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Find all functions $f:\Bbb R^+\to\Bbb R^+$ s.t. for all $x\in \Bbb R^+$ the following is valid:

$$f\bigg(\frac{1}{f(x)}\bigg)=\frac{1}{x}$$


I tried to substitute $\frac{1}{x}$ for $x$ and compare the equations: $$f\bigg(\frac{1}{f(\frac{1}{x})}\bigg)=x $$

From this I found one solution $f(x)=x.$

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    $\begingroup$ Let $g(x):=\log(f(\exp(x))$. Then $x=-g(-g(x)).$ $\endgroup$ – Somos Jul 25 at 20:01
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Using the hint of @Somos we can substitute $$g(x)=-\log{(f(\exp{(x)}))}$$ so that the equation becomes $$g(g(x))=x$$ So we just need $g(x)$ to be an involution (of which there are infinitely many). Choosing any involution $g(x)$ defined over $\mathbb{R}$ gives a solution $$f(x)=\exp{(-g(\log{(x)}))}$$ Some solutions include $$(g(x),f(x))\in\{(x,1/x),(-x,x)\}$$ I believe these are the only continuous solutions but there should be infinitely many discontinuous ones.

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