Originally Posted by
HallsofIvy Unless you have some restriction on the possible values of x (on the domain) this function does not have an inverse!
RRH gets to $\displaystyle y^2= x+ 4$ and then says that $\displaystyle y= \sqrt{x+ 4}$ but there are two values of y whose square is x+ 4.
If the original function is "$\displaystyle f(x)= x^2- 4$ with domain $\displaystyle x\ge 0$", THEN $\displaystyle f^{-1}(x)= \sqrt{x+ 4}$ with domain $\displaystyle x\ge -2$.
If the original function is "$\displaystyle f(x)= x^2- 4$ with domain $\displaystyle x\le 0$, THEN $\displaystyle f^{-1}(x)= -\sqrt{x+ 4}$ with domain $\displaystyle x\ge -2$
RRH is also wrong when he writes $\displaystyle \sqrt{x+ 4}= \sqrt{x}+ 2$. For example, if x= 1, $\displaystyle \sqrt{1+ 4}= \sqrt{5}$ which is certainly not equal to $\displaystyle \sqrt{1}+ 2= 3$.
Follow the pattern RRH used (which is correct). From $\displaystyle y= \frac{1}{x-2}$ (which has natural domain all real numbers except 2), the inverse function is given by $\displaystyle x= \frac{1}{y- 2}$, where I have swapped x and y. Now solve for y.