Explanation Given: A function satisfying the hypothesis of Mean Value Theorem on [ 0 , 4 ] for which there are exactly three values c ∈ ( 0 , 4 ) satisfying the conclusion of the theorem Calculation: If function ‘f’ is continuous on [ a , b ] and differentiable on (a, b), then there exists at least one value c ∈ ( a , b ) such that, f ' ( c ) = f ( b ) − f ( a ) b − a Consider the function, f ( x ) = x 2 − 2 It is a polynomial and therefore, is continuous and differentiable; in particular it is continuous on [0, 2] and differentiable on (0, 2). That is why; the Mean Value Theorem applies to this function on the interval [0, 2]. The slope of the line from (0, f (0)) to (2, f (2)) is f ( 2 ) − f ( 0 ) 2 − 0 = ( 2 2 − 2 ) − ( 0 2 − 2 ) 2 = 4 − 2 + 2 2 = 2 By the Mean Value Theorem, there must exist at least on

14th Edition

ISBN: 9781464151088

Author: Associate Professor Laura Taalman, Peter Kohn

Publisher: W.H. Freeman & Company

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Chapter 3.1, Problem 16C

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The graph for the function satisfying the given description

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Chapter 3.1, Problem 15C

Chapter 3.1, Problem 17C

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