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Solving an Inequality with X in a Denominator in Terms of Intervals

  1. Sep 18, 2012 #1
    I have been tasked with solving the following inequality:

    [itex]\frac{1}{x}[/itex] < 4

    Attached to this thread is my attempted solution. As you can see I begin with simply solving the inequality for x, and I obtain the result x > [itex]\frac{1}{4}[/itex]

    Next, I convert the equation into what I thought was the proper form for a hyperbola. I realize now I should have left the equation alone because it was already in proper form. However, I figure now that graphing at this point in my attempt may have not been the correct thing to do.

    Next I find the roots for the inequality. I find these to be 0, and [itex]\frac{1}{4}[/itex].

    Once the roots are found, I find the possible intervals for the inequality. The intervals I use are the following: x<0, 0<x<[itex]\frac{1}{4}[/itex], and x>[itex]\frac{1}{4}[/itex].

    I then set these up on a chart in order to find which intervals solve the inequality. However, I must have either set this up wrong or am going about this the wrong way. Any tips or guidance on where to go from here would be greatly appreciated.
     

    Attached Files:

  2. jcsd
  3. Sep 18, 2012 #2

    mathman

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    I don't understand you question. However 1/x < 4 has solution in two ranges. For x > 0, then x > 4. For x < 0, all x.
     
  4. Sep 18, 2012 #3
    My textbook states that the solution for this problem to be (-∞, 0) [itex]\cup[/itex] ([itex]\frac{1}{4}[/itex], ∞) (meaning that the roots are 0, and 1/4. I just don't know how to arrive at that answer. Basically I am wondering how to arrive at this solution, because I keep working the problem and getting different answers.
     
  5. Sep 18, 2012 #4
    I don't see what the problem is. Your graph on the right clearly shows that for [itex]x > 1/4[/itex], the value of [itex]1/x - 4[/itex] is less than 0 as required.
     
  6. Sep 18, 2012 #5
    So is graphing it the only way to solve the inequality? Or is there a way to do it arithmetically.
     
  7. Sep 18, 2012 #6
    You knew the function must have roots at 0 and 1/4. These are the only points where it can change positive or negative. All you have to do is plug in one value from each region.

    For [itex]x<0[/itex], pick, say, -1. Clearly [itex]1/-1 - 4 < 0[/itex].

    For [itex]0 < x < 1/4[/itex], pick, say, [itex]1/8[/itex]. Then [itex]1/(1/8) - 4 = 8 - 4 > 0[/itex].

    For [itex]x > 1/4[/itex], pick 1. [itex]1/1 - 4 < 0[/itex].
     
  8. Sep 18, 2012 #7
    Ah thanks, it turns out I was reaching incorrect solutions because in my notebook I was trying to find for >0 instead of <0...

    Thank you very much for your help and for pointing this out for me when I read your last post.
     
  9. Sep 19, 2012 #8

    mathman

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    I can't appreciate why you have a problem for x < 0. If x < 0, then 1/x < 0, so 1/x < 4.
     
  10. Sep 21, 2012 #9
    That's ok.
     
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