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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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