Answer
(a)
Increasing on $(-\infty, -1)\cup(0,1)$
Decreasing on $(-1,0)\cup(1, \infty)$
(b)
No absolute maximum.
Local maxima: $(-1,0.5),\ (1,0.5)$
No absolute minimum.
Local minimum: $(0,0)$
Work Step by Step
$H(t) $is defined everywhere.
$H'(t)=6t^{3}-6t^{5}=6t^{3}(1-t^{2})$
$H'(0)=0$ for $x=-1,0$ and $ 1\quad$... critical points.
$H(-1)=0.5,\qquad H(0)=0,\qquad H(1)=0.5$
The end behavior of a polynomial is dictated by the leading term $(-t^{6})$,
so $ H(t)\rightarrow -\infty$ on the far left and $ H(t)\rightarrow -\infty$ on the far right.
Using testpoints in the intervals between critical points,
$H'(-1)=144 \gt 0$
$H'(0.5)=-0.5625\lt 0$
$H'(1.5)=+0.5625 \gt 0$
$H'(3)=-144 \lt 0$
Tabular view:
$\begin{array}{l}
H':\\
\\
\\
H:
\end{array} \begin{array}{lllllllll}
-\infty & & -1 & & 0 & & 1 & & \infty\\
( & -- & | & ++ & | & -- & | & ++ & )\\
& & & & & & & & \\
& \nearrow & 0.5 & \searrow & & \nearrow & 0.5 & \searrow & \\
-\infty & & & & 0 & & & & -\infty\\
& & & & & & & &
\end{array}$
(a)
Increasing on $(-\infty, -1)\cup(0,1)$
Decreasing on $(-1,0)\cup(1, \infty)$
(b)
No absolute maximum.
Local maxima: $(-1,0.5),\ (1,0.5)$
No absolute minimum.
Local minimum: $(0,0)$
