\
(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
and
and
.
The critical points are ,
and
.
The test intervals are ,
,
and
.
Interval | \Test Value | \Sign of ![]() | \
Conclusion | \
![]() | \
![]() | \
\
| \
Increasing | \
![]() | \
![]() | \
\
| \
Decreasing | \
![]() | \
![]() | \
\
| \
Increasing | \
![]() | \
![]() | \
\
| \
Decreasing | \
The function is increasing on the intervals and
.
The function is decreasing on the intervals and
.
\
(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 minimum at
, because
changes its sign from negative to positive.
Substitute in
.
Local minimum is .
The function has a local maximum at
, because
changes its sign from positive to negative.
Substitute in
.
Local maximum is .
\
(c)
\.
Differentiate on each side with respect to
.
Find the inflection points.
\Equate to zero.
The inflection point is at .
Substitute in
.
Substitute in
.
The test intervals are ,
and
.
\
Interval \ | \
Test Value | \Sign of ![]() | \
Concavity | \
![]() | \
![]() | \
![]() | \
Down | \
![]() | \
![]() | \
![]() | \
\
Up \ | \
![]() | \
![]() | \
![]() | \
Down | \
The graph is concave up on the interval .
The graph is concave down on the intervals and
.
The inflection point are and
.
\
(d)
\Graph :
\Graph the function is
\
(a)
\Increasing on the intervals and
.
Decreasing on the intervals and
.
(b)
\Local maximum are and
.
Local minimum is .
(c)
\Concave up in the interval .
Concave down in the intervals and
.
Inflection point are and
.
(d)
\Graph of the function is