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Math Help - Inequality assistance

  1. #1
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    Inequality assistance

    The question:

    Solve |\frac{x - 1}{x + 1}| < 1

    My attempt:

    |x - 1| < |x + 1|
    x - 1 < x + 1
    x < x + 2

    Help! I'm clearly doing something wrong here. :/
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The question:

    Solve |\frac{x - 1}{x + 1}| < 1

    My attempt:

    |x - 1| < |x + 1|
    x - 1 < x + 1
    This doesn't follow. If x- 1> 0 and x+ 1> 0 (which happens for x> 1), then |x- 1|= x- 1 and |x+ 1|= x+ 1 so x- 1< x+ 1 from which -1< 1 which is always true.

    But if x- 1 and x+ 1 are both negative (which happens for x< -1), then |x- 1|= -(x- 1) and |x+ 1|= -(x+ 1) so -(x- 1)< -(x+ 1) from which x- 1> x+ 1, -1> 1 which is always false.

    If -1< x< 1, then x- 1< 0 and x+ 1> 0 so |x- 1|< |x+ 1| becomes -(x- 1)< x+ 1 or -x+ 1< x+ 1 which reduces to -x< x, 0< 2x, x> 0.

    The inequality is true for all positive x except x= 1. (If the problem had been \frac{|x- 1|}{|x+ 1|}\le 1 then it would be true for all positive x.)

    x < x + 2

    Help! I'm clearly doing something wrong here. :/
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    A rule!

    If you see |f(x)|<t it's : -t<f(x)<t

    And if you see |f(x)|>t then it's:

    f(x)>t or f(x)<-t


    In your case it's the first one when f(x)=(x-1)/(x+1)
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  4. #4
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    Quote Originally Posted by Also sprach Zarathustra View Post
    A rule!

    If you see |f(x)|<t it's : -t<f(x)<t

    And if you see |f(x)|>t then it's:

    f(x)>t or f(x)<-t


    In your case it's the first one when f(x)=(x-1)/(x+1)
    Ahh, thanks. I'll remember that.

    I'm still having trouble simplifying it. I've done:

     -1 < |\frac{x - 1}{x + 1}| < 1
     -(x + 1) <  (x - 1) < (x + 1)
     (-x - 1) <  (x - 1) < (x + 1)
     -x <  x < x + 2
     0 <  2x < 2x + 2
     0 <  x < x + 1

    So x > 0 (which is correct).

    But am I doing this correctly? Or was it just a coincidence I got the answer?
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Glitch View Post
    Ahh, thanks. I'll remember that.

    I'm still having trouble simplifying it. I've done:

     -1 < |\frac{x - 1}{x + 1}| < 1
     -(x + 1) <  (x - 1) < (x + 1)
     (-x - 1) <  (x - 1) < (x + 1)
     -x <  x < x + 2
     0 <  2x < 2x + 2
     0 <  x < x + 1

    So x > 0 (which is correct).

    But am I doing this correctly? Or was it just a coincidence I got the answer?
    Incorrect! (first line)

    Please read again 'my rule'...
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  6. #6
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    So it's this instead?

    -1 < \frac{x-1}{x+1} < 1

    If so, does this mean I have to square the denominator to ensure it's positive before solving?
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Yes!

    And how you solve this?

    In fact you have two inequalities:

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

    AND

    2. (x-1)/(x+1)>-1

    I will start the first one:

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

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

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

    ==> -2/(x+1)<0

    And now like you said , square the denominator.
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  8. #8
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    But I can't do anything with that square, because if I multiply each side by (x + 1)^2 then the RHS is 0(x + 1)^2, which is 0. Doing so gave me x > -1

    Sorry if I'm doing it wrong, I'm having a bad day I think...
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  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
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    Ok. I will continue from


    ==> -2/(x+1)<0

    ==> -2(x+1)/(x+1)^2<0

    ==> -2(x+1)<0 (because (x+1)>0 fo all x in R\(-1) )

    ==> -(x+1)<0

    ==> x+1>0

    ==> x>-1

    Now try the second one.
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  10. #10
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    Ok, the second one gives me x > 0. So since, x > -1 and x > 0, I naturally take x > 0 (since 0 > -1), right?
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  11. #11
    MHF Contributor Also sprach Zarathustra's Avatar
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    Yes!
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  12. #12
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    Thank you good sir. I've just started my calculus course, and these pre-calc questions are worrying me!
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  13. #13
    MHF Contributor Also sprach Zarathustra's Avatar
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    You welcome! And good-luck!
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