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Does there exist a non trivial function $f:\mathbb{R}\to\mathbb{R}$ such that $f^{-1}(x)=\frac{1}{f(x)}$ ?

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  • $\begingroup$ Yeah !! Also the one which @Nilan gave $\endgroup$ – Hirak Aug 14 '14 at 9:58
  • $\begingroup$ That's trivial... $\endgroup$ – Shakespeare Aug 14 '14 at 9:59
  • $\begingroup$ @Nilan what would be the functional inverse of that function. would the inverse even be a function? $\endgroup$ – cirpis Aug 14 '14 at 10:03
  • $\begingroup$ it's not invertible $\endgroup$ – Shakespeare Aug 14 '14 at 10:06
  • $\begingroup$ Yes! you are right. $\endgroup$ – Bumblebee Aug 14 '14 at 10:07
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At least the functional equation can't hold for every $x \in \mathbb{R}$. Note that plugging in $x = 0$ implies $$ f^{-1}(0) = \frac{1}{f(0)} \Rightarrow 0 = f\left(\frac{1}{f(0)}\right) $$ so in particular $f$ attains value $0$ for some $c \in \mathbb{R}$. But now $$ f^{-1}(c) = \frac{1}{f(c)} = \frac{1}{0}, $$ so there is no such function.

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Suppose $f$ continuous. Then $f$, being invertible, is increasing or decreasing. If it is increasing, $f^{-1}(x)$ is increasing, while $[f(x)]^{-1}$ is decreasing. If it is decreasing, $f^{-1}(x)$ is decreasing, while $[f(x)]^{-1}$ is increasing. So there exists no solution among continuous functions. In fact, any solution would have to be everywhere discontinuous, or continuous at only a finite set of points.

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