2
$\begingroup$

I had recently asked the question:
Logistic function passing through two points?
which received a very helpful answer. However, I'd like to ask your help again as the formulation of the problem has slightly changed and i'm lost...

Problem:
Given two points: $(x_l, y_l)$ and $(x_u, y_u)$ with: $x_l < x_u$ and $y_l < y_u$,
and given that the higher asymptote is one ($\lim_{x\to+\infty}f(x)=1$)
what's the logistic function that passes through the two points and the origin $(0, 0)$?

Thanks!

  • 0
    Yes, I understand that. But it is __still__ a logistic function. Just it passes through the origin and two other points, and it has higher asymptote y=1.2010-10-31

1 Answers 1

1

If you do not insist on $y=0$ as lower asymptote, but instead want the graph to pass through $(0,0)$, the general form of logistic function becomes $y=1-\frac{b}{1+b\,e^{ax}}$ where $a,b$ are determined from the two given points, and $b>0$. It's easy to eliminate $b$ from one of two equations: $b= \frac{1-y_\ell}{1+(y_\ell-1)e^{ax_\ell}}$ but plugging this into the second to find $a$, we get something that calls for numeric solution.

  • 0
    Your first equation cannot pass through the origin. $x = 0$ gives $y = 1 - \dfrac{b}{1+b}$2016-12-02