Is this some kind of mean value theorem?

Question

I was reviewing some past doctoral entrance exams, and I stumbled upon an exercise from 2011 that caught my attention. It seems quite similar to the mean value theorem, but I am unable to think of a proof. I have a feeling that I might be missing something not as subtle to prove it. The exercise is as follows:

Let $f$ be a function defined on $[a, b]$ and differentiable on $[a, b]$ such that $f(a) = f(b)$ and $f'(a) = 0$. Prove that $\exists c \in (a, b)$ such that $$ f'(c) = \frac{f(c)-f(a)}{c-a} $$

So far, I have some preliminary observations. I recognize that: $$f'(a) = \lim_{x\to a} \frac{f(x) - f(a)}{x - a} = 0$$ Additionally, there must be some number $e\in (a, b)$ such that $f'(e) = 0$. I was thinking of using a function similar to what is used in the proof of the mean value theorem, but so far, I have not been successful. Could someone please provide a hint, and if possible, a complete proof? Thank you.

Maybe you can consider about Rolle's theorem and Lagrange's Mean Value Theorem. — Itoz Darbien, Jul 01 '23 at 04:39
Following your suggestions, I decided to make some diagrams. The function must exhibit concave or convex behavior within a certain interval. The element c the theorem refers to is the endpoint of this interval, is my understanding correct? — Lehnart, Jul 01 '23 at 05:28
I've finished this problem. Try to prove this quetion for $c\in (a,e)$ by Rolle's theorem. — Itoz Darbien, Jul 01 '23 at 06:07

score 3 · Accepted Answer · answered Jul 01 '23 at 06:28

Lemma: If $f$ is a function defined on $[a,b]$ and differentiable on $[a,b]$ such that $f'(a)=f'(b)=0$, there exist a number $c\in (a,b)$ such that $f'(c)=\frac{f(c)-f(a)}{c-a}$.

Let $F(x)= \begin{cases} \frac{f(x)-f(a)}{x-a}, a<x\leq b\\ 0,x=a \end{cases}$

Then $F(x)$ is continuous and differentiable on $[a,b]$,

and we have $F'(x)=\frac{(x-a)f'(x)-(f(x)-f(a))}{(x-a)^2}, a<x\leq b$,

then $F'(b)=\frac{(b-a)f'(b)-(f(b)-f(a))}{(b-a)^2}=-\frac{F(b)}{b-a}$

Case 1. $F(b)=0$

In this case we have $f(a)=f(b)$, and $F(a)=F(b)=0$. By Rolle's theorem, there must be a number $g\in (a,b)$ such that $F′(g)=0=\frac{(g-a)f'(g)-(f(g)-f(a))}{(g-a)^2}$, which means that $f'(g)=\frac{f(g)-f(a)}{g-a}$.

Case 2. $F(b)\neq 0$

We have $F(b)F'(b)=-\frac{F^2(b)}{b-a}<0$.

If $F(b)>0$, then $F'(b)=lim_{x\to b^-}\frac{F(x)-F(b)}{x-b}<0$.

So there exist a number $\delta$, such that for $x\in (b-\delta,b)$, we have $f(x)>f(b)>0$.

So the maximum of $F(x)$ exist in $(a,b)$.

Let $F(h)$ be the maximum of $F(x)$, then $h\in (a,b)$.

We have $F'(h)=0=\frac{(h-a)f'(h)-(f(h)-f(a))}{(h-a)^2}$, which means that $f'(h)=\frac{f(h)-f(a)}{h-a}$.

Then we proved the lemma.

By Rolle's theorem, there must be some number $e\in (a,b)$ such that $f′(e)=0$. By using the lemma, we know that there exist a number $c\in (a,e)$, such that $f'(c)=\frac{f(c)-f(a)}{c-a}$, hence proved.

(By the way, in the lemma if $f'(a)=f'(b)\neq 0$, the lemma still stands.)

Is this some kind of mean value theorem?

1 Answers1