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Math Help - Solving inequalities with absolute values within absolute values

  1. #1
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    Solving inequalities with absolute values within absolute values

    Solve the inequality:

    |||x-1|-1|-1| \geq 4

    So I can find the first solution:

    x-1-1-1 \geq 4
    x \geq 7

    I can't figure out how to get the other solution correctly. From looking it up I know it should be x \leq -5. I have never had to deal with absolute values like this before so I don't know where to put all the negative signs.
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  2. #2
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    Re: Solving inequalities with absolute values within absolute values

    Use repeatedly the fact that |f(x)| >= a iff f(x) >= a or f(x) <= -a.
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  3. #3
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    Re: Solving inequalities with absolute values within absolute values

    I have attempted that a few times and usually I come up with x \geq 5 which is incorrect. I guess I just don't know how to properly write it out and I end up missing a step or something every time. Would it be possible to get a step by step of how it is correctly done?
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  4. #4
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    Re: Solving inequalities with absolute values within absolute values

    |||x - 1| - 1| - 1| >= 4 iff (a1) ||x - 1| - 1| - 1 >= 4 or (a2) ||x - 1| - 1| - 1 <= -4.

    Case (a1). ||x - 1| - 1| - 1 >= 4 iff ||x - 1| - 1| >= 5 iff (b1) |x - 1| - 1 >= 5 or (b2) |x - 1| - 1 <= -5.

    Case (b1) |x - 1| - 1 >= 5 iff |x - 1| >= 6 iff (c1) x - 1 >= 6 or (c2) x - 1 <= -6.

    Finish analyzing (a2), (b2), (c1) and (c2).
    Thanks from BobRoss and ione
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  5. #5
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    Re: Solving inequalities with absolute values within absolute values

    Sorry, this is really confusing now. I think I see where you go from a1 to b1/b2 and then from that to c1/c2 and from the c1/c2 cases I can solve and find that x \geq 7 and x \leq -5, which is the correct answer for the question. I'm not really following what happens with a2 here. If I do with a2 as you did with a1 I end up with x \leq -1 and x \geq 3, which obviously means I did something wrong there...
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  6. #6
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    Re: Solving inequalities with absolute values within absolute values

    Quote Originally Posted by BobRoss View Post
    If I do with a2 as you did with a1 I end up with x \leq -1 and x \geq 3, which obviously means I did something wrong there...
    Note that (a2) has the form |f(n)| <= a, which is equivalent to -a <= f(n) <= a (and not to a choice between f(n) >= a and f(n) <= -a). In this particular case, (a2) is impossible because it is equivalent to ||x - 1| - 1| <= -3, and an absolute value can't be negative. The case (b2) is also impossible.
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  7. #7
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    Re: Solving inequalities with absolute values within absolute values

    Okay I think I see now what you've done here. That helps a lot, thank you!
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    Re: Solving inequalities with absolute values within absolute values

    I have another absolute value inequality question if you don't mind. I have read the inequality document that is stickied in this forum but it didn't help.

    |x-1|-|x2 -x| < 0

    I can't find any examples like this in my text or any similar examples online so I am lost as to how to properly solve this one too. Another quick walkthrough of this problem would be awesome if you are able.
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  9. #9
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    Re: Solving inequalities with absolute values within absolute values

    We have |ab| = |a| |b| for all a, b. So, |x^2 - x| = |x| |x - 1|. Thus, |x - 1| - |x^2 - x| = |x - 1| (1 - |x|). When is a product negative?
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  10. #10
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    Re: Solving inequalities with absolute values within absolute values

    It is negative when you multiply a positive and a negative. How are you arriving at this last part:

    Quote Originally Posted by emakarov View Post
    |x - 1| (1 - |x|).
    ?
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  11. #11
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    Re: Solving inequalities with absolute values within absolute values

    Quote Originally Posted by BobRoss View Post
    It is negative when you multiply a positive and a negative.
    Yes.

    Quote Originally Posted by BobRoss View Post
    How are you arriving at this last part
    By factoring out |x - 1|.
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  12. #12
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    Re: Solving inequalities with absolute values within absolute values

    Perhaps think of it in terms of distances.
    Let a = ||x-1| - 1|. Then, via |||x-1| - 1| - 1| >=4 (meaning |a-1|>= 4), the distance a to 1 is >=4
    Thus a >=5 or a <= -3.
    But a is an absolute value, hence positive.
    Thus a >=5.
    Let b = |x-1|.
    Then the distance from b to 1 is a (a = ||x-1|-1| = |b-1|), and a >=5.
    Thus the distance from b to 1 is >=5.
    Thus b>= 6 or b <= -4.
    But b is an absolute value, hence positive.
    Thus b >= 6.
    But b = |x-1| = distance from x to 1, and b >=6.
    Thus the distance from x to 1 is >=6.
    Thus x >=7, or x <= -5.
    Last edited by johnsomeone; September 23rd 2012 at 03:02 AM.
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  13. #13
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    Re: Solving inequalities with absolute values within absolute values

    Quote Originally Posted by emakarov View Post
    Yes.

    By factoring out |x - 1|.
    Gah, I'm terrible at this. I can't figure out how you get to that by factoring out |x-1|.
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  14. #14
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    Re: Solving inequalities with absolute values within absolute values

    Quote Originally Posted by BobRoss View Post
    I can't figure out how you get to that by factoring out |x-1|.
    |x - 1| - |x2 - x| < 0 <=>
    |x - 1| - |x(x - 1)| < 0 <=>
    |x - 1| - |x| |x - 1| < 0 <=>
    |x - 1|(1 - |x|) < 0 <=>
    |x - 1| > 0 and 1 - |x| < 0.
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  15. #15
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    Re: Solving inequalities with absolute values within absolute values

    Okay I should have seen that before. So now I need to analyze the cases

    1.) |x-1| is positive and 1-|x| is negative

    and the other case

    2.) |x-1| is negative and 1-|x| is positive

    And since the absolute value cannot be negative, case 2 doesn't exist. And when I solve case 1 I get x>1 and x<-1. Is that correct?
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