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The function is $f:\left(0,\infty\right)\rightarrow\mathbb{R}$, I tried to do it like that, first I saw that: $$f\left(x\right) =2f\left(\frac{1}{x}\right)-\frac{2x^{2}-1}{x^{2}+1}$$ and then decided to try to put $\frac{1}{x}\in\left(0,\infty\right) $ and came out with: $$f\left(\frac{1}{x}\right) =2f\left(x\right)-\frac{2\frac{1}{x^{2}}-1}{\frac{1}{x^{2}}+1} =2f\left(x\right)-\frac{\frac{2-x^{2}}{x^{2}}}{\frac{1+x^{2}}{x^{2}}} =2f\left(x\right)-\frac{2-x^{2}}{1+x^{2}}$$ i used it at the last equation and it came out that: $$f\left(x\right)=\frac{5-4x^{2}}{3\left(1+x^{2}\right)}$$ now i get $f\left(1\right)=\frac{1}{6}$ but i know it should equal 1/2... I have now idea how to move forward, and why it doesn't work, any tips?

Abzikro
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    $f(x) = 4f(x) - \frac{4-2 x^2}{1+x^2} - \frac{2 x^2-1}{1+x^2}=4f(x) - \frac{3}{1+x^2}$, from which. $f(x) = \frac{1}{1+x^2}$. Seems you just did some mistake. – kolobokish Nov 11 '22 at 19:50
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    @kolobokish oh yeah I did a mistake :/ I didn't get why it doesn't work. Thank you very much! – Abzikro Nov 11 '22 at 20:00

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$$f(x)=2f\left(\frac 1x\right)-\frac{2x^2-1}{x^2+1}\\f\left(\frac 1x\right)=2f(x)-\frac{2-x^2}{x^2+1}$$ Adding and simplifying we have $$f(x)+f\left(\frac 1x\right)=1$$ It follows $$\boxed{f(x)= \frac{1}{x^2+1}}$$ and it is easy to verify that in fact $$f(x)=2f\left(\frac 1x\right)-\frac{2x^2-1}{x^2+1}$$

Piquito
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  • Why by derivative i can't get the good answer? I get $f(x)=2-\frac{3}{\left(x^{2}+1\right)}$ by derivating both side by $x$ – X0-user-0X Nov 11 '22 at 22:23