
3
x
2
dx
dt
or
dθ
dt
=−
6
x
2
cos
2
θ
2
dx
dt
.

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.10 209
Note thatdx/dt= 2 in./min=1/6 ft/min. Whenx= 4, we have cosθ/2=4/5 and thus
dθ
dt
=−
6
16
Δ
4
5
θ
2Δ
1
6
θ
=−
1
25
.
The vertex angle decreases at the rate of 0.04 rad/min.
28. tanα=
60
x
; sec
2
α
dα
dt
=−
60
x
2
dx
dt
Now, sec
2
α=
ν
y
x
ω
2
so
dα
dt
=−
60
y
2
dx
dt
.
Withy= 100,and
dx
dt
=−10,we have
dα
dt
=−
60
(100)
2
(−10) = 0.06 rad/min.
29. Find
dx
dt
whenx=1mi
given that
dθ
dt
=2πrad/min.
(∗) tanθ=
x
1/2
=2x
Diﬀerentiation of (∗) with respect totgives
sec
2
θ
dθ
dt
=2
dx
dt
.
Whenx= 1, we get secθ=
√
5 and thus
dx
dt
=5π.
The light is traveling at 5πmi/min.
30.(a) We have tanθ=x,so sec
2
θ
dθ
dt
=
dx
dt
.Switching to radians,
dθ
dt
=4π.
Thus atθ=
π
4
,
dx
dt
=8πmi/min.
(b) Atθ=0,
dx
dt
=4πmi/min.
31.We have tanθ=
x
40
,so sec
2
θ
dθ
dt
=
1
40
dx
dt
,and
dx
dt
=4.
Att=15,x= 60 and secθ=
√
5200
40
,so
dθ
dt
=
2
65
∼
=0.031 rad/sec.
32.We haveV=hπr
2
,soV= 500πcm
3
.Thush=
500
r
2
,and
dh
dt
=
−1000
r
3
dr
dt
.
Atr= 5 and
dr
dt
=
1
2
,we have
dh
dt
=−4 cm/sec.
33.We have sinθ=
4
5
so tanθ=
4
3
=
x
h
.Thusx=
4
3
h. V=12
Δ
3+2x+3
2
h
θ
=36h+16h
2
.
Thus
dV
dt
=(36+32h)
dh
dt
,so at
dV
dt
= 10 andh=2,
dh
dt
=
1
10
ft/min.

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
210 SECTION 4.10
34.tanα=
y
x
; sec
2
α
dα
dt
=
x
dy
dt
−y
dx
dt
x
2
dα
dt
=
x
dy
dt
−y
dx
dt
x
2
·
x
2
x
2
+y
2
=
x
dy
dt
−y
dx
dt
x
2
+y
2
Now
dx
dt
=−30 mph =−44 ft/sec and
dy
dt
=−22.5 mph =−33 ft/sec.
Atx= 300,y= 400,we have
dα
dt
=
300(−33)−400(−44)
(300)
2
+ (400)
2
=0.0308 the angle is increasing 0.0308 rad/sec.
35.Find
dθ
dt
wheny=4ft
given that
dx
dt
= 3 ft/sec.
tanθ=
16
x
,x
2
+ (16)
2
= (16 +y)
2
Diﬀerentiating tanθ=16/xwith respect tot, we obtain
sec
2
θ
dθ
dt
=
−16
x
2
dx
dt
and thus
dθ
dt
=
−16
x
2
cos
2
θ
dx
dt
.
Fromx
2
+ (16)
2
= (16 +y)
2
we conclude thatx= 12, wheny=4.Thus
cosθ=
x
16 +y
=
12
20
=
3
5
and
dθ
dt
=
−16
(12)
2
Δ
3
5
θ
2
(3) =
−3
25
.
The angle decreases at the rate of 0.12 rad/sec.
36.tan(α/2) =
3
x
; sec
2
(α/2)
1
2
dα
dt
=−
3
x
2
dx
dt
,
and
dα
dt
=−
6
y
2
dx
dt
Initially,y=5sox=4.8 seconds laterx=12soy=

(12)
2
+ (3)
2
=
√
153.Therefore,
dα
dt
=−
6
y
2
dx
dt
=−
6
153
(1) =−
6
153
; the angle is decreasing−6/153 rad/sec.
37.Find
dθ
dt
whent= 6 min.
tanθ=
100t
500+75t
=
4t
20+3t
Diﬀerentiation with respect totgives
sec
2
θ
dθ
dt
=
(20+3t)4−4t(3)
(20+3t)
2
=
80
(20+3t)
2
.
Whent=6
tanθ=
24
38
=
12
19
and sec
2
θ=1+

12
19

2
=
505
361

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.10 211
so that
dθ
dt
=
80
(20+3t)
2
·
1
sec
2
θ
=
80
(38)
2
·
361
505
=
4
101
.
The angle increases at the rate of 4/101 rad/min.
38.tanα=
x
H
; sec
2
α
dα
dt
=
1
H
dx
dt
dα
dt
=
1
H
dx
dt
cos
2
α=
H
H
2
+x
2
dx
dt
We haveH= 2 mi. and
dx
dt
= 400 mph.After 2 seconds,x= 400
Δ
2
3600
θ
=
2
9
miles.
dα
dt
=
2
2
2
+(2/9)
2
(400) =
200(81)
82
rad/hr =
9
164
rad/sec.
39.Letxbe the distance from third base to the player. Then the distance from home plate to the player
is given by:y=

(90)
2
+x
2
. Diﬀerentiation with respect totgives
dy
dt
=
x

(90)
2
+x
2
dx
dt
We are given thatdx/dt=−15. Therefore,
dy
dt
=−
15x

(90)
2
+x
2
and
dy
dt

x=10
=
−15
√
82
∼
=1.66
The distance between home plate to the player is decreasing 1.66 ft./sec.
40.The plane is ﬂying at an altitude 6 miles. Ifydenotes the horizontal distance between the radar station
and the plane, then
tanθ=
6
y
=⇒sec
2
θ
dθ
dt
=
−6
y
dy
dt
Therefore,
dy
dt
=
y
2
sec
2
θ
−6
dθ
dt
At the instant the plane is 12 miles from the station,y=
√
108,θ=
1
6
πand the speed is

dy
dt

=

108
−6
4
3
π
360

=
4π
60
mi/sec.
∼
=754 mi/hr.
41.Letxbe the position of the runner on the track, letybe the distance between the runner and the
spectator, and letθbe the central angle indicated in the ﬁgure. The runner’s speed is 5 meters per
second so
50
dθ
dt
= 5 and
dθ
dt
=
1
10
.
By the law of cosines:
Sθ
R
y
2
= (50)
2
+ (200)
2
−2(50)(200) cosθ.
Diﬀerentiating with respect tot, we get
2y
dy
dt
=20,000 sinθ
dθ
dt

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JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
212 SECTION 4.11
dy
dt
=
10,000 sinθ
y
1
10
dy
dt
=
1,000
y
sinθ
Now,
dy
dt

y=200
=5

(200)
2
−(25)
2
200
=5
√
39375
200
=4.96.
The runner is approaching the spectator at the rate of 4.96 meters per second.
42.
dx
dt
=
23
4
43.
dx
dt
=4 44.
dy
dt
=−
1
2
√
3
SECTION 4.11
1. ΔV=(x+h)
3
−x
3
=(x
3
+3x
2
h+3xh
2
+h
3
)−x
3
=3x
2
h+3xh
2
+h
3
,
dV=3x
2
h,
ΔV−dV=3xh
2
+h
3
(see ﬁgure)
2.The area of the ring can be thought of as the increase of the area of a disk as the radius increases from
rtor+h.
A=πr
2
;sodA=2πr dr=2πrh
The exact area is:π(r+h)
2
−πr
2
=π(r
2
+2rh+h
2
)−πr
2
=2πrh+πh
2
.
3.f(x)=x
1/3
,x= 1000,h=2,f
ﬀ
(x)=
1
3
x
−2/3
3
√
1002 =f(x+h)
∼
=f(x)+hf
ﬀ
(x)=
3
√
1000 + 2

1
3
(1000

−2/3
] = 10 +
1
150
∼
=10.0067
4.f(x)=
1
√
x
,x=25,h=−0.5,f
ﬀ
(x)=
−1
2x
3/2
1
√
24.5
=f(x+h)
∼
=f(x)+hf
ﬀ
(x)=
1
5
+(−
1
2
)
−1
2(5)
3
=
101
500
=0.202
5.f(x)=x
1/4
,x=16,h=−0.5,f
ﬀ
(x)=
1
4
x
−3/4
(15.5)
1/4
=f(x+h)
∼
=f(x)+hf
ﬀ
(x) = (16)
1/4
+(−0.5)

1
4
(16)
−3/4

=1
63
64
∼
=1.9844
6.f(x)=x
2/3
,x=27,h=−1,f
ﬀ
(x)=
2
3
x
−1/3
(26)
2/3
=f(x+h)
∼
=f(x)+hf
ﬀ
(x) = (27)
2/3
+(−1)

2
3
(27)
−1/3

=
79
9
∼
=8.778

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.11 213
7.f(x)=x
3/5
,x=32,h=1,f
ﬀ
(x)=
3
5
x
−2/5
(33)
3/5
=f(x+h)
∼
=f(x)+hf
ﬀ
(x) = (32)
3/5
+ (1)

3
5
(32)
−2/5

=8.15
8.f(x)=x
−1/5
,x=32,h=1,f
ﬀ
(x)=−
1
5
x
−6/5
(33)
−1/5
=f(x+h)
∼
=f(x)+hf
ﬀ
(x) = (32)
−1/5
−(1)

1
5
(32)
−6/5

=
159
320
∼
=0.497
9.f(x) = sinx, x=
π
4
,h=
π
180
,f
ﬀ
(x) = cosx
sin 46
◦
=f(x+h)
∼
=f(x)+hf
ﬀ
(x) = sin
π
4
+
π
180
cos
π
4
=
√
2
2
ν
1+
π
180
ω
∼
=0.719
10.f(x) = cosx, x=
π
3
,h=
π
90
,f
ﬀ
(x)=−sinx
cos 62
◦
=f(x+h)
∼
=f(x)+hf
ﬀ
(x) = cos
π
3
+
π
90
ν
−sin
π
3
ω
=
1
2
−
ν
π
90
ω
ﬀ√
3
2
ﬃ
∼
=0.470
11.f(x) = tanx, x=
π
6
,h=
−π
90
,f
ﬀ
(x) = sec
2
x
tan 28
◦
=f(x+h)
∼
=f(x)+hf
ﬀ
(x) = tan
π
6
+
Δ
−π
90
∈√
4
3
θ
=
√
3
3
−
2π
135
∼
=0.531
12.f(x) = sinx, x=
π
4
,h=−
π
90
,f
ﬀ
(x) = cosx
sin 43
◦
=f(x+h)
∼
=f(x)+hf
ﬀ
(x) = sin
π
4
+
ν
−
π
180
ω
cos
π
4
=
√
2
2
−
ν
π
180
ω
ﬀ√
2
2
ﬃ
∼
=0.682
13.f(2.8)
∼
=f(3)+(−0.2)f
ﬀ
(3) = 2 + (−0.2)(2) = 1.6
14.f(5.4)
∼
=f(5)+(0.4)f
ﬀ
(5) = 1 + (0.4)(3) = 2.2
15.V(r)=πr
2
h; volume =V(r+t)−V(r)
∼
=tV
ﬀ
(r)=2πrht
16.Error in diameter = 0.3=⇒error in radius = 0.15.
(a)dS=8πrh=8π(8)(0.15)
∼
=9.6πcm
2
(b)dV=4πr
2
h=4π(8)
2
(0.15)
∼
=38.4cm
3
17.V(x)=x
3
,V
ﬀ
(x)=3x
2
,ΔV
∼
=dV=V
ﬀ
(10)h= 300h
|dV|≤3=⇒|300h|≤3=⇒|h|≤0.01,error≤0.01 feet
18.(a) Letf(x)=
√
x.Thenf
ﬀ
(x)=
1
2
√
x
and
√
x+1−
√
x
∼
=(1)f
ﬀ
(x)=
1
2
√
x
.
Now,
1
2
√
x
<0.01 =
1
100
=⇒
√
x>50 =⇒x>2500.
(b) Letf(x)=x
1/4
.Thenf
ﬀ
(x)=
1
4
x
−3/4
and
4
√
x+1−
4
√
x
∼
=(1)f
ﬀ
(x)=
1
4
x
−3/4
.
Now,
1
4
x
−3/4
<0.002 =
2
1000
=⇒x
3/4
>125 =⇒x>625.

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
214 SECTION 4.11
19.V(r)=
2
3
πr
3
anddr=0.01.
V(r+0.01)−V(r)
∼
=V
ﬀ
(r)(0.01) = 2πr
2
(0.01)
=2π(600)
2
(0.01) (50 ft = 600 in)
= 22619.5in
3
or 98 gallons (approx.)
20.V(r)=
4
3
πr
3
;
dV
dr
=4πr
2
.
Now,V(r+h)−V(r)=8×(10)6
∼
=
dV
dr
h=4πr
2
h.Therefore,
h=
8×(10)
6
4π(4000)
2
∼
=0.0398 (miles)
∼
=210 (feet)
21.P=2π

L
g
impliesP
2
=4π
2
L
g
Diﬀerentiating with respect tot,we have
2P
dP
dt
=
4π
2
g
·
dL
dt
=
P
2
L
·
dL
dt
since
P
2
L
=
4π
2
g
.
Thus
dP
P
=
1
2
·
dL
L
22.
dP
P
=−15 sec/hour =−
15
3600
.By Exercise 21,
1
2
dL
L
=−
15
3600
anddL=−
30
3600
L=−
1
120
L
WithL=90,we havedL=−90/120 =−0.75; the pendulum should be shortened
0.75 cm to 89.25 cm.
23.L=3.26 ft,P= 2 sec, anddL=0.01 ft
dP
P
=
1
2
·
dL
L
dP=
1
2
·
dL
L
·P=
1
2
·
0.01
3.26
·2dP
∼
=0.00307 sec
24.Each edge increases by 0.1%; takeh=0.001x.
S=6x
2
,dS=12xh,and
dS
S
=
12x(0.001x)
6x
2
=0.002 = 0.2%.
A=x
3
,dA=3x
2
h,and
dA
A
=
3x
2
(0.001x)
x
3
=0.003 = 0.3%.
25.A(x)=
1
4
πx
2
,dA=
1
2
πxh,
dA
A
=2
h
x
dA
A
≤0.01⇐⇒ 2
h
x
≤0.01⇐⇒
h
x
≤0.005 within
1
2
%

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.11 215
26.(a) Lety=x
n
.Thendy=nx
n−1
h.
To get
dy
y
=
nx
n−1
h
x
n
<0.01,we need
h
x
<
0.01
n
,that is, within
1
n
%.
(b) Lety=x
1/n
.Thendy=
1
n
x
(1−n)/n
h.
To get
dy
y
=
1
n
x
(1−n)/n
h
x
1/n
<0.01,we need
h
x
<(0.01)n,that is, withinn%.
27.(a) and (b) 28.lim
h→0
g(h) = lim
h→0
g(h)
h
·h=
Δ
lim
h→0
g(h)
h
∈√
lim
h→0
h
θ
=0
29.lim
h→0
g1(h)+g2(h)
h
= lim
h→0
g1(h)
h
+ lim
h→0
g2(h)
h
=0+0=0
lim
h→0
g1(h)g2(h)
h
= lim
h→0
h
g1(h)g2(h)
h
2
=
Δ
lim
h→0
h
∈√
lim
h→0
g1(h)
h
∈√
lim
h→0
g2(h)
h
θ
= (0)(0)(0) = 0
30.(a)g(h)=f(x+h)−f(x)−mh.
(b) lim
h→0
g(h)
h
= lim
h→0

f(x+h)−f(x)
h
−m

=f
ﬀ
(x)−m=0 iﬀm=f
ﬀ
(x).
31.Suppose thatfis diﬀerentiable atx. Then there is a numbermsuch that
lim
h→0
f(x+h)−f(x)
h
=m or lim
h→0
f(x+h)−f(x)
h
−m=0.
Letg(h)=f(x+h)−f(x)−mh. Then
g(h)
h
=
f(x+h)−f(x)
h
−m and lim
h→0
g(h)
h
= lim
h→0
f(x+h)−f(x)
h
−m=0.
Now suppose thatf(x+h)−f(x)−mh=g(h) whereg(h)=o(h). Then
lim
h→0
Δ
f(x+h)−f(x)
h
−m
θ
= lim
h→0
g(h)
h
=0.
Therefore
lim
h→0
f(x+h)−f(x)
h
=m
andfis diﬀerentiable atx.m=f
ﬀ
(x).

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
216 SECTION 4.11
PROJECT 4.11
1.C(x) = 2000 + 50x−
x
2
20
.
Marginal cost:C
ﬀ
(x)=50−
x
10
;C
ﬀ
(20) = 48
Exact cost of the 21st component:C(21)−C(20) = 3027.95−2980 = 47.95.
2.Proﬁt function:P(x)=R(x)−C(x) = 650x−5x
2
−(12,000+30x) = 620x−5x
2
−12000.
Breakeven points:P(x)=0,impliesx
2
−124x+ 2400 = (x−24)(x−100) = 0;x= 24,x= 100
units.
Maximum proﬁt:P
ﬀ
(x) = 620−10x;P
ﬀ
(x)=0 atx= 62 units.
3.(a) Proﬁt function:P(x)=R(x)−C(x)=20x−
x
2
50
−(4x+ 1400) = 16x−
x
2
50
−1400.
Break-even points: 16x−
x
2
50
−1400 = 0 orx
2
−800x+70,000 = 0
Thusx= 100,orx= 700 units.
(b)P
ﬀ
(x)=16−
x
25
;P
ﬀ
(x)=0 atx= 400 units.
(c)
4.(a) (b)
Break-even points atx=81.11 andx= 631.19 Maximum proﬁt at x= 336.11 units
5.(a) (b) P(x)=
10x
1+0.25x
2
−4−0.75x
Breakeven points atx
∼
=0.46,andx
∼
=4.53 Maximum proﬁt at 175 units
6.A(x)=
C(x)
x
,A
ﬀ
(x)=
xC
ﬀ
(x)−C(x)
x
2
.
A
ﬀ
(x) = 0 impliesxC
ﬀ
(x)−C(x)=0 or C
ﬀ
(x)=
C(x)
x
=A(x). The critical points ofAare
the points whereC
ﬀ
(x)=C(x)/x.
A
ﬀﬀ
(x)=
x
2
[xC
ﬀﬀ
(x)]−[xC
ﬀ
(x)−C(x)](2x)
x
4
At the points wherexC
ﬀ
(x)−C(x)=0,A
ﬀﬀ
(x)=
C
ﬀﬀ
(x)
x
.IfC
ﬀﬀ
(x)>0 at such a point, thenA(x)
is a minimum.

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
SECTION 4.12 217
7.B(x)=
R(x)
x
,B
ﬀ
(x)=
xP
ﬀ
(x)−P(x)
x
2
.
B
ﬀ
(x) = 0 impliesxP
ﬀ
(x)−P(x)=0 or P
ﬀ
(x)=
P(x)
x
=B(x). The critical points ofBare
the points whereP
ﬀ
(x)=P(x)/x.
B
ﬀﬀ
(x)=
x
2
[xP
ﬀﬀ
(x)]−[xP
ﬀ
(x)−P(x)](2x)
x
4
At the points wherexP
ﬀ
(x)−P(x)=0,B
ﬀﬀ
(x)=
P
ﬀﬀ
(x)
x
.IfP
ﬀﬀ
(x)<0 at such a point, thenB(x)
is a maximum.
SECTION 4.12
1.(a)xn+1=
1
2
xn+12
Δ
1
xn
θ
(b)x4
∼
=4.89898
2.(a)xn+1=
2x
3
n−1
3x
2
n−4
(b)x4
∼
=1.86081
3.(a)xn+1=
2
3
xn+
25
3
Δ
1
xn
θ
2
(b)x4
∼
=2.92402
4.(a)xn+1=
4
5
xn+6
Δ
1
xn
θ
4
(b)x4
∼
=1.97435
5.(a)xn+1=
xnsinxn+ cosxn
sinxn+1
(b)x4
∼
=0.73909
6.(a)xn+1=
xncosxn−sinxn−x
2
n
cosxn−2xn
(b)x4
∼
=0.87673
7.(a)xn+1=
6+xn
2
√
xn+3−1
(b)x4
∼
=2.30278
8.(a)xn+1=
xnsec
2
xn−tanxn
1 + sec
2
xn
(b)x4
∼
=2.30278
9.Letf(x)=x
1/3
.Thenf
ﬀ
(x)=
1
3
x
−2/3
.The Newton-Raphson method applied to
this function gives:
xn+1=xn−
f(xn)
f
ﬀ
(xn)
=xn−
x
1/3
n
1
3
x
−2/3
n
=−2xn.
Choose anyx1δ=0.Thenx2=−2x1,x3=−2x2=4x1,···,
xn=−2xn−1=(−1)
n−1
2
n
x1,···.
10.xn+1=xnfor alln.
11.(a)fis a continuous function, andf(1) =−2<0,f(2) = 3>0.Thus,fhasarootin(1,2).
(b)f
ﬀ
(x)=6x
2
−6xandf
ﬀ
(1) = 0.Therefore,x1= 1 will fail to generate values that
will approach the root in (1,2).
(c)xn+1=xn−
2x
3
n−3x
2
n−1
6x
2
n−6xn
;
x1=2,x2=1.75,x3=1.68254,x4=1.67768,f(x4)
∼
=0.00020.

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
218 SECTION 4.12
12.(a) Letf(x)=x
4
−2x
2
−
17
16
.Thenf
ﬀ
(x)=4x
3
−4x.The Newton-Raphson method applied to this
function gives:
xn+1=xn−
x
4
n−2x
2
n−
17
16
4x
3
n−4xn
Ifx1=
1
2
,thenx2=−
1
2
,x3=
1
2
,···xn=(−1)
n−11
2
,···.
(b)x1=2,x2=1.71094,x3=1.58569,x4=1.56165;f(x4)=0.00748
13.(a)f(x)=x
2
−a;f
ﬀ
(x)=2x.Substituting into (4.12.1), we have
xn+1=xn−
x
2
n−a
2xn
=
x
2
n+a
2xn
=
1
2
Δ
xn+
a
xn
θ
,n≥1
(b) Leta=5,x1=2,andxn+1=
1
2
Δ
xn+
a
xn
θ
,n≥1.
Thenx2=2.25,x3=2.23611,x4=2.23607,andf(x4)
∼
=0.000009045.
14.(a) Letf(x)=x
k
−a.Thenf
ﬀ
(x)=kx
k−1
.The Newton-Raphson method applied to this function
gives:
xn+1=xn−
x
k
n−a
kx
k−1
n
=xn−
1
k
xn+
1
k
a
x
k−1
n
=
1
k

(k−1)xn+
a
x
k−1
n

(b) Leta=23,k= 3 andx1=3.Then
x1=3,x2=2.85185,x3=2.84389,x4=2.84382;f(x4)=−0.00114
15.(a) Letf(x)=
1
x
−a.Thenf
ﬀ
(x)=−
1
x
2.The Newton-Raphson method applied to
this function gives:
xn+1=xn−
1
xn
−a
−
1
x
2
n
=xn+xn−ax
2
n=2xn−ax
2
n
(b) Leta=2.7153,andx1=0.3.Then
x2=0.35562,x3=0.36785,x4=x5=0.36828,
Thus
1
2.7153
⎨0.36828.
16.(a)f(x)=x
4
−7x
2
−8x−3,f
ﬀ
(x)=4x
3
−14x−8,f
ﬀﬀ
(x)=12x
2
−14.
f
ﬀ
(2) =−4<0 andf
ﬀ
(3)=58>0;f
ﬀ
has a zero in (2,3).
f
ﬀﬀ
(x)=12x
2
−14>0on(2,3). Therefore,f
ﬀ
has exactly one zero in this interval.
(b)xn+1=xn−
4x
3
n−14xn−8
12x
2
n−14
=
4x
3
n+4
6x
2
n−7
;x3
∼
=2.1091. Sincef
ﬀﬀ
(x3)>0,fhas a local mini-
mum atc.
17.(a)f(x) = sinx+
1
2
x
2
−2x, f
ﬀ
(x) = cosx+x−2,f
ﬀﬀ
(x)=−sinx+1.
f
ﬀ
(2) = cos 2
∼
=−0.4161<0 andf
ﬀ
(3) = cos 3 + 1
∼
=0.0100>0;f
ﬀ
has a zero in (2,3).
f
ﬀﬀ
(x)=−sinx+1>0on(2,3). Therefore,f
ﬀ
has exactly one zero in this interval.
(b)xn+1=xn−
cosxn+xn−2
1−sinxn
=
2−xnsinxn−cosxn
1−sinxn
;x3
∼
=2.9883. Sincef
ﬀﬀ
(x3)>0,
fhas a local minimum atc.

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
REVIEW EXERCISES 219
18.(a)xn+1=xn−
sinxn
cos xn
=xn−tanxn,x1=3;x4
∼
=3.14159
(b)x4
∼
=6.28319
19.(a)xn+1=xn−
xn+ tanxn
1+sec
2
xn
,x1=2π/3;r1
∼
=2.029
(b)x1=5π/3;r2
∼
=4.913
REVIEW EXERCISES
1.fis diﬀerentiable on (−1,1) and continuous on [−1,1];f(1) =f(−1) = 0.
f
ﬀ
(x)=3x
2
−1; 3c
2
−1=0 =⇒c=±
√
3
3
.
2.fis diﬀerentiable on (0,2π) and continuous on [0,2π];f(0) =f(2π)=0.
f
ﬀ
(x) = cosx−sinx; cosc−sinc=0 =⇒c=
1
4
π,
5
4
π.
3.fis diﬀerentiable on (−2,3) and continuous on [−2,3].
f
ﬀ
(c)=3c
2
−2=
f(3)−f(−2)
5
=5 =⇒3c
2
=7 =⇒c=±

7
3
4.fis diﬀerentiable on (2,5) and continuous on [2,5].
f
ﬀ
(c)=
1
2
√
c−1
=
f(5)−f(2)
3
=
1
3
=⇒c=
13
4
.
5.fis diﬀerentiable on (2,4) and continuous on [2,4].
f
ﬀ
(c)=−
2
(c−1)
2
=
f(4)−f(2)
2
=−
2
3
=⇒c=1+
√
3.
6.fis diﬀerentiable on (0,16) and continuous on [0,16].
f
ﬀ
(c)=
3
4c
1/4
=
f(16)−f(0)
16
=
1
2
=⇒c=
81
16
.
7.f
ﬀ
(x)=
1+x
2/3
3x
2/3
δ= 0 for allx∈(−1,1).f
ﬀ
(0) does not exist. Thereforefis not diﬀerentiable on
(−1,1).
8.f
ﬀ
(x)=
−3
(x−2)
2
<0 for allx∈(1,4);
f(4)−f(1)
3
=
9
6
>0.f
ﬀ
(2) does not exist. Thereforefis
not diﬀerentiable on (1,4).
9.No. Reason: If such a function did exist, then, by the mean-value theorem, there is a numberc∈(1,4)
such thatf
ﬀ
(c)=−
4
3
<−1.
10.(a)f
ﬀ
(x)=3x
2
−3=3(x
2
−1)<0on(−1,1). Therefore, by Rolle’s theorem,fcan have at most
one zero in [−1,1].
(b) Sincefdecreases on [−1,1], we must havef(−1) =−1+3+k≥0 andf(1) = 1−3+k≤0.
These conditions imply thatk∈[−2,2].

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220 REVIEW EXERCISES
11.f
ﬀ
(x)=6x(x+ 1).
fincreases on (−∞,−1]∪[0,∞) and decreases on [−1,0].
The critical points arex=−1,0;f(−1) = 2 is a local max andf(0) = 1 is a local min.
12.f
ﬀ
(x)=4(x−1)(x
2
+x+ 1).
fincreases on [1,∞) and decreases on (−∞,1]. The critical point isx=1;f(1) = 0 is a local min.
13.f
ﬀ
=(x+ 2)(x−1)
2
(5x+ 4).
fincreases on (−∞,−2]∪[−
4
5
,∞) and decreases on [−2,−
4
5
].
The critical points arex=−2,1,−
4
5
.f(−2) = 0 is a local max andf(−
4
5
)
∼
=−8.981 is a local min.
14.f
ﬀ
(x)=1−
8
x
3
=
x
3
−8
x
3
.
fincreases on (−∞,0)∪[2,∞) and decreases on (0,2].x= 2 is the only critical point; 0 is not a
critical point.f(2) = 3 is a local min.
15.f
ﬀ
(x)=
1−x
2
(1 +x
2
)
2
.
fincreases on [−1,1] and decreases on (−∞,−1]∪[1,∞). The critical points arex=−1,1.f(−1) =
−
1
2
is a local min andf(1) =
1
2
is a local max.
16.f
ﬀ
(x) = sinx+ cosx.
fincreases on [0,
3π
4
]∪[
7π
4
,2π] and decreases on [
3π
4
,
7π
4
]. The critical points arex=
3π
4
,
7π
4
.f(
3π
4
)=
√
2 is a local max andf(
7π
4
)=−
√
2 is a local min.
17.f
ﬀ
(x)=3x
2
+4x+1=(3x+ 1)(x+ 1); critical points:x=−
1
3
,−1.
f(−2) =−1 endpoint and abs. min;f(−1) = 1 local max;f(−
1
3
)=
23
27
local min;f(1) = 5 endpoint
and abs. max.

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REVIEW EXERCISES 221
18.f
ﬀ
(x)=4x
3
−16x=4x(x−2)(x+ 2); critical points:x=0,2.
f(−1) =−5 endpoint min;f(0) = 2 local max;f(2) =−14 local and abs. min;f(3) = 11 endpoint
and abs. max.
19.f
ﬀ
(x)=2x−
8
x
3
=
2(x
4
−4)
x
3
; critical point:x=4
1/4
=
√
2.
f(1) = 5 endpoint max;f(
√
2) = 4 local and abs. min;f(4) =
65
4
endpoint and abs. max.
20.f
ﬀ
(x) = cosx(1−2 sinx); critical points:x=
π
6
,
π
2
,
5π
6
,
3π
2
.
f(0) = 1 endpoint min;f(
π
6
)=5/4 local and abs. max;f(
π
2
) = 1 local min;f(
5π
6
)=
5
4
local and abs.
max;f(
3π
2
)=−1 local and abs. min;f(2π) = 1 endpoint max.
21.f
ﬀ
(x)=
−x
2
√
1−x
+
√
1−x=
2−3x
2
√
1−x
; critical point:x=
2
3
.
f(1) = 0 endpoint min;f(
2
3
)=
2
√
3
9
local and abs. max.
22.f
ﬀ
(x)=
(x−2)2x−x
2
(x−2)
2
=
x(x−4)
(x−2)
2
; critical point:x=4;
f(4) = 8 local and abs. min.
23.f(x)=
3x(x−3)
(x−4)(x+3)
. Vertical asymptotes:x=−3 andx= 4; horizontal asymptote:y=3.
24.f(x)=
(x−2)(x+2)
(x−2)(x−3)
=
x+2
x−3
,xδ= 2. Vertical asymptote:x= 3; horizontal asymptote:y=1.
25.f(x)=x−
x
(x−1)(x
2
+x+1)
. Vertical asymptote:x= 1; oblique asymptote:y=x.
26.f
ﬀ
(x)=
3
5(x−1)
2/5
, vertical tangent.
27.f
ﬀ
(x)=x
2/5
−2x
−3/5
=
x−2
x
3/5
, vertical cusp.
28.f
ﬀ
(x)=2x
−2/3
+4x
1/3
=
2+4x
x
2/3
, vertical tangent.

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222 REVIEW EXERCISES
29.f(x)=6+4x
3
−3x
4
, domain: (−∞,∞)
f
ﬀ
(x)=12x
2
(1−x)
critical pts.x=0,1
f
ﬀﬀ
(x)=12x(2−3x)
f
ﬀ
(x)>0on(−∞,1),
f
ﬀ
(x)<0on(1,∞)
f
ﬀﬀ
(x)>0on(0,2/3)
f
ﬀﬀ
(x)<0on(−∞,0)∪(2/3,∞) −1 1
x
2
4
6
y
30.f(x)=3x
5
−5x
3
+ 1, domain: (−∞,∞)
f
ﬀ
(x)=15x
2
(x−1)(x+1)
critical pts.x=−1,0,1
f
ﬀﬀ
(x)=30x

2x
2
−1

f
ﬀ
(x)>0on(−∞,−1)∪(1,∞)
f
ﬀ
(x)<0on(−1,1)
f
ﬀﬀ
(x)>0on
ν
−
√
2
2
,0
ω
∪
ν√
2
2
,∞
ω
f
ﬀﬀ
(x)<0on
ν
−∞,−
√
2
2
ω
∪
ν
0,
√
2
2
ω
−1 1
x
−1
1
3
5
y
31.f(x)=
2x
x
2
+4
, domain: (−∞,∞)
symmetric with respect to the origin.
f
ﬀ
(x)=
2

4−x
2

(x
2
+4)
2
critical pts.x=2,x=−2
f
ﬀﬀ
(x)=
4x

x
2
−12

(x
2
+4)
3
f
ﬀ
(x)>0on(−2,2),
f
ﬀ
(x)<0on(−∞,2)∪(2,∞)
f
ﬀﬀ
(x)>0on

−
√
12,0

∪
√
12,∞

f
ﬀﬀ
(x)<0on

−∞,−
√
12

∪

0,
√
12

−2 2
x
0.5
−0.5
y

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REVIEW EXERCISES 223
32.f(x)=x
2/3
(x−10), domain: (−∞,∞)
f
ﬀ
(x)=
5(x−4)
3x
1/3
critical pts.x=4,x=0
f
ﬀﬀ
(x)=
10(x+2)
9x
4/3
f
ﬀ
(x)>0on(−∞,0)∪(4,∞),
f
ﬀ
(x)<0on(0,4)
f
ﬀﬀ
(x)>0on(−2,0)∪(0,∞)
f
ﬀﬀ
(x)<0on(−∞,−2)
−2 4 10
x
−10
10
y
33.f(x)=x
√
4−x, domain: (−∞,4]
f
ﬀ
(x)=
8−3x
2
√
4−x
f
ﬀﬀ
(x)=
3x−16
4(4−x)
3/2
f
ﬀ
(x)>0on

−∞,
8
3

,
f
ﬀ
(x)<0on

8
3
,4

f
ﬀﬀ
(x)<0on(−∞,4)
−2 2 4
x
−2
2
y
34.f(x)=x
4
−2x
2
+ 3, domain (−∞,∞);
symmetric with respect to they-axis
f
ﬀ
(x)=4x
3
−4x=4x(x−1)(x+1)
critical pts.x=−1,0,1
f
ﬀﬀ
(x)=12x
2
−4 = 12(x−1/
√
3)(x+1/
√
3)
f
ﬀ
(x)>0on[−1,0]∪[1,∞)
f
ﬀ
(x)<0on(−∞,−1]∪[0,1]
f
ﬀﬀ
(x)>0on(∞,−1/
√
3)∪(1/
√
3,∞)
f
ﬀﬀ
(x)<0on[−1/
√
3,1/
√
3)
-2 -1 1 2
x
1
2
3
y
35.f(x) = sinx+
√
3 cosx, domain [0,2π]
f
ﬀ
(x) = cosx−
√
3 sinx
critical pts.x=
1
6
π, x=
7
6
π
f
ﬀﬀ
(x)=−sinx−
√
3 cosx
f
ﬀ
(x)>0on[0,π/6)∪(7π/6,2π]
f
ﬀ
(x)<0on(π/6,7π/6)
f
ﬀﬀ
(x)>0on(2π/3,5π/3)
f
ﬀﬀ
(x)<0on[0,π/3)∪(5π/3,2π]
π 2π
x
−2
2
y

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224 REVIEW EXERCISES
36.f(x) = sin
2
x−cosx, x∈[0,2π]
f
ﬀ
(x) = 2 sinxcosx+ sinx
critical pts.x=
2
3
π, π,
4
3
π
f
ﬀﬀ
(x) = 4 cos
2
x+ cosx−2
f
ﬀ
(x)>0on

0,
2
3
π

∪

π,
4
3
π

,
f
ﬀ
(x)<0on

2
3
π, π

∪

4
3
π,2π

123456
x
-1
1
y
37.
−1 2
x
y
38.Letrbe the radius of the sphere andxthe side length of the cube. Then the sum of the surface areas
equals constant implies
4πr
2
+6x
2
=C=⇒x=

C−4πr
2
6
.
The sum of the volumes is:
V=
4
3
πr
3
+
Δ
C−4πr
2
6
θ
3/2
,0≤r≤

C
4π
.
Now
V
ﬀ
=4πr
2
+
3
2
Δ
C−4πr
2
6
θ
1/2Δ
−
4
3
πr
θ
=4πr
2
−2πr
Δ
C−4πr
2
6
θ
1/2
SetV
ﬀ
(r)=0:
2r=
Δ
C−4πr
2
6
θ
1/2
=⇒4r
2
=
C−4πr
2
6
=⇒r=

C
24 + 4π
,andx=2

C
24 + 4π
.
Without loss of generality, setC= 1. Then
V(0) = (1/6)
3/2∼
=0.07,V(1/
√
24+4π)
∼
=0.055,V(1/
√
4π)
∼
=0.09.
Thus, the sum of the volumes is a minimum when the side length of the cube equals the diameter of
the sphere. The sum of the volumes is a maximum when the side length of the cube is 0.
39.Letxbe the side length of the square base. Then, since the volume is 27, the heighty=27/x
2
. The
conditions:x
2
≤18,0≤y≤4 imply
3
√
3
2
≤x≤3
√
2. Thus, the surface areaSof the box is

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REVIEW EXERCISES 225
given by
S(x)=2x
2
+
108
x
,
3
√
3
2
≤x≤3
√
2
S
ﬀ
(x)=4x−
108
x
2
;S
ﬀ
(x)=0 =⇒4x
3
=27 =⇒x=3
Now,
S
ν
3
√
3
2
ω
=
27
2
+24
√
3
∼
=55.07,
S(3) = 54
S(3
√
2)=36+18
√
2
∼
=61.46
(a) The minimal surface area is 54 sq. ft. atx=3.
(b) The maximal surface area is 61.46 sq. ft. (approx.) atx=3
√
2.
40.An equation for the line withx-interceptaandy-interceptbis:
x
a
+
y
b
=1
Since the line passes through (1,2),
1
a
+
2
b
=1 =⇒b=
2a
a−1
The area of the right triangleOABis:S=
1
2
ab=
a
2
a−1
Now,
S
ﬀ
(a)=
a
2
−2a
(a−1)
2
andS
ﬀ
(a)=0 ata=0,a=2
The triangleOABwill have minimum area whenA=(2,0) andB=(0,4).
41.Introduce a rectangular coordinate system so that the rectangle is in the ﬁrst quadrant and the base
and left side lie on the coordinate axes. Letxdenote the length of the base, andythe height. Let
P=2x+2ybe the given perimeter. If the rectangle is rotated about they-axis, then the volumeV
is given by:
V(x)=
πP
2
x
2
−πx
3
,0≤x≤P/2.
Now,
V
ﬀ
(x)=πPx−3πx
2
andV
ﬀ
(x)=0 =⇒x=0orx=
1
3
P.
Atx=0,V=0; atx=P/3,y=P/6,V=πP
3
/54; atx=P/2,V= 0. The dimensions that
generate the cylinder of maximum volume arex=P/3,y=P/6.
42.Letxbe the width of the page andythe height. Thenxy= 80. The area available for print is:
A=(x−2)(y−2.5)=(x−2)
Δ
80
x
−2.5
θ
=85−2.5x−
160
x
,0<x<∞.
A
ﬀ
(x)=−2.5+
160
x
2
;A
ﬀ
(x)=0: 2.5x
2
= 160,x
2
=64,x=8

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226 REVIEW EXERCISES
SinceA
ﬀﬀ
(x)<0,Ahas an absolute maximum atx= 8. The dimensions of the page that will maximize
the printed area are: width 8, height 10.
43.x
ﬀ
(t)=1−2 sint, x
ﬀﬀ
(t)=−2 cost.
The object is slowing down whenx
ﬀ
(t)x
ﬀﬀ
(t)<0=⇒t∈[0,
1
6
π)∪(
1
2
π,
5
6
π)∪(
3
2
π,2π].
44.x(t)=(4t−1)(t−1)
2
,t≥0;x
ﬀ
(t)=v(t) = 2(4t−1)(t−1)+4(t−1)
2
= 6(2t
2
−3t+1)
= 6(2t−1)(t−1)
(a) The object is moving to the right:v(t)>0 when 0≤t<
1
2
and whent>1; moving to the
left:v(t)<0 when
1
2
<t<1; the object changes direction att=
1
2
andt=1.
(b) Speed when moving left:ν=|v|= 6(3t−2t
2
−1) on

1
2
,1

.dν/dt= 6(3−4t);dν/dt=0 at
t=3/4. The maximum speed when moving left isν(3/4) = 3/4 units per unit time.
45.(a)x(t)=
√
t+1
x
ﬀ
(t)=v(t)=
1
2
√
t+1
x
ﬀﬀ
(t)=a(t)=−
1
4(t+1)
3/2
=−2[v(t)]
3
,
x
ﬀﬀﬀ
(t)=a
ﬀ
(t)=
3
8(t+1)
5/2
(b) Position:x(17) =x(15+2)
∼
=x(15) + 2x
ﬀ
(15) = 4.25;
Velocity:v(17) =v(15+2)
∼
=v(15) + 2v
ﬀ
(15) =
15
128
∼
=0.1172;
Acceleration:a(17) =a(15+2)
∼
=a(15) + 2a
ﬀ
(15)
∼
=0.0032.
46.The height of the rocket at timetis given by:y(t) = 128t−16t
2
.
(a) At maximum height,v(t)=y
ﬀ
(t) = 128−32t=0 =⇒t= 4. The rocket reaches its
maximum height at 4 seconds; the maximum height is:y(4) = 256 feet.
(b)y(t)=0 =⇒16t(8−t)=0 =⇒t=0,t= 8; the rocket hits the ground att=8
seconds;v(8) =−128 ft/sec.
47.The height at timetis:y(t)=−16t
2
+8t+y0.y(10) = 0 =−1600 + 80 +y0=⇒y0= 1520 ft.
48.The height at timetis:y(t)=−16t
2
+v0t.y(1) = 24 =−16 +v0=⇒v0= 40. Therefore,
y(t)=−16t
2
+40t. The ball achieves maximum height at the instantv(t)=−32t+40=0 = ⇒
t=
5
4
. The maximum height isy(5/4) =−16(5/4)
2
+ 40(5/4) = 25.
49.Letxbe the position of the boy and letybe the length of his shadow. Then
dx
dt
= 168 and
x+y
12
=
y
5
=⇒7y=5xandy=
5
7
x
Diﬀerentiating implicitly with respect tot, we get
7
dy
dt
=5
dx
dt
=⇒
dy
dt
=
Δ
5
7
θ
(168) = 120
The shadow is lengthening at the rate of 120 ft/min.

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REVIEW EXERCISES 227
50.V=
1
3
πr
2
h,Vconstant
0=
d
dt
V=
d
dt
Δ
1
3
πr
2
h
θ
=
2πrh
3
dr
dt
+
πr
2
3
dh
dt
Solving for
dh
dt
we get:
dh
dt
=−
2h
r
·
dr
dt
.
At the instantr= 4 andh=15,
dh
dt
=−
30
4
(0.3) =−2.25; the height is decreasing at the rate of
2.25 in/min.
51.Letxbe the position of the locomotive at timetand letybe the position of the car. By the law of
cosines the distance between the locomotive and the car is given by
s
2
=x
2
+y
2
−2xycos 60
◦
=x
2
+y
2
−xy.
Diﬀerentiating with respect tot,weget
2s
ds
dt
=2x
dx
dt
+2y
dy
dt
−x
dy
dt
−y
dx
dt
.
Settingdx/dt=60, dy/dt=−30 and converting to feet, gives
2s
ds
dt
=2x(60)(5280)−2y(30)(5280) +x(30)(5280)−y(60)(5280) = 5280(150x−120y).
Whenx=y= 500,s= 500 and
2(500)
ds
dt
= 5280(30)(500) =⇒
ds
dt
= 15(5280).
The distance between the locomotive and the car is increasing at the rate of 15 miles per hour.
52.Letrbe the radius of the circle. Thendr/dt= 5. The square has side lengthr
√
2 and areaA=2r
2
.
Diﬀerentiating with respect tot,weget
dA
dt
=4r
dr
dt
=20r; and
dA
dt

r=10
= 200.
53.The cross-section of the water at timetis an isosceles triangle with basexand heightywherexand
yare related by
1
2
x
y
=
1
2
(similar triangles). Thusx=y.
The volume of water in the trough at timetis
V= 12(
1
2
xy)=6y
2
.
Diﬀerentiating with respect totgives
dV
dt
=12y
dy
dt
=⇒
dy
dt
=
1
12y
dV
dt
.
SincedV/dt=−3,
dy
dt
=−
1
4y
and
dy
dt

t=1.5
=−
1
6
. The water level is falling at the rate of
1/6 feet per minute.
54.f(3.8)
∼
=f(4) +f
ﬀ
(4)(−0.2)=2+2(−0.2) = 1.6.

P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-04 JWDD027-Salas-v1 November 25, 2006 15:57
228 REVIEW EXERCISES
55.f(x)=
√
x+1/
√
x, f
ﬀ
(x)=
1
2
x
−1/2
−
1
2
x
−3/2
f(4.2)
∼
=f(4) +f
ﬀ
(4)(0.2) =
5
2
+
3
16
(0.2) = 2.5375.
56.f(x)=x
1/4
,f
ﬀ
(x)=
1
4
x
−3/4
f(83)
∼
=f(81) +f
ﬀ
(81)(2) = 3 +
1
54
∼
=3.01852.
57.f(x) = tanx, f
ﬀ
(x) = sec
2
x
f(43
◦
)=f(
1
4
π−
1
90
π)
∼
=f(π/4) +f
ﬀ
(π/4)(−π/90)=1−2(π/90) = 1−(π/45)
∼
=0.9302.
58.V=
4
3
πr
3
,dV=4πr
2
dr. CalculatedVatr= 10 ft. anddr=0.05 in.
∼
=0.00417 ft.
dV|
r=10,dr=0.00417
=4π(10)
2
(0.00417)
∼
=5.240 cu. ft. = 9055.025 cu. in.;
= 9055.025/231
∼
=39.20 gal.
59.f(x)=x
3
−10
(a)xn+1=xn−
x
3
n−10
3x
2
n
=
2x
3
n+10
3x
2
n
(b)x4
∼
=2.15443;f(2.15443)
∼
=−0.00007
60.f(x)=xsinx−cosx
(a)xn+1=
x
2
ncosxn+xnsinxn+ cosxn
2 sinxn+xncosxn
(b)x4
∼
=0.86057;f(0.86057)
∼
=0.00049

Chapter4 2.pdf微積分。。。。。。。。。。。。。。。。。。。。。。。。。

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Chapter4 2.pdf微積分。。。。。。。。。。。。。。。。。。。。。。。。。

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