IMC 2025 · Day 2 · P6

Observe that if we consider $g(x) = f(x)/x$ and take its derivative, we get $$g'(x) = \frac{x f'(x) - f(x)}{x^2},$$ where the numerator is almost the expression we have in the problem.

Now we apply Cauchy's theorem to the functions $g(x) = f(x)/x$ and $h(x) = 1/x$, well-defined over $[a, b]$. This gives us the existence of a value $\xi \in [a, b]$ such that $$\frac{g(a) - g(b)}{h(a) - h(b)} = \frac{g'(\xi)}{h'(\xi)}.$$ Here, $$\frac{g'(\xi)}{h'(\xi)} = \frac{\frac{\xi f'(\xi) - f(\xi)}{\xi^2}}{-\frac{1}{\xi^2}} = f(\xi) - \xi f'(\xi)$$ and $$\frac{g(a) - g(b)}{h(a) - h(b)} = \frac{\frac{f(a)}{a} - \frac{f(b)}{b}} {\frac1a - \frac1b} = k,$$ which concludes the proof.