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Math Help - Quick absolute value inequality explanation, please

  1. #1
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    Red face Quick absolute value inequality explanation, please

    Hey everyone, I'm struggling with some of the concepts behind absolute value inequalities and found a certain problem I need some assistance with. I'll explain what I tried, please point out my mistake and explain why it is a mistake. Much appreciation.

    Basically (|x|+1)/(|x|-1) < 1

    if x >= 0

    (x+1-x+1) / x-1 < 0

    which gives x < 1


    and if x < 0

    (-x+1)/(-x-1) < 1

    which gives x >= 0

    [0;1)U(1;infinity)
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Squared View Post
    Basically (|x|+1)/(|x|-1) < 1 If x >= 0 (x+1-x+1) / x-1 < 0 which gives x < 1

    Right, if x\geq 0 then:

    \dfrac{|x|+1}{|x|-1}<1 \Leftrightarrow \dfrac{x+1}{x-1}<1\Leftrightarrow 1+\dfrac{2}{x-1}<1\Leftrightarrow \dfrac{2}{x-1}<0\Leftrightarrow x\in[0,1)


    Fernando Revilla
    Last edited by FernandoRevilla; February 9th 2011 at 10:50 AM.
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  3. #3
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    write the inequality as \dfrac2{|x|-1}<0 so |x|<1\implies -1<x<1.

    perhaps is not what the OP intented to ask, but it's the faster way on solving it.
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  4. #4
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    @fernando

    Gracias Um, my question might be a bit confusing, forgive me, what I actually want to know is I did everything like you and got the same answer but when x < 0 should I get (1;infinity), should I make a union between the two and is my final answer correct?

    [0;1)U(1,infinity)
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    If x<0 :


    \dfrac{|x|+1}{|x|-1}<1 \Leftrightarrow \dfrac{-x+1}{-x-1}<1\Leftrightarrow 1-\dfrac{2}{x+1}<1\Leftrightarrow \dfrac{2}{x+1}>0\Leftrightarrow x\in(-1,0)


    Fernando Revilla
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  6. #6
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    Yay, forgot to divide by the -1, works fine now, gracias Fernando, also thanks for the quick method Krizalid.
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  7. #7
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    Quote Originally Posted by Squared View Post
    Hey everyone, I'm struggling with some of the concepts behind absolute value inequalities and found a certain problem I need some assistance with. I'll explain what I tried, please point out my mistake and explain why it is a mistake. Much appreciation.

    Basically (|x|+1)/(|x|-1) < 1

    if x >= 0

    (x+1-x+1) / x-1 < 0

    which gives x < 1

    Good!


    and if x < 0

    (-x+1)/(-x-1) < 1

    which gives

    [0;1)U(1;infinity)
    Setting it up as you did for x\ge0

    to get for x\le0

    \displaystyle\frac{-x+1}{-x-1}<1\Rightarrow\frac{-x+1}{-x-1}-\frac{-x-1}{-x-1}<0

    \displaystyle\frac{-x+1-(-x-1)}{-x-1}<0\Rightarrow\frac{-x+x+1+1}{-x-1}<0

    \displaystyle\frac{-2}{x+1}<0\Rightarrow\ x+1>0\Rightarrow\ x>-1
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  8. #8
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    Hey again everyone, I just wanted to ask, do any of you perhaps have a link to some advanced absolute value inequalities to solve. I googled but kept getting relatively VERY easy ones compared to the ones we are doing. I then tried searching for compound absolute value inequalities and found some who were harder but still too easy?
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  9. #9
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    Quote Originally Posted by Squared View Post
    Hey everyone, I'm struggling with some of the concepts behind absolute value inequalities and found a certain problem I need some assistance with. I'll explain what I tried, please point out my mistake and explain why it is a mistake. Much appreciation.
    Basically (|x|+1)/(|x|-1) < 1
    I have a different take on this altogether.
    Consider the case |x|>1 then |x|-1>0.
    This leads to |x|+1<|x|-1 or 1<-1. Impossible.

    On the other hand, |x|<1 gives |x|+1>|x|-1 which is true.
    So the solution set is \{x:-1<x<1\}
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  10. #10
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    Quote Originally Posted by Squared View Post
    Hey again everyone, I just wanted to ask, do any of you perhaps have a link to some advanced absolute value inequalities to solve. I googled but kept getting relatively VERY easy ones compared to the ones we are doing. I then tried searching for compound absolute value inequalities and found some who were harder but still too easy?
    try this http://www.mathhelpforum.com/math-he...es-132202.html
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  11. #11
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    Thanks, I'll work through it this weekend, last 3 examples are basically what I'm looking for, but still require some that are harder.
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