Answer
(a)
Increasing on $(-2,0)\cup(2, \infty)$.
Decreasing on $(-\infty, -2)\cup(0,2)$
(b)
No absolute maximum.
Local maximum: $(0,16)$
No absolute minimum.
Local minima: $(-2,0)$ and $(2,0)$
Work Step by Step
$f$ is defined everywhere.
$ f'(x)=4x^{3}-16x=4x(x^{2}-4),\quad$ defined everywhere.
$f'(0)=0$ for $x=-2,0$ and $ 2\quad$... critical points.
$f(-2)=0,\qquad f(0)=16,\qquad f(2)=0.$
The end behavior of a polynomial is dictated by the leading term,
so $ f(x)\rightarrow +\infty$ on the far left and $ f(x)\rightarrow +\infty$ on the far right.
Using testpoints in the intervals between critical points,
$f'(-3) \lt 0$
$f'(-1) \gt 0$
$f'(1) \lt 0$
$f'(3) \gt 0$
Tabular view:
$\begin{array}{l}
f':\\
\\
\\
f:
\end{array} \begin{array}{lllllllllll}
-\infty & & -2 & & 0 & & 2 & & \infty & & \\
( & -- & | & ++ & | & -- & | & ++ & ) & & \\
& & & & & & & & & & \\
+\infty & \searrow & & \nearrow & 16 & \searrow & & \nearrow & +\infty & & \\
& & 0 & & & & 0 & & & & \\
& & & & & & & & & &
\end{array}$
(a)
Increasing on $(-2,0)\cup(2, \infty)$.
Decreasing on $(-\infty, -2)\cup(0,2)$
(b)
No absolute maximum.
Local maximum: $(0,16)$
No absolute minimum.
Local minima: $(-2,0)$ and $(2,0)$
