Domain \(\{x\geq-2\}\),      Range\(\{f(x)\geq0\}\)

Find the inverse
\begin{align*} Let\quad y & = f(x)\\ y & = \sqrt{x+2}\\ y^2 & = x+2\\ y^2-2 & = x\\ \end{align*}

Therefore

Domain \(\{x\geq0\}\),      Range\(\{f(x)\geq-2\}\)


Solve for \(f(x)=f^{-1}(x)\)

This is true when \(f(x)=x\)

\begin{align*} \sqrt{x+2} & = x\\ x+2 & = x^2\\ 0 & = x^2-x-2\\ 0 & =(x-2)(x+1)\\ So\\ x & = 2,\; x=-1\\ But\;x & =-1\;is\;impossible\;\\ So\;x & =2 \end{align*}