Answer
$(a)$
Local maximum: $(-3,-9),\ (2,16)$,
Local minimum: $(-2,-16).$
$(b)$
Absolute maximum: $(2,16)$
Absolute minimum: none.
$(c)$
See graph.
Work Step by Step
$f(t)=12t-t^{3},\quad x\in[-3,\infty)$
$f'(t)=12-3t^{2}=-3(x^{2}-4)=-3(x-2)(x+2)$
$f$ and $f'$ are defined on $[-3,\infty)$
$f'(t)=0$ for $ t=\pm 2\qquad$ ... critical point.
$\left[\begin{array}{cccccccccc}
t & -3 & -2.5 & -2 & 0 & 2 & 3\\
interval & & (-3,-2) & & (-2,2) & & (2,\infty)\\
f'(t) & & -14.375 & 0 & 12 & 0 & -15\\
f(t) & -9 & \searrow & -16 & \nearrow & 16 & \searrow\\
& & & & & &
\end{array}\right]$
