# A function $f(x)$ that increases from $0$ to $1$ when $x$ increases from $−\infty$ to $\infty$.

I am looking for a function $f(x)\in[0,1]$ when $x\in(−\infty,+\infty)$.

I bumped up on this question (A function $f(x)$ that increases from $0$ to $1$ when $x$ increases from $-\infty$ to $\infty$.) when I searched for it but the only difference here is the rate of increase.

$f(x)$ should increase slowly when $x$ starts from $−\infty$ until $0$, should increase fast once $x$ crosses zero and then the rate should keep on decreasing fast eventually approaching $1$ when $x$ is $\infty$.

• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Dec 24 '15 at 14:28
• "the rate should keep on decreasing fast": I am afraid this is not possible. – Yves Daoust Dec 24 '15 at 14:35

The common logistic function $$f(x) = \frac{1}{1+e^{-x}}$$ has the desired limits. We can add some parameters to manipulate it's exact growth rates around 0. We can take $$g(x) = \frac{1}{1+e^{-m(x-a)}}$$ and then play around with $m$ and $a$ until we get desired values. The bigger $m$ is the steeper the rise from 0 to 1 will be. And $a$ will be the center point of the rise.
$f(x)=\frac{x}{\sqrt{1+x^2}}$ goes to $(-1,1)$, so you can rescale it to make it go to $(0,1)$: $$g(x)=\frac{1}{2}\left(1+\frac{x}{\sqrt{1+x^2}}\right)$$
The function $f$ has a geometric interpretation. Take the point on the unit circle that intersects the ray from $(0,0)$ to $(x,1)$. Then take the $x$ value of that point. I forget the name of this projection.
Another such function is: $$\arctan x$$ which sends $(-\infty,\infty)$ to $(-\pi/2,\pi/2)$, so again you'd have to rescale.