The function f(x) = 2 + sqrt(x+4) is one to one. Find the common point of the graph of y = f(x) and y = f^-1(x)
1. $\displaystyle D_f = [-4,+\infty)$ and the range is $\displaystyle R_f=[2, +\infty)$
Therefore $\displaystyle D_{f^{-1}}= [2, +\infty)$ and the range of $\displaystyle f^{-1}$ is $\displaystyle R_{f^{-1}} = [-4,+\infty)$
2. The common point of both graphs must be located onn the line y = x (why?)
Therefore solve for x:
$\displaystyle x = 2 + \sqrt{x+4}~,~x\ge 2$
Spoiler: