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Online Notes / Calculus I / Applications of Derivatives / The Mean Value Theorem
Calculus I

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The Shape of a Graph, Part II E-Book   Chapter   Section Optimization

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 the 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.

 

Rolle’s 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  and .  Or, in other words  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.

 

Solution

From basic Algebra principles we know that since  is a 5th degree polynomial there 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  is a polynomial we know that it is continuous everywhere and so by the Intermediate Value Theorem there is a number c such that  and .  In other words  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  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.

 

MeanValueTheorem_G1

 

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  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 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  and that we know that .  What is the largest possible value for ?

 

Solution

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  is 88.