Lesson 15.pdf - 15 The Mean Value Theorem The mean value theorem Theorem 0.1 If f is differentiable on an open interval(a b and if f assumes a maximum

Lesson 15.pdf - 15 The Mean Value Theorem The mean value...

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15. The Mean Value Theorem July 29, 2019 The mean value theorem Theorem 0.1 If f is differentiable on an open interval ( a, b ) , and if f assumes a maximum or minimum value at a point c ( a, b ) , then f 0 ( c ) = 0 . Proof: Suppose f has a maximum at c ( a, b ) with f 0 ( c ) 6 = 0 (We seek a contradiction). Suppose f 0 ( c ) > 0. Since lim x c f ( x ) - f ( c ) x - x > 0, There is a neighborhood N ( c, δ ) on which f ( x ) - f ( c ) x - c > 0. Then on [ c, c + δ ), x - c > 0, f ( x ) - f ( c ) > 0, i.e., f ( x ) > f ( c ) on [ c, c + δ ). This contracts with the fact that f has a maximum at c . Suppose f 0 ( c ) < 0. Since lim x c f ( x ) - f ( c ) x - x < 0, There is a neighborhood N ( c, δ ) on which f ( x ) - f ( c ) x - c < 0. Then on ( c - δ, c ), x - c < 0, f ( x ) - f ( c ) > 0, i.e., f ( x ) > f ( c ) on ( c - δ, c ]. This also contracts with the fact that f has a maximum at c . Therefore, f 0 ( c ) = 0. The minimum case for f has the same proof. Theorem 0.2 (Rolle’s theorem) If f is continuous on [ a, b ] and is differentiable on ( a, b ) with f ( a ) = f ( b ) , then there exists c ( a, b ) , such that f 0 ( c ) = 0 . Proof: Let k := f ( a ) = f ( b ). If f ( x ) = k for all x [ a, b ], then f 0 ( x ) = 0 for any x ( a, b ) so that we can take c to be any point in ( a, b ). If f 6 = constant , then f has a maximum or minimum at some point in ( a, b ). Assume that f has a maximum at some point c [ a, b ]. Then apply Theorem 0.1 to imply f 0 ( c ) = 0. 128
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Theorem 0.3 (The mean value theorem) Let f be continuous on [ a, b ] and differentiable on ( a, b ) . Then there exists at least one number c ( a, b ) such that f 0 ( c ) = f ( b ) - f ( a ) b - a . Proof: Set g ( x ) := f ( x ) - f ( b ) - f ( a ) b - a ( x - a ) + f ( a ) . Then g satisfies the hypothesis of the Rolle’s theorem: g is continuous on [ a, b ] and differ- entiable on ( a, b ) with g ( a ) = g ( b ) = 0. By Rolle’s theorem. there exists a point c ( a, b ) such that g 0 ( c ) = 0, i.e., f 0 ( c ) - f ( b ) - f ( a ) b - a = 0 which implies f ( b ) - f ( a ) b - a = f 0 ( c ) . [Examples and remarks] 1. The conclusion of the Mean Value Theorem can be written equivalently as: there exists at least one number c ( a, b ) such that f ( b ) - f ( a ) = f 0 ( c )( b - a ) .
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  • Fall '08
  • Staff
  • Probability theory, Order theory, Convex function, Suppose F, 6.2.5g

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