Question

Let $f:R\to R$ be a function defined by $f\left(x\right)=\frac{x^2}{x^2+1},\forall x\in R$. Find $f^{-1}$

Answer 1

$f\left(x\right)=\frac{x^2}{x^2+1}$

$f\left(f^{-1}\right(x\left)\right)=x$

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

$\Rightarrow \left(f^{-1}\right(x){)}^2=x+x(f^{-1}{)}^2$

$\Rightarrow f^{-1}\left(x{)}^2\right(1-x)=x$

$\Rightarrow f^{-1}\left(x\right)=\sqrt{\frac{x}{1-x}}$