My Math Forum clarifying absolute values

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 October 6th, 2013, 08:22 PM #1 Member   Joined: Aug 2013 Posts: 69 Thanks: 0 clarifying absolute values Hi, I can't seem to make sense of two inequalities using Absolute Values; can someone explain to me what I'm thinking about wrong? I'm trying to find a constant such that |x - 2|< C |x + 2| < 1 => -3 < x < -1 => -5 < x -2 < -3 => |x - 2| < 5 <-- I can't make sense, and if that is the case, how does |x - 3| < 1 => 2 < x < 4 => 5 < x + 3 < 7 => |x + 3| < 7 ? If someone can help me quick, I have a midterm in 8 hours and I feel like a potato.
 October 8th, 2013, 06:05 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2207 If x + 2 < 0, |x + 2| = -(x + 2), else |x + 2| = x + 2. Thus if x + 2 < 0, |x + 2| < 1 means -(x + 2 ) < 1, which implies x + 2 > -1, so x > -3, and one can write -3 < x < -2. If x + 2 ? 0, |x + 2| < 1 means x + 2 < 1, so x < -1, and one can write -2 ? x < -1. Those results can be combined by writing -3 < x < -1. Or observe that |x + 2| < 1 is equivalent to |x - (-2)| < 1, which means that x differs from -2 by less than 1. That implies x lies strictly between -2 - 1 and -2 + 1, which can be written as -3 < x < -1. Another method: |x + 2| < 1 implies (x + 2)² < 1, so (x + 2)² - 1 < 0. Factorizing gives (x + 2 + 1)(x + 2 - 1) < 0, i.e., (x + 3)(x + 1) < 0. Hence -3 < x < -1. One you have shown that -3 < x < -1, subtracting 2 (throughout) gives -5 < x - 2 < -3, which (as x - 2 must be negative, and so equals -|x - 2|) implies that -5 < -|x - 2| < -3. Multiplying by -1, one then gets 5 > |x - 2| > 3, which means 3 < |x - 2| < 5. In words, if x differs from -2 by less than 1, x must differ from +2 by less than 5 (as a difference of 5 is approached as x nears -3), but more than 3 (as a difference of 3 is approached as x nears -1).

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