Answer
$$\eqalign{
& \left( {\text{a}} \right)t = 2 \cr
& \left( {\text{b}} \right){\text{Decreasing: }}\left( { - \infty ,2} \right),{\text{ Increasing: }}\left( {2,\infty } \right) \cr
& \left( {\text{c}} \right){\text{Relative minimum at }}\left( {2, - 12} \right) \cr} $$
Work Step by Step
$$\eqalign{
& h\left( t \right) = \frac{1}{4}{t^4} - 8t \cr
& \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr
& h'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{1}{4}{t^4} - 8t} \right] \cr
& h'\left( t \right) = {t^3} - 8 \cr
& {\text{Calculating the critical points}}{\text{, set }}h'\left( t \right) = 0 \cr
& {t^3} - 8 = 0 \cr
& {t^3} = 8 \cr
& t = 2 \cr} $$
\[\begin{align}
& \left( \text{b} \right)\text{Set the intervals }\left( -\infty ,2 \right),\left( 2,\infty \right) \\
& \text{Making a table of values } \\
& \begin{matrix}
\text{Interval} & \left( -\infty ,2 \right) & \left( 2,\infty \right) \\
\text{Test Value} & t=-4 & t=4 \\
\text{Sign of }h'\left( t \right) & \text{ }h'\left( -4 \right)=-72<0 & \text{ }h'\left( 4 \right)=56>0 \\
\text{Conclusion} & \text{Decreasing} & \text{Increasing} \\
\end{matrix} \\
\end{align}\]
$$\eqalign{
& \left( {\text{c}} \right){\text{ By Theorem 3}}{\text{.6}} \cr
& h'\left( t \right){\text{ changes from negative to positive at }}t = 2,{\text{ then }}h\left( t \right) \cr
& {\text{has a relative minimum at }}\left( {2,h\left( 2 \right)} \right) \cr
& h\left( 2 \right) = \frac{1}{4}{\left( 2 \right)^4} - 8\left( 2 \right) \cr
& h\left( 2 \right) = - 12 \cr
& {\text{Relative minimum at }}\left( {2, - 12} \right) \cr
& \cr
& \left( {\text{d}} \right){\text{Graph}} \cr} $$
