Answer
(a)
Increasing on $(0,1)\cup(2, \infty)$.
Decreasing on $(-\infty, 0)\cup(1,2)$
(b)
No absolute maximum.
Local maximum: $(1,1)$
No absolute minimum.
Local minima: $(0,0)$ and $(2,0)$
Work Step by Step
$g$ is defined everywhere.
$ g'(x)=4x^{3}-12x^{2}+8x=4x(x^{2}-3x+2)=8x(x-1)(x-2),\quad$ defined everywhere.
$g'(0)=0$ for $x=0,1$ and $ 2\quad$... critical points.
$g(0)=0,\qquad g(1)=1,\qquad g(2)=0.$
The end behavior of a polynomial is dictated by the leading term,
so $ g(x)\rightarrow +\infty$ on the far left and $ g(x)\rightarrow +\infty$ on the far right.
Using testpoints in the intervals between critical points,
$g'(-1)=-24 \lt 0$
$g'(0.5)=1.5 \gt 0$
$g'(1.5)=-1.5 \lt 0$
$g'(3)=24 \gt 0$
Tabular view:
$\begin{array}{l}
g':\\
\\
\\
g:
\end{array} \begin{array}{lllllllllll}
-\infty & & 0 & & 1 & & 2 & & \infty & & \\
( & -- & | & ++ & | & -- & | & ++ & ) & & \\
& & & & & & & & & & \\
+\infty & \searrow & & \nearrow & 1 & \searrow & & \nearrow & +\infty & & \\
& & 0 & & & & 0 & & & & \\
& & & & & & & & & &
\end{array}$
(a)
Increasing on $(0,1)\cup(2, \infty)$.
Decreasing on $(-\infty, 0)\cup(1,2)$
(b)
No absolute maximum.
Local maximum: $(1,1)$
No absolute minimum.
Local minima: $(0,0)$ and $(2,0)$
