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If I have the function f(x) = 2x + 1, is there something I can do to the function that will make it undefined for certain input values, for example input values where x ≥ 1. This would give the graph an endpoint. It is easy enough to just write {2x + 1 | x ≥ 2} but I am looking for a one line equation. The type of answer I am looking for, if you were to graph it, would look like a line that goes from the point (1,3) to infinity to the right and nothing but an endpoint on the left. First, is this even possible and second, how would we achieve this?
Thanks in advance.

Edit: I am looking for an approach that would work in a graphing calculator or a program like Wolfram|Alpha or Symbolab; something generalized.

applemonkey496
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  • Do you need a formula for use in a graphical calculator or some computer program? Or do you want to know the mathematical notation? – Paul Jun 20 '19 at 18:17
  • A formula for use in a computer program that is a graphical calculator. – applemonkey496 Jun 20 '19 at 18:19
  • What program? An approach that works in multiple programs is possible, but maybe there's an easier way for that specific program. – Paul Jun 20 '19 at 18:28
  • I am looking for a generalized approach, so something that I can use for multiple programs. – applemonkey496 Jun 20 '19 at 18:30

3 Answers3

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For the example you've described:

$\frac{(2x+1)\sqrt{x-1}}{\sqrt{x-1}}$ is probably the simplest solution.

You can generalize it to limit the domains of other functions.

RayDansh
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  • This works on paper for me, but for example when I put it into Wolfram|Alpha or Symbolab it cancels the terms out so it doesn't work. Is there something that might work in a program like W|A or Symbolab? – applemonkey496 Jun 20 '19 at 18:34
  • Also, this doesn't work if the program supports complex numbers, good answer otherwise. – Paul Jun 20 '19 at 18:36
  • I graphed it on Wolfram|Alpha and I think it's something to do with the program; if you scroll down it does correctly identify the domain but simply graphs it for all real numbers. – RayDansh Jun 20 '19 at 18:38
  • I graphed it on Symbolab and it seems to work. – RayDansh Jun 20 '19 at 18:41
  • Ok thanks. I think this will work well for me. I agree that it works on symbolab but I didn't check the domain on Wolfram|Alpha; I probably should've. – applemonkey496 Jun 20 '19 at 20:53
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This should work on many systems: $$ (2x+1) \cdot \frac{1-x}{\max(1-x,0)} $$ Whenever $x \ge 1$, there is division by zero.

Unfortunately Wolfram Alpha does not give a plot when entering this formula.

Paul
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Something like:

$g(x) = \frac{f(x)(x-a)(x-b)(x-c)}{(x-a)(x-b)(x-c)} = \begin {cases} f(x)&x\notin\{a,b,c\}\\\text{undef.}&x \in\{a,b,c\}\end{cases} $

user317176
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