hw2sol

# hw2sol - Math 125A Fall 2009 Solutions for Homework...

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Math 125A, Fall 2009 Solutions for Homework 2 (Prepared by Matt Low) 1. Let f ( x ) = x - cos x . By direct computation, we see that f (0) < 0 and f ( π 2 ) > 0. Hence, by the intermediate value theorem (Theorem 18.2), there exists a number c (0 , π 2 ) such that f ( c ) = 0. In other words, c = cos c . 2. (a) True. By Theorem 18.1, there exists a c [ a, b ] such that f ( x ) f ( c ) for all x [ a, b ]. (b) False. For example, the function f : [0 , 2] R defined by: f ( x ) = x, if 0 x < 1 - x + 3 , if 1 x 2 assumes a maximum value, but is not continuous. (c) True. By Theorem 18.1, there exist numbers c, d [ - 2 , 2] for which f ( c ) f ( x ) f ( d ) for all x [ - 2 , 2], and according to the intermediate value theorem, f assumes every value in between, so f ([ - 2 , 2]) = [ f ( c ) , f ( d )]. Remark (by Xia): In the case that f ( c ) = f ( d ), we get only a single point. So, a more suitable answer to this question is “False” as a single point is not an interval. The countex- ample one shall use is a constant function. Nevertheless, for the grading of this problem, the reader will accept both “correct” answers. i.e. “True” using above statements or “false”

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