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!

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?

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.

VietDao29
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 $$y = \frac{1}{\sqrt{-x(x - 1)}}$$ is only defined on the interval (0, 1), and undefined everywhere else.

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.

HallsofIvy
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 $-pi/2$ and $\pi/2$ so $f(x)= tan(-\pi/2+ \pi x)$ should work nicely: when x= 0, $-\pi/2+ \pi (0)= -pi/2$. When x= 1, $-pi/2+ \pi (1)= \pi/2$. Unless I have misunderstood what you are looking for, that should be exactly what you want.
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 $-pi/2$ and $\pi/2$ so $f(x)= tan(-\pi/2+ \pi x)$ should work nicely: when x= 0, $-\pi/2+ \pi (0)= -pi/2$. When x= 1, $-pi/2+ \pi (1)= \pi/2$. Unless I have misunderstood what you are looking for, that should be exactly what you want.