Answer
The Mean Value Theorem can be applied; $c=\dfrac{\sqrt{3}}{3}$ or $c-\dfrac{\sqrt{3}}{3}.$
Work Step by Step
Since $f(x)$ is a polynomial, it is continuous for all values of $x$ and differentiable at every value of $x.$
$a=-1; b=1\to f(a)=-3$ and $f(b)=3.$
Since $f(x)$ is continuous over $[a, b]$ and differentiable over $(a, b)$, applying Mean Value Theorem over the interval $[a , b]$ guarantees the existence of at least one value c such that $a\lt c\lt b$ and $fâ˛(c)=\dfrac{f(b)-f(a)}{b-a}.$
$f'(c)=\dfrac{3-(-3)}{1-(-1)}=3.$
$f'(x)=3x^2+2\to f'(x)=3\to 3x^2+2=3\to c=+\dfrac{\sqrt{3}}{3}$ or $c=-\dfrac{\sqrt{3}}{3}.$
