Cauchy Mean Value Theorem

Theorem
Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.

Suppose:
 * $\forall x \in \openint a b: \map {g'} x \ne 0$

Then:
 * $\exists \xi \in \openint a b: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f b - \map f a} {\map g b - \map g a}$

Proof
First we check $\map g a \ne \map g b$.

$\map g a = \map g b$.

From Rolle's Theorem:


 * $\exists \xi \in \openint a b: \map {g'} \xi = 0$.

This contradicts $\forall x \in \openint a b: \map {g'} x \ne 0$.

Thus by Proof by Contradiction $\map g a \ne \map g b$.

Let $h = \dfrac {\map f b - \map f a} {\map g b - \map g a}$.

Let $F$ be the real function defined on $\closedint a b$ by:


 * $\map F x = \map f x - h \map g x$.

Then:

Also see

 * Mean Value Theorem