Let f x = x + 2 2 - 2 ,   x ≥ - 2 . Then f - 1 x =

Let $f\left(x\right)={\left(x+2\right)}^2-2,x\ge -2$.

Putting $y={\left(x+2\right)}^2-2$

$y=x^2+4x+4-2$

$y=x^2+4x+2$

$x^2+4x+2-y=0$

$a=1, b=4, c=2-y$

$x=\frac{-4\pm \sqrt{4^2-4\times 1\times \left(2-y\right)}}{2\times 1}$

$x=\frac{-4\pm \sqrt{16-8+4y}}{2\times 1}$

Then, $x=-2+\sqrt{2+y}$,  $x\ge -2$

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