Let $f(x)$ is continuous in $[0,1]$, differentiable in $(a,b)$, such that $f(0)=g(0)=f(1)=0$, $g'(x)\neq0$.
Prove:
$\exists\xi,\eta\in(0,1),\xi<\eta,$
$$\frac{f'(\xi)}{g'(\xi)}+\frac{f'(\eta)}{g'(\eta)}=0$$
I have no idea about it, but I can post some link which I think is useful.
This is a question of a real-analysis book which named 'Mathematical analysis course'(Author:Shi Jihuai,Chang Gengzhe).
Book Name:数学分析教程(Mathematical analysis course)
Author: 史济怀 常庚哲
My attempt
When there are two parameters in question,we can express them by a node which associated with both.
And then we can remove the node by some operations.For this question, We can see the answer of I posted.
I want to get a proof,and a way to solve two-parameter($\xi,\eta$) question
By the Intermediate Value Theorem, since $g(0)=0$, $\exists t\in (0,1)$ such that $g(t)=g(1)/2$.
By the Cauchy's Mean Value Theorem, $\exists \xi \in(0,t)$ such that $$\frac{f(t)}{g(t)}=\frac{f(t)-f(0)}{g(t)-g(0)}=\frac{f'(\xi)}{g'(\xi)}.$$ Again, by the same reason as before, $\exists \eta \in(t,1)$ such that $$-\frac{f(t)}{g(1)-g(t)}=\frac{f(1)-f(t)}{g(1)-g(t)}=\frac{f'(\eta)}{g'(\eta)}.$$ Then $0<\xi<\eta<1$ and $$\frac{f'(\xi)}{g'(\xi)}+\frac{f'(\eta)}{g'(\eta)}= \frac{f(t)(g(1)-2g(t))}{g(t)(g(1)-g(t))}=0.$$