math-2011-note9 - Theorem 41 (Thm 30.5, p.828) . Let f : R...

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Unformatted text preview: Theorem 41 (Thm 30.5, p.828) . Let f : R 1 → R 1 be a C 2 function. For any point a < b ∈ R 1 , there is a point c ∈ ( a , b ) such that f ( b ) = f ( a ) + f prime ( a )( b- a ) + 1 2 f primeprime ( c )( b- a ) 2 . Proof. Define M = 2 ( b- a ) 2 bracketleftbig f ( b )- f ( a )- f prime ( a )( b- a ) bracketrightbig . Suppose that f primeprime ( x ) negationslash = M for all x ∈ ( a , b ) . Then either f primeprime ( x ) > M ∀ x ∈ ( a , b ) or f primeprime ( x ) < M ∀ x ∈ ( a , b ) . Suppose that f primeprime ( x ) > M ∀ x ∈ ( a , b ) . Then, we have f prime ( t )- f prime ( a ) = integraldisplay t a f primeprime ( x ) dx > integraldisplay t a M dt = M ( t- a ) , ∀ t ∈ ( a , b ) Hence, f prime ( t ) > f prime ( a )+ M ( t- a ) ∀ t ∈ ( a , b ) . So we have f ( b )- f ( a ) = integraldisplay b a f prime ( t ) dt > f prime ( a )( b- a )+ M integraldisplay b a ( t- a ) dt = f prime ( a )( b- a )+ M ( b- a ) 2 2 , a contradiction to the definition of M , which shows that it cannot be the case that...
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This note was uploaded on 03/26/2012 for the course ECON 205 taught by Professor Mr.lee during the Spring '11 term at Korea University.

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math-2011-note9 - Theorem 41 (Thm 30.5, p.828) . Let f : R...

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