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I'm looking for a simple decay function, where the rate of change of $y$ gradually gets smaller as $x$ grows, like $y=\sqrt{x}$, but it needs to stay less than $y=x$ at all times. You can see how the blue line, $y=\sqrt{x}$, is larger than the red line, $y=x$, between 0 and 1 here:

sqrt(x) vs x

I'd like something like this green line that I drew by hand:

my ideal function vs x

And for clarification, I don't care about negative $x$ values, and want $y$ to be always positive. I would also prefer $y$ to keep growing, not converge to a fixed value. Is there a function like this? It seems simple but I can't seem to find one. Any tips for finding answers to similar questions like this in the future?

lifeformed
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  • Do you only want this property for positive $x$? Or for all $x\in\mathbb{R}$? Have a look at $x\mapsto 1-e^{-x}$. – humanStampedist May 27 '21 at 09:27
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    $\sqrt x$ does not get smaller as $x \to \infty$. It increases to $\infty$. – Kavi Rama Murthy May 27 '21 at 09:27
  • I added some edits clarifying my requirements.

    @Joe $log{x}$ almost works but it is negative when $x<1$ (a new requirement I added), but $2log(x+1)$ is pretty good!

     – lifeformed May 27 '21 at 09:47
  • @humanStampedist Positive x. That's a good solution, although it approaches 1. I think I would prefer it be unbounded. These are caveats I did not think to add, I'll edit my post. – lifeformed May 27 '21 at 10:01
  • You mean $2\log_{10}(x+1)$, right? I prefer $\ln(x+1)$; it is conceptually simpler, and it has the same slope as $y=x$ at $x=0$. – TonyK May 27 '21 at 10:03
  • @lifeformed: Note that my first comment was mistaken. Not all logarithm bases work. However, $y=\ln(x)$ and $y=\log_10(x)$ are both always below $y=x$ for $x>0$. – Joe May 27 '21 at 10:14
  • @RicardoCavalcanti yes! That's the best one for me. It's simple, still grows unbounded, and is easy to scale by switching the $\sqrt{x}$ to $\sqrt{x/a}$ – lifeformed May 28 '21 at 09:09

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