Chapter 12: Applications of the Derivative
Section 12.1 Derivatives and Graphs
1.
The function is increasing on (1,
), and
decreasing on (–
, 1). The lowest point on the
graph has coordinates (1, –4), so the local
minimum of –4 occurs at
x
= 1.
2.
The function is increasing on (–
, 4) and
decreasing on (4,
). The highest point on the
graph has coordinates (4, 1), so the local
maximum of 1 occurs at
x
= 4.
3.
The function is increasing on (–
, –2) and
decreasing on (–2,
). The highest point on the
graph has coordinates
(–2, 3), so a local maximum of 3 occurs at
x
= –2.
4.
The function is increasing on (3,
) and
decreasing on (–
, 3). The lowest point has
coordinates (3, –4), so a local minimum of –4
occurs at
x
= 3.
5.
The function is increasing on (–
, –4) and
(–2,
), and decreasing on (–4, –2). A relatively
high point occurs at (–4, 3) and a relatively low
point occurs at (–2, 1), so a local maximum of 3
occurs at
x
= –4 and a local minimum of 1 occurs
at
x
= –2.
6.
The function is increasing on (1, 5), and
decreasing on (–
, 1) and (5,
). A relatively low
point occurs at (1, –6) and a relatively high point
occurs at (5, 2), so a local minimum of –6 occurs
at
x
= 1 and a local maximum of 2 occurs at
x
= 5.
7.
The function is increasing on (–7, –4) and
(–2,
) and decreasing on (–
, –7) and
(–4, –2). Relatively low points occur at
(–7, –2) and (–2, –2), and relatively high point
occurs at (–4, 3). Thus a local minimum of –2
occurs at
x
= –7 and
x
= –2, and a local maximum of 3 occurs at
x
= –
4.
8.
The function is increasing on (–3, 0) and
(3,
) and decreasing on (–
, –3) and
(0, 3). Relatively low points occur at
(–3, 0) and (3, 0) and a relatively high point
occurs at (0, 4). Thus a local minimum of 0
occurs at
x
= –3 and
x
= 3, and a local maximum
of 4 occurs at
x
= 0.
9.
3
2
2
( )
2
5
4
2
( )
6
10
4
0
0
.
0
0
¡
f x
x
x
x
f
x
x
x
+
,
2
2
6
10
4
0
2 3
5
2
0
2(3
1)(
2)
0
0
0
0
0
.
0
x
x
x
x
x
x
1
or
2
3
0
x
x
Test
( )
¡
f
x
at
x
= –2,
x
= 0,
x
= 3.
2
2
2
( 2)
6( 2)
10( 2)
4
40
0
(0)
6(0)
10(0)
4
4
0
(3)
6(3)
10(3)
4
20
0
0
0
0
0
0
!
¡
0
0
0
?
¡
0
0
!
¡
f
f
f
( )
¡
f
x
is positive on
1
,
3
È
Ø
0¢ 0
É
Ù
Ê
Ú
, so
f
(
x
) is
increasing.
( )
¡
f
x
is negative on
1
, 2
3
È
Ø
0
É
Ù
Ê
Ú
, so
f
(
x
) is
decreasing.
( )
¡
f
x
is positive on (2,
), so
f
(
x
) is increasing.
10.
3
2
2
( )
4
9
12
7
( )
12
18
12
0
0
.
0
0
¡
f x
x
x
x
f
x
x
x
+
,
2
2
12
18
12
0
6 2
3
2
0
6(2
1)(
2)
0
0
0
0
0
.
0
x
x
x
x
x
x
1
or
2
2
0
x
x
Test
( )
¡
f
x
at
x
= –1,
x
= 0,
x
= 3.
2
2
2
( 1)
12( 1)
18( 1)
12
18
0
(0)
12(0)
18(0)
12
12
0
(3)
12(3)
18(3)
12
42
0
0
0
0
0
0
!
¡
0
0
0
?
¡
0
0
!
¡
f
f
f
( )
¡
f
x
is positive on
1
,
2
È
Ø
0¢ 0
É
Ù
Ê
Ú
, so
f
(
x
) is
increasing.
( )
¡
f
x
is negative on
1
, 2
2
È
Ø
0
É
Ù
Ê
Ú
, so
f
(
x
) is
decreasing.
( )
¡
f
x
is positive (2,
), so
f
(
x
) is increasing.
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11.
2
2
1
( )
3
(
3)(1)
(
1)(1)
3
( )
(
3)
(
3)
.
.
.
0
.
¡
.
.
x
f x
x
x
x
f
x
x
x
so
2
3
( )
0
(
3)
¡
.
f
x
x
has no solution.
( )
¡
f
x
does not exist if
2
(
3)
0
.
x
x
= –3.
Test
( )
¡
f
x
at
x
= –4 and
x
= 0.
2
2
3
( 4)
3
( 4
3)
3
3
1
(0)
9
3
(0
3)
0
¡
0
.
¡
.
f
f
( )
¡
f
x
is positive on (–
, –3) and (–3,
), so
f
(
x
)
is increasing on those intervals.
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