Answer :

To determine the domain of the function \( g(x) = \frac{\sqrt{x-2}}{x^2 - x} \), we need to analyze the conditions that ensure the function is defined. The domain is the set of all possible values of \( x \) for which \( g(x) \) is defined.

Step-by-step analysis:

1. Numerator Analysis:
The numerator is \( \sqrt{x-2} \).
- A square root function only has real values when the expression inside the square root is non-negative.
- Thus, \( x - 2 \geq 0 \).

Therefore, we get:
[tex]\[ x \geq 2 \][/tex]

2. Denominator Analysis:
The denominator is \( x^2 - x \).
- For the function to be defined, the denominator cannot be zero because division by zero is undefined.
- Factor the denominator: \( x^2 - x = x(x - 1) \).

Set the denominator equal to zero to find the values that make the denominator zero:
[tex]\[ x(x - 1) = 0 \][/tex]
Solving this equation gives:
[tex]\[ x = 0 \quad \text{or} \quad x = 1 \][/tex]
Therefore, \( x \neq 0 \) and \( x \neq 1 \).

3. Combined conditions:
We combine the conditions from the numerator and the denominator to find the domain.
- From the numerator: \( x \geq 2 \)
- From the denominator: \( x \neq 0 \) and \( x \neq 1 \).

The condition \( x \geq 2 \) already excludes the values \( x = 0 \) and \( x = 1 \) because \( 0 < 2 \) and \( 1 < 2 \).

Therefore, we only need to consider the condition \( x \geq 2 \).

Conclusion:
The domain of the function \( g(x) = \frac{\sqrt{x-2}}{x^2 - x} \) is all \( x \) such that:
[tex]\[ x \geq 2 \][/tex]

So, the domain of [tex]\( g(x) \)[/tex] is [tex]\( x \geq 2 \)[/tex].