# Cauchy Mean Value Theorem

## Contents

## 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:

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

Then:

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

## 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.

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.

$\blacksquare$

## Also known as

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

## Geometrical Interpretation

Consider two functions $f \left({x}\right)$ and $g \left({x}\right)$ continuous on the interval $\left[{a \,.\,.\, b}\right]$ and differentiable on $\left({a \,.\,.\, b}\right)$.

For every $x \in \left[{a \,.\,.\, b}\right]$, we consider the point $\left({f \left({x}\right), g \left({x}\right)}\right)$.

If we trace out the points $\left({f \left({x}\right), g \left({x}\right)}\right)$ over every $x \in \left[{a \,.\,.\, b}\right]$, we get a curve in two dimensions, as shown in the graph:

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

This is because:

- $\dfrac {\Delta y} {\Delta x} = \dfrac {g \left({b}\right) - g \left({a}\right)} {f \left({b}\right) - f \left({a}\right)}$

assuming that the vertical axis, which contains the value of $g \left({x}\right)$, is the $y$-axis.

The slope of the green line is $\dfrac {g' \left({c}\right)} {f' \left({c}\right)}$.

This is because:

- $\left.{\dfrac {\mathrm d g}{\mathrm d f} }\right\vert_{x \mathop = c} = \left.{\dfrac{\mathrm d g / \mathrm d x}{\mathrm d f / \mathrm d x} }\right\vert_{x \mathop = c} = \dfrac{g' \left({c}\right)} {f' \left({c}\right)}$

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:

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

## Example

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:

\(\displaystyle V_a\) | \(=\) | \(\displaystyle \frac {d \left({t_2}\right) - d \left({t_1}\right)} {t_2 - t_1}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {100 \ \mathrm m} {9.63 \ \mathrm s}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 10.384 \ \mathrm {m/s}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 37.38 \ \mathrm {km/h}\) | $\quad$ | $\quad$ |

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:

\(\displaystyle \frac {f' \left({c}\right)} {g' \left({c}\right)}\) | \(=\) | \(\displaystyle \frac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}\) | $\quad$ | $\quad$ | |||||||||

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

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.