Preview Extract

Chapter 2
Applications of Differentiation
2.
Exercise Set 2.1
1.
f x x2 4 x 5
f ‘ x 2 x 4
f ‘ x exists for all real numbers. Solve
f x x 6 x 3
f ‘ x 0
2
First, find the critical points.
f ‘ x 2 x 6
2x 4 0
2 x 4
x 2
The only critical value is 2 . We use 2 to
divide the real number line into two intervals,
A: , 2 and B: 2, .
f ‘ x exists for all real numbers. We solve
f ‘ x 0
2x 6 0
2 x 6
x 3
The only critical value is 3 . We use 3 to
divide the real number line into two intervals,
A: , 3 and B: 3, .
A: Test 3, f ‘ 3 2 3 4 2 0
We see that f x is decreasing on , 2
and increasing on 2, , there is a relative
B
A
f ‘ 0 2 0 4 4 0
B: Test 0,
minimum at x 2 .
f 2 2 4 2 5 1
2
3
Thus, there is a relative minimum at 2,1 . We
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 4, f ‘ 4 2 4 6 2 0
B: Test 0,
sketch the graph.
x
f x
5
4
3
2
1
0
1
f ‘ 0 2 0 6 6 0
We see that f x is decreasing on , 3 and
increasing on 3, , and the change from
decreasing to increasing indicates that a relative
minimum occurs at x 3 . We substitute into
the original equation to find f 3 :
f 3 3 6 3 3 12
2
Thus, there is a relative minimum at 3, 12 .
We use the information obtained to sketch the
graph. Other function values are listed below.
x
f x
6
5
4
3
2
1
0
3
8
11
12
11
8
3
3.
10
5
2
1
2
5
10
f x 2 3x 2 x 2
First, find the critical points.
f ‘ x 3 4 x
f ‘ x exists for all real numbers. We solve
f ‘ x 0
3 4 x 0
3
4
The solution is continued on the next page.
x
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218
Chapter 2: Applications of Differentiation
3
3
. We use to
4
4
divide the real number line into two intervals,
3
3
A: , and B: , .
4
4
The only critical value is
4.
f x 5 x x2
f ‘ x 1 2 x
f ‘ x exists for all real numbers. Solve
f ‘ x 0
1 2 x 0
A
B
2 x 1
x
3
4
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 1, f ‘ 1 3 4 1 1 0
1
1
. We use to
2
2
divide the real number line into two intervals,
1
1
A: , and B: , :
2
2
The only critical value is
f ‘ 0 3 4 0 3 0
B: Test 0,
A: Test 1, f ‘ 1 1 2 1 1 0
3
We see that f x is increasing on ,
4
1
We see that f x is increasing on ,
2
1
and decreasing on , , which indicates
2
there is a relative maximum at x
2
2
1 21
Thus, there is a relative maximum at , .
2 4
We sketch the graph.
x
f x
3 25
Thus, there is a relative maximum at , .
4 8
We use the information obtained to sketch the
graph. Other function values are listed below.
3
2
1
7
0
3
34
25
8
0
1
2
2
3
12
1
.
2
21
1
1 1
f 5
2
2 2
4
25
3
3
3
f 2 3 2
4
4
4
8
f x
f ‘ 0 1 2 0 1 0
B: Test 0,
3
and decreasing on , , and the change
4
from increasing to decreasing indicates that a
3
relative maximum occurs at x . We
4
substitute into the original equation to find
3
f :
4
x
1
2
5.
3
2
1
12
1
3
5
0
1
2
5
3
1
21
4
F x 0.5 x 2 2 x 11
First, find the critical points.
F ‘ x x 2
F ‘ x exists for all real numbers. We solve:
F ‘ x 0
x2 0
x 2
The solution is continued on the next page
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
219
We see that g ‘ x is decreasing on , 1
The only critical value is 2 . We use 2 to
divide the real number line into two intervals,
A: , 2 and B: 2, .
and increasing on 1, , which indicates there
is a relative minimum at x 1 .
B
A
g 1 1 6 1 3 1 2
2
Thus, there is a relative minimum at 1, 2 .
2
We sketch the graph.
x
g x
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 3, F ‘ 3 3 2 1 0
4
3
2
1
0
1
2
F ‘ 0 0 2 2 0
B: Test 0,
We see that F x is decreasing on , 2 and
increasing on 2, , and the change from
decreasing to increasing indicates that a relative
minimum occurs at x 2 . We substitute into
the original equation to find F 2 :
F 2 0.5 2 2 2 11 13
2
Thus, there is a relative minimum at 2, 13 .
We use the information obtained to sketch the
graph. Other function values are listed below.
1 2
x 2x 5
2
First, find the critical points.
g ‘ x 3x 2 x 2
g x x3
g ‘ x exists for all real numbers. We solve
g ‘ x 0
2
3x x 2 0
x
F x
5
3x 2 x 1 0
172
3x 2 0
or
4
3
11
25
2
x
2
3
or
2
1
13
0
1
6.
7.
25
10
1
2
1
10
25
x 1
2
. We use them
3
to divide the real number line into three
intervals,
2
2
A: , 1 , B: 1, , and C: , .
3
3
The critical values are 1 and
25
2
11
172
g x 1 6 x 3x 2
g ‘ x exists for all real numbers. Solve
1
g ‘ x 0
6 6x 0
x 1
The only critical value is 1 . We use 1 to
divide the real number line into two intervals,
A: , 1 and B: 1, :
C
B
A
g ‘ x 6 6 x
2
3
We use a test value in each interval to determine
the sign of the derivative in each interval.
The solution is continued on the next page.
A: Test 2, g ‘ 2 6 6 2 6 0
B: Test 0,
x 1 0
g ‘ 0 6 6 0 6 0
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220
Chapter 2: Applications of Differentiation
G ‘ x exists for all real numbers. We solve
A: Test 2,
g ‘ 2 3 2 2 2 8 0
G ‘ x 0
2
B: Test 0,
2
3x 2 x 1 0
g ‘ 0 3 0 0 2 2 0
3x 1 x 1 0
2
C: Test 1,
g ‘ 1 3 1 1 2 2 0
2
We see that g x is increasing on , 1 ,
2
decreasing on 1, , and increasing on
3
2
, . So there is a relative maximum at
3
x 1 and a relative minimum at x
We find g 1 :
2
.
3
1
2
g 1 1 1 2 1 5
2
1
13
1 2 5
2
2
2
Then we find g :
3
3
3
2
2 2 1 2
2
g 2 5
3 3 2 3
3
13
There is a relative maximum at 1, , and
2
2 113
there is a relative minimum at ,
.
3 27
We use the information obtained to sketch the
graph. Other function values are listed
x
g x
8.
3
5
2
11
x 1 0
3 x 1
or
x 1
1
x
3
or
x 1
1
and 1 . We use them
3
to divide the real number line into three
intervals,
1
1
A: , , B: ,1 , and C: 1, .
3
3
A: Test 1,
The critical values are
G ‘ 1 3 1 2 1 1 4 0
2
B: Test 0,
G ‘ 0 3 0 2 0 1 1 0
2
C: Test 2,
G ‘ 2 3 2 2 2 1 7 0
2
1
We see that G x is increasing on , ,
3
1, . So there is a relative maximum at
x
1
and a relative minimum at x 1 .
3
3
2
59
1 1 1 1
G 2
3 3 3 3
27
G 1 1 1 1 2 1
3
2
1 59
There is a relative maximum at , , and
3 27
there is a relative minimum at 1,1 .
9
2
G x x x x 2
3
or
1
decreasing on ,1 , and increasing on
3
8 2 4
113
5
27 9 3
27
2
0
1
3x 1 0
2
G ‘ x 3x 2 2 x 1
We use the information obtained to sketch the
graph. Other function values are listed below.
x
G x
2
1
0
2
3
8
1
2
4
17
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
9.
221
f x x3 3x 2
10.
f x x3 3 x 6
f ‘ x 3x 2 3
First, find the critical points.
f ‘ x 3x 2 6 x
f ‘ x exists for all real numbers. We solve
f ‘ x 0
f ‘ x exists for all real numbers. We solve
f ‘ x 0
3x 2 3 0
2
3x 6 x 0
3x x 2 0
x0
or
x2
The critical values are 0 and 2 . We use them to
divide the real number line into three intervals,
A: , 0 , B: 0, 2 , and C: 2, .
C
B
A
2
0
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 1, f ‘ 1 3 1 6 1 9 0
2
B: Test 1,
f ‘ 1 3 1 6 1 3 0
C: Test 3,
f ‘ 3 3 3 6 3 9 0
2
2
We see that f x is increasing on , 0 ,
decreasing on 0, 2 , and increasing on 2, .
So there is a relative maximum at x 0 and a
relative minimum at x 2 .
We find f 0 :
f 0 0 3 0 0.
3
2
Then we find f 2 :
f 2 2 3 2 4.
3
2
There is a relative maximum at 0, 0 , and there
is a relative minimum at 2, 4 . We use the
information obtained to sketch the graph. Other
function values are listed below.
x
f x
2
1
1
3
4
20
4
2
0
16
3x 2 3
x2 1
x 1
The critical values are 1 and 1 . We use them to
divide the real number line into three intervals,
A: , 1 , B: 1,1 , and C:1, .
A: Test 3, f ‘ 3 3 3 3 24 0
2
B: Test 0,
f ‘ 0 3 0 3 3 0
C: Test 2,
f ‘ 2 3 2 3 9 0
2
2
We see that f x is increasing on , 1 ,
decreasing on 1,1 , and increasing on 1, .
So there is a relative maximum at x 1 and a
relative minimum at x 1 .
f 1 1 3 1 6 1 3 6 8
3
f 1 1 3 1 6 1 3 6 4
3
There is a relative maximum at 1,8 , and
there is a relative minimum at 1, 4 . We use the
information obtained to sketch the graph. Other
function values are listed below.
x
f x
3
2
0
2
3
12
4
6
8
24
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222
11.
Chapter 2: Applications of Differentiation
We see that f x is increasing on , 1 ,
f x x3 3 x
decreasing on 1, 0 , and increasing on 0, .
First, find the critical points.
f ‘ x 3x 2 3
f ‘ x exists for all real numbers. We solve
So there is a relative maximum at x 1 and a
relative minimum at x 0 .
3x 2 3 0
f 0 3 0 2 0 0
f 1 3 1 2 1 1
2
f ‘ x 0
2
is a relative minimum at 0, 0 . We use the
information obtained to sketch the graph. Other
function values are listed below.
x
f x
f ‘ 0 3 0 3 3 0
2
3
2
We see that f x is increasing on , , and
that there are no relative extrema. We use the
information obtained to sketch the graph. Other
function values are listed below.
x
f x
12.
14
4
0
4
14
1
2
2
13.
27
4
1
28
F x 1 x3
First, find the critical points.
F ‘ x 3x 2
F ‘ x exists for all real numbers. We solve
F ‘ x 0
f x 3 x 2 2 x3
3x 2 0
f ‘ x 6 x 6 x2
f ‘ x exists for all real numbers. We solve
f ‘ x 0
6 x 6 x2 0
x0
The only critical value is 0 . We use 0 to
divide the real number line into two intervals,
A: , 0 , and B: 0, .
6 x 1 x 0
6x 0
3
There is a relative maximum at 1,1 , and there
x 2 1
There are no real solutions to this equation.
Therefore, the function does not have any
critical values.
We test a point
2
1
0
1
2
3
B
A
or
x 1 0
x0
or
x 1
We know the critical values are 1 and 0 . We
use them to divide the real number line into
three intervals,
A: , 1 , B: 1, 0 , and C: 0, .
A: Test 2,
f ‘ 2 6 2 6 2 12 0
2
1
B: Test ,
2
2
3
1
1 1
f ‘ 6 6 0
2
2 2
2
0
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 1, F ‘ 1 3 1 3 0
2
B: Test 1,
F ‘ 1 3 1 3 0
2
We see that F x is decreasing on , 0 and
decreasing on 0, , so the function has no
relative extema. We use the information
obtained to sketch the graph on the next page.
C: Test 1,
f ‘ 1 6 1 6 1 12 0
2
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
223
Using the information from the previous page
and deteriming other function values are listed
below, we sketch the graph.
x
F x
2
1
0
1
2
14.
9
2
1
0
7
A
C
B
0
4
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 1, G ‘ 1 3 1 12 1 15 0
2
g x 2 x3 16
B: Test 1,
First, find the critical points.
g ‘ x 6 x2
G ‘ 1 3 1 12 1 9 0
C: Test 5,
G ‘ 5 3 5 12 5 15 0
g ‘ x exists for all real numbers. We solve
g ‘ x 0
6 x2 0
x0
The only critical value is 0 . We use 0 to
divide the real number line into two intervals,
A: , 0 , and B: 0, .
A: Test 1, g ‘ 1 6 1 6 0
2
g ‘ 1 6 1 6 0
2
B: Test 1,
We see that g x is increasing on , 0 and
increasing on 0, , so the function has no
relative extema. We use the information
obtained to sketch the graph. Other function
values are listed below.
x
g x
2
1
0
1
2
3
15.
The critical values are 0 and 4 . We use them to
divide the real number line into three intervals,
A: , 0 , B: 0, 4 , and C: 4, .
32
18
16
14
0
38
2
2
We see that G x is increasing on , 0 ,
decreasing on 0, 4 , and increasing on 4, .
So there is a relative maximum at x 0 and a
relative minimum at x 4 .
We find G 0 :
G 0 0 6 0 10
3
10
Then we find G 4 :
G 4 4 6 4 10
3
22
There is a relative maximum at 0,10 , and
there is a relative minimum at 4, 22 . We use
the information obtained to sketch the graph.
Other function values are listed below.
x
G x
2
1
1
2
3
22
3
5
6
17
First, find the critical points.
G ‘ x 3x 2 12 x
G ‘ x exists for all real numbers. We solve
G ‘ x 0
x 4x 0
Dividing by 3
x x 4 0
x0
or
x4 0
x0
or
x4
2
64 96 10
G x x3 6 x 2 10
2
2
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224
16.
Chapter 2: Applications of Differentiation
f x 12 9 x 3x 2 x3
17.
f ‘ x 9 6 x 3x 2
f ‘ x exists for all real numbers. Solve
f ‘ x 0
First, find the critical points.
g ‘ x 3 x 2 4 x3
g ‘ x exists for all real numbers. We solve
g ‘ x 0
9 6 x 3x 2 0
x2 2 x 3 0
Dividing by 3
x 3 x 1 0
x3 0
g x x3 x 4
x 1 0
or
x 3
or
x 1
The critical values are 3 and 1 . We use them to
divide the real number line into three intervals,
A: , 3 , B: 3,1 , and C:1, .
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 4,
f ‘ 4 9 6 4 3 4 15 0
2
B: Test 0,
f ‘ 0 9 6 0 3 0 9 0
2
3 x 2 4 x3 0
x 2 3 4 x 0
x2 0
x0
3 4x 0
4 x 3
3
x0
or
x
4
3
The critical values are 0 and .
4
We use the critical values to divide the real
number line into three intervals,
3
3
A: , 0 , B: 0, , and C: , .
4
4
or
or
A
C: Test 2,
0
f ‘ 2 9 6 2 3 2 15 0
2
We see that f x is decreasing on , 3 ,
increasing on 3,1 , and decreasing on 1, .
So there is a relative minimum at x 3 and a
relative maximum at x 1 .
f 3 12 9 3 3 3 3 15
2
3
A: Test 1, g ‘ 1 3 1 4 1 7 0
B: Test
there is a relative maximum at 1,17 . We use
the information obtained to sketch the graph.
Other function values are listed below.
x
f x
5
4
2
1
0
2
3
17
8
10
1
12
10
15
2
3
2
3
1
1
1
1
, g ‘ 3 4
2
2
2
2
1
1 1
3 4 0
4
8 4
3
There is a relative minimum at 3, 15 , and
3
4
We use a test value in each interval to determine
the sign of the derivative in each interval.
f 1 12 9 1 3 1 1 17
2
C
B
C: Test 1,
g ‘ 1 3 1 4 1 1 0
2
3
We see that g x is increasing on , 0 and
3
3
0, , and is decreasing on , . So there
4
4
is no relative extrema at x 0 but there is a
3
relative maximum at x .
4
3
We find g :
4
3
4
27 81
27
3 3 3
g
4 4 4
64 256 256
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
225
We use the information obtained to sketch the
graph. Other function values are listed below.
From the previous page, we determine there is a
3 27
. We use the
relative maximum at ,
4 256
information obtained to sketch the graph. Other
function values are listed below.
18.
x
g x
2
1
0
24
2
0
1
2
1
16
1
2
0
8
19.
f x x 4 2 x3
f ‘ x 4 x 6 x
3
f ‘ x 0
2 x 2 x 3 0
2
2x 3 0
3
2
3
The critical values are 0 and . We use them to
2
divide the real number line into three intervals,
3
3
A: , 0 , B: 0, , and C: , .
2
2
x0
x
or
A: Test 1, f ‘ 1 4 1 6 1 10 0
3
2
B: Test 1,
f ‘ 1 4 1 6 1 2 0
C: Test 2,
f ‘ 2 4 2 6 2 8 0
3
2
3
2
Since f x is decreasing on both , 0 and
3
3
0, , and increasing on , , there is no
2
2
relative extrema at x 0 but there is a relative
3
minimum at x .
2
4
2
1
0
1
2
3
32
3
0
1
0
27
1 3
x 2 x2 4 x 1
3
First, find the critical points.
f ‘ x x2 4 x 4
f x
f ‘ x 0
2
x 4x 4 0
x 2 2 0
4 x3 6 x 2 0
or
f x
f ‘ x exists for all real numbers. We solve
2
f ‘ x exists for all real numbers. Solve
x2 0
x
3
27
3 3
3
f 2
2 2
2
16
3 27
There is a relative minimum at , .
2 16
x2
The only critical value is 2 .
We divide the real number line into two
intervals,
A: , 2 and B: 2, .
A
B
2
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 0, f ‘ 0 0 4 0 4 4 0
2
B: Test 3, f ‘ 3 3 4 3 4 1 0
2
We see that f x is increasing on both , 2
and 2, . Therefore, there are no relative
extrema. We use the information obtained to
sketch the graph. Other function values are
listed below.
x
f x
3
2
40
59
3
1
22
3
0
1
1
2
5
3
3
2
Copyright © 2016 Pearson Education, Inc.
4
3
226
20.
Chapter 2: Applications of Differentiation
1
F x x3 3x 2 9 x 2
3
F ‘ x x2 6 x 9
21.
F ‘ x exists for all real numbers. Solve
F ‘ x 0
f x 3 x 4 15 x 2 12
First, find the critical points.
f ‘ x 12 x3 30 x
f ‘ x exists for all real numbers. We solve
f ‘ x 0
2
x 6x 9 0
12 x3 30 x 0
x2 6 x 9 0
6 x 2×2 5 0
x 32 0
x3 0
x3
The only critical value is 3 . We divide the real
number line into two intervals,
A: ,3 and B: 3, .
A
B
3
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 0, F ‘ 0 0 6 0 9 9 0
2
B: Test 4, F ‘ 4 4 6 4 9 1 0
2
We see that F x is decreasing on both ,3
and 3, . Therefore, there are no relative
extrmea.
We use the information obtained to sketch the
graph. Other function values are listed below.
x
3
2
F x
65
104
3
6x 0
or
x0
or
x0
or
2×2 5 0
5
x2
2
x
10
2
10
10
and
. We
2
2
use them to divide the real number line into four
intervals,
10
10
A: ,
, B: 2 , 0 ,
2
The critical values are 0,
10
10
C: 0,
, and D:
, .
2
2
C
B
A
0
10
2
D
10
2
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 2,
f ‘ 2 12 2 30 2 36 0
3
B: Test 1,
1
43
3
0
1
2
13
3
C: Test 1,
2
20
3
3
7
D: Test 2,
f ‘ 1 12 1 30 1 18 0
3
f ‘ 1 12 1 30 1 18 0
3
f ‘ 2 12 2 30 2 36 0
3
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
227
From the previous page, we see that f x is
10
decreasing on ,
, increasing on
2
10
10
2 , 0 , decreasing again on 0, 2 ,
10
, . Thus, there
and increasing again on
2
10
, a relative
2
maximum at x 0 , and another relative
is a relative minimum at x
10
.
2
10
We find f
:
2
22.
x
f x
3
2
1
1
2
3
120
0
0
0
0
120
g x 2 x 4 20 x 2 18
g ‘ x 8 x3 40 x
g ‘ x exists for all real numbers. We solve
g ‘ x 0
minimum at x
3
8 x 40 x 0
8x x2 5 0
4
2
10
10
10
f
3
15
12
2
2
2
8x 0
or
x2 5 0
x0
or
x2 5
27
4
f
Then we find 0 :
x0
or
x 5
The critical values are 0, 5 and 5 . We use
them to divide the real number line into four
intervals,
f 0 3 0 15 0 12 12
4
2
C: 0, 5 , and D: 5, .
A: , 5 , B: 5, 0 ,
10
Then we find f
:
2
4
2
10
10
10
f
3
15
12
2
2
2
0
5
27
4
10 27
, and
There are relative minima at
4
2
C
B
A
D
5
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 3,
g ‘ 3 8 3 40 3 96 0
3
10 27
2 , 4 .
B: Test 1,
There is a relative maximum at 0,12 .
We use the information obtained above to
sketch the graph. Other function values are
listed at the top of the next column.
g ‘ 1 8 1 40 1 32 0
3
C: Test 1,
g ‘ 1 8 1 40 1 32 0
3
D: Test 3,
g ‘ 3 8 3 40 3 96 0
3
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
228
Chapter 2: Applications of Differentiation
From the previoius page, we see that g x is
We use a test value in each interval to determine
the sign of the derivative in each interval.
1
1
0
A: Test 3, G ‘ 3
2
3
3
3 3 2
decreasing on , 5 , increasing on
5, 0 , decreasing again on 0, 5 , and
increasing again on 5, . Thus, there is a
B: Test 1, G ‘ 1
relative minimum at x 5 , a relative
maximum at x 0 , and another relative
minimum at x 5 .
g 5 2 5
2
4
2
10
3
2
1
6
0,18 We use the information obtained to
sketch the graph. Other function values are
listed below.
23.
x
g x
4
3
1
1
3
4
210
0
0
0
0
210
1
0
3
We use the information obtained to sketch the
graph. Other function values are listed below.
x
G x
2
5 2 5 20 5 18 32
There are relative minima at 5, 32 and
5, 32 . There is a relative maximum at
g
3
, 2 and 2, . Thus, there are no
relative extrema for G x .
g 0 2 0 20 0 18 18
4
3 1 2
2
We see that G x is increasing on both
20 5 18 32
4
1
24.
2
1
0
1
2
F x 3 x 1 x 1 3
1
F ‘ x
2
1
x 1 3 1
3
1
3 x 1 3
2
F ‘ x does not exist when
3 x 1 3 0 , which means that F ‘ x does not
2
exist when x 1 . The equation F ‘ x 0 has
no solution, therefore, the only critical value is
x 1.
We use 1 to divide the real number line into
two intervals,
A: ,1 and B: 1, .
G x 3 x 2 x 2 3
1
First, find the critical points.
2
1
G ‘ x x 2 3 1
3
1
2
3 x 2 3
G ‘ x does not exist when x 2 . The equation
G ‘ x 0 has no solution, therefore, the only
critical value is x 2 .
We use 2 to divide the real number line into
two intervals,
A: , 2 and B: 2, :
A
B
2
We use a test value in each interval to determine
the sign of the derivative in each interval.
1
1
0
A: Test 0, F ‘ 0
2
3
3
3 0 1
B: Test 2, F ‘ 2
1
3 2 1 3
2
1
0
3
We see that F x is increasing on both ,1
and 1, . Thus, there are no relative extrema
for F x . We use the information obtained to
sketch the graph at the top of the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
229
Using the information from the previous page,
we sketch the graph. Other function values are
listed.
x
F x
7
0
1
2
9
25.
2
1
0
1
2
26.
f x 1 x 3
f x
8
1
1
8
3
0
0
3
f x x 3 3 5
2
f ‘ x
2
First, find the critical points.
2 13
f ‘ x
x
3
2
3
3 x
f ‘ x does not exist when
1
2
x 3 3
3
2
3 x 3 3
1
f ‘ x does not exist when x 3 . The equation
f ‘ x 0 has no solution, therefore, the only
3 3 x 0 , which means that f ‘ x does not
exist when x 0 . The equation f ‘ x 0 has
no solution, therefore, the only critical value is
x 0.
We use 0 to divide the real number line into
two intervals,
A: , 0 and B: 0, :
A
x
B
critical value is x 3 .
We use 3 to divide the real number line into
two intervals, A: , 3 and B: 3, :
A: Test 4, f ‘ 4
B: Test 2, f ‘ 2
2
1
1
3 4 3 3
2
3 2 3
3
2
0
3
2
0
3
We see that f x is decreasing on , 3 and
increasing on 3, . Thus, there is a relative
minimum at x 3 .
0
f 3 3 3 3 5 5 :
2
We use a test value in each interval to determine
the sign of the derivative in each interval.
2
2
0
A: Test 1, f ‘ 1 3
3 1 3
2
2
B: Test 1,
f ‘ 1 3 0
3
3 1
We see that f x is increasing on , 0 and
decreasing on 0, . Thus, there is a relative
maximum at x 0 .
We find f 0 :
f 0 1 0 3 1 .
2
Therefore, there is a relative minimum at
3, 5 . We use the information obtained to
sketch the graph. Other function values are
listed below.
x
f x
11
4
2
5
1
4
4
1
Therefore, there is a relative maximum at 0,1 .
We use the information obtained to sketch the
graph. Other function values are listed at the top
of the next column.
Copyright © 2016 Pearson Education, Inc.
230
27.
Chapter 2: Applications of Differentiation
1
x
x 1
First, find the critical points.
G x
3
54
2
85
1
1
2
4
4
85
3
54
8
G x
2
8 x 2 1
2 x
G ‘ x 8 1 x 2 1
2
16 x
x 1
2
2
G ‘ x exists for all real numbers. Setting the
derivative equal to zero, we have:
G ‘ x 0
16 x
x 1
2
2
28.
0
x 1
5 x2 1
A
2 x
0
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 1, G ‘ 1
G ‘ 1
16 1
1 1
2
16 1
12 1
2
16
4 0
4
16
40
2
4
We see that G x is decreasing on , 0 and
increasing on 0, . Thus, a relative minimum
occurs at x 0 .
We find G 0 :
8
0 2 1
2
x 1
2
2
F ‘ x exists for all real numbers. We solve
F ‘ x 0
10 x
x 1
2
B
1
10 x
x0
The only critical value is 0 .
We use 0 to divide the real number line into
two intervals,
A: , 0 and B: 0, :
G 0
2
F ‘ x 5 1 x 2 1
16 x 0
B: Test 1,
5
F x
2
0
x0
The only critical values is 0 .
We use 0 to divide the real number line into
two intervals,
A: , 0 and B: 0, :
A: Test 1,
F ‘ 1
10 1
2
10 5
0
4 2
2
5
10
0
4
2
1 1
2
B: Test 1,
F ‘ 1
10 1
1 1
2
We see that F x is increasing on , 0 and
decreasing on 0, . Thus, a relative maximum
occurs at x 0 .
We find F 0 :
8
5
Thus, there is a relative minimum at 0, 8 .
F 0
We use the information obtained to sketch the
graph. Other function values are listed at the top
of the next column.
Thus, there is a relative maximum at 0,5 .
0 2 1
5
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
231
We use the information obtained on the
previous page to sketch the graph. Other
function values are listed below.
29.
x
F x
3
1
2
2
1
1
1
5
2
2
3
1
A: Test 2, g ‘ 2
5
2
2
2
g 1
x 1
2
2
x 1
2
2
2
4 4 2
2
2
2
2 1
40
12
0
25
0
x2 1 0
2
4
2
2
information obtained to sketch the graph. Other
function values are listed below.
x
g x
g ‘ x 0
4 4 x2 0
4 1
there is a relative maximum at 1, 2 . We use the
g ‘ x exists for all real numbers. We solve
Multiplying by x 2 1
2
Dividing by 4
x2 1
x 1
30.
x 1
The critical values are 1 and 1 . We use them to
divide the real number line into three intervals,
A: , 1 , B: 1,1 , and C:1, .
A
02 1
There is a relative minimum at 1, 2 , and
4 4 x2
x 1
2
1 1
Then we find g 1 :
4 1
4
g 1 2
2
1 1 2
Quotient Rule
4 x2 4 8×2
2
4 4 0
12
0
25
x 1 and a relative maximum at x 1 .
We find g 1 :
2
2
2 1
1, . So there is a relative minimum at
x 1 4 4 x 2 x
g ‘ x
x 1
4 4 x2
2
2
increasing on 1,1 , and decreasing again on
x2 1
First, find the critical points.
2
We see that g x is decreasing on , 1 ,
4x
g ‘ 2
C: Test 2,
1
2
g x
g ‘ 0
B: Test 0,
4 4 2
B
C
1
1
We use a test value in each interval to determine
the sign of the derivative in each interval.
3
65
2
85
0
2
0
3
6
5
g x
g ‘ x
g ‘ x
8
5
x2
x2 1
x 1 2 x x 2 x
x 1
2
2
2
2
2x
x 1
2
2
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
232
Chapter 2: Applications of Differentiation
From the previous page, g ‘ x exists for all real
31.
numbers. We solve
g ‘ x 0
2x
2
x 1
2
x0
The only critical values is 0 .
We use 0 to divide the real number line into
two intervals,
A: , 0 and B: 0, :
f ‘ x does not exist when x 0 . The equation
f ‘ x 0 has no solution, therefore, the only
critical value is x 0 .
We use 0 to divide the real number line into
two intervals,
A: , 0 and B: 0, :
A: Test 1,
2 1
1 1
2
2
1
First, find the critical points.
1 2
f ‘ x x 3
3
1
1
2
3 2
3
3 x
3 x
0
g ‘ 1
f x 3 x x 3
2
1
0
4
2
A
B
B: Test 1,
g ‘ 1
2 1
1 1
2
0
2 1
0
2
4 2
We use a test value in each interval to determine
the sign of the derivative in each interval.
1
1
0
A: Test 1, f ‘ 1
2
3
3 3 1
We see that g x is decreasing on , 0 and
increasing on 0, . Thus, a relative minimum
B: Test 1, f ‘ 1
occurs at x 0 .
We find g 0 :
g 0
02
0
02 1
2
4
5
1
1
2
1
1
2
2
4
5
3
9
10
1
0
3
and 0, . Thus, there are no relative extrema
for f x . We use the information obtained to
use the information obtained to sketch the
graph. Other function values are listed below.
x
g x
9
10
2
3 3 1
We see that f x is increasing on both , 0
Thus, there is a relative minimum at 0, 0 . We
3
1
sketch the graph. Other function values are
listed below.
32.
x
f x
8
1
0
1
8
2
1
0
1
2
f x x 1 3
1
f ‘ x
2
1
x 1 3 1
3
1
3 x 1 3
2
f ‘ x does not exist when x 1 . The equation
f ‘ x 0 has no solution, therefore, the only
critical value is x 1 .
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
233
We use 1 to divide the real number line into
two intervals,
A: , 1 and B: 1, :
Using the information from the previous page,
we use 1 to divide the real number line into
two intervals,
A: , 1 and B: 1, :
1
A: Test 2, f ‘ 2
0
2
3
3
3 2 1
1
We use a test value in each interval to determine
the sign of the derivative in each interval.
A: Test 2,
1
1
B: Test 0, f ‘ 0
0
2
3 0 1 3 3
We see that f x is increasing on both
B: Test 0,
We use the information obtained to sketch the
graph. Other function values are listed below.
33.
f x
9
2
1
0
7
2
1
0
1
2
0 1
1
0
0 2 0 5 5
We see that g x is decreasing on , 1 and
increasing on 1, , and the change from
g ‘ 0
1
g 1
12 2 1 5 4 2
Thus, there is a relative minimum at 1, 2 .
2
First, find the critical points.
12
1
g ‘ x x2 2 x 5
2 x 2
2
2 x 1
1
2
2 x2 2 x 5
We use the information obtained to sketch the
graph. Other function values are listed below.
x 1
2
x 2x 5
The equation x 2 2 x 5 0 has no realnumber solution, so g ‘ x exists for all real
numbers. Next we find out where the derivative
is zero. We solve
g ‘ x 0
x 1
2
0
2
decreasing to increasing indicates that a relative
minimum occurs at x 1 . We substitute into
the original equation to find g 1 :
g x x2 2 x 5 x2 2 x 5
2 1
1
0
2
2 2 2 5 5
g ‘ 2
intervals, Thus, there are no relative extrema for
f x .
x
B
A
1
x 2x 5
x 1 0
x 1
The only critical value is 1 .
34.
x
g x
4
2
0
1
3
3.61
2.24
2.24
2.83
4.47
F x
x 1
1
x2 1
2
12
32
1
F ‘ x x2 1
2 x
2
x
x 1
2
3
2
F ‘ x exists for all real numbers.
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
234
Chapter 2: Applications of Differentiation
Using the derivative from the previous page,
We solve
F ‘ x 0
x
x 1
2
3
2
70. Answers may vary, one such graph is:
0
x0
The only critical value is 0 .
We use 0 to divide the real number line into
two intervals,
A: , 0 and B: 0, :
A: Test 1,
F ‘ 1
1
1 1
2
3
2
1
8
71. Answers may vary, one such graph is:
0
B: Test 1,
F ‘ 1
1
1 1
2
3
2
1
8
0
We see that F x is increasing on , 0 and
decreasing on 0, . Thus, a relative maximum
occurs at x 0 .
1
F 0
1
0 2 1
72. Answers may vary, one such graph is:
Thus, there is a relative maximum at 0,1 . We
use the information obtained to sketch the
graph. Other function values are listed below.
x
F x
3
2
1
1
2
3
0.32
0.45
0.71
0.71
0.45
0.32
35. – 68.
73. Answers may vary, one such graph is:
Left to the student.
69. Answers may vary, one such graph is:
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
235
74. Answers may vary, one such graph is:
79. Answers may vary, one such graph is:
y
x
-5
-4
-3
-2
-1
0
1
2
3
4
5
80. Answers may vary, one such graph is:
y
75. Answers may vary, one such graph is:
x
-5
-4
-3
-2
-1
0
1
2
3
4
5
81. Answers may vary, one such graph is:
f(x)
76. Answers may vary, one such graph is:
x
-5
77. Answers may vary, one such graph is:
-4
-3
-2
-1
0
1
2
3
4
5
82. Answers may vary, one such graph is:
g(x)
x
-9 -8 -7 -6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
7
8
9
83. Answers may vary, one such graph is:
H(x)
78. Answers may vary, one such graph is:
x
-9 -8 -7 -6 -5 -4 -3 -2 -1
Copyright © 2016 Pearson Education, Inc.
0
1
2
3
4
5
6
7
8
9
236
Chapter 2: Applications of Differentiation
84. Answers may vary, one such graph is:
We use them to divide the interval 0, into
K(x)
three intervals:
A: 0,2.108 B: 2.08,3.897 , and C:3.897
85.
0
1
2
3
4
5
6
7
8
9
The critical value of a function f is an
interior value c of its domain at which the
tangent to the graph is horizontal f ‘ c 0 or
the tangent is vertical f ‘ c does not exist .
The critical values for this graph are
x1 , x3 , x4 , x5 , x6 , x7 , x8 , x10 .
86.
The function is increasing on intervals
a, b and c, d . A line tangent to the curve at
any point on either of these intervals has a
positive slope. Thus, the function is increasing
on the intervals for which the first derivative is
positive. Similarly, we see that on the intervals
b, c and d , e the function is decreasing. A
line tangent to the curve at any point on either
of these intervals has a negative slope. Thus, the
function is decreasing on the intervals for which
first derivative is negative.
87. Letting t be years since 2006 and E be thousand
of employees, we have the function:
E t 107.833t 3 971.369t 2
2657.917t 50347.83
First, we find the critical points.
E ‘ t 323.499t 2 1942.738t 2657.917
E ‘ t exists for all real numbers. Solve
E ‘ t 0
323.499t 2 1942.738t 2657.917 0
Using the quadratic formula, we have:
t
1942.738
1942.7382 4 323.499 2657.917
2 323.499
1942.738 334,896.970312
646.998
t 2.108
or
t 3.897
There are two critical values.
C
B
A
x
-9 -8 -7 -6 -5 -4 -3 -2 -1
2.108
3.897
0
Next, we test a point in each interval to
determine the sign of the derivative.
A: Test 1,
E ‘ 1 323.499 1 1942.738 1 2657.917
2
1038.678 0
B: Test 3,
E ‘ 3 323.499 3 1942.738 3 2657.917
2
258.806 0
C: Test 4,
E ‘ 4 323.499 4 1942.738 4 2657.917
2
62.949 0
Since, E t is increasing on 0, 2.108 and
decreasing on 2.108,3.897 and there is a
relative maximum at t 2.108 .
E 2.108 107.833 2.108 971.369 2.108
3
2
2657.917 2.108 50347.83
52, 644.383
There is a relative minimum at
2.108, 52,644.383 .
Since, E t is decreasing on 2.108,3.897 and
increasing 3.897, on and there is a relative
minimum at t 3.897 .
E 3.897 107.833 3.897 971.369 3.897
3
2
2657.917 3.897 50347.83
52,335.73
There is a relative maxnimum at
3.897, 52,335.73 .
We sketch the graph.
t
T t
0
1
3
5
8
50,347
52,142
52, 491
52,832
64, 654
Copyright © 2016 Pearson Education, Inc.
.
Exercise Set 2.1
88.
237
N a a 2 300a 6,
0 a 300
A: Test 20,
N ‘ a exists for all real numbers. Solve,
B: Test 100,
f ‘ 20 0.00518 20 0.457 0.3534 0
N ‘ a 2a 300
f ‘ 100 0.00518 100 0.457 0.061 0
N ‘ a 0
We see that f t is decreasing on 0,88.22
2a 300 0
2a 300
a 150
The only critical value is150 . We divide the
interval 0,300 into two intervals,
and increasing on 88.22, , so there is a
relative minimum at t 88.22 .
f 88.22
0.00259 88.22 0.457 88.22 36.237
2
A: 0,150 and B: 150,300 .
A: Test 100,
16.09
There is a relative minimum at about
88.22,16.08 . Thus, the latitude that is closest
N ‘ 100 2 100 300 100 0
to the equator at which the full eclipse could be
view is 16.08 degrees south and will occur
88.22 minutes after the start of the eclipse.
We use the information obtained above to
sketch the graph. Other function values are
listed below.
B: Test 200,
N ‘ 200 2 200 300 100 0
Since, N a is increasing on 0,150 and
decreasing on 150,300 , there is a relative
maximum at x 150 .
N 150 150 300 150 6 22,506
2
There is a relative maximum at 150, 22,506 .
We sketch the graph.
89.
a
N a
0
100
200
300
6
20, 006
20, 006
6
90.
f x
20
30
40
60
80
28.133
24.858
22.101
18.141
16.253
T t 0.1t 2 1.2t 98.6, 0 t 12
First, we find the critical points.
T ‘ t 0.2t 1.2
f t 0.00259t 2 0.457t 36.237
First, we find the critical points.
f ‘ t 0.00518t 0.457
f ‘ x exists everywhere, so we solve
f ‘ t 0
0.00518t 0.457 0
t 88.22
The only critical value is about 88.22 we use it
to break up the interval 0, into two intervals
A: 0,88.28 and B: 88.22, .
B
A
0
x
T ‘ t exists for all real numbers. Solve
T ‘ t 0
0.2t 1.2 0
t6
The only critical value is 6. We use it to divide
the interval 0,12 into two intervals:
A: 0,6 and B: 6,12
B
A
0
6
12
The solution is continued on the next page.
88.22
Copyright © 2016 Pearson Education, Inc.
238
Chapter 2: Applications of Differentiation
Next, we test a point in each interval found on
the previous page to determine the sign of the
derivative.
A: Test 1, T ‘ 1 0.2 1 1.2 1.0 0
B: Test 7, T ‘ 7 0.2 7 1.2 0.2 0
Since, T t is increasing on 0, 6 and
decreasing on 6,12 , there is a relative
92. The derivative is positive over the interval
, 2 and negative over the interval 2, .
Furthermore, it is equal to zero when x 2 .
This means that the function is increasing over
the interval , 2 , decreasing over the interval
2, and has a horizontal tangent at x 2 . A
possible graph is shown below.
maximum at t 6 .
T 6 0.16 1.2 6 98.6 102.2
2
There is a relative maximum at 6,102.2 . We
sketch the graph.
t
T t
0
3
5
7
8
12
98.6
101.3
102.1
102.1
101.8
98.6
93. The derivative is positive over the interval
,1 and negative over the interval 1, .
Furthermore, it is equal to zero when x 1 .
This means that the function is increasing over
the interval ,1 , decreasing over the interval
91. The derivative is negative over the interval
, 1 and positive over the interval 1, .
Furthermore, it is equal to zero when x 1 .
This means that the function is decreasing over
the interval , 1 , increasing over the
1, and has a horizontal tangent at x 1 . A
possible graph is shown below.
interval 1, and has a horizontal tangent at
x 1 . A possible graph is shown below.
94. The derivative is negative over the interval
, 1 and positive over the interval 1, .
Furthermore, it is equal to zero when x 1 .
This means that the function is decreasing over
the interval , 1 , increasing over the
interval 1, and has a horizontal tangent at
x 1 . A possible graph is shown below.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
239
95. The derivative is positive over the interval
4, 2 and negative over the intervals
97.
f x x 6 4 x5 54 x 4 160 x3 641x 2
828 x 1200
Using the calculator we enter the function into
the graphing editor as follows:
, 4 and 2, . Furthermore, it is equal to
zero when x 4 and x 2 . This means that
the function is decreasing over the interval
, 4 , then increasing over the interval
4, 2 , and then decreasing again over the
interval 2, . The function has horizontal
tangents at x 4 and x 2 . A possible graph
is shown below.
Using the following window:
The graph of the function is:
96. The derivative is negative over the interval
1,3 and intervals and positive over the
intervals , 1 and 3, . Furthermore, it is
We find the relative extrema using the
minimum/maximum feature on the calculator.
There are relative minima at
3.683, 2288.03 and 2.116, 1083.08 .
equal to zero when x 1 and x 3 . This
means that the function is increasing over the
interval , 1 , then decreasing over the
There are relative maxima at
6.262,3213.8 , 0.559,1440.06 , and
interval 1,3 , and then increasing again over
the interval 3, . The function has horizontal
tangents at x 1 and x 3 . A possible graph
is shown below.
5.054, 6674.12.
98.
f x x 4 4 x3 36 x 2 160 x 400
Using the calculator we enter the function into
the graphing editor as follows:
Using the following window:
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
240
Chapter 2: Applications of Differentiation
The graph of the function is:
100. f x x 9 x 2
Using the calculator we enter the function into
the graphing editor as follows:
We find the relative extrema using the
minimum/maximum feature on the calculator.
There are relative minima at
5, 425 and 4, 304 .
Using the following window:
There is a relative maximum at 2,560 .
99.
f x 3 4 x2 1
Using the calculator we enter the function into
the graphing editor as follows:
The graph of the function is:
Using the following window:
Notice, the calculator has trouble drawing the
graph. The graph should continue to the xintercepts at 3, 0 and 3,0 . Fortunately, this
does not hinder our efforts to find the extrema.
We find the relative extrema using the
minimum/maximum feature on the calculator.
There is a relative minimum at 2.12, 4.5 .
The graph of the function is:
There is a relative maximum at 2.12, 4.5 .
101. f x x 2
Using the calculator we enter the function into
the graphing editor as follows:
We find the relative extrema using the
minimum/maximum feature on the calculator.
There are relative minima at 2,1 and 2,1 .
There is a relative maximum at 0, 2.587 .
Using the following window:
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
241
103. f x x 2 1
The graph of the function is:
Using the calculator we enter the function into
the graphing editor as follows:
We find the relative extrema using the
minimum/maximum feature on the calculator.
The graph is decreasing over the interval
, 2 .
Using the following window:
The graph is increasing over the interval 2, .
There is a relative minimum at 2, 0 .
The derivative does not exist at x 2
The graph of the function is:
102. f x 2 x 5
Using the following window:
We find the relative extrema using the
minimum/maximum feature on the calculator.
The graph is decreasing over the interval
, 1 and 0,1 .
The graph is increasing over the interval
1, 0 and 2, .
There are relative mimima at 1, 0 and 1,0 .
The graph of the function is:
There is a relative maximum at 0,1 .
The derivative does not exist at x 1 and
x 1.
104. f x x 2 3x 2
We find the relative extrema using the
minimum/maximum feature on the calculator.
The graph is decreasing over the interval
5
, .
2
The graph is increasing over the interval
5
, .
2
5
There is a relative minimum at , 0 .
2
The derivative does not exist at x
Using the calculator we enter the function into
the graphing editor as follows:
The solution is continued on the next page.
5
2
Copyright © 2016 Pearson Education, Inc.
242
Chapter 2: Applications of Differentiation
The graph of the function is:
Using the following window:
The graph of the function is:
We find the relative extrema using the
minimum/maximum feature on the calculator.
The graph is decreasing over the interval
, 3 and 0,3 .
The graph is increasing over the interval
3, 0 and 3, .
We find the relative extrema using the
minimum/maximum feature on the calculator.
The graph is decreasing over the interval
3
,1 and , 2 .
2
The graph is increasing over the interval
3
1, and 2, .
2
There are relative mimima at 3, 0 and 3, 0 .
There is a relative maximum at 0,9 .
The derivative does not exist at x 3 and
x 3.
106. f x x 2 4 x 4
Enter the function into the graphing editor:
There are relative mimima at 1, 0 and 2, 0 .
3 1
There is a relative maximum at , .
2 4
The derivative does not exist at x 1 and
x 2.
Using the following window:
105. f x 9 x 2
Using the calculator we enter the function into
the graphing editor as follows:
The graph of the function is:
Using the following window:
The solution is continued on the next column.
We find the relative extrema using the
minimum/maximum feature on the calculator.
The graph is decreasing over the interval
, 2 .
The graph is increasing over the interval 2, .
There is a relative minimum at 2, 0 .
The derivative exists for all values of x.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.1
243
107. f x x3 1
The graph of the function is:
Using the calculator we enter the function into
the graphing editor as follows:
We find the relative extrema using the
minimum/maximum feature on the calculator.
The graph is decreasing over the interval
, 1.41 , 1, 0 and 0,1.41 .
Using the following window:
The graph is increasing over the interval
1.41, 0 , 0,1 and 1.41, .
There are relative mimima at
1.41, 0 , 0, 0 and 1.41, 0 .
The graph of the function is:
There are relative maxima at 1, 0 and 1,0 .
The derivative does not exist at x 1.41 and
x 1.41.
109.
We find the relative extrema using the
minimum/maximum feature on the calculator.
The graph is decreasing over the interval
, 1 .
a) We enter the data into the calculator and
run a cubic regression. The calculator
returns
The graph is increasing over the interval 1, .
There is a relative minimum at 1,0 .
The derivative does not exist at x 1.
108. f x x 4 2 x 2
Using the calculator we enter the function into
the graphing editor as follows:
When we try to run a quartic regression, the
calculator returns a domain error. Therefore,
the cubic regression fits best.
b) The domain of the function is the set of
nonnegative real numbers. Realistically,
there would be some upper limit upon daily
caloric intake.
c) The cubic regression model appears to have
a relative minimum at 4316, 77.85 and it
appears to have a relative maximum at
3333, 79.14 . This leads us to believe that
eating too many calories might shorten life
expectancy.
Using the following window:
The solution is continued at the top of the next
column.
Copyright © 2016 Pearson Education, Inc.
244
Chapter 2: Applications of Differentiation
110.
a) The cubic function fits best. In fact some
calculators will return an error message
when an attempt is made to fit a quartic
function to the data.
b) The domain of the function is the set of
nonnegative real numbers. Realistically,
there would be some upper limit upon daily
caloric intake.
c) The cubic regression model does not appears
to have a relative extrema. The greater the
daily caloric intake, the lower the infant
mortality.
b) Answers will vary. When the equations are
somewhat complex, the best way to
determine a viewing window is to use the
table screen on the calculator and observing
appropriate y-values for selected x-values.
You will need to set your table to accept
selected x-values. Enter the table set up
feature on your calculator and turn on the
ask feature for your independent variable.
This will allow you to enter an x-value and
the calculator will return the y-value. You
should make your ranges large enough so
that all the data points will be easily viewed
in the window.
111.
a) Answers will vary. In Exercises 1-16 the
function is given in equation form. The most
accurate way to select an appropriate
viewing window, one should first determine
the domain, because that will help determine
the x-range. For polynomials the domain is
all real numbers, so we will typically select
a x-range that is symmetric about 0. Next,
you should find the critical values and make
sure that your x-range contains them.
Finally, you should determine the xintercepts and make sure the x-range
includes them. To find the y-range, you
should find the y-values of the critical points
and make sure the y-range includes those
values. You should also make sure that the
y-range includes the y-intercept.
To avoid the calculations required to find
the relative extrema and the zeros as
described above, we can determine a good
window by using the table screen on the
calculator and observing the appropriate yvalues for selected x-values.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
245
Next, find the critical points of f x . Since
Exercise Set 2.2
1.
f ‘ x exists for all real numbers x, the only
Next, find the critical points of f x . Since
critical points occur when f ‘ x 0 .
2x 1 0
2 x 1
1
x
2
1
We find the function value at x .
2
critical points occur when f ‘ x 0 .
1 1
5
1 1 1
f 1 1 .
2 2 2
4 2
4
2 x 0
x0
We find the function value at x 0 .
1 5
The critical point is , .
2 4
Next, we apply the Second Derivative test.
f ” x 2
f x 4 x
2
First, find f ‘ x and f ” x .
f ‘ x 2 x
f ” x 2
f ‘ x exists for all real numbers x, the only
2
f ‘ x 0
f 0 4 0 4 .
1
f ” 2 0
2
Next, we apply the Second Derivative test.
f ” x 2
5
1
Therefore, f is a relative minimum.
2
4
2
The critical point is 0, 4 .
f ” 0 2 0
Therefore, f 0 4 is a relative maximum.
2.
f x x2 x
f ‘x 2x 1
f x 5 x2
f ” x 2
f ” x 2
critical points occur when f ‘ x 0 .
critical points occur when f ‘ x 0 .
2x 1 0
f ‘ x exists for all real numbers. The only
f ‘ x 2 x
f ‘ x exists for all real numbers. The only
f ‘ x 0
f ‘ x 0
x
2 x 0
x0
We find the function value at x 0 .
f 0 5 0 5 .
2
The critical point is 0,5 .
1
2
We find the function value at x
1
.
2
2
1
1 1 1 1 1
f .
2 2 2 4 2
4
f ” 0 2 0
1 1
The critical point is , .
2 4
Next, we apply the Second Derivative test.
f ” x 2
f x x x 1
1
f ” 2 0
2
Next, we apply the Second Derivative test.
f ” x 2
Therefore, f 0 5 is a relative maximum.
3.
4.
2
First, find f ‘ x and f ” x .
f ‘ x 2x 1
1
1
Therefore, f is a relative minimum.
2
4
f ” x 2
Copyright © 2016 Pearson Education, Inc.
246
5.
Chapter 2: Applications of Differentiation
f x 4 x 2 3x 1
We find the function value at x
First, find f ‘ x and f ” x .
2
4
4
4
f 5 8 7
5
5
5
16 32 35
.
5
5
5
19
5
4 19
The critical point is , .
5
5
Next, we apply the Second Derivative test.
f ” x 10
f ‘ x 8 x 3
f ” x 8
Next, find the critical points of f x . Since
f ‘ x exists for all real numbers x, the only
critical points occur when f ‘ x 0 .
f ‘ x 0
8 x 3 0
8 x 3
3
x
8
We find the function value at x
4
f ” 10 0
5
3
.
8
19
4
Therefore, f is a relative maximum.
5
5
2
3
3
3
f 4 3 1
8
8
8
9 9
4 1
64 8
7.
9 18 16
16 16 16
7
16
7
3
The critical point is , .
8 16
Next, we apply the Second Derivative test.
f ” x 8
3
f ” 8 0
8
f x 5 x 2 8 x 7
f ‘ x 10 x 8
f ” x 6 x
Next, find the critical points of f x . Since
f ‘ x exists for all real numbers x, the only
critical points occur when f ‘ x 0 .
f ‘ x 0
3x 2 12 0
x2 4
x 4
x 2
There are two critical values.
First, we find the function value at x 2 .
f ” x 10
f 2 2 12 2 1
3
f ‘ x exists for all real numbers. The only
critical points occur when f ‘ x 0 .
10 x 8 0
4
x
5
f ‘ x 3 x 2 12
3x 2 12
7
3
Therefore, f is a relative maximum.
8
16
f ‘ x 0
f x x 3 12 x 1
First, find f ‘ x and f ” x .
.
6.
4
.
5
8 24 1
.
15
The critical point is 2,15 .
Next, we apply the Second Derivative test.
f ” x 6 x
f ” 2 6 2 12 0
Therefore, f 2 15 is a relative maximum.
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
247
Next, we find the function value at x 2 .
1
The critical point is , 1 .
2
Next, we apply the Second Derivative test.
f ” x 48 x
f 2 2 12 2 1
3
.
8 24 1
17
The critical point is 2, 17 .
1
1
f ” 48 24 0
2
2
Next, we apply the Second Derivative test.
f ” x 6 x
1
Therefore, f 1 is a relative minimum.
2
f ” 2 6 2 12 0
Therefore, f 2 17 is a relative minimum.
8.
f x x 3 27 x
a) Find f ‘ x and f ” x .
f x 8×3 6 x 1
f ‘ x 3x 2 27
f ‘ x 24 x 2 6
f ” x 6 x
f ” x 48 x
f ‘ x exists for all real numbers. The only
critical points occur when f ‘ x 0 .
f ‘ x 0
The domain of f is .
b) Find the critical points of f x . Since
f ‘ x exists for all real numbers x, the only
critical points occur when f ‘ x 0 .
24 x 2 6 0
3 x 2 27 0
4×2 1 0
1
x2
4
x
9.
x2 9
x 3
There are two critical values x 3 and
x3.
1
2
1
1
There are two critical values x and x .
2
2
1
We find the function value at x
2
3
1
1
1
f 8 6 1
2
2
2 .
3
1
The critical point is , 3 .
2
Next, we apply the Second Derivative test.
f ” x 48 x
1
1
f ” 48 24 0
2
2
1
Therefore, f 3 is a relative maximum.
2
1
Now, we find the function value at x .
2
3
f 3 3 27 3 54 .
3
The critical point on the graph is 3, 54 .
f 3 3 27 3 54 .
3
The critical point on the graph is 3, 54 .
c) We apply the Second Derivative test to the
critical points.
For x 3
f ” x 6 x
f ” 3 6 3 18 0
The critical point 3, 54 is a relative
maximum.
For x 3
f ” x 6 x
f ” 3 6 3 18 0
The critical point 3, 54 is a relative
minimum.
The solution is continued on the next page.
1
1
1
f 8 6 1
.
2
2
2
1
Copyright © 2016 Pearson Education, Inc.
248
Chapter 2: Applications of Differentiation
If we use the critical values x 3 and
x 3 to divide the real line into three
intervals, , 3 , 3, 3 , and 3, , we
know from the extrema above, that f x is
increasing over the interval , 3 ,
decreasing over the interval 3, 3 and then
increasing again over the interval 3, .
d) Find the points of inflection. f ” x exists
for all real numbers, so we solve the
equation
f ” x 0
6x 0
x0
Therefore, a possible inflection point occurs
at x 0 .
f 0 0 27 0 0 .
A possible inflection point on the graph is
the point 0, 0 .
e) To determine concavity, we use the possible
inflection point to divide the real number
line into two intervals
A : , 0 and B: 0, . We test a point in
each interval.
A: Test 1: f ” 1 6 1 6 0
f ” 1 6 1 6 0
Then, f x is concave down on the interval
, 0 and concave up on the interval
0, , so 0, 0 is an inflection point.
f)
Finally, we use the preceding information to
sketch the graph of the function. Additional
function values can also be calculated as
needed.
x
f x
4
1
1
4
10.
44
26
26
44
f x x 12 x
3
a) Find f ‘ x and f ” x .
f ‘ x 3 x 2 12
f ” x 6 x
The domain of f is .
f ‘x 0
f ‘ x 0
3x 2 12 0
3x 2 12
x2 4
x 2
There are two critical values x 2 and
x 2.
We find the function value at x 2
f 2 2 12 2 16 .
3
The critical point on the graph is 2,16 .
Next, we find the function value at x 2 .
f 2 2 12 2 16 .
3
The critical point on the graph is 2, 16 .
3
B: Test 1:
b) f ‘ x exists for all real numbers. Solve:
c) Apply the Second Derivative test to the
critical points.
For x 2
f ” x 6 x
f ” 2 6 2 12 0
The critical point 2,16 is a relative
maximum.
For x 2
f ” x 6 x
f ” 2 6 2 12 0
We determine the critical point 2, 16 is a
relative minimum.
If we use the critical values x 2 and
x 2 to divide the real line into three
intervals, , 2 , 2, 2 , and 2, , we
know from the extrema above, that f x is
increasing over the interval , 2 ,
decreasing over the interval 2, 2 and then
increasing again over the interval 2, .
d) Find the points of inflection. f ” x exists
for all real numbers, so we solve the
equation
f ” x 0
6x 0
x0
Therefore, a possible inflection point occurs
at x 0 .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
249
We evaluate the function at x 0 .
f 0 0 12 0 0 .
3
Next, we find the function value at x 3 .
This gives the point 0, 0 on the graph.
f 3 2 3 3 3 36 3 28
e) To determine concavity, we use the possible
inflection point to divide the real number
line into two intervals A : , 0 and
54 27 108 28
B: 0, . We test a point in each interval
A: Test 1: f ” 1 6 1 6 0
f ” 1 6 1 6 0
B: Test 1:
Then, f x is concave down on the interval
, 0 and concave up on the interval
0, , so 0, 0 is an inflection point.
f) Finally, we use the preceding information to
sketch the graph of the function. Additional
function values can also be calculated as
needed.
x
f x
3
1
1
3
9
11
11
9
11.
The critical point on the graph is 2, 72 .
3
2
53
The critical point on the graph is 3, 53 .
c) Apply the Second Derivative test to the
critical points.
For x 2
f ” x 12 x 6
f ” 2 12 2 6 30 0
The critical point 2, 72 is a relative
maximum.
For x 3
f ” x 12 x 6
f ” 3 12 3 6 30 0
The critical point 3, 53 is a relative
minimum.
We use the critical values x 2 and x 3
to divide the real line into three intervals,
A: , 2 , B: 2, 3 , and C: 3, , we
know from the extrema above, that f x is
increasing over the interval , 2 ,
f x 2 x 3 3x 2 36 x 28
a) First, find f ‘ x and f ” x .
decreasing over the interval 2, 3 and then
increasing again over the interval 3, .
f ‘ x 6 x 6 x 36
d) Find the points of inflection. f ” x exists
The domain of f is .
for all real numbers, so we solve the
equation
f ” x 0
2
f ” x 12 x 6
b) Find the critical points of f x . Since
f ‘ x exists for all real numbers x, the only
critical points occur when f ‘ x 0 .
f ‘ x 0
6 x 2 6 x 36 0
x2 x 6 0
12 x 6 0
1
2
Therefore, a possible inflection point occurs
1
at x .
2
x
3
2
x 3 x 2 0
1
1
1
1
f 2 3 36 28
2
2
2
2
x 3 or
x 2
There are two critical values x 2 and
x3.
We find the function value at x 2
19
.
2
A possible inflection point on the graph is
1 19
the point , .
2 2
x 3 0 or x 2 0
f 2 2 2 3 2 36 2 28
3
2
16 12 72 28
72
Copyright © 2016 Pearson Education, Inc.
250
Chapter 2: Applications of Differentiation
e) To determine concavity, we use the possible
inflection point to divide the real number
line into two intervals
1
1
A : , and B: , . We test a point
2
2
in each interval
A: Test 0 : f ” 0 12 0 6 6 0
B: Test 1: f ” 1 12 1 6 6 0
Then, f x is concave down on the interval
1
, and concave up on the interval
2
1
1 19
, , so , is an inflection point.
2
2 2
f) We use the preceding information to sketch
the graph of the function. Additional
function values can also be calculated as
needed.
x
f x
3
1
0
1
4
55
59
28
9
36
Next, we find the function value at x 2 .
f 2 3 2 36 2 3 51 .
3
The critical point on the graph is 2, 51 .
c) Apply the Second Derivative test to the
critical points.
For x 2
f ” x 18 x
f ” 2 18 2 36 0
The critical point 2, 45 is a relative
maximum.
For x 2
f ” x 18 x
f ” 2 18 2 36 0
The critical point 2, 51 is a relative
minimum.
We use the critical values x 2 and x 2
to divide the real line into three intervals,
A: , 2 , B: 2, 2 , and C: 2, , we
know from the extrema above, that f x is
increasing over the interval , 2 ,
decreasing over the interval 2, 2 and then
increasing again over the interval 2, .
d) Find the points of inflection. f ” x exists
12.
for all real numbers, so we solve the
equation
f ” x 0
f x 3x 3 36 x 3
a) First, find f ‘ x and f ” x .
18 x 0
f ‘ x 9 x 2 36
x0
Therefore, a possible inflection point occurs
at x 0 .
The domain of f is .
f 0 3 0 36 0 3 3 .
f ” x 18 x
3
b) f ‘ x exists for all real numbers. Solve:
f ‘ x 0
This gives the point 0, 3 on the graph.
e) To determine concavity, we use the possible
inflection point to divide the real number
line into two intervals
A : , 0 and B: 0, . We test a point in
2
9 x 36 0
9 x 2 36
x2 4
x 2
There are two critical values x 2 and
x 2.
We find the function value at x 2 .
f 2 3 2 36 2 3 45 .
3
The critical point on the graph is 2, 45 .
each interval
A: Test 1: f ” 1 18 1 18 0
B: Test 1:
f ” 1 18 1 18 0
Then, f x is concave down on the interval
, 0 and concave up on the interval
0, , so 0, 3 is an inflection point.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
f)
251
The critical point 0,80 is a relative
We use the preceding information to sketch the
graph of the function. Additional function
values can also be calculated as needed.
x
f x
3
1
1
3
24
30
36
30
maximum.
We use the critical values x 6, and x 0
to divide the real line into three intervals,
A : , 6 , B: 6, 0 , and C: 0, , we
know from the extrema above, that f x is
decreasing over the interval , 6 ,
increasing over the interval 6, 0 and then
decreasing again over the interval 0, .
13.
d) Find the points of inflection. f ” x exists
f x 80 9 x 2 x 3
for all real numbers, so we solve the
equation
f ” x 0
a) First, find f ‘ x and f ” x .
f ‘ x 18 x 3x 2
f ” x 18 6 x
18 6 x 0
The domain of f is .
x 3
Therefore, a possible inflection point occurs
at x 3 .
b) Next, find the critical points of f x .
Since f ‘ x exists for all real numbers x, the
f 3 80 9 3 3 26 .
2
only critical points occur when f ‘ x 0 .
A possible inflection point on the graph is
the point 3, 26 .
f ‘ x 0
18 x 3x 2 0
e) To determine concavity, we use the possible
inflection point to divide the real number
line into two intervals
A : , 3 and B: 3, . We test a point
3 x x 6 0
3 x 0 or x 6 0
x 0 or
x 6
There are two critical values x 6 and
x0.
We find the function value at x 6 .
in each interval
A: Test 4 : f ” 4 18 6 4 6 0
B: Test 0 : f ” 0 18 6 0 18 0
Then, f x is concave up on the interval
f 6 80 9 6 6
2
3
, 3 and concave down on the interval
3, , , so 3, 26 is an inflection point.
.
80 324 216
28
The critical point on the graph is 6, 28 .
f) We use the preceding information to sketch
the graph of the function. Additional
function values are also calculated.
Next, we find the function value at x 0.
f 0 80 9 0 0 80 .
2
3
The critical point on the graph is 0,80 .
x
c) We Apply the Second Derivative test to the
critical points.
For x 6
f ” x 18 6 x
f x
9
4
2
0
2
3
80
0
52
80
36
28
f ” 6 18 6 6 18 0
The critical point 6, 28 is a relative
3
minimum.
For x 0
f ” x 18 6 x
f ” 0 18 6 0 18 0
Copyright © 2016 Pearson Education, Inc.
252
14.
Chapter 2: Applications of Differentiation
8 3
1
x 2x
3
3
a) Find f ‘ x and f ” x .
1
2
f ” x 16 x
f x
For x
f ‘ x 8×2 2
1
1
f ” 16 8 0
2
2
The domain of f is .
1 1
The critical point , is a relative
2 3
minimum.
Therefore, f x is increasing over the
f ” x 16 x
b) f ‘ x exists for all real numbers. Solve:
f ‘ x 0
8×2 2 0
1
interval , , decreasing over the
2
8×2 2
x2
1
4
x
1 1
interval , and then increasing again
2 2
1
2
There are two critical values x
x
1
and
2
1
.
2
We find the function value at x
1
.
2
3
1
1
The critical point on the graph is ,1 .
2
1
.
2
3
1 8 1
1 1
f 2
2 32
2 3
1
3
d) We find the points of inflection. f ” x
exists for all real numbers, so we solve the
equation
f ” x 0
16 x 0
1 8 1
1 1
f 2
2 3 2
2 3
Next, we find the function value at x
1
over the interval , .
2
.
1 1
The critical point on the graph is , .
2 3
c) Apply the Second Derivative test to the
critical points.
1
For x
2
f ” x 16 x
x0
Therefore, a possible inflection point occurs
at x 0 .
8 3
1 1
f 0 0 2 0 .
3
3 3
1
This gives the point 0, on the graph.
3
e) To determine concavity, we use the possible
inflection point to divide the real number
line into two intervals A : , 0 and
B: 0, . We test a point in each interval
A: Test 1: f ” 1 16 1 16 0
B: Test 1:
f ” 1 16 1 16 0
Then, f x is concave down on the interval
, 0 and concave up on the interval
1
3
0, , so 0, is an inflection point.
1
1
f ” 16 8 0
2
2
1
The critical point ,1 is a relative
2
maximum.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
253
We use the critical values x 1 and x 1
to divide the real line into three intervals,
A : , 1 , B: 1,1 , and C: 1, , we
f) We use the preceding information to sketch
the graph of the function. Additional
function values can also be calculated as
needed.
x
f x
decreasing over the interval , 1 ,
17
13
2
1
1
2
15.
know from the extrema above, that f x is
increasing over the interval 1,1 and then
decreasing again over the interval 1, .
1
d) Find the points of inflection. f ” x exists
53
3
for all real numbers, so we solve the
equation
f ” x 0
f x x 3 3x 2
a) First, find f ‘ x and f ” x .
6 x 0
f ‘ x 3x 2 3
x0
Therefore, a possible inflection point occurs
at x 0 .
f ” x 6 x
The domain of f is .
f 0 0 3 0 2 2 .
b) Find the critical points of f x . Since
3
f ‘ x exists for all real numbers x, the only
A possible inflection point on the graph is
the point 0, 2 .
critical points occur when f ‘ x 0 .
f) To determine concavity, we use the possible
inflection point to divide the real number
line into two intervals
A : , 0 and B: 0, .
f ‘ x 0
3 x 2 3 0
3 x 2 1 0
We test a point in each interval
A: Test 1: f ” 1 6 1 6 0
2
x 1 0
B: Test 1: f ” 1 6 1 6 0
x 1
There are two critical values x 1 and
x 1.
We find the function value at x 1 .
Then, f x is concave up on the interval
, 0 and concave down on the interval
0, , so 0, 2 is an inflection point.
f 1 1 3 1 2 4 .
3
e) We use the preceding information to sketch
the graph of the function. Additional
function values can also be calculated as
needed.
x
f x
The critical point on the graph is 1, 4 .
We find the function value at x 1 .
f 1 1 3 1 2 0 .
3
The critical point on the graph is 1, 0 .
3
2
2
3
c) We apply the Second Derivative test to the
critical points.
For x 1
f ” x 6 x
f ” 1 6 1 6 0
The critical point 1, 4 is a relative
minimum.
For x 1
f ” x 6 x
f ” 1 6 1 6 0
16.
16
0
4
20
f x x 3 3x 2 4
a) First, find f ‘ x and f ” x .
f ‘ x 3 x 2 6 x
f ” x 6 x 6
The critical point 1, 0 is a relative
The domain of f is .
maximum.
Copyright © 2016 Pearson Education, Inc.
254
Chapter 2: Applications of Differentiation
b) f ‘ x exists for all real numbers. Solve:
line into two intervals A : ,1 and
f ‘ x 0
B: 1, . We test a point in each interval
A: Test 0 : f ” 0 6 0 6 6 0
3x 2 6 x 0
B: Test 2 : f ” 2 6 2 6 6 0
3x x 2 0
3x 0
or
Then, f x is concave up on the interval
x20
,1 and concave down on the interval
1, , so 1, 2 is an inflection point.
x0
or
x2
There are two critical values x 0 and
x 2.
We find the function value at x 0 .
f) We use the preceding information to sketch
the graph of the function. Additional
function values can also be calculated as
needed.
x
f x
f 0 0 3 0 4 4
3
2
The critical point on the graph is 0, 4 .
Next, we find the function value at x 2 .
2
1
3
f 2 2 3 2 4 0 .
3
2
The critical point on the graph is 2, 0 .
c) Apply the Second Derivative test to the
critical points.
For x 0
f ” x 6 x 6
f ” 0 6 0 6 6 0
The critical point 0, 4 is a relative
minimum.
For x 2
f ” x 6 x 6
16
0
4
20
4
17.
f x 3x 4 16 x 3 18 x 2
a) First, find f ‘ x and f ” x .
f ‘ x 12 x 3 48 x 2 36 x
f ” x 36 x 2 96 x 36
The domain of f is .
f ” 2 6 2 6 6 0
The critical point 2, 0 is a relative
maximum.
Therefore, f x is decreasing over the
interval , 0 , increasing over the interval
0, 2 and then decreasing again over the
interval 2, .
d) Find the points of inflection. f ” x exists
for all real numbers, so we solve the
equation
f ” x 0
6 x 6 0
b) Next, find the critical points of f x .
Since f ‘ x exists for all real numbers x, the
only critical points occur when f ‘ x 0 .
f ‘ x 0
3
2
12 x 48 x 36 x 0
12 x x 2 4 x 3 0
12 x x 1 x 3 0
12 x 0 or x 1 0 or x 3 0
x 0 or
x 1 or
x3
There are three critical values x 0 , x 1 ,
and x 3 .
Then
f 0 3 0 16 0 18 0 0
4
x 1
Therefore, a possible inflection point occurs
at x 1 .
f 1 1 3 1 4
3
2
1 3 4
.
2
This gives the point 1, 2 on the graph.
3
2
f 1 3 1 16 1 18 1 5
4
3
2
f 3 3 3 16 3 18 3 27
4
3
2
Thus, the critical points 0, 0 , 1, 5 , and
3, 27 are on the graph.
e) To determine concavity, we use the possible
inflection point to divide the real number
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
255
c) Apply the Second Derivative test to the
critical points.
A: Test 0 :
f ” 0 36 0 96 0 36 36 0
2
f ” 0 36 0 96 0 36 36 0
2
B: Test 1:
The critical point 0, 0 is a relative
f ” 1 36 1 96 1 36 24 0
2
minimum.
C: Test 3:
f ” 1 36 1 96 1 36 24 0
2
f ” 3 36 3 96 3 36 72 0
2
The critical point 1, 5 is a relative
Then, f x is concave up on the interval
maximum.
, 0.451 and concave down on the
interval 0.451, 2.215 and concave up on the
interval 2.215, , so 0.451, 2.321 and
2.215, 13.358 are inflection points.
f ” 3 36 3 96 3 36 72 0
2
The critical point 3, 27 is a relative
minimum.
We use the critical values 0,1, and 3 to
divide the real line into four intervals,
A : , 0 , B: 0,1 , C: 1, 3 and D: 3, ,
we know from the extrema above, that
f x is decreasing over the intervals
f) We use the preceding information to sketch
the graph of the function. Additional
function values can also be calculated as
needed.
x
f x
, 0 and 1, 3 and f x increasing over
the intervals 0,1 and 3, .
d) Find the points of inflection. f ” x exists
1
2
4
37
8
32
for all real numbers, so we solve the
equation f ” x 0 .
f ” x 0
2
18.
36 x 96 x 36 0
12 3x 2 8 x 3 0
f x 3x 4 4 x 3 12 x 2 5
a)
3x 2 8 x 3 0
Using the quadratic formula, we find that
4 7
, so x 0.451 or x 2.215 are
x
3
possible inflection points.
f 0.451 2.321
f 2.215 13.358
f ‘ x 12 x 3 12 x 2 24 x
f ” x 36 x 2 24 x 24
The domain of f is .
b) f ‘ x exists for all real numbers. Solve
f ‘ x 0
12 x 3 12 x 2 24 x 0
12 x x 2 x 2 0
So, 0.451, 2.321 and 2.215, 13.358 are
two more points on the graph.
e) To determine concavity, we use the possible
inflection point to divide the real number
line into three intervals A : , 0.451 ,
B: 0.451, 2.215 , and C: 2.215, .
We test a point in each interval to determine
the sign of the second derivative.
12 x x 2 x 1 0
12 x 0 or x 2 0
x 1 0
or
x 0 or
x 2 or
x 1
There are three critical values x 2 ,
x 0 , and x 1 . Then
f 2 3 2 4 2 12 2 5
4
3
2
27
f 0 3 0 4 0 12 0 5 5
4
3
2
f 1 3 1 4 1 12 1 5 0
4
3
2
Thus, the critical points 2, 27 , 0,5 ,
and 1, 0 are on the graph.
Copyright © 2016 Pearson Education, Inc.
256
Chapter 2: Applications of Differentiation
f) We use the preceding information to sketch
the graph of the function. Additional
function values can also be calculated as
needed.
c) Apply the Second Derivative test to the
critical points.
f ” 2 36 2 24 2 24 72 0
2
The critical point 2, 27 is a relative
minimum.
f ” 0 36 0 24 0 24 24 0
2
The critical point 0,5 is a relative
maximum.
x
f x
3
1
2
32
8
37
f ” 1 36 1 24 1 24 36 0
2
The critical point 1, 0 is a relative
minimum.
Then f x is decreasing over the intervals
, 2 and 0,1 and f x increasing
over the intervals 2, 0 and 1, .
d) Find the points of inflection. f ” x exists
for all real numbers, so we solve the
equation
f ” x 0
36 x 2 24 x 24 0
Using the quadratic formula, we find that
1 7
, so x 1.215 or x 0.549 are
x
3
possible inflection points.
f 1.215 13.358
f 0.549 2.321
So, 1.215, 13.358 and 0.549, 2.321 are
two more points on the graph.
e) To determine concavity, we use the possible
inflection points to divide the real number
line into three intervals A : , 1.215 ,
B: 1.215, 0.549 ,and C: 0.549, .
19.
f x x4 6×2
a) First, find f ‘ x and f ” x .
f ‘ x 4 x 3 12 x
f ” x 12 x 2 12
The domain of f is .
b) Find the critical points of f x . Since
f ‘ x exists for all real numbers x, the only
critical points occur when f ‘ x 0 .
f ‘ x 0
3
4 x 12 x 0
4x 0
or
x0
or
4 x x2 3 0
C: Test 1: f ” 1 36 0
Then, f x is concave up on the interval
, 1.215 and concave down on the
interval 1.215, 0.549 and concave up on
the interval 0.549, , so
1.215, 13.358 and 0.549, 2.321 are
inflection points.
x 3
There are three critical values 3, 0, and
3.
Then
f 3 3
We test a point in each interval
A: Test 2 : f ” 2 72 0
B: Test 0 : f ” 0 24 0
x2 3 0
6 3
4
2
9 6 3
9
f 0 0 6 0 0
4
f
2
3 3 6 3
4
2
9 6 3
9
Thus, the critical points 3, 9 , 0, 0 ,
3, 9 and are on the graph.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
257
A: Test 2 :
c) Apply the Second Derivative test to the
critical points.
f ” 3 12 3
f ” 2 12 2 12 36 0
2
12
2
B: Test 0 :
12 3 12 24 0
f ” 0 12 0 12 12 0
2
The critical point 3, 9 is a relative
C: Test 2:
f ” 2 12 2 12 36 0
2
minimum.
f ” 0 12 0 12 12 0
2
Then, f x is concave up on the intervals
The critical point 0, 0 is a relative
, 1 and 1, and concave down on
the interval 1,1 , so 1, 5 and 1, 5
maximum.
f ”
3 12 3 12
2
are inflection points.
f) We use the preceding information to sketch
the graph of the function. Additional
function values can also be calculated as
needed.
12 3 12 24 0
The critical point
3, 9 is a relative
minimum.
If we use the critical values 3, 0, and 3
to divide the real line into four intervals,
A : , 3 , B: 3, 0 , C: 0, 3 ,
and D:
3,
Then f x is decreasing over the intervals
, 3 and 0, 3 , and f x
20.
increasing over the intervals
3,0 and 3, .
for all real numbers, so we solve the
equation
f ” x 0
12 x 2 12 0
f x
3
2
2
3
27
8
8
27
f x 2 x2 x4
a)
d) Find the points of inflection. f ” x exists
x
f ‘ x 4 x 4 x3
f ” x 4 12 x 2
The domain of f is .
b) f ‘ x exists for all real numbers. Solve
f ‘ x 0
4 x 4 x3 0
x2 1 0
4 x 1 x2 0
x2 1
x 1
So x 1 or x 1 are possible inflection
points.
f 1 1 6 1 1 6 5
4
2
f 1 1 6 1 1 6 5
4
2
.
So, 1, 5 and 1, 5 are two more points
on the graph.
e) To determine concavity, we use the possible
inflection point to divide the real number
line into three intervals A : , 1 ,
B: 1,1 , and C: 1, . We test a point
in each interval
4 x 1 x 1 x 0
4x 0
or 1 x 0
or 1 x 0
x 0 or
x 1 or
x 1
There are three critical values x 1 , x 0 ,
and x 1 .
Then
f 1 2 1 1 1
2
4
f 0 2 0 0 0
2
4
f 1 2 1 1 1
2
4
Thus, the critical points 1,1 , 0, 0 , and
1,1 are on the graph.
Copyright © 2016 Pearson Education, Inc.
258
Chapter 2: Applications of Differentiation
Therefore, f x is concave down on the
c) Apply the Second Derivative test to the
critical points.
1
1
, and
intervals ,
and
3
3
f ” 1 4 12 1 8 0
2
The critical point 1,1 is a relative
f x is concave up on the interval
maximum.
1 5
1 1
1 5
,
, and
,
, so
3 9
3 3
3 9
are inflection points.
f) We use the preceding information to sketch
the graph of the function. Additional
function values can also be calculated as
needed.
x
f x
f ” 0 4 12 0 4 0
2
The critical point 0, 0 is a relative
minimum.
f ” 1 4 12 1 8 0
2
The critical point 1,1 is a relative
maximum.
Then f x is increasing over the intervals
for all real numbers, so we solve the
equation
f ” x 0
4 12 x 2 0
1
x2
3
1 5
1 5
, and
,
are two more
So,
3 9
3 9
points on the graph.
e) To determine concavity, we use the possible
inflection points to divide the real number
1
line into three intervals A : ,
,
3
1 1
1
B:
,
, and C:
, .
3
3 3
We test a point in each interval
A: Test 1: f ” 1 8 0
C: Test 1: f ” 1 8 0
21.
f x x3 6 x2 9 x 1
a) First, find f ‘ x and f ” x .
f ‘ x 3x 2 12 x 9
1
1
x
3
3
1
1
there exists
So at x
and x
3
3
possible inflection points.
1 5
f
3 9
.
1 5
f
3 9
B: Test 0 : f ” 0 4 0
8
1
0
1
8
2
1
0
1
2
, 1 and 0,1 , and f x is decreasing
over the intervals 1, 0 and 1, .
d) Find the points of inflection. f ” x exists
f ” x 6 x 12
The domain of f is .
b) f ‘ x exists for all values of x, so the only
critical points of f are where f ‘ x 0 .
f ‘ x 0
2
3 x 12 x 9 0
3 x2 4 x 3 0
3 x 1 x 3 0
x 1 0 or x 3 0
x 1 or
x3
The critical values are 1 and 3 .
Then,
f 1 1 6 1 9 1 1 5
3
2
f 3 3 6 3 9 3 1 1
3
2
So, 1, 5 and 3,1 are on the graph.
c) Applying the Second Derivative Test, we
have:
f ” 1 6 1 12 6 0
f ” 3 6 3 12 6 0
Therefore, 1, 5 is a relative maximum and
3,1 is a relative minimum.
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
259
If we use the points 1 and 3 to divide the real
number line into three intervals,
,1 , 1,3 , and 3, . Using the
extrema, we know f x is increasing on the
intervals ,1 and 3, , and f x is
decreasing on the interval 1, 3 .
d) Find the points of inflection. f ” x exists
for all real numbers. Solve
f ” x 0
6 x 12 0
6 x 12
f 2 2 6 2 9 2 1 3
2
Thus, the point 2, 3 on the graph is a
possible inflection point.
e) To determine concavity, we use 2 to divide
the real number line into two intervals,
A: , 2 and B: 2, . Then test a point
in each interval.
A: Test 0, f ” 0 6 0 12 12 0
B: Test 3, f ” 3 6 3 12 6 0
Thus, f x is concave down on the interval
, 2 and concave up on the interval
2, and 2, 3 is an inflection point.
f) We sketch the graph. Additional points may
be found as necessary.
22.
x
f x
2
1
0
4
5
49
15
1
5
21
2
2 2
2
2
f 2 4 3
3 3
3
3
8 8 8
3
27 9 3
121
27
f 2 2 2 2 4 2 3 5
2
2 121
The critical points ,
and 2, 5
3 27
are on the graph.
c) Applying the Second Derivative Test, we
have:
2
2
f ” 6 4 4 4 8 0
3
3
2 121
,
is a relative maximum.
3 27
f ” 2 6 2 4 12 4 8 0
2, 5 is a relative minimum.
2
Therefore f x is increasing on ,
3
and on 2, and f x is decreasing on
2
, 2 .
3
inflection points occur when f ” x 0 .
6x 4 0
4 2
x
6 3
There is a possible inflection point at x
3
2
2 2
2
2
f 2 4 3
3 3
3
3
f ” x 6 x 4
The domain of f is .
b) f ‘ x exists for all real numbers. Solve
3x 2 x 2 0
3
2
and 2 .
3
for all values of f x , so the only possible
f ‘ x 3x 2 4 x 4
3x 2 4 x 4 0
The critical values are
d) Find the points of inflection. f ” x exists
f x x3 2 x2 4 x 3
a)
or x 2 0
2
x
or x 2
3
3
x2
There is a possible inflection point at x 2 .
3
3x 2 0
8 8 8
3
27 9 3
7
27
7
2
Another point on the graph is , .
3 27
Copyright © 2016 Pearson Education, Inc.
2
.
3
260
Chapter 2: Applications of Differentiation
2
to divide
3
the real number line into two intervals,
2
2
A : , and B: , . Then test a
3
3
point in each interval.
A: Test 0, f ” 0 6 0 4 4 0
e) To determine concavity we use
B: Test 1, f ” 1 6 1 4 2 0
2
We see that f is concave down on ,
3
7
2
2
and concave up on , , so , is
3
3 27
an inflection point.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
23.
x
f x
2
1
0
1
3
4
5
4
3
2
0
19
2
The Second Derivative Test fails, we will
have to use the First Derivative Test for
3
x 0 . Divide , into two intervals,
2
3
A: ,0 and B: 0, , and test a point
2
in each interval.
A: Test 1,
f ‘ 1 4 1 6 1 10 0
3
2
B: Test 1,
f ‘ 1 4 1 6 1 2 0
Since, f is decreasing on both intervals,
0, 0 is not a relative extremum.
3
2
2
3
3
3
f ” 12 12 9 0
2
2
2
3 27
Therefore, , is a relative minimum.
2 16
a) First, find f ‘ x and f ” x .
Thus, f x is increasing on the interval
f ‘ x 4 x3 6 x2
3
, , and f x is decreasing on the
2
f ” x 12 x 2 12 x
The domain of f is .
b) f ‘ x exists for all values of x, so the only
critical points of f are where f ‘ x 0 .
f ‘ x 0
4 x3 6 x2 0
12 x 0
The critical values are 0 and
3
.
2
f 0 0 2 0 0
3
4
d) Find the points of inflection. f ” x exists
for all real numbers. Solve
f ” x 0
12 x x 1 0
or 2x 3 0
3
or
x
2
4
3
intervals ,0 and 0, .
2
12 x 2 12 x 0
2 x 2 2 x 3 0
x0
f ” 0 12 0 12 0 0
We use the Second Derivative Test for
3
x .
2
f x x4 2 x3
2 x2 0
3 27
So, 0, 0 and , are on the graph.
2 16
c) Applying the Second Derivative Test, we
have:
3
or
x 0 or
x 1
There are a possible inflection points at
x 0 and x 1 .
f 0 0
found earlier
f 1 1 2 1 1
4
27
3 3
3
f 2
2 2
2
16
x 1 0
3
Thus, the points 0, 0 and 1, 1 on the
graph are possible inflection points.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
261
e) Use 0 and 1 to divide the real number line
into two intervals, A: , 0 , B: 0,1 ,
and C: 1, . Then test a point in each
interval.
A: Test 1,
f ” 1 12 1 12 1 24 0
2
1
,
2
B: Test
2
1
1
1
f ” 12 12 3 0
2
2
2
C: Test 2,
f ” 2 12 2 12 2 24 0
2
Thus, f x is concave up on the intervals
, 0 and 1, and concave down on the
interval 0,1 . Also, the points 0, 0 and
1, 1 are inflection points.
f) Using the preceding information, we sketch
the graph. Additional points may be found
as necessary.
24.
x
f x
2
1
2
3
32
3
0
27
f ” 1 36 1 24 1 36 24
2
12 0
So 1, 1 is a relative minimum.
f ” 0 36 0 24 0 0
2
The test fails. We will use the First
Derivative Test. We use 0 to divide the
interval 1, into two intervals,
A: 1, 0 and B: 0, , and test a point in
each interval.
1
A: Test ,
2
3
1
1
1
f ‘ 12 12
2
2
2
2
3
0
2
B: Test 2,
f ‘ 2 12 2 12 2 144 0
f is increasing on both intervals
1, 0 and 0, . Therefore, 0, 0 is not a
3
2
relative extremum. Since 1, 1 is a
relative minimum, we know that f is
decreasing on , 1 .
d) Find the points of inflection. f ” x exists
for all values of x, so the only possible
inflection points occur when f ” x 0 .
f x 3x 4 4 x 3
a)
c) Applying the Second Derivative Test, we
have:
f ” x 0
f ‘ x 12 x 3 12 x 2
36 x 2 24 x 0
12 x 3x 2 0
f ” x 36 x 2 24 x
12 x 0
The domain of f is .
b) f ‘ x exists for all real numbers. Solve
12 x 3 12 x 2 0
12 x 2 x 1 0
12 x 2 0
or x 1 0
or
3x 2 0
2
3
There are a possible inflection points at
2
x and x 0 .
3
x0
x
or
4
3
x 0 or x 1
The critical values are 1 and 0 .
16
2
2
2
f 3 4
3
3
3
27
f 1 3 1 4 1 1
f 0 3 0 4 0 0
f 0 3 0 4 0 0
2 16
This gives one additional point ,
3 27
on the graph.
4
4
3
3
The critical points 1, 1 and 0, 0 are on
the graph.
4
Copyright © 2016 Pearson Education, Inc.
3
262
Chapter 2: Applications of Differentiation
2
and 0 to
3
divide the real number line into three
2
2
intervals, A : , , B: ,0 , and
3
3
C: 0, . Then test a point in each interval.
e) To determine concavity we use
A: Test 1,
f ” 1 36 1 24 1 12 0
2
1
B: Test ,
2
2
1
1
1
f ” 36 24
2
2
2
3 0
C: Test 1,
f ” 1 36 1 24 1 60 0
2
We see that f is concave up on the intervals
2
, and 0, , and concave down on
3
2
2 16
the interval , 0 , so both ,
3
3 27
and 0, 0 are inflection points.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
2
1
2
25.
16
7
80
3
2
125 150 675
400
f 9 9 6 9 135 9
3
2
729 486 1215
972
The critical points 5, 400 and 9, 972
are on the graph.
c) Applying the Second Derivative Test, we
have:
f ” 5 6 5 12 30 12
42 0
The critical point 5, 400 is a relative
maximum.
f ” 9 6 9 12 54 12
42 0
The critical point 9, 972 is a relative
minimum.
If we use the points 5 and 9 to divide the
real number line into three intervals
, 5 , 5, 9 , and 9, we see that
f x is increasing on the intervals
, 5 and 9, and f x is
decreasing on the interval 5, 9 .
d) Find the points of inflection. f ” x exists
set the second derivative equal to zero and
find possible inflection points.
6 x 12 0
6 x 12
x2
The only possible inflection point is 2 .
f ‘ x 3x 2 12 x 135
f ” x 6 x 12
f 2 2 6 2 135 2
3
The domain of f is .
b) f ‘ x exists for all values of x, so the only
critical points of f are where f ‘ x 0 .
3x 2 12 x 135 0
2
8 24 270
286
The point 2, 286 is a possible inflection
point on the graph.
e) To determine concavity we use 2 to divide
the real number line into two intervals,
A : , 2 and B: 2, , Then test a point
x 2 4 x 45 0
x 9 x 5 0
x9 0
f 5 5 6 5 135 5
for all values of x, so the only possible
inflection points occur when f ” x 0 . We
f x x 3 6 x 2 135 x
a)
The function values are
or x 5 0
x9
or
x 5
The critical values are 5 and 9 .
in each interval at the top of the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
263
Therefore, f x is increasing on the
Using the information from the previous
page, we test a point in each interval.
A: Test 0, f ” 0 6 0 12 12 0
intervals , 6 and 8, and f x is
decreasing on the interval 6,8 .
B: Test 3, f ” 3 6 3 12 6 0
We see that f is concave down on the
interval , 2 and concave up on the
d) Find the inflection points. f ” x exists for
all real numbers. Solve:
f ” x 0
interval 2, , Therefore 2, 286 is an
6x 6 0
inflection point.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
The possible inflection point 1, 286 is on
the graph.
e) To determine concavity, use 1 to divide the
real number line into two intervals,
A : ,1 and B: 1, and test a point in
572
250
324
0
286
700
0
400
11
10
3
0
2
5
15
16
26.
x 1
f 1 286
each interval.
A: Test 0, f ” 0 6 0 6 6 0
B: Test 2, f ” 2 6 2 6 6 0
Therefore, f x is concave down on the
interval ,1 and concave up on the
f ‘ x x 3 3x 2 144 x 140
a)
interval 1, . The point 1, 286 is an
f ‘ x 3x 6 x 144
2
inflection point.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
f ” x 6 x 6
The domain of f is .
b) f ‘ x exists for all real numbers. Solve:
f ‘ x 0
12
10
3
0
4
14
15
2
3 x 6 x 144 0
x 2 2 x 48 0
x 6 x 8 0
x6 0
or x 8 0
x 6
or
f 6 400
x8
Critical
Values
f 8 972
So the critical points 6, 400 and
8, 972 are on the graph.
c) Using the Second Derivative Test, we have:
f ” 6 42 0
27.
572
0
238
140
700
0
400
f x x 4 4 x 3 10
a)
f ‘ x 4 x 3 12 x 2
f ” x 12 x 2 24 x
The domain of f is .
f ” 8 42 0
So, 6, 400 is a relative maximum and
8, 972 is a relative minimum.
Copyright © 2016 Pearson Education, Inc.
264
Chapter 2: Applications of Differentiation
b) f ‘ x exists for all values of x, so the only
Possible points of inflection occur at
x 0 and x 2.
critical points of f are where f ‘ x 0 .
f 0 0 4 0 10 10
4
4 x 3 12 x 2 0
f 2 2 4 2 10 6
4
4 x 2 x 3 0
4 x2 0
inflection points on the graph.
e) To determine concavity we use 0 and 2 to
divide the real number line into three
intervals,
A : , 0 , B: 0, 2 , and C: 2, , Then
x0
or
x3
The critical values are 0 and 3 .
f 0 0 4 0 10 10
3
f 3 3 4 3 10 17
4
3
The points 0,10 and 2, 6 are possible
or x 3 0
4
3
3
test a point in each interval.
A: Test 1,
The critical points 0,10 and 3, 17 are on
the graph.
c) Applying the Second Derivative Test, we
have:
f ” 1 12 1 24 1 36 0
2
B: Test 1,
f ” 0 12 0 24 0 0
2
f ” 1 12 1 24 1 12 0
2
The test fails, we will use the First
Derivative Test.
Divide , 3 into two intervals,
C: Test 3,
f ” 3 12 3 24 3 36 0
We see that f is concave up on the intervals
, 0 and 2, and concave down on
2
A: , 0 and B: 0, 3 , and test a point in
each interval.
A: Test 1,
the interval 0, 2 . Therefore both 0,10
and 2, 6 are inflection points.
f ‘ 1 4 1 12 1 16 0
3
2
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
B: Test 1, f ‘ 1 4 1 12 1 8 0
3
2
Since, f is decreasing on both intervals,
0,10 is not a relative extremum.
We use the Second Derivative Test for
x3.
2
1
1
4
5
f ” 3 12 3 24 3 36 0
2
The critical point 3, 17 is a relative
minimum.
When we applied the First Derivative Test,
we saw that f x was decreasing on the
intervals , 0 and 0, 3 . Since 3, 17
is a relative minimum, we know that f x
is increasing on 3, .
d) Find the points of inflection. f ” x exists
for all values of x, so the only possible
inflection points occur when f ” x 0 .
12 x 2 24 x 0
12 x x 2 0
12 x 0 or x 2 0
x 0 or
x2
28.
58
15
7
10
135
4 3
x 2 x2 x
3
f ‘ x 4 x2 4 x 1
f x
a)
f ” x 8 x 4
The domain of f is .
b) f ‘ x exists for all real numbers. Solve:
f ‘ x 0
2
4x 4x 1 0
2 x 12 0
2x 1 0
1
x
2
Copyright © 2016 Pearson Education, Inc.
Critical Value
Exercise Set 2.2
265
Therefore, f x is concave down on the
We evaluate the function at the critical
value.
1 1
f
2 6
1
interval , and concave up on the
2
1
1 1
interval , . The point , is an
2
2 6
inflection point.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
1 1
So the critical point , is on the graph.
2 6
c) Using the Second Derivative Test, we have:
1
f ” 0 and the test fails. We will use
2
the First Derivative Test.
Divide the real number line into two
1
1
intervals, A: , and B: , , and
2
2
test a point in each interval.
A: Test 0, f ‘ 0 4 0 4 0 1 1 0
2
B: Test 1, f ‘ 1 4 1 4 1 1 1 0
2
Since, f is increasing on both intervals,
1 1
, is not a relative extremum.
2 6
However, we now know that f x is
increasing over the intervals
1
1
, and , .
2
2
d) Find the inflection points. f ” x exists for
all real numbers. Solve:
f ” x 0
29.
x
f x
6
3
1
366
57
13
3
0
1
0
3
6
21
222
1
3
f x x 3 6 x 2 12 x 6
a)
f ‘ x 3x 2 12 x 12
f ” x 6 x 12
The domain of f is .
b) f ‘ x exists for all values of x, so the only
critical points of f are where f ‘ x 0 .
8x 4 0
3x 2 12 x 12 0
1
2
1 1
f
2 6
x2 4 x 4 0
x
Dividing by 3
x 2 0
2
x20
1 1
The possible inflection point is , .
2 6
1
to divide the
2
real number line into two intervals,
1
1
A: , and B: , and test a point
2
2
in each interval.
A: Test 0, f ” 0 8 0 4 4 0
e) To determine concavity, use
B: Test 1, f ” 1 8 1 4 4 0
x2
The critical value is 2 .
f 2 2 6 2 12 2 6 2
3
2
The critical point 2, 2 is on the graph.
c) Applying the Second Derivative Test, we
have:
f ” 2 6 2 12 0
The test fails, we will use the First
Derivative Test.
Divide the real line into two intervals,
A: , 2 and B: 2, , and test a point in
each interval on the next page.
Copyright © 2016 Pearson Education, Inc.
266
Chapter 2: Applications of Differentiation
b) f ‘ x exists for all real numbers, and the
Testing a point in each interval, we have:
A: Test 0,
equation f ‘ x 0 has no real solution.
f ‘ 0 3 0 12 0 12 12 0
2
3x 2 3 0 for all x . Thus, f x has no
critical points.
c) Since f ‘ x 0 for all real numbers, f x
B: Test 3,
f ‘ 3 3 3 12 3 12 3 0
Since, f is increasing on both intervals,
2, 2 is not a relative extremum.
2
is increasing over the entire domain , .
d) Find the points of inflection. Since f ” x
When we applied the First Derivative Test,
we saw that f x was increasing on the
exists for all real numbers, solve:
f ” x 0
intervals , 2 and 2, .
6x 0
d) Find the points of inflection. f ” x exists
x0
f 0 1
for all values of x, so the only possible
inflection points occur when f ” x 0 .
6 x 12 0
6 x 12
x2
We have already seen that f 2 2 ,so
So the point 0,1 is a possible inflection
point on the graph.
e) To determine concavity, we use 0 to divide
the real number line into two intervals.
A : , 0 and B: 0, , Then test a point
the point 2, 2 is a possible inflection point
in each interval.
A: Test 1, f ” 1 6 1 6 0
on the graph.
e) To determine concavity we use 2 to divide
the real number line into two intervals,
A : , 2 and B: 2, , Then test a point
B: Test 1,
We see that f x is concave down on the
interval , 0 and concave up on the
in each interval.
A: Test 0, f ” 0 6 0 12 12 0
interval 0, . Therefore, the point 0,1 is
an inflection point.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
B: Test 3, f ” 3 6 3 12 6 0
We see that f x is concave down on the
interval , 2 and concave up on the
interval 2, . Therefore, the point 2, 2
is an inflection point.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
1
0
1
3
4
25
6
1
3
10
f ” 1 6 1 6 0
31.
x
f x
2
1
1
2
13
3
5
15
f x 5 x 3 3x 5
a)
f ‘ x 15 x 2 15 x 4
f ” x 30 x 60 x 3
30.
f x x 3 3x 1
a)
The domain of f is .
f ‘ x 3x 2 3
f ” x 6 x
The domain of f is .
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
267
b) f ‘ x exists for all values of x, so the only
critical points of f are where f ‘ x 0 .
15 x 2 15 x 4 0
decreasing on , 1 . Since 1, 2 is a
or 1 x 2 0
x0
or
x 1
The critical values are 1, 0, and 1 .
f 1 5 1 3 1 2
3
5
3
5
f 1 5 1 3 1 2
5
are on the graph.
c) Applying the Second Derivative Test, we
have:
f ” 1 30 1 60 1 30 0
3
So, the critical point 1, 2 is a relative
minimum.
3
The test fails, we will use the First
Derivative Test.
Divide 1,1 into two intervals, A: 1, 0
and B: 0,1 , and test a point in each
interval.
1
A: Test ,
2
2
1
1
1
f ‘ 15 15
2
2
2
4
45
0
16
1
,
2
2
1
1
1
f ‘ 15 15
2
2
2
4
45
0
16
Since, f is increasing on both intervals,
0, 0 is not a relative extremum.
We use the Second Derivative Test for x 1 .
f ” 1 30 1 60 1 30 0
3
The critical point 1, 2 is a relative
maximum.
d) Find the points of inflection. f ” x exists
30 x 0
or
1 2×2 0
x0
or
x2
x0
or
x
30 x 1 2 x 2 0
1
2
1
1
2
2
3
f ” 0 30 0 60 0 0
B: Test
decreasing on 1, .
30 x 60 x 3 0
The critical points 1, 2 , 0, 0 and 1, 2
relative maximum, we know that f x is
for all values of x, so the only possible
inflection points occur when f ” x 0 .
f 0 5 0 3 0 0
3
intervals 1, 0 and 0,1 . Since 1, 2 is
a relative minimum, we know that f x is
15 x 2 1 x 2 0
15 x 2 0
When we applied the First Derivative Test,
we saw that f x was increasing on the
1
1
1
f
5
3
2
2
2
5
1.237
f 0 5 0 3 0 0
3
5
3
1
1
1
f
5
3
2
2
2
5
1.237
1
, 1.237 , 0, 0 and
The points
2
1
,1.237 are possible inflection points
2
on the graph.
e) To determine concavity we use
1
1
to divide the real number
, 0, and
2
2
1
line into four intervals, A : ,
,
2
1
1
B:
,0 , C: 0,
, and
2
2
1
D:
, .
2
Then test a point in each interval on the next
page.
Copyright © 2016 Pearson Education, Inc.
268
Chapter 2: Applications of Differentiation
Testing the point in each interval, we have
f ” 1 30 1 60 1
3
A: Test 1,
30 0
3
15
0
2
3
15
0
2
3
30 0
We see that f is concave up on the intervals
1
1
,
and 0,
and concave
2
2
down on the intervals
1
1
,0 and
, . Therefore, the
2
2
1
, 1.237 , 0, 0 and
points
2
1
,1.237 are inflection points.
2
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
32.
2
56
f x 20 x 3 3x 5
a)
x 2
f ‘ x 60 x 2 15 x 4
f ” x 120 x 60 x 3
The domain of f is .
So the critical points 2, 64 , 0, 0 , and
2, 64 are on the graph.
c) Applying the Second Derivative Test, we
have:
3
f ” 1 30 1 60 1
17
32
x 0 or
f 2 64
f ” 2 120 2 60 2 240 0
D: Test 1 ,
1
2
or 4 x 2 0
f 2 64
1
1
1
f ” 30 60
2
2
2
56
17
32
15 x 2 0
f 0 0
1
C: Test ,
2
2
12
15 x 2 4 x 2 0
1
1
1
f ” 30 60
2
2
2
f ‘ x 0
60 x 2 15 x 4 0
1
B: Test ,
2
b) f ‘ x exists for all real numbers. Solve:
So, the critical point 2, 64 is a relative
minimum.
Testing the remaining critical values we
have:
f ” 0 120 0 60 0 0
3
The test fails, we will use the First
Derivative Test.
Divide 2, 2 into two intervals,
A: 2, 0 and B: 0, 2 , and test a point in
each interval.
A: Test 1,
f ‘ 1 60 1 15 1 45 0
2
4
B: Test 1,
f ‘ 1 60 1 15 1 45 0
Since, f is increasing on both intervals,
0, 0 is not a relative extremum.
We use the Second Derivative Test for
x 2.
2
4
f ” 2 120 2 60 2 240 0
3
The critical point 2, 64 is a relative
maximum.
c) When we applied the First Derivative Test,
we saw that f x was increasing on the
intervals 2, 0 and 0, 2 . Since 2, 64
is a relative minimum, we know that f x
is decreasing on , 2 . Since 2, 64 is a
relative maximum, we know that f x is
decreasing on 2, .
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
269
d) Find the inflection points. f ” x exists for
all real numbers. Solve:
120 x 60 x 3 0
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
60 x 2 x 2 0
3
1
1
3
2
60 x 0
or
2 x 0
x0
or
x2 2
x0
or
x 2
f 2 28 2
33.
f 0 0
2 28 2
The points 2 , 28 2 , 0, 0 and
2, 28 2 are possible inflection points on
f
the graph.
e) To determine concavity we use
2, 0, and 2 to divide the real number
line into four intervals,
2, .
f ” 2 120 2 60 2
3
240 0
B: Test 1, f ” 1 120 1 60 1
3
60 0
C: Test 1,
f ‘ x 2 x 6 x2 4 x3
f ” x 2 12 x 12 x 2
b) f ‘ x exists for all values of x, so the only
critical points of f are where f ‘ x 0 .
2 x 6 x 2 4 x3 0
2x 0
or 1 2 x 0
x0
or
x
or 1 x 0
1
2
or
The critical values are 0,
f 0 0 1 0 0
x 1
1
, and 1 .
2
2
2
2
1
1 1 1
f 1
2 2 2
16
2
2
3
f ” 2 120 2 60 2
a)
f 1 1 1 1 0
f ” 1 120 1 60 1
60 0
D: Test 2 ,
2
2 x 1 x 1 2 x 0
Then test a point in each interval.
A: Test 2,
f x x 2 1 x x 2 2 x 3 x 4
2 x 1 3 x 2 x 2 0
A : , 2 , B: 2, 0 , C: 0, 2 ,
and D:
189
17
17
189
3
240 0
We see that f is concave up on the intervals
, 2 and 0, 2 and concave down
on the intervals 2, 0 and 2, .
2
1 1
The critical points 0,0 , , , and
2 16
1,0 are on the graph.
c) Applying the Second Derivative Test, we
have:
f ” 0 2 12 0 12 0 2 0
2
So, the critical point 0,0 is a relative
minimum.
Therefore, the points 2 , 28 2 , 0, 0
and
2, 28 2 are inflection points.
2
1
1
1
f ” 2 12 12 1 0
2
2
2
1 1
So, the critical point , is a relative
2 16
maximum.
f ” 1 2 12 1 12 1 2 0
2
So, the critical point 1,0 is a relative
minimum. The solution is continued.
Copyright © 2016 Pearson Education, Inc.
270
Chapter 2: Applications of Differentiation
Therefore, the points 0.211,0.028 and
1
, and 1 to divide the
2
real number line into four intervals,
1
1
,0 , 0, , ,1 , and 1, , we
2
2
We use the points 0,
0.789,0.028 are inflection points.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
know that f x is decreasing on the
2
1
2
3
1
intervals ,0 and ,1 , and f x is
2
increasing on the intervals
1
0, 2 and 1, .
d) Find the points of inflection. f ” x exists
for all values of x, so the only possible
inflection points occur when f ” x 0 .
34.
f x x 2 3 x
2
1 6 x 6 x2 0
Dividing by 2
Using the quadratic formula we have:
a)
3
4
9x 6x x
f ‘ x 18 x 18 x 2 4 x 3
f ” x 18 36 x 12 x 2
3 3
x
6
x 0.211 or x 0.789
f 0.211 0.028
The domain of f is .
b) f ‘ x exists for all values of x, so the only
critical points of f are where f ‘ x 0 .
f 0.789 0.028
18 x 18 x 2 4 x 3 0
The points, 0.211,0.028 and
0.789,0.028 are possible inflection points
on the graph.
e) To determine concavity we use
0.211 and 0.789 to divide the real number
line into three intervals,
A : ,0.211 , B: 0.211, 0.789 ,
and C: 0.789,
Then test a point in each interval.
2x 0
or 3 2 x 0
x0
or
2 x 3 2 x 3 x 0
x
or 3 x 0
3
2
or
The critical values are 0,
x3
3
, and 3 .
2
2
2
2
81
3 3 3
f 3
2 2 2
16
2
20
2
2
20
We see that f is concave up on the intervals
,0.211 and 0.789, and concave
down on the interval 0.211,0.789 .
2
f ” 0 2 12 0 12 0
f ” 1 2 12 1 12 1
2 x 9 9 x 2 x2 0
f 0 0 3 0 0
1
1
1
1
B: Test , f ” 2 12 12
2
2
2
2
1 0
C: Test 1,
2
x2 9 6x x2
2 12 x 12 x 2 0
A: Test 0,
36
4
4
36
f 3 3 3 3 0
2
2
3 81
The critical points 0, 0 , , ,and
2 16
3, 0 are on the graph.
c) Applying the Second Derivative Test, we
have:
f ” 0 18 36 0 12 0 18 0
2
So, the critical point 0, 0 is a relative
minimum.
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
271
Then test a point in each interval.
A: Test 0,
Applying the Second Derivative test:
2
3
3
3
f ” 18 36 12 9 0
2
2
2
f ” 0 18 36 0 12 0
3 81
So, the critical point , is a relative
2 16
maximum.
18 0
B: Test 1,
f ” 3 18 36 3 12 3 18 0
6 0
C: Test 3,
f ” 1 18 36 1 12 1
2
So, the critical point 3, 0 is a relative
3
, and 3 to divide the
2
real number line into four intervals,
3
3
,0 , 0, , ,3 , and 3, . We
2
2
know that f x is decreasing on the
down on the interval 0.634,2.366 .
Therefore, the points 0.634,2.250 and
2.366,2.250 are inflection points.
3
intervals ,0 and ,3 , and f x is
2
increasing on the intervals
3
0, and 3, .
2
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
d) Find the points of inflection. f ” x exists
for all values of x, so the only possible
inflection points occur when f ” x 0 .
18 36 x 12 x 2 0
35.
x
f x
2
1
1
2
4
100
16
4
4
16
f x x 1 3
2
a)
f ‘ x
1
2
2
x 1 3
1
3
3 x 1 3
f 2.366 2.250
f ” x
2.366,2.250 are possible inflection points
The domain of f is .
The points, 0.634,2.250 and
on the graph.
e) To determine concavity we use
0.634and 2.366 to divide the real number
line into three intervals,
A : ,0.634 , B: 0.634,2.366 ,
and C: 2.366,
2
18 0
We see that f is concave up on the intervals
,0.634 and 2.366, and concave
We use the points 0,
3 6 x 2×2 0
Dividing by 6
Using the quadratic formula we have:
3 3
x
2
x 0.634 or x 2.366
f 0.634 2.250
2
f ” 3 18 36 3 12 3
minimum.
2
4
2
2
x 1 3
4
9
9 x 1 3
b) f ‘ x does not exist for x 1 . The equation
f ‘ x 0 has no solution, therefore, x 1 is
the only critical point.
f 1 1 1 3 0 .
2
So, the critical point, 1,0 is on the graph.
c) We apply the First Derivative Test. We use
1 to divide the real number line into two
intervals A : ,1 and B: 1, and then
we test a point in each interval at the top of
the next page.
Copyright © 2016 Pearson Education, Inc.
272
Chapter 2: Applications of Differentiation
Testing a point in each interval, we have
2
2
0
A: Test 0, f ‘ 0
1
3
3
3 0 1
B: Test 2, f ‘ 2
2
3 2 1
1
3
2
0
3
1,0 is a relative minimum. f x is
decreasing on the interval ,1 and
increasing on the interval 1, .
d) Find the points of inflection. f ” x does not
exist when x 1 . The equation f ” x 0
has no solution, so x 1 is the only possible
inflection point. We know that f 1 0 .
e) To determine concavity, we divide the real
number line into two intervals,
A : ,1 and B: 1, and then we test a
point in each interval.
A: Test 0, f ” 0
B: Test 2, f ” 2
2
9 0 1
4
3
2
9 2 1
4
3
2
0
9
2
0
9
Thus, f x is concave down on the intervals
,1 and 1, . Therefore, the point
1,0 is not an inflection point.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
7
0
2
9
36.
4
1
1
4
equation f ‘ x 0 has no solution,
therefore, x 1 is the only critical value.
f 1 1 1 3 0 .
2
So, the critical point, 1,0 , is on the
graph.
c) We apply the First Derivative Test. We use
1 to divide the real number line into two
intervals A : , 1 and B: 1, and
then we test a point in each interval.
A: Test 2,
2
2
0
f ‘ 2
1
3
3
3 2 1
B: Test 0,
f ‘ 0
2
3 0 1
1
3
2
0
3
Thus, 1,0 is a relative minimum. We also
know that f x is decreasing on the interval
, 1 and increasing on the interval
1, .
d) Find the points of inflection. f ” x does not
exist when x 1 . The equation f ” x 0
has no solution, so x 1 is the only
possible inflection point. We know that
f 1 0 .
e) To determine concavity, we divide the real
number line into two intervals, A : , 1
and B: 1, and then we test a point in
each interval.
A: Test 2,
f ” 2
2
2
2
1
f ‘ x x 1 3
1
3
3 x 1 3
2
2
4
f ” x x 1 3
4
9
9 x 1 3
The domain of f is .
2
9 2 1
4
3
2
0
9
B: Test 0,
f x x 1 3
a)
b) f ‘ x does not exist for x 1 . The
f ” 0
2
9 0 1
4
3
2
0
9
Thus, f x is concave down on the interval
, 1 and on the interval 1, .
Therefore, the point 1,0 is not an
inflection point.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
273
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
37.
x
f x
9
2
0
7
4
1
1
4
A: Test 2, f ” 2
B: Test 4, f ” 4
3
2
9 4 3
5
3
2
0
9
2
0
9
, 3 and f x is concave down on the
interval 3, . Therefore, the point 3, 1
is an inflection point.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
f x x 3 3 1
f ‘ x
9 2 3
5
Thus, f x is concave up on the interval
1
a)
2
2
1
1
x 3 3
2
3
3 x 3 3
2
2
5
f ” x x 3 3
5
9
9 x 3 3
The domain of f is .
b) f ‘ x does not exist for x 3 . The equation
f ‘ x 0 has no solution, therefore, x 3 is
the only critical value.
f 3 3 3 3 1 1 .
1
So, the critical point, 3, 1 is on the graph.
c) We apply the First Derivative Test. We use
3 to divide the real number line into two
intervals A : , 3 and B: 3, and then
we test a point in each interval.
1
1
A: Test 2, f ‘ 2
0
2
3
3
3 2 3
B: Test 4, f ‘ 4
1
3 4 3
2
3
1
0
3
f x is increasing on both intervals
, 3 and 3, , therefore 3, 1 is not a
relative extremum.
d) Find the points of inflection. f ” x does not
exist when x 3 . The equation f ” x 0
has no solution, so at x 3 is the only
possible inflection point. We know that
f 3 1 .
e) To determine concavity, we divide the real
number line into two intervals,
A : , 3 and B: 3, and then we test a
point in each interval.
38.
x
f x
5
2
4
11
3
2
0
1
f x x 2 3 3
1
a)
f ‘ x
2
1
1
x 2 3
2
3
3 x 2 3
f ” x
5
2
2
x 2 3
5
9
9 x 2 3
The domain of f is .
b) f ‘ x does not exist for x 2 . The equation
f ‘ x 0 has no solution, therefore, x 2 is
the only critical point.
f 2 2 2 3 3 3 .
1
So, the critical point, 2, 3 is on the graph.
c) We apply the First Derivative Test. We use
2 to divide the real number line into two
intervals A : , 2 and B: 2, and then
we test a point in each interval.
1
1
A: Test 1, f ‘ 1
0
2
3
3
3 1 2
B: Test 3, f ‘ 3
1
3 3 2
2
3
1
0
3
f x is increasing on both intervals
, 2 and 2, , therefore 2, 3 is not a
relative extremum.
Copyright © 2016 Pearson Education, Inc.
274
Chapter 2: Applications of Differentiation
d) Find the points of inflection. f ” x does not
A: Test 3, f ‘ 3
exist when x 2 . The equation f ” x 0
3
4
0
3
B: Test 5, f ‘ 5
determine concavity, we divide the real
number line into two intervals, A : , 2
Thus, 4, 5 is a relative maximum.
each interval.
the interval , 4 and decreasing on the
A: Test 1, f ” 1
B: Test 3, f ” 3
2
9 1 2
5
3
2
9 3 2
5
3
e) To determine concavity, we divide the real
number line into two intervals, A : , 4
and B: 4, and then we test a point in
each interval:
A: Test 3, f ” 3
B: Test 5, f ” 5
1
2
4
5
4
9 3 4 3
4
4
9 5 4
4
3
4
0
9
4
0
9
Thus, f x is concave up on both intervals
, 4 and 4, Therefore, the point
4, 5 is not an inflection point.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
f x 2 x 4 3 5
2
f ” x
4
0
3
has no solution, so x 4 is the only possible
inflection point. We know that f 4 5 .
f x
f ‘ x
exist when x 4 . The equation f ” x 0
2
0
9
an inflection point.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
6
1
3
10
3
d) Find the points of inflection. f ” x does not
, 2 and f x is concave down on the
interval 2, . Therefore, the point 2, 3 is
x
3 5 4
1
interval 4, .
2
0
9
4
We also know that f x is increasing on
Thus, f x is concave up on the interval
a)
3 3 4
1
has no solution, so x 2 is the only possible
inflection point. We know that f 2 3 . To
and B: 2, and then we test a point in
39.
4
1
4
4
x 4 3
1
3
3 x 4 3
4
3
5
12
4
4
4
x 4 3
4
9
9 x 4 3
3
3
3
3
The domain of f is .
b) f ‘ x does not exist for x 4 . The equation
f ‘ x 0 has no solution, therefore, x 4 is
the only critical point.
f 4 2 4 4 5 5 .
2
3
So, the critical point 4, 5 , is on the graph.
c) We apply the First Derivative Test. We use
4 to divide the real number line into two
intervals A : , 4 and B: 4, and then
40.
f x 3 x 2 3 3
2
a)
f ‘ x
f ” x
1
6
2
x 2 3
1
3
x 2 3
4
6
2
x 2 3
4
9
3 x 2 3
The domain of f is .
we test a point in each interval at the top of
the next column:
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
275
b) f ‘ x does not exist for x 2 . The equation
f ‘ x 0 has no solution; therefore, x 2 is
f 2 3 2 2 3 3 .
3
So, the critical point, 2, 3 is on the graph.
c) We apply the First Derivative Test. We use
2 to divide the real number line into two
intervals A : , 2 and B: 2, and then
we test a point in each interval.
2
A: Test 1, f ‘ 1
1 2 0
1 2 3
B: Test 3, f ‘ 3
2
3 2
1
3
2 0
Thus, 2, 3 is a relative maximum. We also
know that f x is increasing on the interval
, 2 and decreasing on the interval
2, .
d) Find the points of inflection. f ” x does not
exist when x 2 . The equation f ” x 0
has no solution, so x 2 is the only possible
inflection point. We know that f 2 3 .
e) To determine concavity, we divide the real
number line into two intervals,
A : , 2 and B: 2, and then we test a
point in each interval.
2
2
A: Test 1, f ” 1
0
4
3
3
3 1 2
B: Test 3, f ” 3
f x x 1 x2 x 1 x2
a)
the only critical point.
2
41.
2
3 3 2
4
3
2
0
3
Thus, f x is concave up on both intervals
, 2 and 2, Therefore, the point
2, 3 is not an inflection point.
f ‘ x x
1 x2
f x
6
1
3
10
9
0
0
9
1
2
12
1
1 x2
2 x
2
1
1
2
2 x2 1
2 x2 1
1 x 1 x
1
f ” x 2 x 1 1 x 2 x
2
1
2
2
2
2 2
3
2
1 x 4 x
2 2
1
2 x 3 3x
1 x
2
3
2
The domain of f x is 1,1 .
b) f ‘ x does not exist when x 1 .
However, the domain of f x is 1,1 .
Therefore, relative extrema cannot occur at
x 1 or x 1 because there is not an open
interval containing 1 or 1 on which the
function is defined. The other critical points
occur where f ‘ x 0 .
2 x2 1
1 x2
0
2 x2 1 0
x2
1
2
x
1
2
The critical values are
1
1
.
and
2
2
1
1
1
1
f
2
2
2
2
1
1
1
2
2
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
1 1
2 2
1
2
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
276
Chapter 2: Applications of Differentiation
f ” x 0
Evaluating the second critical value, we have:
1
1
1
f
1
2
2
2
2
2 x 3 3 x
1
1
1 2
2
3
2
0
3
2 x 3 x 0
x 2 x 2 3 0
1 1
2 2
1 x2
1
2
1 1
1
1
, and
, are
Therefore,
2 2
2 2
critical points on the graph.
c) We use the Second Derivative Test.
1
f ”
4 0
2
1 1
, is a relative
The critical point
2 2
maximum.
1
f ”
40
2
1 1
,
.
2 2
d) Find the points of inflection. f ” x does
not exist when x 1 and x 1 . However,
inflection points cannot occur at those
values because the domain of the function is
1,1 . The remaining possible inflection
x0
or
x0
or
2x 2 3 0
3
2
6
x
2
x2
6
.
2
Therefore, the only possible inflection point is
x 0 . Evaluating the function we have
f 0 0 1 0 0 .
2
Therefore, 0, 0 is a possible inflection point on
e)
the graph.
To determine concavity, we use 0 to divide the
interval 1,1 into two intervals, A: 1, 0
and B: 0,1 and then we test a point in each
interval.
1
1 10
A: Test , f ” 3 0
2 3 2
2
B: Test
1 10
f ” 3 0
2 3 2
1
,
2
Thus, f x is concave down on the interval
1 1 1
1
,
,1 ,
1,
,
, and
2
2 2
2
we see that f x is increasing on the
f x is decreasing on the interval
or
Note that f x is not defined for x
1
1
, is a relative
The critical point
2 2
minimum.
1
1
to divide
If we use the points
and
2
2
the interval 1,1 into three intervals
1
1
and
,1 and
intervals 1,
2
2
x0
1, 0 and f x is concave up on the interval
0,1 . Therefore, the point 0, 0 is an inflection
f)
point.
We sketch the graph using the preceding
information. Additional function values may
also be calculated as necessary.
x
f x
1
12
0
1
2
43
1
0
3
4
points occur when f ” x 0 . We set the
second derivative equal to zero and solve for
the possible inflection points.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
42.
277
f x x 4 x2 x 4 x2
a)
4 x2
1
1
2
12
1
4 x2
2 x
2
f ‘ x x
1
4 2×2
2
Next, we simplify the derivative.
1
x2
2 2
f ‘ x
1 4 x
2
4 x2
x2 4 x2
4 x
2
1
1
2
2 2
2
1
f ” x
0
4 2 x2 0
x 2
The critical values are 2 and 2 .
2 2 2
f 2 2 4 2 2 2 2
Therefore, 2, 2 and 2, 2 are critical
1
4 2 x2 4 x2
2
4 x 4 x
x 4 2 x
4x
4 x 4 x
4 x 2 x 4 x 4 x
4 x
2 x
3
2
2 2
1
points on the graph.
c) We use the Second Derivative Test.
12 2
f ” 2
4 2
2
2
2
3
2
3
4 x 2 x 16 x 4 x
4 x
2
3
2
3
2
2
8 2
40
2 2
minimum.
2 12 2
f ” 2
4 2
3
2
2
3
2
2 x 3 12 x
4 x
2
3
2
3
4 2 12 2
3
The critical point 2, 2 is a relative
2
2
2
1
3
2 2
2
3
2
2
4 x
4 2 x 4 x
2
4 x2
f 2 2 4 2
2
4 2 x2
For this reason, we do not consider 2 or 2
in our discussion of relative extrema.
The other critical points occur where
f ‘ x 0
4 2 12 2
2
The critical point
2
The domain of f x is 2, 2 .
b) First, we find the critical points.
f ‘ x does not exist when 4 x 2 0 .
Solve:
4 x2 0
x2 4
x 2
Since the domain of f x is 2, 2 ,
3
2
3
2
8 2
4 0
2 2
2, 2 is a relative
maximum.
If we use the points 2 and 2 to divide
the interval 2, 2 into three intervals
2, 2 , 2, 2 , and 2, 2 , we
see that f x is decreasing on the intervals
2, 2 and 2, 2 and f x is
increasing on the interval 2 , 2 .
relative extrema cannot occur at
x 2 or x 2 because there is not an open
interval containing 2 or 2 on which the
unction is defined.
Copyright © 2016 Pearson Education, Inc.
278
Chapter 2: Applications of Differentiation
d) Find the points of inflection. f ” x does
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
not exist where 4 x 2 0 . We know that
this occurs at x 2 and x 2 . However,
just as relative extrema cannot occur at
2, 0 and 2,0 , they cannot be inflection
points either. Inflection points could occur
where f ” x 0 .
x
f x
1
3
1
3
We set the second derivative equal to zero
and solve for x.
f ” x 0
2 x 3 12 x
4 x
2
3
2
0
43.
8x
x 1
f x
2
x 1 8 2 x8x Quotient
a) f ‘ x
Rule
x 1
2
2 x 3 12 x 0
2x 0
or
x2 6 0
x0
or
x2 6
x0
or
x 6
2
2
2 x x2 6 0
8 x 2 8 16 x 2
Therefore, the only possible inflection point
is x 0 .
2
Therefore, 0, 0 is a possible inflection
point on the graph.
e) To determine concavity, we use 0 to divide
the interval 2, 2 into two intervals,
f ” x
2
2
8 8×2
Note that f x is not defined for x 6 .
f 0 0 4 0 0 .
x 1
x 1
2
2
Next we find the second derivative
2
1
x 2 1 16 x 8 8 x 2 2 x 2 1 2 x
x2 1
2 2
A: 2, 0 and B: 0, 2 and then we test a
x 1 16 x x 1 4 x 8 8x
x 1
point in each interval.
A: Test 1,
f ” 1
2 1 12 1
3
4 12
3
2
10
3
32
0
B: Test 1,
f ” 1
2 1 12 1
3
10
3 0
3
32
4 12 2
Thus, f x is concave up on the interval
2
2
2
2
4
16 x3 16 x 32 x 32 x3
x 1
2
3
16 x3 48 x
x 1
3
2
The domain of f is .
b) f ‘ x exists for all real numbers. Solve:
f ‘ x 0
8 8×2
0
2, 0 and f x is concave down on the
interval 0, 2 . Therefore, the point 0, 0 is
x 1
an inflection point.
x2 1
x 1
The two critical values are x 1 and x 1 .
The solution is continued on the next page.
2
2
8 8×2 0
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
279
We will test a point in each interval.
32
0
A: Test 2, f ” 2
125
B: Test 1, f ” 1 4 0
Evaluating the function at the critical values,
we have:
8 1
8
f 1
4
2
1 1 2
f 1
8 1
1 1
2
C: Test 1, f ” 1 4 0
8
4
2
32
0
125
We see that f is concave down on the
D: Test 2, f ” 2
The critical points 1, 4 and 1, 4 are on
the graph.
c) We use the Second Derivative Test.
f ” 1 4 0
, 3 and 0, 3 and
intervals
concave up on the intervals
3,0 and 3, . Therefore the points
3, 2 3 , 0, 0 , and 3, 2 3 are
So the point 1, 4 is a relative minimum.
f ” 1 4 0
So the point 1, 4 is a relative maximum.
inflection points.
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
f x is decreasing on the intervals
,1 and 1, , and f x is increasing
on the interval 1,1 .
d) Find the points of inflection. f ” x exists
x
f x
for all real numbers, so the only possible
points of inflection occur when f ” x 0 .
3
125
2
16 x 3 48 x
16
5
2
16
5
12
5
x 1
3
2
0
3
16 x 3 48 x 0
44.
16 x x 2 3 0
or x 2 3 0
16 x 0
x0
x 1 1 x 2 x Quotient
a) f ‘ x
Rule
x 1
f 3 2 3
f 0 0
3 2 3
The points 3, 2 3 , 0, 0 , and
3, 2 3 are three possible inflection
f
points on the graph.
e) To determine concavity we use 3, 0, and
f ” x
1 x2
x 1
2
3,
x2 1
2 2
2 x3 6 x
x 1
2
3
The domain of f is .
intervals, A : , 3 , B: 3, 0 ,
C: 0, 3 ,and D:
2
Next, we find the second derivative.
2
1
x 2 1 2 x 1 x 2 2 x 2 1 2 x
3 to divide the real number line into four
2
2
x 0 or
x 3
There are three possible inflection points
x 3, 0, and 3 .
x
x 1
2
2
x2 3
or
f x
Copyright © 2016 Pearson Education, Inc.
280
Chapter 2: Applications of Differentiation
b) Since f ‘ x exists for all real numbers, the
f ” x 0
only critical values are where f ‘ x 0 .
1 x2
x2 1
0
2
x2 1
3
2x 6x 0
Multiplying
by x 2 1
2
x2 1
x 1 1
The two critical values are x 1 and x 1 .
1
1 1
2
1
1
2
2x x2 3 0
2x 0
or x 2 3 0
x0
or
f 3
1
1
The critical points 1, and 1, are
2
2
on the graph.
c) We use the Second Derivative Test.
f 0
1 1
2
f ” 1
2 1 6 1
3
3
4 1
0
8 2
12 1
1
So the point 1, is a relative
2
minimum.
f ” 1
2 1 6 1
3
3
1
4
0
8
2
12 1
1
So the point 1, is a relative maximum.
2
We use 1 and 1 to divide the real number
line into three intervals , 1 , 1,1 ,
and 1, . f x is decreasing on the
intervals ,1 and 1, , and f x is
increasing on the interval 1,1 .
d) Find the points of inflection. f ” x exists
for all real numbers, so the only possible
points of inflection occur when f ” x 0 .
We find the possible points of inflection at
the top of the next column.
x2 3
x 0 or
x 3
There are three possible inflection points at
x 3, 0, and 3 .
1
2
f 1
0
3
1 x2 0
f 1
2 x3 6 x
f
3
3 1
2
0
0 1
2
3
3
3 1
2
3
4
0
0
1
3
4
3
The points 3,
, 0, 0 , and
4
3
3, 4 are three possible inflection
points on the graph.
e) To determine concavity we use
3, 0, and 3 to divide the real number
line into four intervals,
A : , 3 , B: 3, 0 , C: 0, 3 ,
and D:
3, .
We test a point in each interval.
4
A: Test 2, f ” 2
0
125
1
B: Test 1, f ” 1 0
2
1
C: Test 1, f ” 1 0
2
4
0
D: Test 2, f ” 2
125
We see that f is concave down on the
intervals
, 3 and 0, 3 and
concave up on the intervals
3,0 and 3, .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
281
Using the information from the previous
3
,
page, we determine the points 3,
4
3
0, 0 , and 3, are inflection points.
4
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
3
103
2
25
2
2
5
3
10
3
45.
2
Extended
Power Rule
2
x 1
2
8 8x 2 x2 1 2 x
2
2
2
x 1
2
2
2
4
8 x 2 8 32 x 2
x 1
3
8 24 x 2
x 1
2
2
8x 0
x0
The critical value is x 0 .
4
f 0
4
0 2 1
f ” x 0 .
8 24 x 2
x2 1
3
0
Multiplying
8 24 x 2 0
2
2
x 1 x 1 8 8x22 x
x 1
by x 2 1
d) f ” x exists for all real numbers. Solve
8x
2
Multiplying
0
and f x is increasing on the interval 0, .
Next, we find the second derivative.
f ” x
2
f x is decreasing on the interval , 0 ,
2
x 1
2
So the point 0, 4 is a relative minimum.
2
8x
c) We use the Second Derivative Test.
f ” 0 8 0
a) f ‘ x 4 1 x 1 2 x
f ‘ x 8 x x 1
x2 1
f ‘ x 0
The critical point 0, 4 is on the graph.
1
4
4 x 2 1
f x 2
x 1
b) f ‘ x exists for all real numbers Solve:
3
The domain of f is .
x2
by x2 1
3
1
3
x
1
3
There are two possible inflection points
1
1
.
and
3
3
1
4
4
4
f
3
2
1
4
3 1
1
1 3
3
3
1
1
, 3 and
, 3 are
The points
3
3
possible inflection points on the graph.
1
e) To determine concavity we use
and
3
1
to divide the real number line into three
3
1 1
1
, B:
,
intervals, A : ,
,
3
3 3
1
, . The solution is continued.
and C:
3
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282
Chapter 2: Applications of Differentiation
Then test a point in each interval.
A: Test 1, f ” 1 2 0
B: Test 0, f ” 0 8 0
1 1
,
concave up on the interval
.
3 3
1
, 3 and
Therefore the points
3
1
, 3 are inflection points.
3
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
46.
4
4
17
2
45
1
1
2
2
2
45
4
4
17
1
2
2
6 x
x 1
2
2
2
2
2
2
2
2
4
18 x 2 6
x 1
2
2
The critical point 0, 3 is on the graph.
c) We use the Second Derivative Test.
18 0 6 6
f ” 0
6 0
2
2
1
0 1
So the point 0, 3 is a relative maximum.
We use 0 to divide the real number line into
two intervals , 0 and 0, . f x is
x2 1
3
0
2
18 x 6 0
18 x 2 6
1
x2
3
1
3
There are two possible inflection points at
1
1
.
x
and
3
3
x
2
x 1 6 6 x 2 x 1 2 x
x 1
x 1 x 1 6 6 x 22 x
x 1
x0
The critical value is x 0 .
3
f 0
3
0 2 1
18 x 2 6
2
2
Multiplying
by x 2 1
for all real numbers, so the only possible
points of inflection occur when f ” x 0 .
2
2
6 x 0
d) Find the points of inflection. f ” x exists
3
3 x2 1
x 1
f ” x
0
f x is decreasing on the interval 0, .
2
2
increasing on the interval , 0 , and
a) f ‘ x 3 1 x 1 2 x
6 x x 1
f x
x 1
2
We see that f is concave down on the
1
1
, and
intervals ,
and
3
3
f x
only critical values are where f ‘ x 0 .
6 x
C: Test 1, f ” 1 2 0
x
b) Since f ‘ x exists for all real numbers, the
3
1
3
f
3 1 2
1
3
3
3 9
1
4 4
1
3
3
The solution is continued on the next page.
The domain of f is .
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
283
Evaluating the function at the second possible
inflection point, we have:
1
3
f
3 1 2
1
3
3
3 9
1
4 4
1
3
3
1 9
1 9
, and
, are
The points
3 4
3 4
possible inflection points on the graph.
e) To determine concavity we use
1
1
to divide the real number
and
3
3
line into three intervals,
1 1
1
A : ,
,
, B:
, and
3
3 3
1
C:
,
3
Then test a point in each interval.
A: Test 1, f ” 1
B: Test 0, f ” 0
C: Test 1, f ” 1
18 1 6
2
1 1
2
18 0 6
3
3
0
2
f) We sketch the graph using the preceding
information. Additional function values
may also be calculated as necessary.
x
f x
3
1
1
3
3
10
3
2
3
2
3
10
47. Answers may vary, one possible graph is:
48. Answers may vary, one possible graph is:
2
02 1
6 0
3
18 1 6
2
1 1
2
3
3
0
2
We see that f is concave up on the intervals
1
1
, and concave
,
and
3
3
49. Answers may vary, one possible graph is:
1 1
,
down on the interval
.
3 3
1 9
, and
Therefore the points
3 4
1 9
, are inflection points.
3 4
Copyright © 2016 Pearson Education, Inc.
284
Chapter 2: Applications of Differentiation
50. Answers may vary, one possible graph is:
55. Answers may vary, one possible graph is:
7y
6
5
4
3
2
1
-1
-1 0
-2
-3
-4
x
1
2
3
4
5
56. Answers may vary, one possible graph is:
3
51. Answers may vary, one possible graph is:
y
2
1
-3
-2
-1
-1
x
0
1
2
3
-2
-3
57.
R x 50 x 0.5 x 2
C x 4 x 10
P x R x C x
52. Answers may vary, one possible graph is:
50 x 0.5 x 2 4 x 10
0.5 x 2 46 x 10
We will restrict the domains of all three
functions to x 0 since a negative number of
units cannot be produced and sold.
First graph R x 50 x 0.5 x 2
R ‘ x 50 x
53. Answers may vary, one possible graph is:
R ” x 1
Since R ‘ x exists for all x 0 , the only
critical points are where R ‘ x 0 .
50 x 0
50 x
Critical Value
Find the function value at x 50 .
R 50 50 50 0.5 50
2
2500 1250
54. Answers may vary, one possible graph is:
1250
This critical point 50,1250 is on the graph.
We use the Second Derivative Test:
R ” 50 1 0
The point 50,1250 is a relative maximum.
We use 50 to divide the interval 0, into two
intervals, 0,50 and 50, . We know that R
is increasing on 0,50 and decreasing on
50, .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
285
Next, find the inflection points. Since R ” x
exists for all x 0 , and R ” x 1 , there are
Sketch the graph using the preceding
information.
no possible inflection points.
Furthermore, since R ” x 0 for all x 0 , R
is concave down over the interval 0, .
Sketch the graph using the preceding
information. The x-intercepts of R are found by
solving R x 0.
50 x 0.5 x 2 0
0.5 x 100 x 0
0.5 x 0 or 100 x 0
100 x
x 0 or
The x-intercepts are 0, 0 and 100,0 .
Next, we graph C x 4 x 10 . This is a linear
function with slope 4 and y-intercept 0,10 .
C x is increasing over the entire domain x 0
58. R x 50 x 0.5 x 2
C x 10 x 3
P x R x C x
50 x 0.5 x 2 10 x 3
2
and has no relative extrema or points of
inflection.
Finally, we graph P x 0.5 x 2 46 x 10
0.5 x 40 x 3
We will restrict the domains of all three
functions to x 0 .
First graph R x 50 x 0.5 x 2 as in Exercise
P ” x 1
57.
Next, we graph C x 10 x 3 . This is a linear
P ‘ x x 46
Since P ‘ x exists for all x 0 , the only
critical points occur when P ‘ x 0 .
x 46 0
46 x
Critical Value
Find the function value at x 46 .
P 46 0.5 46 46 46 10
2
1058 2116 10
1048
The critical point 46,1048 is on the graph.
We use the Second Derivative Test:
P ” 46 1 0
The point 46,1048 is a relative maximum.
We use 46 to divide the interval 0, into two
intervals, 0, 46 and 46, , we know that P is
increasing on 0, 46 and decreasing on 46, .
Next, find the inflection points. Since P ” x
exists for all x 0 , and P ” x 1 , there are
no possible inflection points. Furthermore, since
P ” x 0 for all x 0 , P is concave down
over the interval 0, .
function with slope 10 and y-intercept 0, 3 .
C x is increasing over the entire domain x 0
and has no relative extrema or points of
inflection.
Finally, we graph P x 0.5 x 2 40 x 3
P ‘ x x 40
P ” x 1
Since P ‘ x exists for all x 0 , the only
critical points occur when P ‘ x 0 .
x 40 0
40 x
P 40 797
Critical Value
The critical point 40, 797 is on the graph.
We use the Second Derivative Test:
P ” 40 1 0
The point 40, 797 is a relative maximum.
We know that P is increasing on 0, 40 and
decreasing on 40, .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
286
Chapter 2: Applications of Differentiation
Next, find the inflection points. Since P ” x
exists for all x 0 , and P ” x 1 , there are
no possible inflection points. Furthermore, since
P ” x 0 for all x 0 , P is concave down
over the interval 0, .
Sketch the graph using the preceding
information.
The point 16.20, 7.17 is a relative minimum.
If we use the point 16.20 to divide the domain
into two intervals, 0,16.20 and 16.20,40 ,
we know that p is decreasing on 0,16.20 and
increasing on 16.20, 40 .
Next, we find the inflection points. p ” x exists
for all real numbers, so the only possible
inflection points are where p ” x 0
78 x 480
0
75, 000
78 x 480 0
78 x 480
59.
p x
13x 3 240 x 2 2460 x 585,000
75,000
39 x 2 480 x 2460
p ‘ x
75,000
78 x 480
p ” x
75,000
Since p ‘ x exists for all real numbers, the only
p 6.15
x 6.15
13 6.15 240 6.15 2460 6.15 585, 000
3
2
75, 000
7.52
The point 6.15, 7.52 is a possible inflection
critical points are where p ‘ x 0 .
point.
To determine concavity, we use 6.15 to divide
the domain into two intervals
A : 0, 6.15 and B: 6.15,40 and test a point in
39 x 2 480 x 2460
0
75,000
A: Test 1, p ” 1
39 x 2 480 x 2460 0
Using the quadratic formula, we have:
x
b b 2 4ac
2a
480
4802 4 39 2460
2 39
480 614,160
78
x 3.89 or x 16.20
Critical values
Since the domain of the function is 0 x 40 ,
we consider only x 16.20
p 16.20
13 16.20 240 16.20 2460 16.20 585, 000
3
2
75, 000
7.17
The critical point 16.20, 7.17 is on the graph.
each interval.
78 1 480
0.005 0
75,000
78 7 480
0.00088 0
75, 000
Then p is concave down on 0, 6.15 and
B: Test 7, p ” 7
concave up on 6.15, 40 and the point
6.15, 7.52 is a point of inflection.
Sketch the graph for 0 x 40 using the
preceding information. Additional function
values may be calculated if necessary.
x
p x
0
8
12
20
24
32
40
7.8
7.42
7.25
7.25
7.57
9.15
12.46
We use the Second Derivative Test:
78 16.20 480
p ” x
0.01 0
75,000
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
60.
287
f x 0.025 x 2 0.71x 20.44
These are two possible points of inflection.
R 3.26 18.36
f ‘ x 0.05 x 0.71
R 14.48 62.69
The two points of inflection are
3.26,18.36 and 14.48, 62.69 . However,
f ” x 0.05
f ‘ x exists for all real numbers. Solve
f ‘ x 0
t 14.48 is not in the domain of this function.
The left most inflection point 3.26,18.36
0.05 x 0.71 0
0.05 x 0.71
x 14.2
implies that rate of change of the amount of
rainfall is decreasing the fastest at this point.
f 14.2 0.025 14.2 0.71 14.2 20.44
2
15.399
f ” 14.2 0.05 , so 14.2,15.399 is a relative
V ‘ r k 40r 3r
62. V r k 20r 2 r 3 ,
2
minimum. Then f x is decreasing on
V ” r k 40 6r
0,14.2 and increasing on 14.2, 30 .
V ‘ r exists for all r in 0, 20 , so the only
Next, find the points of inflection. Since
f ” x 0.05 exists for all real numbers and is
critical points occur where V ‘ r 0 .
interval 0, 30 .
40r 3r 2 0
r 40 3r 0
r 0 or 40 3r 0
r 0 or
40 3r
40
r 0 or
r
3
Using the Second Derivative Test:
V ” 0 k 40 6 0 40k 0
Sketch the graph of f x using the information
above. Additional values may be calculated as
necessary.
f x
0
5
10
15
20
25
30
20.44
17.52
15.84
15.42
16.24
18.32
21.64
40
Since V ” 0 , we know that there is a
3
0.615t 27.745
To find the inflection points we find the first
and second derivatives.
R ‘ t 0.024t 3 0.639t 2 3.404t 0.615
R ” t 0.072t 2 1.278t 3.404
Since R ” t exists for all real values, we solve
R ” t 0 By the quadratic formula.
1.278
t 3.26
k 0
40
40
V ” k 40 6 40k 0
3
3
relative maximum at x
61. The monthly rainfall is approximated by
R t 0.006t 4 0.213t 3 1.702t 2
t
k 40r 3r 2 0
always positive, f x is concave up on the
x
0 r 20
or
40
. Thus, for an
3
40
mm or 13.33 mm , the
3
maximum velocity is needed to remove the
object.
object whose radius is
63.
Observe that h is increasing for all values of
x for which g is positive and h is decreasing for
all values of x for which g is negative.
Furthermore, for all values of x for which g=0,
h has a horizontal tangent. Therefore, g=h’.
1.278 4 0.072 3.404
2 0.072
2
t 14.48
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288
64.
65.
Chapter 2: Applications of Differentiation
Observe that g is increasing for all values of
x for which h is positive and g is decreasing for
all values of x for which h is negative.
Furthermore, for all values of x for which h=0,
g has a horizontal tangent. Therefore, h g ‘ .
f x ax 2 bx c,
70. False, consider f x 3 x 3 x 2 . This
function has points of inflection at x 0 , x 1
and x 2 without any critical values between
the points of inflection.
71. True.
a0
f ‘ x 2ax b
72. False: The function does not switch concavity at
the extreme value.
Since f ‘ x exists for all real numbers, the only
73. True.
f ” x 2a
critical points occur when f ‘ x 0 . We solve:
2ax b 0
2ax b
b
x
2a
b
So the critical value will occur at x
.
2a
Applying the second derivative test, we see that
f ” x 2a 0, for a 0
f ” x 2a 0, for a 0
Therefore, a relative maximum occurs at
b
when a 0 and a relative minimum
x
2a
b
when a 0 .
occurs at x
2a
74.
The rate of change is maximized at the
points of inflection. Looking at the graph, we
estimate the points of inflection to be 75 days
after January first, and 270 days after January
first. Therefore, the number of hours of daylight
are increasing most rapidly approximately 75
days after January 1st or approximately March
16th and the number of hours of daylight are
decreasing most rapidly approximately 270 days
after January 1st or approximately September
27th.
75.
f x 4x 6x 3
Graphing the function on the calculator we
have:
2
66. The point of inflection will occur when the
second derivative is equal to zero or undefined.
The derivatives of the function are:
g x ax 3 bx 2 cx d
g ‘ x 3ax 2 2bx c
Using the minimum/maximum feature on the
calculator, we estimate a relative maximum at
0, 0 and a relative minimum at 1, 2 .
g ” x 6ax 2b.
Setting the second derivative equal to zero and
solving for x, we have:
6ax 2b 0
2b
b
x
.
6a
3a
Therefore, the point of inflection must occur at
b
x
.
3a
67. True.
68. True.
76.
f x 3x 3 2 x
2
We estimate a relative maximum at 1,1 and a
relative minimum at 0, 0 .
69. True.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.2
77.
289
f x x 2 1 x
3
80.
f x x x
Graphing the function on the calculator we
have:
We estimate a relative minimum at 0.25, 0.25 .
Using the minimum/maximum feature on the
calculator, we estimate a relative maximum at
0.4, 0.035 and a relative minimum at 0, 0 .
78.
f x x2 x 2
3
We estimate a relative maximum at 0, 0 and a
4
relative minimum at , 1.106 .
5
79.
81.
a) The cubic and quartic functions appear to be
equally good fits. The cubic function will
ease the computations, where as the quartic
function will be a little better fit to the data.
Also they quartic function declines sharply
after 2011 so the cubic function is probably
a better fit.
b) The domain of is the set of non-negative real
numbers.
c) The graphs show that the cubic function
does not have relative extrema on a
reasonable domain. The quartic function has
a relative maximum around x 10.3 .
However, judging by the data, it appears that
the percentage of households headed by
someone with a bachelor’s degree or higher
is increasing with time.
f x x 1 3 x 1 3
2
2
Graphing the function on the calculator we
have:
Using the minimum/maximum feature on the
calculator, we estimate a relative maximum at
1,1.587 and a relative minimum at
1, 1.587 .
Copyright © 2016 Pearson Education, Inc.
290
Chapter 2: Applications of Differentiation
Once the expression is in simplified form, we
set the denominator equal to zero and solve.
x x 1 x 1 0
Exercise Set 2.3
1.
2.
3.
x4
x2
The expression is in simplified form. We set the
denominator equal to zero and solve.
x2 0
x2
The vertical asymptote is the line x 2 .
x 0 or x 1 0 or x 1 0
x 0 or
x 1 or
x 1
The vertical asymptotes are the lines
x 0, x 1, and x 1 .
f x
6.
2x 3
x5
The expression is in simplified form.
The vertical asymptote is the line x 5 .
f x
f x
x 25
First, we write the function in simplified form.
5x
f x
x 5 x 5
7.
3x
2
3x
x 3 x 3
x 9
Once the expression is in simplified form, we
set the denominator equal to zero and solve.
x 3 x 3 0
x 3 0 or x 3 0
x 3 or
x 3
The vertical asymptotes are the lines
x 3 and x 3 .
5.
f x
x3
x x 1 x 1
x2
2
x 6x 8
First, we write the function in simplified form.
x2
f x
x
2 x 4
1
Dividing common
, x 2 factors
x4
Once the expression is in simplified form, we
set the denominator equal to zero and solve.
x4 0
x 4
The vertical asymptote is the line x 4 .
8.
x6
x 7x 6
First, we write the function in simplified form.
x6
1
f x
, x 6
x 6 x 1 x 1
f x
2
The vertical asymptote is the line x 1 .
9.
x3
x3 x
First, we write the function in simplified form.
x3
f x
x x2 1
f x
x 5 0 or x 5 0
x 5 or
x 5
The vertical asymptotes are the lines
x 5 and x 5 .
f x
The vertical asymptotes are the lines
x 0, x 2, and x 4 .
Once the expression is in simplified form, we
set the denominator equal to zero and solve.
x 5 x 5 0
4.
3
5x
2
x2
x 6×2 8x
First, we write the function in simplified form.
x2
x2
f x
2
x x 4 x 2
x x 6x 8
f x
10.
7
x 49
The function is in simplified form. The equation
x 2 49 0 has no real solution; therefore, the
function does not have any vertical asymptotes.
f x
2
6
x 36
The function is in simplified form. The equation
x 2 36 0 has no real solution; therefore, the
function does not have any vertical asymptotes.
f x
2
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
11.
291
6x
8x 3
To find the horizontal asymptote, we consider
lim f x . To find the limit, we will use some
f x
13.
x
4x
x 3x
To find the horizontal asymptote, we consider
lim f x . To find the limit, we will use some
f x
x
b
algebra and the fact that as x , n 0 for
ax
any positive integer n.
6x
lim f x lim
x
x 8 x 3
1
6x
Multiplying by a
lim
x
form of 1
x 8 x 3 1
x
6x
x
lim
3
x 8 x
x x
6
lim
3
x
8
x
6
as x , b 0
n
80
ax
algebra and the fact that as x ,
1
2
4x
Multiplying by a
lim 2
x
1 form of 1
x x 3 x
x2
4x
2
lim 2 x
x x
3
x2 x2
4
lim x
3
x
1 2
x
0
as x , b 0
n
1 0
ax
6 3
.
8 4
In a similar manner, it can be shown that
3
lim f x .
x
4
x
3x 2
3
3
1
lim 2
lim
.
1 60 2
x 6 x x
x
6
x
In a similar manner, it can be shown that
1
lim f x .
x
2
1
The horizontal asymptote is the line y .
2
b
0 for
ax n
any positive integer n.
4x
lim f x lim 2
x
x x 3 x
3
The horizontal asymptote is the line y .
4
3x 2
12. f x 2
6x x
Find lim f x .
2
0.
In a similar manner, it can be shown that
lim f x 0.
x
The horizontal asymptote is the line y 0 .
14.
2x
3x x 2
Find lim f x .
f x
3
x
2
2
2x
0
0.
lim 3 2 lim x
1 3 0
x 3x x
x
3
x
In a similar manner, it can be shown that
lim f x 0.
x
The horizontal asymptote is the line y 0 .
Copyright © 2016 Pearson Education, Inc.
292
15.
Chapter 2: Applications of Differentiation
In a similar manner, it can be shown that
lim f x .
2
x
To find the horizontal asymptote, we consider
lim f x . To find the limit, we will use the
f x 4
x
The function increases without bound as
x and decreases without bound as
x . Therefore, the function does not have
a horizontal asymptote.
x
fact that as x ,
b
0 for any positive
ax n
integer n.
2
lim f x lim 4
x
x
x
40
18.
as x , b 0
.
n
ax
8 x 4 5x 2
2 x3 x2
Find lim f x .
f x
x
5
5
lim 8 x
x
x
x .
lim
lim
1
x 2 x 3 x 2
x
20
2
x
In a similar manner, it can be shown that
lim f x .
4
In a similar manner, it can be shown that
lim f x 4.
8x
8×4 5×2
x
The horizontal asymptote is the line y 4 .
x
3
16. f x 5
x
Find lim f x .
The function increases without bound as
x and decreases without bound as
x . Therefore, the function does not have
a horizontal asymptote.
x
3
5 0 5.
x
In a similar manner, it can be shown that
lim f x 5.
lim 5
x
x
The horizontal asymptote is the line y 5 .
6 x3 4 x
17. f x
3x 2 x
To find the horizontal asymptote, we consider
lim f x . To find the limit, we will use some
x
algebra and the fact that as x ,
b
0 for
ax n
any positive integer .
6 x3 4 x
lim f x lim
x
x 3 x 2 x
1
6 x3 4 x x 2
lim
1
x 3 x 2 x
x2
4
6x
x
lim
1
x
3
x
4
lim 6 x
x
x
3 0
Multiplying by a
form of 1
19.
4 x 3 3x 2
x3 2 x 4
To find the horizontal asymptote, we consider
lim f x . To find the limit, we will use some
f x
x
algebra and the fact that as x ,
b
0 for
ax n
any positive integer n.
4 x3 3 x 2
lim f x lim 3
x
x x 2 x 4
1
4 x 3 x 2 x3
lim 3
1
x x 2 x 4
x3
3
2
4 2 3
x
x
lim
2
4
x
1 2 3
x
x
b
4
as x ,
0
n
1
ax
4
In a similar manner, it can be shown that
lim f x 4.
3
x
The horizontal asymptote is the line y 4 .
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
20.
293
6 x4 4 x2 7
2 x5 x 3
Find lim f x .
f x
22.
x
5x 4 2 x3 x
x5 x 3 8
Find lim f x .
f x
x
6 4
7
6 x4 4 x2 7
x x3 x5
lim
lim
1
3
x 2 x5 x 3
x
2 4 5
x
x
0
0
20
In a similar manner, it can be shown that
lim f x 0.
5 2
1
2 5
5×4 2 x3 x
x
x
x 0 0.
lim
lim
1
8
x x 5 x 3 8
x
1
1 2 5
x
x
In a similar manner, it can be shown that
lim f x 0.
x
The horizontal asymptote is the line y 0 .
x
The horizontal asymptote is the line y 0 .
21.
2 x3 4 x 1
4 x3 2 x 3
To find the horizontal asymptote, we consider
lim f x . To find the limit, we will use some
f x
x
algebra and the fact that as x ,
b
0 for
ax n
any positive integer n.
2 x3 4 x 1
lim f x lim 3
x
x 4 x 2 x 3
1
2 x 4 x 1 x3
lim 3
x 4 x 2 x 3 1
x3
4
1
2 2 3
x
x
lim
2
3
x
4 2 3
x
x
200
as x , b 0
n
400
ax
2 1
4 2
In a similar manner, it can be shown that
1
lim f x .
x
2
1
The horizontal asymptote is the line y .
2
3
23.
4
4 x 1
x
a) Intercepts. Since the numerator is the
constant 4, there are no x-intercepts. The
number 0 is not in the domain of the
function, so there are no y-intercepts.
b) Asymptotes.
Vertical. The denominator is 0 for x 0 , so
the line x 0 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
4
f ‘ x 4 x 2 2
x
8
3
f ” x 8 x 3
x
The domain of f is , 0 0, as
f x
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
x except 0, but 0 is not in the domain of the
function, so x 0 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 0 to divide the real number line into
two intervals A: , 0 and B: 0, , and
we test a point in each interval.
4
A: Test 1, f ‘ 1
4 0
12
B: Test
1, f ‘ 1
4
12
4 0
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
294
Chapter 2: Applications of Differentiation
Then f x is decreasing on both intervals.
d) Critical Points. f ‘ x exists for all values of
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 0,
x except 0, but 0 is not in the domain of the
function, so x 0 is not a critical value. The
equation f ‘ x 0 has no solution, so there
but because 0 is not in the domain of the
function, there cannot be an inflection point
at 0. The equation f ” x 0 has no
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 0 to divide the real number line into
two intervals A: , 0 and B: 0, , and
solution; therefore, there are no inflection
points.
g) Concavity. We use 0 to divide the real
number line into two intervals
A: , 0 and B: 0, , and we test a
point in each interval.
A: Test 1, f ” 1
B: Test
1, f ” 1
B: Test
8
13
8
1
3
8 0
80
Therefore, f x is concave down on
, 0 and concave up on 0, .
h) Sketch.
24.
5
5 x 1
x
a) Intercepts. Since the numerator is the
constant 5 , there are no x-intercepts. The
number 0 is not in the domain of the
function, so there are no y-intercepts.
b) Asymptotes.
Vertical. The denominator is 0 for x 0 , so
the line x 0 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
5
f ‘ x 5 x 2 2
x
10
f ” x 10 x 3 3
x
The domain of f is , 0 0, as
f x
we test a point in each interval.
5
A: Test 1, f ‘ 1
5 0
12
1, f ‘ 1
5
12
5 0
Then f x is increasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 0,
but because 0 is not in the domain of the
function, there cannot be an inflection point
at 0. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 0 to divide the real
number line into two intervals
A: , 0 and B: 0, , and we test a
point in each interval.
10
A: Test 1, f ” 1
10 0
13
B: Test
1, f ” 1
10
13
10 0
Therefore, f x is concave up on , 0
and concave down on 0, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
determined in step (b).
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
25.
295
g) Concavity. We use 5 to divide the real
number line into two intervals
A: ,5 and B: 5, , and we test a
2
1
2 x 5
x5
a) Intercepts. Since the numerator is the
constant 2 , there are no x-intercepts. To
find the y-intercepts we compute f 0 .
f x
f 0
2
0 5
point in each interval.
A: Test 4, f ” 4
2
5
2
The point 0, is the y-intercept.
5
b) Asymptotes.
Vertical. The denominator is 0 for x 5 , so
the line x 5 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
2
2
f ‘ x 2 x 5
x 52
f ” x 4 x 5
3
4
x 53
The domain of f is ,5 5, as
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
x except 5, but 5 is not in the domain of the
function, so x 5 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 5 to divide the real number line into
two intervals A: ,5 and B: 5, , and
we test a point in each interval.
2
2
A: Test 4, f ‘ 4
20
2
1
4 5
B: Test 6, f ‘ 6
2
6 5
2
2
20
1
Then f x is increasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 5,
but because 5 is not in the domain of the
function, there cannot be an inflection point
at 5. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
B: Test 6, f ” 6
4
4 5
3
4
6 5
3
4
40
1
4
4 0
1
Therefore, f x is concave up on ,5
and concave down on 5, .
h) Sketch.
26.
1
1
x 5
x5
a) Intercepts. Since the numerator is the
constant 1, there are no x-intercepts. To find
the y-intercepts we compute f 0 .
f x
f 0
1
1
0
5
5
1
The point 0, is the y-intercept.
5
b) Asymptotes.
Vertical. The denominator is 0 for x 5 , so
the line x 5 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
1
2
f ‘ x x 5
x 5 2
f ” x 2 x 5
3
2
x 53
The domain of f is ,5 5, as
determined in step (b).
Copyright © 2016 Pearson Education, Inc.
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Chapter 2: Applications of Differentiation
d) Critical Points. f ‘ x exists for all values of
x except 5, but 5 is not in the domain of the
function, so x 5 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 5 to divide the real number line into
two intervals A: ,5 and B: 5, , and
we test a point in each interval.
1
1
A: Test 4, f ‘ 4
1 0
2
1
4 5
B: Test 6, f ‘ 6
1
6 5
2
1
1 0
1
Then f x is decreasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 5,
but because 5 is not in the domain of the
function, there cannot be an inflection point
at 5. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 5 to divide the real
number line into two intervals
A: ,5 and B: 5, , and we test a
point in each interval.
2
2
A: Test 4, f ” 4
2 0
3
1
4 5
B: Test 6, f ” 6
2
2
20
1
27.
1
1
x 3
x3
a) Intercepts. Since the numerator is the
constant 1, there are no x-intercepts. To find
the y-intercepts we compute f 0 .
f x
f 0
1
0 3
1
3
1
The point 0, is the y-intercept.
3
b) Asymptotes.
Vertical. The denominator is 0 for x 3 , so
the line x 3 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
1
2
f ‘ x x 3
x 32
f ” x 2 x 3
3
2
x 33
The domain of f is , 3 3, as
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
x except 3, but 3 is not in the domain of the
function, so x 3 is not a critical value. The
equation f ‘ x 0 has no solution, so there
Therefore, f x is concave down on
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 3 to divide the real number line into
two intervals A: , 3 and B: 3, , and
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
we test a point in each interval.
1
A: Test 2, f ‘ 2
1 0
2
2 3
6 5
3
,5 and concave up on 5, .
B: Test 4, f ‘ 4
1
4 3
2
1 0
Then f x is decreasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 3,
but because 3 is not in the domain of the
function, there cannot be an inflection point
at 3. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
297
g) Concavity. We use 3 to divide the real
number line into two intervals
A: , 3 and B: 3, , and we test a
point in each interval.
A: Test 2, f ” 2
B: Test 4, f ” 4
2
2 3
3
2
4 3
3
2 0
20
Therefore, f x is concave down on
, 3 and concave up on 3, .
d) Critical Points. f ‘ x exists for all values of
x except 2 , but 2 is not in the domain of
the function, so x 2 is not a critical
value. The equation f ‘ x 0 has no
solution, so there are no critical points.
e) Increasing, decreasing, relative extrema.
We use 2 to divide the real number line
into two intervals
A: , 2 and B: 2, , and we test a
point in each interval.
A: Test 3, f ‘ 3
h) Sketch.
1
3 2
B: Test 1, f ‘ 1
2
1
1 2
2
1 0
1 0
Then f x is decreasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 2 ,
1
1
28. f x
x 2
x2
a) Intercepts. Since the numerator is the
constant 1, there are no x-intercepts. To find
the y-intercepts we compute f 0 .
f 0
1
1
0 2 2
1
The point 0, is the y-intercept.
2
b) Asymptotes.
Vertical. The denominator is 0 for x 2 ,
so the line x 2 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
1
2
f ‘ x x 2
x 2 2
f ” x 2 x 2
3
but because 2 is not in the domain of the
function, there cannot be an inflection point
at 2 . The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 2 to divide the real
number line into two intervals
A: , 2 and B: 2, , and we test a
point in each interval.
A: Test 3, f ” 3
B: Test 1, f ” 1
2
3 2
3
2
1 2
3
2 0
20
Therefore, f x is concave down on
, 2 and concave up on 2, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
2
x 2 3
The domain of f is , 2 2, as
determined in step (b).
Copyright © 2016 Pearson Education, Inc.
298
29.
Chapter 2: Applications of Differentiation
2
1
2 x 5
x 5
a) Intercepts. Since the numerator is the
constant 2 , there are no x-intercepts. To
find the y-intercepts we compute f 0 .
f x
f 0
2
0 5
f) Inflection points. f ” x does not exist at 5 ,
but because 5 is not in the domain of the
function, there cannot be an inflection point
at 5 . The equation f ” x 0 has no
2
2
5
5
solution; therefore, there are no inflection
points.
g) Concavity. We use 5 to divide the real
number line into two intervals
A: , 5 and B: 5, , and we test a
2
The point 0, is the y-intercept.
5
b) Asymptotes.
Vertical. The denominator is 0 for x 5 , so
the line x 5 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
2
2
f ‘ x 2 x 5
x 52
f ” x 4 x 5
3
point in each interval.
A: Test 6, f ” 6
B: Test 4, f ” 4
4
6 5
3
4
4 5
3
40
4 0
Therefore, f x is concave up on , 5
and concave down on 5, .
h) Sketch.
4
x 5 3
The domain of f is , 5 5, as
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
x except 5 , but 5 is not in the domain of
the function, so x 5 is not a critical value.
The equation f ‘ x 0 has no solution, so
there are no critical points.
e) Increasing, decreasing, relative extrema.
We use 5 to divide the real number line
into two intervals
A: , 5 and B: 5, , and we test a
point in each interval.
A: Test 6, f ‘ 6
2
6 5
B: Test 4, f ‘ 4
20
2
2
4 5
2
20
Then f x is increasing on both intervals.
Since there are no critical points, there are
no relative extrema.
30.
3
1
3 x 3
x3
a) Intercepts. Since the numerator is the
constant 3 , there are no x-intercepts. To
find the y-intercepts we compute f 0 .
f x
f 0
3
3
1
0 3 3
The point 0,1 is the y-intercept.
b) Asymptotes.
Vertical. The denominator is 0 for x 3 , so
the line x 3 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
299
c) Derivatives and Domain.
f ‘ x 3 x 3
2
f ” x 6 x 3
3
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
3
x 32
6
x 33
The domain of f is , 3 3, as
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
x except 3, but 3 is not in the domain of the
function, so x 3 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 3 to divide the real number line into
two intervals A: , 3 and B: 3, , and
we test a point in each interval.
3
A: Test 2, f ‘ 2
3 0
2
2 3
B: Test 4, f ‘ 4
3
4 3
2
3 0
Then f x is increasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 3,
but because 3 is not in the domain of the
function, there cannot be an inflection point
at 3. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 3 to divide the real
number line into two intervals
A: , 3 and B: 3, , and we test a
point in each interval.
A: Test 2, f ” 2
B: Test 4, f ” 4
6
2 3
3
6
4 3
3
60
6 0
Therefore, f x is concave up on , 3
and concave down on 3, .
31.
2x 1
x
a) Intercepts. To find the x-intercepts, solve
f x 0 .
f x
2x 1
0
x
2x 1 0
1
2
This value does not make the denominator
1
0; therefore, the x-intercept is , 0 .
2
x
The number 0 is not in the domain of f x
so there are no y-intercepts.
b) Asymptotes.
Vertical. The denominator is 0 for x 0 , so
the line x 0 is a vertical asymptote.
Horizontal. The numerator and the
denominator have the same degree, so
2
y , or y 2 is the horizontal asymptote.
1
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
1
f ‘ x 2
x
2
f ” x 2 x 3 3
x
The domain of f is , 0 0, as
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
x except 0, but 0 is not in the domain of the
function, so x 0 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
Copyright © 2016 Pearson Education, Inc.
300
Chapter 2: Applications of Differentiation
e) Increasing, decreasing, relative extrema.
We use 0 to divide the real number line into
two intervals A: , 0 and B: 0, , and
we test a point in each interval.
1
1 0
A: Test 1, f ‘ 1
12
B: Test 1, f ‘ 1
1
12
1 0
Then f x is decreasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 0,
but because 0 is not in the domain of the
function, there cannot be an inflection point
at 0. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 0 to divide the real
number line into two intervals
A: , 0 and B: 0, , and we test a
point in each interval.
A: Test 1, f ” 1
B: Test 1, f ” 1
2
1
3
2
1
3
2 0
20
Therefore, f x is concave down on
, 0 and concave up on 0, .
h) Sketch.
1
does not make the denominator
3
1
0, the x-intercept is ,0 .
3
The number 0 is not in the domain of f x
so there are no y-intercepts.
b) Asymptotes.
Vertical. The denominator is 0 for x 0 , so
the line x 0 is a vertical asymptote.
Horizontal. The numerator and the
denominator have the same degree, so
3
y , or y 3 is the horizontal asymptote.
1
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
1
f ‘ x 2
x
2
f ” x 2 x 3 3
x
The domain of f is , 0 0, as
Since x
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
x except 0, but 0 is not in the domain of the
function, so x 0 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 0 to divide the real number line into
two intervals A: , 0 and B: 0, , and
we test a point in each interval.
1
A: Test 1, f ‘ 1
1 0
12
B: Test 1, f ‘ 1
32.
3x 1
x
a) Intercepts. To find the x-intercepts, solve
f x 0 .
f x
3x 1
0
x
3x 1 0
3x 1
x
1
3
1
12
1 0
Then f x is increasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 0,
but because 0 is not in the domain of the
function, there cannot be an inflection point
at 0. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
301
g) Concavity. We use 0 to divide the real
number line into two intervals
A: , 0 and B: 0, , and we test a
point in each interval.
A: Test 1, f ” 1
B: Test 1, f ” 1
2
13
2
13
20
d) Critical Points. f ‘ x exists for all values of
x except 0, but 0 is not in the domain of the
function, so x 0 is not a critical value. The
critical points will occur when f ‘ x 0 .
1
2 0
Therefore, f x is concave up on , 0
and concave down on 0, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
2
0
x2
2
1 2
x
2
x 2
x 2
Thus, 2 and 2 are critical values.
2 2 2 , so the
critical points 2, 2 2 and 2, 2 2
f 2 2 2 and f
are on the graph.
e) Increasing, decreasing, relative extrema.
We use 2 , 0, and 2 to divide the real
number line into four intervals
B: 2,0 , C: 0, 2 ,
and D: 2, .
A: , 2
33.
2 x2 2
x
x
a) Intercepts. The equation f x 0 has no
f x x
real solutions, so there are no x-intercepts.
The number 0 is not in the domain of f x
so there are no y-intercepts.
b) Asymptotes.
Vertical. The denominator is 0 for x 0 , so
the line x 0 is a vertical asymptote.
Horizontal. The degree of the numerator is
greater than the degree of the denominator,
so there are no horizontal asymptotes.
Slant. The degree of the numerator is exactly
one greater than the degree of the
denominator. As x approaches ,
2
approaches x. Therefore,
x
y x is the slant asymptote.
c) Derivatives and Domain.
2
f ‘ x 1 2 x 2 1 2
x
4
f ” x 4 x 3 3
x
The domain of f is , 0 0, as
f x x
determined in step (b).
2
A: Test 2, f ‘ 2 1
2
B: Test 1, f ‘ 1 1
C: Test 1, f ‘ 1 1
D: Test 2, f ‘ 2 1
2
12
2
2
2
2
2
12
1
0
2
1 0
1 0
1
0
2
Then f x is increasing on
, 2 and 2, and is decreasing on
2, 0 and 0, 2 . Therefore,
2, 2 2 is a relative maximum, and
2, 2 2 is a relative minimum.
f) Inflection points. f ” x does not exist at 0,
but because 0 is not in the domain of the
function, there cannot be an inflection point
at 0. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
Copyright © 2016 Pearson Education, Inc.
302
Chapter 2: Applications of Differentiation
g) Concavity. We use 0 to divide the real
number line into two intervals
A: , 0 and B: 0, , and we test a
point in each interval.
A: Test 1, f ” 1
B: Test 1, f ” 1
4
13
4
13
4 0
40
Therefore, f x is concave down on
, 0 and concave up on 0, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
d) Critical Points. f ‘ x exists for all values of
x except 0, but 0 is not in the domain of the
function, so x 0 is not a critical value. The
solution to f ‘ x 0 is x 3 . Thus,
3and 3 are critical values.
f 3 6 and f 3 6 , so the critical
points 3, 6 and 3, 6 are on the graph.
e) Increasing, decreasing, relative extrema.
We use 3, 0, and 3 to divide the real
number line into four intervals
A: , 3 B: 3,0 , C: 0, 3 ,
and D: 3, .
7
0
16
B: Test 1, f ‘ 1 8 0
A: Test 4, f ‘ 4
C: Test 1, f ‘ 1 8 0
7
0
16
Then f x is increasing on
D: Test 4, f ‘ 4
34.
9
x
a) Intercepts. The equation f x 0 has no
f x x
real solutions, so there are no x-intercepts.
The number 0 is not in the domain of f x
so there are no y-intercepts.
b) Asymptotes.
Vertical. The denominator is 0 for x 0 , so
the line x 0 is a vertical asymptote.
Horizontal. The degree of the numerator is
greater than the degree of the denominator,
so there are no horizontal asymptotes.
9
Slant. As x approaches , f x x
x
approaches x. Thus, y x is the slant
asymptote.
c) Derivatives and Domain.
9
f ‘ x 1 9 x 2 1 2
x
18
3
f ” x 18 x 3
x
The domain of f is , 0 0, as
, 3 and 3, and is decreasing on
3, 0 and 0,3 . Therefore, 3, 6 is a
relative maximum, and 3, 6 is a relative
minimum.
f) Inflection points. f ” x does not exist at 0,
but because 0 is not in the domain of the
function, there cannot be an inflection point
at 0. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 0 to divide the real
number line into two intervals
A: , 0 and B: 0, , and we test a
point in each interval.
A: Test 1, f ” 1 18 0
B: Test 1, f ” 1 18 0
Therefore, f x is concave down on
, 0 and concave up on 0, .
determined in step (b).
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
303
Then f x is decreasing on , 0 and is
h) Sketch.
increasing on 0, . Since there are no
critical points, there are no relative extrema.
f) Inflection points. f ” x does not exist at 0,
but because 0 is not in the domain of the
function, there cannot be an inflection point
at 0. The equation f ” x 0 has no
35.
1
x 2
x2
a) Intercepts. Since the numerator is the
constant 1 , there are no x-intercepts. The
number 0 is not in the domain of the
function, so there are no y-intercepts.
b) Asymptotes.
Vertical. The denominator is 0 for x 0 , so
the line x 0 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
2
f ‘ x 2 x 3 3
x
6
4
f ” x 6 x 4
x
The domain of f is , 0 0, as
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
solution; therefore, there are no inflection
points.
g) Concavity. We use 0 to divide the real
number line into two intervals
A: , 0 and B: 0, , and we test a
f x
x except 0, but 0 is not in the domain of the
function, so x 0 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 0 to divide the real number line into
two intervals A: , 0 and B: 0, , and
we test a point in each interval.
2
A: Test 1, f ‘ 1
2 0
13
B: Test
1, f ‘ 1
2
13
20
point in each interval.
A: Test 1, f ” 1
B: Test
1, f ” 1
6
14
6
14
6 0
6 0
Therefore, f x is concave down on both
intervals.
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
36.
2
2 x 2
x2
a) Intercepts. Since the numerator is the
constant 2, there are no x-intercepts. The
number 0 is not in the domain of the
function, so there are no y-intercepts.
b) Asymptotes.
Vertical. The denominator is 0 for x 0 , so
the line x 0 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
f x
Copyright © 2016 Pearson Education, Inc.
304
Chapter 2: Applications of Differentiation
c) Derivatives and Domain.
4
f ‘ x 4 x 3 3
x
12
4
f ” x 12 x 4
x
The domain of f is , 0 0, as
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
x except 0, but 0 is not in the domain of the
function, so x 0 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 0 to divide the real number line into
two intervals A: , 0 and B: 0, , and
we test a point in each interval.
A: Test 1, f ‘ 1 4 0
B: Test
1, f ‘ 1 4 0
Then f x is increasing on , 0 and is
decreasing on 0, . Since there are no
critical points, there are no relative extrema.
f) Inflection points. f ” x does not exist at 0,
but because 0 is not in the domain of the
function, there cannot be an inflection point
at 0. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 0 to divide the real
number line into two intervals
A: , 0 and B: 0, , and we test a
point in each interval.
A: Test 1, f ” 1
B: Test
1, f ” 1
12
1
12
14
4
12 0
12 0
Therefore, f x is concave up on both
intervals.
h) Sketch.
37.
x
x3
a) Intercepts. To find the x-intercepts, solve
f x 0 .
f x
x
0
x3
x0
Since x 0 does not make the denominator
0, the x-intercept is 0, 0 . f 0 0 , so the
y-intercept is 0, 0 also.
b) Asymptotes.
Vertical. The denominator is 0 for x 3 , so
the line x 3 is a vertical asymptote.
Horizontal. The numerator and the
denominator have the same degree, so
1
y , or y 1 is the horizontal asymptote.
1
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
3
f ‘ x
x 32
f ” x 6 x 3
3
6
x 33
The domain of f is , 3 3, as
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
x except 3, but 3 is not in the domain of the
function, so x 3 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 3 to divide the real number line into
two intervals A: , 3 and B: 3, , and
we test a point in each interval.
3
A: Test 2, f ‘ 2
3 0
2
2 3
B: Test 4, f ‘ 4
3
4 3
2
3 0
Then f x is decreasing on both intervals.
Since there are no critical points, there are
no relative extrema.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
305
f) Inflection points. f ” x does not exist at 3,
but because 3 is not in the domain of the
function, there cannot be an inflection point
at 3. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 3 to divide the real
number line into two intervals
A: , 3 and B: 3, , and we test a
point in each interval.
A: Test 2, f ” 2
B: Test 4, f ” 4
6
2 3
3
6
6 0
60
3
4 3
Therefore, f x is concave down on
, 3 and concave up on 3, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
c) Derivatives and Domain.
2
f ‘ x
x 2 2
f ” x 4 x 2
3
4
x 23
The domain of f is , 2 2, as
determined in step (b).
d) Critical Points. f ‘ x exists for all values of
x except 2 , but 2 is not in the domain of
the function, so x 2 is not a critical
value. The equation f ‘ x 0 has no
solution, so there are no critical points.
e) Increasing, decreasing, relative extrema.
We use 2 to divide the real number line
into two intervals A: , 2 and
B: 2, , and we test a point in each
interval.
A: Test 3, f ‘ 3 2 0
B: Test 1, f ‘ 1 2 0
Then f x is increasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 2 ,
38.
x
x2
a) Intercepts. The numerator is 0 for x 0 and
since this value of x does not make the
denominator 0, the x-intercept is 0, 0 .
f x
f 0 0 , so the y-intercept is 0, 0 also.
b) Asymptotes.
Vertical. The denominator is 0 for x 2 ,
so the line x 2 is a vertical asymptote.
Horizontal. The numerator and the
denominator have the same degree, so
1
y , or y 1 is the horizontal asymptote.
1
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
but because 2 is not in the domain of the
function, there cannot be an inflection point
at 2 . The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 2 to divide the real
number line into two intervals
A: , 2 and B: 2, , and we test a
point in each interval.
A: Test 3, f ” 3 4 0
B: Test 1, f ” 1 4 0
Therefore, f x is concave up on , 2
and concave down on 2, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
Copyright © 2016 Pearson Education, Inc.
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39.
Chapter 2: Applications of Differentiation
f) Inflection points. f ” x exists for all real
1
1
x2 3
x 3
a) Intercepts. Since the numerator is the
constant 1, there are no x-intercepts.
1
1
f 0
, so the y-intercept is
2
0 3 3
f x
2
numbers. Solve f ” x 0 .
6 x2 6
x 2
6 x2 6
x2 1
x 1
1
1
f 1 and f 1
4
4
1
1
So, 1, and 1, are possible points
4
4
of inflection.
g) Concavity. We use 1 and 1 to divide the
real number line into three intervals
A: , 1 B: 1,1 , and C: 1,
18
0
343
2
B: Test 0, f ” 0 0
9
18
C: Test 2, f ” 2
0
343
Therefore, f x is concave up on , 1
A: Test 2, f ” 2
6 x2 6
f ” x
x 3
2
3
The domain of f is as determined in step
(b).
d) Critical Points. f ‘ x exists for all real
and 1, , and concave down on 1,1 .
numbers. Solve f ‘ x 0
2x
x2 3
2
1
1
Thus the points 1, and 1, are
4
4
points of inflection.
h) Sketch.
0
2x 0
x0
The critical value is 0. From step (a) we
1
found 0, is on the graph.
3
e) Increasing, decreasing, relative extrema.
We use 0 to divide the real number line into
two intervals A: , 0 and B: 0, , and
we test a point in each interval.
1
A: Test 1, f ‘ 1 0
8
1
B: Test 1, f ‘ 1 0
8
Then f x is increasing on , 0 and is
1
decreasing on 0, . Thus 0, is a
3
relative maximum.
0
6 x2 6 0
1
0, .
3
b) Asymptotes.
Vertical. x 2 3 0 has no real solution, so
there are no vertical asymptotes.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
2
2x
f ‘ x 2 x x 2 3
2
x2 3
3
2
40.
1
1
x2 2
x 2
a) Intercepts. Since the numerator is the
constant 1 , there are no x-intercepts.
1
1
f 0 , so the y-intercept is 0, .
2
2
b) Asymptotes.
Vertical. x 2 2 0 has no real solution, so
there are no vertical asymptotes.
f x
2
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
307
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
2
2x
f ‘ x 2 x x2 2
2
2
x 2
f ” x
2
2
to divide
and
3
3
the real number line into three intervals
2 2
2
A: ,
B:
,
,
3
3 3
g) Concavity. We use
2
and C:
,
3
A: Test 1, f ” 1
2
6 x 4
x 2
2
3
2
0
27
1
0
2
2
C: Test 1, f ” 1
0
27
Therefore, f x is concave down on
0, f ” 0
B: Test
The domain of f is .
d) Critical Points. f ‘ x exists for all real
numbers. f ‘ x 0 for x 0 , so 0 is a
critical value. From step (a) we already
1
know 0, is on the graph.
2
e) Increasing, decreasing, relative extrema.
We use 0 to divide the real number line into
two intervals A: , 0 and B: 0, , and
2
2
, 3 and 3 , and concave up
2 2
on ,
. Therefore the points
3 3
2 3
2 3
3 , 8 and 3 , 8 are points of
inflection.
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
we test a point in each interval.
2
A: Test 1, f ‘ 1 0
9
2
B: Test 1, f ‘ 1 0
9
Then f x is decreasing on , 0 and is
1
increasing on 0, . Thus 0, is a
2
relative minimum.
f) Inflection points. f ” x exists for all real
numbers. f ” x 0 for x
2
and
3
2
, so
3
2
are possible inflection points.
3
2
2
3
3
f
and f
8
8
3
3
2 3
2 3
, and
, are points of
So,
3 8
3 8
inflection.
41.
x3
x3
1
, x 3
x 2 9 x 3 x 3 x 3
We write the expression in simplified form
noting that the domain is restricted to all real
numbers except for x 3.
a) Intercepts. f x 0 has no solution. x 3
f x
is not in the domain of the function.
Therefore, there are no x-intercepts. To find
the y-intercepts we compute f 0 .
f 0
1
1
0 3 3
1
The point 0, is the y-intercept.
3
Copyright © 2016 Pearson Education, Inc.
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Chapter 2: Applications of Differentiation
f) Inflection points. f ” x does not exist at 3,
b) Asymptotes.
Vertical. In the original function, the
denominator is 0 for x 3 or x 3 ,
however, x 3 also made the numerator
equal to 0. We look at the limits to
determine if there are vertical asymptotes at
these points.
x3
1
1
1
lim
lim
.
x 3 x 2 9
x 3 x 3
3 3
6
Because the limit exists, the line x 3 is
not a vertical asymptote. Instead, we have a
removable discontinuity, or a “hole” at the
1
point 3, .
6
but because 3 is not in the domain of the
function, there cannot be an inflection point
at 3. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 3 and 3 to divide the
real number line into three intervals
A: , 3 B: 3, 3 and C: 3, and
we test a point in each interval.
A: Test 4, f ” 4
B: Test 2, f ” 2
1
An open circle is drawn at 3, to show
6
that it is not part of the graph.
The denominator is 0 for x 3 and the
numerator is not 0 at this value, so the line
x 3 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
1
2
f ‘ x x 3
x 32
f ” x 2 x 3
3
C: Test 4, f ” 4
2
4 3
2
2 3
3
2
4 3
3
3
2
343
0
2 0
20
Therefore, f x is concave down on
, 3 and 3, 3 and concave up on
3, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
2
x 33
The domain of f as determined in step (b) is
, 3 3, 3 3, .
d) Critical Points. f ‘ x exists for all values of
x except 3, but 3 is not in the domain of the
function, so x 3 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 3 and 3 to divide the real number
line into three intervals
A: , 3 B: 3, 3 and C: 3, .
We notice that f ‘ x 0 for all real
numbers, f x is decreasing on all three
intervals , 3 , 3, 3 , and 3, .
Since there are no critical points, there are
no relative extrema.
42.
x 1
1
, x 1
2
x 1 x 1
We write the expression in simplified form
noting that the domain is restricted to all real
numbers except for x 1.
a) Intercepts. f x 0 has no solution, so
there are no x-intercepts. To find
the y-intercepts we compute f 0
f x
f 0
1
1
0 1
The point 0,1 is the y-intercept.
b) Asymptotes.
Vertical. In the original function, the
denominator is 0 for x 1 or x 1 ,
however, x 1 also made the numerator
equal to 0.
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
309
g) Concavity. We use 1 and 1 to divide the
real number line into three intervals
A: , 1 B: 1,1 and C: 1, .
We look at the limits to determine if there
are vertical asymptotes at these points.
x 1
1
1
lim 2
lim
. Because the limit
x 1 x 1
x 1 x 1
2
exists, the line x 1 is not a vertical
asymptote. Instead, we have a removable
1
discontinuity, or a “hole” at the point 1, .
2
An open circle is drawn at this point to show
that it is not part of the graph.
The denominator is 0 for x 1 and the
numerator is not 0 at this value, so the line
x 1 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
1
2
f ‘ x x 1
x 12
f ” x 2 x 1
3
and we test a point in each interval.
A: Test 2, f ” 2 2 0
B: Test 0, f ” 0 2 0
2
0
27
Therefore, f x is concave down on
C: Test 2, f ” 2
, 1 and concave up on 1,1 and
1, .
h) Sketch.
43.
2
x 13
The domain of f is
, 1 1,1 1, as determined
above.
d) Critical Points. f ‘ x exists for all values of
x except 1 , but 1 is not in the domain of
the function, so x 1 is not a critical value.
The equation f ‘ x 0 has no solution, so
there are no critical points.
e) Increasing, decreasing, relative extrema.
We use 1 and 1 to divide the real number
line into three intervals
A: , 1 B: 1,1 and C: 1, .
We notice that f ‘ x 0 for all real
numbers, f x is decreasing on all three
intervals , 1 , 1,1 , and 1, . Since
there are no critical points, there are no
relative extrema.
f) Inflection points. f ” x does not exist at 1 ,
but because 1 is not in the domain of the
function, there cannot be an inflection point
at 1 . The equation f ” x 0 has no
x 1
x2
a) Intercepts. To find the x-intercepts, solve
f x 0 .
f x
x 1
0
x2
x 1
Since x 1 does not make the denominator
0, the x-intercept is 1, 0 .
0 1
1
, so the y-intercept is
02
2
1
0, .
2
b) Asymptotes.
Vertical. The denominator is 0 for x 2 ,
so the line x 2 is a vertical asymptote.
Horizontal. The numerator and the
denominator have the same degree, so
1
y , or y 1 is the horizontal asymptote.
1
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
f 0
solution; therefore, there are no inflection
points.
Copyright © 2016 Pearson Education, Inc.
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Chapter 2: Applications of Differentiation
c) Derivatives and Domain.
3
f ‘ x
x 2 2
f ” x
44.
x2
x 1
a) Intercepts. f x 0 for x 2 and this value
f x
does not make the denominator 0, the xintercept is 2, 0 . f 0 2 , so the y-
6
x 2
3
The domain of f is , 2 2, as
determined in part (b).
d) Critical Points. f ‘ x exists for all values of
x except 2 , but 2 is not in the domain of
the function, so x 2 is not a critical
value. The equation f ‘ x 0 has no
solution, so there are no critical points.
e) Increasing, decreasing, relative extrema.
We use 2 to divide the real number line
into two intervals
A: , 2 and B: 2, , and we test a
point in each interval.
A: Test 3, f ‘ 3 3 0
B: Test 1, f ‘ 1 3 0
Then f x is increasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 2 ,
but because 2 is not in the domain of the
function, there cannot be an inflection point
at 2 . The equation f ” x 0 has no
intercept is 0, 2 .
b) Asymptotes.
Vertical. The denominator is 0 for x 1 , so
the line x 1 is a vertical asymptote.
Horizontal. The numerator and the
denominator have the same degree, so
1
y , or y 1 is the horizontal asymptote.
1
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
3
f ‘ x
x 12
f ” x
6
x 13
The domain of f is , 1 1, as
determined in part (b).
d) Critical Points. f ‘ x exists for all values of
x except 1 , but 1 is not in the domain of
the function, so x 1 is not a critical value.
The equation f ‘ x 0 has no solution, so
solution; therefore, there are no inflection
points.
g) Concavity. We use 2 to divide the real
number line into two intervals
A: , 2 and B: 2, , and we test a
point in each interval.
A: Test 3, f ” 3 6 0
there are no critical points.
e) Increasing, decreasing, relative extrema.
We use 1 to divide the real number line
into two intervals
A: , 1 and B: 1, . We notice that
Therefore, f x is concave up on , 2
there are no critical points, there are no
relative extrema.
f) Inflection points. f ” x does not exist at 1 ,
B: Test 1, f ” 1 6 0
and concave down on 2, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
f ‘ x 0 for all real numbers; therefore,
f x is increasing on both intervals. Since
but because 1 is not in the domain of the
function, there cannot be an inflection point
at 1 . The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
311
g) Concavity. We use 1 to divide the real
number line into two intervals
A: , 1 and B: 1, , and we test a
point in each interval.
A: Test 2, f ” 2 6 0
B: Test 0, f ” 0 6 0
Therefore, f x is concave up on , 1
and concave down on 1, .
h) Sketch.
As x approaches , f x approaches
x 1 , so y x 1 is the slant asymptote.
c) Derivatives and Domain.
x2 2 x 9
f ‘ x
x 12
f ” x
16
x 13
The domain of f is , 1 1, as
determined in part (b).
d) Critical Points. f ‘ x exists for all values of
x except 1 , but 1 is not in the domain of
the function, so x 1 is not a critical value.
f ‘ x 0 has no real solution, so there are
45.
x2 9
x 1
a) Intercepts. To find the x-intercepts, solve
f x 0 .
f x
x2 9
0
x 1
x2 9 0
x 3
Neither of these values make the
denominator 0, so the x-intercepts are
3, 0 and 3, 0 .
f 0 9 , so the y-intercept is 0, 9 .
b) Asymptotes.
Vertical. The denominator is 0 for x 1 , so
the line x 1 is a vertical asymptote.
Horizontal. The degree of the numerator is
greater than the degree of the denominator,
so there are no horizontal asymptotes.
Slant. Divide the numerator by the
denominator.
1
x
2
9
x 1 x
x2 x
x9
x 1
no critical points.
e) Increasing, decreasing, relative extrema.
We use 1 to divide the real number line
into two intervals
A: , 1 and B: 1,
We test a point in each interval.
A: Test 2, f ‘ 2 9 0
B: Test
0, f ‘ 0 9 0
Then f x is increasing on both intervals.
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 1 ,
but because 1 is not in the domain of the
function, there cannot be an inflection point
at 1 . The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 1 to divide the real
number line into two intervals
A: , 1 and B: 1, , and we test a
point in each interval.
A: Test 2, f ” 2 16 0
B: Test
0, f ” 0 16 0
Therefore, f x is concave up on , 1
and concave down on 1, .
h) Sketch.
8
By dividing, we get
8
f x x 1
x 1
Copyright © 2016 Pearson Education, Inc.
312
46.
Chapter 2: Applications of Differentiation
x2 4
x3
a) Intercepts. The numerator is 0 for x 2 or
x 2 and neither of these values make the
denominator 0, so the x-intercepts are
02 4
4
, so
2, 0 and 2, 0 . f 0
03
3
4
the y-intercept is 0, .
3
b) Asymptotes.
Vertical. The denominator is 0 for x 3 , so
the line x 3 is a vertical asymptote.
Horizontal. The degree of the numerator is
greater than the degree of the denominator,
so there are no horizontal asymptotes.
Slant. By dividing the numerator by the
denominator, we get
5
f x x 3
x3
As x approaches , f x approaches
f x
x 3 , so y x 3 is the slant asymptote.
c) Derivatives and Domain.
x2 6x 4
f ‘ x
x 32
f ” x
10
x 33
The domain of f is , 3 3, as
determined in part (b).
d) Critical Points. f ‘ x exists for all values of
x except 3 , but 3 is not in the domain of
the function, so x 3 is not a critical value.
Solve f ‘ x 0 .
x2 6x 4
x 32
We test a point in each interval.
4
A: Test 6, f ‘ 6 0
9
B: Test 4, f ‘ 4 4 0
C: Test 2, f ‘ 2 4 0
4
0
9
Then f x is increasing on the intervals
D: Test
0, f ‘ 0
, 5.236 and 0.764, , and is
decreasing on the intervals
5.226, 3 and 3, 0.764 . Therefore,
5.236, 10.472 is a relative maximum
and 0.764, 1.528 is a relative minimum.
f) Inflection points. f ” x does not exist at 3 ,
but because 3 is not in the domain of the
function, there cannot be an inflection point
at 3 . The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 3 to divide the real
number line into two intervals
A: , 3 and B: 3, , and we test a
point in each interval.
A: Test 4, f ” 4 10 0
B: Test 2, f ” 2 10 0
Therefore, f x is concave down on
, 3 and concave up on 3, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
0
x2 6x 4 0
Using the
x 3 5 Quadratic
Formula
x 5.236 or x 0.764
f 5.236 10.472 and
f 0.764 1.528 , so 5.236, 10.472
and 0.764, 1.528 are on the graph.
e) Increasing, decreasing, relative extrema.
We use 5.236, 3, and 0.764 to divide
the real number line into four intervals
A: , 5.236 , B: 5.236, 3 ,
C: 3, 0.764 , and D: 0.764,
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
47.
f x
313
x3
2
x 2 x 15
x3
1
,
x 3 x 5 x 5
x 3.
We write the expression in simplified form
noting that the domain is restricted to all real
numbers except for x 5 and x 3 .
a) Intercepts. f x 0 has no solution. x 3
is not in the domain of the function.
Therefore, there are no x-intercepts. To find
the y-intercepts we compute f 0 :
f 0
03
02 2 0 15
1
5
1
The point 0, is the y-intercept.
5
b) Asymptotes.
Vertical. In the original function, the
denominator is 0 for x 5 or x 3 ;
however, x 3 also made the numerator
equal to 0.
We look at the limits to determine if there
are vertical asymptotes at these points.
x3
1
1
lim 2
lim
x 3 x 2 x 15
x 3 x 5
8
Because the limit exists, the line x 3 is not
a vertical asymptote. Instead, we have a
removable discontinuity, or a “hole” at the
1
point 3, . An open circle is drawn at this
8
point to show that it is not part of the graph.
The denominator is 0 for x 5 and the
numerator is not 0 at this value, so the line
x 5 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
1
2
f ‘ x x 5
x 5 2
f ” x 2 x 5
3
d) Critical Points. f ‘ x exists for all values of
x except 5 , but 5 is not in the domain of
the function, so x 5 is not a critical value.
The equation f ‘ x 0 has no solution, so
there are no critical points.
e) Increasing, decreasing, relative extrema.
We use 5 and 3 to divide the real number
line into three intervals A: , 5 ,
B: 5, 3 , and C: 3, . We notice that
f ‘ x 0 for all real numbers, f x is
decreasing on all three intervals
, 5 , 5, 3 , and 3, . Since there
are no critical points, there are no relative
extrema.
f) Inflection points. f ” x does not exist at 5 ,
but because 5 is not in the domain of the
function, there cannot be an inflection point
at 5 . The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 5 and 3 to divide the
real number line into three intervals
A: , 5 B: 5, 3 and C: 3, , and
we test a point in each interval.
A: Test 6, f ” 6 2 0
B: Test 4, f ” 4 2 0
2
0
729
Therefore, f x is concave down on
C: Test 4, f ” 4
, 5 and concave up on
5, 3 and 3, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
2
x 53
The domain of f as determined in part (b) is
, 5 5, 3 3, .
Copyright © 2016 Pearson Education, Inc.
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48.
Chapter 2: Applications of Differentiation
d) Critical Points. f ‘ x exists for all values of
x 1
x 1
1
,
x 2 x 3 x 3 x 1 x 3
x 1.
We write the expression in simplified form
noting that the domain is restricted to all real
numbers except for x 1 and x 3
a) Intercepts. f x 0 has no solution. x 1
f x
2
x except 3, but 3 is not in the domain of the
function, so x 3 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 1 and 3 to divide the real number
line into three intervals
A: , 1 B: 1, 3 and C: 3, .
is not in the domain of the function.
Therefore, there are no x-intercepts. To find
the y-intercepts we compute f 0 :
We notice that f ‘ x 0 for all real
0 1
numbers, so f x is decreasing on all three
1
f 0
2
0 2 0 3 3
intervals , 1 , 1, 3 , and 3, .
1
The point 0, is the y-intercept.
3
b) Asymptotes.
Vertical. In the original function, the
denominator is 0 for x 1 or x 3 ,
however, x 1 also made the numerator
equal to 0. We look at the limits to
determine if there are vertical asymptotes at
these points.
x 1
1
1
1
lim 2
lim
x 1 x 2 x 3
x 1 x 3
1 3
4
Because the limit exists, the line x 1 is
not a vertical asymptote. Instead, we have a
removable discontinuity, or a “hole” at the
1
point 1, . An open circle is drawn at
4
this point to show that it is not part of the
graph.
The denominator is 0 for x 3 and the
numerator is not 0 at this value, so the line
x 3 is a vertical asymptote.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
1
2
f ‘ x x 3
x 32
f ” x 2 x 3
3
Since there are no critical points, there are
no relative extrema.
f) Inflection points. f ” x does not exist at 3,
but because 3 is not in the domain of the
function, there cannot be an inflection point
at 3. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 1 and 3 to divide the
real number line into three intervals
A: , 1 B: 1, 3 and C: 3, , and
we test a point in each interval.
A: Test 2, f ” 2
2
125
0
B: Test 2, f ” 2 2 0
C: Test 4, f ” 4 2 0
Therefore, f x is concave down on
, 1 and 1, 3 and concave up on
3, .
h) Sketch.
2
x 33
The domain of f as determined in part (b) is
, 1 1, 3 3, .
49.
2 x2
x 2 16
a) Intercepts. The numerator is 0 for x 0 and
this value does not make the denominator 0,
so the x-intercept is 0, 0 .
f x
f 0 0 , so the y-intercept is 0, 0 also.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
315
f) Inflection points. f ” x does not exist at
b) Asymptotes.
Vertical. The denominator is 0 when
x 2 16 0
4 and 4 , but because 4 and 4 are not in
the domain of the function, there cannot be
an inflection point at 4 or 4 . The equation
x 2 16
f ” x 0 has no real solution; therefore,
x 4
So the lines x 4 and x 4 are vertical
asymptotes.
Horizontal. The numerator and the
denominator have the same degree, so
2
y , or y 2 is the horizontal asymptote.
1
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
64 x
f ‘ x
2
x 2 16
f ” x
there are no inflection points.
g) Concavity. We use 4 and 4 to divide the
real number line into three intervals
A: , 4 B: 4, 4 and C: 4, , and
we test a point in each interval.
5824
A: Test 5, f ” 5
0
729
1
B: Test 0, f ” 0 0
4
5824
0
C: Test 5, f ” 5
729
Therefore, f x is concave up on the
192 x 2 1024
intervals , 4 and 4, and concave
x 16
2
3
down on the interval 4, 4 .
The domain of f is
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
, 4 4, 4 4, as determined in
part (b).
d) Critical Points. f ‘ x exists for all values of
x except x 4 and x 4 , but 4 and 4 are
not in the domain of the function, so
x 4 and x 4 are not critical values.
f ‘ x 0 for x 0 , so 0, 0 is the only
critical point.
e) Increasing, decreasing, relative extrema.
We use 4, 0, and 4 to divide the real
number line into four intervals
A: , 4 B: 4, 0 , C: 0, 4 ,
and D: 4, .
We test a point in each interval.
320
A: Test 5, f ‘ 5
0
81
64
B: Test 1, f ‘ 1
0
225
64
C: Test 1,
f ‘ 1
0
225
320
D: Test 5, f ‘ 5
0
81
Then f x is increasing on the intervals
50.
x 2 x 2 x 2 x 1
x2
2 x 1 x 1 2 x 1
2×2 2
We write the expression in simplified form
noting that the domain is restricted to all real
numbers except for x 1.
a) Intercepts. f x 0 for x 2 ; therefore,
f x
the x-intercept is 2, 0 .
0 2 0 2 1
2
2 0 2
The point 0,1 is the y-intercept.
f 0
, 4 and 4, 0 , and is decreasing on
the intervals 0, 4 and 4, . Thus, there is
a relative maximum at 0, 0 .
Copyright © 2016 Pearson Education, Inc.
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Chapter 2: Applications of Differentiation
f) Inflection points. f ” x does not exist at 1 ,
b) Asymptotes.
Vertical. In the original function, the
denominator is 0 for x 1 or x 1 ,
however, x 1 also made the numerator
equal to 0. We look at the limits to
determine if there are vertical asymptotes at
these points.
x2 x 2
x2
3
lim
lim
. Because
x 1 2 x 2 2
x 1 2 x 1
4
the limit exists, the line x 1 is not a vertical
asymptote. Instead, we have a removable
3
discontinuity, or a “hole” at the point 1, .
4
An open circle is drawn at this point to show
that it is not part of the graph.
The denominator is 0 for x 1 and the
numerator is not 0 at this value, so the line
x 1 is a vertical asymptote.
Horizontal. The numerator and the
denominator have the same degree, so
1
y is the horizontal asymptote.
2
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
1
2
f ‘ x x 1
2
2 x 1
f ” x 2 x 1
3
1
x 1
The domain of f is , 1 1,1 1,
3
as determined in part (b).
d) Critical Points. f ‘ x exists for all values of
x except 1 , but 1 is not in the domain of
the function, so x 1 is not a critical value.
The equation f ‘ x 0 has no solution, so
there are no critical points.
e) Increasing, decreasing, relative extrema.
We use 1 and 1 to divide the real number
line into three intervals
A: , 1 B: 1,1 and C: 1, .
We notice that f ‘ x 0 for all real
numbers, f x is decreasing on all three
intervals , 1 , 1,1 , and 1, . Since
there are no critical points, there are no
relative extrema.
but because 1 is not in the domain of the
function, there cannot be an inflection point
at 1 . The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 1 and 1 to divide the
real number line into three intervals
A: , 1 B: 1,1 and C: 1, .
and we test a point in each interval.
A: Test 2, f ” 2 1 0
B: Test 0, f ” 0 1 0
1
0
27
Therefore, f x is concave down on the
C: Test 2, f ” 2
interval , 1 and concave up on the
intervals 1,1 and 1, .
h) Sketch.
51.
10
x 4
a) Intercepts. Since the numerator is the
constant 10, there are no x-intercepts.
5
5
f 0 , so the y-intercept is 0, .
2
2
b) Asymptotes.
Vertical. x 2 4 0 has no real solution, so
there are no vertical asymptotes.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
2
20 x
f ‘ x 20 x x 2 4
2
2
x 4
f x
2
f ” x
2
20 3 x 4
x 4
2
3
The domain of f is .
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
317
d) Critical Points. f ‘ x exists for all real
Therefore, f x is concave up on
numbers. f ‘ x 0 for x 0 , so 0 is a
2
2
, and concave
,
and
3
3
critical value. From step (a) we already
5
know 0, is on the graph.
2
e) Increasing, decreasing, relative extrema.
We use 0 to divide the real number line into
two intervals A: , 0 and B: 0, , and
2 2
,
down on
. Thus, the points
3 3
2 15
2 15
, and
,
are points of
3 8
3 8
inflection.
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
we test a point in each interval.
4
A: Test 1, f ‘ 1 0
5
4
B: Test 1, f ‘ 1 0
5
Then f x is increasing on , 0 and is
5
decreasing on 0, . Thus 0, is a
2
relative maximum.
f) Inflection points. f ” x exists for all real
2
, which
3
are possible points of inflection. We have:
2 15
2 15
f
and f
.
3 8
3 8
numbers. f ” x 0 for x
2 15
2 15
, and
,
are inflection
So,
3 8
3 8
points.
2
2
g) Concavity. We use
to divide
and
3
3
the real number line into three intervals
2 2
2
A: ,
B:
,
, and
3
3 3
2
C:
, .
3
5
0
16
5
0, f ” 0 0
4
5
0
2, f ” 2
16
A: Test 2, f ” 2
B: Test
C: Test
52.
1
x2 1
a) Intercepts. Since the numerator is a constant
1, there are no x-intercepts.
1
f 0
1
0 2 1
f x
The point 0, 1 is the y-intercept.
b) Asymptotes.
Vertical. The denominator
x 2 1 x 1 x 1 is 0 for
x 1 or x 1 , so the lines x 1 and x 1
are vertical asymptotes.
Horizontal. The degree of the numerator is
less than the degree of the denominator, so
y 0 is the horizontal asymptote.
Slant. There is no slant asymptote since the
degree of the numerator is not one more than
the degree of the denominator.
c) Derivatives and Domain.
2 x
f ‘ x
2
x2 1
f ” x
x 1
2
2 3x 1
2
3
The domain of f is
, 1 1,1 1, as determined in
part (b).
Copyright © 2016 Pearson Education, Inc.
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Chapter 2: Applications of Differentiation
d) Critical Points. f ‘ x exists for all values of
h) Sketch.
x except 1 and 1 , but these values are not
in the domain of the function, so x 1 and
x 1 are not critical values. f ‘ x 0 for
x 0 . From step (a) we know f 0 1 ,
so the critical point is 0, 1 .
e) Increasing, decreasing, relative extrema.
We use 1, 0, and 1 to divide the real
number line into four intervals
A: , 1 , B: 1, 0 , C: 0,1 , and
D: 1, , and we test a point in each
interval.
4
A: Test 2, f ‘ 2 0
9
1
1 16
B: Test , f ‘
0
2 9
2
1
16
1
, f ‘ 0
2
2
9
4
f ‘ 2 0
D: Test 2,
9
We see that f x is increasing on the
C: Test
intervals , 1 and 1, 0 , and is
decreasing on the intervals 0,1 and 1, .
Therefore, 0, 1 is a relative maximum.
f) Inflection points. f ” x does not exist at
1 and 1 , but because these values are not
in the domain of the function, there cannot
be an inflection point at 1 or 1 . The
equation f ” x 0 has no real solution;
therefore, there are no inflection points.
g) Concavity. We use 1 and 1 to divide the
real number line into three intervals
A: , 1 B: 1,1 and C: 1, .
and we test a point in each interval.
26
A: Test 2, f ” 2
0
27
B: Test 0, f ” 0 2 0
26
0
27
Therefore, f x is concave up on , 1
C: Test 2, f ” 2
and 1, , and concave down on 1,1 .
53.
x2 1
x
a) Intercepts. The equation f x 0 has no
f x
real solutions, so there are no x-intercepts.
The number 0 is not in the domain of f x
so there are no y-intercepts.
b) Asymptotes.
Vertical. The denominator is 0 for x 0 , so
the line x 0 is a vertical asymptote.
Horizontal. The degree of the numerator is
greater than the degree of the denominator,
so there are no horizontal asymptotes.
Slant. The degree of the numerator is exactly
one greater than the degree of the
denominator. When we divide the numerator
by the denominator we have
x2 1
1
f x
x . As x approaches
x
x
1
, f x x approaches x. Therefore,
x
y x is the slant asymptote.
c) Derivatives and Domain.
x2 1
f ‘ x
x2
2
f ” x 3
x
The domain of f is , 0 0, as
determined in part (b).
d) Critical Points. f ‘ x exists for all values of
x except 0, but 0 is not in the domain of the
function, so x 0 is not a critical value. The
critical points will occur when f ‘ x 0 .
x2 1
0
x2
x2 1 0
x2 1
x 1
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
319
From the previous page, we determined
1 and 1 are critical values, thus
f 1 2 and f 1 2 , so the critical
points 1, 2 and 1, 2 are on the graph.
e) Increasing, decreasing, relative extrema.
We use 1, 0, and 1 to divide the real
number line into four intervals
A: , 1 B: 1, 0 , C: 0,1 , and
D: 1, . We test a point in each interval.
3
f ‘ 2 0
A: Test 2,
4
1
1
B: Test , f ‘ 3 0
2
2
C: Test
1
,
2
1
f ‘ 3 0
2
3
0
4
Then f x is increasing on
D: Test
2,
f ‘ 2
, 1 and 1, and is decreasing on
1, 0 and 0,1 . Therefore, 1, 2 is a
relative maximum, and 1, 2 is a relative
minimum.
f) Inflection points. f ” x does not exist at 0,
but because 0 is not in the domain of the
function, there cannot be an inflection point
at 0. The equation f ” x 0 has no
solution; therefore, there are no inflection
points.
g) Concavity. We use 0 to divide the real
number line into two intervals
A: , 0 and B: 0, , and we test a
point in each interval.
A: Test 1, f ” 1 2 0
B: Test 1,
f ” 1 2 0
Therefore, f x is concave down on
, 0 and concave up on 0, .
h) Sketch. Use the preceding information to
sketch the graph. Compute additional
function values as needed.
54.
x3
x 1
a) Intercepts. The numerator is 0 for x 0 and
this value does not make the denominator 0,
the x-intercept is 0, 0 .
f x
2
f 0 0 , so the y-intercept is 0, 0 also.
b) Vertical. The denominator
x 2 1 x 1 x 1 is 0 for
x 1 or x 1 , so the lines x 1 and x 1
are vertical asymptotes.
Horizontal. The degree of the numerator is
greater than the degree of the denominator,
so there are no horizontal asymptotes.
Slant. The degree of the numerator is exactly
one greater than the degree of the
denominator.
When we divide the numerator by the
denominator we have
x3
x
f x 2
x 2
. As x
x 1
x 1
x
approaches , f x x 2
approaches
x 1
x. Therefore, y x is the slant asymptote.
c) Derivatives and Domain.
x 4 3x 2
f ‘ x
2
x2 1
f ” x
3
2x 6x
x 1
2
3
The domain of f as determined in part (b)
is , 1 1,1 1, .
d) Critical Points. f ‘ x exists for all values of
x except 1 and 1 , but these values are not
in the domain of the function, so x 1 and
x 1 are not a critical values. f ‘ x 0 for
x 3 , x 0, and x 3 . Therefore,
3 3
, f 0 0 , and
2
3 3
. The critical points
f 3
2
3 3
3 3
3, 2 , 0, 0 and 3, 2 are
on the graph.
f 3
Copyright © 2016 Pearson Education, Inc.
320
Chapter 2: Applications of Differentiation
e) Increasing, decreasing, relative extrema.
We use 3, 1, 0, 1, and 3 to divide the
real number line into six intervals
1
208
1
, f ”
0
2
2
27
28
D: Test 2, f ” 2
0
27
Therefore, f x is concave down on
C: Test
B: 3, 1 , C: 1,0 ,
D: 0,1 , E: 1, 3 , and F: 3, . We
A: , 3
, 0 and 0,1 and concave up on
1, 0 and 1, . Thus, the point 0, 0 is a
test a point in each interval.
4
A: Test 2,
f ‘ 2 0
9
3
27
3
B: Test , f ‘
0
2
2
25
1
C: Test ,
2
11
1
f ‘ 0
2
9
D: Test
1
,
2
11
1
f ‘ 0
2
9
E: Test
3
,
2
27
3
0
f ‘
2
25
4
0
9
Then f x is increasing on the intervals
F: Test
2,
f ‘ 2
, 3 and 3, and is decreasing on
the intervals 3, 1 , 1, 0 ,
0,1 , and 1, 3 . Therefore, 0, 0 is not a
3 3
is a
relative extremum, 3,
2
3 3
relative maximum, and 3,
is a
2
relative minimum.
f) Inflection points. f ” x does not exist at
1 and 1 , but because these values are not
in the domain of the function, there cannot
be an inflection point at 1 or 1 . f ” x 0
when x 0 . We know that f 0 0 , so
there is a possible inflection point at 0, 0 .
g) Concavity. We use 1, 0, and 1 to divide
the real number line into four intervals
A: , 1 B: 1, 0 , C: 0,1 , and
D: 1, , and we test a point in each
interval.
28
A: Test 2, f ” 2
0
27
1
1 208
B: Test , f ”
0
2 27
2
point of inflection.
h) Sketch.
55.
x 2 16 x 4 x 4
x 4, x 4
x4
x4
Notice that f x x 4 for all values of x
f x
except x 4 , where it is undefined. The
graph of f x will be the graph of
y x 4 except at the point x 4 .
a) Intercepts. f x 0 when x 4 , so the xintercept is 4, 0 .
f 0 4.
The point 0, 4 is the y-intercept.
b) Asymptotes.
In simplified form f x x 4 , a linear
function everywhere except x 4 . So
there are no asymptotes of any kind.
In the original function, the denominator is 0
for x 4 ; however, x 4 also made the
numerator equal to 0. We look at the limits
to determine if there are vertical asymptotes
at these points.
x 2 16
lim
lim x 4 8 . Because
x 4 x 4
x 4
the limit exists, the line x 4 is not a
vertical asymptote. Instead, we have a
removable discontinuity, or a “hole” at the
point 4, 8 . An open circle is drawn at
4, 8 to show that it is not part of the
graph.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
321
c) Derivatives and Domain.
f ‘ x 1, x 4
f 0
f ” x 0, x 4
d) Critical Points. There are no critical points.
e) Increasing, decreasing, relative extrema.
We use 4 to divide the real number line
into two intervals
A: , 4 and B: 4, .
We notice that f ‘ x 0 for all real
numbers in the domain, f x is increasing
on both intervals. Since there are no critical
points, there are no relative extrema.
f) Inflection points. f ” x is constant;
therefore, there are no points of inflection.
g) Concavity. f ” x is 0; therefore, there is no
concavity.
h) Sketch. Use the preceding information to
sketch the graph.
0 2 9 9
3
0 3 3
The point 0, 3 is the y-intercept.
b) Asymptotes.
In simplified form f x x 3 , a linear
function everywhere except x 3 . So there
are no asymptotes of any kind.
In the original function, the denominator is 0
for x 3 ; however, x 3 also made the
numerator equal to 0. We look at the limits
to determine if there is a vertical asymptote
at this point.
x2 9
lim
lim x 3 6 .
x3 x 3
x3
The limit exists; therefore, the line x 3 is
not a vertical asymptote. Instead, we have a
removable discontinuity, or a “hole” at the
point 3, 6 . An open circle is drawn at this
point to show that it is not part of the graph.
c) Derivatives and Domain.
f ‘ x 1, x 3
f ” x 0, x 3
The domain of f is , 3 3, as
Note: In the preceding problem, we could
have noticed that the graph of
x 2 16
is the graph of f x x 4
f x
x4
with the exception of the point 4, 8
which is a removable discontinuity. We
simply need to graph f x x 4 with a
hole at the point 4, 8 and determine all
other aspects of the graph of f x from the
linear graph.
56.
x2 9
x3
We write the expression in simplified form:
x 3 x 3 x 3, x 3
f x
x3
Note that the domain is restricted to all real
numbers except for x 3.
a) Intercepts. The numerator is 0 when x 3
or x 3 however, x 3 is not in the
domain of f x , so the x-intercept is
f x
determined in part (b).
d) Critical Points. f ‘ x exists for all values of
x except 3, but 3 is not in the domain of the
function, so x 3 is not a critical value. The
equation f ‘ x 0 has no solution, so there
are no critical points.
e) Increasing, decreasing, relative extrema.
We use 3 to divide the real number line into
two intervals A: , 3 and B: 3, .
We notice that f ‘ x 0 for all real
numbers in the domain. f x is increasing
on both intervals. Since there are no critical
points, there are no relative extrema.
f) Inflection points. f ” x is constant;
therefore, there are no points of inflection.
g) Concavity. f ” x is 0; therefore, there is no
concavity.
h) Sketch.
3, 0 .
Copyright © 2016 Pearson Education, Inc.
322
57.
Chapter 2: Applications of Differentiation
C x 3x 2 80
C x 3 x 2 80
80
3x
x
x
x
b) Using the techniques of this section we find
the following information. We will only
consider the values of x is 0, .
a) A x
Intercepts. None.
Asymptotes. x 0 is the vertical asymptote.
There is no horizontal asymptote. As x
80
approaches
x
3x. Therefore, y 3 x is the slant asymptote.
Increasing, decreasing, relative extrema.
80
A ‘ x 3 2 . A ‘ x is not defined for
x
x 0 , however that value is outside the
domain of the function. A ‘ x 0 when
approaches , A x 3x
x
80
80
, and A
2 240 .
3
3
Using x
80
to divide the interval 0,
3
80
80
and
, ,
into two intervals, 0,
3
3
and testing a point in each interval, we find
80
that A x is decreasing on 0,
and
3
80
, . Therefore, the
increasing on
3
80
, 2 240 is a relative minimum.
point
3
Inflection points, concavity.
160
A ” x 3 exists for all values of t in
x
0, . The equation A ” x 0 has no real
solution, so there are no possible points of
inflection. Furthermore, A ” x 0 for all x
in the domain, so A x is concave up on
0, .
We use this information to sketch the graph.
Additional values may be computed as
necessary.
c) The degree of the numerator is exactly one
greater than the degree of the denominator.
When we divide the numerator by the
denominator we have
3 x 2 80
80
. As x
A x
3x
x
x
80
approaches , A x 3x
approaches
x
3x. Therefore, y 3 x is the slant asymptote.
This means that when a large number of
pairs of rocket skates are produced, the
average cost can be estimated by
multiplying the number of pairs produced by
3 thousand dollars.
58. V t 50
25t 2
t 2 2
a) V 0 50
25 0
2
0 2
2
50 0 50
The inventory’s value after 0 months is $50
hundreds or $5000.
V 5 50
25 5
2
5 2
2
50
625
37.24
49
The inventory’s value after 5 months is
$37.24 hundreds or $3724.
V 10 50
25 10
2
10 2
2
50
2500
32.64
144
The inventory’s value after 10 months is
$32.64 hundreds or $3264.
V 70 50
25 70
2
70 2
2
50
122, 500
5184
26.37
The inventory’s value after 70 months is
$26.37 hundreds or $2637.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
323
b) Find V ‘ t and V ” t .
V ‘ t
Inflection points, concavity. V ” t exist for
all values of t in 0, . V ” t 0 , when
t 22 50t 25t 2 2 t 21
t 1 . We use this to split the domain into
two intervals. A: 0,1 and B: 1, .
t 2
2 2
Testing points in each interval, we see V t
t 2 t 250t 25t 2 2
t 24
V ” t
is concave down on 0,1 and concave up on
1, .
100t
We use this information to sketch the graph.
Additional values may be computed as
necessary.
t 23
t 23 100 100t 3 t 22 1
t 2
3 2
t 22 100 t 2 100t 3
t 26
200t 200
t 2
this from the horizontal asymptote on the
graph.
200t 200
t 24
59.
Solve V ‘ t 0 .
100t
t 23
Yes; The value below which V will
never fall is lim V t 25 . We observed
t
4
V ‘ t exists for all values of t in 0, .
d)
0
100t 0
t0
Since t 0 is an endpoint of the domain,
there cannot be a relative extrema at t 0 .
We notice that V ‘ t 0 for all x in the
domain, therefore, V t is decreasing over
the interval 0, . Since V t is decreasing,
the absolute maximum value of the
inventory will be $5000 when t 0 .
c) Using the techniques of this section, we find
the following additional information:
Intercepts. There are no t-intercepts in
0, . The V-intercept is the point 0,50
Asymptotes. There are no vertical
asymptotes in 0, .
The line V 25 is a horizontal asymptote.
There are no slant asymptotes.
Increasing, decreasing, relative extrema.
We have already seen that V t is
C x 5000 600 x
1 2
x 1000 x
2
a) Profit is revenue minus cost, therefore, the
total profit function is:
P x R x C x
R x
1 2
x 1000 x 5000 600 x
2
1
x 2 400 x 5000
2
P x
b) A x
x
1 2
x 400 x 5000
2
x
1
5000
x 400
2
x
c) As x approaches , A x approaches
1
1
x 400 . Therefore, y x 400 is
2
2
the slant asymptote. This represents the
average profit for x items, when x is a large
number of items.
decreasing over the interval 0, . There
are no relative extrema.
Copyright © 2016 Pearson Education, Inc.
324
Chapter 2: Applications of Differentiation
d) Using the techniques of this section we find
the following additional information.
Intercepts. The x-intercepts are 12.70, 0
c) Using the techniques of this section we find
the following additional information.
Intercepts. No p-intercepts. The point
0, 480 is the C-intercept.
and 787.30, 0 . There is no P-intercept.
Vertical:
p 100
Horizontal : C 0
Slant:
None
Increasing, decreasing, relative extrema.
C p is increasing over the interval 0,100 .
There are no relative extrema.
Inflection points, concavity. C p is
Asymptotes.
x0
Asymptotes. Vertical:
Horizonta: None
1
x 400
2
Increasing, decreasing, relative extrema.
A x is increasing over the interval 0,100
Slant:
y
and decreasing over the interval 100, .
concave up on the interval 0,100 . there
The point 100, 300 is a relative maximum.
are no inflection points.
We use this information and compute other
function values as necessary to draw the
graph.
Inflection points, concavity. A x is concave
down on the interval 0, . There are no
inflection points. We use this information
and compute other function values as
necessary to sketch the graph.
d)
60.
C p
48, 000
100 p
48,000 48, 000
480
100 0
100
The cost of removing 0% of the pollutants
from a chemical spill is $480.
48,000
48,000
C 20
600
100 20
80
The cost of removing 20% of the pollutants
from a chemical spill is $600.
48,000
48, 000
C 80
2400
100 80
20
The cost of removing 80% of the pollutants
from a chemical spill is $2400.
48, 000
48,000
C 90
4800
100 90
10
The cost of removing 90% of the pollutants
from a chemical spill is $4800.
b) The domain of C is 0 p 100 since it is
not possible to remove less than 0% or more
than 100% of the pollutants, and C p is not
a) C 0
61.
From the result in part (b) we see that
there is a vertical asymptote at p 100 .
This means that the cost of cleaning up the
spill increases without bound as the amount
of pollutants removed approaches 100%.
The company will not be able to afford to
clean up 100% of the pollutants.
P x
a)
1
1 0.0362 x
P 5
1
0.84674
1 0.0362 5
In 1995, the purchasing power of a dollar
was $0.85.
1
P 10
0.73421439
1 0.0362 10
In 2000, the purchasing power of a dollar
was $0.73.
1
P 25
0.5249343
1 0.0362 25
In 2015, the purchasing power of a dollar
was $0.52.
defined for p 100 .
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
325
b) Solve P x 0.50
1
0.50
1 0.0362 x
1 0.50 1 0.0362 x
1 0.50 0.0181x
0.50 0.0181x
27.624 x
27.6 years after 1990, or in 2017, the
purchasing power of a dollar will be $0.50.
c) Find lim P x .
x
1
lim P x lim
x
x 1 0.0362 x
1
1
lim
x
x 1 0.0362 x 1
x
1
x
lim
x 1
0.0362
x
0
0
0 0.0362
lim P x 0 .
x
62.
A0
A t 2
t 1
If we assume 100 cc is the initial amount
100
.
injected, then A t 2
t 1
100
a) A 0
100
0 2 1
After 0 hours (at the time of injection) there
are 100 cc’s of medication in the
bloodstream.
100
A 1 2
50
1 1
After 1 hour, there are 50 cc’s of the
medication in the bloodstream.
100
A 2
20
2
2 1
After 2 hours, there are 20 cc’s of the
medication in the bloodstream.
100
A 7
2
7 2 1
After 7 hours, there are 2 cc’s of the
medication in the bloodstream.
100
A 10
102 1
0.99009901
After 10 hours, there is approximately
0.9901 cc’s of the medication in the
bloodstream.
200t
. Notice that A ‘ t 0 for
b) A ‘ t
2
t2 1
all t in the interval 0, , so there are no
relative extrema, and A t is decreasing on
the interval 0, . Therefore, the maximum
value of A t is the initial value of 100 cc at
the time of injection.
c) Using the techniques of this section we find
the following additional information.
Intercepts. There are no t-intercepts. The
A-intercept is 0,100 .
Asymptotes. Vertical:
None
Horizontal: y 0
Slant:
None
Increasing, decreasing, relative extrema.
A t is decreasing over the interval 0, .
There are no relative extrema.
Inflection points, concavity.
A ” t
exists for all t in the
200 3t 2 1
t 1
2
3
interval 0, . A ” t 0 when 3t 2 1 0 .
The only solution to the equation on 0, is
t
1
. We divide the interval 0, into
3
1
1
and
, and test
two intervals 0,
3
3
a point in each interval. A t is concave
1
down on the interval 0,
and is concave
3
1
, . The point
up on the interval
3
1
,75 is an inflection point.
3
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
326
Chapter 2: Applications of Differentiation
We use the information on the previous page
and compute other function values as
necessary to draw the graph.
d)
4
36
lim
n 0
n 0
n n 0 n
If the pitcher gives up one or more runs but
gets no one out, the pitcher would be
credited with zero innings pitched.
b) lim E n lim 9
64.
Vertical asymptotes occur at values of the
variable for which the function is undefined.
Thus, they cannot be part of the graph.
65.
Asymptotes can be thought of as “limiting
lines” for the graph of a function. The graphs
and limits in Examples 1, 3 and 6 in section 2.3
of the text book illustrate vertical, horizontal
and slant asymptotes.
No, the medication does not completely
leave the bloodstream. We notice that
lim A t 0 . This means that as t
t
approaches , A approaches 0, but does
not actually reach the value 0. We also
notice that the equation A t 0 has no
solution, so according to this model, the
medication will never completely leave the
bloodstream.
63.
4
n
a) Calculate each value for the given n.
4
E 9 9 4.00
9
4
E 6 9 6.00
6
4
E 3 9 12.00
3
4
E 1 9 36.00
1
4
2
3
E 9 9 4 54.00
2
3
2
3
4
1
3
E 9 9 4 108.00
1
3
1
3
We complete the table.
Earned-Run
Innings
average E
Pitched n
66.
E n 9
9
6
3
1
2
3
1
3
4.00
6.00
12.00
36.00
54.00
108.00
67.
5
3x
3x 2 5
x
lim
lim
x 2 x
x 2
1
x
lim 3 x 0
x
lim 3 x
x
1
lim
x
x 0 x
x, for x 0
, we have
Using x
x , for x 0
x
x
lim 1 and lim 1 .
x 0 x
x 0 x
x
Therefore, lim does not exist.
x 0 x
68.
x 2 x 2 x 4
x3 8
lim
2
x 2 x 4
x 2
x 2 x 2
lim
2
x2 2 x 4
x 2
x2
lim
22 2 2 4
.
2 2
3
69.
7
6 x
6 x 3 7 x
x
lim
lim
3 10
x 2 x 2 3 x 10
x
2 2
x x
lim 6 x 0
x
200
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
70.
327
x 1 x x 1
x3 1
lim
2
x 1 x 1
x 1
x 1 x 1
lim
2
75.
f x
x 3 2 x 2 3x
x 2 25
76.
f x
1
2
x
77.
f x
x2 3
2x 4
x2 x 1
x 1
x 1
lim
12 1 1
1 1
71.
f x x2
72.
f x
3
2
1
x2
x
x2 1
a) Solve f x 0.
73.
f x
x3 4 x2 x 6
x2 x 2
x2 3
0
2x 4
x2 3 0
x
1
2
x2 3
x 3 1.732
The x-intercepts are:
1.732, 0 and 1.732, 0 .
74.
x 3 2 x 2 15 x
f x 2
x 5 x 14
b) Evaluate at x 0.
02 3 3
2 0 4 4
The y-intercept is 0, 0.75 .
f 0
Copyright © 2016 Pearson Education, Inc.
328
Chapter 2: Applications of Differentiation
c) Vertical.
x2
Horizontal.
None
Slant. Dividing the numerator by the
denominator we have
x2 3
f x
2x 4
1
1
x 1
2
2x 4
As x approaches ,
We verify this algebraically.
lim
x 2
lim
x 1
79.
x 2 3x 2
x3
15 y
6 x3 0 x2 x
5
-5
x
2
4
6
-15
Therefore, the nonlinear asymptote is
y x2 6 .
Using a calculator, we graph the function, f, and
the asymptote, y, below.
a) lim f x 1; lim f x 1
x
As x increases without bound,
f x
+36 x
37x 9
Therefore,
x5 x 9
37 x 9
f x 3
x2 6 3
.
x 6x
x 6x
As x gets large, f x approaches x 2 6 x .
8
-10
x
x5 x 9
x3 6 x
Using long division we have:
x2
6
3
5
x 6x x
x9
f x
6 x 3
0
12 3 1 2
1 3
x5 6 x 3
10
-8 -6 -4 -2
x 2 3x 2
x3
0
x
1 0.5 x 1.
2
The slant asymptote is
y 0.5 x 1
f x
2 2 3 2 2
2 3
0
f x approaches
78.
x 2 3x 2
x3
x 2 3x 2
approaches
x3
x
x2
1.
x
x
As x decreases without bound,
f x
x 2 3x 2
approaches
x3
x
x2
1.
x
x
b) The domain of the function appears to be
, 2 1, 3 3, . We must throw
out values that make the radicand negative,
or the denominator 0.
c) From the graph it appears that
lim f x 0 and lim f x 0 .
x 2
x 1
80.
a) Visually inspecting the graph, there appears
1
to be relative maxiumum at 0, .
6
However, noticing the graph dips below the
horizontal asymptote around x 5 , we
would think there is a relative minimum for
some value x 5. This graph does not give
us enough detail to visually determine that
point.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
329
b) Calculating the first derivative, we have:
x 2 10 x 1
f ‘ x
.
2
x2 x 6
f ‘ x does not exist when x 3 or x 2 ;
however, those values are not in the domain
of f x . So we set f ‘ x 0 and solve for
x.
x 2 10 x 1
x x 6
2
2
0
x 2 10 x 1 0
By the quadratic formula, we have:
x
10
102 4 11
2 1
81. One possible rational function would be
2 x
f x
.
x2
Using the techniques in this section, we sketch
the graph.
Intercepts. f x 0 when x 0 . It turns out
the x-intercept and the y-intercept is 0, 0 .
Asymptotes. x 2 is the vertical asymptote.
The degree of the numerator equals that of the
denominator, the line y 2 is the horizontal
asymptote.
Increasing, decreasing, relative extrema.
4
. f ‘ x is not defined for
f ‘ x
x 2 2
x 2 , however that value is outside the domain
of the function. f ‘ x 0 so f x is increasing
10 100 4
2
10 96
2
Using the calculator, f ‘ x 0 when
on the intervals , 2 and 2, . There are
x 0.10 and x 9.90 .
Substititing these values back into the
function, we find the critical values occur
approximately at 0.101, 0.168 and
The equation f ” x 0 has no real solution, so
9.899, 0.952 .
c) Entering the graph in to the calculator and
using the maximum feature we see:
no relative extrema.
Inflection points, concavity.
8
does not exist when x 2 .
f ” x
x 23
there are no possible points of inflection.
Furthermore, f ” x 0 for all x in , 2 , so
f x is concave up on , 2 , and ,
f ” x 0 for all x in 2, , so f x is
concave down on 2, .
We use this information to sketch the graph.
Additional values may be computed as
necessary.
8
6
4
2
The result is confirmed.
d) Entering the graph in to the calculator and
using the minimum feature we see:
e)
The result is confirmed.
Our estimates from part (a) are not very
close. Even the “obvious” maximum was
not correct. It is doubtful that we would
have identified the relative minimum
without calculus.
-5
-4
-3
-2
-1
-2
-4
-6
-8
f(x)
x
0
1
2
3
4
5
82. One possible rational function would be
3x 1
f x
.
x
Using the techniques in this section, we sketch
the graph.
1
Intercepts. f x 0 when x . The x3
1
intercept is ,0 . When x 0 , f x is
3
undefined, so there is no y-intercept.
Copyright © 2016 Pearson Education, Inc.
330
Chapter 2: Applications of Differentiation
Asymptotes. x 0 is the vertical asymptote.
The degree of the numerator equals that of the
denominator, the line y 3 is the horizontal
asymptote.
Increasing, decreasing, relative extrema.
1
f ‘ x 2 . f ‘ x is not defined for x 0 ,
x
however that value is outside the domain of the
function. f ‘ x 0 so f x is increasing on
the interval , 0 and 0, . There are no
relative extrema.
Inflection points, concavity.
2
f ” x 3 does not exist when x 0 . The
x
equation f ” x 0 has no real solution, so
there are no possible points of inflection.
Furthermore, f ” x 0 for all x in , 0 , so
f x is concave up on , 0 , and ,
f ” x 0 for all x in 0, , so f x is
-4
-3
-2
-1
-2
-4
-6
-8
x
1
2
3
4
5
so there is a critical value at 0, 2 .
We notice that g ‘ x 0 when x 1 so g x
is decreasing on the interval , 1 .
g ‘ x 0 when 1 x so g x is increasing on
the interval 1, . g ‘ x 0 when 1 x 0
so g x is decreasing on the interval 1, 0 .
g ‘ x 0 when 0 x 1 so g x is increasing
on the interval 0,1 . There are is a relative
minimum at 0, 2 .
Inflection points, concavity.
2 3x 2 1
. g ” x does not exist
g ” x
3
x2 1
g x is concave up on 1,1 , and , g ” x 0
for all x in , 1 and 1, , so g x is
concave down on , 1 and 1, .
83. One possible rational function would be
x2 2
g x 2
.
x 1
Using the techniques in this section, we sketch
the graph.
Intercepts. g x 0 when x 2 . The x-
intercepts are 2,0 and
x 1 , however that value is outside the
domain of the function. g ‘ x 0 , when x 0 ,
real solution, so there are no possible points of
inflection.
Furthermore, g ” x 0 for all x in 1,1 , so
f(x)
0
when x 1 . The equation g ” x 0 has no
We use this information to sketch the graph.
Additional values may be computed as
necessary.
-5
concave down on 0, .
8
6
4
2
Increasing, decreasing, relative extrema.
2x
g ‘ x
. g ‘ x is not defined for
2
x2 1
We use this information to sketch the graph.
Additional values may be computed as
necessary.
8
6
4
2
-5
-4
-3
-2
2,0 When
x 0 , g x 2 , so the y-intercept is 0, 2 .
Asymptotes. x 1 and x 1 are the vertical
asymptote.
The degree of the numerator equals that of the
denominator, the line y 1 is the horizontal
asymptote.
-1
-2
-4
-6
-8
g(x)
x
0
1
2
3
4
5
84. One possible rational function would be
3x 2 15
g x 2
.
x 2x
Using the techniques in this section, we sketch
the graph.
Intercepts. g x 0 when x 5 . The x-
5, 0. When x 0,
intercepts are 5, 0 and
g x is undefined, so there is no y-intercept.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
331
Asymptotes. x 2 and x 0 are the vertical
asymptote.
The degree of the numerator equals that of the
denominator, the line y 3 is the horizontal
asymptote.
Increasing, decreasing, relative extrema.
g ‘ x
6 x 2 5 x 5
x x 2
2
2
. g ‘ x is not defined
for x 2 or x 0 , however that value is
outside the domain of the function. g ‘ x 0 ,
5 5
5 5
and x
, so there
2
2
5 5 3 5 5
are critical values at
,
2
2
when x
5 5 3
and
,
2
5 5
g ” x
6 2 x 3 15 x 2 30 x 20
3
3
We find that g ” x 0 when x 4.816 and
g ” x 0 when 4.816 x 2 , so a point of
inflection exists at 4.816, 4.025
Furthermore, g ” x 0 for all x in 0, , so
g x is concave up on 0, , and , g ” x 0
for all x in 2, 0 , so g x is concave down on
2, 0 .
We use this information to sketch the graph.
Additional values may be computed as
necessary.
g(x)
5
x
5 5
x 2 so g x is
2
5 5
, 2 .
increasing on the interval
2
g ‘ x 0 when
5 5
so g x is
2
5 5
increasing on the interval 2,
.
2
There is a relative minimum at
5 5 3 5 5
.
,
2
2
g ‘ x 0 when 2 x
5 5
x0
2
so g x is increasing on the interval
We find that g ‘ x 0 when
-18-16-14-12-10 -8 -6 -4 -2 0 2 4 6 8 10 12 14 16 18
-5
-10
-15
85. One possible rational function would be
8
h x 2
.
x x6
Using the techniques in this section, we sketch
the graph.
Intercepts. h x 0 has no real solution. The
are no x-intercepts. When x 0 , h x
4
, so
3
4
the y-intercept is 0, .
3
Asymptotes. x 3 and x 2 are the vertical
asymptotes.
The degree of the numerator is less than that of
the denominator, the line y 0 is the horizontal
asymptote.
5 5
2 ,0 . There is a relative maximum at
. g ” x does
x x 2
not exist when x 2 and x 0 . The equation
g ” x 0 has a solution when x 4.816 .
2
5 5
so g x is
g ‘ x 0 when x
2
5 5
decreasing on the interval ,
.
2
Inflection points, concavity.
5 5 3 5 5
.
,
2
2
Copyright © 2016 Pearson Education, Inc.
332
Chapter 2: Applications of Differentiation
Increasing, decreasing, relative extrema.
8 2 x 1
h ‘ x
. h ‘ x is not defined
x 2 2 x 32
for x 3 or x 2 , however those values are
outside the domain of the function. h ‘ x 0 ,
1
so there is a critical value at
2
when x
are no x-intercepts. When x 0 , h x 3 , so
the y-intercept is 0, 3 .
1 32
,
2 25
Asymptotes. x
1
so h x is
2
1
decreasing on the interval 3, .
2
h ‘ x 0 when 3 x
1
x 2 so h x is
2
1
increasing on the interval , 2 . There is a
2
We find h ‘ x 0 when
1 32
relative minimum at , . Furthermore,
2 25
h ‘ x 0 when x 3 so h x is decreasing
on the interval , 3 . h ‘ x 0 when 2 x
so h x is increasing on the interval 2, .
Inflection points, concavity.
h ” x
86. One possible rational function would be
3
h x 2
.
4x 1
Using the techniques in this section, we sketch
the graph.
Intercepts. h x 0 has no real solution. The
16 3x 2 3x 7
x 2 x 3
3
3
. h ” x does not
exist when x 2 and x 0 . The equation
h ” x 0 has no real solution. Therefore h x
has no points of inflection. h ” x 0 when
asymptotes.
The degree of the numerator is less than that of
the denominator, the line y 0 is the horizontal
asymptote.
Increasing, decreasing, relative extrema.
24 x
h ‘ x
. h ‘ x is not defined for
2
4 x2 1
intervals , 3 and 2, . h ” x 0 when
3 x 2 , so h x is concave up on 3, 2 .
We use this information to sketch the graph.
Additional values may be computed as
necessary.
h(x)
5
1
1
x or x , however those values are
2
2
outside the domain of the function. h ‘ x 0 ,
when x 0 so there is a critical value at 0, 3
1
x 0 so h x is
2
1
increasing on the interval ,0 . h ‘ x 0
2
h ‘ x 0 when
1
so h x is decreasing on the
2
1
interval 0, . There is a relative maximum at
2
when 0 x
1
so h x
2
1
is increasing on the interval , and
2
1
h ‘ x 0 when x so h x is decreasing on
2
1
the interval , .
2
Furthermore, h ‘ x 0 when x
x
-6
-4
-2
0
2
4
6
0, 3 .
x 3 and x 2 .
Therefore, h x is concave down on the
-8
1
1
and x are the vertical
2
2
8
-5
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.3
333
Inflection points, concavity.
h ” x
. h ” x does not exist
24 12 x 2 1
4 x 1
2
3
1
1
and x . The equation
2
4
h ” x 0 has no real solution.
when x
Therefore h x has no points of inflection.
1
1
x and x 2 .
2
2
Therefore, h x is concave down on the interval
h ” x 0 when
1
1 1
, . h ” x 0 when x and when
2 2
2
1
x , so h x is concave up on the intervals
2
1
1
, and , .
2
2
We use this information to sketch the graph.
Additional values may be computed as
necessary.
h(x)
5
x
-4
-2
0
2
4
-5
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
334
c) The critical value and the endpoints are:
1
0, , and 2 .
2
d) Evaluate f x for each value in step (c).
Exercise Set 2.4
1.
2.
a) The absolute maximum gasoline mileage is
obtained at a speed of 55 mph.
b) The absolute minimum gasoline mileage is
obtained at a speed of 5 mph.
c) At 70 mph, the fuel economy is 25 mpg.
f 0 4 0 0 4
2
2
17
1
1 1
f 4
4.25
2
2 2
4
f 2 4 2 2 2
2
Over the interval 30, 70 , the vehicle’s
On the interval 0, 2 , the absolute
absolute maximum fuel economy is 30 mpg at
55 mph. The vehicle’s absolute minimum fuel
economy is 25 mpg at 70 mph.
3.
f x 5 x x2 ;
a) Find f ‘ x
0, 2
f ‘x 1 2x
5.
b) Find the Critical values. The derivative
exists for all real numbers. Thus, we solve
f ‘ x 0
1 2x 0
1 2x
1
x
2
c) List the critical values and endpoints. These
1
values are 0, , and 2 .
2
d) Evaluate f x at each value in step (c).
f 0 5 0 0 5
2
2
21
1
1 1
f 5
5.25
2
2 2
4
f 2 5 2 2 3
2
21
, is the
4
1
absolute maximum, it occurs at x . The
2
smallest of these values, 3, is the absolute
minimum, it occurs at x 2 .
The largest of these values,
4.
f x 4 x x2 ;
a)
f ‘x 1 2x
17
1
, which occurs at x .
4
2
The absolute minimum is 2, which occurs at
x 2.
maximum is
0, 2
2,1
1 2
x 2 x 5;
2
a) Find f ‘ x
f x x3
f ‘ x 3x 2 x 2
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘ x 0
3x 2 x 2 0
3x 2 x 1 0
3x 2 0 or x 1 0
2
x or x 1
3
c) List the critical values and endpoints. These
2
values are are: 2, , and 1 .
3
d) Evaluate f x for each value in step (c).
f 2 2
3
3
1
22 2 2 5 9
2
2
85
2 2 1 2
2
f 2 5
3.148
3 3 2 3
3
27
f 1 1
3
1 2
15
1 2 1 5 7.5
2
2
On the interval 2,1 , the absolute
b) f ‘ x exists for all real numbers. Solve:
1 2x 0
1
x
2
15
, which occurs at x 1 . The
2
absolute minimum is 9 , which occurs at
x 2.
maximum is
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
6.
335
f x x 3 x 2 x 2;
a)
f ‘ x 3x 2 2 x 1
1, 2
f 0 0
3
maximum is 5.5 , which occurs at x 1 .
The absolute minimum is 2, which occurs at
x 2.
f ‘ x 0
3x 2 2 x 1 0
8.
1
x
or
x 1
3
c) The critical value and the endpoints are:
1
1, , 1, and 2 .
3
d) Evaluate f x for each value in step (c).
f ‘ x 0
2
3x 2 x 1 0
3x 1 x 1 0
2
3
1
or
x 1
3
c) List the critical values and endpoints. The
critical value x 1 is not in the interval, so
we exclude it. We will test the values
1
1, , and 0 .
3
d) Evaluate f x for each value in step (c).
x
2
f 1 1 1 1 2 1
3
2
f 2 2 2 2 2 4
3
2
On the interval 1, 2 , the absolute
maximum is 4 , which occurs at x 2 . The
absolute minimum is 1, which occurs at
x 1 and x 1.
1 2
x 2 x 4;
2
a) Find f ‘ x
f x x3
f 1 1 1 1 3 2
3
3
2, 0
f 0 0 0 0 3 3
3
3x 2 x 1 0
3 x 2 0 or x 1 0
2
x
x 1
or
3
2
c) The critical value x is not in the
3
interval, so we exclude it. We will test the
values: 2, 1, and 0 .
d) Evaluate f x for each value in step (c).
f 1 1
1
22 2 2 4 2
2
1
2
2
On the interval 1, 0 , the absolute
86
1
, which occurs at x .
27
3
The absolute minimum is 2, which occurs at
x 1.
maximum is
3x 2 x 2 0
3
2
86
1 1 1 1
3
3.2
3 3 3 3
27
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f 2 2
2
f
f ‘ x 3x 2 x 2
3
f ‘ x 3x 2 2 x 1
1, 0
b) f ‘ x exists for all real numbers. Solve:
59
1 1 1 1
f 2
2.2
3 3 3 3
27
7.
f x x 3 x 2 x 3;
a)
f 1 1 1 1 2 1
3
2
02 2 0 4 4
On the interval 2, 0 , the absolute
b) f ‘ x exists for all real numbers. Solve:
3x 1 x 1 0
1
12 2 1 4
11
2
9.
f x 2 x 4;
a) Find f ‘ x
1,1
f ‘ x 2
b) and c)
The derivative exists and f ‘ x 2 for all
real numbers. Note that the derivative is
never 0. Thus, there are no critical values for
f x , and the absolute maximum and
absolute minimum will occur at the
endpoints of the interval.
5.5
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
336
d) Evaluate f x at the endpoints.
12.
f 1 2 1 4 2
a)
f 1 2 1 4 6
maximum is 6, which occurs at x 1 . The
absolute minimum is 2, which occurs at
x 1.
f x 5 x 7;
a)
f ‘x 5
f 10 2 3 10 28
2, 3
f 10 2 3 10 32
On the interval 10,10 , the absolute
b) and c)
The derivative exists and is 5 for all real
numbers. Note that the derivative is never 0.
Thus, there are no critical values for f x ,
and the absolute maximum and absolute
minimum will occur at the endpoints of the
interval.
d) Evaluate f x at the endpoints.
maximum is 28, which occurs at x 10 .
The absolute minimum is 32 , which occurs
at x 10.
13.
absolute maximum is 5 for 1 x 1 and the
absolute minimum is 5 for 1 x 1 .
f 3 5 3 7 8
On the interval 2, 3 , the absolute
14.
maximum is 8, which occurs at x 3 . The
absolute minimum is 17 , which occurs at
x 2.
f x 7 4 x;
a) Find f ‘ x
f ‘ x 2x 6
1,5
b) f ‘ x exists for all real numbers. Solve:
2x 6 0
2x 6
x3
c) The critical value and the endpoints are
1, 3, and 5 .
d) Evaluate f x for each value in step (c).
f ‘ x 4
b) and c)
The derivative exists and is 4 for all real
numbers. Note that the derivative is never 0.
Thus, there are no critical values for f x ,
f 1 1 6 1 3 4
2
f 3 3 6 3 3 12
2
and the absolute maximum and absolute
minimum will occur at the endpoints of the
interval.
d) Evaluate f x at the endpoints.
f 5 5 6 5 3 8
2
On the interval 1, 5 , the absolute
maximum is 4, which occurs at x 1 . The
absolute minimum is 12 , which occurs at
x 3.
f 2 7 4 2 15
f 5 7 4 5 13
maximum is 15, which occurs at x 2 .
The absolute minimum is 13 , which
occurs at x 5.
f x x 2 6 x 3;
a)
2, 5
On the interval 2,5 , the absolute
1,1
f x 5;
Note for all values of x, f x 5 . Thus, the
f 2 5 2 7 17
11.
f ‘ x 3
b) and c)
f ‘ x 3 for all real numbers; therefore,
there are no critical points. The absolute
maximum and minimum values occur at the
endpoints.
d) Evaluate f x at the endpoints.
On the interval 1,1 , the absolute
10.
10,10
f x 2 3 x ;
15.
g x 24;
4,13
Note for all values of x, g x 24 . Thus, the
absolute maximum is 24 for 4 x 13 and the
absolute minimum is 24 for 4 x 13 .
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
16.
337
f x 3 2 x 5x 2 ;
a)
f ‘ x 2 10 x
b) f ‘ x exists for all real numbers. Solve:
3,3
3x 2 6 x 0
3x x 2 0
3x 0
or
x20
x 0 or
x2
c) The critical value and the endpoints are
0, 2, and 5 . Note, since 0 is an endpoint of
the interval, x 0 is included in this list as
an endpoint, not as a critical value.
d) Evaluate f x for each value in step (c).
b) f ‘ x exists for all real numbers. Solve:
2 10 x 0
1
x
5
c) The critical value and the endpoints are
1
3, , and 3 .
5
d) Evaluate f x for each value in step (c).
f 0 0 3 0 0
3
f 3 3 2 3 5 3 36
2
f 2 2 3 2 4
3
2
16
1
1
1
f 3 2 5
3.2
5
5
5
5
3
2
is 50, which occurs at x 5 . The absolute
minimum is 4 , which occurs at x 2.
maximum is
f x x 2 4 x 5;
a)
f ‘ x 2x 4
19.
f 0 1 6 0 3 0 1
2
f 1 1 6 1 3 1 4
2
f 1 1 4 1 5 10
2
f 4 1 6 4 3 4 23
2
f 2 2 4 2 5 1
2
On the interval 0, 4 , the absolute maximum
f 3 3 4 3 5 2
2
is 4, which occurs at x 1 . The absolute
minimum is 23 , which occurs at x 4.
On the interval 1, 3 , the absolute
f x x 3 3x 2 ;
a)
f ‘ x 3x 2 6 x
0,5
f ‘ x 6 6x
b) f ‘ x exists for all real numbers. Solve:
6 6x 0
x 1
c) The critical value and the endpoints are
0, 1, and 4 . Note, since 0 is an endpoint of
the interval, x 0 is included in this list as
an endpoint, not as a critical value.
d) Evaluate f x for each value in step (c).
b) f ‘ x exists for all real numbers. Solve:
2x 4 0
2x 4
x2
c) The critical value and the endpoints are
1, 2, and 3 .
d) Evaluate f x for each value in step (c).
18.
0, 4
f x 1 6 x 3x 2 ;
a)
1,3
maximum is 10, which occurs at x 1 . The
absolute minimum is 1, which occurs at
x 2.
2
On the interval 0,5 , the absolute maximum
On the interval 3, 3 , the absolute
17.
2
f 5 5 3 5 50
f 3 3 2 3 5 3 48
16
1
, which occurs at x .
5
5
The absolute minimum is 48 , which
occurs at x 3.
2
20.
f x x 3 3x 6;
a)
f ‘ x 3x 3
1,3
2
b) f ‘ x exists for all real numbers. Solve:
3x 2 3 0
x2 1 0
x 1
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
338
d) Evaluate f x for each value in step (c).
c) The critical value and the endpoints are
1, 1, and 3 . Note, since 1 is an endpoint
of the interval, x 1 is included in this list
as an endpoint, not a critical value.
d) Evaluate f x for each value in step (c).
f 5 3 5 2 5 325
2
f 0 3 0 2 0 0
2
2
f 1 1 3 1 6 8
3
maximum is 325, which occurs at x 5.
The absolute minimum is 0, which occurs at
x 0.
f 3 3 3 3 6 24
3
On the interval 1, 3 , the absolute
f x x 3 3 x;
a)
f ‘ x 3x 2 3
23.
f ‘ x 3x 2
b) f ‘ x exists for all real numbers. Solve:
5,1
3 x 2 0
x0
c) The critical value and the endpoints are
8, 0, and 8 .
d) Evaluate f x for each value in step (c).
3x 2 3 0
x2 1 0
f 8 1 8 513
3
x 1
c) The critical value and the endpoints are
5, 1, and 1 . Note, since 1 is an endpoint
of the interval, x 1 is included in this list
as an endpoint, not a critical value.
d) Evaluate f x for each value in step (c).
f 0 1 0 1
3
f 8 1 8 511
3
On the interval 8,8 , the absolute
maximum is 513, which occurs at x 8.
The absolute minimum is 511 , which
occurs at x 8.
f 5 5 3 5 110
3
f 1 1 3 1 2
3
24.
f 1 1 3 1 2
3
On the interval 5,1 , the absolute
maximum is 2, which occurs at x 1. The
absolute minimum is 110 , which occurs at
x 5.
f x 3x 2 2 x 3 ;
a)
8,8
f x 1 x3;
a)
b) f ‘ x exists for all real numbers. Solve:
22.
3
On the interval 5,1 , the absolute
f 1 1 3 1 6 4
21.
3
f 1 3 1 2 1 1
3
maximum is 24, which occurs at x 3. The
absolute minimum is 4, which occurs at
x 1.
3
f ‘ x 6x 6×2
5,1
b) f ‘ x exists for all real numbers. Solve:
6x 6×2 0
10,10
f x 2 x3;
a)
f ‘ x 6x
2
b) f ‘ x exists for all real numbers. Solve:
6×2 0
x0
c) The critical value and the endpoints are
10, 0, and 10 .
d) Evaluate f x for each value in step (c).
f 10 2 10 2000
3
f 0 2 0 0
3
6 x 1 x 0
x 0 or x 1
c) The critical value and the endpoints are
5, 0, and 1 . Note, since 1 is an endpoint of
the interval, x 1 is included in this list as
an endpoint, not a critical value.
f 10 2 10 2000
3
On the interval 10,10 , the absolute
maximum is 2000, which occurs at x 10.
The absolute minimum is 2000 , which
occurs at x 10.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
25.
339
0, 4
f x x 3 6 x 2 10;
a)
f ‘ x 3 x 12 x
27.
2
a)
b) f ‘ x exists for all real numbers. Solve:
x 2 3 4 x 0
x2 0
3 4x 0
3
x 0 or
x
4
c) The critical values and the endpoints are
3
1, 0, , and 1 .
4
d) Evaluate f x for each value in step (c).
3
2
3
2
On the interval 0, 4 , the absolute maximum
3
4
f 1 1 1 0
3
4
e) On the interval 1,1 , the absolute
3,1
f ‘ x 9 6 x 3x 2
4
27
3 3 3
f
0.105
4 4 4
256
is 10, which occurs at x 0. The absolute
minimum is 22 , which occurs at x 4.
a)
4
f 0 0 0 0
f 4 4 6 4 10 22
f x 12 9 x 3 x 2 x 3 ;
or
f 1 1 1 2
f 0 0 6 0 10 10
26.
3
3x 2 4 x 3 0
3x x 4 0
3x 0 or x 4 0
x 0 or
x4
c) The critical values and the endpoints are
0 and 4 . Note, since the possible critical
values are the endpoints of the interval, they
are included in this list as endpoints, not as
critical values.
d) Evaluate f x for each value in step (c).
3
f ‘ x 3x 4 x
2
b) f ‘ x exists for all real numbers. Solve:
3x 2 12 x 0
3
1,1
f x x3 x4 ;
27
3
, which occurs at x .
256
4
The absolute minimum is 2 , which occurs
at x 1 .
maximum is
b) f ‘ x exists for all real numbers. Solve:
9 6 x 3×2 0
x2 2 x 3 0
28.
x 3 x 1 0
x 3 or
x 1
c) The critical values and the endpoints are
3 and 1 . Note, since the possible critical
values are the endpoints of the interval, they
are included in this list as endpoints, not as
critical values.
d) Evaluate f x for each value in step (c).
f 3 12 9 3 3 3 3 15
2
3
f 1 12 9 1 3 1 1 17
2
3
On the interval 3,1 , the absolute
maximum is 17, which occurs at x 1. The
absolute minimum is 15 , which occurs at
x 3.
2, 2
f x x4 2 x3;
a)
f ‘ x 4x 6x
3
2
b) f ‘ x exists for all real numbers. Solve:
4 x3 6 x2 0
2 x 2 2 x 3 0
3
2
c) The critical values and the endpoints are
3
2, 0, , and 2 .
2
d) Evaluate f x for each value in step (c).
x0
x
or
f 2 2 2 2 32
4
3
f 0 0 2 0 0
4
3
4
3
27
3 3
3
f 2
1.6875
2 2
2
16
f 2 2 2 2 0
The solution is continued on the next page.
4
Copyright © 2016 Pearson Education, Inc.
3
Chapter 2: Applications of Differentiation
340
From the previous page, we determine on
the interval 2, 2 , the absolute maximum is
f 2 2 8 2 3 13
4
f 3 3 8 3 3 12
4
32, which occurs at x 2. The absolute
27
3
minimum is
, which occurs at x .
16
2
29.
a)
f ‘ x 4 x3 4 x
b) f ‘ x exists for all real numbers. Solve:
maximum is 12, which occurs at x 3 and
x 3. The absolute minimum is 13 , which
occurs at x 2 and x 2.
31.
8,8
f ‘ x
2 13
2
x 1
3
3x 3
a)
4 x x2 1 0
b) f ‘ x does not exist for x 0 . The
4x 0
or
equation f ‘ x 0 has no solution, so x 0
x2 1 0
is the only critical value.
c) The critical values and the endpoints are
8, 0, and 8 .
d) Evaluate f x for each value in step (c).
f 8 1 8 3 3
2
f 2 2 2 2 5 13
4
2
f 1 1 2 1 5 4
f 0 1 0 3 1
f 0 0 2 0 5 5
f 8 1 8 3 3
4
2
2
4
2
2
On the interval 8,8 , the absolute
f 1 1 2 1 5 4
4
2
maximum is 1, which occurs at x 0. The
absolute minimum is 3 , which occurs at
x 8 and x 8.
f 2 2 2 2 5 13
4
2
On the interval 2, 2 , the absolute
maximum is 13, which occurs at x 2 and
x 2. The absolute minimum is 4 , which
occurs at x 1 and x 1.
3, 3
f x x 4 8 x 2 3;
a)
2
f x 1 x 3 ;
4 x3 4 x 0
x 0 or
x 1
c) The critical values and the endpoints are
2, 1, 0, 1, and 2 .
d) Evaluate f x for each value in step (c).
30.
2
On the interval 3, 3 , the absolute
2, 2
f x x 4 2 x 2 5;
2
f ‘ x 4 x 16 x
3
b) f ‘ x exists for all real numbers. Solve:
4 x 3 16 x 0
4x 0
or
4 x x2 4 0
x2 4 0
f 3 3 8 3 3 12
2
f 2 2 8 2 3 13
4
2
f 0 0 8 0 3 3
4
2
4,5
f x x 3 3 5;
2
a)
f ‘ x
1
2
2
x 3 3
1
3
3 x 3 3
b) f ‘ x does not exist for x 3 . The
equation f ‘ x 0 has no solution, so
x 3 is the only critical value.
c) The critical values and the endpoints are
4, 3, and 5 .
d) Evaluate f x for each value in step (c).
f 4 4 3 3 5 4
2
x 0 or
x 2
c) The critical values and the endpoints are
3, 2, 0, 2, and 3 .
d) Evaluate f x for each value in step (c).
4
32.
f 3 3 3 3 5 5
2
f 5 5 3 3 5 1
2
On the interval 4,5 , the absolute
maximum is 1 , which occurs at x 5. The
absolute minimum is 5 , which occurs at
x 3.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
33.
f x x
341
4
;
x
d) Evaluate f x for each value in step (c).
8, 1
1
f 1 1 2
1
1 401
f 20 20
20.05
20 20
On the interval 1, 20 , the absolute
4
a) f ‘ x 1 4 x 1 2
x
b) f ‘ x does not exist for x 0 . However,
2
x 0 is not in the interval. Solve f ‘ x 0 .
1
4
0
x2
4
1 2
x
2
x 4
35.
x 2
The only critical value in the interval is at
x 2 .
c) The critical values and the endpoints are
8, 2 and 1 .
d) Evaluate f x for each value in step (c).
f 8 8
f 2 2
f 1 1
4
2
4
1
;
x
2
2
x 1 2 x x 2 x Quotient Rule
a) f ‘ x
x 1
2
2
2 x3 2 x 2 x3
x 1
2
2
2x
x 1
2
2x
1,20
f ‘ x 1 x 2 1
x 0 is not in the interval. Solve f ‘ x 0 .
1
2, 2
2
f ‘ x 0
4
5
1
1
x2
b) f ‘ x does not exist for x 0 . However,
a)
x2
;
x 1
2
b) f ‘ x exists for all real numbers. Solve:
maximum is 4 , which occurs at x 2.
17
The absolute minimum is
, which
2
occurs at x 8.
f x x
f x
4
17
8.5
8
2
On the interval 8, 1 , the absolute
34.
maximum is 20.05, which occurs at x 20.
The absolute minimum is 2, which occurs at
x 1.
1
0
x2
1
1 2
x
x2 1
x 1
2
2
0
2x 0
x0
c) The critical values and the endpoints are
2, 0, and 2 .
d) Evaluate f x for each value in step (c).
2 2 4
2 2 1 5
0 2 0
f 0
0 2 1
2 2 4
f 2
2 2 1 5
f 2
On the interval 2, 2 , the absolute
4
, which occurs at x 2 and
5
x 2. The absolute minimum is 0, which
occurs at x 0.
maximum is
x 1
The critical value x 1 is not in the
interval, and the other critical value is an
endpoint.
c) The endpoints are 1 and 20 .
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
342
36.
f x
a)
4x
;
x 1
d) Evaluate f x for each value in step (c).
3,3
2
f 2 2 1 3 1
1
x 1 4 4 x 2 x
x 1
2
f ‘ x
f 1 1 1 3 0
1
2
2
f 26 26 1 3 3
1
4 4×2
x 1
2
On the interval 2, 26 , the absolute
2
minimum is 1 , which occurs at x 2.
The absolute maximum is 3, which occurs at
x 26.
b) f ‘ x exists for all real numbers. Solve:
4 4 x2
x 1
2
0
2
38.
2
4 4x 0
x 1
c) The critical values and the endpoints are
3, 1, 1, and 3 .
d) Evaluate f x for each value in step (c).
4 3
1
On the interval 8, 64 , the absolute
maximum is 4, which occurs at x 64. The
absolute minimum is 2, which occurs at
x 8.
39. – 48.
49.
maximum is 2, which occurs at x 1. The
absolute minimum is 2 , which occurs at
x 1.
1
a)
f ‘ x
1
f 64 64 3 4
On the interval 3, 3 , the absolute
f x x 1 3 ;
f ‘ x
f 8 8 3 2
6
5
3 1
4 1
2
f 1
12 1
4 1
2
f 1 2
1 1
4 3
6
f 3
2
3 1 5
2
8,64
1 23
1
x 2
3
3x 3
b) There are no critical values in the interval.
c) The endpoints are 8 and 64 .
d) Evaluate f x for each value in step (c).
x 1 0
37.
1
a)
2
f 3
f x 3 x x 3 ;
2, 26
2
1
1
x 1 3
2
3
3 x 1 3
b) f ‘ x does not exist for x 1 . The
equation f ‘ x 0 has no solution, so
x 1 is the only critical value.
c) The critical values and the endpoints are
2, 1, and 26 .
Left to the student.
f x 30 x x 2
When no interval is specified , we use the real
line , .
a) Find f ‘ x
f ‘ x 30 2 x
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
f ‘ x 0
30 2 x 0
2 x 30
x 15
The only critical value is x 15.
c) Since there is only one critical value, we can
apply Max-Min Principle 2. First we find
f ” x .
f ” x 2
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
343
Since the second derivative is positive at 10,
we have a minimum at x 10. Next, we
find the function value at x 10.
On the previous page, we determined the
second derivative is constant, so
f ” 15 2 . Since the second derivative is
f 10 2 10 40 10 270 70
2
negative at 15, we have a maximum at
x 15. Next, we find the function value at
x 15.
Therefore, the absolute minimum is 70,
which occurs at x 10. The function has no
maximum value.
f 15 30 15 15 225
2
Therefore, the absolute maximum is 225,
which occurs at x 15. There is no
minimum value.
50.
f x 12 x x 2 ;
a)
52.
a)
f ‘ x 0
4 x 20 0
b) f ‘ x exists for all real numbers. Solve
f ‘ x 0
c)
12 2 x 0
Thus, we have a minimum at x 5.
f 5 2 5 20 5 340 290
2
Therefore, the absolute minimum is 290,
which occurs at x 5. There is no maximum
value.
f ” 6 2 0 .
Thus, we have a maximum at x 6.
f 6 12 6 6 36
2
51.
x5
f ” x 4
f ” 5 4 0
6 x
The only critical value is x 6 .
c) f ” x 2 .
Therefore, the absolute maximum is 36,
which occurs at x 6. The function has no
minimum value.
f ‘ x 4 x 20
,
b) f ‘ x exists for all real numbers. Solve
,
f ‘ x 12 2 x
f x 2 x 2 20 x 340;
53.
f x 16 x
4 3
x ;
3
0,
a) Find f ‘ x
f ‘ x 16 4 x 2
f x 2 x 2 40 x 270
When no interval is specified , we use the
real line , .
a) Find f ‘ x
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
16 4 x 2 0
f ‘ x 4 x 40
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
4 x 40 0
4 x 40
x 10
The only critical value is x 10. .
c) Since there is only one critical value, we can
apply Max-Min Principle 2. First we find
f ” x .
f ” x 4 .
The second derivative is constant, so
f ” 10 4 .
4 x 2 16
x2 4
x 2
There are two critical values; however,
x 2 is the only critical value on the
interval 0, .
c) Since there is only one critical value in the
interval, we can apply Max-Min Principle 2.
First we find f ” x .
f ” x 8 x .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
344
Thus, we solve f ‘ x 0 .
Next evaluate the second derivative at
x 2.
f ” 2 16 0
Since the second derivative is negative at 2 ,
we have a maximum at x 2.
4 3 64
f 2 16 2 2
3
3
64
Therefore, the absolute maximum is
,
3
which occurs at x 2. There is no minimum
value.
4
54. f x x x 3 ;
3
a) f ‘ x 1 4 x 2
f ‘ x 0
60 2 x 0
60 2 x
30 x
The only critical value is x 30 .
c) Since there is only one critical value, we can
apply Max-Min Principle 2. First we find
f ” x .
f ” x 2 .
The second derivative is constant, so
f ” 30 2 . Since the second derivative is
negative at 30, we have a maximum at
x 30. Next, we find the function value at
x 30.
f 30 30 60 30 900
Therefore, the absolute maximum is 900,
which occurs at x 30. The function has no
minimum value.
0,
b) f ‘ x exists for all real numbers. Solve
1 4 x2 0
4 x 2 1
x2
1
4
1
x
2
There are two critical values; however, the
1
only critical value in 0, is x .
2
c) f ” x 8 x .
56.
a)
f ‘ x 0
25 2 x 0
c)
1
.
2
25
f ” 2 0
2
3
1
1 1 4 1
f
2 2 3 2
3
Therefore, the absolute maximum is
Thus, we have a maximum at x
1
,
3
1
. The function has no
2
Therefore, the absolute maximum is
which occurs at x
f x x 60 x 60 x x 2
25
. There is no
2
minimum value.
When no interval is specified , we use the real
line , .
57.
f ‘ x 60 2 x .
b) Find the critical values. The derivative exists
for all real numbers.
25
.
2
25 625
25 25
f 25
156.25
2 2
2
4
minimum value.
a) Find f ‘ x .
25
2
f ” x 2
x
Thus, we have a maximum at x
55.
f ‘ x 25 2 x
,
b) f ‘ x exists for all real numbers. Solve
1
f ” 4 0 .
2
which occurs at x
f x x 25 x 25 x x 2 ;
1 3
x 5 x;
3
a) Find f ‘ x .
f x
f ‘ x x2 5 .
Copyright © 2016 Pearson Education, Inc.
3, 3
625
,
4
Exercise Set 2.4
345
b) Find the critical values. The derivative
exists for all real numbers. Thus, we solve
f ‘x 0 .
1
10
23 3 2 3.33
3
3
3
1
f 3 3 3 3 2 3 3.464
3
3
1
f 3
3 3 3 2 3 3.464
3
1 3
10
f 2 2 3 2 3.33
3
3
Thus, the absolute maximum over the
interval 2, 2 , is 2 3 , which occurs at
f 2
x2 5 0
x2 5
x 5 2.236
Both critical values are in the interval 3, 3 .
c) The interval is closed and there is more than
one critical value, so we use Max-Min
Principle 1.
The critical points and the endpoints are
3, 5, 5, and 3.
Next, we find the function values at these
points.
1
3
f 3 3 5 3 6
3
3
1
10 5
f 5 5 5 5
7.454
3
3
3
1
10 5
f 5
5 5 5
7.454
3
3
1 3
f 3 3 5 3 6
3
Thus, the absolute maximum over the
10 5
interval 3, 3 , is
, which occurs at
3
x 5 , and the absolute minimum over
3, 3 is
58.
a)
x 3 , and the absolute minimum over
2, 2 is 2 3 , which occurs at x 3 .
59.
f x 0.001x 2 4.8 x 60
When no interval is specified , we use the real
line , .
a) Find f ‘ x
f ‘ x 0.002 x 4.8
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
0.002 x 4.8 0
0.002 x 4.8
x 2400
The only critical value is x 2400.
c) Since there is only one critical value, we can
apply Max-Min Principle 2. First we find
f ” x .
10 5
, which occurs at x 5 .
3
1 3
x 3 x;
3
f ‘ x x2 3
f x
f ” x 0.002 .
The second derivative is constant, so
f ” 2400 0.002 .
Since the second derivative is negative at
2400, we have a maximum at x 2400.
Next, we find the function value at
x 2400.
2, 2
b) f ‘ x exists for all real numbers. Solve
f ‘ x 0
x2 3 0
f 2400 0.001 2400 4.8 2400 60
2
x 3 1.732.
Both critical values are in the interval
2, 2 .
c) The critical points and the endpoints are
2, 3, 3, and 2.
Next, we find the function values at these
points.
5700
Therefore, the absolute maximum is 5700,
which occurs at x 2400. The function has
no minimum value.
60.
f x 0.01x 2 1.4 x 30
a)
f ‘ x 0.02 x 1.4
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
346
b) f ‘ x exists for all real numbers. Solve
f ‘ x 0
62.
0.02 x 1.4 0
c)
1
f x x 3 6 x 2 11x 50;
3
a) f ‘ x x 2 12 x 11
b) f ‘ x exists for all real numbers. Solve
x 70
f ” x 0.02
f ‘ x 0
f ” 70 0.02 0
2
x 12 x 11 0
Thus, we have a maximum at x 70.
x 2 12 x 11 0
f 70 0.01 70 1.4 70 30 19
Therefore, the absolute maximum is 19,
which occurs at x 70. There is no
minimum value.
2
61.
f x x 3 x 2 5 x 1;
x 11 x 1 0
x 11 0
c)
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
3 x 2 2 x 5 0
63.
5
x
or
x 1
3
5
x is the only critical value on the
3
interval 0, .
c) The interval 0, is not closed. The only
5
critical value in the interval is x .
3
Therefore, we can apply Max-Min Principle
2. First, we find the second derivative.
f ” x 6 x 2
5
5
f ” 6 2 8 0
3
3
Since the second derivative is negative when
5
5
x , there is a maximum at x .
3
3
Next, find the function value:
2
148
5
5 5
5
f 5 1
3
3 3
3
27
Thus, the absolute maximum over the
148
, which occurs at
interval 0, is
27
5
x . The function has no minimum value.
3
x 1 0
x 11 or
f ” x 2 x 12
x 1
Therfore, there is a minimum at x 1.
1 3
166
2
f 1 1 6 1 111 50
3
3
Thus, the absolute minimum over the
166
, which occurs at
interval 0, 3 is
3
x 1.
f ‘ x 3x 2 2 x 5 .
3 x 5 x 1 0
3 x 5 0 or x 1 0
or
f ” 1 2 1 12 10 0 .
0,
a) Find f ‘ x .
3
0,3
1 3
x ;
2
a) Find f ‘ x .
f x 15 x 2
0, 30
3 2
x .
2
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
f ‘ x 30 x
3 2
x 0
2
60 x 3x 2 0
30 x
3x 20 x 0
3x 0
or
20 x 0
x 0 or
x 20
Both critical values are in the interval
0, 30 .
c) Since the interval is closed and there is more
than one critical value, we apply the MaxMin Principle 1.
The critical values and the endpoints are
0, 20, and 30 .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
347
Next, we find the function values.
1 3
2
f 0 15 0 0 0
2
1
2
3
f 20 15 20 20 2000
2
1
2
3
f 30 15 30 30 0
2
The largest of these values, 2000, is the
maximum. It occurs at x 20. The smallest
of these values, 0, is the minimum. It occurs
at x 0 and x 30.
Thus, the absolute maximum over the
interval 0, 30 , is 2000, which occurs at
65.
72
;
x
f x 2 x 72 x 1
f x 2x
a) Find f ‘ x .
72
.
x2
b) Find the critical values. f ‘ x does not exist
for x 0 ; however, 0 is not in the interval
0, . Therefore, we solve
f ‘ x 2 72 x 2 2
f ‘ x 0
72
0
x2
72
2 2
x
2 x 2 72
2
x 20 , and the absolute minimum over
0, 30 is 0, which occurs at
x 0 and x 30.
64.
1 3
x ;
0,8
2
3
a) f ‘ x 8 x x 2
2
b) f ‘ x exists for all real numbers. Solve:
f x 4 x2
x 6
c) The interval 0, is not closed. The only
critical value in the interval is x 6.
Therefore, we can apply Max-Min Principle
2. First, we find the second derivative.
144
f ” x 144 x 3 3
x
Evaluating the second derivative at x 6 ,
we have:
144 2
f ” 6
0.
63 3
3 2
x 0
2
16 x 3 x 2 0
8x
x 16 3 x 0
x0
or 16 x 3 0
3
x0
or
x
16
c) The critical values and the endpoints are
3
0,
, and 8.
16
d) Find the function values.
1 3
2
f 0 4 0 0 0
2
2
2
Multiplying by x , since x 0.
x 2 36
f ‘ x 0
Since the second derivative is positive when
x 6 , there is a minimum at x 6.
Next, find the function value at x 6.
72
f 6 2 6
24
6
Thus, the absolute minimum over the
interval 0, is 24, which occurs at x 6.
3
1 16
1024
16
16
f 4
37.93
3
3
2 3
27
1 3
2
f 8 4 8 8 0
2
Therefore, the absolute maximum over the
1024
25
, which
or 37
interval 0,8 is
27
27
16
. The absolute minimum is
occurs at x
3
0, which occurs at x 0 and x 8.
0,
The function has no maximum value over
the interval 0, .
66.
3600
;
x
3600
f ‘ x 1 2
x
f x x
a)
Copyright © 2016 Pearson Education, Inc.
0,
Chapter 2: Applications of Differentiation
348
b) Find the critical values. f ‘ x does not
First, we find the second derivative.
864
f ” x 2 864 x 3 2 3
x
Evaluating the second derivative at x 6 ,
we have:
864
f ” 6 2
6 0.
63
Since the second derivative is positive when
x 6 , there is a minimum at x 6.
Next, find the function value at x 6.
432
2
f 6 6
108
6
Thus, the absolute minimum over the
interval 0, is 108, which occurs at x 6.
exists for x 0 ; however, 0 is not in the
interval 0, . Therefore, we solve
f ‘ x 0
1
3600
0
x2
x 2 3600
x 60
c) The interval 0, is not closed. The only
critical value in the interval is x 60.
7200
f ” x 3
x
7200
1
f ” 60
0.
3
30
60
Since the second derivative is positive when
x 60 , there is a minimum at x 60.
3600
f 60 60
120
60
Thus, the absolute minimum over the
interval 0, is 120, which occurs at
x 60. The function has no maximum
value over the interval 0, .
67.
432
;
x
f x x 2 432 x 1
f x x2
250
;
0,
x
250
a) f ‘ x 2 x 2
x
b) Find the critical values. f ‘ x does not exist
f x x2
for x 0 ; however, 0 is not in the interval
0, . Therefore, we solve
2x
250
x2
0
2x
432
f ‘ x 2 x 432 x 2 x 2 .
x
b) Find the critical values. f ‘ x does not exist
2
for x 0 ; however, 0 is not in the interval
0, . Therefore, we solve
f ‘ x 0
432
0
x2
432
2x 2
x
3
2 x 432
68.
f ‘ x 0
0,
a) Find f ‘ x .
2x
The function has no maximum value over
the interval 0, .
2
Multiplying by x , since x 0.
x 3 216
x6
c) The interval 0, is not closed. The only
critical value in the interval is x 6.
Therefore, we can apply Max-Min Principle
2.
250
x2
x3 125
x5
c) The interval 0, is not closed. The only
critical value in the interval is x 5.
500
f ” x 2 3
x
500
f ” 5 2 3 6 0 .
5
There is a minimum at x 5.
250
2
f 5 5
75
5
Thus, the absolute minimum over the
interval 0, is 75, which occurs at x 5.
The function has no maximum value over
the interval 0, .
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
69.
349
1,1
f x 2 x 4 x;
a) Find f ‘ x .
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
f ‘ x 8×3 1 .
8×3 1 0
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
8×3 1
1
x3
8
1
x
2
8×3 1 0
8 x 3 1
1
x3
8
1
x
2
The only critical value x
interval 1,1 .
c) The interval is closed, and we are looking
for both the absolute maximum and absolute
minimum values, so we use Max-Min
Principle 1.
The critical values and the endpoints are
1
1, , and 1.
2
Next, we find the function values at these
points.
1
The only critical value x is in the
2
interval 1,1 .
c) The interval is closed, and we are looking
for both the absolute maximum and absolute
minimum values, so we use Max-Min
Principle 1.
The critical points and the endpoints are
1
1, , and 1.
2
Next, we find the function values at these
points.
f 1 2 1 1 3
4
4
3
1
1 1
f 2
2
2 2
8
f 1 2 1 1 1
4
f 1 2 1 1 1
4
The largest of these values, 3, is the
maximum. It occurs at x 1.
3
The smallest of these values, , is the
8
1
minimum. It occurs at x .
2
Thus, the absolute maximum over the
interval 1,1 , is 3, which occurs at x 1 ,
4
3
1
1 1
f 2
2
2 2
8
f 1 2 1 1 3
4
The largest of these values, 3, is the
maximum. It occurs at x 1. The smallest of
3
these values, , is the minimum. It occurs
8
1
at x .
2
Thus, the absolute maximum over the
interval 1,1 , is 3, which occurs at x 1 ,
and the absolute minimum over 1,1 is
3
1
, which occurs at x .
8
2
70.
f x 2 x x;
4
a)
f ‘ x 8×3 1
1,1
1
is in the
2
and the absolute minimum over 1,1 is
3
1
, which occurs at x .
8
2
71.
f x 3 x x 3 ;
1
a) Find f ‘ x
f ‘ x
0,8
1 2 3
1
x
3 2
3
3 x
b) Find the critical values. f ‘ x does not exist
for x 0 . The equation f ‘ x 0 has no
solution, so the only critical value is 0,
which is also an endpoint.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
350
c) The interval is closed, and we are looking
for both the absolute maximum and absolute
minimum values, so we use Max-Min
Principle 1.
The only critical value is an endpoint. The
endpoints are 0 and 8 .
Next, we find the function values at these
points.
f 0 3 0 0
73.
a) Find f ‘ x .
f ‘ x 3 x 1 .
2
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
3 x 1 0
2
x 1 0
x 1
The only critical value is x 1.
c) Since there is only one critical value, we can
apply Max-Min Principle 2. First we find
f ” x .
and the absolute minimum over 0,8 is 0,
which occurs at x 0.
f x x x 2 ;
1
a)
f ‘ x
f ” x 6 x 1 .
Now,
f ” 1 6 1 1 0 , so the Max-Min
0,4
1 12
1
x
2
2 x
Principle 2 fails. We cannot use Max-Min
Principle 1, because there are no endpoints.
b) Find the critical values. f ‘ x does not exist
We note that f ‘ x 3 x 1 is never
2
for x 0 . The equation f ‘ x 0 has no
negative. Thus, f x is increasing
solution, so the only critical value is 0,
which is also an endpoint.
c) The interval is closed, and we are looking
for both the absolute maximum and absolute
minimum values, so we use Max-Min
Principle 1.
The only critical value is an endpoint. The
endpoints are 0 and 4 .
Next, we find the function values at these
points.
f 0 0 0
f 4 4 2
The largest of these values, 2, is the
maximum. It occurs at x 4. The smallest
of these values, 0, is the minimum. It occurs
at x 0. Thus, the absolute maximum over
the interval 0, 4 is 2, which occurs at x 4 ,
and the absolute minimum over 0, 4 is 0,
which occurs at x 0.
3
When no interval is specified , we use the real
line , .
f 8 3 8 2
The largest of these values, 2, is the
maximum. It occurs at x 8. The smallest of
these values, 0, is the minimum. It occurs at
x 0.
Thus, the absolute maximum over the
interval 0,8 , is 2, which occurs at x 8 ,
72.
f x x 1
everywhere except at x 1 . Therefore, the
function has no maximum or minimum over
the interval , .
Notice
f ” 0 6 0
f ” 2 6 0
and
f 1 0
Therefore, there is a point of inflection at
1, 0 .
74.
f x x 1
3
When no interval is specified , we use the real
line , .
a)
f ‘ x 3 x 1
2
b) f ‘ x exists for all real numbers. Solve:
f ‘ x 0
3 x 1 0
2
x 1
The only critical value is x 1.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
c)
351
f ” x 6 x 1
Therefore, the absolute maximum over the
interval 10,10 is 59, which occurs at
f ” 1 6 1 1 0 .
x 10 , and the absolute minimum over
the interval 10,10 is 41 , which occurs
The Max-Min Principle 2 fails. We cannot
use Max-Min Principle 1, because there are
no endpoints. We note that
f ‘ x 3 x 1 is always positive, except
at x 10.
2
at x 1 . Thus, f x is increasing
77.
a) Find f ‘ x
everywhere except at x 1 . Therefore, the
function has no maximum or minimum over
the interval , .
f ‘ x 2
b) and c)
The derivative exists and is 2 for all real
numbers. Therefore, f ‘ x is never 0. Thus,
Notice
f ” 2 6 0
there are no critical values. We apply the
Max-Min Principle 1. There is only one
endpoint, x 1 . We find the function
value at the endpoint.
f 1 2 1 3 5
f ” 0 6 0
and
f 1 0
Therefore, there is a point of inflection at
1, 0 .
75.
f x 2 x 3;
a) Find f ‘ x .
We know f ‘ x 0 over the interval, so the
function is increasing over the interval
1, 5 . Therefore, the minimum value will
1,1
be the left hand endpoint. The absolute
minimum over the interval 1, 5 is 5 ,
f ‘ x 2 .
.
which occurs at x 1. Since the right
endpoint is not included in the interval, the
function has no maximum value over the
interval 1, 5 .
b) and c)
The derivative exists and is 2 for all real
numbers. Therefore, f ‘ x is never 0. Thus,
there are no critical values. We apply the
Max-Min Principle 1. The endpoints are
1 and 1. We find the function values at the
endpoints.
f 1 2 1 3 5
f 1 2 1 3 1
Therefore, the absolute maximum over the
interval 1,1 is 1 , which occurs at x 1 ,
and the absolute minimum over the interval
1,1 is 5 , which occurs at x 1.
76.
f x 9 5 x;
a)
f ‘ x 5
1,5
f x 2 x 3;
10,10
b) and c)
The derivative exists for all real numbers
and is never 0. There are no critical values,
so the maximum and minimum occur at the
endpoints, 10 and 10. We find the function
values at the endpoints.
f 10 9 5 10 59
f 10 9 5 10 41
78.
1,1
f x x 3 ;
2
a)
f ‘ x
2 13
2
x 3
3
3 x
b) Find the critical values. f ‘ x does not exist
for x 0 . The equation f ‘ x 0 has no
solution, so the only critical value is 0.
c) The interval is closed, and we are looking
for both the absolute maximum and absolute
minimum values, so we use Max-Min
Principle 1.
The critical value and the endpoints are
1, 0, and 1.
Next, we find the function values at these
points.
f 1 1 3 1
2
f 0 0 3 0
2
f 1 1 3 1
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
2
Chapter 2: Applications of Differentiation
352
From the previous page, the largest of these
values, 1, is the maximum. It occurs at
x 1 and x 1. The smallest of these
values, 0, is the minimum. It occurs at x 0.
Thus, the absolute maximum over the
interval 1,1 , is 1, which occurs at x 1
and x 1 , and the absolute minimum over
1,1 is 0, which occurs at x 0.
79.
f x 9 5 x;
a) Find f ‘ x .
We determine that the function has no
absolute extrema over the interval , .
2, 3
f ‘ x 5 .
81.
b) and c)
The derivative exists for all real numbers
and is never 0. There are no critical values.
The only endpoint is the left endpoint 2 .
f ‘ x 0 over the interval, so the function
g x x 3
When no interval is specified , we use the real
line , .
2
a) Find g ‘ x .
g ‘ x
is decreasing and a maximum occurs at
x 2 . We find the function value at
x 2 .
f 2 9 5 2 19 .
b) Find the critical values. g ‘ x does not exist
for x 0 . The equation g ‘ x 0 has no
solution, so the only critical value is 0.
c) We apply the Max-Min Principle 2.
2
g ” x 4
9x 3
g ” 0 does not exist.
The absolute maximum over the interval
2, 3 is 19, which occurs at x 2 . The
function has no minimum value over the
interval 2, 3 .
Note that g ‘ x 0 for x 0 and g ‘ x 0
1
2
80. f x x 3 x
3
3
When no interval is specified , we use the real
line , .
a)
for x 0 , so g x is decreasing on , 0
and increasing on 0, .
Therefore, the absolute minimum over the
interval , is 0, which occurs at x 0.
The function has no maximum value.
f ‘ x x 1
2
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
2
x 1 0
x 1
There are two critical values 1 and 1.
c) The interval , is not closed, so the
Max-Min Principle 1 does not apply. Since
there is more than one critical value, the
Max-Min Principle 2 does not apply.
A quick sketch of the graph at the top of the
next column will help us determine whether
absolute or relative extrema occur at the
critical values.
2 13
2
x 3 .
3
3 x
82.
1 3
x 2 x 2 x;
3
f ‘ x x2 4 x 1
f x
a)
0, 4
b) f ‘ x exists for all real numbers. Solve:
f ‘ x 0
x2 4x 1 0
We use the quadratic formula to solve the
equation.
x
4
42 4 11
2 1
4 12
2 3
2
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
353
c) The interval , is not closed, so the
Both critical values x 2 3 0.27 and
Max-Min Principle 1 does not apply. Since
there is more than one critical value, the
Max-Min Principle 2 does not apply.
A quick sketch of the graph will help us
determine absolute or relative extrema occur
at the critical values.
x 2 3 3.73 are in the closed interval
0, 4 .
c) The interval is closed, and there is more than
one critical value in the interval, so we use
Max-Min Principle 1.
The critical values and the endpoints are
0, 2 3, 2 3, and 4.
Next, we find the function values at these
points.
1 3
2
f 0 0 2 0 0 0
3
13 2 3 2 2 3 2 3
3
f 2 3
2
10
2 3 0.131
3
3
10
3
2
2 3 6.797
1 3
4 2 4 2 4
3
20
6.666
3
Thus, the absolute maximum over the
10
interval 0, 4 , is 2 3 , which occurs
3
at x 2 3 , and the absolute minimum
10
over 0, 4 is 2 3 , which occurs at
3
x 2 3 .
f 4
83.
We determine that the function has no
absolute extrema over the interval , .
3 2 3 2 2 3 2 3
f 2 3
1
1 3 1 2
x x 2x 1
3
2
When no interval is specified , we use the real
line , .
f x
a) Find f ‘ x
f ‘ x x2 x 2
b) Find the critical values. The derivative
exists for all real numbers. Thus, we solve
f ‘ x 0 .
x2 x 2 0
x 2 x 1 0
x 2 or x 1
There are two critical values 1 and 2.
84.
1 3
x 2 x 2 x;
3
a) g ‘ x x 2 4 x 1
g x
4, 0
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
g ‘ x 0 .
x2 4 x 1 0
We use the quadratic formula to solve the
equation.
The solution to the quadratic equation is
x 2 3 .
Both critical values x 2 3 3.73
and x 2 3 0.27 are in the closed
interval 4, 0 .
c) The interval is closed, and there is more than
one critical value in the interval, so we use
Max-Min Principle 1.
The critical values and the endpoints are
4, 2 3, 2 3, and 0.
Next, we find the function values at the
critical values and the endpoints.
20
g 4
6.666
3
10
g 2 3 2 3 6.797
3
10
g 2 3 2 3 0.131
3
g 0 0
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
354
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
f ‘x 0 .
From the previous page, we determine the
absolute maximum over the interval 4, 0 ,
10
2 3 , which occurs at x 2 3 ,
3
and the absolute minimum over 4, 0 is
is
8×3 8x 0
10
2 3 , which occurs at x 2 3 .
3
85.
t x x 2x
4
x0
or
8x x2 1 0
x2 1 0
x 0 or x 1
There are three critical values 1, 0, and 1.
c) The interval , is not closed, so the
2
When no interval is specified , we use the real
line , .
Max-Min Principle 1 does not apply. Since
there is more than one critical value, the
Max-Min Principle 2 does not apply.
A quick sketch of the graph will help us
determine absolute or relative extrema occur
at the critical values.
a) Find t ‘ x
t ‘ x 4 x3 4 x
b) Find the critical values. The derivative exists
for all real numbers. Thus, we solve
t ‘ x 0 .
4 x3 4 x 0
x0
or
4 x x2 1 0
x2 1 0
x 0 or x 1
There are three critical values 1, 0, and 1.
c) The interval , is not closed, so the
Max-Min Principle 1 does not apply. Since
there is more than one critical value, the
Max-Min Principle 2 does not apply.
A quick sketch of the graph will help us
determine whether absolute or relative
extrema occur at the critical values.
We determine that the function has no
absolute maximum over the interval
, . The function’s absolute minimum
is 0, which occurs at x 1 and x 1.
87. – 96.
97.
Left to the student.
M t 2t 2 100t 180,
0 t 40
a) M ‘ t 4t 100
b) M ‘ t exists for all real numbers. We solve
M ‘ t 0.
4t 100 0
4t 100
We determine that the function has no
absolute maximum over the interval
, . The function’s absolute minimum
is 1 , which occurs at x 1 and x 1.
86.
f x 2 x4 4 x2 2
When no interval is specified , we use the real
line , .
a) Find f ‘ x
f ‘ x 8×3 8x
t 25
c) Since there is only one critical value, we
apply the Max-Min Principle 2. First, we
find the second derivative.
M ” t 4 . The second derivative is
negative for all values of t in the interval,
therefore, a maximum occurs at t 25.
M 25 2 25 100 25 180 1430
2
The maximum productivity for 0 t 40 is
1430 units per month, which occurs after
t 25 years of employment.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
98.
355
N a a 2 300a 6,
0 a 300
a) N ‘ a 2 a 300
b) N ‘ a exists for all real numbers. Solve:
N ‘ a 0
2a 300 0
a 150
c) Since there is only one critical value, we
apply the Max-Min Principle 2. First, we
find the second derivative.
N ” a 2 . The second derivative is
negative for all values of t in the interval,
therefore, a maximum occurs at a 150.
N 150 150 300 150 6 22, 506
2
The maximum number of units that can be
sold is 22,506. In order to achieve this
maximum, $150,000 must be spent on
advertising.
99.
p x 0.039 x3 0.594 x 2 1.967 x 7.555
The absolute maximum occurs when
x 8.07.
According to this model, the maximum
percentage of unemployed workers in
service occupations occurred in 2011.
100. f x 0.0135 x 2 0.265 x 74.6
We restrict our attention to the years 1992 to
2012. That is, we will look at the x-values
0 x 20 .
a) f ‘ x 0.027 x 0.265
b) f ‘ x exists for all real numbers. Solve:
f ‘ x 0
0.027 x 0.265 0
x 9.81
The critical value 9.81 is in the interval.
c) The critical value and the endpoints are
0, 9.81, and 20.
d) Find the function values.
f 0 0.0135 0 0.265 0 74.6
2
We restrict our attention to the years 2003 to
2013. That is, we will look at the x-values
0 x 10 .
a) Find p ‘ x .
f 9.81 0.0135 9.81 0.265 9.81 74.6
p ‘ x 0.117 x 1.188 x 1.967 .
f 20 0.0135 20 0.265 20 74.6
2
b) Find the critical values. p ‘ x exists for all
real numbers. Therefore, we solve
p ‘ x 0 .
0.117 x 2 1.188 x 1.967 0
Using the quadratic formula, we have:
x
1.188
1.1882 4 0.1171.967
2 0.117
1.188 0.490788
0.234
x 2.08 or x 8.07
The critical values are x 2.08 and x 8.07
is in the interval 0,10 .
c) The critical values and the endpoints are
0, 2.08, 8.07, and 10.
d) Using a calculator, we find the function
values.
p 0 7.555
p 2.08 5.683
p 8.07 9.869
p 10 8.285
74.6
2
75.9
2
74.5
The maximum occurs when x 9.81.
According to this model for the period 1992
to 2012, the percentage of women aged 2154 in the U.S. Civilian labor forced was at
maximum in 2001.
101. We use the model
P t 2.69t 4 63.941t 3 459.895t 2
688.692t 24,150.217
We consider the interval 0, , where
t 0 corresponds to the year 2000.
a) Find P ‘ t .
P ‘ t 10.76t 3 191.823t 2 919.79t 688.692
b) P ‘ t exists for all real numbers. Solve:
P ‘ t 0
10.76t 3 191.823t 2 919.79t 688.692 0
Using a graphing calculator, we approximate
the zeros of P ‘ t . We find the solutions:
t 0.914
t 7.232
t 9.681
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
356
c) The critical values and the endpoints are:
0, 0.914, 7.232, and 9.681.
d) Find the function values.
P 0 24,150
P 0.914 23,858
P 9.681 39,533
The absolute minimum production of world
wide oil was 23,858,000 barrels. The world
achieved this production 0.914 years after
2000, or approximately the year 2001.
1500
x 6 x 10
We will restrict our analysis to the nonnegative
real numbers 0, , since you cannot produce
and sell a negative number of amplifiers.
3000 x 3
a) P ‘ x
2
x 2 6 x 10
2
b) P ‘ x exists for all real numbers. Solve
P ‘ x 0
3000 x 3
x 6 x 10
2
2
0
x3
c) Since the interval is open, we apply the
Max-Min Principle 2.
P ” x
3000 3x 2 18 x 26
x 6 x 10
2
P ‘ x exists for all real numbers. Solve:
P ‘ x 0
x 400 0
P 7.232 26,396
102. P x
b) First, we find the critical values.
P ‘ x x 400
3
x 400
The critical value is 400 and the endpoints
are 0 and 600. Using the Max-Min Principle
1. We evaluate the function at the endpoints
and critical values:
1 2
P 0 0 400 0 5000
2
5000
1
2
P 400 400 400 400 5000
2
75, 000
1
2
P 600 600 400 600 5000
2
55, 000
The total profit is maximized when 400
items are produced.
104. From Exercise 103, we know that
1
P x x 2 400 x 5000
2
P x
1
5000
x 400
,
a) A x
x
x
2
0 x 600 .
1 5000
b) A ‘ x 2
2
x
A ‘ x exists everywhere on 0 x 600 .
P ” 3 3000
Solve A ‘ x 0 .
Therefore, a maximum occurs at x 3 . We
find the function value at x 3 .
1500
P 3
1500
2
3 6 3 10
Producing and selling 3 amplifiers will
result in a maximum weekly profit of $1500.
1 5000
2 0
2
x
103. C x 5000 600 x
1 2
x 1000 x , 0 x 600
2
a) P x R x C x
R x
1 2
x 1000 x 5000 600 x
2
1
x 2 400 x 5000
2
x 2 10,000
x 100
x 100 is the only critical value in the
interval 0 x 600 .
The critical value and the endpoint are 100
and 600.
A 100 300
A 600 91.67
The average profit is maximized when 100
items are produced.
105. B x 305 x 2 1830 x 3 ,
a) B ‘ x 610 x 5490 x
Copyright © 2016 Pearson Education, Inc.
0 x 0.16
2
Exercise Set 2.4
357
b) B ‘ x exists for all real numbers. Solve:
We sketch a graph of the function.
B ‘ x 0
610 x 5490 x 2 0
610 x 1 9 x 0
x 0 or 1 9 x 0
1
0.11
9
c) The critical points and the endpoints are
1
0, , and 0.16.
9
d) We find the function values.
x 0 or
x
B 0 305 0 1830 0 0
2
3
2
1
1
1
B 305 1830
9
9
9
g 2 2 4
2
B 0.16 305 0.16 1830 0.16
3
0.312
The maximum blood pressure is
approximately 1.255, which occurs at a dose
1
of x cc , or about 0.11 cc of the drug.
9
106.
x 0 , which is also the endpoint of this part of
the domain. For 0 x 2 , g ‘ x 5 .
Therefore, there are no critical values of g x .
The absolute extrema will occur at one of the
endpoints. The function values are:
3
305
1.255
243
2
108. We look at the derivative on each piece of the
function to determine any critical values.
For 2 x 0 , g ‘ x 2 x . g ‘ x 0 when
g 0 0 0
2
g 2 5 2 10
On the interval 2, 2 , the absolute minimum
is 0 , which occurs at x 0. The absolute
maximum is 10, which occurs at x 2
A sketch of the graph is shown below.
In finding the absolute extrema of a
function, the second derivative is used when
there is exactly one critical value interior to an
interval. If there is exactly one critical value, the
second derivative will determine the concavity
of the function on the interval, and thus
determine if there is an absolute maximum or
absolute minimum on the interval.
107. We look at the derivative on each piece of the
function to determine any critical values.
For 3 x 1 , f ‘ x 2 so there are no
critical values for this part of the function. For
1 x 2 , f ‘ x 2 x . f ‘ x 0 when x 0 ,
which is outside the domain of this piece of the
function. Therefore, there are no critical values
of f x . The absolute extrema will occur at
one of the endpoints. The function values are:
f 3 2 3 1 5
f 1 2 1 1 3
f 2 4 2 0
2
109. We look at the derivative on each piece of the
function to determine any critical values.
For 4 x 0 , h ‘ x 2 x . h ‘ x 0 when
x 0 , which is also the endpoint of this part of
the domain. For 0 x 1 , h ‘ x 1 .
Therefore, there are no critical values of h x
on this part of the domain. For 1 x 2 ,
h ‘ x 1 . Therefore, there are no critical values
of h x on this part of the domain. The
absolute extrema will occur at one of the
endpoints. The function values are determined
on the next page.
On the interval 3, 2 , the absolute minimum
is 5 , which occurs at x 3. The absolute
maximum is 3, which occurs at x 1.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
358
We determine the function values from the
previous page:
111. a) The sketch of the graph is shown below:
h 4 1 4 15
2
h 0 1 0 1
h 1 1 1 0
h 2 2 1 1
On the interval 4, 2 , the absolute minimum
is 15 , which occurs at x 4. The absolute
maximum is 1, which occurs at x 0 and x 2.
A sketch of the graph is shown below.
110. We look at the derivative on each piece of the
function to determine any critical values.
For 2 x 0 , F ‘ x 2 x . F ‘ x 0 when
x 0 , which is also the endpoint of this part of
the domain. For 0 x 3 , F ‘ x 1 .
Therefore, there are no critical values of F x
on this part of the domain. For 3 x 67 ,
1
. Therefore, there are no
F ‘ x
2 x2
critical values of F x on this part of the
domain. The absolute extrema will occur at one
of the endpoints. The function values are:
F 2 2 4 8
2
b) From the graph, the absolute maximum is 6
and occurs at x 2 .
c) The absolute minimum value for this
function is 2. This value occurs over the
range 0 x 4 .
112. a) According to the graph, the maximum
elevation is 2290 feet, this elevation occurs
at 6.5 miles from the western edge of Rogers
Dry Lake.
b) According to the graph, the minimium
elevation occurs at 2269 feet. This elevation
occurs over the range of 0.5 miles to 5.5
miles from the western edge of Rogers Dry
Lake.
3, 3
113. g x x x 3;
a) Find g ‘ x .
1
1
1
g ‘ x x x 3 2 1 1 x 3 2
2
1
x
2
1 x 3
2
2 x 3
x 3 2 x 3
1
2 x 3
2 x 3
1
x
1
F 0 4 0 4
2
2
1
1
2
2
Multiplying by a form of 1
F 3 3 2 1
F 67 67 2 65 8.062
On the interval 2, 67 , the absolute minimum
is 1 , which occurs at x 3. The absolute
maximum is approximately 8.062, which occurs
at x 67.
A sketch of the graph is shown below.
x
2 x 3 2
1
3x 6
2 x 3 2
Copyright © 2016 Pearson Education, Inc.
1
2 x 3
2 x 3 2
, or
1
3x 6
2 x3
Exercise Set 2.4
359
b) Find the critical values. g ‘ x exists for all
d) Find the function values.
values in 3, 3 except 3 . This is a
h 0 0 1 0 0
critical value as well as an endpoint. To find
the other critical values, we solve
g ‘ x 0
2
2 2
2 2 1
h 1
3 3
3 3 3 3 3
3x 6
0
2 x3
3x 6 0
3 x 6
x 2
The second critical value on the interval is
2 .
c) On a closed interval, the Max-Min Principle
1 can always be used. The critical values
and the endpoints are 3, 2, and 3 .
d) Find the function value at each value in step
(c)..
g 3 3
3 3 0
g 2 2 2 3 2
g 3 3 3 3 3 6
Thus, the absolute maximum over the
interval 3, 3 is 3 6 , which occurs at
x 3 , and the absolute minimum is 2 ,
which occurs at x 2.
114. h x x 1 x ;
a) Find h ‘ x .
h 1 1 1 1 0
Thus, the absolute maximum over the
2
, which occurs at
interval 0,1 is
3 3
2
x , and the absolute minimum over the
3
interval is 0, which occurs at x 0 and
x 1.
2
115. C x 2 x 4
,
x 6
2 x 4 2 x 6
1
1
1
h ‘ x x 1 x 2 1 11 x 2
2
x
1 x
2 1 x
2 1 x
x
2 1 x 2 1 x
3 x 2
2 1 x
b) h ‘ x does not exist for x 1. Solve:
h ‘ x 0
3x 2
0
2 1 x
3 x 2 0
2
x
3
c) The critical values and the endpoints are
2
0, , and 1.
3
1
a) Find C ‘ x .
C ‘ x 2 2 x 6
2
1 2
2
x 6 2
b) Find the critical values.
C ‘ x does not exist for x 6 ; however,
this value is not in the domain interval, so it
is not a critical value. Solve C ‘ x 0 .
2
0,1
x6
2
x 6 2
0
2
2
x 6 2
2 x 6 2
2
Multiplying by x 6
Since x 6.
2
x 6 2 1
x 6 1
Taking the square root
of both sides.
x 6 1
x 5 or x 7
The only critical value in 6, is 7.
c) Since there is only one critical value, we
apply the Max-Min Principle 2.
4
3
C ” x 4 x 6
x 6 3
C ” 7
4
7 63
40
Therefore, since C ” 7 0 , there is a
minimum at x 7 .
The Katie’s Clocks should use 7 “quality
units” to minimize its total cost of service.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
360
116. y x a x b
2
118.
2
dy
2 x a 2 x b 4 x 2 a 2b
dx
b) The derivative exists for all real numbers.
Solve:
dy
0
dx
4 x 2 a 2b 0
4 x 2 a 2b
ab
x
2
c) There is only one critical value, we apply
the Max-Min Principle 2.
d2y
4 0 for all values of x.
dx 2
ab
Thus, y is a minimum for x
.
2
117. From exercise 101, we know that the first
derivative is
R t P ‘ t 10.76t 3 191.823t 2
a)
919.79t 688.692
a) We find the maximum rate of change, by
finding the critical values of the derivative.
0,8 . Taking the derivative we have:
P ” t 32.28t 2 383.646t 919.79
b) P ” t exists for all real numbers. Solve:
P ” t 0
2
32.28t 383.646t 919.79 0
Using the quadratic formula, we find the
zeros of P ” t .
t
383.646
383.646 4 32.28 919.79
2 32.28
2
The two solutions are t 3.33 and t 8.554.
Only one of the critical values, t 3.33 , is
in the interval.
c) Since there is only one critical value, we
apply the Max-Min Principle 2.
P ”’ t 64.56t 383.646
P ”’ 3.33 64.56 3.33 383.646 168.6 0
Therefore, the absolute maximum over the
interval 0,8 occurs at t 3.33 .
P ‘ 3.33 644.427 640.
In the year 2003 – 2004, worldwide oil
production was increasing most rapidly. It
was increasing at a rate of approximately
640,000 barrels per year.
The first derivative is used to find the critical
values of the function on an interval. For closed
intervals, we know that absolute extrema occur
at a critical value in the interval, or at an
endpoint of the interval. For open intervals,
absolute extrema, if they exists, will occur at a
critical value in the interval.
119. P t 0.0000000219t 4 0.0000167t 3
0.00155t 2 0.002t 0.22,
a) Find P ‘ t .
0 t 110
P ‘ t 0.0000000876t 3 0.0000501t 2
0.0031t 0.002
P ‘ t exists for all real numbers. Solve
P ‘ t 0 . We use a calculator to find the
zeros of P ‘ t . We estimate the solutions to
be:
x 0.639
x 71.333
x 501.223
Only one of the critical values, x 71.333 ,
is in the interval 0,110 . We apply the
Max-Min Principle 1, to find the absolute
maximum.
The critical values and the endpoints are 0,
71.333, and 110.
The function values at these points are
P 0 0.22
P 71.333 2.755
P 110 0.174
Thus, the absolute maximum oil production
for the U.S. after 1910 was 2.755 billion
barrels per year. This production level
occurred 71.333 year after 1910, or in 1981.
b) In 2010, t 2010 1910 100 .We plug
this value into the first derivative to obtain:
P ‘ 100
0.0000000876 100 0.0000501 100
3
2
0.0031 100 0.002
0.1014.
The rate of oil was declining at
approximately 0.1014 billion of barrels per
year.
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.4
361
In 2015, t 2015 1910 105 .We plug this
value into the first derivative to obtain:
Then, using the window:
P ‘ 105
0.0000000876 105 0.0000501 105
3
2
0.0031 105 0.002
0.1234.
The rate of oil was declining at
approximately 0.1234 billion of barrels per
year.
120. f x x 3 x 5 ;
2
We get the graph:
1, 4
Using a calculator, we enter the equation into
the graphing editor:
Using the table feature, we locate the extrema.
We estimate the absolute minimum to be 0,
which occurs at x 1. There is no absolute
maximum.
4
x
122. f x x 5 ;
2
Using a calculator, we enter the equation into
the graphing editor:
Then, using the window:
We get the graph:
Then, using the window:
Using the table feature, we locate the extrema.
We estimate the absolute maximum to be
2.520 , which occurs at x 4 , and the
absolute minimum to be 4.762 , which occurs
at x 2.
2
3 2
3
12 ,
x 1 ;
4
Using a calculator, we enter the equation into
the graphing editor:
121. f x
We get the graph:
Looking at the graph, it is clear to see that there
are no absolute extrema over the real numbers.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
362
123. a) Using a graphing calculator, we fit the linear
equation y x 8.857 . This corresponds to
the model P t t 8.857 . Where P is the
pressure of the contractions and t is the time
in minutes. We substitute 7 for t to find the
pressure at 7 minutes.
P 7 7 8.857 15.857 .
The pressure at 7 minutes is 15.857 mm of
Hg.
b) Rounding the coefficients to 3 decimal
places, we find the quartic regression
y 0.117 x 4 1.520 x 3 6.193 x 2
7.018 x 10.009
Changing the variables we get the model
P t 0.117t 4 1.520t 3 6.193t 2
7.018t 10.009
Using the table feature, when x 7 ,
y 24.857 . So the pressure at 7 minutes is
24.86 mm of mercury. (If we use the
rounded coefficients above, we get 23.897
mm of Hg.)
Using the trace feature, we estimate the
smallest contraction on the interval 0,10
was about 7.62 mm of Hg. This occurred
when x 0.765 min.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
363
3.
Exercise Set 2.5
1.
Express Q xy as a function of one variable.
First, we solve x y 70 for y.
x y 70
y 70 x
Next, we substitute 70 x for y in Q xy .
Q xy
Q x 70 x
Substituting
2
70 x x
Now that Q is a function of one variable we can
find the maximum. First, we find the critical
values.
Q ‘ x 70 2 x . Since Q ‘ x exists for all real
numbers, the only critical value will occur when
Q ‘ x 0 . We solve:
70 2 x 0
70 2 x
35 x
There is only one critical value. We use the
second derivative to determine if the critical
value is a maximum. Note that:
Q ” x 2 0 . The second derivative is
Q x x x 6 x 2 6 x .
Finding the deriviative, we have:
Q ‘ x 2x 6
The derivative exists for all values of x; thus,
the only critical values are where Q ‘ x 0 .
2x 6 0
2x 6
x3
There is only one critical value. We can use the
second derivative to determine whether we have
a minimum.
Q ” x 2 0 for all values of x. Therefore, a
negative for all values of x. Therefore, a
maximum occurs at x 35.
minimum occurs at x 3.
Q 3 3 6 3 9
2
Now, Q 35 70 35 35 1225
2
Therefore, the maximum product is 1225, which
occurs when x 35. If x 35 , then
y 70 35 35 . The two numbers are 35 and
35.
2.
x y 50, so y 50 x.
Q x xy x 50 x 50 x x 2
Q ‘ x 50 2 x
Q ‘ x exists for all real numbers. Solve:
Q ‘ x 0
Let x be one number and y be the other number.
Since the difference of the two numbers must be
6, we have x y 6.
The product, Q, of the two numbers is given by
Q xy , so our task is to minimize Q xy ,
where x y 6.
First, we express Q xy as a function of one
variable.
Solving x y 6 for y, we have:
x y 6
y 6 x
y x6
Next, we substitute x 6 for y in Q xy.
Thus, the minimum product is 9 when x 3 ,
and y 3 6 3 .
4.
Let x be one number and y be the other number.
The product, Q, of the two numbers is given by
Q xy , so our task is to minimize Q xy ,
where x y 4.
First, we express Q xy as a function of one
variable.
Solving x y 4 for y, we have: y x 4
Next, we substitute x 4 for y in Q xy.
Q x x 4 x 2 4 x
50 2 x 0
x 25
Q ” x 2 0 for all values of x, so a
maximum occurs at x 25 .
Q 25 50 25 25 625
2
Thus, the maximum product is 625 when
x 25 and y 50 25 25.
Q ‘ x 2x 4
The derivative exists for all values of x; thus,
the only critical values are where Q ‘ x 0 .
2x 4 0
2x 4
x2
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
364
There is only one critical value. We can use the
second derivative to determine whether we have
a minimum.
Q ” x 2 0 for all values of x. Therefore, a
2 y2 4
y2 4 2
y2 2
minimum occurs at x 2.
y 2
Q 2 2 4 2 4
2
Thus, the minimum product is 4 when x 2.
Substitute 2 for x in y x 4 to find y.
y 2 4 2 .
The two numbers which have the minimum
product are 2 and 2 .
5.
2
Maximize Q xy , where x and y are positive
2
numbers such that x y 4 .
Express Q xy 2 as a function of one variable
2
2
First, we solve x y 4 for y .
2
x y 4
y2 4 x
Next, we substitute 4 x for y 2 in Q xy 2 .
Q xy 2
Q x 4 x
4x x2
Now that Q is a function of one variable we can
find the maximum. First, we find the critical
values.
Q ‘ x 4 2x .
Since Q ‘ x exists for all real numbers, the only
critical value will occur when Q ‘ x 0 . We
set Q ‘ x 0 and solve for x:
4 2x 0
2 x 4
x 2.
There is only one critical value. We use the
second derivative to determine if the critical
value is a maximum. Note that:
Q ” x 2 0 . The second derivative is
negative for all values of x. Therefore, a
maximum occurs at x 2 .
Now,
Q 2 4 2 2 4
2
Substitute 2 in for x in x y 2 4 and solve for
y.
y 2
x and y are positive
Then Q is a maximum when x 2 and y 2 .
6.
Maximize Q xy 2 , where x and y are positive
numbers such that x y 2 1 .
Express Q xy 2 as a function of one variable.
x y2 1
y2 1 x
Q xy 2
Q x 1 x x x 2 Substituting
Now that Q is a function of one variable we can
find the maximum. First, we find the critical
values.
Q ‘ x 1 2 x . Since Q ‘ x exists for all real
numbers, the only critical value will occur when
Q ‘ x 0 . We solve:
1 2x 0
1
x.
2
There is only one critical value. We use the
second derivative to determine if the critical
value is a maximum. Note that:
Q ” x 2 0 . The second derivative is
negative for all values of x. Therefore, a
1
maximum occurs at x .
2
2
1
1 1 1
Now, Q
2 2 2
4
1
When x , we have:
2
1
y2 1
2
1
1
y
2
2
1
y
x and y must be positive
,
2
1
1
The maximum value of Q is when x and
4
2
1
.
y
2
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
7.
365
There is one critical value, we use the second
derivative to determine if it is a minimum.
Q ” x 10 0 for all values of x, therefore,
Minimize Q x 2 2 y 2 , where x y 3 .
Express Q as a function of one variable. First,
solve x y 3 for y.
x y 3
y 3 x
2
Q ” 3 0 and a minimum occurs when x 3 .
When x 3 , Q 3 2 3 3 5 3 30.
2
2
Then substitute 3 x for y in Q x 2 y .
Q x 2 3 x
x2 2 9 6 x x2
3x 2 12 x 18
Find Q ‘ x , where Q x 3x 2 12 x 18 .
Q ‘ x 6 x 12
This derivative exists for all values of x; thus
the only critical values are where
Q ‘ x 0
6 x 12 0
6 x 12
x2
Since there is only one critical value, we can use
the second derivative to determine whether we
have a minimum. Note that:
Q ” x 6 , which is positive for all real
numbers. Thus Q ” 2 0 , so a minimum
occurs when x 2. The value of Q is
Q 2 2 2 2 3 2
2
42
6.
Substitute 2 for x in y 3 x to find y.
y 3 x
y 3 2
y 1
Thus, the minimum value of Q is 6 when
x 2 and y 1.
8.
When x 3 , y 5 3 2 .
Therefore, the minimum value of Q is 30 when
x 3 and y 2.
2
2
9.
Maximize Q xy , where x and y are positive
4 2
y 1.
3
Express Q as a function of one variable. First,
4
solve x y 2 1 for x.
3
4 2
x y 1
3
4
x 1 y2
3
4
Then substitute 1 y 2 for x in Q xy .
3
4 2
Q xy 1 y y
3
4
y y3
3
4
Find Q ‘ y , where Q y y y 3 .
3
Q ‘ y 1 4 y2
numbers such that x
This derivative exists for all values of y; thus
the only critical values are where
Q ‘ y 0
1 4 y2 0
4 y 2 1
y2
x y 5 , so y 5 x.
1
4
Q 2 x 2 3 y 2 2 x 2 3 5 x
y
5 x 2 30 x 75
Q ‘ x 10 x 30
y
2
Q ‘ x exists for all real numbers. Solve:
Q ‘ x 0
10 x 30 0
x3
2
1
4
1
2
1
y must be positive
2
Since there is only one critical value, we can use
the second derivative to determine whether we
have a maximum.
y
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
366
Only 2 is in the domain of Q.
Q ” x 8 x
Note that:
Q ” y 8 y
and
1
1
Q ” 8 4 0 .
2
2
When x 2 , Q ” 2 8 2 16 , so a
1
Since Q ” is negative, a maximum occurs at
2
y
1
.
2
Evaluating the function at y
1 1 4 1
Q
2 2 3 2
1
we have:
2
x 2 and y
1 4 1
2 3 8
1 1
2 6
1
.
3
1
4
for y in x 1 y 2 to find x.
Substitute
2
3
4 2
x 1 y
3
2
x 1
Thus, the maximum value of Q is
x
10.
1
when
3
2
1
and y .
3
2
4 2
4
x y 16 , so y 16 x 2 .
3
3
4 2
4
Q xy x 16 x 16 x x3
3
3
Q ‘ x 16 4 x .
2
Q ‘ x exists for all real numbers.
Solve:
Q ‘ x 0
16 4 x 2 0
x 2
x2
Therefore, the maximum value of Q is
3
4 1
3 2
1
x 1
3
2
x
3
maximum occurs when x 2.
When x 2 ,
4 2
16 32
y 16 2 16
.
3
3
3
and
32 64
Q 2 2
.
3
3
x must be positive
64
when
3
32
.
3
11. Let x represent the width and y represent the
length of the area. It is helpful to draw a picture.
y
x
x
y
x
y
River
Since the rancher has 240 yards of fence, the
perimeter is 3 x y 240 . Solving this equation
for y, we have y 240 3 x . Since x and y must
be positive, we are restricted to the interval
0 x 80 .
The objective is to maximize area, which is
given by
A lw
Substituting y 240 3 x for the width and x for
the length, we have:
A x x 240 3x 240 x 3x 2 .
We will maximize the area over the restricted
interval by first finding the derivative of the
area function.
A x 240 x 3 x 2
A ‘ x 240 6 x
The derivative exists for all values of x in the
interval 0,80 . The only critical values are
where:
A ‘ x 0
240 6 x 0
6 x 240
x 40
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
367
Since there is only one critical value in the
interval, we use the second derivative to
determine whether we have a maximum. Note,
A ” x 6 0 for all values of x, so there is a
maximum at x 40.
Next we find the dimensions and the area.
When x 40 ,
y 240 3 x 240 3 40 120
and
A 40 240 40 3 40 4800.
2
2
Therefore, the maximum area is 4800 yd
when the overall dimensions are 40 yd by 120
yd.
12. Let x represent the length and y represent the
width of the swimming area.
Since the life guard has 180 yd of rope and
floats, the perimeter of the swimming area is
x 2 y 180 , or x 180 2 y
The objective is to maximize area, which is
given by
A lw
Substituting x 180 2 y for the length and y for
the width, we have:
A 180 2 y y 180 y 2 y 2 .
We will maximize the area over the interval
0 y 90 , because y is the length of one side,
and cannot be negative. We find the derivative:
A ‘ y 180 4 y .
This derivative exists for all values of y in
0, 90 . Thus the only critical values are where
A ‘ y 0
180 4 y 0
y 45
There is one critical value on the interval.
A ” y 4 0 for all values of y, so a
maximum occurs at y 45 .
Next, find the dimensions and the area.
When y 45 , we have
x 180 2 45 90
and
A 45 180 45 2 45 4050.
2
Therefore, the maximum area is 4050 yd 2 when
the overall dimensions are 45 yd by 90 yd.
13. Let x represent the width and y represent the
length of the area. It is helpful to draw a picture.
y
x
x
x
x
y
Since 1200 yards of fence is available, the
perimeter is 4 x 2 y 1200 . Solving this
equation for y, we have y 600 2 x . Since x
and y must be positive, we are restricted to the
interval 0 x 300 .
The objective is to maximize area, which is
given by
A lw
Substituting y 600 2 x for the width and x for
the length, we have:
A x xy x 600 2 x 600 x 2 x 2 .
We will maximize the area over the restricted
interval by first finding the derivative of the
area function.
A x 600 x 2 x 2
A ‘ x 600 4 x
The derivative exists for all values of x in the
interval 0,300 . The only critical values are
where:
A ‘ x 0
600 4 x 0
4 x 600
x 150
Since there is only one critical value in the
interval, we use the second derivative to
determine whether we have a maximum. Note,
A ” x 4 0 for all values of x, so there is a
maximum at x 150.
Next we find the dimensions and the area.
When x 150 ,
y 600 2 x 600 2 150 300
and
A 150 600 150 2 150
2
45, 000.
Therefore, the maximum area is 45, 000 yd 2
when the overall dimensions are 150 yd by 300
yd.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
368
14. a) Let x represent the width and y represent the
length of the area. If the three areas are
parallel to the stone wall, only one of the
areas will utilize the stone wall as a side.
With 600 ft of fencing, the perimeter of the
total area will be
3
3 x 2 y 600 , or y 300 x . Since x
2
and y must be positive, we are restricted to
the interval 0 x 200 . Using this
information, we find the area as a function
of x.
3
3
A x xy x 300 x 200 x x 2 .
2
2
Therefore,
A ‘ x 300 3 x
This derivative exists for all values of x.
Solve:
A ‘ x 0
300 3 x 0
x 100
There is one critical value on the interval.
A ” x 3 0 for all values of x. So a
maximum occurs at x 100.
When x 100 ,
3
y 300 100 150
2
and
3
2
A 100 300 100 100 15, 000.
2
Therefore, the maximum area is 15,000 ft 2
when the overall dimensions are 100 ft by
150 ft.
b) Let x represent the width and y represent the
length of the area. If the three areas are
perpendicular to the stone wall, all three
areas will utilize the stone wall as a side.
With 600 ft of fencing, the perimeter of the
total area will be
4 x y 600 , or y 600 4 x . Since x and
y must be positive, we are restricted to the
interval 0 x 150 . Using this information,
we find the area as a function of x.
A x xy x 600 4 x 600 x 4 x 2 .
Therefore,
A ‘ x 600 8 x
This derivative exists for all values of x.
Solve:
A ‘ x 0
600 8 x 0
x 75
There is one critical value on the interval.
A ” x 8 0 for all values of x. So a
maximum occurs at x 75.
When x 100 ,
y 600 4 75 300
and
A 75 600 75 4 75 22, 500.
2
Therefore, the maximum area is 22,500 ft 2
when the overall dimensions are 75 ft by
300 ft.
15. Let x represent the length and y represent the
width. It is helpful to draw a picture.
x
y
y
x
The perimeter is found by adding up the length
of the sides. Since it is fixed at 42 feet, the
equation of the perimeter is 2 x 2 y 42 .
The area is given by A xy .
First, we solve the perimeter equation for y.
2 x 2 y 42
2 y 42 2 x
y 21 x
Then we substitute for y into the area formula.
A xy
x 21 x 21x x 2
We want to maximize the area on the interval
0, 21 . We consider this interval because x is
the length of the rectangle and cannot be
negative.
Since the perimeter cannot exceed 42 feet, x
cannot be greater than 21, also if x is 21 feet, the
width of the rectangle would be 0 feet. We
begin by finding A ‘ x .
A ‘ x 21 2 x .
This derivative exists for all values of x in
0, 21 . Thus, the only critical values occur
where A ‘ x 0 . We solve the equation on the
next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
369
Solving A ‘ x 0 , we have:
21 2 x 0
2 x 21
21
x
10.5
2
Since there is only one critical value in the
interval, we can use the second derivative to
determine whether we have a maximum. Note
that
A ” x 2 0 for all values of x. Thus,
17. When squares of length h on a side are cut out
of the corners, we are left with a square base of
length x. A picture will help.
x
x
x
A 10.5 2110.5 10.5
2
110.25
The maximum area is 110.25 ft2 .
Note: when x 10.5 , y 21 10.5 10.5 , so
the overall dimensions that will achieve the
maximum area are 10.5 ft by 10.5 ft.
16. The perimeter is 2l 2 w 54 , so l 27 w
Since l and w must be positive, we are restricted
to the interval 0 w 27.
A w l w 27 w w 27 w w2
A ‘ w 27 2 w
This derivative exists for all values of w. Solve:
A ‘ x 0
27 2 w 0
27
13.5
2
There is one critical value on the interval.
We find the second derivative.
A ” w 2 0 for all values of w. So a
w
maximum occurs at w 13.5 .
When w 13.5 ,
l 27 13.5 13.5
and
A 13.5 13.5 27 13.5 182.25.
Therefore, the maximum area is 182.25 ft 2
when the overall dimensions are
13.5 ft by 13.5 ft.
h
h
A ” 10.5 0 , so a maximum occurs at
x 10.5.
Now,
A x 21x x 2
h
x
x
x
The resulting volume of the box is
V lwh x x h x 2 h .
We want to express V in terms of one variable.
Note that the overall length of a side of the
cardboard is 20 in. We see from the drawing,
that h x h 20, or x 2 h 20. Solving for h
we get:
2 h 20 x
1
1
h 20 x 10 x.
2
2
Substituting h into the volume equation, we
have:
1
1
V x 2 10 x 10 x 2 x 3 . The objective
2
2
is to maximize V x on the interval 0, 20 .
First, we find the derivative.
3
V ‘ x 20 x x 2
2
This derivative exists for all x in the interval
0, 20 . We set the derivative equal to zero and
solve for the critical values
V ‘ x 0
3 2
x 0
2
3
x 20 x 0
2
20 x
3
x0
2
3
x 20
x 0 or
2
40
13 13
x 0 or
x
3
The solution is continued on the next page.
x 0 or 20
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
370
From the previous page, the only critical value
40
or about 13.33. Therefore, we
in 0, 20 is
3
can use the second derivative V ” x 20 3 x
to determine if we have a maximum. We have
40
40
V ” 20 3 20 0 .
3
3
40
Therefore, there is a maximum at
.
3
2
1 40
40
40
V 10
3
3
2 3
3
100
, or
3
about 33.33. Therefore, we can use the second
derivative V ” x 50 3 x to determine if we
The only critical value in 0,50 is
have a maximum. We have
100
100
V ”
50 3
50 0 .
3
3
100
Therefore, there is a maximum at
.
3
2
1 100
100
100
V
25
3
3
2 3
3
16, 000
592 16
27
27
Now, we find the height of the box.
1 40 10
h 10
3 13 .
2 3 3
Therefore, a box with dimensions
13 13 in. by 13 13 in. by 3 13 in. will yield a
250, 000
7
9259 27
27
Now, we find the height of the box.
1 100 25
h 25
8 13 .
2 3 3
Therefore, a box with dimensions
33 13 cm by 33 13 cm by 8 13 cm will yield a
in 3.
maximum volume of 592 16
27
7
cm3 .
maximum volume of 9259 27
18. Using the picture drawn in Exercise 17, the
resulting volume of the box is
V lwh x x h x 2 h .
We want to express V in terms of one variable.
Note that the overall length of a side of the
aluminum is 50 cm. We see from the drawing,
that h x h 50, or x 2 h 50. Solving for h
we get:
1
1
h 50 x 25 x.
2
2
Substituting h into the volume equation, we
have:
1
1
V x 2 25 x 25 x 2 x 3 . The objective
2
2
is to maximize V x on the interval 0,50 .
First, we find the derivative.
3
V ‘ x 50 x x 2
2
This derivative exists for all x in the interval
0,50 , so the critical values will occur when
V ‘ x 0 . Solving this equation, we have:
3 2
x 0
2
3
x 50 x 0
2
50 x
x 0 or x
100
33 13 .
3
19. First, we make a drawing.
y
x
x
The surface area of the open-top, square-based,
rectangular tank is found by adding the area of
the base and the four sides. x 2 is the area of the
base, xy is the area of one of the sides and there
are four sides, therefore the surface area is given
by S x 2 4 xy .
The volume must by 32 cubic feet, and is given
by V l w h x 2 y 32 .
To express S in terms of one variable, we solve
x 2 y 32 for y:
y
32
.
x2
Substituting, we have:
32
S x x2 4x 2
x
128
x 2 128 x 1
x
Now S is defined only for positive numbers, so
we minimize S on the interval 0, .
x2
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
371
First, we find S ‘ x .
We restrict the analysis to the interval 0, .
S ‘ x 2 x 128 x 2
S ‘ x 2 x 250 x 2 2 x
2x
128
x2
Since S ‘ x exists for all x in 0, , the only
critical values are where S ‘ x 0 . We solve
the following equation:
S ‘ x 0
128
0
x2
x 3 64
x4
Since there is only one critical value, we use the
second derivative to determine whether we have
a minimum. Note that
256
S ” x 2 256 x 3 2 3 .
x
256
S ” 4 2 3 6 0 . Since the second
4
derivative is positive, we have a minimum at
x 4 . We find y when x 4 .
32
y 2
x
32
2
4
2
The surface area is minimized when x 4 ft
and y 2 ft. We find the minimum surface
area by substituting these values into the surface
area equation.
S x 2 4 xy
2x
4 4 42
2
16 32
48
S 4 42 4 4 2 48.
Therefore, when the dimensions are 4 ft by 4 ft
by 2 ft, the minimum surface area will be 48 ft2.
20. Using the drawing in Exercise 19, we see that
S x 2 4 xy and V l w h x 2 y 62.5 .
62.5
, and S x x 2 4 x 2 , or
x
x2
250
S x x2
x 2 250 x 1 .
x
Then y
62.5
250
x2
S ‘ x exists for all values of x in 0, . Solve:
S ‘ x 0
2x
250
x2
0
x3 125
x5
There is one critical value in the interval. We
use the second derivative to determine if it is a
minimum.
500
S ” x 2 500 x 3 2 3
x
500
S ” 5 2 3 6 0 , so a minimum occurs
5
when x 5 .
62.5
When x 5 , y 2 2.5 .
5
Therefore,
S 5 52 4 5 2.5 75.
Therefore, when the dimensions are 5 in by 5 in
by 2.5 in, the minimum surface area will be 75
in2.
21. First, we make a drawing.
y
x
2x
The surface area of the open-top, rectangular
dumpster is found by adding the area of the base
and the four sides. 2x 2 is the area of the base,
xy is the area of two of the sides, while 2 xy is
the area of the other two sides. Therefore the
surface area is given by
S 2 x 2 2 xy 2 2 xy 2 x 2 6 xy .
The volume must by 12 cubic yards, and is
given by V l w h 2 x x y 2 x 2 y 12 .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
372
To express S in terms of one variable, we solve
2 x 2 y 12 for y:
6
x2
Then
y
6
S x 2×2 6x 2
x
22. Let x be the width of the container, y be the
length of the container and 2x be the height of
the container.
Since we are including the top and the bottom
of the container, the surface area is given by:
S 4 x 2 6 xy and the volume is given by
V y x 2 x 2 x 2 y 18
36
2 x 2 36 x 1
x
Now S is defined only for positive numbers, so
we minimize S on the interval 0, .
Then, y
First, we find S ‘ x .
9
, and
x2
9
S 4 x 2 6 x 2 4 x 2 54 x 1 .
x
We are restricted to the interval 0, .
S ‘ x 4 x 36 x 2
S ‘ x 8 x 54 x 2 8 x
36
x2
Since S ‘ x exists for all x in 0, , the only
S ‘ x 0
2×2
54
x2
S ‘ x exists for all x in the interval. Solve:
4x
54
0
x2
8 x 3 54
critical values are where S ‘ x 0 . We solve
8x
the following equation:
36
4x 2 0
x
36
4x 2
x
3
x 9
27
4
x 1.89
Since there is only one critical value, we use the
second derivative to determine if it is a
minimum.
108
S ” x 8 3
x
S ” 1.89 24 0
x3
x 3 9 2.08
Since there is only one critical value, we use the
second derivative to determine whether we have
a minimum.
72
Note that S ” x 4 72 x 3 4 3 .
x
72
S ” 3 9 4
12 0 . Since the second
3
3
9
Therefore, there is a minimum at x 1.89 .
The height is twice that of the width, therefore,
the height is 2 1.89 3.78 . We solve for the
length y using:
9
y
2.52
1.892
Therefore, the dimensions of the compost
container with minimal surface area are 1.89 ft.
by 2.52 ft. by 3.78 ft.
derivative is positive, we have a minimum at
x 3 9 2.08 . The width is 2.08 yd.;
therefore, the length is 2 2.08 4.16 . We find
the height y
6
y 2
x
6
2.082
23.
1.387
The overall dimensions of the dumpster that
will minimize surface area are 2.08 yd by 4.16
yd by 1.387 yd.
R x 50 x 0.5 x 2 ; C x 4 x 10
Profit is equal to revenue minus cost.
P x R x C x
50 x 0.5 x 2 4 x 10
0.5 x 2 46 x 10
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
373
Because x is the number of units produced and
sold, we are only concerned with the
nonnegative values of x. Therefore, we will find
the maximum of P x on the interval 0, .
25.
Profit is equal to revenue minus cost.
P x R x C x
2 x 0.01x 2 0.6 x 30
First, we find P ‘ x .
P ‘ x x 46
2
0.01x 1.4 x 30
Because x is the number of units produced and
sold, we are only concerned with the
nonnegative values of x. Therefore, we will find
the maximum of P x on the interval 0, .
The derivative exists for all values of x in 0, .
Thus, we solve P ‘ x 0.
x 46 0
x 46
x 46
There is only one critical value. We can use the
second derivative to determine whether we have
a maximum.
P ” x 1 0
The second derivative is less than zero for all
values of x. Thus, a maximum occurs at x 46.
First, we find P ‘ x .
P ‘ x 0.02 x 1.4
The derivative exists for all values of x in 0, .
Thus, we solve P ‘ x 0.
0.02 x 1.4 0
0.02 x 1.4
x 70
There is only one critical value. We can use the
second derivative to determine whether we have
a maximum.
P ” x 0.02 0
The second derivative is less than zero for all
values of x. Thus, a maximum occurs at x 70.
P 46 0.5 46 46 46 10
2
1058 2116 10
1048
The maximum profit is $1048 when 46 units are
produced and sold.
24.
R x 2 x; C x 0.01x 2 0.6 x 30
R x 50 x 0.5 x 2 ; C x 10 x 3
P 70 0.01 70 1.4 70 30
2
P x R x C x
49 98 30
50 x 0.5 x 10 x 3
19
The maximum profit is $19 when 70 units are
produced and sold.
2
0.5 x 2 40 x 3,
P ‘ x x 40
0 x
The derivative exists for all values of x in 0, .
Thus, we solve P ‘ x 0.
x 40 0
x 40
There is only one critical value.
P ” x 1 0
The second derivative is less than zero for all
values of x. Thus, a maximum occurs at x 40.
P 40 0.5 40 40 40 3 797
2
The maximum profit is $797 when 40 units are
produced and sold.
26.
R x 5 x; C x 0.001x 2 1.2 x 60
P x R x C x
5 x 0.001x 2 1.2 x 60
0.001x 2 3.8 x 60,
P ‘ x 0.002 x 3.8
0 x
The derivative exists for all values of x in 0, .
Thus, we solve P ‘ x 0.
0.002 x 3.8 0
x 1900
There is only one critical value.
P ” x 0.002 0
The second derivative is less than zero for all
values of x. The maximum occurs at x 1900.
P 1900 0.001 1900 3.8 1900 60 3550
2
The maximum profit is $3550 when 1900 units
are produced and sold.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
374
27.
R x 9 x 2 x2
5
thousand is approximately
3
148
1.667 thousand or 1667 units, and that
27
thousand is approximately 5.481 thousand or
5481.
Thus, the maximum profit is approximately
$5481 when approximately 1667 units are
produced and sold.
Note that x
C x x 3 3x 2 4 x 1
R x and C x are in thousands of dollars and x
is in thousands of units.
Profit is equal to revenue minus cost.
P x R x C x
9 x 2 x 2 x 3 3x 2 4 x 1
x3 x 2 5x 1
Because x is the number of units produced and
sold, we are only concerned with the
nonnegative values of x. Therefore, we will find
the maximum of P x on the interval 0, .
First, we find P ‘ x .
P ‘ x 3 x 2 2 x 5
The derivative exists for all values of x in 0, .
Thus, we solve P ‘ x 0.
3x 2 2 x 5 0
3x 5 x 1 0
or
or
is in thousands of units.
P x R x C x
1
100 x x 2 x3 6 x 2 89 x 100
3
0 x
Thus, we solve P ‘ x 0.
5
or
x 1
3
There is only one critical value in the interval
0, . We can use the second derivative to
determine whether we have a maximum.
P ” x 6 x 2
Therefore,
5
5
P ” 6 2 10 2 8 0
3
3
The second derivative is less than zero for
5
5
x . Thus, a maximum occurs at x .
3
3
2
5
5 5
5
P 5 1
3
3 3
3
125 25 25
1
27 9
3
125 75 225 27
27 27 27 27
148
27
1 3
x 6 x 2 89 x 100
3
R x and C x are in thousands of dollars and x
C x
The derivative exists for all values of x in 0, .
x 1 0
x 1
x
3
R x 100 x x 2
1
x3 5 x 2 11x 100,
3
P ‘ x x 2 10 x 11
3 x 2 2 x 5 0
3x 5 0
3x 5
28.
x 2 10 x 11 0
x 2 10 x 11 0
x 11 x 1 0
x 11 0
or x 1 0
x 11 or
x 1
There is only one critical value in the interval
0, . We can use the second derivative to
determine whether we have a maximum.
P ” x 2 x 10
Therefore,
P ” 11 2 11 10 12 0
The second derivative is less than zero for
x 11. Thus, a maximum occurs at x 11.
1
3
2
P 11 11 5 11 11 11 100
3
182.333
Note that x 11 thousand is 11,000 units, and
that 182.333 thousand is 182,333.
Thus, the maximum profit is approximately
$182,333 when 11,000 units are produced and
sold.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
29.
375
p 280 0.4 x Price per unit.
30.
C x 5000 0.6 x 2 Cost per unit.
a) Revenue is price times quantity. Therefore,
revenue can be found by multiplying the
number of unit sold, x, by the price of the
unit, p . Substituting 280 0.4x for p, we
have:
R x x p
x 280 0.4 x
p 150 0.5 x , C x 4000 0.25 x 2
a) R x x p x 150 0.5 x 150 x 0.5 x 2
b) P x R x C x
150 x 0.5 x 2 4000 0.25 x 2
P x 0.75 x 2 150 x 4000,
0 x
c) P ‘ x 1.5 x 150
P ‘ x exists for all real numbers in the
interval 0, . Solve:
R x 280 x 0.4 x 2
b) Profit is revenue minus cost. Therefore,
P x R x C x
280 x 0.4 x 2 5000 0.6 x 2
0x
x 2 280 x 5000,
Since x is the number of units produced and
sold, we will restrict the domain to the
interval 0 x .
c) To determine the number of suits required to
maximize profit, we first find P ‘ x .
P ‘ x 2 x 280 .
The derivative exists for all real numbers in
the interval 0, . Thus, we solve
P ‘ x 0
2 x 280 0
2 x 280
x 140
Since there is only one critical value, we can
use the second derivative to determine
whether we have a maximum.
P ” x 2 0
The second derivative is negative for all
values of x; therefore, a maximum occurs at
x 140.
Riverside Appliances must sell 140
refrigerators to maximize profit.
d) The maximum profit is found by substituting
140 for x in the profit function.
P 140 140 280 140 5000
2
19, 600 39, 200 5000
14, 600
The maximum profit is $14,600.
e) The price per refrigerator is given by:
p 280 0.4 x
Substituting 140 for x, we have:
p 280 0.4 140 224
1.5 x 150 0
x 100
Since there is only one critical value, we can
use the second derivative to determine
whether we have a maximum.
P ” x 1.5 0
The second derivative is negative for all
values of x; therefore, a maximum occurs at
x 100.
Raggs, Ltd. must sell 100 suits to maximize
profit.
d) Substitute 100 for x in the profit function.
P 100 0.75 100 150 100 4000
2
3500
The maximum profit is $3500.
e) p 150 0.5 100 150 50 100
The price per suit will be $100.
31. Let x be the amount by which the price of $80
should be increased. First, we express total
revenue R as a function of x. There are two
sources of revenue, revenue from tickets and
revenue from concessions.
Number of Price of
R x
Rooms Rooms
Note, when the price increases x dollars, the
number of rooms occupied falls by x rooms.
Thus, the number of rooms is 300 x when
price increases x dollars. Therefore, the total
revenue function is
R x 300 x 80 x
x 2 220 x 24, 000
The solution is continued on the next page.
The price per refrigerator will be $224.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
376
The cost of maintining each occupied room is
$22. Therefore the total cost function is:
Number of
C x
22
Rooms
= 300 x 22
6600 22 x.
Now we can find the profit function for the
hotel.
P x R x C x
x 2 220 x 24, 000 6600 22 x
x 2 242 x 17, 400.
To find x such that P x is a maximum, we first
find P ‘ x :
P ‘ x 2 x 242
This derivative exists for all real numbers x.
thus, the only critical values are where
P ‘ x 0 ; so we solve that equation:
P ‘ x 0
10, 000
R x 40, 000
x 18 x
3
10, 000
x 4.50
40, 000
3
10, 000 2
x 35, 000 x 900, 000
3
Therefore, the total revenue function is
10,000 2
R x
x 35,000 x 900,000
3
20,000
R ‘ x
x 35,000
3
This derivative exists for all real numbers x.
thus, the only critical values are where
R ‘ x 0 ; so we solve that equation:
20,000
x 35,000 0
3
20,000
x 35,000
3
x 5.25
20, 000
0
3
is negative, for all values of x, therefore a
maximum occurs at x 5.25 . In order to
maximize revenue, the university should charge
$18 $5.25 , or $12.75.
We can find the attendance using
10, 000
40,000
5.25 57,500
3
The average attendance when ticket price is
$12.75 is 57,500 people.
The second derivative R ” x
2 x 242 0
2 x 242
x 121
The second derivative P ” x 2 0 is
negative for all values of x, therefore a
maximum occurs at x 121 .
The charge per unit should be $80 + $121, or
$201.
32. Let x be the amount by which the price of $18
should be decreased (if x is negative, the price
would be increased to maximize revenue). First,
we express total revenue R as a function of x.
There are two sources of revenue, revenue from
tickets and revenue from concessions.
Revenue from Revenue from
R x
concessions
tickets
Number of Ticket Number of
4.50
People Price People
33. Let x be the amount the number of new officers
Oak Glen should place on patrol. The total
number of parking tickets written per day is
Number of Avg. Tickets
p x
officers per day
8 x 24 4 x
4 x 2 8 x 192.
To find x such that p x is a maximum, we first
find p ‘ x :
Note, the increase in ticket sales is 10,000 x,
when price drops 3x dollars. Therefore, the
10, 000
x when price
increase in ticket sales is
3
drops x dollars.
p ‘ x 8 x 8.
This derivative exists for all real numbers x.
thus, the only critical values are where
P ‘x 0 .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
377
We solve the equation:
P ‘ x 0
b) First, we express total revenue R as a
function of x. Revenue is price times
quantity demanded, therefore,
R x x q x
8 x 8 0
8 x 8
x 1
The second derivative P ” x 8 0 is
negative for all values of x, therefore a
maximum occurs at x 1 .
This means that Oak Glean should place one
fewer officer on patrol in order to maximize the
number of parking tickets written in a day.
34. Let x equal the number of additional trees per
acre which should be planted. The total yield
per acre is equal to the yield per tree times the
number of trees so, we have:
Y x 30 x 20 x
600 10 x x 2
Find Y ‘ x :
Y ‘ x 10 2 x.
This derivative exists for all real numbers x.
Thus, the only critical values are where
Y ‘ x 0 ; so we solve that equation:
10 2 x 0
x5
This corresponds to planting 5 trees.
Since this is the only critical value, we can use
the second derivative,
R ” x 2 0, which is negative for all values
of x. A maximum occurs at x 5.
Therefore, in order to maximize yield, the apple
farm should plant 20 + 5, or 25 trees per acre.
35. a) First find the slope of the line.
1.12 1
12
m
0.59 1
41
12
y 1 x 1
41
12
53
y x
41
41
So the demand function is:
12
53
q x x
41
41
53
12
R x x x
41
41
12 2 53
x x
41
41
To find x such that R x is a maximum, we
first find R ‘ x :
24
53
x
41
41
R ‘ x exists for all real numbers. Solve:
R ‘ x
R ‘ x 0
24
53
x
0
41
41
53
x
24
x 2.21
24
0,
41
is negative, for all values of x, therefore a
maximum occurs at x 2.21 .
To maximize revenue, the price of nitrogen
should increase 221% from the January
2001 price.
The second derivative R ” x
36. a) When x 25 , q 2.13 . When x 25 1 ,
or 26, then q 2.13 0.04 2.09. We use
the points 25, 2.13 and 26,2.09 to find
the linear demand function q x . First, we
find the slope:
2.13 2.09 0.04
m
0.04 .
25 26
1
Next, we use the point-slope equation:
q 2.13 0.04 x 25
q 2.13 0.04 x 1
q 0.04 x 3.13
Therefore, the linear demand function is:
q x 0.04 x 3.13.
b) Revenue is price times quantity; therefore,
the revenue function is:
R x x q x
x 0.04 x 3.13
0.04 x 2 3.13x
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
378
Find R ‘ x :
38. The volume of the box is given by
V x x y x 2 y 320.
R ‘ x 0.08 x 3.13 .
This derivative exists for all values of x. So
the only critical values occur when
R ‘ x 0 ; so we solve that equation:
0.08 x 3.13 0
0.08 x 3.13
x 39.125
Since this is the only critical value, we can
use the second derivative,
R ” x 0.08 0,
to determine whether we have a maximum.
Since R ” 39.125 is negative, R 39.125 is
The area of the base is x 2 . The cost of the base
is 15x 2 cents.
The area of the top is x 2 . The cost of the top is
10x 2 cents.
The area of each side is xy . The cost of the four
a maximum.
In order to maximize revenue, the State of
Maryland should charge $39.125 or
rounding up to $39.13 per license plate.
y
37. Let x be the number of $0.10 increase that
should be made. Then,
R x Attendance Admission Price
180 x 5 0.1x
0.1x 2 13x 900
To find x such that R x is a maximum, we first
find R ‘ x :
R ‘ x 0.2 x 13
R ‘ x exists for all real numbers. Solve:
R ‘ x 0
0.2 x 13 0
x 65
Since there is only one critical value, we can
use the second derivative,
R ” x 0.2 0,
to determine whether we have a maximum.
Since R ” 65 is negative, a maximum occurs at
x 65 .
When x 65 the admission price that will
maximize revenue is given by:
$5 0.165 11.50
Therefore, the theater owner should charge
$11.50 per ticket.
sides is 2.5 4xy cents.
The total costs in cents is given by
C 15 x 2 10 x 2 2.5 4 xy 25 x 2 10 xy.
To express C in terms of one variable, we solve
x 2 y 320 for y:
320
.
x2
Then,
3200
320
C x 25 x 2 10 x 2 25 x 2
x
x
The function is defined only for positive
numbers, so we are minimizing C on the
interval 0, .
First, we find C ‘ x .
3200
x2
Since C ‘ x exists for all x in 0, , the only
C ‘ x 50 x 3200 x 2 50 x
critical values are where C ‘ x 0 . Thus, we
solve the following equation:
3200
50 x 2 0
x
3200
50 x 2
x
x3 64
x4
This is the only critical value, so we can use the
second derivative to determine whether we have
a minimum.
6400
C ” x 50 6400 x 3 50 3
x
Note that the second derivative is positive for all
positive values of x, therefore we have a
minimum at x 4. We find y when x 4.
320 320 320
y 2
20
x
4 2 16
The cost is minimized when the dimensions are
4 ft by 4 ft by 20 ft.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
379
We find y when x 79.93.
5000
5000
y
62.55.
x
79.93
39. The area of the parking area is given by
A x y 5000.
Since three sides are chain link fencing, the total
cost of the chain link fencing is given by
4.50 2x y dollars.
The cost is minimized when the dimensions are
62.55 ft by 79.92 ft. (Note, the wooden fence
side should be 62.55 feet.)
Substituting these dimensions into the cost
function we have:
C 9 x 11.5 y
One side is wooden fencing, the cost of the
wooden fence is 7 y dollars.
The total costs in dollars is given by
C 4.50 2 x y 7 y
9 79.93 11.5 62.55
9 x 11.5 y.
To express C in terms of one variable, we solve
xy 5000 for y:
5000
.
x
Then,
y
40.
57,500
x
The function is defined only for positive
numbers, so we are minimizing C on the
interval 0, .
9x
First, we find C ‘ x .
57,500
x2
Since C ‘ x exists for all x in 0, , the only
critical values are where C ‘ x 0 . Thus, we
solve the following equation:
57,500
0
9
x2
57,500
9
x2
9 x 2 57,500
Cost of Wood 28 2 y
Therefore the total cost function is:
C 70 x 56 y.
To express C in terms of one variable, we solve
xy 1200 for y:
1200
.
x
Then,
y
67, 200
1200
70 x
C x 70 x 56
x
x
The function is defined only for positive
numbers, so we are minimizing C on the
interval 0, .
First, we find C ‘ x .
C ‘ x 70
67, 200
x2
Since C ‘ x exists for all x in 0, , the only
critical values are where C ‘ x 0 . Thus, we
57,500
x
9
2
57,500
9
x 79.93
This is the only critical value, so we can use the
second derivative to determine whether we have
a minimum.
172,500
C ” x 172,500 x 3
x3
Note that the second derivative is positive for all
positive values of x, therefore we have a
minimum at x 79.93.
x
A x y 1200
Cost of Stone 35 2 x
5000
C x 9 x 11.5
x
C ‘ x 9 57,500 x 2 9
1438.695 1439.
The total cost of fencing the parking area is
approximately $1439.
solve the following equation:
67, 200
0
70
x2
70 x 2 67, 200
x 30.98
This is the only critical value, so we can use the
second derivative to determine whether we have
a minimum.
134, 400
C ” x
x3
Note that the second derivative is positive for all
positive values of x, therefore we have a
minimum at x 30.98.
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
380
We find y when x 30.98.
1200
y
38.73.
30.98
Therefore, the cost will be minimized when the
stone wall is 30.98 yards long and the wooden
fencing is 38.73 yards long. The minimum cost
is given by:
C 70 30.98 56 38.73 4337.48.
The minimum cost will be approximately
$4337.
41. Let x and y represent the outside length and
width, respectively.
0.75
x 1.5
x
0.5
0.5
109.6875 0 , so A 109.6875 is a
maximum.
When
x 109.6875 10.47,
73.125
6.98.
109.6875
Therefore, the outside dimensions should be
approximately 10.47 in. by 6.98 in. to maximize
the print area.
y
C x
y 1
y
73.125
.
x
We want to maximize the print area:
A x 1.5 y 1
We know that xy 73.125, so y
xy x 1.5 y 1.5.
Substituting for y, we get:
73.125
73.125
A x x
x 1.5
1.5
x
x
109.6875
1.5
x
109.6875
74.625 x
x
109.6875
A ‘ x 1 109.6875 x 2 1
x2
A ‘ x exists for all values in the domain of A.
73.125 x
1
A ”
42. Let x equal the lot size. Now the inventory costs
are given by:
0.75
Solve:
Therefore, we can use the second derivative to
determine if it is a maximum.
219.375
A ” x 219.375 x 3
.
x3
Yearly carrying costs:
x
Cc x 20
2
10 x
Yearly reorder costs:
100
Cr x 40 16 x
x
4000
1600
x
Hence, the total inventory cost is:
C x Cc x Cr x
4000
1600.
x
We want to find the minimum value of C on the
interval 1,100 . First, we find C ‘ x :
10 x
4000
.
x2
The derivative exists for all x in 1,100 , so the
C ‘ x 10 4000 x 2 10
only critical values are where C ‘ x 0 .
C ‘ x 0
10
A’ x 0
109.6875
0
x2
x 2 109.6875
x 109.6875
The only critical value in the domain of A is
x 109.6875 ;
Yearly Carrying Yearly reorder
Cost
Cost
4000
0
x2
4000
10 2
x
2
10 x 4000
x 2 400
x 20
The only critical value in the interval is x 20 .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
381
We can use the second derivative to determine
whether we have a minimum.
8000
C ” x 8000 x 3 3
x
Notice that C ” x is positive for all values of x
in 1,100 , we have a minimum at x 20. Thus,
to minimize inventory costs, the store should
100
5 times per year. The
order pool tables
20
lot size will be 20 tables.
43. Let x be the lot size.
Yearly Carrying Yearly reorder
C x
Cost
Cost
We consider each cost separately.
Yearly carrying costs, C c x : Can be found by
multiplying the cost to store the items by the
number of items in storage. The average
x
amount held in stock is , and it cost $4 per
2
bowling ball for storage. Thus:
x
Cc x 4
2
2x
Yearly reorder costs, C r x : Can be found by
multiplying the cost of each order by the
number of reorders. The cost of each order is
1 0.5x , and the number of orders per year is
200
. Therefore,
x
200
Cr x 1 0.5 x
x
200
100
x
The total inventory cost is given by:
C x Cc x Cr x
200
2x
100, 1 x 200
x
We want to find the minimum value of C on the
interval 1, 200 . First, we find C ‘ x :
200
x2
C ‘ x exists for all x in 1, 200 . 1,100 , so the
C ‘ x 2 200 x 2 2
only critical values are where C ‘ x 0 .
Solving the equation, we have:
C ‘ x 0
200
0
x2
x 10
The only critical value in the domain is x 10 .
Therefore, we use the second derivative,
400
C ” x 400 x 3 3
x
to determine whether we have a minimum.
C ” 10 0.4 0 , so C 10 is a minimum.
2
In order to minimize inventory costs. The store
200
should order
20 times per year. The lot
10
size will be 10 bowling balls.
44. Let x equal the lot size.
C x
Yearly Carrying Yearly reorder
Cost
Cost
Yearly carrying costs, C c x :
x
x
2
Yearly reorder costs, C r x :
Cc x 2
720 3600
1800
Cr x 5 2.50 x
x
x
Hence, the total inventory cost is:
C x Cc x Cr x
x
3600
1800.
x
Then,
3600
.
x2
The derivative exists for all x in 1, 720 . Solve:
C ‘ x 1 3600 x 2 1
C ‘ x 0
1
3600
x2
0
x 2 3600
x 60
The only critical value in the interval is x 60 ,
so we can use the second derivative to
determine whether we have a minimum.
7200
C ” x 7200 x 3 3 .
x
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
382
Notice that C ” x is positive for all values of x
in 1, 720 , we have a minimum at x 60. Thus,
to minimize inventory costs, the store should
720
12 times per year. The
order calculators
60
lot size will be 60 calculators.
45. Let x equal the lot size. Now the inventory costs
are given by:
C x
Yearly Carrying Yearly reorder
Cost
Cost
We consider each cost separately.
Yearly carrying costs, C c x : Can be found by
multiplying the cost to store the items by the
number of items in storage. The average
x
amount held in stock is , and it cost $2 per
2
calculator for storage. Thus:
x
Cc x 8
2
4x
Yearly reorder costs, C r x : Can be found by
multiplying the cost of each order by the
number of reorders. The cost of each order is
10 5x , and the number of orders per year is
360
. Therefore,
x
360
Cr x 10 5 x
x
3600
1800
x
Hence, the total inventory cost is:
C x Cc x Cr x
3600
4x
1800, 1 x 360
x
We want to find the minimum value of C on the
interval 1, 360 . First, we find C ‘ x :
3600
C ‘ x 4 3600 x 4 2
x
The derivative exists for all x in 1, 360 , so the
2
only critical values are where C ‘ x 0 . We
solve the equation at the top of the next column.
C ‘ x 0
4
3600
x2
0
4
3600
x2
4 x 2 3600
x 2 900
x 30
The only critical value in the domain is x 30 .
Therefore, we use the second derivative,
7200
C ” x 7200 x 3 3
x
to determine whether we have a minimum.
C ” 30 0.27 0 , so C 30 is a minimum.
In order to minimize inventory costs. The store
360
12 times per year. The lot
should order
30
size will be 30 surf boards.
46. Let x equal the lot size.
Yearly carrying costs:
x
Cc x 2
2
x
Yearly reorder costs:
256
Cr x 4 2.50 x
x
1024
640
x
Hence, the total inventory cost is:
C x Cc x Cr x
1024
640.
x
We want to find the minimum value of C on the
interval 1, 256 . First, we find C ‘ x :
x
1024
.
x2
The derivative exists for all x in 1, 256 , so the
C ‘ x 1 1024 x 2 1
only critical values are where C ‘ x 0 . We
solve that equation:
1024
1 2 0
x
1024
1 2
x
2
x 1024
x 32
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
383
The only critical value in the interval is x 32 ,
so we can use the second derivative to
determine whether we have a minimum.
2048
C ” x 2048 x 3 3
x
Notice that C ” x is positive for all values of x
in 1, 256 , so we have a minimum at x 32.
Thus, to minimize inventory costs, the store
256
should order calculators
8 times per year.
32
The lot size will be 32 calculators.
47. Let x be the lot size.
x
4x
2
360
Cr x 10 6 x
x
Yearly carrying cost: Cc x 8
Yearly reorder cost:
3600
2160
x
Therefore,
C x Cc x Cr x
3600
2160, 1 x 360
x
3600
C ‘ x 4 3600 x 2 4 2
x
C ‘ x exists for all x in 1, 360 . Solve:
4x
C ‘ x 0
3600
0
x2
x 30
The only critical value in the domain is x 30 .
Therefore, we use the second derivative,
7200
C ” x 7200 x 3 3
x
to determine whether we have a minimum.
C ” 30 0.27 0 , so C 30 is a minimum.
4
In order to minimize inventory costs. The store
360
12 times per year. The lot
should order
30
size will be 30 surf boards.
48. The volume of the container must be 250 in3.
Therefore, we use formula for the volume
cylinder to obtain 250 r 2 h .
Solving for h we have:
250
h 2 .
r
The container consists of a circular top and a
circular bottom. The area for each of the top and
bottom of the container is given by A r 2 .
The side material that when laid out is a
rectangle of height h and whose length is the
same as the circumference of the circular ends,
2 r . Therfore, the surface area of the side
material is A 2 rh . Therefore, the total
surface area is the sum of the area of the top and
the bottom plus the the side material:
A 2 r 2 2 rh .
Substituting for h, we have area as a function
of the radius, r.
250
A r 2 r 2 2 r 2
r
500
A r 2 r 2
r
The nature of this problem requires r 0 . We
differentiate the area function with respect to r:
500
A ‘ r 4 r 2 .
r
We find the critical values by setting the
derivative equal to zero and solving for r.
Remember, r 0 .
A ‘ r 0
500
0
r2
500
4 r 2
r
3
4 r 500
4 r
r3
r
500
4
125
3
r 3.414
This is the only critical value in the interval
r 0 . We will calculate the second derivative
to determine the concavity of the function.
1000
A ” r 4 3 .
r
Evaluating the second derivative at the critical
value, we have:
1000
A ” 3.414 4
3.4143
37.697 0.
Since the second derivative is positive at the
critical value, the critical value represents a
relative minimum.
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
384
We determine the height of the container by
substituting back into
250
h 2 .
r
250
6.828
h
2
3.414
Therefore, the dimensions of the container that
will minimize the surface area are a height of
6.828 inches and a radius of 3.414 inches.
49. The volume of the container must be 400 cm3.
Therefore, we use formula for the volume
cylinder to obtain 400 r 2 h .
Solving for h we have:
400
h 2 .
r
Therefore, the total surface area is the sum of
the area the bottom plus the the side material:
A r 2 2 rh .
Substituting for h in area formula we have area
as a function of the radius, r.
400
A r r 2 2 r 2
r
800
r
The nature of this problem requires r 0 . We
differentiate the area function with respect to r:
800
A ‘ r 2 r 2 .
r
We find the critical values by setting the
derivative equal to zero and solving for r.
Remember, r 0 .
A ‘ r 0
Ar r2
2 r
800
0
r2
800
2 r 2
r
800
r3
2
400
r3
A ” 5.03 2
1600
5.033
18.855 0.
Since the second derivative is positive at the
critical value, the critical value represents a
relative minimum. We determine the height of
the container by substituting back into
400
h 2 .
r
400
5.03
h
2
5.03
Therefore, the dimensions of the container that
will minimize the surface area are a height of
5.03 cm and a radius of 5.03 cm.
50. The cost function for the container would be is
given by:
area of the
C 0.005
top and bottom
area of the
0.003
side material
Using the information from problem 48, we
know the total area for the top and bottom of the
can was given by 2 r 2 and the area of the side
500
material was given by r . Therefore, the cost
r
function is:
500
C r 0.005 2 r 2 0.003
r
1.5
.
r
The nature of the problem still requires r 0.
Calculating the derivative of the cost function
we have:
0.01 r 2
C ‘ r 0.01 2 r 2 1 1.5 1r 11
0.02 r
1.5
r2
Find the critical values by setting the derivative
equal to zero and solving for r.
C ‘ r 0
r 5.03
This is the only critical value in the interval
r 0 . We will calculate the second derivative
to determine the concavity of the function.
1600
A ” r 2 3 .
r
We evaluate the second derivative at the critical
value at the top of the nexg column.
0.02 r
1.5
r2
0
r3
75
r3
75
r 2.879
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
385
From the previous page, we determine
r 2.879 is the only critical value in the
interval r 0.
Next, we find the second derivative:
3
C ” r 0.02 3 .
r
Substituting the critical value into the second
derivative, we have:
3
C ” 2.879 0.022
2.8793
0.19 0.
Since the second derivative is positive at the
critical value, the critical value represents a
relative minimum. We substitute the critical
250
value back into h 2 to determine the
r
value for h. Notice, we substitute before the
rounding of r.
250
h
2
2.879411
9.598
The dimensions of the container that will
minimize the cost of the container are a radius
of 2.879 in and a height of 9.589 in.
51. The cost function for the container would be is
given by:
area of
C 0.0015
the base
area of
0.0008
the side
Using the information from problem 49, the
cost function is:
800
C r 0.0015 r 2 0.0008
r
0.64
0.0015 r 2
.
r
The nature of the problem still requires r 0.
Calculating the derivative of the cost function
we have:
d
0.64
C ‘ r
0.0015 r 2
dr
r
0.64
0.003 r 2
r
Find the critical values by setting the derivative
equal to zero and solving for r.
C ‘ r 0
0.64
0
r2
0.64
0.003 r 2
r
0.64
r3
0.003
0.64
r3
0.003
r 4.08
This is the only critical value in the interval
r 0.
Next, we find the second derivative:
d
0.64
C ” r
0.003 r 2
dr
r
1.28
0.003 3 .
r
Substituting the critical value into the second
derivative, we have:
1.28
C ” 4.08 0.003
4.083
0.003 r
0.028 0.
Since the second derivative is positive at the
critical value, the critical value represents a
relative minimum. We substitute the critical
400
value back into h 2 to determine the
r
value for h. Notice, we substitute before the
rounding of r.
400
h
2
4.08
7.65
The dimensions of the container that will
minimize the cost of the container are a radius
of 4.08 cm and a height of 7.65 cm.
52. Case I.
If y is the length, the girth is
x x x x, or 4 x.
Case II.
If x is the length, the girth is
x y x y , or 2 x 2 y.
Case I.
The combined length and girth is
y 4 x 84.
The volume is V x x y x 2 y .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
386
We want express V in terms of one variable.
We solve y 4 x 84 , for y.
y 4 x 84
y 84 4 x
Thus,
V x 2 84 4 x 84 x 2 4 x 3 .
To maximize V x we first find V ‘ x .
V ‘ x 168 x 12 x
2
This derivative exists for all x, so the critical
values will occur when V ‘ x 0 ; therefore,
9 2
x 0
2
3
3x 28 x 0
2
84 x
3x 0
or
x0
or
x0
or
3
x0
2
3
x 28
2
56
x
3
28
56
3
we solve that equation.
Since x 0 , the only critical value is x
168 x 12 x 2 0
We can use the second derivative,
V ” x 84 9 x ,
to determine whether we have a maximum.
56
56
V ” 84 9 84 0
3
3
56
Therefore, we have a maximum at x
.
3
56
3 56
If x
18.67 , then y 42 14.
3
2 3
Therefore, the dimensions that will maximize
the volume of the package are 18.67 in. by
18.67 in. by 14 in. The volume is
56 56
14 4878.2 in 3 .
3
3
Comparing Case I and Case II, we see that the
maximum volume is 5488 in3 when the
dimensions are 14 in. by 14 in. by 28 in.
12 x 14 x 0
12 x 0 or 14 x 0
x 0 or
x 14
Since x 0 , the only critical value is x 14.
We can use the second derivative,
V ” x 168 24 x ,
to determine whether we have a maximum.
V ” 14 168 24 14 168 0
Therefore, we have a maximum at x 14 .
If x 14 , then y 84 4 14 28.
Therefore, the dimensions that will maximize
the volume of the package are 14 in. by 14 in.
by 28 in. The volume is 14 14 28 5488 in 3 .
Case II.
The combine length and girth is
x 2 x 2 y 3 x 2 y 84.
The volume is V x x y x 2 y .
We want express V in terms of one variable.
We solve 3 x 2 y 84 , for y.
3 x 2 y 84
2 y 84 3 x
3
y 42 x
2
Thus,
3
3
V x 2 42 x 42 x 2 x 3 .
2
2
To maximize V x we first find V ‘ x .
9 2
x
2
This derivative exists for all x, so the critical
values will occur when V ‘ x 0 . We solve the
V ‘ x 84 x
equation at the top of the next column.
53. Let y represent the dimension on the lot line and
let x represent the other dimension. Then the
length of fencing that the person must pay for is
1
3
y x y x 2x y .
2
2
48
We know xy 48 ; therefore, y .
x
The length of fencing as a function of x is:
3 48
72
F x 2x 2x , 0 x .
2 x
x
72
x2
F ‘ x exists for all x in 0, . Solve:
F ‘ x 2 72 x 2 2
F ‘ x 0
72
0
x2
x 6
The solution is continued on the next page.
2
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
387
Only the critical value x 6 is in 0, . Since
Then,
there is only one critical value, we use the
second derivative to determine whether we have
a minimum.
144
F ” x 144 x 3 3
x
2
F ” 6 0 , so F 6 is a minimum.
3
48
8. The dimensions that
When x 6, y
6
minimize the cost are 6 yd by 8 yd. Where the
longer side of the lot is adjacent to the
neighbor’s yard.
A x
54. Use the figure in the text book. Since the radius
of the window is x, the diameter of the window
is 2x , which is also the length of the base of the
window.
The circumference of a circle whose radius is x
is given by:
C 2 x.
C 2 r
Therefore, the perimeter of the semicircle is
1
2 x
x.
C
2
2
The perimeter of the three sides of the rectangle
which form the remaining part of the total
perimeter of the window is given by:
2 x y y 2 x 2 y.
The total perimeter of the window is:
x 2 x 2 y 24 .
Maximizing the amount of light is the same as
maximizing the area of the window. The area of
the circle with radius x is:
A x2 ,
A r .
1 2
x 2 x 12 x x
2
2
1
x 2 24 x 2 x 2 x 2
2
1
2 x 2 24 x
2
We maximize A on the interval 0, 24 . We first
find A ‘ x .
A ‘ x 4 x 24 .
Since A ‘ x exists for all x in 0, 24 , the only
critical points are where A ‘ x 0 . Thus, we
solve the following equation:
A’ x 0
4 x 24 0
4 x 24
x
24
4
x
24
4
24
3.36
4
This is the only critical value, so we can use the
second derivative to determine whether we have
a maximum.
A ” x 4 0
x
Since A ” x is negative for all values of x, we
have a maximum at x
2
Therefore, the area of the semicircle is:
1
1
A x2.
2
2
The area of the rectangle is 2 x y .
The total area of the Norman window is
1
A x 2 2 xy.
2
To express A in terms of one variable, we solve
x 2 x 2 y 24 for y:
x 2 x 2 y 24
2 y 24 2 x x
y 12 x x
2
x
24
4
y 12
24
4
. We find y when
:
2
x x
24 24
2 4 4
24
4 12
12
4 4 4
12 48 12 24
4
12
24
3.36
4
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
388
To maximize the amount of light through the
24
ft
window, the dimensions must be x
4
24
ft, or approximately
and y
4
x 3.36 ft and y 3.36 ft.
55. Since the stained glass transmits only half as
much light as the semicircle in Exercise 54, we
express the function A as:
1 1
A x 2 2 xy
2 2
1
x 2 2 xy
4
The perimeter is still the same, so we can
substitute 12
2
x x for y to get:
1 2
x 2 x 12 x x
4
2
1
x 2 24 x 2 x 2 x 2
4
3
2 x 2 24 x, 0 x 24.
4
A x
y 12
12
2
xx
48
48
2 3 8 3 8
12 48
4.92
3 8
To maximize the amount of light through the
48
window, the dimensions must be x
ft
3 8
12 48
ft or approximately
and y
3 8
x 2.75 ft and y 4.92 ft.
56-110. Left to the Student.
1
is
x
the reciprocal of the number, and x 2 is the
square of the number. The sum, S, of the
reciprocal and five times the square is given by:
1
S x 5x 2 .
x
We want to minimize S x on the interval
111. Let x represent a positive number. Then,
Find A ‘ x .
0, . First, we find S ‘ x
3
A ‘ x 4 x 24
2
1
10 x
x2
Since S ‘ x exists for all values of x in 0, ,
S ‘ x x 2 10 x
Since A ‘ x exists for all x in 0, 24 , the only
critical points are where A ‘ x 0 . Thus, we
solve the following equation:
3
4 x 24 0
2
3 8 x 48 0
48
2.75
3 8
This is the only critical value, so we can use the
second derivative to determine whether we have
a maximum.
3
A ” x 4 0
2
Since A ” x is negative for all values of x, we
x
have a maximum at x
when x
48
. We find y
3 8
48
at the top of the next column.
3 8
the only critical values occur when S ‘ x 0 .
We solve the following equation:
1
2 10 x 0
x
1
10 x 2
x
10 x 3 1
1
x3
10
1
1
x3
10 3 10
Since there is only one critical value, we use the
second derivative,
2
S ” x 2 x 3 10 3 10 ,
x
to determine whether it is a minimum. The
second derivative is positive for all x in 0, ;
therefore, the sum is a minimum when x 3
Copyright © 2016 Pearson Education, Inc.
1
.
10
Exercise Set 2.5
389
112. Let x represent a positive number. The sum, S,
of the reciprocal and four times the square is
given by:
1
S x 4 x2.
x
We want to minimize S x on the interval
0, . First, we find S ‘ x
1
8x
x2
S ‘ x exists for all values of x in 0, . Solve:
S ‘ x x 2 8 x
S ‘ x 0
We solve the following equation:
0.18k 2ki 0
2ki 0.18k
0.18k
i
2 k
i 0.09
Since there is only one critical point, we can use
the second derivative to determine whether we
have a minimum. Notice that P ” i 2 k ,
which is a negative constant k 0 . Thus,
P ” 0.09 is negative, so P 0.09 is a
maximum. To maximize profit, the bank should
pay 9% on its savings accounts.
1
8x 0
x2
1 1
8 2
Since there is only one critical value, we use the
second derivative,
2
S ” x 2 x 3 8 3 8 ,
x
to determine whether it is a minimum. The
second derivative is positive for all x in 0, ;
x3
therefore, the sum is a minimum when x
1
.
2
113. Let A represent the amount deposited in savings
account and i represent the interest rate paid on
the money deposited. If A is directly
proportional to i , then there is some positive
constant k such that A ki . The interest earned
by the bank is represented by 18% A, or 0.18 A .
The interest paid by the bank is represented by
iA . Thus the profit received by the bank is
given by
P 0.18 A iA.
We express P as a function of the interest the
bank pays on the money deposited, i , by
substituting ki for A.
P 0.18 ki i ki
0.18ki ki 2
We maximize P on the interval 0, . First, we
find P ‘ i .
P ‘ i 0.18k 2 ki
Since P ‘ i exists for all i in 0, , the only
critical values are where P ‘ i 0 .
114. The circumference of the circle is x 2 r .
x
.
Solving the equation for r we get, r
2
The area of the circle is A r 2 , thus,
2
x2
x
.
substituting for r we have A
2
4
The perimeter of the square is 24 x .
24 x
The length of a side of the square is
.
4
Therefore, the area of the square is:
2
2
x 48 x 576
24 x
.
As
4
16
The total area is:
A Ac As
x 2 x 2 48 x 576
4
16
x2 1 2
x 3x 36
4 16
1
1
x 2 3x 36
4 16
4 2
x 3x 36
16
We minimize the area on the interval 0, 24 .
First, we find A ‘ x .
4
A’ x 2
x3
16
4
x3
8
A ‘ x exists for all x in 0, 24 ; therefore, the
only critical value occurs when A ‘ x 0.
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
390
Setting the derivative equal to zero, we solve for
the critical value.
4
x3 0
8
8 24
10.56
x 3
4 4
Since there is only one critical value, we use the
second derivative,
4
A ” x
0
8
to determine whether we have a minimum. The
second derivative is positive for all values of x
in the interval; therefore a minimum occurs at
24
10.56 .
x
4
The wire should be cut to x 10.56 in. in order
to form the circle and 24 10.56 13.44 in. in
order to form the square.
There is no maximum if the string is to be cut.
One would interpret the maximum to be at the
endpoint, with the string uncut and used to form
a circle.
115. Using the drawing in the text, we write a
function that gives the cost of the power line.
The length of the power line on the land is given
by 4 x , so the cost of laying the power line
underground is given by:
C L x 3000 4 x 12, 000 3000 x.
The length of the power line that will be under
water is 1 x 2 , so the cost of laying the
power line underwater is given by:
CW x 5000 1 x
2
Therefore, the total cost of laying the power line
is:
C x C L x CW x
12, 000 3000 x 5000 1 x 2 .
We want to minimize C x over the interval
0 x 4. First, we find the derivative.
12
1
C ‘ x 3000 5000 1 x 2
2 x
2
3000 5000 x 1 x 2
3000
5000 x
1 x2
12
Since the derivative exists for all x, we find the
critical values by solving the equation:
C ‘ x 0
5000 x
3000
0
1 x2
3000 1 x 2 5000 x 0
5000 x 3000 1 x 2
5
x 1 x2
3
2
5
x
3
1 x
2
2
25 2
x 1 x2
9
16 2
x 1
9
9
x2
16
9
x
16
3
x
4
The only critical value in the interval 0, 4 is
3
, so we can use the second derivative to
4
determine if we have a minimum.
x
1 x 5000 5000 x 12 1 x 2 x
2
C ” x
1
2
2
5000 1 x 2
1 x 2 12
2
2
5000 x 2
1 x
1 x2
2
5000
5000 x 2
1 x
5000 1 x 5000 x
1 x
1 x2
2
2
2
1
3
3
2
2
2
5000
1 x
2
3
2
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
391
C ” x is positive for all x in 0, 4 ; therefore, a
minimum occurs at x
3
3
. When x ,
4
4
3 13
3.25 .
4 4
Therefore, S should be 3.25 miles down shore
from the power station.
4
Note: since we are minimizing cost over a
closed interval, we could have used Max-Min
Principle 1 to determine the minimum, and
avoided finding the second derivative. The
3
critical value and the endpoints are 0, , and 4.
4
The function values at these three points are:
2
C 0 12, 000 3000 0 5000 1 0
17, 000
2
3
3
3
C 12, 000 3000 5000 1
4
4
4
16, 000
2
C 4 12, 000 3000 4 5000 1 4
20,615.53
Therefore, the minimum occurs when x
3
, or
4
when S is 3.25 miles down shore from the
power station.
116. The distance from point C to point S is given by
9 x 2 , and the distance from point S to point
A is 8 x . Therefore, the total energy expended
by the pigeon is:
E x 1.28r 9 x 2 r 8 x , where r is a
positive constant measuring the rate of energy
the pigeon uses.
We want to minimize E x on the interval
0,8 .
E ‘ x
1.28rx
9 x2
1
2
r
E ‘ x exists for all x in 0,8 . Solve:
E ‘ x 0
1.28rx
9 x
2
1
2
r0
x 3.755
The only critical value in 0,8 is x 3.755 , so
we can use the second derivative to determine
whether we have a maximum.
11.52r
and that
Note that E ” x
3
2
9 x2
E ” x 0 for all x in the interval 0,8 .
Therefore, a minimum occurs when x 3.755 .
The pigeon should reach land about 8 3.755
or 4.245 miles down shore from A.
117. The objective is to maximize the parallel areas
subject to the constraint of having k units of
fencing. Using the figure in the book, we can let
x equal the length of the two parallel sides and y
equal the length of the three parallel sides. Since
the two areas are the identical, we can either
maximize the area of one of the smaller areas or
maximize the area of the entire field. That is, we
can either maximize
A1 x y or A2 2 x y.
First we will maximize A1. The perimeter of the
fields is given by P 4 x 3 y . Since we have k
units of fencing, we know that
4 x 3 y k.
Solve for y and substitute into the area function
to get the area as a function of one variable.
4x 3 y k
3y k 4x
k 4x
3
1
4
y k x
3
3
Substituting into the area function we have:
4 1
4
1
A1 x k x kx x 2 .
3
3 3
3
We find the first derivative with respect to x.
dA1 1
8
k x.
dx 3
3
The deriviative exists for all values of x. We
find the critical values by solving:
dA1
0
dx
1
8
k x0
3
3
8
1
x k
3
3
1
x k.
8
The solution is continued on the next page.
y
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
392
There is only one critical value, so we can use
the second derivative to determine if it is a
maximum. Notice that the second derivative,
d 2 A1
8
, is negative for all values of x.
2
3
dx
Therefore the maximum area will occur when
1
the width of the fields are k units long. We
8
substitute in to find the length of the fields y.
1
4
y k x
3
3
1
4 1
k k
3
3 8
1
1
k k
3
6
1
k
k .
6
6
Therefore the length of the rectangular areas
that will maximize the area given k units of
k
fencing is y units.
6
Notice, if we maximized A2 2 x y . The
constraint is the same:
1
4
4 x 3 y k y k x.
3
3
4 2
8
1
A2 2 x k x kx x 2 .
3
3 3
3
dA2 2
16
k x.
3
3
dx
dA2
0
dx
2
16
k x0
3
3
1
x k.
8
There is only one critical value, so we can use
the second derivative to determine if it is a
maximum. Notice that the second derivative,
d 2 A1
16
, is negative for all values of x.
2
3
dx
Therefore the maximum area will occur when
1
the width of the fields are k units long. We
8
substitute in to find the length of the fields y.
1
4 1 k
y k k .
3
3 8 6
Therefore the length of the rectangular areas
that will maximize the area given k units of
k
fencing is y units as determined
6
previously.
118. Using the drawing in the text, we write a
function which gives the total distance between
the cities.
The distance from C1 to the bridge can be given
a 2 p x . The distance over the bridge
2
by
is r. The distance from the bridge to C2 can be
given by b2 x 2 . Therefore, the total
distance between the two cities is given by:
D x a 2 p x r x 2 b2
2
To minimize the distance, we find the derivative
of the function first.
12
1
2
D ‘ x a 2 p x 2 p x 1
2
1
1 2
2
b x 2 2 x
2
x p
x
2
2
b x2
a2 p x
The derivative exists for all values of x in the
interval 0, p . Therefore, the only critical
values occur when D ‘ x 0. We solve this
equation.
x p
a p x
2
2
x
2
b x2
0.
The solution to this equation is
bp
bp
x
or x
.
ba
ba
bp
is in 0, p .
Only x
ba
Since there is only one critical value, we can use
the second derivative to determine if there is a
minimum. The second derivative is given by:
a2
b2
D ” x
.
3
3
a 2 p x 2 2 x 2 b 2 2
D ” x 0 for all values of x; therefore, a
bp
. The bridge
ba
should be located such that the distance x is
bp
units.
ba
minimum occurs at x
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
393
x3
100
a) To determine the average cost, we divide the
total cost function by the number of units
produced:
C x
A x
.
x
x3
8 x 20
100
A x
x
119. C x 8 x 20
20 x 2
x 100
b) Taking the derivative of the total cost
function and the average cost function we
have:
d
x3
C ‘ x
8
x
20
dx
100
8
3x 2
.
100
d
20 x 2
A ‘ x
8
dx
x 100
8
Find the function value when x 10 :
20 102
11.
10 100
The minimum average cost is $11 when 10
units are produced.
3
C ‘ 10 8
102 11 .
100
The marginal cost is $11 when 10 units are
produced.
d) The two values are equal:
A 10 C ‘ 10 11.
A 10 8
C x
x
a) Taking the derivative of A x we have:
120. A x
A’ x
d
1 2
1
x
8 20 x
dx
100
1
20 x 2
2 x
100
20 x
2 .
50
x
c) The derivative exists for all x in 0, ;
therefore, the critical values occur when
A ‘ x 0 . Solve:
20
x
2
0
50
x
x
20
50 x 2
d C x
dx x
x C ‘ x C x 1
x2
x C ‘ x C x
Quotient Rule
x2
b) The derivative exists for all x in 0, ,
therefore, the critical values will occur when
A ‘ x0 0 . We solve the equation:
x0 C ‘ x0 C x0
x02
0
x0 C ‘ x0 C x0 0 Multiplying by x02 0.
x0 C ‘ x0 C x0
C ‘ x0
C x0
x0
A x0
121. Express Q as a function of one variable. First,
solve x y 1 for y. We have:
y 1 x .
Substituting we have:
x x 2 20 50
x3 1000
x 10
There is only one critical value, so we use
the second derivative to determine whether
we have a minimum.
40 1
A ” x 3
x 50
3
A ” 10
0. Thus A 10 is a minimum.
50
Q x 3 2 1 x
3
x 3 2 1 3x 3x 2 x 3
x3 6 x 2 6 x 2
Next, we find Q ‘ x .
Q ‘ x 3x 2 12 x 6
The derivative exists for all values of x in the
interval 0, .
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
394
Note the constraint y 1 x actually limits us
to look at the interval 0,1 .
The only critical values are where Q ‘ x 0 .
The derivative exists for all values of x in the
interval 2, 2 ; thus, the only critical
values are where Q ‘ x 0 . We solve:
Q ‘ x 0
We solve the equation:
3 x 2 12 x 6 0
2
x 4x 2 0
Using the quadratic formula, we have:
x 2 2 .
3 3x 2 x 2 0
3 3 x 2 x 2
1 x 2 x2
When x 2 2 , y 1 2 2 1 2.
When x 2 2 , y 1 2 2 1 2.
x4 2 x2 1 0
x 1 0
Q ” 2 2 6 2 2 12 8.48 0 , so we
have a minimum at x 2 2 .
The minimum value of Q is found by
substituting.
Q x3 2 y3
2 2
2 1 2
3
2
1 2 x2 x4
Note, Q ” x 6 x 12 and
1 x2 2 x2
Since x and y must be positive, we only consider
x 2 2 and y 1 2.
12 x 2 x 2
3
2
2
x2 1 0
x2 1
x 1
We notice that x 1 is an extraneous solution
which does not work.
3 3 1 2 1 3 3 6 0 .
64 2
2
122. Express Q as a function of one variable. First,
solve x 2 y 2 2 for y. We have:
y 2 2 x2
y 2 x2
y is a real number if x is in the interval
2, 2 .
Therefore, the only critical value is x 1 . The
critical value and the endpoints are
2, 1, and 2.
3
2 2
Q 2 3 2 2 2 3 2
Q 1 3 1 2 1
2
4
3
2
3
2 3 2 2 2 3 2
2
2
If y 2 x 2 , we substitute for y to get:
Q
Q 3x y 3
The minimum value of Q is 3 2 and occurs
Q 3x 2 x
3x 2 x 2
3
2
3
when x 2 and y 2 2
2
Next, we find Q ‘ x .
1
2
3
Q ‘ x 3 2 x 2 2 x
2
3 3x 2 x
2
3 3x 2 x 2
1
2
0
2
Next, we repeat the process for y 2 x 2 .
We notice that:
Q 3x y 3
Q 3x
2 x2
3x 2 x 2
3
3
2
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
395
Next, we find Q ‘ x .
123. Let x be the lot size.
3 3x 2 x 2
1
2
3 3x 2 x 2
The derivative exists for all values of x in the
interval 2, 2 ; thus, the only critical
values are where Q ‘ x 0 . We solve the
equation:
3 3x 2 x 2 0
3 3x 2 x 2
1 x 2 x2
12 x 2 x 2
2
1 x 2 x
2
2
C ‘ x 0
a bQ
0
2 x2
a bQ
2 x2
x4 2 x2 1 0
x 1 0
2
x2 1 0
x2 1
x 1
We notice that x 1 is an extraneous solution
which does not work.
3 3 1 2 1 3 3 6 0 .
2
Therefore, the only critical value is x 1 . The
critical value and the endpoints are
2, 1, and 2. We evaluate the function:
3
2 2
Q 2 3 2 2 2 3 2
Q 1 3 1 2 1
Q
2 3
2
4
2 2
3
2
3
2 3 2
2 2
The minimum value of Q is 3 2 occurs when
ax bQ
cQ, 1 x Q
2
x
To find the minimum, we take the first
deriviative:
a
a bQ
C ‘ x bQx 2 2
2
2 x
C ‘ x exists for all x in 1,Q . So the only
critical values occur when C ‘ x 0 . Solve:
1 2 x2 x4
2
x ax
2 2
Q
Cr x b cx
x
Yearly reorder cost:
bQ
cQ
x
Then,
C x Cc x Cr x
Yearly carrying cost: Cc x a
1
3
2
Q ‘ x 3 2 x 2 2 x
2
x 2 and y 2 2
0.
2
Regardless of what value of y we chose, we see
that the minimum of Q, is 3 2 , when
x 2 and y 0.
ax 2 2bQ
2bQ
x2
a
2bQ
x
a
The only critical value in the domain is
2bQ
.
a
Since there is only one critical value in the
domain, we use the second derivative,
2bQ
C ” x 2bQx 3 3
x
to determine whether we have a minimum.
C ” x 0 for all x in 1,Q , so a minimum
x
2bQ
.
a
In order to minimize inventory costs. The store
Q
aQ
times per year.
should order
2b
2bQ
a
occurs at x
The lot size will be
Copyright © 2016 Pearson Education, Inc.
2bQ
units.
a
Chapter 2: Applications of Differentiation
396
124. From Exercise 123, we know that the store
2bQ
aQ
should order a lot size of
units,
a
2b
times per year.
When Q 2500, a 10, b 20, c 9 , the
store should order:
126. Using a spreadsheet we numerically estimate
the maximum:
Starting
Value:
Step Size:
10 2500
aQ
25 times per year.
2b
2 20
x
3.22
3.23
3.24
3.25
3.26
3.27
3.28
3.29
3.3
The lot size of each order should be:
2 20 2500
100 units.
10
125-128. The starting value and step size were
chosen to limit the amount of space used .
Approaches can vary.
125. Using a spreadsheet we numerically estimate
the maximum:
Starting
Value:
Step Size:
x
24
24.25
24.5
24.75
25
25.25
25.5
25.75
26
3.22
0.01
y 16 x 2
Q x2 y
2.373099239
2.35947028
2.345719506
2.331844763
2.317843826
2.303714392
2.289454083
2.275060439
2.260530911
24.60524215
24.61611748
24.62442508
24.63011031
24.63311704
24.63338762
24.63086281
24.6254817
24.61718162
We determine the maximum of Q 24.63 to
occur when x 3.27 and y 2.30.
24
0.25
127. Using a spreadsheet we numerically estimate
the maximum:
100 2 x
y
3
q x y
17.33333333
17.16666667
17
16.83333333
16.66666667
16.5
16.33333333
16.16666667
16
416
416.2917
416.5
416.625
416.6667
416.625
416.5
416.2917
416
We determine the maximum of Q 416.67 to
occur when x 25 and y 16.67.
Starting
Value:
Step Size:
2.2
0.01
y 10 0.5 x 2
x
2.2
2.21
2.22
2.23
2.24
2.25
2.26
2.27
2.28
2.75317998
2.749172603
2.745141162
2.741085551
2.737005663
2.732901389
2.728772618
2.724619239
2.720441141
Q x y3
45.91202934
45.91962105
45.92477319
45.92748186
45.92774328
45.9255538
45.92090987
45.91380804
45.90424501
We determine the maximum of Q 45.93 to
occur when x 2.24 and y 2.74.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.5
397
128. Using a spreadsheet we numerically estimate
the maximum:
Starting
Value:
Step Size:
4.6
0.1
x
4.6
4.7
4.8
4.9
5
5.1
5.2
5.3
5.4
y 50 x 2
Q x2 y2
5.370288633
5.282991577
5.192301994
5.098038839
5
4.897958759
4.79165942
4.680811895
4.565084884
610.2544
616.5319
621.1584
624.0199
625
623.9799
620.8384
615.4519
607.6944
We determine the maximum of Q 625 to
occur when x 5 and y 5.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
398
2.
Exercise Set 2.6
1.
a) Total profit is revenue minus cost.
P x R x C x
R x 50 x 0.5 x 2 ; C x 4 x 10
5 x 0.001x 2 1.2 x 60
a) Total profit is revenue minus cost.
P x R x C x
5 x 0.001x 1.2 x 60
0.001x 2 3.8 x 60
b) R 100 5 100 500
50 x 0.5 x 2 4 x 10
The total revenue from the sale of the first
100 units is $500.
0.5 x 46 x 10
b) Substituting 20 for x into the three
functions, we have:
2
C 100 0.001 100 1.2 100 60 190
2
The total cost of producing the first 100
units is $190.
R 20 50 20 0.5 20 800
2
P 100 0.001 100 3.8 100 60 310
The total revenue from the sale of the first
20 units is $800.
C 20 4 20 10 90
2
The total profit is $310 when the first 100
units are produced and sold.
c) R ‘ x 5
The total cost of producing the first 20 units
is $90.
P 20 R 20 C 20
C ‘ x 0.002 x 1.2
P ‘ x 0.002 x 3.8
800 90
710
The total profit is $710 when the first 20
units are produced and sold.
Note, we could have also used the profit
function, P x , from part (a) to find the
d) R ‘ 100 5
Once 100 units have been sold, the
approximate revenue for the 101st unit is $5.
C ‘ 100 1.4
Once 100 units have been produced, the
approximate cost for the 101st unit is $1.40.
P ‘ 100 0.002 100 3.8 3.6
profit.
P 20 0.5 20 46 20 10 710
2
Once 100 units have been produced and
sold, the approximate profit from the sale of
the 101st unit is $3.60.
e)
In part (b), we are observing the total
revenue, cost and profit from the production
and sale of the first 100 items. In part (d),
we are observing the approximate revenue,
cost and profit from the production and sale
of the 101st unit only. These quantities are
also known as the marginal revenue,
marginal cost and marginal profit.
c) Finding the derivative for each of the
functions, we have:
R ‘ x 50 x
C ‘ x 4
P ‘ x x 46
d) Substituting 20 for x in each of the three
marginal functions, we have:
R ‘ 20 50 20 30
Once 20 units have been produced, the
approximate cost for the 21st unit is $4.
P ‘ 20 20 46 26
Once 20 units have been produced and sold,
the approximate profit from the sale of the
21st unit is $26.
2
P x 50 x 0.5 x 2 4 x 10
Once 20 units have been sold, the
approximate revenue for the 21st unit is $30.
C ‘ 20 4
R x 5 x; C x 0.001x 2 1.2 x 60
3.
C x 0.002 x 3 0.1x 2 42 x 300
a) Substituting 40 for x into the cost function,
we have:
C 40 0.002 40 0.1 40 42 40 300
3
2
2268 (hundreds of dollars)
The current daily cost of producing 40
security systems is $226,800.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.6
399
c) C ‘ x 0.003 x 2 0.14 x 19.
b) In order to find the additional cost of
producing 41 chairs monthly, we first find
the total cost of producing 41 security
systems in a day.
C ‘ 25 0.003 25 0.14 25 19
2
24.375
The marginal cost when 25 daypacks have
been produced is $24.38.
d) Using the marginal cost from part (c), the
additional cost required to produce 2
additional daypacks monthly is:
2 24.375 48.75.
C 41 0.002 41 0.1 41 42 41 300
3
2
2327.94 (hundreds of dollars)
Next, we subtract the cost of producing 40
security systsems daily found in part (a)
from the cost of producing 41 security
systems daily.
C 41 C 40 2327.94 2268 59.94
Therefore, the difference in cost between
producing 25 and 27 daypacks per month is
approximately $48.75.
e) In part (a) we found that it cost $1234.38 to
produce 25 daypacks per month. In part (d)
we found that the difference in cost between
25 daypacks and 27 daypacks per month
was $48.75. Therefore, the approximate total
cost of producing 27 daypacks per month is
C 27 1234.38 48.75 1283.13.
The additional daily cost of increasing
production to 41 security systems daily is
$5994.
c) First, we find the marginal cost function,
C ‘ x 0.006 x 2 0.2 x 42
Next, substituting 40 for x, we have:
C ‘ 40 59.6
The marginal cost when 40 security systems
are produced daily is $5960.
d) In part (a) we found that it cost $2268
hundred to produce 40 security systems per
day. The additional cost of producing 2
additional security systems is
2($59.60) $119.20 hundreds. Therefore,
the estimated daily cost of producing 42
security systems per day is
C 42 $2268 $119.20 $2387.20
hundreds or $238,720 per day.
4.
C x 0.001x 3 0.07 x 2 19 x 700
a) Substituting 25 for x into the cost function,
we have:
C 25 0.001 25 0.07 25 19 25 700
3
2
1234.375
The current monthly cost of producing 25
daypacks is $1234.38.
b) We first find the total cost of producing 26
daypacks in a month.
C 26 0.001 26 0.07 26 19 26 700
3
2
1258.896
Therefore,
C 26 C 25 1258.896 1234.375
24.521
The additional cost of increasing production
to 26 daypacks monthly is $24.52.
We predict the cost of producing 27
daypacks monthly will be $1283.13.
5.
R x 0.005 x 3 0.01x 2 0.5 x
a) Substituting 70 for x, we have:
R 70 0.005 70 0.01 70 0.5 70
3
2
1715 49 35
1799
The currently daily revenue from selling 70
lawn chairs per day is $1799.
b) Substituting 73 for x, we have:
R 73 0.005 73 0.01 73 0.5 73
3
2
2034.875
2034.88
Therefore, the increase in revenue from
increasing sales to 73 chairs per day is:
R 73 R 70 2034.88 1799
235.88
Revenue will increase $235.88 per day if the
number of chairs sold increases to 73 per
day.
c) First we find the marginal revenue function
by finding the derivative of the revenue
function.
R ‘ x 0.015 x 2 0.02 x 0.5
Substituting 70 for x, we have:
R ‘ 70 0.015 70 0.02 70 0.5
2
75.40
The marginal revenue when 70 lawn chairs
are sold daily is $75.40.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
400
b) First we find the total montly revenue for
selling 28 suitcases
3
2
R 28 0.007 28 0.5 28 150 28
d) In part (a) we found that selling 70 lawn
chairs per day resulted in a revenue of
$1799. In part (c) we found that the
marginal revenue when 70 chairs were sold
is $75.40. Using these two numbers, we
estimate the daily revenue generated by
selling 71 chairs is
R 71 R 70 R ‘ 70
3961.66
The difference in monthly revenue from
selling 26 suitcases and 28 suitcases a month
is:
R 28 R 26 $3961.66 $3685.03
$1799 $75.40 $1874.40.
Similarly, the daily revenue generated by
selling 72 chairs, or 2 additional chairs,
daily is approximately
R 72 R 70 2 R ‘ 70
$276.63.
If sales increased from 26 to 28 suitcases,
monthly revenue would increase $276.63
c) First we find marginal revenue by taking the
derivative of the revenue function.
R ‘ x 0.021x 2 x 150
$1799 2 $75.40 $1949.80.
The daily revenue generated by selling 73
chairs, or 3 additional chairs, daily is
approximately
R 73 R 70 3 R ‘ 70
Next we substitute 26 in for x.
R ‘ 26 0.021 26 26 150 138.196
2
Marginal revenue is 138.20 when 26
suitcases are sold.
d) From part (a), we know that when 26
suitcases are sold, total monthly revenue is
$3685.03. From part (c), we know that when
26 suitcases are sold, marginal revenue is
$138.20. Therefore, we estimate:
R 27 R 26 R ‘ 26
$1799 3 $75.40 $2025.20.
6.
P x 0.006 x 3 0.2 x 2 900 x 1200
a) P 60 $50,784
b) P 60 P 59 50, 784 49, 971.53 812.47
The dealership would lose $812.47 per week
if it were only able to sell 59 cars weekly.
c) P ‘ x 0.018 x 2 0.4 x 900
P ‘ 60 0.018 60 0.4 60 900
R 27 $3685.03 $138.20 $3823.23
We estimate the revenue from selling 27
suitcases per month to be $3823.23.
2
811.20
The marginal profit is $811.20 when 60 cars
are sold each week.
d) P 61 P 60 P ‘ 60 $51,595.20
The estimated profit of selling 61 cars per
week is $51,595.20.
7.
R x 0.007 x 3 0.5 x 2 150 x
a) Substituting 26 for x, we have:
R 26 0.007 26 0.5 26 150 26
3
2
123.032 338 3900
3685.03
The current monthly revenue is $3685.03.
8.
P x 0.004 x 3 0.3 x 2 600 x 800
a) Substituting 9 for x, we have:
P 9 0.004 9 0.3 9 600 9 800
3
2
4572.784
The currently weekly profit is $4572.78.
b) The difference in weekly profit from selling
8 laptops and 9 laptops per week is
P 9 P 8 4572.78 3978.75
594.03
Therefore, Crawford Computing would lose
$594.03 each week if 8 laptops were sold
each week instead of 9.
c) P ‘ x 0.012 x 2 0.6 x 600
P ‘ 9 0.012 9 0.6 9 600
2
593.628
The marginal profit is $593.63 when 9
laptops are sold weekly.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.6
401
d) We estimate:
P 10 P 9 P ‘ 9
12.
C C x x C x
4572.78 593.63
C C 70 1 C 70
5166.41
The total weekly profit is approximately
$5166.41 when 10 laptops are built and sold
weekly.
9.
C 71 C 70
0.01 71 0.6 71 30
2
0.01 70 2 0.6 70 30
2.01
The additional cost of producing the 71st unit is
$2.01.
C ‘ x 0.02 x 0.6
N 1000 500,000 means that 500,000
computers will be sold annually when the price
of the computer is $1000. N ‘ 1000 100
means that when the price is increased $1 to
$1001, sales will decrease by 100 computers per
year.
10.
C ‘ 70 0.02 70 0.6 2.00
The marginal cost when 70 units are produced is
$2.00.
N 1025 N 1000 25 N ‘ 1000
N 1025 500,000 25 100 497,500
13.
R x 2 x
R R x x R x
We estimate that 497,500 computers will be
sold annually if the price is increased to $1025.
11.
C x 0.01x 2 0.6 x 30
Substituting x 70, and x 1 we have
R R 70 1 R 70
C x 0.01x 1.6 x 100
2
R 71 R 70
C C x x C x
2 71 2 70
Substituting x 80, and x 1 we have
C C 80 1 C 80
2
The additional revenue from selling the 71st
unit is $2.00.
Finding the derivative of R x we have:
0.01 81 1.6 81 100
2
0.01 80 2 1.6 80 100
3.21
The additional cost of producing the 81st unit is
$3.21.
Finding the derivative of C x we have:
R ‘ x 2 .
The derivative is constant; therefore,
R ‘ 70 2
The marginal revenue when 70 units are
produced is $2.00.
C ‘ x 0.02 x 1.6
Substituting 80 for x, we have:
C ‘ 80 0.02 80 1.6 3.20
The marginal cost when 80 units are produced is
$3.20.
14.
R x 3x
R R x x R x
R R 80 1 R 80
3 81 3 80
3
The additional revenue from selling the 81st
unit is $3.00.
R ‘ x 3 .
R ‘ 80 3
The marginal revenue when 80 units are
produced is $3.00.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
402
15.
C x 0.01x 2 1.6 x 100; R x 3 x
The additional profit of producing and
selling the 71st unit is $0.01.
P ‘ x 0.02 x 1.4
a) Finding the profit function we have:
P x R x C x
3 x 0.01x 2 1.6 x 100
P ‘ 70 0.02 70 1.4 0.00
The marginal profit when 70 units are
produced and sold is $0.00.
0.01x 2 1.4 x 100
b) P P x x P x
Substituting x 80, and x 1 we have
P P 80 1 P 80
17.
0.01 81 1.4 81 100
2
0.01 80 2 1.4 80 100
0.21
The additional profit of producing and
selling the 81st unit is $0.21.
Finding the derivative of P x we have:
S 0.007 25 0.5 25 150 25
3
Substituting 80 for x, we have:
P ‘ 80 0.02 80 1.4 0.20
The marginal profit when 80 units are
produced and sold is $0.20.
Note: We notice that P R C and
P ‘ x R ‘ x C ‘ x . We could have used
This result implies when the price is $25, a
$1 increase in price will lead to an increase
in supply of approximately 138 pens.
d)
We would expect the rate of change of
quantity with respect to price to be positive.
All things being equal, it is reasonable to
assume as the price of a good or service
increases, the supply for that good or service
will increase.
this knowledge and our work from Exercises
11 and 14 to simplify our work. We have:
P R C
3 3.21 0.21
P ‘ 80 R ‘ 80 C ‘ 80
3 3.20
0.20
C x 0.01x 2 0.6 x 30; R x 2 x
a) Finding the profit function we have:
P x R x C x
2 x 0.01x 2 0.6 x 30
0.01x 2 1.4 x 30
b) P P x x P x
P P 70 1 P 70
18.
13 x 100
x
x 13 13 x 100 1
A ‘ x
x2
100
2
x
Therefore,
A A ‘ x x
A x
A ‘ 100 x
P 71 P 70
0.01 71 1.4 71 30
2
0.01 70 2 1.4 70 30
0.01
2
109.375 312.50 3750
3546.875
Producers will want to supply 3547 units
when price is $25 per unit.
c)
Substituting 25 for p into the answer
from part (a) we have:
dS
2
0.021 25 25 150 138.125
dp p 25
P ‘ x 0.02 x 1.4
16.
S 0.007 p 3 0.5 p 2 150 p
a) We take the derivative of the supply
function with respect to price.
dS
0.021 p 2 p 150
dp
b) Substituting 25 for p in the supply function
we have:
100
100
2
1
x 100
x 1
0.01
The average cost changes by about $0.01 .
(We see an approximate decrease in average
cost of one cent.)
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.6
19.
403
M t 2t 2 100t 180
21.
a) Substituting for t, we have:
567 36 x 104 x
x is the number of years since 1960; therefore,
the year 2014 corresponds to
x 2014 1960 54 , and the year 2015
corresponds to x 2015 1960 55 . To
estimate the increase in gross domestic product
from 2014 to 2015, we establish that
x 54 and x 1 . Next, we find the derivative
of P x :
2
M 10 2 10 100 10 180 980
2
M 25 2 25 100 25 180 1430
2
M 45 2 45 100 45 180 630
b) To find marginal productivity, we take the
derivative of the productivity function:
M ‘ t 4t 100 .
2
P ‘ x 36 1.6 x 0.6 104 57.6 x 0.6 104 .
Substituting for t, we have:
M ‘ 5 4 5 100 80
Therefore,
P P ‘ x x
M ‘ 10 4 10 100 60
P ‘ 54 x
M ‘ 25 4 25 100 0
M ‘ 45 4 45 100 80
We see that the additional monthly output
per years of service decreases each year the
employee is with the company.
d)
The employees marginal productivity is
at its highest point when the employee is
new to the company. The employee is still
learning how to do the job and will make the
greatest gains. As the employee gains
experience, the marginal productivity begins
to decrease. The employee is still being
more productive each month, but just
doesn’t increase total output as much as the
previous month’s increase. Eventually, age
catches up to the employee and they cannot
produce the output they once did. Marginal
productivity becomes negative as total
output starts to fall.
20.
S p 0.08 p3 2 p2 10 p 11
p 18.00, p 18.20 18.00 0.20
S ‘ p 0.24 p2 4 p 10
S S ‘ p p
S ‘ 18.00 p
0.24 18 4 18 10 0.2
2
31.952
The supplier will supply approximately 32 more
units.
1.6
M 5 2 5 100 5 180 630
c)
P x 567 x 36 x 0.6 104
57.6 54
0.6
x 54
104 x
526.7521497 1
x 1
526.75
The gross domestic product should increase
about $526.75 billion between 2014 and 2015.
22.
N x x 2 300 x 6
x is in thousands of dollars so, x 100, x 1 .
N ‘ x 2 x 300
N N ‘ x x
N ‘ 100 x
2 100 300 1
100
Norris will sell approximately 100 more units
by increasing its advertising expenditure from
$100,000 to $101,000.
23.
Yes, the taxation in 2014 was progressive.
The 25,001st dollar is taxed at a rate of 15%, the
80,001st dollar is taxed at a rate of 25%, and the
140,001st dollar is taxed at a rate of 33%.
24. Marcy’s marginal tax rate is 28%, while Tyrone
marginal tax rate is also 28%. However, an
increase in $5000 will push Tyrone into the next
tax bracket putting his marginal tax at
33%.Therefore, since Marcy is in a lower
marginal tax bracket, she will keep more of the
$5000 after taxes.
25. Alan’s marginal tax rate is currently 25%. If he
earns another $10,000, dollars, this will push
him into the 28% tax bracket and he will pay
about $0.28 per dollar earned in taxes.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
404
26. The marginal tax rate at $50,000 is 25%. By
earning an extra $3000 it will not push her out
of the 25% tax bracket. Therefore, her tax
liability will not grow if she takes the extra
work.
27.
30.
y f x x f x
f 3 0.02 f 3
2
2
3.02 3.02 3 3
0.1004
f ‘ x x 1 2 x x
y f x x 3 , x 2, and x 0.01
y f x x f x
f ‘ 3 x 1 2 3 0.02
2 for x and
f 2 0.01 f 2 Substituting
0.01 for x
f 2.01 f 2
2.01 2
3
0.10
3
0.1206
f ‘ x x 3x 2 x f x x3 ; f ‘ x 3x 2
f ‘ 2 x 3 2 0.01
2
12 0.01
31.
f 1.2 f 1
1 1
1.2 1
0.1667
y f x x 2 , x 2, and x 0.01
y f x x f x
f ‘ x x x 2 x
f 2 0.01 f 2
2.01 2
2
2
f ‘ 1 x 1
0.0401
f ‘ x x 2 x x
32.
0.04
y f x x x 2 , x 3, and x 0.04
y f x x f x
f 3.04 f 3
Substituting 3 for x and
0.04 for x
2
2
3.04 3.04 3 3
12.2816 12
0.2816
f ‘ x x 1 2 x x
0.2
f x x 1 ;
2
f ‘ x x
Substituting 1 for x and
0.2 for x
0.2
4 0.01
f 3 0.04 f 3
2
Substituting 1 for x and
0.2 for x
1 0.2
f ‘ 2 x 2 2 0.01
29.
1
x 1 , x 1, and x 0.2
x
y f x x f x
y f x
f 1 0.2 f 1
Substituting 2 for x and
0.01 for x
0.12
28.
y f x x x 2 , x 3, and x 0.02
f x x x 2 ;
f ‘ x 1 2 x
f ‘ 3 x 1 2 3 0.04
1
x 2 , x 1, and x 0.5
x2
y f x x f x
y f x
f 1 0.5 f 1
1 1
2
2
1.5 1
0.5556
f ‘ x x 2 x 3 x
f ‘ 1 x 2 1 0.5
2 0.5
3
1
Substituting 3 for x and
0.04 for x
7 0.04
0.28
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.6
33.
405
y f x 3 x 1, x 4, and x 2
y f x x f x
f 4 2 f 4
f 6 f 4
6
f ‘ x x 3 x f x 3 x 1; f ‘ x 3
Substituting 4 for x and
2 for x
6
y f x x f x
f 8 0.5 f 8
2 8.5 3 2 8 3
1
f ‘ x x 2 x
1
2 x
1
x
1
2 9
1
0.167
6
We can now approximate
8;
3 y
3 0.167 2.833
To five decimal places 8 2.82843. Thus,
our approximation is reasonably accurate.
1
35. We first think of the number closest to 26 that is
a perfect square. This is 25. What we will do is
approximate how y x , changes when x
changes from 25 to 26. Let
y f x x x 2
1
1
1 12
1
x
2
2 x
y f ‘ x x
x
37. We first think of the number closest to 102 that
is a perfect square. This is 100. What we will do
is approximate how y x , changes when x
changes from 100 to 102. Let
y f x x x 2
Using, y f ‘ x x , we have
1
y f ‘ x x
8 9 y
f ‘ 8 x 2 0.5
36. Let y f x x , x 8, x 1
y f x 2 x 3, x 8, and x 0.5
Then f ‘ x
5 0.100
5.100
To five decimal places 26 5.09902. Thus,
our approximation is reasonably accurate.
17 11
34.
26 ;
5 y
Substituting 4 for x and
2 for x
3 6 1 3 4 1
f ‘ 4 x 3 2
We can now approximate
26 25 y
Then f ‘ x
1 12
1
x
2
2 x
Using, y f ‘ x x , we have
y f ‘ x x
2 x
We are interested in y as x changes from 25 to
26, so
1
y
x
2 x
1
1 Replacing x with 25 and x with 1
2 25
1
25
1
0.100
10
1
x
2 x
We are interested in y as x changes from 100
to 102, so
1
y
x
2 x
1
2 Replacing x with 100 and x with 2
2 100
1
2
2 10
1
0.100
10
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
406
We can now approximate
102 100 y
We are interested in y as x changes from 1000
to 1005, so
1
y
x
3 2
3 x
102 ;
10 y
10 0.100
Replacing x with 1000 and x with 5, we have
10.100
To five decimal places 102 10.09950. Thus,
our approximation is reasonably accurate.
1
x
3
3
2 100
3
0.150
20
We can now approximate
10.017
To five decimal places 3 1005 10.01664 Thus,
our approximation is reasonably accurate.
103 ;
40. Let y f x 3 x , x 27, x 1
10.150
y f ‘ x x
To five decimal places 103 10.14889. Thus,
our approximation is reasonably accurate.
39. We first think of the number closest to 1005 that
is a perfect cube. This is 1000. What we will do
is approximate how y 3 x , changes when x
changes from 1000 to 1005. Let
1 23
1
x
3 2
3
3 x
Using, y f ‘ x x , we have
y f ‘ x x
3 x2
3 x2
1
x
3 3 27
2
1
1
0.037
27
We can now approximate 3 28 ;
3
28 3 27 y
3 0.037
3.037
To five decimal places 3 28 3.03659 Thus,
our approximation is reasonably accurate.
1
1
1
3
y f x 3 x x 3
3
1005 3 1000 y
10 0.017
10 0.15
5
10 y
103 100 y
Then f ‘ x
2
y f ‘ x x
2 x
1
3 3 1000
1
5
3 100
1
0.017
60
We can now approximate 3 1005 ;
38. Let y f x x , x 100, x 3
1
x
41.
y 3 x 2 3 x 2 2
1
First, we find
dy
:
dx
dy 1
3
1
3 x 2 2 3
.
dx 2
2 3x 2
Then
3
dy
dx .
2 3x 2
Note that the expression for dy contains two
variables x and dx.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.6
42.
407
y x 1 x 1 2
1
First, we find
45.
dy
:
dx
dy
dx
dy
1
dy 9 x 2 2 x 3 1 dx .
Note that the expression for dy contains two
variables x and dx.
First, we find
46.
2
2
2
.
2
x4 8×2 4 x 3
x 3
2
2
dx .
y 5 x 27 x 27 5
1
dy
:
dx
dy 1
1
4
x 27 5 1
.
4
dx 5
5 5 x 27
2
Then
dy
:
dx
4 x 3 2 x 5 3 x 2 2 x 5
dy
2
4 x 6 x 15 x 2 x 5
10 x 15 x 2 x 5
4 x 3 x 2 x 5 2 x 5
2
3
3
x 3
First, we find
dy
2
x 3 2 2 x 5 2 3 x 2 2 x 5
dx
3
4
Note that the expression for dy contains two
variables x and dx.
9 x 2 2 x 3 1.
Then
y x 2 x 5
2
Then
9 x2 2 x3 1 2
3
2
x4 8×2 4 x 3
2
3
2
By the extended power rule
3
4
y 2 x3 1 2
dy
. By the quotient rule we have:
dx
x2 3 3×2 1 x3 x 2 2 x
x 3
3x 10 x 3 2 x 2 x 4 x .
x 3
dy
:
First, we find
dx
1
dy 3
2 x3 1 2 6 x2
dx 2
44.
x3 x 2
x2 3
First, we find
dy 1
1
1
x 1 2 1
.
dx 2
2 x 1
Then
1
dy
dx .
2 x 1
43.
y
47.
.
1
5 5 x 27
4
dx .
y x 4 2 x3 5×2 3x 4
First, we find
2
dy
:
dx
dy
4 x 3 6 x 2 10 x 3 .
dx
Then
dy 4 x 3 6 x 2 10 x 3 dx .
2
5 x 2 2 x 3 2 x 5 .
Then
dy 5 x 2 2 x 3 2 x 5 dx .
Note that the expression for dy contains two
variables x and dx.
48.
y 7 x
8
First, we find
dy
:
dx
dy
7
7
8 7 x 1 8 7 x .
dx
Then
dy 8 7 x dx .
7
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
408
53. Let y f x x 4 x 2 8
49. From Exercise 48, we know:
dy 8 7 x dx ,
7
when x 1 and dx 0.01 , we have:
dy 8 7 1 0.01
7
7
22,394.88.
To approximate f 5.1 , we will use
When x 2 and dx 0.1 we have:
dy 4 2 6 2 10 2 3 0.1
2
31 0.1
3.1
51.
x 5 and dx 0.1 to determine the differential
dy.
Substituting 5 for x and 0.1 for dx we have:
dy f ‘ 5 dx
y 3 x 10
490 0.1
dy
:
dx
15 3 x 10
Then
49
Next, we find
f 5 5 5 8
4
4
When x 4 and dx 0.03 we have:
Now,
f 5.1 f 5 f ‘ 5 dx
608 49
dy 15 3 4 10 0.03
4
657
15 2 0.03
4
54. Let y f x x 3 5 x 9
7.2
First we find f ‘ x :
y x5 2 x3 7 x
f ‘ x 3x 2 5 .
dy
:
First, we find
dx
dy
5×4 6 x2 7 .
dx
Then
dy 5 x 4 6 x 2 7 dx .
Then
dy f ‘ x dx
To approximate f 3.2 , we will use
When x 3 and dx 0.02 we have:
dy 5 3 6 3 7 0.02
2
3 x 2 5 dx
4
2
625 25 8
608
4
dy 15 3 x 10 dx .
6.88.
3
500 10 0.1
5
dy
4
5 3 x 10 3 By the extended power rule
dx
4 5 2 5 0.1
4 125 10 0.1
First, we find
52.
4 x 3 2 x dx
50. From Exercise 47, we know:
dy 4 x 3 6 x 2 10 x 3 dx .
3
f ‘ x 4 x3 2 x .
Then
dy f ‘ x dx
8 6 0.01
First we find f ‘ x :
x 3 and dx 0.2 to determine the differential
dy.
dy f ‘ 3 dx
3 3 5 0.2
2
27 5 0.2
22 0.2
4.4
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.6
409
Next, we find
3
f 3 3 5 3 9
The differential is:
dN N ‘ t dt
27 15 9
21
Now,
f 3.2 f 3 f ‘ 3 dx
25.4
S 0.02235h0.42246 w0.51456
We begin by noticing that we are wanting to
estimate the change in surface area due to a
change in weight w; therefore, we will first find
dS
. Since h 160 , we have:
dw
S 0.02235 160
0.42246
0.02235 8.53399783 w 0.51456
Now that we have the differential, we can use
her weight of 60 kg to approximate how much
her surface area changes when her weight drops
1 kg.
We substitute 60 for w and 1 for dw to get:
dS 0.09814452 60
0.48544
1
0.01345
The patient’s surface area will change by
0.01345 m 2 .
56.
0.8t 1000
5t 4
First we find N ‘ t by the quotient rule.
N t
N ‘t
5t 4 0.8 0.8t 1000 5
5t 4 2
4t 3.2 4t 5000
5t 4 2
4996.8
5t 4 2
dt
4996.8
9
2
0.1
6.16889
Next, we approximate the change in bodily
concentration from 2.8 hr to 2.9 hr by using 2.8
for t and 0.1 for dt.
4996.8
0.1
dN
2
5 2.8 4
w 0.51456
0.19073485w 0.51456
Now we can take the derivative of S with
respect to w.
dS
0.19073485 0.51456 w 0.48544
dw
0.09814452w 0.48544
Therefore,
dS 0.09814452 w 0.48544 dw
5t 4 2
We approximate the change in bodily
concentration from 1.0 hr to 1.1 hr by using 1.0
for t and 0.1 for dt.
4996.8
0.1
dN
2
5 1.0 4
21 4.4
55.
4996.8
4996.8
18
2
0.1
1.54222
The concentration changes more from 1.0 hr to
1.1 hr.
57.
p x 0.06 x3 0.5 x 2 1.64 x 24.76
First we find p ‘ x .
p ‘ x 0.06 3x 2 0.5 2 x 1.64
0.18 x 2 x 1.64.
The differential is:
dp p ‘ x dx
0.18 x 2 x 1.64 dx
Since x is the number of years since 2008, we
have 2010 implies x 2 and 2012 implies
x 4 . To estimate the change in ticket prices
from 2010 and 2012, we substitute 2 for x and 2
for dx.
dp 0.18 2 2 1.64 2
2
0.36 2 0.72
To estimate the change in ticket prices from
2014 and 2016, we substitute 6 for x and 2 for
dx.
dp 0.18 6 6 1.64 2
2
2.12 2 4.24
Ticket prices will increase more between 2014
and 2016.
Copyright © 2016 Pearson Education, Inc.
410
Chapter 2: Applications of Differentiation
58. The circumference of the earth, which is the
original length of the rope, is given by
C r 2 r , where r is the radius of the earth.
b) The possible extra area is 628 square feet.
Since each additional can will cover 300
square feet, they will need to bring 3 extra
cans to account for any extra area.
c) 3 additional cans will cost the painters $90.
We need to find the change in the length of the
radius, r , when the length of the rope is
increased 10 feet.
61. The volume of the spherical cavern is given by:
V r
4 3
r
3
First we find the derivatve of the volume
function.
r
4
V ‘ r 3r 2 4 r 2 .
3
r
Therefore, the differential is:
dV 4 r 2 dr
Substituting the given information, we have:
dV 4 3.14400 2 4, 019, 200.
2
Using differentials, C C ‘ r r represents
the change in the length of the rope. Therefore,
C 10 , and we have:
10 C ‘ r r
Noticing that C ‘ r 2 , we have:
The enlarged cavern will contain an additional
4,019,200 cubic feet.
62.
R x p x
10 2 r
10
r
2
Therefore, the rope is raised approximately
5
r 1.59 feet above the earth.
59.
A x
3
100 x x 2 .
3 1
3 x
.
R ‘ x 100 x 2 100
2
2
x
To find Marginal Average Cost, we take the
derivative of the average cost function. By the
quotient rule we have:
x C ‘ x C x 1
A ‘ x
x2
x C ‘ x C x
.
x2
60. a) The surface area of the water tank is given
by A 2 r .
Calculating the differential we have:
2
dA A ‘ r dr 4 rdr.
The tolerance is in feet is 0.5 .
dA 4 3.14100 0.5 628
100 x x
63.
C x
p 100 x
p 400 x
Since revenue is price times quantity, the
revenue function is given by:
R x p x
400 x x
400 x x 2
To find the marginal revenue, we take the
derivative of the revenue function. Thus:
R ‘ x 400 2 x
64.
p 500 x
R x p x
500 x x
500 x x 2
R ‘ x 500 2 x .
The approximate difference in surface area
when the tolerance is taken into
consideration is 628 square feet.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.6
65.
411
4000
3
x
Since revenue is price times quantity, the
revenue function is given by:
R x p x
p
4000
3 x
x
4000 3 x
To find the marginal revenue, we take the
derivative of the revenue function. Thus:
R ‘ x 3
66.
3000
5
x
R x p x
p
3000
5 x
x
3000 5 x .
R ‘ x 5.
67.
Answers will vary. Calculus in its present
form was essentially developed independently
in the 17th century by Isaac Newton and
Gottfried Wilhelm von Leibniz. During the 18th
century calculus was challenged by some
philosophers and religious leaders who argued
that the infinitely small quantities represented
by differentials were meaningless. These critics
were silenced when the concept of “limit” was
introduced. In it, a differential was not thought
of as an infinitely small quantity; rather, a
derivative was considered to be the limit
approached by two differentials as each
becomes infinitely small.
68.
For a function y f x , the differential, dy,
can be used to approximate the true change in
the value of f x when a small change is made
in the value of x.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
412
c) Solve E x 1
Exercise Set 2.7
1.
x
1
500 x
x 500 x
2 x 500
x 250
A price of $250 will maximize total revenue.
a) The demand function is
q D x 400 x .
The definition of the elasticity of demand is
x D ‘ x
. In order to
given by: E x
D x
find the elasticity of demand, we need to
find the derivative of the demand function
first.
dq
d
D ‘ x
400 x 1 .
dx
dx
Next, we substitute 1 for D ‘ x , and
3.
a) D ‘ x 4
E x
x
50 x
b) Substituting x 46 into the expression
found in part (a) we have:
46
46 23
E 46
11.5
50 46 4
2
elasticity.
x 1
x
400 x 400 x
b) Substituting x 125 into the expression
found in part (a) we have:
125 125 5
E 125
400 125 275 11
E x
Since E 46 1 , demand is elastic.
c) We solve E x 1
x
1
50 x
x 50 x
2 x 50
x 25
A price of $25 will maximize total revenue.
5
Since E 125 is less than one, the
11
demand is inelastic.
c) The values of x for which E x 1 will
maximize total revenue. We solve:
E x 1
2.
q D x 500 x; x 38
a) D ‘ x 1
E x
x D ‘ x
x 1
x
D x
500 x 500 x
38
19
b) E 38
500 38 231
E 38 1 , so demand is inelastic.
x D ‘ x
x 4
4x
D x
200 4 x 200 4 x
400 x for D x into the expression for
x
1
400 x
x 400 x
2 x 400
x 200
A price of $200 will maximize total revenue.
q D x 200 4 x; x 46
4.
q D x 500 2 x; x 57
a) D ‘ x 2
E x
x D ‘ x
D x
x 2
500 2 x
2x
500 2 x
x
250 x
57
57
b) E 57
250 57 193
E 57 1 , so demand is inelastic.
c) Solve E x 1
x
1
250 x
x 250 x
2 x 250
x 125
A price of $125 will maximize total revenue.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.7
5.
413
Next, we take the derivative of the demand
function, using the Chain Rule:
1
1
d
D ‘ x 600 x 2 600 x
dx
2
1
2 600 x
Making the appropriate substitutions into the
elasticity function, we have
1
x
x D ‘ x
2 600 x
E x
D x
600 x
400
; x 50
x
a) First, we rewrite the demand function.
400
D x
400 x 1 .
x
Next, we take the derivative of the demand
function, using the Power Rule.
D ‘ x 400 1 x 2 400 x 2
q D x
Making the appropriate substitutions into the
elasticity function, we have
x 400 x 2
x D ‘ x
E x
D x
400 x 1
x
x
2 600 x
2 600 x
600 x
400 x 1
1
400 x 1
Therefore, E x 1 for all values of x .
x
1200 2 x
b) Substituting x 100 into the expression
found in part (a) we have:
100
100
1
E 100
1200 2 100 1000 10
b) E 50 1 , so demand is unit elastic.
c) E x 1 for all values of x . Therefore, total
revenue is maximized for all values of x .
In other words, total revenue is the same
regardless of the price.
6.
c) Solve E x 1
3000
; x 60
x
3000
a) D x
3000 x 1
x
D ‘ x 3000 x 2
q D x
x
1
1200 2 x
x 1200 2 x
3x 1200
Making the appropriate substitutions into the
elasticity function, we have
x 3000 x 2
x D ‘ x
E x
D x
3000 x 1
3000 x 1
1
3000 x 1
Therefore, E x 1 for all values of x .
b) E 60 1 , so demand is unit elastic.
c) E x 1 for all values of x . Therefore, total
revenue is maximized for all values of x .
In other words, total revenue is the same
regardless of the price.
7.
Since E 100 1 , demand is inelastic.
q D x 600 x ; x 100
a) First rewrite the demand function:
D x 600 x 2 .
1
x 400
A price of $400 will maximize total revenue.
8.
q D x 300 x ; x 250
a) D x 300 x 2 .
1
1
1
d
300 x 2 300 x
2
dx
1
D ‘ x
2 300 x
E x
x D ‘ x
D x
1
x
2 300 x
300 x
x
x
x
2
300
2 300 x
300 x
x
600 2 x
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
414
b)
E 250
250
5
600 2 250 2
10.
Since E 250 1 , demand is elastic.
c) Solve E x 1
x
1
600 2 x
x 600 2 x
x 200
A price of $200 will maximize total revenue.
q D x
100
x 32
; x8
2 x 12 2
2
a) D x 500 2 x 12 .
3
D ‘ x 500 2 2 x 12 2
3
2000 2 x 12
3x 600
9.
500
q D x
2000
2 x 123
Making the appropriate substitutions into the
elasticity function, we have
2000
x
3
x D ‘ x
2 x 12
E x
500
D x
; x 1
a) First, we rewrite the demand function:
2 x 12 2
D x 100 x 3 .
2
2000 2 x 12 2
2x
x
3
500
x6
2 x 12
Next, we take the derivative of the demand
function, using the Chain Rule:
3 d
D ‘ x 100 2 x 3 x 3
dx
200 x 3
b) E 8
3
2 8 16 8
8 6 14 7
Since E 8 1 , demand is elastic.
200
c) Solve E x 1
x 33
Making the appropriate substitutions into the
elasticity function, we have:
200
x
3
x D ‘ x
x 3
E x
100
D x
x 32
200 x 32
x
3
x 3 100
2x
x3
b) Substituting x 1 into the expression
found in part (a) we have:
2 1 2 1
E 1
1 3 4 2
Since E 1 1 , demand is inelastic.
c) Solve E x 1
2x
1
x6
2x x 6
x6
A price of $6 will maximize total revenue.
11.
q D x 50, 000 300 x 3 x 2
a) D ‘ x 300 6 x
E x
x D ‘ x
D x
x 300 6 x
50, 000 300 x 3 x 2
300 x 6 x 2
50, 000 300 x 3 x 2
6 x 2 300 x
50, 000 300 x 3 x 2
2x
1
x3
2x x 3
x3
A price of $3 will maximize total revenue.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.7
415
f) Substituting the answer found in part (e) into
the demand function we have:
b) Substituting in to the elasticity of demand
we have:
E 75
6 75 300 75
D 114.98 50, 000 300 114.98 3 114.98
2
50, 000 300 75 3 75
2
44,832.80.
The demand for oil is about 44,832 million
barrels per day or 44.8 billion barrels per
day at a price of $114.98 a barrel. At the
time this solution was created, the price of
oil was $100.83 per barrel. Thus according
to this model, by increasing the price, oil
producers could increase revenue.
g) The demand for oil is inelastic at $110 a
barrel. Therefore an increase in price will
result in an increase in total revenue.
11, 250
55, 625
0.20.
Since E 75 1 , the demand for oil is
inelastic at $75 a barrel.
c) Substituting in to the elasticity of demand
we have:
E 100
6 100 300 100
2
50, 000 300 100 3 100
2
30, 000
50, 000
0.6.
Since E 100 1 , the demand for oil is
inelastic at $100 a barrel.
d) Substituting in to the elasticity of demand
we have:
E 125
6 125 300 125
2
50, 000 300 125 3 125
2
56, 250
40, 625
1.38.
Since E 125 1 , the demand for oil is
elastic at $125 a barrel.
e) Revenue will be maximized when E x 1 .
Therefore, we solve:
6 x 2 300 x
1
50, 000 300 x 3 x 2
6 x 2 300 x 50, 000 300 x 3 x 2
9 x 2 600 x 50, 000 0
Using the quadratic formula, we find that the
solutions to the equation.
x
600
6002 4 950000
2 9
600 2,160, 000
18
x 48.316 or
x 114.983.
The only solution that is feasible is
x 114.98 . Thus, oil revenues will be
maximized when price is $114.98 a barrel.
x
2
12.
q D x 967 25 x
a) D ‘ x 25
E x
x D ‘ x
x 25
D x
967 25 x
25 x
967 25 x
b) We set E x 1 and solve for x.
25 x
1
967 25 x
25 x 967 25 x
50 x 967
x 19.34
Demand is unitary elastic when price is
19.34 cents.
c) Demand is elastic when E x 1 .
Testing a value on each side of 19.34 cents,
we have:
25 19
E 19
0.97 1
967 25 19
25 20
E 20
1.07 1
967 25 20
Therefore, the demand for cookies is elastic
for prices greater than 19.34 cents.
d) Demand is inelastic when E x 1 .
Using the calculations from part (c), we see
that the demand for cookies is inelastic for
prices less than 19.34 cents.
e) Total revenue is maximized when E x 1 .
In part (b) we showed that E x 1 when
price was 19.34 cents. Therefore, revenue
will be maximized when price is 19.34
cents.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
416
f) We have shown that the demand for cookies
is elastic when the price of cookies is 20
cents. Therefore a small increase in price
will cause total revenue to decrease.
13.
14.
q D x 200 x 3
2 x 300
10 x 11
a) Using the Quotient Rule, we have:
10 x 11 2 2 x 300 10
D ‘ x
10 x 112
q D x
a) First, we rewrite the demand function to
make it easier to find the derivative.
D x 200 x 3
.
1
b) E 3
200 x
3
3x 3
2 200 x
3
x D ‘ x
D x
x
2978
10 x 112
2 x 300
10 x 11
2978 x
10 x 11
10 x 112 2 x 300
b) E 3
2
1489 x
10 x 11 x 150
1489 3
10 3 11 3 150
1489
2091
0.7121
Since E 3 1 , the demand for tomato
3x
400 2 x 3
plants is inelastic when the price is $3 per
plant.
c) We determined in part (b) that the demand
for tomato plants was inelastic at $3.
Therefore, an increase in the price of tomato
plants will lead to an increase in revenue.
3
400 2 3
10 x 112
E x
3
3 3
2978
Therefore,
3x 3
2
10 x 112
2
Next, using the Chain Rule, we have:
1
1
2
D ‘ x 200 x 3
3 x 2
2
3 x 2
2 200 x 3
Now, substituting in the elasticity function
we get:
3 x 2
x
3
x D ‘ x
2 200 x
E x
D x
200 x 3
20 x 22 20 x 3000
3
81
346
0.2341
Since E 3 1 , the demand for computer
games is inelastic when price is $3.
c) From part (b) we know that the demand for
computer games is inelastic at a price of $3.
Therefore an increase in the price of
computer games will lead to an increase in
the total revenue.
15.
q D x 180 10 x
a) First we find the derivative of the demand
function.
D ‘ x 10
Therefore, the elasticity of demand is:
x D ‘ x
E x
D x
x 10
180 10 x
10 x
180 10 x
10 x
180 10 x
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.7
417
b) Substituting in to the elasticity of demand
we have:
10 8
E 8
180 10 8
Substituting into the elasticity function we
have:
80
100
0.8.
c) Revenue will be maximized when E x 1 .
Therefore, we solve:
10 x
1
180 10 x
10 x 180 10 x
20 x 180
x 9.
When the price of sunglass cases is $9 a
case, revenue will be maximized.
d)
A 20% increase in the price would
increase the price from $8 a case to $9.60 a
case.
At $8 a case, demand is:
q D 8 180 10 8 100 cases.
Therefore revenue is price times quantity or
R p q $8 100 $800.
At $9.60 a case, demand is:
q D 9.60 180 10 9.60 84 cases.
Therefore revenue is price times quantity or
R p q $9.60 84 $806.40. Therefore,
revenue will increase if Tipton Industries
raises the price 20%.
We know increasing from $8 to $9 would
increase revenue, and increasing from $9 to
$9.60 would decrease revenue, we just
didn’t know how much the increase and
decrease would be. Therefore, the answer to
part (d) does not contradict the answers to
parts (b) and (c). In fact it confirms them.
16.
n 1
x nkx
x D ‘ x
E x
D x
kx n
nk x x n 1
kx
n
nk x n
kx n
n
b) No, the elasticity of demand is constant for
all prices. E x n .
c) Total revenue is maximized when
E x 1 . Since E x n , total revenue
will be maximized when n 1 .
17.
Answers will vary. The elasticity of demand
is a measure of the responsiveness of
quantity demanded to changes in price. This
measure allows economists to determine
how sensitive quantity demanded is to price
changes and help predict the effect of price
changes on total revenue.
18.
Answers will vary. In general, the greater
the availability of substitutes and the better
the substitutes are will cause goods to have a
higher elasticity of demand. For example,
the demand for tea is relatively elastic
because of the wide range of substitutes
available, such as coffee, soda, or water.
However, a diabetic’s demand for insulin is
very inelastic because there are no close
substitutes for insulin.
k
xn
a) First, we rewrite the demand function to
make it easier to find the derivative.
D x k xn
Using the Power Rule, we have:
D ‘ x k n x n 1 nkx n 1
q D x
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
418
3.
Exercise Set 2.8
1.
Differentiate implicitly to find
dy
.
dx
We have 3 x 3 y 2 8.
Differentiating both sides with respect to x
yields:
d
d
3×3 y 2
8
dx
dx
d 3 d 2 d
3x
8
y
dx
dx
dx
dy
dy
Next, we isolate
9 x2 2 y
.
0
dx
dx
dy
2 y
9 x 2
dx
dy 9 x 2
dx 2 y
Find the slope of the tangent line to the curve at
2, 4 .
2
dy
3x 2
dx
4y
Find the slope of the tangent line to the curve at
2, 1 .
Replacing x with 2 and y with 1 , we have:
dy
3 x 2 3 2
12
3.
dx
4y
4 1
4
2
The slope of the tangent line to the curve at
2, 1 is 3.
4.
The slope of the tangent line to the curve at
9
2, 4 is .
2
2.
x3 2 y3 6
Differentiating both sides with respect to x
yields:
d 3
d
x 2 y3
6
dx
dx
dy
dy
0
Next, we isolate
3×2 6 y 2
dx
dx
dy
3x 2
6 y2
dx
dy x 2
dx 2 y 2
Find the slope of the tangent line to the curve at
2, 1 .
dy 6 x 2
dx
8y
Replacing x with 2 and y with 4 , we have:
dy 9 x 2 9 2
36 9
.
dx 2 y
2 4
8 2
2 x 3 4 y 2 12
Differentiating both sides with respect to x
yields:
d
d
2 x3 4 y 2
12
dx
dx
dy
dy
0
Next, we isolate
6 x2 8 y
.
dx
dx
dy
6 x 2
8y
dx
2 x2 3 y3 5
Differentiating both sides with respect to x
yields:
d
d
2 x2 3 y3
5
dx
dx
dy
dy
0
Next, we isolate
4x 9 y2
.
dx
dx
dy
9 y 2
4 x
dx
dy
4x
dx 9 y 2
Find the slope of the tangent line to the curve at
2,1 .
dy 4 2 8
8
.
2
dx 9 1
9
9
The slope of the tangent line to the curve at
8
2,1 is .
9
4
dy 2
2 .
dx 2 12
2
2
The slope of the tangent line to the curve at
2, 1 is 2.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.8
5.
419
x2 y2 1
Differentiating both sides with respect to x
yields:
d 2
d
x y2
1
dx
dx
dy
dy
2x 2 y
.
Next, we isolate
0
dx
dx
dy
2y
2 x
dx
dy 2 x
2y
dx
7.
4 x3 y
dy
3y
dx
2x
Find the slope of the tangent line to the curve at
1, 3 .
1
3
and y with
, we have:
2
2
x2 y2 1
Differentiating both sides with respect to x
yields:
d 2
d
x y2
1
dx
dx
dy
dy
0
2x 2 y
.
Next, we isolate
dx
dx
dy
2 y
2 x
dx
dy x
dx y
Find the slope of the tangent line to the curve at
3, 2 .
3
3
dy
.
2
dx
2
The slope of the tangent line to the curve at
3
3, 2 is
.
2
dy
dy
6 x 2 y 2 0 Next, we isolate .
dx
dx
dy
3
2 2
6 x y
4x y
dx
dy 6 x 2 y 2
dx
4 x3 y
1
dy
x
1
2
.
dx
y
3
3
2
The slope of the tangent line to the curve at
1 3
1
.
2 , 2 is
3
6.
dy
x
dx
y
Find the slope of the tangent line to the curve at
1 3
2, 2 .
Replacing x with
2 x 3 y 2 18
Differentiating both sides with respect to x
yields:
d
d
2 x3 y 2
18
dx
dx
d 2
d
y y2
Product Rule
2 x3
2 x3 0
dx
dx
dy
2 x3 2 y y 2 6 x 2 0
dx
Replacing x with 1 and y with 3, we have:
3 3 9
dy
3y
dx
2x
2 1 2
The slope of the tangent line to the curve at
9
1, 3 is .
2
3 x 2 y 4 12
Differentiating both sides with respect to x
yields:
d
d
3x 2 y 4
12
dx
dx
dy
Product Rule
3 x 2 4 y 3 y 4 3 2 x 0
dx
8.
dy
dy
6 xy 4 0 Next, we isolate .
dx
dx
dy
12 x 2 y 3
6 xy 4
dx
dy
y
2x
dx
Find the slope of the tangent line to the curve at
2, 1 .
12 x 2 y 3
1 1
dy
.
dx
2 2 4
The slope of the tangent line to the curve at
1
2, 1 is .
4
Copyright © 2016 Pearson Education, Inc.
420
Chapter 2: Applications of Differentiation
x 4 x 2 y 3 12
Differentiating both sides with respect to x
yields:
d 4
d
x x2 y3
12
dx
dx
d 4
d 2 3
d
x
x y
12
dx
dx
dx
dy
4 x3 x 2 3 y 2 y 3 2 x 0 Product Rule
dx
Find the slope of the tangent line to the curve at
3, 2 .
9.
4 x3 3 x 2 y 2
dy 3 3 2 2
98
1
.
dx
12
2 3 2
12
The slope of the tangent line to the curve at
1
3, 2 is .
12
2
11.
dy
2 xy 3 0
dx
Next, we isolate
3 x 2 y 2
dy
dx
.
dy
2 xy 3 4 x3
dx
dy 2 xy 3 4 x3
dx
3 x 2 y 2
x
Replacing x with 2 and y with 1, we have:
dy 4 x 2 2 y 3 4 2 2 1
7
2
dx
3
3 xy 2
3 21
3
The slope of the tangent line to the curve at
7
2,1 is .
3
10.
x 3 x 2 y 2 9
Differentiating both sides with respect to x
yields:
d 3
d
x x2 y2
9
dx
dx
dy
3 x 2 x 2 2 y y 2 2 x 0
dx
Replacing x with 1 and y with 2 , we have:
2 2
dy
2 y
4
4
dx x 2 y 1 2 2 3
3
The slope of the tangent line to the curve at
4
1, 2 is .
3
dy
2 xy 2 0
dx
dy
2 x 2 y
2 xy 2 3 x 2
dx
3x 2 2 x 2 y
dy 2 xy 2 3 x 2
dx
2 x 2 y
dy 3 x 2 y
2 xy
dx
2
dy
dy
y 2y 2 0
dx
dx
dy
dy
x 2y
2 y
dx
dx
dy
x 2 y 2 y
dx
2 y
dy
dx x 2 y
Find the slope of the tangent line to the curve at
1, 2 .
dy 4 x 2 2 y 3
dx
3 xy 2
Find the slope of the tangent line to the curve at
2,1 .
2
xy y 2 2 x 0
Differentiating both sides with respect to x
yields:
d
d
xy y 2 2 x
0
dx
dx
d
d
d
d
xy y 2 2 x 0
dx
dx
dx
dx
dy
dy
x y 1 2 y 2 1 0
dx
dx
12.
xy x 2 y 3
Differentiating both sides with respect to x
yields:
d
d
xy x 2 y 3
dx
dx
dy
dy
0
x y 1 1 2
dx
dx
dy
dy
y 1 2
0
dx
dx
dy
at the top of the next page.
We isolate
dx
x
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Exercise Set 2.8
421
Continued from the previous page.
dy
x 2 1 y
dx
dy 1 y
dx x 2
Find the slope of the tangent line to the curve at
2
5, .
3
x2 y 2 x3 y 3 1 0
Differentiating both sides with respect to x
yields:
d 2
d
x y 2 x3 y 3 1
0
dx
dx
2 dy
dy
2
2
0
x y 2 x 2 3 x 3 y
dx
dx
14.
x2
2
1
1
3
1
dy
3 .
dx 5 2 3
9
The slope of the tangent line to the curve at
2
1
5, is .
3
9
4 x3 y 4 3 y 5x 1 0
Differentiating both sides with respect to x
yields:
d
d
4 x3 y 4 3 y 5 x 1
0
dx
dx
d
d 4 d
d
d
y 3 y 5x 1 0
4 x3
dx
dx
dx
dx
dx
dy
dy
12 x 2 4 y 3 3 5 0 0
dx
dx
dy
dy
4 y 3 3
12 x 2 5
dx
dx
dy
4 y 3 3
12 x 2 5
dx
dy 12 x 2 5
dx
4 y 3 3
dy 12 x 2 5
dx 4 y 3 3
Find the slope of the tangent line to the curve at
1, 2 .
Replacing x with 1 and y with 2 , we have:
17
dy 12 x 2 5 12 1 5
29
dx 4 y 3 3 4 23 3
2
The slope of the tangent line to the curve at
17
1, 2 is .
29
Find the slope of the tangent line to the curve at
2, 3 .
dy 6 2 2 2 3 24 12 36
36
2
2
dx
4
27
23
23
2 3 3
2
dy
dy
2 xy 6 x 2 3 y 2
0
dx
dx
dy
6 x 2 2 xy
x2 3 y 2
dx
dy 6 x 2 2 xy
2
dx
x 3y2
13.
The slope of the tangent line to the curve at
36
2, 3 is .
23
15.
x 2 2 xy 3 y 2
Differentiating both sides with respect to x
yields:
d 2
d
3y2
x 2 xy
dx
dx
d 2 d
d
3y2
x 2 xy
dx
dx
dx
dy
dy
2 x 2 x y 1 3 2 y
dx
dx
dy
dy
2y 6y
dx
dx
dy
dy
2x 6 y
2 x 2 y
dx
dx
dy
2 x 6 y 2 x y
dx
dy 2 x y
dx
2x 6 y
2x 2x
2 x y
dy
dx 2 x 3 y
dy
x y
dx 3 y x
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
422
16.
2 xy 3 0
Differentiating both sides with respect to x
yields:
d
d
2 xy 3 0
dx
dx
dy
2 x y 1 0 0
dx
2x
17.
dy
2y 0
dx
dy
2x
2 y
dx
dy
y
dx
x
18.
x 2 y 2 25
Differentiating both sides with respect to x
yields:
d 2
d
x y2
25
dx
dx
dy
0
2x 2 y
dx
dy
2 x
2y
dx
dy 2 x
dx 2 y
dy
x
dx
y
dy 5 x 4
dx 3 y 2
y5 x3
Differentiating both sides with respect to x
yields:
d 5
d 3
y
x
dx
dx
dy
3x 2
5 y4
dx
dy x
dx y
y 3 x5
Differentiating both sides with respect to x
yields:
d 3
d 5
y
x
dx
dx
dy
5x 4
3y2
dx
20.
x 2 y 2 16
Differentiating both sides with respect to x
yields:
d 2
d
x y2
16
dx
dx
d 2 d 2 d
x
y
16
dx
dx
dx
dy
0
2x 2 y
dx
dy
2 y
2 x
dx
dy 2 x
dx 2 y
19.
dy 3x 2
dx 5 y 4
21.
x 2 y 3 x 3 y 4 11
Differentiating both sides with respect to x
yields:
d 2 3
d
x y x3 y 4
11
dx
dx
d 2 3
d 3 4
x y
x y 0
dx
dx
Notice:
d 2 3
dy
x y x2 3 y 2 y 3 2 x
dx
dx
3x 2 y 2
dy
2 xy 3
dx
and
d 3 4
dy
x y x 3 4 y 3 y 4 3x 2
dx
dx
dy
4 x 3 y 3 3x 2 y 4
dx
Therefore,
d 2 3
d 3 4
x y
x y 0
dx
dx
dy
dy
3 x 2 y 2 2 xy 3 4 x 3 y 3 3 x 2 y 4 0
dx
dx
dy
, on the next page.
We isolate
dx
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.8
423
Continued from the previous page.
dy
4 x 3 y 3 3x 2 y 2
3 x 2 y 4 2 xy 3
dx
dy 3x 2 y 4 2 xy 3
dx 4 x 3 y 3 3x 2 y 2
3
3 p 1 dp
dx
2
dp
3
dx 3 p 2 1
2
2
dy xy 3 xy 2 y
dx
xy 2 4 x 2 y 3x
25.
2 y 3xy 2
dy
2
dx
4 x y 3x
22.
x 3 y 2 x 5 y 3 19
Differentiating both sides with respect to x
yields:
d 3 2
d
x y x5 y 3
19
dx
dx
2 x3 y
dy
dx
3x 2 y 2 3x 5 y 2
5 2
3xp 2
dy
dx
5x 4 y 3 0
5 x y 3x y
2 x y 3x y dy
dx
3
xp 3 24
Differentiating both sides with respect to x
yields:
d
d
xp 3
24
dx
dx
dp
Product Rule
x 3 p 2 p 3 1 0
dx
4 3
2 2
4 3
2 2
dy
5 x y 3x y
5 2
dx
3x y 2 x 3 y
26.
x 3 p 2 108
Differentiating both sides with respect to x
yields:
dy 5 x 2 y 2 3 y
dx
3x 3 y 2 x
23.
p 2 p 2 x 40
Differentiating both sides with respect to x
yields:
d
d
p2 p 2 x
40
dx
dx
dp dp
2p
2 1 0
dx dx
dp
2 p 1 2
dx
2
dp
dx 2 p 1
24.
p 3 p 3 x 50
Differentiating both sides with respect to x
yields:
d 3
d
p p 3x
50
dx
dx
dp dp
3 p2
3 1 0
dx dx
dp
We isolate
at the top of the next column.
dx
d 3 2
d
x p
108
dx
dx
dp
x 3 2 p p 2 3x 2 0
dx
2 x3 p
dp
p3
dx
dp p 3
dx 3 xp 2
dp
p
3x
dx
27.
dp
3x 2 p 2
dx
dp 3x 2 p 2
dx
2 x3 p
dp
3p
dx
2x
x 2 p xp 1
1
2x p
Multiply both sides by 2 x p to clear the
fraction.
x 2 p xp 1
2 x p
1 2 x p
2x p
x 2 p xp 1 2 x p
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
424
Differentiating both sides of the equation on the
previous page with respect to x yields:
d 2
d
x p xp 1
2 x p
dx
dx
d 2
d
d
d
d
x p
xp 1
p
2x
dx
dx
dx
dx
dx
dp
dp
dp
p 2x x
p 1 0 2
x2
dx
dx
dx
dp
dp
2 dp
x
1
2 2 xp p
x
dx
dx
dx
dp
x2 x 1
2 2 xp p
dx
dp 2 2 xp p
2
dx
x x 1
xp
28.
2
x p
Multiply both sides by x p to clear the
fraction.
xp
2 x p
x p
x p
xp 2 x 2 p
Differentiate both sides with respect to x.
d
d
xp 2 x 2 p
dx
dx
dp
dp
p 1 2 1 2
x
dx
dx
dp
dp
2
2 p
x
dx
dx
dp
x 2 2 p
dx
dp 2 p
dx x 2
29.
p 4 x 3 48
Expanding the left hand side of the equation we
have:
px 3 p 4 x 12 48
px 3 p 4 x 36
Differentiating both sides with respect to x
yields:
d
d
px 3 p 4 x 36
dx
dx
dp
dp
p 1 x
3
4 1 0
dx
dx
dp
x 3 p 4
dx
dp p 4
dx
x3
30. 1000 300 p 25 p 2 x
Differentiating both sides with respect to x
yields:
d
d
1000 300 p 25 p 2
x
dx
dx
dp
dp
300
25 2 p
1
dx
dx
dp
50 p 300 1
dx
1
dp
dx 50 p 300
31.
G 2 H 2 25
We differentiate both sides with respect to t.
d
d
G 2 H 2 25
dt
dt
dG
dH
2G
2H
0
dt
dt
dH
dG
2H
2G
dt
dt
dH
G dG
dt
H dt
We find H when G 0 :
02 H 2 25
H 2 25
H 5,
H is nonnegative
Next, we substitute 5 in for H , 0 in for G , and
dG
dH
to determine
3 in for
.
dt
dt
dH
G dG
dt
H dt
0
3 0
5
We find H when G 1 :
12 H 2 25
H 2 24
H 24 2 6,
H is nonnegative
Next, we substitute 2 6 in for H , 1 in for G ,
dG
dH
and 3 in for
to determine
.
dt
dt
dH
G dG
dt
H dt
1
3
3
2 6
2 6
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.8
425
We find H when G 3 :
Next, we substitute 10 for x and 5 for dx/dt.
dR
50 10 5 40 5 200
dt
The rate of change of total revenue with respect
to time is $200 per day.
C x 10 x 3
Differentiating with respect to time we have:
d
d
C x 10 x 3
dt
dt
dC
dx
10
dt
dt
Next, we substitute 10 for x and 5 for dx/dt.
dC
10 5 50
dt
The rate of change of total cost with respect to
time is $50 per day.
Profit is revenue minus cost. Therefore;
P x R x C x
3 H 25
2
2
H 2 16
H 4,
H is nonnegative
Next, we substitute 4 in for H , 3 in for G , and
dG
dH
to determine
3 in for
dt
dt.
dH
G dG
dt
H dt
3
9
3
4
4
32.
A3 B 3 9
We differentiate both sides with respect to t.
d 3
d
A B 3 9
dt
dt
dA
dB
3B 2
0
3 A2
dt
dt
dA
dB
3 B 2
3 A2
dt
dt
50 x 0.5 x 2 10 x 3
0.5 x 2 40 x 3
Differentiating with respect to time we have:
d
d
P x
0.5 x 2 40 x 3
dt
dt
dP
dx
dx
x 40
dt
dt
dt
dP
dx
40 x
dt
dt
Next, we substitute 10 for x and 5 for dx/dt.
dP
40 10 5 305 150
dt
The rate of change of total profit with respect to
time is $150 per day.
dA 3B 2 dB
dt
3 A2 dt
dA B 2 dB
2
dt
dt
A
We find B when A 2 :
A3 B 3 9
2 3 B 3 9
8 B3 9
B3 1
B 1
Next, we substitute 2 for A, 1 for B, and 3 for
dB
dA
into the formula for
:
dt
dt
dA 1
3
3
2
4
dt
2
2
33.
R x 50 x 0.5 x 2
Differentiating with respect to time we have:
d
d
R x
50 x 0.5 x 2
dt
dt
dR
dx
dx
50 x
dt
dt
dt
dR
dx
50 x
dt
dt
34.
R x 50 x 0.5 x 2
Differentiating with respect to time we have:
d
d
R x
50 x 0.5 x 2
dt
dt
dR
dx
dx
50 x
dt
dt
dt
dR
dx
50 x
dt
dt
Next, we substitute 30 for x and 20 for dx/dt.
dR
50 30 20 20 20 400
dt
The rate of change of total revenue with respect
to time is $400 per day.
The solutioin is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
426
Next, we substitute 200 for x and 300 for dx/dt.
dC
1.2 200 300 72, 000
dt
The rate of change of total cost with respect to
time is $72,000 per day.
Profit is revenue minus cost. Therefore;
P x R x C x
Looking at the cost function
C x 4 x 10 .
Differentiating with respect to time we have:
d
d
C x 4 x 10
dt
dt
dC
dx
4
dt
dt
Next, we substitute 30 for x and 20 for dx/dt.
dC
4 20 80
dt
The rate of change of total cost with respect to
time is $80 per day.
Profit is revenue minus cost. Therefore;
P x R x C x
280 x 0.4 x 2 5000 0.6 x 2
x 280 x 5000
Differentiating with respect to time we have:
d
d
P x
x 2 280 x 5000
dt
dt
dP
dx
dx
2 x 280
dt
dt
dt
dP
dx
280 2 x
dt
dt
Next, we substitute 200 for x and 300 for dx/dt.
dP
280 2 200 300 36,000
dt
The rate of change of total profit with respect to
time is $36, 000 per day.
50 x 0.5 x 2 4 x 10
0.5 x 2 46 x 10
Differentiating with respect to time we have:
d
d
P x
0.5 x 2 46 x 10
dt
dt
dP
dx
dx
x 46
dt
dt
dt
dP
dx
46 x
dt
dt
Next, we substitute 30 for x and 20 for dx/dt.
dP
46 30 20 16 20 320
dt
The rate of change of total profit with respect to
time is $320 per day.
35.
R x 280 x 0.4 x 2
Differentiating with respect to time we have:
d
d
R x
280 x 0.4 x 2
dt
dt
dR
dx
dx
280 0.8 x
dt
dt
dt
dR
dx
280 0.8 x
dt
dt
Next, we substitute 200 for x and 300 for dx/dt.
dR
280 0.8 200 300 36,000
dt
The rate of change of total revenue with respect
to time is $36,000 per day.
C x 5000 0.6 x 2
Differentiating with respect to time we have:
d
d
C x
5000 0.6 x 2
dt
dt
dC
dx
1.2 x
dt
dt
2
36.
R x 2x
Differentiating with respect to time we have:
d
d
R x 2 x
dt
dt
dR
dx
2
dt
dt
Next, we substitute 20 for x and 8 for dx/dt.
dR
2 8 16
dt
The rate of change of total revenue with respect
to time is $16 per day.
C x 0.01x 2 0.6 x 30
Differentiating with respect to time we have:
d
d
C x
0.01x 2 0.6 x 30
dt
dt
dC
dx
dx
0.02 x 0.6
dt
dt
dt
dC
dx
0.02 x 0.6
dt
dt
Next, we substitute 20 for x and 8 for dx/dt.
dC
0.02 20 0.6 8 8
dt
The rate of change of total cost with respect to
time is $8 per day.
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.8
427
Profit is revenue minus cost. Therefore;
P x R x C x
2
2 x 0.01x 0.6 x 30
38.
From Exercise 37, we know that
0.01x 2 1.4 x 30
Differentiating with respect to time we have:
d
d
P x
0.01x 2 1.4 x 30
dt
dt
dP
dx
dx
0.02 x 1.4
dt
dt
dt
dP
dx
1.4 0.02 x
dt
dt
Next, we substitute 20 for x and 8 for dx/dt.
dP
1.4 0.02 20 8 8
dt
The rate of change of total profit with respect to
time is $8 per day.
37.
Total revenue is changing at a rate of $1.39
per day.
39.
5 p 4 x 2 px 60
First, we take the derivative of both sides of the
equation with respect to t.
d
d
5 p 4 x 2 px 60
dt
dt
dp
dx
dx dp
5 4 2 p
x 0
dt
dt
dt
dt
Product Rule
dx 5 2 3
1.5
dt
4 2 5
11
1.5
14
16.5
14
1.18
Sales are changing at a rate of 1.18 sales per
day.
Constant Multiple Rule
Chain Rule
dr
4.7
dt
miles per year. Substituting these values into the
derivative, we have:
dA
2 792 4.7
dt
23,388.52899
23,389
Therefore, in 2013 the Arctic ice cap was
changing at a rate of 23, 389 mi 2 per year.
Another way of stating this is to say that the
Arctic ice cap was shrinking at a rate of 23,389
mi2/yr.
In 2013, the r was 792 miles, and
dp
dx
dx
dp
4 2 p 2x
0
dt
dt
dt
dt
dx
Next, we solve for
.
dt
dx
dx
dp
dp
5 2 x
4 2p
dt
dt
dt
dt
dx
dp
4 2 p 5 2 x
dt
dt
dx 5 2 x dp
dt
4 2 p dt
have:
A r2
To find the rate of change of the area of the
Arctic ice cap with respect to time, we take the
derivative of both sides of the equation with
respect to t.
d
d
A r 2
dt
dt
dA
d
r 2
dt
dt
dA
dr
2r
dt
dt
dA
dr
2 r
dt
dt
5
dp
, we
dt
dx 16.5
dt
14
dR
dp
dx
x
p
dt
dt
dt
Substituting the appropriate values, we have:
dR
16.5
31.5 5
1.39
14
dt
Substituting 3 for x¸ 5 for p, and 1.5 for
R xp
40.
A r2
d
d
A r 2
dt
dt
dA
dr
2 r
dt
dt
Substituting, we have:
dA
2 25 1 50 157.0796
dt
The area of the wound is decreasing at a rate of
157.08 mm2/day.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
428
41.
b) Using the derivative found in part (a), and
substituting the values for R and dR/dt, we
have:
dV 500 0.075
0.0002
0.48
dt
0.0156
The speed of the blood is changing at a rate
of 0.0156 mm / sec 2 .
hw
60
First, we substitute 165 for h, and then we take
the derivative of both sides with respect to t.
165w
165 1 2
w
S
60
60
d
d 165 1
S dt 60 w 2
dt
S
dS
165 1 1 2 dw
w
dt
60 2
dt
165 1 dw
1
120 w 2 dt
p
dR
0
2R
4 Lv
dt
pR dR
2 Lv dt
Substituting 70 for L, 400 for p and 0.003
for v , we have:
400 R
dV
dR
dt 2 70 0.003 dt
400 R dR
0.42 dt
dR
952.38 R
dt
b) Using the derivative in part (a), we
substitute 0.00015 for dR/dt and 0.1 for R to
get:
dV
952.38 0.1 0.00015
dt
0.0143
The speed of the person’s blood will be
increasing at a rate of 0.0143 mm/sec2.
p
dR
2R
4 Lv
dt
pR dR
2 Lv dt
Substituting, we have:
500 R
dV
dR
dt 2 80 0.003 dt
500 R dR
0.48 dt
p
R2 r2
4 Lv
We assume that r, p, L and v are constants.
a) Taking the derivative of both sides with
respect to t, we have:
dV d p
R2 r2
dt dt 4 Lv
p d 2 d 2
R r
4 Lv dt
dt
165 dw
120 w dt
Now, we will substitute 70 for w and
dw
2 for
.
dt
dS
165
2
dt 120 70
0.0256
Therefore, Kim’s surface area is changing at a
rate of 0.0256 m2 /month. We could also say
that Kim’s surface area is decreasing by 0.0256
m2/month.
42. V
p
R2 r2
4 Lv
We assume that r, p, L and v are constants.
a) Taking the derivative of both sides with
respect to t, we have:
dV d p
R2 r2
dt dt 4 Lv
p d 2 d 2
R r
4 Lv dt
dt
43. V
44.
D2 x2 y2
After 1 hour,
D 2 252 602
D 2 4225
D 65
dx
dy
25 and
60
dt
dt
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.8
429
Differentiating both sides of the distance
equation on the previous page with respect to t,
we have:
d 2 d 2
D x y 2
dt
dt
dD
dx
dy
2D
2x 2 y
dt
dt
dt
dx
dy
x y
dD
dt
dt
dt
D
Substituting the appropriate information, we
have:
dD 25 25 60 60
65
dt
65
One hour after the cars leave, the distance
between the two cars is increasing at a rate of 65
mph.
45. Since the ladder forms a right triangle with the
wall and the ground, we know that:
x 2 y 2 262
x 2 y 2 676
dy
.
dt
Differentiating both sides of the equation with
respect to t, we have:
d 2
d
x y 2 676
dt
dt
dx
dy
2x 2 y
0
dt
dt
dy
dx
2y
Subtracting
2 x
dt
dt
dy 2 x dx
Dividing by 2 y
2 y dt
dt
dy x dx
dt
y dt
The lower end of the wall is being pulled away
from the wall at a rate of 5 feet per second;
dx
5.
therefore,
dt
When the lower end is 10 feet away from the
wall, x 10 , we substitute and solve for y at the
top of the next column.
We are looking for
102 y 2 676
100 y 2 676
y 2 676 100
y 2 576
y 576
y 24
y 24
y must be positive
We substitute 10 for x, 24 for y and 5 for
dx
dt
into the derivative to get:
dy x dx
dt
y dt
10 5
24
25
12
2 121
When the lower end of the ladder is 10 feet
from the wall, the top of the ladder is moving
down the wall at a rate of 2 121 feet per second.
46. First we draw a picture.
D
x
2x
First, from the picture we know that:
x2 2 x D2
2
5x 2 D2
When x is 440 m wide, we have
D 2 5 440
2
D 2 968,000
D 440 5
Differentiating both sides of 5x 2 D 2 with
respect to t, we have:
dx
dD
10 x
2D
dt
dt
dx D dD
dt 5 x dt
The solution is continued on the next page.
Copyright © 2016 Pearson Education, Inc.
Chapter 2: Applications of Differentiation
430
dD
90 , x 440 , and
dt
D 440 5 we have:
Now, when
48.
5
90
5
18 5
We are looking to find how fast the area is
changing.
A 2 x2
dA
dx
4x
dt
dt
Substituting, we have:
dA
4 440 18 5
dt
dy x 2
1
dx
y 2
1
1
dy y 2 y
1
dx
x
x2
31,680 5
70,838.63
The area is changing at a rate of 70,838.63
m2/yr.
49.
1
1
2 5
2
x
y
x 2 y 2 5
Differentiating both sides with respect to x, we
have:
d 2 d 2 d
x
y 5
dx dx
dx
dy
2 x 3 2 y 3
0
dx
dy
2 y 3
2 x 3
dx
4 3
r
3
Differentiating both sides with respect to t, we
have:
dV d 4 3
r
dt dt 3
4
d
r 3
3 dt
47. V
4 2 dr
3r
3
dt
dr
4 r 2
dt
Next, substituting 0.7 for dr/dt and 7.5 for r, we
have:
dV
2
4 7.5 0.7
dt
4 56.25 0.7
dy
x 3
3
dx
y
dy
y3
3
dx
x
157.5
494.8
The cantaloupe’s volume is changing
approximately at the rate of 494.8 cm3/week.
1
d 12
d
x y 2 1
dx
dx
1 1 2 1 1 2 dy
x y
0
dx
2
2
1 1 2 dy
1 1
y
x 2
dx
2
2
1 12
dy 2 x
1 12
dx
y
2
440 5
dx
90
dt
5 440
x y 1
50.
y3
x 1
x 1
d 3 d x 1
y
dx dx x 1
3y2
dy x 11 x 11
dx
x 12
3y2
2
dy
dx x 12
2
dy
dx 3 y 2 x 12
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.8
51.
431
dy
dy
2
2
3 x y 1 3 x y 1
dx
dx
x2 1
x2 1
Differentiating both sides with respect to x, we
have:
d 2 d x2 1
y
dx dx x 2 1
y2
5×4 5 y 4
3 x y 3 x y
2
x2 1 2 x x2 1 2 x
dy
2y
2
dx
x2 1
2y
2
54.
2
dy
.
dx
d
d
xy x 2 y dx 4
dx
dy
dy
x
y 1 1 2
0
dx
dx
dy
x 2 1 y
dx
dy 1 y 1 y
2 x
dx
x2
d2y
dy
implicitly to find
.
Differentiate
dx
dx 2
d dy d 1 y
dx dx dx 2 x
2
d 32
d
x y 3
1
dx
dx
3 1 2 2 13 dy
x y
0
dx
2
3
2 13 dy
3 1
y
x 2
dx
3
2
3 12
x
dy
2
2 1 3
dx
y
3
1
dy 9 x 2
9 1 2 1 3
1
x y
4 y 3
4
dx
53.
2
xy x 2 y 4
Differentiate implicitly to find
x 2 y 3 1
d2y
x y 3 x y 3 x 5 y 5
dx 2
Differentiating both sides with respect to x, we
have:
d
d
x y 3 x y 3 x5 y 5
dx
dx
d
d
d
d
3
3
x y x y x5 y5
dx
dx
dx
dx
2 d
2 d
3 x y x y 3 x y
x y
dx
dx
dy
5×4 5 y 4
dx
Continued at the top of the next column.
dy
dx
2
Simplification will yield:
dy 5 x 4 6 x 2 6 y 2
dx
12 xy 5 y 4
2x
dy
dx y x 2 1 2
3
dx
5×4 5 y 4
4
dy 5 x 3 x y 3 x y
dx 3 x y 2 3 x y 2 5 y 4
4x
dy
dx 2 y x 2 1 2
52.
2 dy
2
2
5×4 3 x y 3 x y
3x y
3 x y 3 x y 5 y 4 dy
dx
2
4x
dy
2y
2
2
dx
x 1
dx
3 x y
dy 2 x 3 2 x 2 x 3 2 x
2
dx
x2 1
2 dy
dy
dx
2 x
dy
1 y 1
dx
2 x 2
2 x
1 y
1 y
2 x
2 x 2
1 y 1 y
2 x 2
2 2y
2 x 2
Copyright © 2016 Pearson Education, Inc.
Substituting
dy
1 y
for
dx
2 x
2
Chapter 2: Applications of Differentiation
432
55.
y 2 xy x 2 5
56.
dy
.
dx
Differentiate implicitly to find
Differentiate implicitly to find
Differentiate
d2y
dy
implicitly to find
.
dx
dx 2
d y
dx 2
2 y x
dy
dy
2 y 2 x 2 1
dx
dx
2 y x 2
d y
dx 2
57.
dy
dx
2 y x
3 y 3 x
y 2x
dy
for
2y x
dx
y 2x
2y x
2 y x 2
3 y
2y x
y 2x
3x
2y x
2y x
2 y x 2
2
6 y 3xy 3xy 6 x
2 y x 3
6 y 2 6 xy 6 x 2
2 y x
2
6 y 2 xy x 2
2 y x
3
dy
dx
y2 x2
y3
x3 y3 8
dy
.
dx
d 3
d
x y 3 8
dx
dx
dy
3x 2 3 y 2
0
dx
dy
3 y 2
3x 2
dx
dy 3x 2
dx 3 y 2
dy x 2
dx y 2
The solution is continued on the next page.
3
Substituting for
Differentiate implicitly to find
2
Substituting
d2y
dx 2
y2
dy
dx
y 2 x2
y
y
2
y
dy
dy
2
y 4 x 2 x 2 y x
2 y
dx
dx
3 y 3 x
y 1 x
x
y x
y
2
y
Simplifying the numerator we have:
d 2 y dy
dy
2 y
4 y x 2x
2
dx
dx
dx
2
d2y
dy
implicitly to find
dx
dx 2
d2y d x
dx 2 dx y
d dy d y 2 x
dx dx dx 2 y x
2
dy
.
dx
d 2
d
x y 2 5
dx
dx
dy
0
2x 2 y
dx
dy x
dx y
d 2
d
y xy x 2 5
dx
dx
dy dy
x y 1 2 x 0
2y
dx dx
dy
2 y x y 2 x 0
dx
dy
2 y x y 2 x
dx
dy y 2 x
dx 2 y x
Differentiate
x2 y2 5
Copyright © 2016 Pearson Education, Inc.
Exercise Set 2.8
433
60. Using the calculator, we have:
d2y
dy
implicitly to find
.
dx
dx 2
d 2 y d x2
dx 2 dx y 2
dy
y 2 2 x x2 2 y
dx
Differentiate
y
2 2
x2
2 xy 2 2 x 2 y 2
y
4
y
2 xy 2
61. Using the calculator, we have:
Substituting for
dy
dx
2 x4
y
y4
Simplifying the derivative, we have:
2 xy 2 y 2 x 4
2
d y
1
y
y
2
4
dx
y
62. Using the calculator, we have:
2 xy 3 2 x 4
y
y4
58.
59.
2 xy 3 2 x 4
y5
3
2x y x
y
5
3
63. Using the calculator, we have:
Given an equation in x and y where y is a
function of x but where it is difficult or
impossible to express y in terms of x, implicit
differentiation allows us to find the derivative of
y with respect to x.
64. Using the calculator, we have:
One dictionary defines “implicit” as
“capable of being understood from something
else though unexpressed” or “involved in the
nature or essence of something though not
revealed, expressed, or developed.” When a
function y of x is defined implicitly it is written
as an equation in x and y where y is not
expressed in terms of x but where it is
understood or implied that y is indeed a function
of x.
Copyright © 2016 Pearson Education, Inc.

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