Question

Find the domain and the inverse of the following functions:
$f\left(x\right)=\frac{x^2}{1+x^2}$

Answer 1

The function $f\left(x\right)=\frac{x^2}{1+x^2}$ is only defined when $1+x^2\ne 0x^2\ne -1x\in R$

Let $y=\frac{x^2}{1+x^2}{x}^2=y(1+x^2)x^2=y+x^{2}y{x}^2-x^{2}y=y{x}^2(1-y)=y{x}^2=\frac{y}{1-y}x=\sqrt{\frac{y}{1-y}}implies{f}^{-1}\left(x\right)=\sqrt{\frac{x}{1-x}}$