\
(a)
\Thu function is
.
Differentiate
on each side with respect to
.

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

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

Local minimum is
.
\
(c)
\
.
Differentiate
on each side with respect to
.

Find the inflection points.
\Equate
to zero.

and 
and
.
The inflection point is at
and
.
Substitute
in
.

Substitute
in
.

The test intervals are
,
and
.
| \
Interval \ | \
Test Value | \Sign of ![]() | \
Concavity | \
![]() | \
![]() | \
![]() | \
Up | \
![]() | \
![]() | \
![]() | \
\
Down \ | \
![]() | \
![]() | \
![]() | \
Up | \
The graph is concave up on the intervals
and
.
The graph is concave down on the interval
.
The inflection point are
and
.
\
(d)
\Graph :
\Graph the function
:

\
(a)
\Decreasing on the interval
.
Increasing on the interval
.
(b)
\Local minimum is
.
(c)
\Concave up in the intervals
and
.
Concave down in the interval
.
Inflection point are
and
.
(d)
\Graph of the function
is
.