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Suppose $f(x)$ is differentiable on $[0,1]$, and $f(0)=0$, $f(x)\ne 0,\forall x\in(0,1)$ , Prove for every $n,m\in\mathbb{N^+}$, there exists $\xi=\xi_{n,m}\in(0,1)$ such that $$n\cdot\frac{f'(\xi)}{f(\xi)}=m\cdot\frac{f'(1-\xi)}{f(1-\xi)}$$

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  • $\begingroup$ Actually I think this form is about the application of Mean Value Theorem. $\endgroup$ – Jaqen Chou Aug 22 '18 at 4:40
  • $\begingroup$ Yeah, you are right, but what have you tried? $\endgroup$ – xbh Aug 22 '18 at 5:05
  • $\begingroup$ Apply Cauchy's Mean Value Theorem. $\endgroup$ – Anik Bhowmick Aug 22 '18 at 6:08
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MVT method:

Consider an auxiliary function $$ F(x) =f(x)^n f(1-x)^m, \quad x \in [0,1], $$ then Rolle's theorem achieves the goal.

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Without loss of generality, we may assume $f(x) > 0$ for all $x$, since it cannot change sign, without violating the intermediate value theorem.

Let $h(x) = \ln(f(x)) + \frac{m}{n} \ln(f(1 - x))$, defined over $(0, 1)$. Then, $$h'(x) = \frac{f'(x)}{f(x)} - \frac{m}{n} \frac{f(1 - x)}{f'(1 - x)}$$ so $h'(x) = 0$ if and only if $x$ is a suitable choice of $\xi$. Since $\lim_{x \to 0^+} f(x) = 0$ and $\lim_{x \to 1^-} f(x) = f(1) > 0$, we have that $$\lim_{x \to 0^+} h(x) = \lim_{x \to 1^-} h(x) = -\infty.$$ It follows therefore that $h$ achieves a maximum somewhere in $(0, 1)$. This point will be the $\xi$ you're looking for.

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