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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.

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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:

enter image description here

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

enter image description here

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

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  • $\begingroup$ Great! Does anything come to your mid that is not piecewise defined? It's not really an issue but more examples are welcome. $\endgroup$ – Kplusn Sep 22 '18 at 14:08
  • $\begingroup$ 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. $\endgroup$ – Adrian Keister Sep 22 '18 at 14:15
  • $\begingroup$ 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. $\endgroup$ – Kplusn Sep 22 '18 at 14:21
  • $\begingroup$ I've added another possibility, inspired by Stork's answer. Don't know if it's correct, but it looks good! $\endgroup$ – Adrian Keister Sep 22 '18 at 14:25
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    $\begingroup$ I plotted it's derivative and it seems that it has a unique zero. $\endgroup$ – Kplusn Sep 22 '18 at 14:31
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How about $(\tanh (x) + 1) + {e^{-x^2/2} \over \sqrt{2 \pi}}$?

enter image description here

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  • $\begingroup$ That isn't decreasing on some interval $(-\infty, y)$. Also it's nowhere negative as the function clearly must be. $\endgroup$ – Kplusn Sep 22 '18 at 12:58
  • $\begingroup$ It seems that the picture isn't even the same function you have drawn? $\endgroup$ – Kplusn Sep 22 '18 at 13:56
  • $\begingroup$ @Kplusn: Ooops... inserted the missing "+" sign. $\endgroup$ – David G. Stork Sep 22 '18 at 17:14
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Here's mine $$ f(x) = \frac{e^x(e^{2x}-1)}{e^{2x}+1} $$ picture

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