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 $c$?
Best Answer
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.