Function

The function $f$ satisfies $f(\sqrt{2x} + x) = \frac{1}{x}$ for all $x \ge \frac{1}{2}$, find $f(4)$.

$f(4) = f(\sqrt{2x}+x)={1\over{x}}$

$\sqrt{2x}+x=4$; $\sqrt{2x} = -x + 4$; $2x = x^2 - 8x + 16$; $x^2-10x+16=0$; $(x-8)(x-2)=0$

Note that x = 8 i s an extraneous solution, because $\sqrt{2*8}$ does not equal to $-8 + 4 = -4$, for a square root can't end up negative.

Thus, our only solution, x = 2. $f(\sqrt{2(2)}+2)={1\over2}=f(4)$, so our final answer is 1/2.