Rolle's theorem for a equation

Question

Let $f:R \to(0,+\infty)$ be a differtiable function, and $f(1)=1$. Prove that there is at least on $ξ \in (0,1)$ s.t. $$f(ξ)=\frac{1}{e^\frac{ξf'(ξ)}{f(ξ)}}$$ My try was to maybe use Rolle's theorem. Now $$e^\frac{xf'(x)}{f(x)}=\frac{1}{f(x)} $$ But I have no clue as on how to continue. $$\frac{d}{dx}(\frac{xf'(x)}{f(x)})= \frac{f'(x)+xf''(x)+x(f^{'}(x))^2}{f^2(x)}$$ which is not that close to $\frac{1}{f(x)}$. Can anyone help?

Answer 1

This is equivalent to proving there exists some $\xi$ such that $f(\xi) e^{\xi\frac{f'(\xi)}{f(\xi)}} = 1$. Since the range of $f$ is strictly positive, we may take a logarithm of it. \begin{align} f(x)e^{x \frac{f'(x)}{f(x)}} =& e^{\ln(f(x)) + x \frac{f'(x)}{f(x)}} \\ =& e^{\ln(f(x) + x \frac{d}{dx}\ln(f(x))} \end{align} So we have to show that this exponent equals $0$ somewhere in $(0,1)$. Define $g(x) := \ln(f(x))$. Then we know $g$ is a differentiable function such that $g(1) = 0$. This is exactly the $\xi$ you were looking for.

Let us look at the following integral. \begin{align} \int_0^1 g(x) + xg'(x) dx =& \int_0^1 g(x)dx + \int_0^1 x dg(x) \\ =& \int_0^1 g(x) dx + [xg(x)]_0^1 - \int_0^1 g(x)dx \\ =&0 \end{align} Hence, by Rolle's theorem on the function $t \mapsto \int_0^t g(x)+xg'(x)dx$, there must be some value $\xi \in (0,1)$ such that $g(\xi) + \xi g'(\xi) = 0$.

Answer 2

Thanks to @Lukas Rollier for the help, I managed to solve it in a different way,(using his help),probably a bit easier.For the record:

Using $$f(x)e^{x\frac{f'(x)}{f(x)}}=1 \Rightarrow e^{ln(f(x))(x)'+x(ln(f(x))'}=1 \Rightarrow (xln(f(x))'=0 $$

And if we set $t(x)=xln(f(x))$, we have $t(0)=0, t(1)=0$ and apply Rolle's theorem.