Answer
$c=\pm\frac{2\sqrt 3}{3}$
Work Step by Step
The Mean Value Theorem states that if $f$ is a continuous function over the closed interval $[a,b]$ and differentiable over the open interval $(a,b)$, then there exists at least one point $c\in(a,b)$ so that
$$f′(c)=\dfrac{f(b)−f(a)}{b−a}.$$
The function $f(x)$ = $x^{3}-3x+2$ is a polynomial function, therefore continuous on $[-2, 2]$ and differentiable on $(-2, 2)$ since polynomials are continuous and differentiable on $R$.
According the the Mean Value Theorem, there is at least one point $c\in(-2,2)$ so that we have:
$f'(c)=\dfrac{f(2)-f(-2)}{2-(-2)}$
$f'(c)=\dfrac{[(2^3)-3(2)+2]-[(-2)^3-3(-2)+2]}{4}$
$f'(c)=\dfrac{4-0}{4}$
$f'(c)=1$
But $f'(x)=3x^2-3$, so we have:
$3c^2-3=1$
$x^2=\frac{4}{3}$
$x=\pm\dfrac{2}{\sqrt 3}=\pm\dfrac{2\sqrt 3}{3}$
