Graphing function with asymptotes at y=0,1

Question

Hi everyone. I'm searching for a formula to describe the above function piecewise. I've tried using things like translations of $1/x$ and $\sqrt{x}$. $1/x$ gives a nice asymptotic behavior, but I'm not sure how to neutralize the side of it to the left of the $x$-axis. Any advice? Thanks.

Answer 1

I don't think $1/x$ is what you want, since it goes asymptotically to zero, not to one (and it is negative when $x$ is negative, not positive like you want). How about a scaled version of the arctangent on the right side (which constant would you use to scale it?), $-1/x$ on the left, and linear function in between?

Another possibility for the function on the right is $x/(x+1)$.

Answer 2

One possibility is:

$$f(x)=\left\{\begin{array}\\ -\frac{1}{x}&\text{if}\, x\leq-1\\ -x&\text{if}\, -1\leq x\leq0\\ \tanh(x)&\text{if}\, x\geq0\\ \end{array}\right.$$

Answer 3

(@Eric Auld) Do you mean $\pi$ as part of the argument for the arctan? (i'm interested). The above graph, $(2/3)arctan(x)$, looks to do the job... Is this incorrect?