#### Answer

$c= 3\sqrt{2}-2 $

#### Work Step by Step

Given $$y=\frac{x}{x+2}, \quad[1,4]$$
Since
$$f'(x)=\frac{\frac{d}{dx}\left(x\right)\left(x+2\right)-\frac{d}{dx}\left(x+2\right)x}{\left(x+2\right)^2}= \frac{2}{\left(x+2\right)^2} $$
Then by MVT, there exists a constant $c\in (1,4)$ such that
\begin{align*}
f'(c)&= \frac{f(b)-f(a) }{b-a}\\
\frac{2}{(c+2)^2}&=\frac{2/3-1/3}{3}\\
\frac{2}{(c+2)^2}&=\frac{1}{9}\\
(c+2)^2&= 18\\
c+2&=\pm \sqrt{18}\\
c&= \pm 3\sqrt{2}-2
\end{align*}
Choose $c= 3\sqrt{2}-2\in (1,4)$