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

In summary: I am glad that you were able to see the mistake I was making and that you were able to find the proper solutions.
  • #1
EcKoh
13
0
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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  • #2
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
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
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
So is graphing it the only way to solve the inequality? Or is there a way to do it arithmetically.
 
  • #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].
 
  • #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.
 
  • #8
I can't appreciate why you have a problem for x < 0. If x < 0, then 1/x < 0, so 1/x < 4.
 
  • #9
That's ok.
 

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

1. How do I solve an inequality with x in the denominator in terms of intervals?

To solve an inequality with x in the denominator in terms of intervals, you can follow these steps:

  • Isolate the fraction with x in the denominator on one side of the inequality.
  • Multiply both sides by the denominator to eliminate it from the inequality.
  • Determine the interval(s) where the fraction will be positive or negative based on the sign of the numerator.
  • Use the interval(s) to write the solution in interval notation or as a compound inequality.

2. Can I solve an inequality with x in the denominator algebraically?

Yes, you can solve an inequality with x in the denominator algebraically by following the steps mentioned in the previous answer. However, it may be easier to graph the inequality and use the visual representation to find the solution.

3. Why is it important to determine the intervals in an inequality with x in the denominator?

Determining the intervals in an inequality with x in the denominator is important because it helps you understand the behavior of the fraction and identify where it is positive or negative. This information is crucial in writing the solution in interval notation or as a compound inequality.

4. Are there any restrictions when solving an inequality with x in the denominator in terms of intervals?

Yes, there are restrictions when solving an inequality with x in the denominator. The denominator cannot be equal to zero, as it would result in a division by zero error. Additionally, any values that would make the denominator equal to zero must be excluded from the solution.

5. Can you give an example of solving an inequality with x in the denominator in terms of intervals?

Sure, let's say we have the inequality 5/(x-3) < 2. To solve this, we would first isolate the fraction by subtracting 2 from both sides, giving us 5/(x-3) - 2 < 0. Next, we multiply both sides by the denominator (x-3) to get rid of it in the fraction: 5 - 2(x-3) < 0. Simplifying, we get -2x + 11 < 0. Now, we can determine the intervals by setting the numerator equal to zero and solving for x, giving us x = 11/2. This creates two intervals: x < 11/2 and x > 11/2. Finally, we can write the solution in interval notation as (-∞, 11/2) and (11/2, ∞) or as a compound inequality, x < 11/2 or x > 11/2.

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