TITLE: Is there a continuous function $f(x)$ such that the inverse function is $1/f(x)$?
QUESTION [16 upvotes]: A student came to me with this question and I cracked my head for one hour but I couldn't unambiguously prove that it exists or it doesn't exist. Continuous and invertible function such that $f^{-1}(x) =1/f(x)$ on its domain of definition.

REPLY [7 votes]: Given

$$
f^{-1}(x) = \frac{1}{f(x)}. \tag 1
$$

Whence

$$
x = f^{-1}(f(x)) = \frac{1}{f(f(x))}. \tag 2
$$

Thus

$$
f(f(x)) = \frac{1}{x}.\tag 3
$$


Consider the function

$$
f(x) = \frac{1 + a x}{b + c x}. \tag 4
$$

Then

$$
f(f(x)) = \frac
    {\displaystyle 1 + a \frac{1 + a x}{b + c x}}
    {\displaystyle b + c \frac{1 + a x}{b + c x}} = \frac
    {[b + a] + [c + a^2] x}
    {[b^2 + c] + [bc + ac] x}. \tag 5
$$

So we get $f(f(x)) = x^{-1}$ for

$$
\left[
\begin{array}{rcl}
c + a^2 &=& 0\\
b^2 + c &=& 0\\
\displaystyle \frac{b+a}{bc+ac} &=& 1
\end{array} \tag 6
\right.
$$

Therefore $c=1$, $a=b=\pm \mathbf{i}$.
Whence

$$
\bbox[16px,border:2px solid #800000] {
   f(x) = \frac{1 \pm \mathbf{i} x}{x \pm \mathbf{i}}.} \tag 7
$$


Consider the function

$$
f(x) = x^\alpha. \tag 8
$$

Then

$$
f(f(x)) = \big( x^\alpha \big)^\alpha = x^{\alpha^2}. \tag 9
$$

So we get $f(f(x)) = x^{-1}$ for $\alpha = \pm \mathbf{i}$.
Whence

$$
\bbox[16px,border:2px solid #800000] {
   f(x) = x^{\displaystyle \pm \mathbf{i}}. } \tag {10}
$$

I have no idea if there are other possible function.