International Mathematics Competition
for University Students
2026

IMC2026: Day 2, Problem 6

Problem 6.

(a) Is there a differentiable function \(\displaystyle f\colon \mathbb{R} \to \mathbb{R}\) such that f'(f(x)) = x for every \(\displaystyle x \in \mathbb{R}?\)

(b) Is there a differentiable function \(\displaystyle f\colon \mathbb{R} \to \mathbb{R}\) such that f'(f(x)) = |x| for every \(\displaystyle x \in \mathbb{R}?\)

Nikolaos Kolliopoulos, University of Cyprus

Solution.

(a) There is no such function. The given identity implies that the function is injective. Since it is differentiable and thus continuous, it is also strictly monotone, meaning that \(\displaystyle f'(x)\) cannot admit both negative and positive values. The latter clearly contradicts the given identity, meaning that such a function does not exist.

(b) We will show an example for such a function. It is reasonable to look for a function of the form \(\displaystyle f(x) = k|x|^a\) for some \(\displaystyle k > 0\), which is differentiable for \(\displaystyle a > 1\) with \(\displaystyle f'(x) = (ka)\text{sign}(x)|x|^{a-1}\) for any \(\displaystyle x \in \mathbb{R}\). Since \(\displaystyle f(x) \geq 0\) for each \(\displaystyle x \in \mathbb{R}\), we see that

\(\displaystyle f'(f(x)) = (ka)(k|x|^a)^{a-1} = ak^a|x|^{a^2 - a}\)

for all \(\displaystyle x \in \mathbb{R}\), so all we need is to pick \(\displaystyle a = \frac{\sqrt{5}+1}{2} > 1\) which solves the quadratic equations \(\displaystyle a^2 - a = 1\) and thus \(\displaystyle |x|^{a^2 - a} = |x|\) for all \(\displaystyle x \in \mathbb{R}\), and then \(\displaystyle k = \frac{1}{a^{\frac{1}{a}}}\) so that \(\displaystyle ak^a = 1\).