Answer
$c=-\frac23$
Work Step by Step
Here, $f(x)=x^3-2x^2-4x+2$ in interval $[-2,2]$.
1. Since $f(x)$ is a polynomial, it is continuous over the given interval $[-2,2]$.
2. Similarly, since $f(x)$ is a polynomial, it is differentiable on the open interval $(-2,2)$.
3. $f(-2)=(-2)^3-2(-2)^2-4(-2)+2=-6$. Similarly,$f(2)=-6$. Thus, $f(-2)=f(2)$.
Thus, all three criteria of Rolle's theorem satisfied. This implies that there exists a number $c$ in the open interval $(-2,2)$ such that $f'(c)=0$. Let's differentiate $f(x)$, $$f'(x)=3x^2-4x-4\implies3c^2-4c-4=0$$ $$c=-\frac23,2$$ Thus, $c=-\frac23$ satisfies the required condition.
