Calculus: Rolle's Theorem Verification

1. $f(x)$ is a polynomial function, so it is continuous on $[-2, 2]$. 2. $f'(x) = 3x^2 - 4$ exists for all $x \in (-2, 2)$, so $f$ is differentiable on $(-2, 2)$. 3. Evaluate endpoints: $f(-2) = (-2)^3 - 4(-2) = -8 + 8 = 0$; $f(2) = 2^3 - 4(2) = 8 - 8 = 0$. Thus $f(-2) = f(2)$.

By Rolle's Theorem, there exists at least one $c \in (-2, 2)$ such that $f'(c) = 0$: $3c^2 - 4 = 0 \implies c^2 = \frac{4}{3} \implies c = \pm \frac{2}{\sqrt{3}}$

Since $\frac{2}{\sqrt{3}} \approx 1.155 \in (-2, 2)$ and $-\frac{2}{\sqrt{3}} \approx -1.155 \in (-2, 2)$, both points lie strictly within $(-2, 2)$. Rolle's Theorem is verified.