If $f:R\to R$  be given by$f\left(x\right)={\left(x+1\right)}^2-1,x\ge -1$  then the set of values of x for which $f\left(x\right)=f^{-1}x$  is given by

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

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

$\left(x+1\right)=\sqrt{y+1}$

$x=\sqrt{y+1}-1$

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

Here $f=f^{-1}$

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

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