Darboux Theorem Question Regarding:
Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})<0$ for every $0<x \ne 1$.
Then $f'(1)=0$.
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So I am trying to prove this, and I said:
We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0< \delta < 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ .
So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})<0$
So by Darboux Theorem, there exists a $x_{0}<c< \frac{1}{x_{0}}$ such that $f'(c)=0$.
But now, I do not know how to link $c$ with $1$...
Any hints?
Thanks! :)
 A: Note if $c\neq 1$ then $f'(c)f'(1/c)<0$ by hypothesis. This implies $f'(c)\neq 0$. Hence there is only one possible value left for $c\in (x_0,1/x_0)$, which is $c=1$.
A: As you noted, we have $$f'(1-\delta_n)f'\left(\frac {1}{1-\delta_n} \right)<0$$ for any positive sequence $\{\delta_n\}$ such that $\lim\delta_n = 0$. Take $1+\varepsilon_n = \dfrac{1}{1-\delta_n}>1$, where $\lim \varepsilon_n = 0$. Then by the given condition,
$$f'(1-\delta_n)f'\left( 1+\varepsilon_n \right)<0$$
WLOG, assume that $f'(1-\delta_n)<0$ and $f'\left( 1+\varepsilon_n \right)>0$.
By sending $n$ to infinity and remembering Darboux's theorem, we see that $f'(1)=0$.
A: You can also make it more simple without using $\delta, x_0$ if you want. But your method is better practice for general cases.
We know that $f$ is differentiable at $(0, \infty)$, thus $f$ is differentiable at $\left[\frac{1}{1.1},1.1\right]$
By hypostasis: $$f'\left(\frac{1}{1.1}\right)\cdot f'\left(1.1\right)<0$$
By Darboux's Theorem, there exists a $0.91\approx\frac{1}{1.1}<c<1.1$  such that:
$$f'\left(c\right)=0$$
Assume towards a contradiction that $f\left(c\right)\ne1$
Then By hypostasis:
$$0=f'\left(c\right)\cdot f'\left(\frac{1}{c}\right)<0$$
Contradiction!
