Answer
(a)
Increasing on $(-3,0)\cup(0,3)$
Decreasing on $(-\infty, -3)\cup(3, \infty)$
(b)
No absolute maximum.
Local maximum: $(3,162),\ $
No absolute minimum.
Local minimum: $(-3,-162)$
Work Step by Step
$K(t) $is defined everywhere.
$K'(t)=45t^{2}-5t^{4}=5t^{2}(9-t^{2})$
$K'(0)=0$ for $x=-3,0$ and $ 3\quad$... critical points.
$K(-3)=-162,\qquad K(0)=0,\qquad K(3)=162$
The end behavior of a polynomial is dictated by the leading term $(-t^{5})$,
so $ K(t)\rightarrow +\infty$ on the far left and $ K(t)\rightarrow -\infty$ on the far right.
Using testpoints in the intervals between critical points,
$K'(-4)=-560 \lt 0$
$K'(-1)=40\gt 0$
$K'(1)=40 \gt 0$
$K'(4)=-560 \lt 0$
Tabular view:
$\begin{array}{l}
K':\\
\\
\\
K:
\end{array} \begin{array}{lllllllll}
-\infty & & -3 & & 0 & & 3 & & \infty\\
( & -- & | & ++ & | & ++ & | & -- & )\\
& & & & & & & & \\
+\infty & \searrow & & \nearrow & & \nearrow & 162 & \searrow & \\
& & -162 & & 0 & & & & -\infty\\
& & & & & & & &
\end{array}$
(a)
Increasing on $(-3,0)\cup(0,3)$
Decreasing on $(-\infty, -3)\cup(3, \infty)$
(b)
No absolute maximum.
Local maximum: $(3,162),\ $
No absolute minimum.
Local minimum: $(-3,-162)$
