Any self-respecting study of differential calculus must include the mean value theorem, which has ubiquitous uses in all of calculus. We will state it here and prove it in two steps: the easier step, followed by the harder step.
Theorem (Mean Value Theorem). Let be continuous, and
be differentiable on
. Then there exists
such that
Proof. We will first prove the special case when , so that
. This is known as Rolle’s theorem. Suppose for simplicity that
is non-constant.
Without loss of generality, there exists such that
. Since
is continuous, by the extreme value theorem, there exists
such that for any
,
In particular, so that
. Since
is differentiable at
, we will compute its derivative
in two steps.
For the first step, by considering -values to the left of
, since
and
,
For the second step, by considering -values to the right of
, since
and
,
Since , we have
, as desired.
Now that we have established Rolle’s theorem, we prove the mean value theorem. Define the continuous function that is differentiable on
by
This helps us ensure that . By Rolle’s theorem, there exists
such that
. Alternatively, by computing
directly,
Setting yields
as required.
The mean value theorem is incredibly useful in calculus, and we will be using it repeatedly as we slowly but surely define rigorously the usual functions that we have been taking for granted.
—Joel Kindiak, 20 Oct 24, 1611H
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