I am having difficulties to find the form of the equation of the curve $y=f(x)$.
$f(x)$ has the all of the following properties:
it is continuous (and not piecewise),
always increasing,
having at most one inflection point, (depends on our requirement, the constants of $f$).
it has upper bound.
For example, the common equation of a sigmoid is invalid,
say the equation is $y=a+\frac{b}{1+e^{cx}}$, now when $c=0$, then the second condition is not valid because we are told that $f$ is always increasing. Otherwise, it will always have one inflection point, this makes the third condition invalid because if we require no inflection point, we can not fix the constants.
Another example of invalid form; $y=A-Be^{Cx}$, because if we require one inflection point, then we can not have it by changing $A,B,C$
See the following to illustrate my problem:
The red curve: No inflection point, and the rate (or the slope) is always decreasing
The purple curve: There is one inflection point, and the rate (the slope) is increasing slowly, but then decreasing rapidly.
The green curve: There is one inflection point, the rate (or the slope) is increasing slowly, then increasing rapidly, and then decreasing slowly.
The blue curve: There is one inflection point, the rate (or the slope) is increasing slowly, then decreasing slowly.
The equations of these curves, I believe, have a common form. I tried to guess, but I did not reach to any answer.
I also tried to use more than $3$ constants.
Any help would be appreciated. THANKS!
$\tan^{-1} x$.
It is continuous on $\mathbb R$.
$\dfrac d{dx} \tan^{-1}x = \dfrac1{1+x^2} > 0$ for all $x \in \mathbb R$.
$\dfrac {d^2}{dx^2} \tan^{-1}x = -\dfrac{2x}{(1+x^2)^2}$ which is equal to $0$ only at $x=0$.
$\displaystyle \lim_{x \to \infty} \tan^{-1}x = \dfrac\pi2$ is the upper bound.