Answer
$f$ satisfies the three hypotheses of Rolle's Theorem on the interval $[-2, 2]$
On the interval $(-2, 2)$, $f'(c) = 0$ when $c = -\frac{2}{3}$
Work Step by Step
$f(x) = x^3-2x^2-4x+2$
1. The function is a polynomial and all polynomials are continuous on the interval $(-\infty, \infty)$. Therefore, the function is continuous on the closed interval $[-2,2]$
2. $f'(x) = 3x^2-4x-4$ which exists for all numbers in the interval $(-\infty, \infty)$. Therefore, $f'(x)$ exists for all numbers on the interval $(-2, 2)$.
Thus $f$ is differentiable on the open interval $(-2, 2)$
3. $f(-2) = (-2)^3-2(-2)^2-4(-2)+2 = -6$
$f(2) = (2)^3-2(2)^2-4(2)+2 = -6$
Thus $f(-2) = f(2)$
$f$ satisfies the three hypotheses of Rolle's Theorem on the interval $[-2, 2]$
Therefore, according to Rolle's Theorem, there is a number $c$ in the interval $(-2,2)$ such that $f'(c) = 0$
We can find $c$:
$f'(x) = 3x^2-4x-4 = 0$
$(3x+2)(x-2) = 0$
$x = -\frac{2}{3}~~$ or $~~x=2$
Note that we do not include $x=2$ as a number that satisfies the conclusion of Rolle's Theorem because the point $x=2$ is not included in the open interval $(-2,2)$
On the interval $(-2, 2)$, $f'(c) = 0$ when $c = -\frac{2}{3}$
