Answer
The maximum is $f(-1)=f(1)=2$; the minimum is $f(-2)=f(2)=-160$
Work Step by Step
Given $$f(x)=6 x^{4}-4 x^{6}, \quad[-2,2]$$
Since
$$ f'(x)= 24x^3 -24x^5 $$
Find the critical points
\begin{align*}
f'(x)&=0\\
24x^3 -24x^5&=0\\
24x^3(1-x^2)
\end{align*}
Then $f(x)$ has critical points at $x= 0,\ x=-1,\ x=1$. We check the values at the critical points and end points:
\begin{align*}
f(-2)&= -160 \\
f(-1)&= 2\\
f(0)&= 0 \\
f(1)&= 2\\
f(2)&=-160
\end{align*}
Then the maximum is $f(-1)=f(1)=2$ and the minimum is $f(-2)=f(2)=-160$.
