The mean value theorem states that given a planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints.

## Rigorous definition of the theorem[edit]

Let $f: \mathbb{R} \supset \left[ a, b \right] \to \mathbb{R}$ be a function that is both continuous and differentiable on the interval $\left] a, b \right[$. Then a $\xi \in \left] a, b \right[$ exists so that

- [math]f'\left( \xi \right) = \frac{ f \left( b \right) - f \left( a \right) }{ b - a }.[/math]

## Prerequisite information[edit]

## The proof[edit]

Lets look at the funtion $f$ from above a little closer. We can calculate the secant $s \left( x \right) $ of that function's graph over $\left[ a, b \right]$ like this:
[math]s \left( x \right) = f \left( a \right) + \frac{ f\left( b \right) - f \left( a \right) }{ b - a} \cdot \left( x - a \right)[/math]

Now we can take the secant and calulate the height of the graph of $f$ above (or below) the secant: $h \left( x \right) = f \left( x \right) - s \left( x \right) $.
Now we just have to search for $\xi \in \left] a, b \right[ $ where $h$ has an extremum, because we know that if $h$ has an extremum in $\xi$, we have:

[math]0 = h' \left( \xi \right) = f' \left( \xi \right) - s' \left( \xi \right) = f' \left( \xi \right) - \frac{ f\left( b \right) - f \left( a \right) }{ b - a}[/math]

$h$ is continuous in $\left[ a, b \right]$, so we know it reaches its maximum and minimum somewhere in $\left[ a, b \right]$. If the maximum and minimum of $h$ are equal then $h$ must be constant and the statement is true for everey $\xi \in \left] a, b \right[$. Otherwise we know that $h \left( a \right) = h \left( b \right) $ and then at least one of the two must be an inside point $\xi \in \left] a, b \right[$.

## All proofs that use this theorem[edit]