The function is
in the interval
.
Find the inflection points.
\Differentiate
with respect to
.
Differentiate
with respect to
.
Find the inflection points by equating
to zero.

\
are not in the interval
.
Thus, the inflection points are
.
Substitute
in
.
.
Therefore the inflection point is
.
Consider the test intervals
and
.
| \
Interval \ | \
Test Value | \Sign of ![]() | \
Concavity | \
![]() | \
![]() | \
\
| \
Down | \
![]() | \
![]() | \
\
| \
Up | \
Concavity test:
\(a) If
for all
in
, then the graph of
is concave upward on
.
(b) If
for all
in
, then the graph of
is concave downward on
.
Thus, the graph is concave up in the interval
.
The graph is concave down in the interval
.
Inflection point is
.
The function
is concave up in the interval
.
The graph is concave down in the interval
.