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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)� < 1, so (x + 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). Tags absolute, clarifying, values Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shiseonji Algebra 2 September 24th, 2013 08:36 AM Kagiso Algebra 9 September 10th, 2013 03:37 AM matemego Calculus 3 October 25th, 2012 03:03 AM BackY Algebra 4 September 28th, 2010 04:03 PM Aurica Calculus 1 June 10th, 2009 05:24 PM

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