In this section we want to take a look at the Mean Value
Theorem. In most traditional textbooks
this section comes before the sections containing the First and Second
Derivative Tests because many of the proofs in those sections need the Mean
Value Theorem. However, we feel that
from a logical point of view it’s better to put the Shape of a Graph sections
right after the absolute extrema section.
So, if you’ve been following the proofs from the previous two sections
you’ve probably already read through this section.
Before we get to the Mean Value Theorem we need to cover the
To see the proof of Rolle’s Theorem see the Proofs From Derivative
Applications section of the Extras chapter.
Let’s take a look at a quick example that uses Rolle’s
The reason for covering Rolle’s Theorem is that it is needed
in the proof of the Mean Value Theorem.
To see the proof see the Proofs From Derivative
Applications section of the Extras chapter.
Here is the theorem.
Note that the Mean Value Theorem doesn’t tell us what c is.
It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.
Also note that if it weren’t for the fact that we needed
Rolle’s Theorem to prove this we could think of Rolle’s Theorem as a special
case of the Mean Value Theorem. To see
that just assume that and then the result of the Mean Value Theorem
gives the result of Rolle’s Theorem.
Before we take a look at a couple of examples let’s think
about a geometric interpretation of the Mean Value Theorem. First define and and then we know from the Mean Value theorem
that there is a c such that and that
Now, if we draw in the secant line connecting A and B then we can know that the slope of the secant line is,
Likewise, if we draw in the tangent line to at we know that its slope is .
What the Mean Value Theorem tells us is that these two
slopes must be equal or in other words the secant line connecting A and B and the tangent line at must be parallel. We can see this in the following sketch.
Let’s now take a look at a couple of examples using the Mean
Example 2 Determine
all the numbers c which satisfy the
conclusions of the Mean Value Theorem for the following function.
There isn’t really a whole lot to this problem other than
to notice that since is a polynomial it is both continuous and
differentiable (i.e. the derivative
exists) on the interval given.
First let’s find the derivative.
Now, to find the numbers that satisfy the conclusions of
the Mean Value Theorem all we need to do is plug this into the formula given
by the Mean Value Theorem.
Now, this is just a quadratic equation,
Using the quadratic formula on this we get,
So, solving gives two values of c.
Notice that only one of these is actually in the interval
given in the problem. That means that
we will exclude the second one (since it isn’t in the interval). The number that we’re after in this problem
Be careful to not assume that only one of the numbers will
work. It is possible for both of them
It is completely possible to generalize the previous example
significantly. For instance if we know
that is continuous and differentiable everywhere
and has three roots we can then show that not only will have at least two roots but that will have at least one root. We’ll leave it to you to verify this, but the
ideas involved are identical to those in the previous example.
We’ll close this section out with a couple of nice facts
that can be proved using the Mean Value Theorem. Note that in both of these facts we are
assuming the functions are continuous and differentiable on the interval [a,b].
This fact is very easy to prove so let’s do that here. Take any two x’s in the interval ,
say and . Then since is continuous and differential on [a,b]
it must also be continuous and differentiable on . This means that we can apply the Mean Value
Theorem for these two values of x. Doing this gives,
where . But by assumption for all x
in an interval and so in particular we must have,
Putting this into the equation above gives,
Now, since and where any two values of x in the interval we can see that we must have for all and in the interval and this is exactly what it
means for a function to be constant on the interval and so we’ve proven the
This fact is a direct result of the previous fact and is
also easy to prove.
If we first define,
Then since both and are continuous and differentiable in the
interval then so must be . Therefore the derivative of is,
However, by assumption for all x
in an interval and so we must have that for all x
in an interval . So, by Fact 1 must be constant on the interval.
This means that we have,
which is what we were
trying to show.