# 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?

Hint: You're about half-way there. If you plug in $x=\sqrt{2}$, you get $$f(\sqrt{2})+2f\left(\frac{1}{\sqrt{2}}\right)=4.$$ Now, if you plug in $x=\frac{1}{\sqrt{2}}$, then you get, instead $$f\left(\frac{1}{\sqrt{2}}\right)+2f(\sqrt{2})=1.$$ You can use these two equations to solve for $f(\sqrt{2})$: in the second equation, solve for $f\left(\frac{1}{\sqrt{2}}\right)$ in terms of $f(\sqrt{2})$ and then plug this into the first 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 $f(\sqrt{2})=-\frac{2}{3}$.
Presumably the equation holds for all $x\neq0$.
You were given $$f(x)+2f(\frac1x)=2x^2.\qquad(1)$$ Plugging in $1/x$ in place of $x$ gives $$f(\frac1x)+2f(x)=\frac2{x^2}.\qquad(2)$$ You can treat $A=f(x)$ and $B=f(1/x)$ as unknowns in the linear system you get by combining $(1)$ and $(2)$. So $$\begin{cases}A+2B&=2x^2,\\ 2A+B&=\dfrac2{x^2}. \end{cases}$$ You can solve for $A$ and $B$ any which way you want. For example multiplying latter equation by $2$ and subtracting the resulting equation from the former eliminates $B$ and gives $$-3A=2x^2-\frac4{x^2},$$ or $$f(x)=A=-\frac23x^2+\frac4{3x^2}.$$ Plugging in $x=\sqrt2$ gives $f(\sqrt2)=-2/3$.
The key was that the substitution $x\mapsto 1/x$ is an order two rational transformation (= do it twice and you get what you started with). Replacing $x\mapsto -x$ occurs more frequently (say, when you look at even/odd functions), but often any order two substitution can be handled with the same trick. The transformationion $x\mapsto g(x)=-1/(x+1)$ has order three: $g(g(x))=-1-1/x$, $g(g(g(x)))=x$. Then you would need to combine three values, $f(x)$, $f(-1/(x+1))$ and $f(-1-1/x)$, to take advantage.