# A function with certain shape

I'm trying to simulate a certain type of condition with a continuous and twice differentiable function $$f(x)$$ that has the following shape:

1. The limit of $$f(x)$$ at $$-\infty$$ is finite but less or equal to zero, i.e. $$-\infty < \lim_{x \to -\infty}f(x) \leq 0$$
2. There exists a unique point $$y \in \mathbb{R}$$ such that $$f(x)$$ is decreasing on $$(-\infty,y)$$ and increasing on $$(y,\infty)$$.
3. The limit at $$\infty$$ is positive or infinity i.e. $$0<\lim_{x \to \infty} f(x) \leq \infty.$$

However I'm struggling to find a good example of such a function that is simple enough to make the point clear. In other words I'm looking for a explicit examples of a function with the properties that I described above.

Here's one that will work: $$f(x)=\begin{cases}-e^{-x^2},\;&x<0 \\ 1-2e^{-x^2/2},\;&x\ge 0. \end{cases}$$ Here's a plot:

Here's another that might work (inspired by David G. Stork's answer): $$f(x)=\operatorname{tanh}(x)-e^{-x^2/4}.$$ Plot:

The thing is, I haven't double-checked that this function is monotonically decreasing before some $$x,$$ and monotonically increasing after.

• Great! Does anything come to your mid that is not piecewise defined? It's not really an issue but more examples are welcome. – Kplusn Sep 22 '18 at 14:08
• My guess is that David G. Stork's answer could be massaged to be correct. The $\operatorname{tanh}$ or $\arctan$ functions would be useful here, but they'd have to be modified to get the local minimum your conditions require. – Adrian Keister Sep 22 '18 at 14:15
• Yes I tried to plot a bit different versions of his answers but I didn't manage to get one that I was looking for yet. – Kplusn Sep 22 '18 at 14:21
• I've added another possibility, inspired by Stork's answer. Don't know if it's correct, but it looks good! – Adrian Keister Sep 22 '18 at 14:25
• I plotted it's derivative and it seems that it has a unique zero. – Kplusn Sep 22 '18 at 14:31

How about $$(\tanh (x) + 1) + {e^{-x^2/2} \over \sqrt{2 \pi}}$$?

• That isn't decreasing on some interval $(-\infty, y)$. Also it's nowhere negative as the function clearly must be. – Kplusn Sep 22 '18 at 12:58
• It seems that the picture isn't even the same function you have drawn? – Kplusn Sep 22 '18 at 13:56
• @Kplusn: Ooops... inserted the missing "+" sign. – David G. Stork Sep 22 '18 at 17:14

Here's mine $$f(x) = \frac{e^x(e^{2x}-1)}{e^{2x}+1}$$