L2RolleMean - ROLLES THEOREM AND THE MEAN VALUE THEOREM...

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Unformatted text preview: ROLLES THEOREM AND THE MEAN VALUE THEOREM WILLIAM A. LAMPE Recall the Theorem on Local Extrema. If f ( c ) is a local extremum, then either f is not differentiable at c or f ( c ) = 0 . That is, at a local max or min f either has no tangent, or f has a horizontal tangent there. We will use this to prove Rolles Theorem. Let a < b . If f is continuous on the closed interval [ a,b ] and differen- tiable on the open interval ( a,b ) and f ( a ) = f ( b ) , then there is a c in ( a,b ) with f ( c ) = 0 . That is, under these hypotheses, f has a horizontal tangent somewhere between a and b . Rolles Theorem, like the Theorem on Local Extrema, ends with f ( c ) = 0. The proof of Rolles Theorem is a matter of examining cases and applying the Theorem on Local Extrema. Proof. We seek a c in ( a,b ) with f ( c ) = 0. That is, we wish to show that f has a horizontal tangent somewhere between a and b . Since f is continuous on the closed interval [ a,b ], the Extreme Value Theorem says that f has a maximum value f ( M ) and a minimum value f ( m ) on the closed interval [ a,b ]. Either f ( M ) = f ( m ) or f ( M ) negationslash = f ( m ). Case 1. We suppose the maximum value f ( M ) = f ( m ), the minimum value. So all values of f on [ a,b ] are equal, and f is constant on [ a,b ]. Then f ( x ) = 0 for all x in ( a,b ). So one may take c to be anything in ( a,b ); for example, c = a + b 2 would suffice. Case 2. Now we suppose f ( M ) negationslash = f ( m ). So at least one of f ( M ) and f ( m ) is not equal to the value f ( a ) = f ( b ). Case 2.a We first consider the case where the maximum value f ( M ) negationslash = f ( a ) = f ( b ). (See the figure to the right.) So M is neither a nor b . But M is in [ a,b ] and not at the end points. So M must be in the open interval (...
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L2RolleMean - ROLLES THEOREM AND THE MEAN VALUE THEOREM...

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