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Unformatted text preview: Rolle’s Theorem: Let f : [a, b] → R be continuous on [a, b] and diﬀerentiable on (a, b), and let f (a) = f (b). Then there exists c ∈ (a, b) such that f (c) = 0. Proof: If f is constant, then f (c) = 0 for every c ∈ (a, b). Now suppose f is not constant. By the Extreme Value Theorem, f achieves its absolute maximum and minimum on [a, b]. Since f is not constant, it has to go above or below f (a) and f (b). Therefore there is a point c ∈ (a, b) where f attains an absolute extreme value. Since f (c) must then also be a local extreme value and f is diﬀerentiable at c, Fermat’s Theorem tells us that f (c) = 0. 1 ...
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This note was uploaded on 06/26/2010 for the course MATH 1A taught by Professor Wilkening during the Fall '08 term at Berkeley.
 Fall '08
 WILKENING

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