# hw8 - Math 3110 Homework 8 Solutions Exercise 9.2.3 Since f...

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Unformatted text preview: Math 3110 Homework 8 Solutions Exercise 9.2.3 Since f ( x ) = 1 f ( x ) , we must have ( f ( x )) 2 = 1, and so f ( x ) = ± 1. Clearly any function that has either f ( x ) = 1 or f ( x ) =- 1 for each real x satisfies the equation. This must hold for each real x , but different inputs could give different outputs. That is, there could be distinct real x and y such that f ( x ) = 1 and f ( y ) =- 1. We can formalize this by saying that there is some set S that is a subset of the real numbers such that f ( x ) = 1 if x ∈ S and f ( x ) =- 1 if x 6∈ S . Exercise 9.3.5 We show a contrapositive, namely, that a periodic function that is not constant is not increasing. Let f be a periodic function that is not constant. Since f is not constant, there are real a and b such that f ( a ) < f ( b ). Since f is periodic, let its period be c . If a > b , then these two points demonstrate that f is not increasing. Otherwise, a < b . Define d = a + b- a c c . Since f has period c , f ( d ) = f ( a )...
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## This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell.

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hw8 - Math 3110 Homework 8 Solutions Exercise 9.2.3 Since f...

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