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Unformatted text preview: The Mean Value Theorem and the Extended Mean Value Theorem Willard Miller September 21, 2006 0.1 The MVT Recall the Extreme Value Theorem (EVT) from class: If the function f is defined and continuous on a closed bounded interval [ a, b ] then there is some point c [ a, b ] where it takes on its maximum value M = f ( c ) and some point d [ a, b ] where it takes on its minimum value m = f ( d ). Thus M = f ( c ) f ( x ) f ( d ) = m for all x [ a, b ]. Theorem 1 Fermats Theorem. (Not his last one but a very useful observa- tion that is easy to prove.) Suppose the function f is defined and continuous on a closed bounded interval [ a, b ] and takes on an extreme value (either its maximum M or its minimum m ) at an interior point c of the interval, so a < c < b . If the derivative f ( c ) exists, then f ( c ) = 0 . PROOF: To be definite, we assume that f ( c ) = M . (The proof for the case f ( c ) = m is virtually the same.) By definition f ( c ) = lim x f ( c + x ) f ( c ) x . (1) Since f ( c ) = M is a maximum, it follows that for x > 0 we have f ( c + x ) f ( c ) x , 1 since the numerator is nonpositive and the denominator is positive. On the other hand, for x < 0 we have f ( c + x ) f ( c ) x , since the numerator is nonpositive and the denominator is negative. However by assumption the limit (1) exists and the same value is obtained as x through postitive or negative values. Thus we must have f ( c ) = 0. Q.E.D....
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