Conditions of Cauchy's Mean Value Theorem 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 $c$?
 A: 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. 
A: 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!
A: 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
