Mean-value Theorem $f(x)=\sqrt{x+2}; [4,6]$ 
Verify that the hypothesis of the mean-value theorem is satisfied for the given function on the indicated interval. Then find a suitable value for $c$ that satisfies the conclusion of the mean-value theorem.

$$f(x)=\sqrt{x+2}; [4,6]$$
So,
$$f'(x) = {1 \over 2} (x+2)^{-{1\over 2}}$$
$f(x)$ is differentiable for all x. Now,
$$f'(c) = {f(b)-f(a)\over b- a} \\
= {f(6) - f(4)\over 6-4} \\
= {2\sqrt{2} - \sqrt{6}\over2}$$
Since $f'(c) = {2\sqrt{2} - \sqrt{6}\over2}$,
$${1\over 2}(c+2)^{-{1\over 2}} = {2\sqrt{2} - \sqrt{6}\over2}$$
Then after, this will give me $c$ right?
I just want to know if I did it right so far. I tried simplifying the last equation, but it wasn't right. Please let me know if there is anything wrong in my steps, if not could anyone help me solve for $c$ at the end? Thank you.

You did everything correctly, let's solve for $c$ together.
You have
$$
\frac{1}{2} (c+2)^{-1/2} = \frac{a}{2}\\
(c+2)^{-1/2} = a \\
\frac{1}{\sqrt{c+2}} = a \\
c+2 = \frac{1}{a^2}
$$
so
$$
\begin{split}
c &= \frac{1}{a^2} - 2 = \frac{1}{\left(2 \sqrt{2} - \sqrt{6}\right)^2} - 2 \\
  &= \frac{1}{8 + 6 - 4\sqrt{2}\sqrt{6}} - 2 \\
  &= \frac{1}{14 - 8\sqrt{3}} - 2 \\
  &= \frac{14 + 8 \sqrt{3}}{14^2 - 3 \cdot 8^2} - 2 \\
  &= \frac{14 + 8 \sqrt{3}}{4} - 2 \\
  &= \frac{7}{2} +2\sqrt{3} - 2 \\
  &= \frac{3}{2} +2\sqrt{3}
\end{split}
$$
