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 .