Rolle's Theorem and a Constant Function

Even though $f(-2)=f(3)$, there exists a unique point in $(-2,3)$ where $f'(x)$ reaches its maximum value.

Since $f(x)=5$ is constant with $f(-2)=f(3)$, Rolle’s Theorem guarantees that every point in $(-2,3)$ is a point where $f'(x)=0$.

The constant function $f(x)=5$ does not satisfy the differentiability condition required by Rolle’s Theorem, so the theorem does not apply.

Rolle’s Theorem cannot be applied because constant functions do not have any critical points.