Let $f(x)$ is continuous in $[0,1]$, differentiable in $(a,b)$, such that $f(0)=g(0)=f(1)=0$, $g'(x)\neq0$.




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

  • $\begingroup$ do you have a source for this problem? In case of having no idea about the solution, while the mention of similar questions was nice you can also mention the source. Also, see if any of the questions whose links you gave, are special cases of the question above, and see what you can modify in that answer to get the answer in the general case. $\endgroup$ – Teresa Lisbon Feb 18 '19 at 11:55
  • $\begingroup$ This is a question of a real-analysis book.I think the three questions that I posted are similar.They all have two-parameter.And I'm trying to solve the question I ask. $\endgroup$ – Li Taiji Feb 18 '19 at 12:00
  • $\begingroup$ @梦里年华似烟花 If you don't mind can you please mention the name of book. Thanks. $\endgroup$ – Aditya Prasad Feb 18 '19 at 12:05
  • $\begingroup$ @StammeringMathematician It's a Chinese book,named 《数学分析教程》(常庚哲,史济怀著). $\endgroup$ – Li Taiji Feb 18 '19 at 12:25

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 $$\frac{f'(\xi)}{g'(\xi)}+\frac{f'(\eta)}{g'(\eta)}= \frac{f(t)(g(1)-2g(t))}{g(t)(g(1)-g(t))}=0.$$

  • $\begingroup$ For the two-parameter question, I find it is solved by a node which associated with both usually.For example, $\frac{f'(\xi)}{g'(\xi)}+\frac{f'(\eta)}{g'(\eta)}$,and then $\frac{f(t)}{g(t)}$ is removed. $\endgroup$ – Li Taiji Feb 18 '19 at 12:18
  • $\begingroup$ Why don't you edit your question and write down your attempt? $\endgroup$ – Robert Z Feb 18 '19 at 12:22
  • $\begingroup$ I think your answer has introduced the way.@Robert Z $\endgroup$ – Li Taiji Feb 18 '19 at 12:27
  • $\begingroup$ In order to avoid downvotes, it is better that you say something about the background (english title of the book, authors?) and your attempt. $\endgroup$ – Robert Z Feb 18 '19 at 13:13
  • $\begingroup$ Ok,Thanks for your suggestions. $\endgroup$ – Li Taiji Feb 18 '19 at 13:50

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