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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. Rolle’s Theorem Suppose is a function that satisfies all of the following. is continuous on the closed interval [a,b]. is differentiable on the open interval (a,b). 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 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.