Simple inverse function of $\frac{1-2x}{1+x}$

Question

Just started learning about inverse functions, and got stuck on this one:

$$f(x) = \frac{1-2x}{1+x}$$

So I tried multiplying by $(1+x)$ on both sides and got $y+yx = 1-2x$ but that doesn't seem to lead anywhere. Any help is appreciated.

Thanks in advance

Answer 1

$$y+yx=1-2x\\yx+2x=1-y\\x(y+2)=1-y\\x=\frac{1-y}{y+2}$$

Answer 2

$$ y=\frac{1-2x}{1+x}=\frac{3}{1+x}-2\implies y+2=\frac{3}{1+x}\implies x=\frac{3}{y+2}-1. $$

Answer 3

$$f(x) = \frac{1-2x}{1+x}\\x = \frac{1-2f^{-1}(x)}{1+f^{-1}(x)}\\x+xf^{-1}(x)=1-2f^{-1}(x)\\xf^{-1}(x)+2f^{-1}(x)=f^{-1}(x)(x+2)=1-x\\\therefore f^{-1}(x)=\frac{1-x}{x+2}$$