Answer
The Mean Value Theorem can be applied; $c=-\dfrac{1}{4}.$
Work Step by Step
$f(x)$ is continuous for all values of $x\le2$ and differentiable for at all values of $x\lt2/.$
$a=-7; b=2\to f(a)=3$ and $f(b)=0.$
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 $fâ˛(c)=\dfrac{f(b)-f(a)}{b-a}.$
$f'(c)=\dfrac{0-3}{2-(-7)}=-\dfrac{1}{3}.$
$f'(x)=-\dfrac{1}{2\sqrt{2-x}}\to f'(c)=-\dfrac{1}{3}\to -\dfrac{1}{2\sqrt{2-c}}=-\dfrac{1}{3}\to$
$\sqrt{2-c}=\dfrac{3}{2}\to c=-\dfrac{1}{4}.$
