# A function $f:\mathbb{R}\to\mathbb{R}$ such that $f^{-1}(x)=\frac{1}{f(x)}$

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

• Yeah !! Also the one which @Nilan gave – creative Aug 14 '14 at 9:58
• That's trivial... – ShakesBeer Aug 14 '14 at 9:59
• @Nilan what would be the functional inverse of that function. would the inverse even be a function? – cirpis Aug 14 '14 at 10:03
• it's not invertible – ShakesBeer Aug 14 '14 at 10:06
• Yes! you are right. – Bumblebee Aug 14 '14 at 10:07

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.
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.