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Unformatted text preview: Math 3110 Homework 11 Solutions Exercise 12.1.1 Let f be a continuous function whose values are always rational numbers. Assume for contradiction that f is not a constant function. Then there exist x 1 , x 2 with x 1 < x 2 such that f ( x 1 ) 6 = f ( x 2 ). Without loss of generality, assume f ( x 1 ) < f ( x 2 ). Then by Theorem 2.5, we can find an irrational number k such that f ( x 1 ) < k < f ( x 2 ). Since f is continuous, then by the Intermediate Value Theorem (Corollary 12.1), there exists a number c with x 1 < c < x 2 such that k = f ( c ). But this contradicts that the values of f are always rational. So f must be constant. Exercise 12.1.3 Let f ( x ) = x 2 ( x- 1), which has roots x = 0 and x = 1. Bolzanos Theorem cannot find the root x = 0 because f ( x ) < 0 immediately to both sides of the root, so f ( x ) does not change sign on a small neighborhood of x = 0. Newtons method can still find this root, however, if we start close enough to it....
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This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).