Applying the Mean Value Theorem

Since $f(x) = x^3 - x$ is a polynomial, it is continuous on $[-1,2]$ and differentiable on $(-1,2)$, so the Mean Value Theorem applies. Compute $f(2) = 8 - 2 = 6$ and $f(-1) = -1 + 1 = 0$. The average rate of change is $\dfrac{f(2)-f(-1)}{2-(-1)} = \dfrac{6-0}{3} = 2$.

Next, $f'(x) = 3x^2 - 1$. Setting $f'(c) = 2$ gives $3c^2 - 1 = 2$, so $3c^2 = 3$, meaning $c^2 = 1$ and $c = \pm 1$. Since we need $c$ in the open interval $(-1, 2)$, the value $c = -1$ is excluded (it's an endpoint), leaving only $c = 1$.