Cauchy Mean Value Theorem

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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:

$\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$


$\exists \xi \in \left({a \,.\,.\, b}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$


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.

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

We need to make:

$f \left({a}\right) + h g \left({a}\right) = f \left({b}\right) + h g \left({b}\right)$

We have that:

$\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$

So, by Rolle's Theorem:

$g \left({a}\right) \ne g \left({b}\right)$

Thus we can let:

$h = - \dfrac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$

So, by Rolle's Theorem:

$\exists \xi \in \left({a \,.\,.\, b}\right): 0 = F' \left({\xi}\right) = f' \left({\xi}\right) + h g' \left({\xi}\right)$

That is:

$h = - \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)}$

Hence the result, from the definition of the derivative.


Also known as

The Cauchy mean value theorem is also known as the generalized mean value theorem.

Geometrical Interpretation

Consider two functions $\map f x$ and $\map g x$:

[Definition:Continuous Real Function|continuous]] on the closed interval $\closedint a b$
differentiable on $\openint a b$.

For every $x \in \closedint a b$, we consider the point $\tuple {\map f x, \map g x}$.

If we trace out the points $\tuple {\map f x, \map g x}$ over every $x \in \closedint a b$, we get a curve in two dimensions, as shown in the graph:

Cauchy's Mean Value Theorem.png

In the drawing, the slope of the red line is $\dfrac {\map g b - \map g a} {\map f b - \map f a}$.

This is because:

$\dfrac {\Delta y} {\Delta x} = \dfrac {\map g b - \map g a} {\map f b - \map f a}$

assuming that the vertical axis, which contains the value of $\map f x$, is the $y$-axis.

The slope of the green line is $\dfrac {\map {g'} c} {\map {f'} c}$.

This is because:

$\valueat {\dfrac {\d g} {\d f} } {x \mathop = c} = \valueat {\dfrac {\d g / \d x} {\d f / \d x} } {x \mathop = c} = \dfrac {\map {g'} c} {\map {f'} c}$

The drawing illustrates that for the value of $c$ chosen in the pictures, the slopes of the red line and green line are the same.

That is:

$\dfrac {\map g b - \map g a} {\map f b - \map f a} = \dfrac {\map {g'} c} {\map {f'} c}$


In the 2012 Olympics Usain Bolt won the 100 metres gold medal with a time of $9.63$ seconds.

By definition, his average speed was the total distance travelled divided by the total time it took:

\(\ds V_a\) \(=\) \(\ds \frac {d \left({t_2}\right) - d \left({t_1}\right)} {t_2 - t_1}\)
\(\ds \) \(=\) \(\ds \frac {100 \ \mathrm m} {9.63 \ \mathrm s}\)
\(\ds \) \(=\) \(\ds 10.384 \ \mathrm {m/s}\)
\(\ds \) \(=\) \(\ds 37.38 \ \mathrm {km/h}\)

The Mean Value Theorem gives:

$f' \left({c}\right) = \dfrac {f \left({b}\right) - f \left({a}\right)} {b - a}$

Hence, at some point Bolt was actually running at the average speed of $37.38 \ \mathrm {km/h}$

Asafa Powell was participating in that same race.

He achieved a time of $11.99 \ \mathrm s = 1.245 \times 9.63 \ \mathrm s$.

So Bolt's average speed was $1.245$ times the average speed of Powell.

The Cauchy Mean Value Theorem gives:

\(\ds \frac {f' \left({c}\right)} {g' \left({c}\right)}\) \(=\) \(\ds \frac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}\)
\(\ds \) \(=\) \(\ds \frac {\frac {f \left({b}\right) - f \left({a}\right)} {b - a} } {\frac {g \left({b}\right) - g \left({a}\right)} {b - a} }\)

Hence, at some point, Bolt was actually running at a speed exactly $1.245$ times that of Powell's.

Source of Name

This entry was named for Augustin Louis Cauchy.