Cauchy Mean Value Theorem

Theorem

Let $f$ and $g$ be a real functions which are continuous on the closed interval $\left[{a .. b}\right]$ and differentiable on the open interval $\left({a .. b}\right)$.

Suppose that $\forall x \in \left({a .. b}\right): g^{\prime} \left({x}\right) \ne 0$.

Then:

$\displaystyle \exists \xi \in \left({a .. b}\right): \frac {f^{\prime} \left({\xi}\right)} {g^{\prime} \left({\xi}\right)} = \frac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$

Proof

Let $F$ be the real function defined on $\left[{a .. b}\right]$ by $F \left({x}\right) = f \left({x}\right) + h g \left({x}\right)$, where $h \in \R$ is a constant.

Let us choose the constant $h$ such that $F \left({a}\right) = F \left({b}\right)$ and so apply Rolle's Theorem.

We need to make $f \left({a}\right) + h g \left({a}\right) = f \left({b}\right) + h g \left({b}\right)$.

Since $\forall x \in \left({a .. b}\right): g^{\prime} \left({x}\right) \ne 0$, by Rolle's Theorem, $g(a) \neq g(b)$.

Thus we can let:

$\displaystyle h = - \frac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$.

So, by Rolle's Theorem, $\exists \xi \in \left({a .. b}\right): 0 = F^{\prime} \left({\xi}\right) = f^{\prime} \left({\xi}\right) + h g^{\prime} \left({\xi}\right)$.

That is:

$\displaystyle h = - \frac {f^{\prime} \left({\xi}\right)} {g^{\prime} \left({\xi}\right)}$

Hence, from the definition of the derivative, the result.

$\blacksquare$

Source of Name

This entry was named for Augustin Louis Cauchy.

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 11.8 \ (2)$