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This question is about an interesting observation that comes up whenever one solves an inequality with a variable in the denominator.

eg.

1/x < 1

Why is it that, in the solution to these inequalities, x can be a values that are 1) on the outsides of the two critical values or 2) in between the two critical values?

I have never seen inequalities where x can take values in between both values AND on the outside of one critical value.

Another query: what is the formal, purely algebraic definition of a critical value (if there is one)?

thetrueembodimentofstupidity
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  • Hi, welcome to MSE. Please use MathJax to format equations. I have attached a link to a tutorial. https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference?page=2&tab=scoredesc#tab-top – Cheese Cake Sep 22 '22 at 03:23
  • What do you mean by critical values? – Karl Sep 22 '22 at 03:29
  • critical point? – thetrueembodimentofstupidity Sep 22 '22 at 03:29
  • Could you elaborate on your example in the question? What are the critical values? – Karl Sep 22 '22 at 03:32
  • in the inequality, the critical value would be x = 0 and x = 1. The way I see it, critical values are values for x of which all the numbers on one side of x cannot be solutions to the inequality. for instance in the example, x = 0 is a critical value because when you use a test points between 0 to 1 (which are numbers on the right side of 0) they do not work in the inequality. So we say x=0 is a critical value. – thetrueembodimentofstupidity Sep 22 '22 at 03:36
  • Well, it is an inequality. It can't be both bigger and smaller than something at the same time. I suppose you can do $f(x) = x$ if $x\ge 1$ and $f(x) = \frac 1x$ if $0 < x < 1$. THen the inequality $\frac 1{f(x)}<1$ will do what you want. – fleablood Sep 22 '22 at 03:37
  • @fleablood. what are you talking about? I don't follow – thetrueembodimentofstupidity Sep 22 '22 at 03:40
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    Well, you are asking a rather meaningless question. If $\frac 1x < 1$ then $x<0;x>1$ and there isn't any reason to ask why we can't have an $x$ between $0$ and $1$. That's just not the answer. I figured you wanted an inequality $f(x)< 1$ with two critical points where $x < 1$ and $x>1$ are solutions. Define $f(x) = \begin{cases}x& x>1\ \frac 1{1-x}& 0<x<1\end{cases}$. Then the inequality $\frac 1{f(x)} < 1$ will have solutions $0< x < 1$ and $x > 1$. – fleablood Sep 22 '22 at 03:54
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    The premise of your question, i.e. the observation you make for such inequalities- is itself an issue. for e.g. $1/(x^2+1)<1$ has the variable in denominator, but the solution does not conform to your observation. OTOH, $x<x^2$ has no denominator, but its solutions are outside a closed interval. So the pattern you claim itself is questionable. – Macavity Sep 22 '22 at 06:37

1 Answers1

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I'm a little unclear about your question, as it mentions two critical values (I'm assuming you mean restrictions or non-permissible values) but the inequality you included only has one. However, here are two examples that might address what you're asking:

$$\frac{1}{(x+1)(x-1)^2} > 0 \quad \text{and} \quad \frac{1}{(x+1)^2(x-1)^2} > 0$$

Both the above inequalities have the restrictions $x \neq \pm 1\,$.

The solution for the first is $\,-1 < x < 1\,$ or $\,x>1\,$, i.e. in between the two non-permissible values and to the right of the larger one.

The solution to the second is $\,x<-1\,$ or $\,-1 < x < 1\,$ or $\,x>1\,$, i.e. in between the two non-permissible values and outside both.

You'll no doubt notice that the difference between the two examples is the power of the $\,(x+1)\,$ factor in the denominator. In general, for inequalities involving polynomial or rational functions, 'the solutions "change" around zeroes that come from factors with odd exponents (or zeroes with odd 'multiplicity', if you're familiar with that term), whereas they don't "change" around zeroes that come from factors with even exponents.

A.J.
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  • so my observation is incorrect. There can be inequalities where the solution ranges over values on one side of a restriction and also in between the two restriction (critical value) points. Cool, thanks! – thetrueembodimentofstupidity Sep 22 '22 at 08:49