Homework Assignment 15 - Solutions
1. Page 227: 36*, 37*, 46*, Extra points: 47, 52.
Know s U Ÿt "240 mph
Find x U Ÿt when h 4.
Relation of sŸt and xŸt :
ŸsŸt 2 6 2 ŸxŸt 2 36 ŸxŸt 2
Relation of s U Ÿt and x U Ÿt :
36.
2sŸt s U Ÿt 2xŸt x U Ÿt , x U Ÿt x U Ÿt | when h4 x U Ÿt | when h4
sŸt s U Ÿt xŸt 40 2 6 2 Ÿ"240 "242. 68 mph
40
"242. 68 mph
dy
50 mph, dx 0. Find dD | y 1 , x 1 .
2
4
dt
dt
dt
dy
dD 1 Ÿx 2 y 2 "1/2 2x dx 2y
2
dt
dt
dt
1
dD | 1 1 0 2 1 Ÿ50 20 5 44. 721 mph
2
2
2
dt y 2 , x 4
1
1
2
2
4
37. Know that D 46.
x2 y2 ,
Know dx "3 ft/s
dt
ds
|
.
Find
dt when x6
Relation of s and x : x s s , 6x 6s 18s
6
18
1
6x 12s, s x
2
dx
ds
ds
and
:
Relation of
1 dx
2 dt
dt
dt
dt
1
ds |
Ÿ"3 "1. 5 ft/s
2
dt when x12
U
A dr
47. Know that V 4 =r 3 , and A 4=r 2 . Show that V Ar U dV
dt
dt
3
d ¡V¢ d 4 =r 3 , dV 4 =3r 2 dr 4=r 2 dr A dr .
3
dt
dt
dt
dt
dt 3
dt
52. Know that V 1 =r 2 h, h d 2r, or r 1 h, dV 5 m 3 /s. Find dh | h2.
2
3
dt
dt
2
dV 1 = 3h 2 dh 1 =h 2 dh
V 1 = 1 h h = h3,
3
12
12
4
2
dt
dt
dt
1
4 Ÿ5 5 1. 59 m/s
dh | h2 4 dV =
dt
=h 2 dt
=Ÿ2 2
2. State Rolle’s Theorem and the Mean Value Theorem.
Rolle’s Theorem: Suppose that fŸx is continuous on the interval a, b and is differentiable on the
U
interval a, b and fŸa fŸb . Then there is a number c in a, b such that f Ÿc 0.
Mean Value Theorem: Suppose that f is continuous on the interval a, b and differentiable on the
interval a, b . Then there exists a number c in a, b such that
U
fŸb " fŸa f Ÿc .
b"a
3. *Find graphically all possible values of c in the interval "5, 4. 7 satisfying the conclusion of Rolle’s
Theorem for fŸx whose graph is given below.
4
Approximately,
3
x "3. 8, x "1,
x 1. 2, x 3. 5
1
-4
0
-2
2 x
-1
4
-2
y fŸx 4. *Find graphically all possible values of c in the interval "4, 4 satisfying the conclusion of the Mean
Value Theorem for fŸx whose graph is given below.
4
Approximately,
c "3. 2, c "1,
c 1. 4, c 3. 5
-4
-3
-2
-1
0
-2
y fŸx 2
1
2x
3
4
5. Page 236:
S 5, 8* (hint: study first the graph of f), 11, 12*, 13, 14*
S For each of the following problems, determine algebraically if the function is increasing, decreasing or
neither.
19, 20* 21, 22*, 23, 24*
S Extra points: 26, 30
5. fŸx 1x , "1, 1
f is not continuous at x 0. Hence, the conclusion given in the Mean-Value Theorem may not be true.
In this case, the conclusion given in the Mean-Value
10
Theorem is not true. Observe the following.
fŸ1 " fŸ"1 1 " Ÿ"1 1
1 " Ÿ"1 1 " Ÿ"1 U
f Ÿx " 12 0 p 1.
x
So, there is no c in "1, 1 .
5
0
-1 -0.8 -0.6 -0.4 -0.2
0.2 0.4 x 0.6 0.8
1
-5
-10
8. fŸx 3
x,
"1, 1
U
f is continuous on "1, 1 but f Ÿx is not defined at x 0. Hence, the conclusion given in the
Mean-Value Theorem may not be true. For this example, actually the conclusion given in the
Mean-Value Theorem is true.
fŸ1 " fŸ"1 2 1
2
1 " Ÿ"1 1
U
f Ÿx 0.5
-1
-0.8 -0.6 -0.4 -0.2 0
-0.5
-1
11. fŸx x 3 x 2 ,
3
0, 1
x 2/3 0.2 0.4 x 0.6 0.8
1
1
3
1
3
x "2/3 1, x "2/3 3
, xo
1
3
3/2
fŸ0 0, fŸ1 2,
2
1.8
U
1.6
f Ÿx 3x 2 2x 2, 3x 2 2x " 2 0
1.4
1.2
x
1
0.8
"2 o 4 " 4Ÿ3 Ÿ"2 "2 o 2 7
6
2Ÿ3 "1 o 7
3
"1 7
"1 " 7
, x
3
3
"1 7
In 0, 1 ,
c
3
x
0.6
0.4
0.2
0
0.2
0.4
12. fŸx x 3 x 2 ,
0.6
x
0.8
1
"1, 1
fŸ"1 0, fŸ1 2,
U
1
x
0.5
-1
-0.8 -0.6 -0.4 -0.2
13. fŸx sin x,
"2 o 4 " 4Ÿ3 Ÿ"1 "2 o 4 ,
6
2Ÿ3 0
0.2 0.4 x 0.6 0.8
1
0, =
2
fŸ0 0, f = 2
f = " fŸ0 2
= " Ÿ0 2
U
2,
f Ÿx cos x =
0.8
0.6
0.4
0.2
c 0. 88
0.2
14. fŸx sin x,
x "1
x 1
3
In "1, 1 , c 1
3
1
0
fŸ1 " fŸ"1 2"0 1
1 " Ÿ"1 1 " Ÿ"1 f Ÿx 3x 2 2x 1, 3x 2 2x " 1 0
1.5
4
fŸ1 " fŸ"1 2"0 2
1 " Ÿ0 1 " Ÿ0 0.4
0.6
"=, 0
0.8
x
1
1.2
1.4
1
1"0
2
= " Ÿ0 =
2
2
x cos "1 =
0. 88
1
fŸ"= 0, fŸ0 0
0.5
-3
-2
-1
0
1
2
x
3
U
f Ÿx cos x 0, x = o n=
2
=
c"
2
-0.5
-1
19. fŸx "x 3 " 3x 1
U
f Ÿx "3x 2 " 3 "3Ÿx 2 1 0 since x 2 1 u 1 0
Hence f is always decreasing.
20. fŸx x 4 2x 2 1
0, if x 0
U
f Ÿx 4x 3 4x 4xŸx 2 1 0, if x 0
Hence, fŸx is increasing on 0, . and is decreasing on "., 0 .
U
21. fŸx e x , f Ÿx e x 0, so, f is always increasing.
22. fŸx e "x , f” Ÿ "e "x 0, hence f is always decreasing.
U
f Ÿx 1x 0, if x in D f . So, f is always increasing in 0, 2
23. fŸx lnŸx , D f 0, . ,
24. fŸx ln x 2 , D f "., 0 : 0, . ,
U
f Ÿx 12 Ÿ2x 2x
x
0, if x 0
0, if x 0
,
Hence, fŸx is increasing on 0, . and is decreasing on "., 0
26. fŸx x 3 4x 3.
Since lim xv. fŸx . and lim xv" . fŸx " ., the graph of f crosses the x-axis at least once. Since
U
f Ÿx 3x 2 4 u 0 for all x
f is always increasing, the graph of f crosses the x-axis only once.
30. fŸx x 4 ax 2 " b 0 has exactly 2 solutions where a 0, b 0.
Since lim xv. fŸx ., fŸ0 "b 0 and lim xv" . fŸx ., fŸx 0 has at least one solution.
U
U
U
U
f Ÿx 4x 3 2ax, since lim xv. f Ÿx . and lim xv" . f Ÿx " ., f Ÿx 0 at least once.
UU
U
f Ÿx 12x 2 2a 2a 0, f Ÿx 0 only once and hence fŸx 0 has exactly 2 solutions.
5
6