Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $c$?

share|cite|improve this question
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? – David Mitra May 3 '12 at 19:36
@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. – EuYu May 4 '12 at 5:31
up vote 2 down vote accepted

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.

share|cite|improve this answer

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



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

share|cite|improve this answer

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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.