How do you solve a Mean Value Theorem problem that involves multiple square roots? For the equation k(x) = $4\sqrt{x}+\frac{2}{\sqrt{x}}$ on the interval [$\frac{1}{4}$,1], I need to find the value c as stated in the Mean Value Theorem. I've done other problems like this before and know how to use $\frac{f(b)-f(a)}{b-a}$ and set the derivative of k(x) equal to the second equation, but after that I can't get the x's on one side to compute the value of x.
 A: Typically to solve something like
$$
a \sqrt{x} + bx = d
$$
for instance, you square both sides, to get
$$
a^2 x + 2ab x\sqrt{x} + b^2x^2 = d
$$
and you can now move everything but the square root to the other side:
$$
2ab x\sqrt{x}= d - a^2 - b^2 x^2
$$
Now you can square again, and get 
$$
4a^2 b^2 x^3 = (d - a^2 - b^2 x^2)^2
$$
which is a polynomial that you might hope to solve (esp. if some coefficients happen to be zeroes, etc.). At any rate, there are no square roots left. 
There's a caution here. Suppose you're asked to solve
$$
\sqrt{x} = -\sqrt{x}
$$
and you square both sides. You get
$$
x = x
$$
whose solution is "all numbers $x$". But the original equation has only the solution $x = 0$. In general, squaring an equation can introduce extraneous solutions (because $u^2 = v^2$ doesn't tell you that $u = v$, only that $u = \pm v$). So when you find a solution to that polynomial equation at the end, you have to PLUG IT BACK INTO THE ORIGINAL EQUATION and verify that it's also a solution to the original equation. 
This should help, I hope, with your problem. 
A: At first compute $$f(1)-f\left(\frac{1}{4}\right)$$, that is $$f(1)=4+2=6$$ and $$f\left(\frac{1}{4}\right)=6$$ thus $$\frac{f(b)-f(a)}{b-a}=0$$
Now you must solve $$2x^{-1/2}-x^{-3/2}=0$$ (this is $$f'(x)=0$$) Here we get $$x=\frac{1}{2}$$
