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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?

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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.

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

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  • $\begingroup$ Sorry about reproducing the early parts of your answer in detail. I wanted to get to the order of this rational transformation pronto, and didn't check the other answers in detail. +1 was of course there. $\endgroup$ – Jyrki Lahtonen Jan 18 '17 at 14:46
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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.

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