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Question about a function

  1. Jan 28, 2008 #1
    I'm just wondering if anyone can think of a function that is contained in (0,1). I've been brainstorming for awhile and I can't think of anything particularly. Furthermore, it would be fantastic if it was not piecewise defined as well. Help would be appreciated!
  2. jcsd
  3. Jan 28, 2008 #2
    what do u mean it is contained, that for any values of x the values of the function varie from y=0 to y=1 or what?
  4. Jan 28, 2008 #3
    I'm essentially looking for a function where x is contained in the open interval of (0,1). Someone suggested tan(x) might work, but I would rather not use trigonometric functions.
  5. Jan 28, 2008 #4


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    Homework Helper

    You can give any function you think of, and then restrict the domain of the function to be (0, 1), like this:

    f: (0, 1) --> R
    x |--> x2 + x - 5

    That means, your f now, is only defined in (0, 1), and undefined everywhere else.

    Or, if you want some other nicer function, then square root is a function to look for:

    In the reals, the function [tex]y = \frac{1}{\sqrt{-x(x - 1)}}[/tex] is only defined on the interval (0, 1), and undefined everywhere else.
  6. Jan 28, 2008 #5
    Sorry, I should have clarified. I'm trying to prove that there exists a function between (0,1) that is 1-1 and onto, therefore the range of this function must span all y values, and the domain is contained in (0,1).

    For example, between (-1,1) f(x) = x/(x^2-1) works, but when I tried converting that into a form that is simply between (0,1), the inverse function looks exceptionally nasty! Using the inverse, I'm basically proving the function is onto, which in turn is proving that the (0,1) ~ R. Proving the function is 1-1 simply requires that the derivative is either always positive or always negative.
  7. Jan 28, 2008 #6


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    Science Advisor

    One to one and onto? Onto what? If I were talking about a function with domain between 0 and 1 and said it was "onto" the reasonable assumption would be it was "onto" (0, 1). The function y= x does that nicely!

    As for a function that is one-to-one and onto from (0, 1) to the set of all real numbers, I would think a variation on tan(x) would work nicely. tan(x) covers all real numbers for x between [itex]-pi/2[/itex] and [itex]\pi/2[/itex] so [itex]f(x)= tan(-\pi/2+ \pi x)[/itex] should work nicely: when x= 0, [itex]-\pi/2+ \pi (0)= -pi/2[/itex]. When x= 1, [itex]-pi/2+ \pi (1)= \pi/2[/itex]. Unless I have misunderstood what you are looking for, that should be exactly what you want.
  8. Jan 28, 2008 #7
    You are pretty much the man! Thanks a whole heap. And yes, I was looking for a 1-1 correspondence between (0,1) and R. I might have neglected to mention that.
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