Applying the Mean Value Theorem

Since $f(x) = x^4 - 2x^2$ is a polynomial, it is continuous on $[-2,2]$ and differentiable on $(-2,2)$, so the Mean Value Theorem applies. Compute $f(-2) = 16 - 8 = 8$ and $f(2) = 16 - 8 = 8$, so the average rate of change is $\frac{f(2)-f(-2)}{2-(-2)} = \frac{8-8}{4} = 0$.

The derivative is $f'(x) = 4x^3 - 4x$. Setting $f'(c) = 0$ gives $4c^3 - 4c = 0$, i.e. $4c(c^2-1)=0$, so $c = 0, 1, -1$. All three values lie in the open interval $(-2,2)$, so all three satisfy the conclusion of the MVT.

The tangent lines at c equals negative one, zero, and one must be horizontal. This is because the secant line connecting the endpoints has slope zero, since f of negative two equals f of two. The Mean Value Theorem guarantees a tangent parallel to the secant, so each guaranteed tangent line has the same slope as the secant, namely zero, making it horizontal.