The given function is:

$f(x) = \frac{3}{\sqrt{x - 1}}$

To find the domain of this function, we need to identify all values of $x$ for which the function is defined.

Steps:

1. Square Root Constraint: The square root is only defined for non-negative numbers. Therefore, the expression under the square root, $x - 1$, must be non-negative: $x - 1 \geq 0$ Solving this inequality: $x \geq 1$ Thus, $x$ must be greater than or equal to 1.

2. Denominator Constraint: Since the function involves a denominator, we need to ensure that the denominator is not zero. The denominator is $\sqrt{x - 1}$, and this equals zero when $x = 1$. Therefore, $x \neq 1$.

Conclusion:

Combining both conditions, the function is defined for $x > 1$. Therefore, the domain of the function is: $x \in (1, \infty)$

Would you like more details on this, or do you have any questions?

Here are five related questions for further exploration:

1. How would the domain change if the function was $f(x) = \frac{3}{\sqrt{2x - 5}}$?
2. What is the range of the given function $f(x) = \frac{3}{\sqrt{x - 1}}$?
3. Can we find the inverse of this function? If so, what is it?
4. What happens if the square root is replaced with a cube root?
5. How does this function behave as $x$ approaches infinity?

Tip: When dealing with square root functions, always consider the conditions under which the expression inside the root is non-negative.