Let f : [−1, ∞) → [−1, ∞) be given by

Injectivity : Let $\mathrm{x}$ and $\mathrm{y} \in[-1, \infty)$, such that

$f(x)=f(y)$

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

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

$\Rightarrow(x+1)=(y+1)$

$\Rightarrow x=y$

So, $f$ is a injection.

Surjectivity : Let $\mathrm{y} \in[-1, \infty)$.

Then, $\mathrm{f}(\mathrm{x})=\mathrm{y}$

$\Rightarrow(\mathrm{x}+1)^{2}-1=\mathrm{y}$

$\Rightarrow \mathrm{x}+1=\sqrt{\mathrm{y}+1}$

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

Clearly, $\mathrm{x}=\sqrt{\mathrm{y}+1}-1$ is real for all $\mathrm{y} \geq-1$.

Thus, every element $y \in[-1, \infty)$ has its pre-image $x \in[-1, \infty)$ given by $x=\sqrt{y+1}-1$.

$\Rightarrow f$ is a surjection.

So, $f$ is a bijection.

Hence, $f$ is invertible.

Let $f^{-1}(x)=y$                    ...(1)

$\Rightarrow f(y)=x$

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

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

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

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

$\Rightarrow f^{-1}(x)=\pm \sqrt{x+1}-1 \quad[$ from $(1)]$

$f(x)=f^{-1}(x)$

$\Rightarrow(x+1)^{2}-1=\pm \sqrt{x+1}-1$

$\Rightarrow(x+1)^{2}=\pm \sqrt{x+1}$

$\Rightarrow(x+1)^{4}=x+1$

$\Rightarrow(x+1)\left[(x+1)^{3}-1\right]=0$

$\Rightarrow x+1=0$ or $(x+1)^{3}-=0$

$\Rightarrow x=-1$ or $(x+1)^{3}=1$

$\Rightarrow x=-1$ or $x+1=1$

$\Rightarrow x=-1$ or $\mathrm{x}=0$

$\Rightarrow S=\{0,-1\}$