Answer
The domain of $g(x)$ is $[0, \infty)$ while the domain of $g(x-k)$ $[k, \infty)$.
The domain of $g(x-k)$ is equal to the domain of $g(x)$ shifted $k$ units to the right.
Work Step by Step
Step 1. The domain of $g(x)=\sqrt x$ is $[0,\infty)$
Step 2. $g(x-k)$ can be obtained by shifting $g(x)$ to the right $k$ units ($k\ge0$).
This means the domain of $g(x-k)$ is equal to the domain of $g(x)$ shifted $k$ units to the right.
Step 3. Thus the domain of $g(x-k)$ is $[0+k,\infty)=[k, \infty)$.
