WA 2 - Nicola Hodges David Walnut MATH 290-002 September...

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Nicola Hodges David Walnut MATH 290-002 September 21, 2010 Writing Assignment 2 1. Prove that for every real number x > 0, x + (1/x) ≥ 2 and that x + (1/x) = 2 if and only if x = 1. Proof 1. x > 0 then x + (1/x) ≥ 2. Proof by contradiction. Let x be a real number. Assume x > 0 and x + (1/x) < 2. x + (1/x) < 2 is equivalent to x + (1/x) – 2 < 0. x + (1/x) – 2 < 0 is equivalent to (x 2 +1-2x)/x < 0. This inequality factors to, (x-1) 2 / x < 0. Solving the inequality, the critical values are 0 and 1. When x < 0, f(x) is negative. When 0 < x < 1, f(x) is positive. When x > 1, f(x) is positive. (x-1) 2 / x < 0, implies that f(x) must be less than zero. Therefore, the interval that satisfies this inequality is x < 0. However, this is a contradiction since we assumed x > 0. Therefore, when x > 0, x + (1/x) ≥ 2. Proof 2. x + (1/x) = 2 if and only if x=1. ( ) Let x be a real number. Assume, x + (1/x) = 2. We will show x=1. Note, x + (1/x) = 2 is equivalent to (1/x)=2 – x. Which is equivalent to 1 = 2x – x 2 . Which is equivalent to x 2 -2x + 1 = 0. This factors into (x-1)(x-1)=0.

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This note was uploaded on 03/05/2011 for the course MATH 290 taught by Professor Alligood,k during the Fall '08 term at George Mason.

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WA 2 - Nicola Hodges David Walnut MATH 290-002 September...

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