Unformatted text preview: The Mean Value Theorem 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 following theorem.
Suppose is a function that satisfies all of the following. 1. is continuous on the closed interval [a,b]. 2. is differentiable on the open interval (a,b). 3.
Then there is a number c such that
. Or, in other words and
has a critical point in (a,b). 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 Theorem.
Example 1 Show that
has exactly one real root.
From basic Algebra principles we know that since
is a 5th degree polynomial it
will have five roots. What we’re being asked to prove here is that only one of those 5 is a real
number and the other 4 must be complex roots.
First, we should show that it does have at least one real root. To do this note that and that and so we can see that
. Now, because
a polynomial we know that it is continuous everywhere and so by the Intermediate Value
Theorem there is a number c such that
. In other words is and
has at least one real root. We now need to show that this is in fact the only real root. To do this we’ll use an argument that
is called contradiction proof. What we’ll do is assume that
has at least two
real roots. This means that we can find real numbers a and b (there might be more, but all we
need for this particular argument is two) such that
. But if we do this then we know from Rolle’s Theorem that there must then be
another number c such that . This is a problem however. The derivative of this function is, Because the exponents on the first two terms are even we know that the first two terms will
always be greater than or equal to zero and we are then going to add a positive number onto that
and so we can see that the smallest the derivative will ever be is 7 and this contradicts the
statement above that says we MUST have a number c such that . We reached these contradictory statements by assuming that
has at least two
roots. Since this assumption leads to a contradiction the assumption must be false and so we can
only have a single real root. 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.
Mean Value Theorem
Suppose is a function that satisfies both of the following. 1. is continuous on the closed interval [a,b]. 2. is differentiable on the open interval (a,b). Then there is a number c such that a < c < b and Or, 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
Mean Value theorem that there is a c such that and then we know from the
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
know that its slope is at we . 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 Value Theorem.
Example 2 Determine all the numbers c which satisfy the conclusions of the Mean Value
Theorem for the following function. Solution
There isn’t really a whole lot to this problem other than to notice that since
polynomial it is both continuous and differentiable (i.e. the derivative exists) on the interval
given. is a 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 is, Be careful to not assume that only one of the numbers will work. It is possible for both of them
to work. Example 3 Suppose that we know that is continuous and differentiable on [6, 15]. Let’s also suppose that we know that
that and that we know . What is the largest possible value for
Let’s start with the conclusion of the Mean Value Theorem. Plugging in for the known quantities and rewriting this a little gives, Now we know that so in particular we know that . This gives us the following, All we did was replace with its largest possible value. This means that the largest possible value for Example 4 Suppose that we know that
everywhere. Let’s also suppose that we know that
that must have at least one root. is 88. is continuous and differentiable
has two roots. Show Solution
It is important to note here that all we can say is that
will have at least one
root. We can’t say that it will have exactly one root. So don’t confuse this problem with the first
one we worked.
This is actually a fairly simple thing to prove. Since we know that has two roots let’s suppose that they are a and b. Now, by assumption we know that
continuous and differentiable everywhere and so in particular it is continuous on [a,b] and
differentiable on (a,b). is Therefore, by the Mean Value Theorem there is a number c that is between a and b (this isn’t
needed for this problem, but it’s true so it should be pointed out) and that, But we now need to recall that a and b are roots of Or, has a root at and so this is, . Again, it is important to note that we don’t have a value of c. We have only shown that it exists.
We also haven’t said anything about c being the only root. It is completely possible for
to have more than one root. 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].
If for all x in an interval
is constant on then . This fact is very easy to prove so let’s do that here. Take any two x’s in the
. Then since
continuous and differential on [a,b] it must also be continuous and differentiable
. 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
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 fact.
If for all x in an interval then in this interval we have
constant. where c is some This fact is a direct result of the previous fact and is also easy to prove.
If we first define, Then since both and in the interval
derivative of then so must be all x in an interval
on the interval. . Therefore the is, However, by assumption
interval are continuous and differentiable for all x in an
and so we must have that for . So, by Fact 1 must be constant This means that we have, which is what we were trying to show. ...
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