Here's my awesome drawing:

enter image description here

Basically it's a function that takes a high (or infinite) value at $0$, then falls off logarithmically for a while before falling off exponentially.

It doesn't need to be symmetric, it would be ok if the negative $x$ values were reversed in sign or something like that. I really only care about $x\ge0$.

Edit: I would prefer if the function is not periodic.

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    $\begingroup$ Awesome drawing, dude. $\endgroup$ – anakhro Aug 13 '15 at 15:00
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    $\begingroup$ It's surprising to me how many upvotes this got when you consider the lack of rigor in the question. I've seen so many questions shot down because they are of the form: " I know the first 4 numbers of the sequence... what is it? ". I fail to see how this question is any different... I guess an 'awesome' drawing makes all of the difference $\endgroup$ – jameselmore Aug 13 '15 at 17:46
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    $\begingroup$ @jameselmore I fail to see the similarities between the sequence questions you mention and this question. The question is about finding a function that looks like this image shown, not implying there is a single function that they are looking for. People (like myself) may find these kind of questions interesting, thus up-voting it. $\endgroup$ – anakhro Aug 14 '15 at 13:42

You almost literally describe a function satisfying your description: e.g.

$$ f(x) = -(\ln(x) + \exp(x)) $$

Tweak the constants to taste. Absolute value to make it symmetric: i.e. play with $A,B,C,D$ to shape it how you like:

$$ f(x) = A - B \ln |x| - C \exp(D |x|) $$

Here's a sample function with $A=20$, $B=5$, $C=1/20$ and $d=20$.

sample plot

generated by wolfram alpha

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    $\begingroup$ This is almost certainly the right answer; how could it not be, given the first sentence? $\endgroup$ – pjs36 Aug 13 '15 at 18:16
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    $\begingroup$ :tear: it's beautiful $\endgroup$ – rhombidodecahedron Aug 14 '15 at 6:59
  • $\begingroup$ By the way, why no coefficient inside the \ln? $\endgroup$ – rhombidodecahedron Aug 14 '15 at 9:35
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    $\begingroup$ @rhombidodecahedron: I put one in at first, but quickly remembered $\ln |Ex| = \ln |E| + \ln |x|$, so it would be redundant with $A$. However, $\ln(|x| + F)$ could be useful if you don't want the peak at $0$ to go all the way to $+\infty$. $\endgroup$ – Hurkyl Aug 14 '15 at 9:42

Motivation: "Falls off logarithmically to the right" and "falls off exponentially to the right" means that it increases exponentially in one direction and decreases exponentially in the other.

This sounds like a bit like a logistic function, which models exponential growth as we approach the lower $y$-bound and upside-down exponential growth near the upper $y$-bound (like $y=e^x$ and $y=-e^{-x}$ respectively), but it's rotated inconveniently.

So I looked up its inverse (the logit function), which has the form $\log(\frac{x}{1-x})$. After some modifications to reflect across the $x$-axis and $y$-axis, I got the plot: $$f(x)=-\log{\frac{|x|}{1-|x|}}$$ (Try $f(ax), af(x), f(x)+a,$ etc. to scale/translate as needed)


For instance: $\arccos(|x| -1)$ Plot

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    $\begingroup$ If you alter it to $y=\cos^{-1} (0.4|x|-1)$, it looks almost identical to OP's graph. $\endgroup$ – Trogdor Aug 13 '15 at 14:55

How about $\frac{x^2-1}{x^2(x-2)(x+2)}$


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