Looking for monotonic function example that satisfies several conditions I need an example of any monotonic function that:
1) grows asymptotically to 8/3:  f(infinity)=8/3 and f(x)<=8/3 for any x,
2) f(1)=1,
3) easy to write in programming (JavaScript language)
4) pretty fast to calculate the function value
I've found a function that is OK for 1)..4). For example it could be
f(x)=8/3 -(5/3)*(1/x);
However, I need 5) as well:
5) for small x (from 1 to 5) it doesn't grow so fast, but is around 2.0 for f(6). Let's say it is something like:
f(1)=1, f(2)<=1.2, f(3)<=1.4,  f(6)>1.8, f(6)<2.0.
For my example above f(2) = 1.8(3), that is too much.
I appreciate if somebody helps me with minimum changes of my function so it looks tricky and smart and 1)-5) are met.  (the less changes or the shorter function, the better).
Thank you.
 A: This might be more fitting as a comment but i don't yet have enough reputation to post it as such:
A straightforward solution would be to handle the small x-es with a polynomial function and then use your function for x>6 (or from wherever the polynomial and your function intersects).
I imagine polynomials would be pretty fast to calculate, and if you don't know how to interpolate, wolfram alpha can do it for you. 
(this answer completely fails at looking "tricky and smart" though, and the function won't be short or simple by any means) 
A: You are looking for a sigmoid function. Common examples are
$\phi(x)=0.5+\arctan(x)/\pi$, $\phi(x)=1/(1+\exp(-x))$, $\phi(x)=0.5+0.5\,x/(1+|x|))$. 
They all grow from 0 to 1 and have their turning point at 0. 
By translating and scaling you can adapt them to your situation, using $f(x)=c*\phi(ax+b)$. Obvious conditions are that $c=8.3$ and $a>0$. 
You want the turning point to be larger than 6, i.e., $6a+b<0$, for instance by putting the turning point at 10 to 20, e.g. at 15 by demanding $15a+b=0$. 
$f(1)=1$ one has to solve directly. In the first function example, this leads to the equation $a+b=\tan(\pi*(1/8.3-0.5))\approx -2.5$. With the turning point at 15, this gives $a=2.5/14\approx 0.2$ and thus $b=-2.7$.
With (a,b,c)=(0.2,-2.7,8.3) you get $f(1)\approx 1$, $f(8)\approx 2$, the turning point with value 4.15 at about x=13, $f(23)\approx 7$.
To be able to claim the homework as your own effort, I would suggest that you determine the constants in a similar fashion for the second example function, the logistic function.
