How to find inverse function of function with parameter?

Question

I'm trying to find the inverse function of this function: $$ f_a(x) = \frac{x^a}{x^a + (1-x)^a} $$

• $a$ - parameter
• $x$ - variable

Is it possible to find if there is a parameter here?
Hope for your help! Thanks in advance!

Answer 1

$$\require{cancel} f_a\circ f_b(x)=\dfrac{\left(\dfrac{x^b}{\cancel{x^b+(1-x)^b}}\right)^a}{\left(\dfrac{x^b}{\cancel{x^b+(1-x)^b}}\right)^a+\left(1-\dfrac{x^b}{\cancel{x^b+(1-x)^b}}\right)^a}\\ =\dfrac{x^{ab}}{x^{ab}+(\cancel{x^b}+(1-x)^b-\cancel{x^b})^a}\\ =\dfrac{x^{ab}}{x^{ab}+(1-x)^{ab}} =f_{ab}(x)$$

So we have $\quad f_a\circ f_b = f_{ab}$

You can verify that since $f_1(x)=\dfrac{x}{x+(1-x)}=x$ then $f_a^{-1}=f_{\frac 1a}$

Answer 2

$$f_a(x) = \frac{x^a}{x^a + (1-x)^a} \implies \frac 1{f_a(x)}=1+\frac{(1-x)^a } {x^a}\implies $$ $$\frac 1{f_a(x)}-1=\Bigg[\frac {1-x}x\Bigg]^a\implies\Bigg[\frac 1{f_a(x)}-1\Bigg]^{\frac 1a}=\frac {1-x}x\implies x=???$$

Answer 3

On the domain you mentioned $(0,1)$, I think you just need to invert the parameter $a$ itself, and it works. $$f^{-1}_a(x) = \frac{x^\frac{1}{a}}{x^\frac{1}{a} + (1-x)^\frac{1}{a}}$$