Visualize $f(x)$. It is a horizontal shift of $y=\frac{1}{x}$, which has a vertical asymptote at $x = 0$. $f(x)$ also has a vertical asymptote. Where is it, and how will the vertical asymptote restrict the domain?

Visualize $g(x)$. It is a horizontal shift of 
$y=\sqrt{x}$, which does not exist for negative values of $x$. What can you conclude about $g(x)$?

Find the domain of $\left ( \frac{1}{\sqrt{x-4}+2} \right )$

Both the square root and the denominator have restricted domains. Combine.

Refer to hints in part (c).