Single Variable Calculus, Chapter 4, 4.2, Section 4.2, Problem 14

Verify that the function $f(x) = \frac{x}{x+2}$ satisfies the hypothesis of the Mean Value Theorem on the interval $[1,4]$. Then find all the numbers $c$ that satisfy the conclusion of the Mean Value Theorem.
We know that $f(x)$ is a rational function that is continuous everywhere except on the values of $x$ that will make its denominator equal to 0. In this case, that value is $x= -2$.But its not included in the interval. Hence, $f(x)$ is continuous on the interval $[1,4]$. Also the derivative, $\displaystyle f'(x) = \frac{2}{(x+2)^2}$ is differentiable everywhere except at $x = -2$. Hence, $f(x)$ is differentiable on the open interval $(1,4)$.

Now solving for $c$, we have...

$
\begin{equation}
\begin{aligned}
f'(c) &= \frac{f(b) - f(a)}{b-a}\\
\\
f'(c) &= \frac{\left[\frac{4}{4+2} \right] - \left[ \frac{1}{1+2} \right]}{4-1}\\
\\
f'(c) &= \frac{1}{9}
\end{aligned}
\end{equation}
$

but $\displaystyle f'(x) = \frac{2}{(x+2)^2}$, so $f'(c) = \frac{2}{(c+2)^2}$

$
\begin{equation}
\begin{aligned}
\frac{2}{(c+2)^2} &= \frac{1}{9}\\
\\
18 &= (c+2)^2\\
\\
18 &= c^2 + 4c + 4\\
\\
0 &= c^2 + 4c - 14
\end{aligned}
\end{equation}
$

Using Quadratic Formula, we get...
$c = 2.2426$ and $c = -6.2426$
We got two values of $c$. However, the function is defined only at interval $[1,4]$. Therefore, $c = 2.2426$