Answer
The Mean Value Theorem can be applied; $c=\sqrt[3]{2}.$
Work Step by Step
Since $f(x)$ is a polynomial, it is continuous for all values of $x$ and differentiable at every value of $x.$
$a=0; b=2\to f(a)=0$ and $f(b)=0$
Since $f(x)$ is continuous over $[a, b]$ and differentiable over $(a, b)$, applying the 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 $fâ˛(c)=\dfrac{f(b)-f(a)}{b-a}.$
$f'(c)=\dfrac{0-0}{2-0}=0.$
$f'(x)=4x^3-8\to x^3=2.\to c=\sqrt[3]{2}.$
