inverse of $y(x) = 1 - \exp( - (\alpha x + \beta x^2 )) $

Question

I'd like to compute the inverse of $y(x) = 1 - \exp( - (\alpha x + \beta x^2 )) $. I used to know a method but I can't remember how to do it. I am stuck at the step where I have:

$$- \ln( 1 - y) = x ( \alpha + \beta x ) $$

The function is not itself invertible, but I am quite positive about the idea that it is if one sets some conditions on $\alpha$ and $\beta$. The function is 2-piecewise monotone and since I am only interested about $x > 0$, if the extremum is below zero, my function is invertible on the positive axis.

How can I find such inverse ?

Cheers.

Answer 1

As suggested in the comments we simply have

$$ \beta x^2+\alpha x+ \ln( 1 - y)=0 $$

$$\implies x=\frac{-\alpha+\sqrt{\alpha^2-4\beta\ln(1-y)}}{2\beta}\: x=\frac{-\alpha-\sqrt{\alpha^2-4\beta\ln(1-y)}}{2\beta}$$

Answer 2

Alternatively, when you get

$${\alpha x + \beta x^2 = -\ln(1-y)}$$

Divide through by ${\beta}$:

$${\Rightarrow x^2 + \frac{\alpha }{\beta }x = \frac{-1}{\beta}\ln(1-y)}$$

Then complete the square:

$${\left(x + \frac{\alpha}{2\beta}\right)^2 - \frac{\alpha^2}{4\beta^2} = \frac{-1}{\beta}\ln(1-y)}$$

$${\Rightarrow \left(x + \frac{\alpha}{2\beta}\right)^2 = \frac{-1}{\beta}\ln(1-y) + \frac{\alpha^2}{4\beta^2}}$$

$${\Rightarrow x + \frac{\alpha}{2\beta} = \pm\sqrt{\frac{-1}{\beta}\ln(1-y) + \frac{\alpha^2}{4\beta^2}}}$$

And so finally

$${x = -\frac{\alpha}{2\beta}\pm\sqrt{\frac{-1}{\beta}\ln(1-y) + \frac{\alpha^2}{4\beta^2}}}$$

Now this isn't technically a "function", since it maps one input value to multiple output values - so saying

$${f^{-1}(x)=-\frac{\alpha}{2\beta}\pm\sqrt{\frac{-1}{\beta}\ln(1-x) + \frac{\alpha^2}{4\beta^2}}}$$

is not really technically correct - you'd have to pick which branch you are interested in.