Find a decreasing function $f:[0,\infty)\to(0,1]$ with one inflection point satisfying certain properties

Question

Define a function $f:[0,\infty)\to(0,1]$. Define three constants: $j\in(0,\infty)$, $\alpha\in(0,1)$, and $\beta\in(-\infty, 0)$.

$f(0)=1$

$f'(0)=0$

$f(j)=\alpha$

$f'(j)=\beta$

$j$ is the only inflection point of $f$

$f'(x)<0 \space \space \space \space \forall x \in (0, \infty)$

Can you find such a function $f$? Piecewise definitions are fine as long as these properties are satisfied for all $x$, $j$, $\alpha$, and $\beta$ in their respective domains.

Essentially, I'm looking for a decreasing continuous function that maps the non-negative real numbers to positive numbers less than or equal to one, and I want to be able to choose the inflection point, the value of the function at the inflection point, as well as the slope of the line tangent to the function at the inflection point.

Answer 1

$f(x)=e^{-kx^2}$ or something like it seems to be what you want.

Answer 2

We first make a function for all values before the inflection point and proceed to a function for after it.

Our $\textbf{left}$ function is of the form $f(x)=ae^{kx^2}+b$. To follow the first rule, $$f(0)=1=a+b \Rightarrow \fbox{$b = a-1$}.$$ This follows the second rule since $f'(x)=2akxe^{kx^2}$ and if $x=0$, $f'(0)=0$.

Following the third requires $$f(j)=ae^{kj^2}+b=\alpha$$ $$\Rightarrow ae^{kj^2}+1-a=\alpha$$ by substituting in what we found from the first rule. With some more manipulation we find $$\fbox{$k = \frac{\ln\bigl(\frac{\alpha-1}{a} + 1 \bigr)}{j^2}$}$$

For the final rule, $$f'(j)=2akje^{kj^2}=\beta$$ Substituting in what we just found for $k$ we get $$\fbox{$\ln\Bigl(\frac{\alpha-1}{a}+1\Bigr)=\frac{\beta j}{2 (\alpha + a - 1)}$}.$$ This requires numerical approximation so you will need to choose your favorite method to do so in order to find $a$. Once $a$ is found, just work your way back up the totem pole and you will get all necessary coefficients.

Our $\textbf{right}$ function is of the form $f(x)=\frac{r}{x-s}$. It is much simpler since we have fewer rules to follow. To stay continuous (and to follow the third rule), $$f(j)=\frac{r}{j-s}=\alpha$$ $$\Rightarrow \fbox{$r = \alpha(j-s)$}$$ To keep the derivative continuous, $$f'(j)=-\frac{r}{(j-s)^2}=\beta$$ With some substitution and reorganization this gives $$\fbox{$s = \frac{\alpha}{\beta}+j$}$$ Once again you can go back and find the remaining variable, $r$.

The total piecewise function is

$$ f(x) = \begin{cases} ae^{kx^2}+b, & \text{if $0 \leq x \leq j$} \\ \frac{r}{x-s}, & \text{if $j < x$} \end{cases} $$ where $a$, $b$, $k$, $r$, and $s$ are found using the previous equations.

I got inspiration for the left function from herb steinberg's answer. Thank you!

The solution I came up with is computationally intensive so if you are trying to use this as an activation function for a neural network or something of the sort, approximation via power series may be better. I hope that this helps a little!