Let $f:R\to R$ be given by $f\left(x\right)={\left(x+1\right)}^2-1,x\ge -1$ Then $f^{-1}\left(x\right)=$

• A

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

• B

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

• C

  does not exist because f is not one - one

• D

  does not exist because f is not onto