Answer
We can't apply the theorem because f is not continuous on the closed interval $[-2,2]$.
Work Step by Step
Rolle's Theorem
Let $f$ be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ .
If $f(a)=f(b)$ , then there is at least one number $c$ in $(a, b)$ such that $f^{\prime}(c)=0.$
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f is not defined when
$1-x^{2}=0,$
$x=\pm 1\in[-2,2]$
So, f is not continuous on the closed interval $[-2,2]$.
A premise of the theorem is not satisfied,
so we can't use its conclusion.
