## Calculus (3rd Edition)

$t=-2,\ 0,\ 2$ $s(t)$ has local maxima at $t=0$ and has local minima at $t= -2,\ \ t=2$
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 $t= -2,\ \ t=2$