Answer
$t=-2,\ 0,\ 2$
$ s(t) $ has local maxima at $t=0 $ and has local minima at $t= -2,\ \ t=2$
Work Step by Step
Given $$s(t)=t^{4}-8 t^{2} $$
Since
\begin{align*}
s'(t) &= 4t^3-16t\\
s''(t)&= 12t^2-16
\end{align*}
Since $f(x)$ has critical point when $s'(t)= 0$
\begin{align*}
s'(t)&=0\\
4t^3-16t&=0\\
4t(t^2-4)&=
0\end{align*}
Then $s(t)$ has critical points at $t=0,\ t=2,\ \ t=-2$, and
\begin{align*}
s''(-2)&= 32>0\\
s''(0)&=-16<0\\
s''(2)&=32 >0
\end{align*}
Then $ s(t) $ has a local maxima at $t=0 $ and it has a local minima at $t= -2,\ \ t=2$
