# hw14 - Math 3110 Homework 14 Solutions Exercise 15.1.3 If f...

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Math 3110 Homework 14 Solutions Exercise 15.1.3 If f ( x ) changes sign from negative to positive on [ a, b ], then there are some h, k [ a, b ] with h < k , f ( h ) < 0, and f ( k ) > 0. Since f ( x ) is diﬀerentiable on [ a, b ] and [ h, k ] [ a, b ], f ( x ) is diﬀerentiable on [ h, k ]. Then by Theorem 15.1, there is some c [ h, k ] such that f ( k ) - f ( h ) = f 0 ( c )( k - h ), from which f 0 ( c ) = f ( k ) - f ( h ) k - h . Since f ( k ) > 0 > f ( h ) and k > h , f 0 ( c ) is a quotient of two positive numbers, and so f 0 ( c ) > 0. Exercise 15.2.1 (a) We prove the contrapositive, namely, that if f ( x ) is not bounded on I , then neither is f 0 ( x ). Suppose that f ( x ) is not bounded on I . Pick some a I . Let the length of the interval I be ` . For any m > 0, since f ( x ) is not bounded, there is some b I such that | f ( b ) | > m` + | f ( a ) | . Hence, | f ( b ) - f ( a ) | ≥ | f ( b ) |-| f ( a ) | > m` + | f ( a ) |-| f ( a ) | = m` . By Theorem 15.1, there is some c I (as I contains all points between a and b , regardless of whether the interval is [ a, b

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## This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).

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hw14 - Math 3110 Homework 14 Solutions Exercise 15.1.3 If f...

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