a. Note that
is continuous for all
. If
attains a maximum at
, then
. Compute the derivative of
.
Evaluate this at
and solve for
.
To ensure that a maximum is reached for this value of
, we need to check the sign of the second derivative at this critical point.
The second derivative at
is negative, which indicate the function is concave downward, which in turn means that
is indeed a (local) maximum.
b. When
, we have derivatives
Inflection points can occur where the second derivative vanishes.
Then we have three possible inflection points when
,
, or
.
To decide which are actually inflection points, check the sign of
in each of the intervals
,
,
, and
. It's enough to check the sign of any test value of
from each interval.
The sign of
changes to either side of
and
, but not
. This means only
and
are inflection points.