EXERCISE 51
137
26.
f
(
x
) = 2
3
+ 3
2
+ 120
'(
) = 6
2
+ 6
+ 120 which is continuous for all
.
'(
) = 6(
2

 20) = 6(
+ 4)(
 5) = 0
= 4, 5
Thus,
= 4 and
= 5 are partition numbers for
'.
Next, we construct a sign chart for
':
   + + + + + + + +   
Increasing
Decreasing
'(
)
(
)
5
4
0
5 6
Test Numbers
'(
)
!
5
!
60(
!
)
0
120(
+
)
6
!
60(
!
)
Therefore,
is decreasing on (
!
, 4) and (5,
!
); increasing on
(4, 5);
has a local minimum at
= 4 and a local maximum at
= 5.
28.
(
) =
4
+ 2
3
+ 5
'(
) = 4
3
+ 6
2
which is continuous for all
.
'(
) = 4
3
+ 6
2
= 2
2
(2
+ 3) = 0
= 
3
2
, 0
Thus,
= 
3
2
and
= 0 are partition numbers for
'.
Next, we construct a sign chart for
:
'(
)
(
)
2
1
0
1
3
2

Incr. Increasing
   + + + + +
Test Numbers
)
!
2
!
8(
!
)
!
1
2(
+
)
1
10(
+
)
Therefore,
is decreasing on
!"
,
!
3
2
#
$
%
&
; increasing on
!
3
2
,
"
#
$
%
&
;
(1.5) = 3.3125 is a local minimum.
30.
(
) =
ln
x – x
,
> 0
f’
(
) = (1)ln
+
1
"
#
$
%
&
’
 1 = ln
+ 1 – 1 = ln
(
) = ln
= 0 for
= 1.
(
) < 0 for
< 1 and
(
) > 0 for
> 1.
Therefore,
is decreasing on (0, 1) and increasing on (1,
!
).
has
a local minimum at
= 1.
32.
(
) = (
2
– 9)
2/3
(
) =
2
3
(
2
– 9)
1/3
(2
) =
3(
2
"
9)
(
) = 0 at
= 0. We have to determine the sign of
(
) in the
intervals (
!
, 3), (3, 0) and (0,
!
).
For
< 3,
2
> 9 and hence (
2
– 9)
1/3
> 0.
Thus,
(
) < 0 on the interval (
!
, 3).