The problem provided asks for the domain of continuity of the function:

$f(x) = \sqrt{x - 1}$

To determine where $f(x)$ is continuous, we need to find where the function is defined since a function is continuous wherever it is defined and not interrupted.

Step-by-step Solution:

1. Identify the domain of $f(x)$:

   • The square root function $\sqrt{x}$ is defined only when its argument is non-negative.
   • Therefore, for $\sqrt{x - 1}$ to be defined, the expression inside the square root, $x - 1$, must be non-negative:

   $x - 1 \geq 0$

   Solving this inequality:

   $x \geq 1$

2. Determine the domain in interval notation:

   • The domain of $f(x)$ is all $x$ values where $x \geq 1$.

   In interval notation, this is written as:

   $[1, \infty)$

3. Conclusion:

   • The function $f(x) = \sqrt{x - 1}$ is continuous for all $x$ in the interval $[1, \infty)$.

So, the domain of continuity for $f(x)$ is $[1, \infty)$.

Would you like further explanation on any step? Here are five related questions that you might find helpful:

1. What is the general rule for the domain of a square root function?
2. How do you determine continuity for functions involving fractions?
3. How do you solve inequalities involving square roots?
4. What is the significance of continuity in calculus?
5. How would the domain change if the function were $f(x) = \sqrt{1 - x}$?

Tip: When working with square root functions, always check the expression inside the square root to ensure it's non-negative, as it defines the domain of the function.