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.
Therefore the function have inflection points at
and
.
Consider the test intervals
,
and
.
| \
Interval \ | \
Test Value | \Sign of ![]() | \
Concavity | \
![]() | \
![]() | \
\
| \
Down | \
![]() | \
![]() | \
\
| \
Up | \
![]() | \
![]() | \
\
| \
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 intervals
and
.
The graph is concave down in the intervals
and
.
The function have inflection points at
and
.
The function
is concave up in the interval
and
.
The graph is concave down in the intervals
and
.