If we take $f(x)=4 x^{4}$, then
(i) $f(x)$ is continuous in $(-2,2)$
(ii) $f(x)$ is differentiable in $(-2,2)$
(iii) $f(-2)=f(2)$
So, $f(x)=4 x^{4}$ satisfies all the conditions of Rolle's theorem therefore $\exists$ a point $c$ such that $f'(c)=0$
$\Rightarrow 16 c^{3}=0 \Rightarrow c=0 \in(-2,2)$