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

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  • #1
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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.
 

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Answers and Replies

  • #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.
 
  • #3
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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.
 
  • #4
834
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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.
 
  • #5
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So is graphing it the only way to solve the inequality? Or is there a way to do it arithmetically.
 
  • #6
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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].
 
  • #7
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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.
 
  • #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.
 
  • #9
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That's ok.
 

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