Answer
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[-2, 2]$
On the interval $(-2, 2)$, $f'(c) = 1$ when $c = -\frac{2\sqrt{3}}{3}~~$ or $~~c=\frac{2\sqrt{3}}{3}$
Work Step by Step
$f(x) = x^3-3x+2$
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. $f'(x) = 3x^2-3$ 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 $(-2, 2)$
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[-2, 2]$
Therefore, according to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(-2,2)$ such that $f'(c) = \frac{f(2)-f(-2)}{2-(-2)}$
$\frac{f(2)-f(-2)}{2-(-2)} = \frac{4-0}{4} = 1$
We can find $c$:
$f'(x) = 3x^2-3 = 1$
$3x^2=4$
$x^2 =\frac{4}{3}$
$x = -\frac{2}{\sqrt{3}}~~$ or $~~x=\frac{2}{\sqrt{3}}$
$x = -\frac{2\sqrt{3}}{3}~~$ or $~~x=\frac{2\sqrt{3}}{3}$
On the interval $(-2, 2)$, $f'(c) = 1$ when $c = -\frac{2\sqrt{3}}{3}~~$ or $~~c=\frac{2\sqrt{3}}{3}$
