Function with limit constraints

Question

In working with predator-prey population modeling, I ended up requiring a function $f(x)$ that satisfies the following conditions:

• $f$ is real and continuous, and so is $\frac{df}{dx}$;
• $\frac{df}{dx}\geq 0, \forall x$;
• $\lim\limits_{x\to\infty}{f(x)}=f_0$, where $f_0$ is a positive constant;
• $\lim\limits_{x\to-\infty}{f(x)}=-\infty$;
• $f(x)=0$ for a single and positive value of $x$, namely $x_0$.

I've tried finding functions that satisfy this set of rules, but I can't seem to find one that fits all of them (even though it's probably very easy). Any function will do, but the simpler and more "mathematically elegant" the better. Does anyone have any suggestions?

Note:

If it helps, $f(x)$ represents the growth rate of the predator population as a function of the prey population $x$. I'm doing this to find a function that will yield a maximum growth rate ($f_0$) when the population of prey, $x$ is infinite, and its minimum positive value when there are no prey ($x=0$).

Much appreciated.

Answer 1

Hint try for example a function like this: $-e^{-x}+x_0$

Answer 2

Try $f(x) = a-b^{-cx}$ where $b,c>0$. For example $a=b=c=1$, then $f(x) = 1-e^{-x}$. Note that

1) $f(x)$ and $f'(x) = e^{-x}$ are smooth,

2) $f'(x) = e^{-x} >0$,

3) $\lim_{x\rightarrow \infty} f(x) = 1$ and $\lim_{x\rightarrow -\infty} f(x) = -\infty$,

4) $f(0)=0$.

Answer 3

have you considered this?

$$ f(x) = \left\{ \begin{array}{lr} \frac{\sqrt{\pi}}{2}\text{erf}(x-1), \quad x\geq1\\ (x-1), \quad x<1 \end{array} \right. $$

$f(x),\,f'(x)$ are real and continuous, $\quad$ $f'(x)>0$,$\quad$ $f=0$ at $x=1$,$\quad$ $f(-\infty)=-\infty$, $\quad$ $f(\infty)=\frac{\sqrt{\pi}}{2}\approx 0.886$