(10)
\Step 1:
\The function is
.
Differentiate
on each side with respect to
.

Find the critical points.
\Since it is a polynomial it is continuous at all the point.
\Thus, the critical points exist when
.
Equate
to zero.

and
.
The critical points are
and
.
The test intervals are
.
| Interval | \Test Value | \Sign of ![]() | \
Conclusion | \
![]() | \
| \
\
| \
Increasing | \
| \
![]() | \
\
| \
Decreasing | \
| \
![]() | \
\
| \
Increasing | \
Therefore the function is increasing on the intervals
and
.
The function is decreasing on the interval
.
in the interval of
.
So the function
is increasing at
.
\
\
\
(b)
\Find the local maximum and local minimum.
\The function
has a local maximum at
, because
changes its sign from positive to negative.
Substitute
in
.

Local maximum is
.
The function
has a local maximum at
, because
changes its sign from positive to negative.

Local minimum is
.
\
(c)
\
.
Differentiate
on each side with respect to
.
Find the inflection points.
\Equate
to zero.

The inflection point is
.
Substitute
in
.

The test intervals are
and
.
| \
Interval \ | \
Test Value | \Sign of ![]() | \
Concavity | \
![]() | \
| \
![]() | \
Down | \
![]() | \
![]() | \
![]() | \
\
Up \ | \
The graph is concave up on the interval
.
The graph is concave down on the interval
.
The inflection point is
.
\
(a)
\Increasing on the intervals
and
.
Decreasing on the interval
.
(b)
\Local maximum is
.
Local minimum is 
(c)
\Concave up on the interval
.
Concave down on the interval
.
Inflection point is
.
\
\