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?
 A: 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.
A: 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 $f(\sqrt{2})=-\frac{2}{3}$.
A: 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.
