An alternative proof of Cauchy's Mean Value Theorem

Or we could just consider the function:

$F(x)=f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}(g(x)-g(a))$

Clearly $F(x)$ is defined iff $g(b) $ is different from $g(a)$ , otherwise if $g(a)=g(b)$ then $F(x)$ would satisfy Role's theorem's conditions hence there would exist a point $c$ $ \epsilon$ $(a,b)$ such that $g'(c)=0$ this is a contradiction to our third condition :

3) $g'$ does not vanish at any point in $(a, b)$

Then from the MVT we have :

There exists a point $c$ $\epsilon$ $(a,b)$ such that $F'(c)=0$

thus we get :

$f'(c)=\frac{f(b)-f(a)}{g(b)-g(a)}g'(c)$

or

$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$