Calculus - Chapter 3 Solutions
1.
2.
See graph at right.
a.
The slope gradually decreases from large and positive
to level at 0 a little after x = 2 . It becomes a little
negative before leveling off again around x = 4.3 and
then gradually increases thereafter.
b.
m ! 0.2
c.
Fastest at the beginning and end, slowest at the max and min. Steep slope means fast.
d.
The tangent lines help approximate the slope.
b.
d.
x
m
–3
–6
–2
–4
–1
–2
0
0
1
2
2
4
3
6
f !(x) = 2x ; a line
3.
It is 3x 2 , a parabola.
4.
f !(x) = 1 , a constant
5.
The slope function is always one degree less than the original function.
6.
nx n!1 = f "(x)
7.
g!(x) = an(x " h)n"1
8.
h!(x) = f !(x) + g!(x)
5
9.
" 13 g ( !1 + 3i )
i=0
10.
% (x + 2! )2 for
x < "2!
'
Sample answer: f (x) = & sin(x)
for " 2! # x < 0
'
for
x$0
( x
Chapter 3 Solutions
Calculus
11.
x
m
–2
24
–1
6
1
6
2
24
This is a cubic function, so the slope function will be quadratic: f !(x) = 6x 2
12.
a.
b.
lim
h!0
(2(x+h)" 3)"(2 x" 3)
h
= lim
h!0
2 x+2h" 3"2 x+ 3
h
2h
h!0 h
=2
((x+h)2 +(x+h))"(x 2 + x)
((x 2 +2 xh+h 2 )+(x+h))"(x 2 + x)
= lim
=
h
h
h! 0
h! 0
2
lim 2 xh+h +h = lim (2x + h + 1) = 2x + 1
h
h! 0
h! 0
lim
13.
–4, –2: Continuous
0, 2: Not continuous; 0: limit does not exist
2: Not continuous; f (2) and limit do not exist
14.
D : (!", 2) ! (2, ") , R : (!", !2.5] ! (!2, ")
15.
a.
9x 8
16.
a.
Stevie is traveling back towards class.
b.
This counts all area as positive: A =
c.
This is A(v, 0 ! t ! 6) =
d.
10 ft min 2
e.
Between 3 and 4 minutes.
17.
= lim
b.
13x12
20"2
2
+
c.
20+15 + 15 + 15 ! 1
2
2 2
20+15 + 15 + 0 # 15 = 45ft
2
2
20!2
2
+
2
d.
0
+ 15
= 67.5 ft
2
See graph at right.
" log(!x) for x < 0
a.
f (x) = #
for x > 0
$ log(x)
b.
D = { x : x ! 0}
Chapter 3 Solutions
Calculus
18.
a.
b.
c.
x 2 " x"2
x!2 x 2 " 3x+2
lim
lim
x!5
x" 5
x"5
(x"2)(x+1)
x!2 (x"2)(x"1)
= lim
x" 5
x!5 ( x " 5 )( x + 5 )
(2 x+1)(x"5)6
7
x!2 " (x"2) 9" x
(2 x+1)(x"5)6
=
x+1
x!2 x"1
= lim
= lim
x!5
2+1
2"1
=3
1
x+ 5
=
1
2 5
=
5
10
lim
: There is no algebraic simplification, therefore x = 2 is a V.A.
lim
!
7
x!2 " (x"2) 9" x
d.
= lim
2
lim 4 x +2 x+9
x!" #(3x#6 x 2 +2)
=
(2(1)+1)(1"5)6
(1"2)7 9"1
=
(+)(+)
!!or!!y !
(")(+)
4 x2 + 2 x + 9
2
2
2
lim x x 2 x
x!" # $ 3x # 6 x + 2 '
&% 2
)
x
x2 x2 (
=
"#
4+ 2 + 9
x 2
lim 3 x 2
x!" # +6#
x
x2
=
4
6
=
2
3
19.
All of them except (d) according to problem 3-5.
20.
a.
g!(x) = 6x 2
b.
g!(x) = 8x 7 " 2x
c.
g!(x) = "12x 2 " 2
d.
g!(x) = 24(x + 2)3
e.
g!(x) = 10(x + 7)9 " 60x 4
e.
g!(x) = 6(x " 3)2 + 8(x + 1)
21.
A tangent will intersect a curve only once while a secant will intersect twice.
22.
a.
See graph at right.
b.
See graph at right.
c.
i.
m = 160!90
= 70
4! 3
ii.
m=
96.1!90
3.1! 3
iii.
m=
90.601!90
3.01! 3
23.
=
6.1
0.1
=
= 61
0.601
0.01
= 60.1
a.
(3 + h) ! 3 = h
b.
10(3 + h)2 ! 90 = (90 + 60h + 10h 2 ) ! 90
= 60h + 10h 2
c.
60h+10h 2
h
Chapter 3 Solutions
= 60 + 10h ft/sec
Calculus
24.
25.
a.
b.
The instantaneous rate of change of the object.
a.
By finding the slope of the tangent line.
b.
26.
lim (60 + 10h) = 60 ft/sec
h!0
lim
h!0
s(12+h)" s(12)
h
a.
y = 0 is the slope function.
b.
The slope function is quadratic.
c.
(A) is the slope function.
d.
y = 2x
e.
(C) is the slope function.
27.
No, we were not told that f (x) is continuous.
28.
a.
b.
29.
f !(x) = 14x
f !(x) = 8(x " 2)3 + 18
d.
f !(x) = 0
f !(x) = 2x 5 + 8x 3
b.
–4 → Yes
–3 → No, f (!3) " lim f (x) .
x#!3
0 → No, f (0) does not exist or equal lim f (x) .
x!0
2 → No, lim f (x) does not exist or equal f (2) .
x!2
30.
D = (!", 0) ! (0, ") , R!=!(!", 0) ! (0, ")
31.
a.
3 ! 4 2 = 48
b.
6!6=0
c.
!
d.
0
32.
ft
v(3) ! 13.8 sec
Chapter 3 Solutions
Calculus
33.
34.
35.
36.
37.
a.
3 ! 2 + 3 ! 3 + 3 ! 4 + 3 ! 5 + 3 ! 6 = 60
b.
( 5 + 12 ) + ( 5 + 14 ) + ( 5 + 16 ) + ( 5 + 18 ) + … + ( 5 + 201 ) ! 51.46
a.
They both describe the difference between x and another point.
f (x+ "x)$ f (x)
"x
b.
f !(x) = lim
a.
m
sec
because the slope measures distance (in meters) per unit of
time (seconds).
b.
Each method uses a different pair of points, i.e., a different
secant, to estimate slope.
c.
See graph at right below.
d.
Yes, Hanah’s slope is the average. Hana’s method will not
always give the best approximation—it depends on the
concavity of the graph.
a.
"x#0
f (2+h)! f (2)
(2+h!2)
=
0.25(2+h)2 !0.25"4
h
=
h+(h 2 + 4)
h
b.
Hana’s
c.
We should let h approach 0: lim 1 +
d.
lim 1 +
h!0
h!0
h
4
= 1+
h
4
h
4
=1
e.
No, h can be negative since the secant can be on the other side.
a.
The slope is:
f (3+h)! f (3!h)
(3+h)!(3!h)
=
(3+h)2 !11(3+h)+2![(3!h)2 !11(3!h)+2)]
2h
=
!1!2h!h 2 +1!2h+h 2 2h
2h
=
!4h
2h
= !2
This is also the slope of the tangent.
38.
b.
f (3+h)! f (3!h)
(3+h)!(3!h)
c.
g!(x) = 2x " 11
(3+h)2 !11(3+h)+2![(3!h)2 !11(3!h)+2]
2h
f (x+h)! f (x)
h
f (x+h)! f (x!h)
Method:
2h
Hana’s Method:
Hanah’s
=
Chapter 3 Solutions
= 12h!22h
=
2h
!10h
2h
Anah’s Method:
= !5
f (x)! f (x!h)
h
Calculus
39.
40.
f (x)# f (x#h)
h
h" 0
Anah’s Method: f !(x) = lim
f (x+h)# f (x)
h
h"0
f !(x) = lim
f (x+h)# f (x#h)
2h
h" 0
, Hanah’s Method: f !(x) = lim
[4(x+h)2 # 3]#(4 x 2 # 3)
h
h"0
= lim
Therefore f !(11) = 88 and f !(1000) = 8000 .
41.
a.
f (x) = 2x 3
b.
f !(x) = lim
=
42.
h"0
2(x+h)3 #2 x 3
2(x 3 + 3x 2 h+ 3xh 2 +h 3 )#2 x 3
= f !(x) = lim
h
h
h" 0
h" 0
2
2 3
lim 6 x h+6hxh +h = lim 6x 2 + 6xh + h 2 = 6x 2
h" 0
h" 0
a.
x1/2
b.
x !3
c.
x !1/2
d.
x 5/3
9
43.
= lim 8x + 4h = 8x
Sample answer: 10 midpoint rectangles:
# 15
i=0
f (!0.9 + 5i ) " !0.273
It is negative, the function is non-positive on this interval.
44.
a.
lim
x 2 +2 x"8
x!0 x 2 "2 x
=
b.
x 2 +2 x"8
x!2 x 2 "2 x
= lim
c.
2
lim x 2+2 x#8
x!" x #2 x
=
d.
x 2 +2 x"8
x!0 x 2 "2 x
= lim
lim
lim
36+12"8
36"12
x!2
=
40
24
=
(x+ 4)(x"2)
x(x"2)
1+ 2 # 8
x x2
lim
x!" 1# 2 x
x2
5
3
= lim
x!2
x+ 4
x
=3
=1
(x+ 4)(x"2)
x!0 x(x"2)
= lim
x+ 4
x!0 x
is !" from the left and ! from the right,
so it does not exist.
45.
The slopes are linear: f !(x) = 4x .
This is twice the derivative of x 2 so f (x) is 2x 2
(as verified by the graph).
46.
See graph at right. A =
Chapter 3 Solutions
1
2
(8)(4) = 16!un 2
Calculus
47.
a.
There is not enough information.
b.
v(6) !
c.
Use the derivative o estimate velocity: v(t) = d !(t) " 0.789 # 2t + 0.703 " 1.578t + 0.703 ,
therefore v(6) = d !(6) " 1.578 # 6 + 0.703 " 10.171 m/sec and
v(10) = d !(10) " 1.578 #10 + 0.703 " 16.482 m/sec .
32.7"9.1
6" 3.1
= 8.138!m/sec
48.
Slope is extremely large and positive, decreasing to zero, then abruptly becomes very
negative and increases to zero then becomes positive.
49.
Slope should be about 2 and it goes through
50.
a.
mph
b.
It is 0.
c.
That is when the slope is large,
followed by a brief flatness
when he is pulled over.
d.
David is driving toward his
home when slope is negative.
a.
DNE (V.A.), but y ! "#
c.
2+2 x
x!" 2#2 x
x
lim 2+2 x
x!" 2#2
51.
( !4 , 1) : y ! 2x " 0.571 .
The ranges may vary; e.g., for x on the interval ( !4 " 0.1, !4 + 0.1) .
d.
52.
lim
!
2+"
2#"
b.
(x+2)2 "2
x
x!2
lim
=
(2+2)2 "2
2
= 14
=7
2
Since we do not know how to evaluate this, look at the graph.
= #1
π is a constant, therefore lim " = "
x!" +
f (x+h)# f (x)
(x+h)2 + 4(x+h)+2#(x 2 + 4 x+2)
=
lim
h
h
h"0
h"0
2
lim 2 xh+hh + 4h = lim 2x + h + 4 = 2x + 4
h"0
h"0
a. f !(x) = lim
=
b.
c.
f !(x) = 2x + 4
For h = 0.01 ,
Chapter 3 Solutions
f (x+h)! f (x)
h
= 2x + h + 4 = 2x + 4.01 " f #(x)
Calculus
53.
For n = 0 , f (x) = x 0 = 1 has a slope of 0 = 0 ! x "1 . For negative and non-integer values
d
(x !1 ) = !1x !2 and
the Power Rule also works: e.g., dx
seems to work for any real number n.
d
dx
(x1/2 ) =
1
2
x !1/2 . The Power Rule
54.
f (x) = 2x 3 + 4x ! 3x 2 ! 6 = 2x 3 ! 3x 2 + 4x ! 6 , f !(x) = 6x 2 " 6x + 4
For x = 3 , f (3) = (2 ! 3 " 3)(32 + 2) = 33 and f !(3) = 6 " 32 # 6 " 3 + 4 = 40 .
The tangent is y ! 33 = 40(x ! 3) .
55.
a.
f (x) = x 2 ! 1
b.
a=9
c.
f !(x) = 2x
d.
f !(9) = 18
e.
y ! 80 = 18(x ! 9)
a.
x2 " 4
x!2 x"2
56.
lim
= lim
x!2
(x+2)(x"2)
x"2
= lim (x + 2) = 4
x!2
Using the Power Rule, f !(x) = 2x!" f !(2) = 4 , so yes, Ana’s method works.
57.
b.
This is not possible because Ana’s method will only find the derivative at a point, not
a derivative function.
c.
She does not have an “h” in her name, or her definition.
See graph at right.
a.
No it is not of the form x n .
b.
c.
( "6 ) # 1.39
h! ( "6 ) = "6 sin ( 2 # "6 ) $ 0.453 ;
h!
(
)
So the tangent line is roughly y ! 0.453 = 1.39 x ! "6 .
58.
Graph B is distance and Graph A is velocity, because Graph A is the slope function for
Graph B. Graph A has its peak at x = 0 , when the slope of Graph B is highest.
59.
a.
dv
dt
60.
a.
d
dx
( 9x ) = dxd (9x !1 ) = !9x !2
b.
d
dx
(!3x 7 ! 6x) = !21x 6 ! 6
c.
d
dt
(5t !4 ) = !20t !5
d.
d
dm
(m) = 1
b.
Chapter 3 Solutions
dv
dr
c.
dA
dp
Calculus
61.
Answers vary but must be an odd function with horizontal
asymptotes at y = ±4 . See sample graph at right.
62.
a.
f (x) = x 2 ! 3
c.
f !(0) = 0 , f !(1) = 2
63.
64.
f !(x) = 2x
b.
a.
The runner gradually speeds up and then gradually slows down and starts running in
the negative direction, speeding up and then continuing to run in the negative
direction at a steady speed for a bit before gradually coming to a stop.
b.
At p the runner has stopped.
c.
Area A gives the distance the runner travels in the positive direction, and Area B gives
the distance the runner subsequently runs in the negative direction.
d.
A + B represents the runner’s total displacement. B is negative because the graph is
below the x-axis there, representing negative velocity.
1 " 1
(x+h)2 x 2
lim
h
h!0
=
=
1 " 1
x 2 (x+h)2 (x+h)2 x 2
x 2 "(x+h)2
lim 2
#
= lim 2
2
h
h!0 x (x+h)
h!0 x (x+h)2
(2 x+h)h
x
lim "
= lim " 2 2 x+h 2 = "2
x 2 #x 2
h!0 x 2 (x+h)2 h
h!0 x (x+h) h
"2 xh"h 2
h!0 x 2 (x+h)2 h
= lim
= "2x "3
2 x3
x(x+ 3)(x!5)
65.
Possible function: y =
66.
a.
x 2 +2 x" 3
2
x!1+ x "2 x+1
b.
2 x+5
lim x +6
x!" 3x 2 + 4
c.
lim x " 3
x!9 x"9
d.
x 2 + x+1
x+ 7
x!2
a.
sin x cos y + cos x sin y
b.
cos x cos y ! sin x sin y
c.
sin x cos y ! cos x sin y
d.
cos x cos y + sin x sin y
67.
lim
lim
Chapter 3 Solutions
(x+ 3)(x"1)
(x"1)(x"1)
x!1+
= lim
=
x2 + 6 x + 5
2
2
2
lim x 2x x
x!" 3x + 4
x2 x2
x "3
x!9 ( x + 3)( x " 3)
= lim
=
2 2 +2+1
2+ 7
=
= lim
x+ 3
x"1
+
x!1
=
! vertical asymptote: y ! "
1+ 6 + 5
x x2
lim
x!" 3+ 4
x2
= lim
x!9
1
x +3
=
=
1
3
1
6
7
3
Calculus
68.
The Power Rule, the definition of derivative as a limit, examining the slopes of tangent
lines, graphing secants such as
tangents.
69.
Recall: lim
x!0
sin x
x
= 1 and lim
x!0
sin(x+h)"sin x
h
h!0
sin(x) :! lim
= lim (sin x)
h!0
f (x+h)! f (x)
h
1"cos x
x
for small h, or using a table of slopes of
=0
= lim
sin x cos h+cos x sin h"sin x
h
h!0
sin x(cos h"1)+cos x sin h
h
h!0
= lim
lim (cos x) ( sinh h ) = 0 + cos x = cos x
( coshh"1 ) + h!0
cos(x+h)"cos x
cos x(cos h"1)"sin x sin h
= lim cos x cos h"sinh x sin h"cos x = lim
h
h
h!0
h!0
h!0
(cos h"1)
= lim (cos x) h
" lim (sin x) sinh h = 0 " sin x = " sin x
h!0
h!0
cos(x) :! lim
f (20) (x) = f (x) = sin x and f (101) (x) = cos x
70.
f.
71.
y = x !1
72.
1 1
#
x+h
lim h x
h"0
a.
y! =
b.
y! = "1x "2 = "
=
x#(x+h)
x(x+h)
lim
h
h"0
x#(x+h)
h"0 x(x+h)h
= lim
#h
h"0 x(x+h)h
= lim
#1
h"0 x(x+h)
= lim
= #x #2
1
x2
We must take the derivative and solve for f !(x) = 5 :
f (!1) = 7,!! f (7) = !209
f !(x) = 3x 2 " 18x " 16 (Power Rule)
!( !1, 7 ) and (7, !209)
5 = 3x 2 " 18x " 16
0 = 3x 2 " 18x " 21
0 = x 2 " 6x " 7
0 = (x " 7)(x + 1)
x = "1, 7
73.
a.
f !(x) = "3x 2
b.
f !(x) = "2x "3
c.
f !(x) = 0
d.
f !(x) = 3 cos(x + " )
Chapter 3 Solutions
Calculus
74.
Approximation using 16 midpoint rectangles:
15
#
i=0
1
2
(
f !3.75 +
i
2
15
) = # 12
(
16 !3.75 +
i=0
)
i 2
2
Geometrically, it is a semi circle with area
75.
76.
= 8! # 25.13!un 2
0
b.
c.
!9t !10
d.
a.
f !(x) = lim
b.
c.
78.
! "4 2
2
a.
=
77.
" 25.25
6x14 !
3
2
3 m !1/4
4
x
2
2
3
2(x+h)3 #(x+h)#(2 x 3 # x)
= lim 6 x h+6 xhh +2h #h
h
h"0
h"0
lim 6x 2 + 6xh + 2h 2 # 1 = 6x 2 # 1
h"0
f !(x) = 6x 2 " 1
It should basically match.
f (x)# f (a)
x#a
x"a
a.
f !(a) = lim
b.
f (x) = x 3 + x and a = 3
c.
f (x+h)! f (x)
h
=
0.01
Yes, the graph should look the same as the answer to part (a).
a.
Approximately 0.354, 0.204, 0.151, and 0.0500 m/sec , respectively.
b.
It approaches 0.
c.
Position approaches ! . It does make sense, although it is a bit paradoxical.
37.8m
30 sec
2(x+0.01)3 !(x+0.01)!(2 x 3 ! x)
m
= 0.756 sec
a.
80.
a.
The ball is traveling straight up, so the derivative or slope of its height function will
give us its velocity.
b.
At Point A since it has the greatest slope (Point C has negative velocity).
c.
At Point B, with velocity 0.
d.
v(t) = s!(t) = "32t + 76.8
e.
m v(3) = !19.2 m v(4) = !51.2 m
v(2) = 12.8 sec
,
,
sec
sec
f.
Negative velocity means the ball is falling, i.e., traveling in the downward direction.
Chapter 3 Solutions
b.
m
0.756 sec
79.
Calculus
81.
3x"21
2 " x" 42
+
x
x!7
b.
3x 2
lim 2"
2
x!"# 2 x +5 x" 7
c.
d.
82.
lim
3(x" 7)
(x" 7)(x+6)
+
x!7
= lim
a.
=
2 "3
2
x
lim
5
x!"# 2+ " 7
x x2
= lim
3
x+6
+
x!7
="
3
= 13
3
2
(x+2)"2
x+2 " 2 ( x+2 + 2 )
= lim
x
(
x+2
+
2
)
x(
x+2 + 2 )
x!0
x!0
1+ 3
2
x 2
0 =0
lim x 2+ 3x3 = lim 1 x1 = #1
# #1
x!" 1# x # x
x!"
x3 x3
lim
= lim
1
(
x+2
+ 2)
x!0
=
1
2 2
a.
There are many possible shapes, e.g. exponential or logarithmic.
b.
See graph at right.
A (negative) cubic is a good example.
c.
Sinusoidal functions, e.g. sin x and cos x .
83.
A is the slope of B. For example, when B is level, A is 0.
84.
a.
At about –3, 0, and 2, at which f !(x) = 0 .
b.
[!3, 0] and [2, !) : f !(x) must be positive within these intervals.
c.
See graph at right.
85.
a.
d.
b.
=
2
4
c.
The slopes are not the same from both sides, so the derivative does not exist.
Chapter 3 Solutions
Calculus
86.
a.
Possible answers:
Negative,
decreasing
slope.
Positive,
decreasing
slope.
Negative,
increasing
slope.
Answers vary, one possible curve is shown at right.
c.
Yes, as long as the beginning and end points of the arcs
are not forced to have vertical and horizontal tangents.
87.
positive: (!", !4) # (1, ")
negative: (!4, 1)
88.
a.
–1
b.
0
c.
1
2
d.
9 + cos x
89.
a.
E
b.
C, D
90.
Velocity (m/sec)
b.
(x ! 2)2
Positive,
increasing
slope.
zero: x = !4, 1
c.
C
Time (sec)
91.
See graph at right.
3
a.
" 12 ( f (i) + f (i + 1)) = 12 (0 + 2 ! 6 + 2 ! 8 + 2 ! 6 + 0) = 20 !un2
i=0
3
b.
! f (i) = 20 !un2 , the same answer.
i=0
c.
No, sometimes for less symmetric functions the approximations are quite different.
92.
6x 2 ! 5x + 3 : The limit is the derivative
93.
x 3 + x 2 ! 5x + C will be an answer for any real number C: students should have two
different constants (but not “C”).
Chapter 3 Solutions
d
dx
(6x 2 ! 5x + 3) = 12x ! 5 (using the Power Rule).
Calculus
94.
If f !(x) is even, then f !(x) = f !("x) for all x, so f (x) will have rotational symmetry about
(0, f (0)) = (0, 0) , so f (x) is odd, f !(x) = " f ("x) for all x and f (x) has reflective
symmetry about the y-axis, and since f (0) = 0 , f (x) must be even.
95.
a.
c.
96.
97.
x 2 + 3x"10
x"2
x!0
2 3x"10
lim x +x"2
x!2
lim
=
"10
"2
= lim
x!2
=5
b.
(x"2)(x+5)
x"2
1+ 3 # 10
x 2
lim 1 2x
#
x!"
x x2
x 2 + 3x"10
x"2
x!"5
lim
=
25"15"10
"7
=0
= lim (x + 5) = 7
x!2
d.
x 2 + 3x#10
x#2
x!"
e.
The graph is linear with a hole at (2, 7) there are no horizontal or vertical asymptotes.
a.
f increases, slopes decrease
b.
f decreases, slopes increase
c.
f increases, slopes decrease
d.
f decreases then increases, slopes increase
e.
f decreases, slopes decrease
f.
f increases, slopes increase then decrease
lim
=
$
1
0
! DNE, but y ! "
a.
b.
c.
d.
98.
(B) and (D) are concave up; (A), (C), and (E) are concave down; (F) is mixed.
99.
a.
At x = !3, 1,!and!4 f (x) has a flat, horizontal tangent. At –3 and 4 and it changes
from negative to positive slope, and at 1 it changes from positive to negative slope.
b.
This is when f !(x) is positive (and endpoints may be included): in [!3, 1] " [4, #) .
c.
This happens when f !(x) changes from negative to positive, which is at x = !3, 4 .
d.
! ("#, "1.3] ! [2.8, #) ; f is concave up.
100. f (x) is increasing means the y-values get higher, therefore the slopes are positive and
f !(x) is positive. f !(x) is increasing means the slopes of f (x) are increasing. The graph
will be concave up.
Chapter 3 Solutions
Calculus
101. Sample graph:
102. a. f !(x) = 2(x " 5)4 + 2x
f !(x) = 0
c.
103. a.
x "2/3
b.
f !(x) = (x1/3 )! =
d.
f !(x) = "2(x + 1)"3
1
3
It starts small and positive and gradually increases over the first 5 hours. Over the
next 5 hours it gradually decreases almost to zero (staying positive).
= 10 mph
b.
100 miles
10 hours
c.
! 36!mph
g(x) = 2x ! 5, f (x) = x 3
104. a.
b.
g(x) = 3x ! 1, f (x) = sin x
c.
g(x) = tan x, f (x) = 5 x
105. Students should give two functions of the form ! sin x + c . There are infinite solutions.
106.
a. 12
b.
3 and 6
107. He differentiated f (2) instead of differentiating f (x) and substituting x = 2 afterward.
108. a
b.
f(x) maintains
a slope of 0.
Chapter 3 Solutions
c.
f(x) maintains
a slope of –1.
The slope starts small,
positive and increases.
Calculus
109. a.
b.
110. a.
lim 1" x
x!0 x"1
= lim
= lim
b.
1" x
x!1 x"1
= lim
= lim
c.
lim
lim
h!0
1" x
x!0 "(1" x )(1+ x )
1" x
"(1"
x )(1+ x )
x!1
25+h "5
h
d.
(h+2)"2
2
h!"2
e.
(b) and (c)
111. d.
e.
112. a.
lim
(
25+h +5
25+h +5
=
("2+2)"2
2
1
x!0 "(1+ x )
) = lim
1
"(1+
x)
x!1
25+h"25
h!0 h( 25+h +5)
= "1
="
= lim
1
2
h!0
1
25+h +5
1
= 10
= "1
Zeros on f ! represent the maxima and minima of f .
Zeros on f !! represent points of inflection on f and maxima and minima of f ! .
A and B. This is where f !(x) = 0 and therefore the slope of f (x) = 0 .
b.
At x = A, the slope changes from positive to negative. This is a maximum.
At x = B, the slope changes from negative to positive. This is a minimum.
c.
Kat is correct. The slope of f (x) goes from positive to zero, then positive again.
113. f !(x) = 3x 2 + 3x " 6,! f !!(x) = 6x + 3,! f !!(0) = 3 .
Since f !!(0) > 0 , f (x) is concave up when x = 0 .
y
114. Sample graph:
115.
Chapter 3 Solutions
x
Calculus
23
116. a.
(
i
8
1
8
f !2 +
A( f , !2 " x " 1) # $
1
2
f !2 +
i=0
23
b.
(
i
8
A(f , !2 " x " 1) # $
i=0
)
23
= $ 18 i2(!2+i /8) # 2.417
i=0
23
) = $ 18
i
8
i=0
# 3.347
y
117. See sample graph at right.
There should be an inflection point at x = !3 .
x
118. e
They will both compute the derivative of f (x) = (x + 1)(x + 2) . Part (i) uses Hana’s
method to find f !(x) while part (ii) uses Hanah’s method.
119. a.
b.
120.
f !(x) = 2x + 3
(5+h)(6+h)" 30
h
h!0
(x+1)(x+2)" 30
lim
x" 4
x!4
a.
lim
b.
30+11h+h 2 " 30 = lim (11 + h) = 11
h
h!0
h!0
2
30 = lim (x+ 7)(x" 4) = lim (x +
lim x + 3x+2"
x" 4
x" 4
x!4
x!4
x!4
= lim
=
7) = 11
Using trapezoids, D ! 49 + 47 + 44.5 + 41.5 + 38.5 + 35.5 + 31.5 + 26 + 17.5 + 7
121.
! 338 miles
" seconds " 3600hours
! 0.0939 miles
hour
seconds
122. f !(x) = x 2 " 5x " 4 = 2
!!!!!!!!!!!! x 2 " 5x " 6 = 0
!!!!!!!!!!(x " 6)(x + 1) = 0
!!!!!!!!!!!!!!!x = 6, "1
123. a.
b.
c.
d.
3 #3
lim x 3# x #3
x!" 5 x + x
=
x 3 # x #3
3
3
lim x 3 x #3
x!" 5 x + x
x3 x3
=
1# 1
6
lim x1
x!" 5+
x6
=
1
5
2" x 2+ x
4" x
i
= lim
= "1 = " 14
x"
4
2+
x
(x"
4)(2+
x
)
2+ x
x!4
x!4
2
2 x(x"6)
2 x = 12
lim 22x "12 x = lim (x"6)(x+ 7) = lim (x+
7)
13
x!6 x + x" 42
x!6
x!6
x3
3
x3
lim x 2 = lim
= lim 11 = lim x =
x!"# 4+ x
x!"# 4 + x 2
x!"#
x!"#
x
x3 x3
lim
Chapter 3 Solutions
"# or DNE
Calculus
126. a.
b.
c.
f !!(x) is the rate of change of f !(x) .
Concavity gives the sign (+ or –) of the slope of the slope (which is f !!(x) ).
Symbols
Description
Increasing,
concave up
Decreasing,
concave up
Decreasing,
concave down
Increasing,
concave down
f !(x)
Pos: f (x) inc
f !!(x)
Pos: slope inc
Neg: f (x) dec
Pos: slop inc
Neg: f (x) dec
Neg: slope dec
Pos: f (x) inc
Neg: slope dec
d.
When f (x) increases, f !(x) is positive, and when f (x) decreases, f !(x) is negative.
e.
When f (x) is concave up, f ''(x) is positive, and when f (x) is concave down,
f !!(x) is negative.
f.
No.
f !(x) = 3x 2 " 12
127.
!3x 2 " 12 > 0
3x 2 > 12
So, f (x) increases in (!", !2] # [2, ") , and decreases in [!2, 2] .
f !!(x) = 6x , so f (x) is concave up in [0, !) and concave down
in (!", 0] . (The interval ends may be open or closed.)
x2 > 4
x > 2,!x < "2
128. a.
b.
(0, 0)
At the inflection point, y!! is zero.
129. This is the same graph as problems 3-127 and 3-128, shifted up 4 units.
Therefore the maximum occurs at x = –2, y = (!2)3 ! 12(!2) + 4 = 20 , or (–2, 20).
The minimum occurs at x = 2, y = (2)3 ! 12(2) + 4 = !12 , or (2, –12).
The point of inflection is at x = 0, y = 0 3 ! 12(0) + 4 = 4 , or (0, 4).
See the solution to problem 3-127 for inc., dec., and concavity.
130. f !(x) > 0 " f (x) is increasing.
f !(x) < 0 " f (x) is decreasing.
f !(x) = 0 " f (x) has a max or min.
f !!(x) > 0 " f (x) is concave up.
f !!(x) < 0 " f (x) is concave down.
f !!(x) = 0 " f (x) has an inflection point.
Chapter 3 Solutions
Calculus
0, f !(4),
131.
f (3)" f (1)
,
3"1
f !(1)
132. g(z) = z 5 + 5z 4 ! z,! g"(z) = 5z 4 + 20z 3 ! 1,! g"(1) = 5 + 20 ! 1 = 24
g(1) = 5 → tangent line: y ! 5 = 24(z ! 1)
133. Functions should be of the form !4x 3/2 + cos x + C
134. a.
b.
c.
d.
d2
dx 2
(8x 99 ) =
d
dx
(8 ! 99x 98 ) = 8 ! 99 ! 98x 97 = 77616x 97
( !3 x ) = dxd ( !3 " 12 x ! 12 ) = !3 " 12 " ( ! 12 ) x !3/2 = 43 x !3/2
d 2 2 x ! 6x !2 = d 2 ! 6(!2)x !3 = !6(!2)(!3)x !4 = !36x !4
(
) dx ( 3
)
dx 2 3
d2
dx 2
d2
dx 2
(7 ! 2 cos x) =
d
dx
(2 sin x) = 2 cos x
135. f !(x) must be greater than or equal to zero.
f !(x) = 6x " 3 # 0, or on "# 12 , !
!!!!!!!!!!!!!!!!!!6x # 3
!!!!!!!!!!!!!!!!!!!!x #
1
2
19
136.
A( f , 2 ! x ! 2) " (
A( f , 2 ! x ! 3) =
error =
)
20.5!20.275
20.5
i=0
16+25
2
1
20
(
f 2+
i
20
19
) = ( 201 $% 9 ( 2 + 20i ) # 2 &' = 20.275
i=0
= 20.5
" 1.10%
137. Two “tangents” may be drawn, but neither is valid.
Chapter 3 Solutions
Calculus
138.
a.
b.
The slope is negative
and constant.
The slope starts positive
and decreases to about –1.
Then it increases to about 1.
c.
The slope starts large and positive, decreases to zero and
then becomes large and negative.
139. a.
b.
m
Speed is always positive. At t = 4 , speed is 32 sec
The drop is A(v, 0 ! t ! 4) =
32.4
2
= 64 meters , so it ends at a height of 36 meters.
140. a. lim (2x 2 " 6x + 5)2 = (2("2)2 " 6("2) + 5)2 = 25 2 = 625
x!"2
e x + e# x
x
x
lim e x e# x
x!" e # e
ex ex
1+ 1
2x
lim e 1
x!" 1#
e2 x
b.
x #x
lim e x +e# x
x!" e #e
c.
(x"1)(x"2)
x+1
x!2 "
=
(0"1)(2"2)
(2+1)
d.
x+2
(x"1)(x"2)
"
x!2
#
(1.99+2)
(1.99"1)(1.99"2)
=
lim
lim
=
=1
=0
=
+
(+)(")
! "$ or DNE
141. To get the real slope of the tangent we must take the limiting slope of secant lines, which
gave us our definition of a derivative.
142. The slope of the tangent (i.e., f(2)) does not exist because x = 2 is vertical.
143. a.
b.
It looks like it is approaching ! or is undefined.
f !(x) =
Chapter 3 Solutions
1
3
x "2/3 !# f !(0) is undefined, confirming the graph.
Calculus
144. Koy is correct; the derivative is defined by a limit which does not exist at c.
145. a.
b.
c.
146. a.
b.
147. a.
i.
iii.
Vertical tangent
There is no tangent.
ii.
iv.
i.
ii.
iii.
iv.
f (c) does not exist.
The limit which gives the slope does not exist.
f(x) needs to be continuous and limits from both sides must lead to the same slope.
2x 4 ! 5x 2 ! 5x + c for any c
Any constant may be added to get another antiderivative, because adding a constant
has no effect on the derivative.
F(x) = !x 6 + 4x 3 + C
b.
F(x) = 3 sin x + C
148. Sample answer graphs:
149. a.
b.
f !(x) = 2x + 5 → decreasing: (!", !2.5] , increasing: [!2.5, ") ;
f !!(x) = 2 → concave up: (!", ")
f !(x) = x 2 + 6x " 1 → decreasing: [!6.16, 0.162] , increasing: (!", !6.16] # [0.162, ")
f !!(x) = 2x + 6 → inflection point: (!3, 26) , concave up: (!3, ") ,
concave down: (!", !3)
Chapter 3 Solutions
Calculus
150.
(
3 (x+h)2 "11(x+h)+ 34" 3 x 2 "11x+ 34
4
4
lim
h
h!0
) = lim 23 xh+ 43 h2 "11h = lim 3 x + 3 h " 11 = 3 x " 11
h!0
h
h!0 2
4
2
This agrees with the Power Rule.
151. a.
b.
D: {x : x ! "1} , R:!{y : y is real}
D:!{x : x ! "2} , R:!{y : y ! "5}
152. z! = 6x + 5,!! z!("2) = "7
z("2) = 12 " 10 + 1 = 3
tangent : y " 3 = "7(x + 2)
y
153. See sample graph at right.
a.
1
b.
It is between 0 and 2.
c.
This limit must be !" .
d.
f !(1) may be 1 because f !(x) is monotonically decreasing ( f (x) is concave down)
and f !(2) = 12 . So f !(1) cannot be 14 .
x
154. See graph at right.
f !(x) = 3 " 13 (x # 2)#2/3 = (x # 2)#2/3 !!$!! f !(3) = 1
f (3) = 3 ! 3 3 " 2 = 3
tangent: y ! 3 = x ! 3!!or!!y = x
At x = 1 , the tangent line has the same slope.
155. a.
b.
The tangent should be approximately vertical.
f !(2) = (2 " 2)"2/3 =
1
0 2/3
which is undefined
156. Typical response: If f(c) does not exist, there is no point of tangency for the tangent. If
both sides of the limit of f(x) as x → c do not agree, then there can not be a tangent at f(c).
Chapter 3 Solutions
Calculus
157. a.
b.
c.
d.
#% 2 + (x ! 0.1)2 , x " 0
See graph at right. f (x) = $
2
&% 2 + (x + 0.1) , x < 0
Not when you zoom in close enough. It is actually a cusp.
# 2x " 0.2, x > 0
f !(x) = $
% 2x + 0.2, x < 0
The left hand limit is 2(0) + 0.2 = 0.2 and the right hand limit is 2(0) ! 0.2 = !0.2
(from part (c)), so they do not agree.
4(x+h)2 # 3(x+h)#(4 x 2 # 3)
4(x 2 +2 xh+h 2 )# 3#(4 x 2 # 3)
= lim
h
h
h"0
h"0
2
lim 8 xh+h4h = lim (8x + 4h) = 8x
h"0
h"0
158. f !(x) = lim
=
f !(11) = 8(11) = 88,! f !(1000) = 8(1000) = 8000
159. a.
b.
See graph at right.
dy
dx
is undefined at the vertex, jumping from –1 to 1.
c.
It may give an incorrect slope of 0 there.
d.
For h = 0.1 , f !(0) =
x+0.1 " x"0.1
2(0.1)
=
For h = !0.1 , f !(0) =
x"0.1 " x+0.1
2(0.1)
For h = 0.01 , f !(0) =
x+0.01 " x"0.01
2(0.01)
0.2
0.2
=
= 1.
"0.2
0.2
=
= "1 .
0.02
0.02
= 1.
Due to symmetry, the difference quotient will always result in zero for this function
when x = 0. Therefore, since the limit as is 0, the calculator will falsely state that
f !(0) = 0 .
160. a.
There are 16: (1)
(6)
(12)
(7)
(13)
(2)
(8)
(3)
(4)
(5)
(9)
(10)
(11)
(14)
(15)
b.
There are 10: 1, 3, 4, 5, 7, 9, 12, 13, 15, and 16 above.
c.
There are 8: 2, 4, 6, 7, 10, 12, 14, and 16 above.
(16)
Solution continues on next page. →
Chapter 3 Solutions
Calculus
160. Solution continued from previous page.
d.
Cusps
Points of
Inflection
Neither
161. y! = 3x 2 " 4, y!(0) = "4, y(0) = 0 , tangent line: y = !4x
y(2) = 0, y!(2) = 8, tangent line: y = 8(x ! 2)
intersection: 12x ! 16 = 0, x =
4
3
, y = ! 163 ;
( 43 , ! 163 )
162. a.
See graph at right above.
The slope starts large and negative and gradually
increases thereafter, becoming 0 at x = 2 .
b.
See graph at right.
The slope starts large and negative and gradually
increases thereafter, passing 0 at x = 2 .
c.
See graph at right below.
The slope starts large and negative and gradually
increases to 0 at about x = 2 , after which it
gradually decreases and becomes large and
negative.
9
163.
!
i=0
7
10
(
f 1+
7
10
" 4 %
7
i = ! 10
$ 1+ 7 i ' ( 9.70
# 10 &
i=0
)
9
164. See sample graph at right.
Chapter 3 Solutions
Calculus
( )
=
d
dx
( x !2/3 ) = ! 23 x !5/3
d
dx
(x x ) =
d
dx
( x 3/2 ) = 23 x1/2
d
dx
(sin 2 x + cos2 x) =
165. a.
d
dx
b.
c.
3 1
x2
d
dx
(1) = 0
166. a.
F(x) = 2x 3/2 !
167. a.
f (x) = cos x,!g(x) = 3x ! 11
c.
168. a.
x2 + C
lim
x!0
3" x " 3 3" x + 3
i
x
3" x + 3
=
x+ 3
x"
4
x!2
c.
x 2 x"2
x"2
x!2 +
lim
2+ 3
2" 4
lim
="
b.
f (x) = 3 x ,!g(x) = 2 x ! 1
= lim
3" x" 3
x!0 x( 3" x + 3)
= lim
x!0
"1
3" x + 3
=
"1
2 3
= lim x 2 = 4
x!2 +
lim e# x + 1 = e#" + 1 = 0 + 1 = 1
x!"
2(x+h)+9#(2 x+9)
h
h"0
c.
F(x) = sin x ! 2 cos x + C
5
2
169. f !(x) = lim
170. a.
b.
f (x) = 3x ,!g(x) = 5 ! 2x
b.
d.
7
2
= lim
2h
h"0 h
=2
Not differentiable: hole at f (0) .
b.
Not differentiable: vertical tangent
Not differentiable
171. See graph at right.
a.
Yes: the limits from both sides agree with f (2) = 0 .
b.
No: on one side it is steep and on the other side it is flat.
172. See sample graph at right.
Chapter 3 Solutions
Calculus
d !(t) = 12t + 2,! d !(1) = 14,! d !(3) = 38,! d !(10) = 122 , using the Power Rule and
derivative of a sum (and because velocity is the derivative of distance).
173. a.
80
32
2
80!
= 100ft is the area of the
See graph at right. max height=
triangle under the velocity curve. The antiderivative of velocity
gives the height. The antiderivative of v(t) = !32t + 80 is
b.
d(t) = !16t 2 + 80t .
c.
The derivative of distance is velocity, and the area under the
velocity curves gives the distance traveled.
174. f (x) = x(x ! 1) = x 2 ! x
f "(x) = 2x ! 1
2(x+h)2 # 3(x+h)+ 4#(2 x 2 # 3x+ 4)
h
h"0
175. f !(x) = lim
4 xh+2h 2 # 3h
h
h"0
= lim
= lim 4x + 2h # 3 = 4x # 3
h"0
(agrees with power rule), f !(3) = 12 " 3 = 9,! f !("2) = "8 " 3 = "11
5x 4
176. a.
n!1
177.
b.
n!1
& ( 2n i ) = & 2n "# ! ( 2in )
i=0
2
n
g
i=0
19
a.
' 101 "# ! ( 20i2 )
2
i=0
b.
178. a.
2
x !1/2
+ 4$
%
+ 4 $ & 5.53
%
Since g(x) is even, A(g, !2 " x " 0) is also ! 5.53 and A(g, !2 " x " 2) is twice as
much: ! 11.06 .
f (x) =
b.
sin x
c.
0
Chapter 3 Solutions
2 x 2 ! 3x+1
x+2
=
(x+2)(2 x! 7)+15
x+2
= 2x ! 7 +
15
x+2
, so it is 2x ! 7
Calculus
179. (A) and (C) have inverses because they satisfy the horizontal line test (only one input for
any output).
a.
f !1 (x) =
c.
f !1 (x) = 3 x ! 2
x!8
!10
b.
g !1 (x) = x ! 4
d.
h !1 (x) = sin !1
x
3
180. f !(x) = 3x 2 " 6x " 24 = 3(x " 4)(x + 2) is positive in (!", !2) ! (4") so f (x) is
increasing in (!", !2] and [4, ") .
181. The shape is a sphere with a cylinder cut out of it.
Radius of the sphere = 6 2 + 4 2 = 52 = 2 13
V= 43 ! (2 13)3 " 8! (6)2 # 665.9
182. a.
From the four intercepts we know f (x) = a(x + 3)x(x ! 1)(x ! 3) for some a.
Patrol one is at (!2.1, 2.32) , so 2.32 = a(0.9)(!2.1)(!3.1)(!5.1)
a=
2.32
0.9(!2.1)(!3.1)(!5.1)
" !0.776
f (x) ! "0.0776(x + 3)x(x " 1)(x " 3)
! "0.0776(x 4 " x 3 " 9x 2 + 9x)
b.
f !(x) " #0.0776(4x 3 # 3x 2 # 18x + 9)
f !(#2.1) " #0.0776(4(#2.1)3 # 3(2.1)2 # 18(#2.1) + 9) " 0.270
First tangent: y ! 0.232 = 0.270(x + 2.1)
f (2.5) ! 0.801, f '(2.5) ! "0.602
Second tangent: y ! 0.801 = !0.602(x ! 2.5)
c.
Subtracting the two equations: !1.519 = 0.872x ! 0.938
x=
!1.519+0.938
0.872
" !0.667
y ! 0.232 = 0.270(!0.667 + 2.1), y " 2.71
It is about 0.667 miles west and 2.71 miles north of the lighthouse, at a distance of
0.667 2 + 2.712 ! 2.79 miles.
183.
dy
dx
= 6x 2 ! 10x,!
d2 y
dx 2
= 12x ! 10,! y""(0) = !10,! y""(2) = 14 ,
so it is concave down at x = 0 and concave up at x = 2 .
184. a.
F(x) = !2x !2 +
Chapter 3 Solutions
x2
4
! 3x + C
b.
!6x !1 + C
Calculus
185. h(x) = 3 !
n"1
186.
#
i=o
8
n
1
3! x
(
, j(x) =
1
3!(3! x)
= 1x ; they are not equal
n"1
n"1
i=0
i=0
) # 8n ! 4 3 5 + 8n i " 5 = # 32n 3 85 i : 42.92, 46.08, 47.26
f 5 + 8n i =
The last is more accurate because it uses more rectangles.
187. a.
b.
188. a.
lim
h!0
3(x+h)"2"(3(x"h)"2)
2h
= lim
h!0
3h+ 3h
2h
=3
See graph at right. The graph looks like y = cos x .
See graph at right below.
b.
Left limit as x ! "# is –1, right limit as x ! +" is 1. Therefore, y! does not exist at
x = 0.
c.
Most calculators will incorrectly provide a slope of 0 at
the vertex.
d.
For h = 0.1 , f !(0) =
(0+0.1)2/3 "(0"0.1)2/3
2(0.1)
=
For h = !0.1 , f !(0) =
(0"0.1)2/3 "(0+0.1)2/3
2(0.1)
For h = 0.01 , f !(0) =
(0"0.01)2/3 "(0+0.01)2/3
2(0.01)
0
0.2
=
=0
0
0.2
=
=0
0
0.02
y = x 2/3
dy
dx
=0
Due to symmetry, the difference quotient will always result in zero for this function
when x = 0. Therefore, since the limit as is 0, the calculator will falsely state that
f !(0) = 0 .
189. g(0) is undefined, so there is no tangent line.
190. y! = 3x 2 + 3x " 6 = 3(x 2 + x " 2) = 3(x + 2)(x " 1) is negative in (!2, 1) so y is decreasing
in [!2, 1] . Where the derivative is negative, the function decreases.
Chapter 3 Solutions
Calculus