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# hw13 - Math 3110 Homework 13 Solutions Exercise 14.1.5 Let...

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Math 3110 Homework 13 Solutions Exercise 14.1.5 Let > 0 be given, and let δ = . If | x | < δ , then | f ( x ) - f (0) | = x sin 1 x ≤ | x | < δ = . This shows that f is continuous at 0. In order for f ( x ) to be differentiable at 0, we would need for f (0) = lim x 0 f ( x ) - f (0) x - 0 = lim x 0 x sin ( 1 x ) x = lim x 0 sin 1 x , but this limit does not exist because x = 0 is an essential discontinuity of sin ( 1 x ) (by page 164 of the book), so f is not differentiable at 0. Exercise 14.2.5 This is true. If f ( x ) has period c , then f ( x + c ) = lim h 0 f ( x + c + h ) - f ( x + c ) h = lim h 0 f ( x + h ) - f ( x ) h = f ( x ) . Therefore, f ( x ) is periodic, with a period that divides c . The relevant limits exist because we are given that f ( x ) is differentiable. Exercise 14.3.3 (a) If f ( x ) has degree n , then f ( x ) has degree n - 1, so it has at most n - 1 roots by Problem 15-1 (b) (which is conveniently due on the same day). Therefore, f ( x ) has at most n - 1 critical points. It can have exactly n - 1 critical points if f ( x ) = x 0 ( t - 1)( t - 2) . . . ( t - ( n - 1)) dt , which gives f ( x ) = ( x - 1)( x - 2) . . . ( x - ( n - 1)), which has as roots x = 1 , 2 , . . . , n - 1. (b) If n is odd, f ( x ) need not have any critical points at all. For example, if f ( x ) = x n + x , then f ( x ) = nx n - 1 + 1, which has no real roots.

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