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I sometimes see Cauchy's Mean Value Theorem stated as follows:

Let $f,\ g:\mathbb{R}\rightarrow\mathbb{R}$ be continuous on $[a,\ b]$ and differentiable on $(a,\ b)$. Suppose that $g(b) \neq g(a)$. Then there exists $c\in(a,\ b)$ such that $g'(c)\neq 0$ and such that $$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$$

I have never once seen a proper proof of the bolded fact and I'm beginning to wonder about the validity of it. Is the assumption $g(b) \neq g(a)$ really enough to prove the existence of such a $c$?

Edit: I think my question is being misunderstood. I am not asking for a standard proof of the Cauchy Mean Value Theorem. The proofs I see assume that $g'(x) \neq 0\ \forall\ x\in(a,\ b)$. This version also claims $g'(c) \neq 0$ when $g(b) \neq g(a)$ (along with the standard continuity/differentiably conditions of course). How can we guarentee there exists such a $c$?

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  • $\begingroup$ I found this version on Wolfram here. They give a reference: Hille, E. Analysis, Vol. 1. New York: Blaisdell, 1964. I could find no other statement of the theorem as you phrased it (they either require $g'$ is never 0, or that $f'$ and $g'$ are never simultaneously 0) . Where have you seen this? $\endgroup$ May 3, 2012 at 19:36
  • $\begingroup$ @David Mitra It shows up occasionally through scattered sources. Wolfram is the primary source I got this version from. Wolfram seemed like a legitimate enough source that I considered this version to be true. But I've never seen a proof for this version before which prompted me to ask this question. $\endgroup$
    – EuYu
    May 4, 2012 at 5:31

3 Answers 3

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You are correct, this isn't true.

Take $f(x)=x^2$ and $g(x)=x^3$ on $[-1,1]$. Then $f(-1)-f(1)=0$ and $g(-1)-g(1)=-2$, so $$ {f(-1)-f(1)\over g(-1)-g(1)}={0\over-2}=0. $$ But $f'(x)=2x$ and $g'(x)=3x^2$; and so there is no number $c$ with ${f'(c)\over g'(c)}={2\over3c}=0$.

It seems the hypothesis that $g'\ne0$ on $[a,b]$ (or that $f'\ne g'$ on $[a,b]$) is needed.

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you can define $h(x):=f(x)[g(b)-g(a)]-g(x)[f(b)-f(a)]$ functions $f$ and $g$ are continuous and differentiable on $(a, b)$ and hence $h$ is continuous and differentiable.we have

$h(a)=f(a)g(b)-f(b)g(a)$

$h(b)=f(a)g(b)-f(b)g(a)$

By Rolle's theorem, there exists $c$ such that $a<c<b$ and $h'(c)=0$. we done

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Answers previous version of the original post -- Will be savaged at a later time.

This is a standard fact.

The assumption is not only $g(b) \neq g(a)$ but also continuity of $f$, $g$ in $[a,b]$ and differentiability of the functions in $(a,b)$

Proof -- hint

Consider $F:[a,b] \to \Bbb{R}$ defined by $$F(x)=f(x)-f(a)-(g(x)-g(a))\frac{f(b)-f(a)}{g(b)-g(a)}$$ and verify $F$ satisfies the hypothesis of Rolle's theorem and use this to conclude!

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