Answer
The Mean Value Theorem can be applied; $c=-\dfrac{1}{2}.$
Work Step by Step
Since $f(x)$ is a polynomial, it is continuous for all values of $x$ and differentiable at every value of $x.$
$a=-2; b=1\to f(a)=4$ and $f(b)=1.$
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 $fâ˛(c)=\dfrac{f(b)-f(a)}{b-a}.$
$f'(c)=\dfrac{1-4}{1-(-2)}=-1.$
$f'(x)=2x\to f'(x)=-1\to c=-\dfrac{1}{2}$
