if $f(x)+2f(\frac{1}{x})=2x^2$ what is $f(\sqrt{2})$ Question
if $f(x)+2f(\frac{1}{x})=2x^2$  what is $f(\sqrt{2})$
My steps
I tried to plug in $\sqrt{2}$ into the equation but that didn't get me anywhere because then i would have $2f(\frac{1}{\sqrt{2}})$ in the way. I was wondering on how to solve this equation?

If $f(x)+2f(\frac{1}{x})=2x^2$, then by substitution $f(\frac{1}{x})+2f(x)=\frac{2}{x^2}$ and so $f(\frac{1}{x})=\frac{2}{x^2}-2f(x)$.  Substituting $f(\frac{1}{x})$ into the first equation yields $f(x)+2[\frac{2}{x^2}-2f(x)]=2x^2$.  Solve for $f(x)$: $f(x)=\frac{2}{3}[\frac{2}{x^2}-x^2]$.  So $f(\sqrt{2})=-\frac{2}{3}$.
