Let $f(x)=(x+1)^{2}-1, x\ge -1$. Then the set $\left\{x: f(x)=f^{-1}(x)\right\}$ is-

$\begin{array}{l}f(x)=(x+1)^{2}-1 \\ \text { Let, } y=f(x) \\ \Rightarrow x=f^{-1}(y) \\ y=(x+1)^{2}-1 \\ \Rightarrow(x+1)^{2}=y+1 \\ \Rightarrow x+1= \pm \sqrt{y+1} \\ \Rightarrow x=1 \pm \sqrt{y+1} \\ \Rightarrow f^{-1}(y)=1 \pm \sqrt{y+1}\end{array}$

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

$\begin{array}{l} f(x)=f^{-1}(x) \\ \Rightarrow(x+1)^{2}-1=1 \pm \sqrt{x+1} \\ \Rightarrow(x+1)^{2}-1-1= \pm \sqrt{x+1} \\ \Rightarrow(x+1)^{2}-2= \pm \sqrt{x+1} \\ \Rightarrow\left\{(x+1)^{2}-2\right\}^{2}=x+1\end{array}$

$\begin{array}{l}\Rightarrow(x+1)^{4}-2(x+1)^{2}.2+4=x+1 \\ \Rightarrow(x+1)^{4}-4(x+1)^{2}+4=x+1 \\ \Rightarrow(x+1)^{4}-4(x+1)^{2}-(x+1)+4=0\end{array}$

Let,

$\begin{array}{c} t=x+1 \\ t^{4}-4 t^{2}-t+4=0 \\ f(t)=t^{4}-4 t^{2}-t+4 \end{array}$

$\begin{array}{l}\Rightarrow f(1)=1-4-1+4=0 \\ \therefore t^{4}-4 t^{2}-t+4=0 .\end{array}$

$\begin{array}{l}\Rightarrow t^{3}(t-1)+t^{2}(t-1)-3 t(t-1)-4(t-1)=0 \\ \Rightarrow(t-1)\left(t^{3}+t^{2}-3 t-4\right)=0\end{array}$

$\begin{array}{l}t-1=0 \\ \Rightarrow t=1 \\ \Rightarrow x+1=1 \Rightarrow x=0\end{array}$