Mean value theorem

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]