The domain of the real valued function `f(x)=sqrt((x-2)/(3-x))` is . .

Options

A. (2,3]
B. [2,3)
C. (2,3)
D. [2,3]

Explanation

To find the domain of the function \( f(x) = \sqrt{\frac{x-2}{3-x}} \), we need to ensure that the expression inside the square root is non-negative (greater than or equal to zero) and that the denominator is not zero.

### Step 1: Set the expression inside the square root greater than or equal to zero
We need to solve the inequality:
\[ \frac{x-2}{3-x} \geq 0 \]

### Step 2: Determine where the numerator and denominator are zero
1. **Numerator**:
\[ x - 2 = 0 \implies x = 2 \]
2. **Denominator**:
\[ 3 - x = 0 \implies x = 3 \]

### Step 3: Analyze the sign of the expression in intervals
We will analyze the sign of \( \frac{x-2}{3-x} \) in the intervals determined by the critical points \( x = 2 \) and \( x = 3 \). The intervals are:
- \( (-\infty, 2) \)
- \( (2, 3) \)
- \( (3, \infty) \)

**Interval 1: \( (-\infty, 2) \)**
- Choose a test point, e.g., \( x = 0 \):
\[ \frac{0-2}{3-0} = \frac{-2}{3} < 0 \]

**Interval 2: \( (2, 3) \)**
- Choose a test point, e.g., \( x = 2.5 \):
\[ \frac{2.5-2}{3-2.5} = \frac{0.5}{0.5} = 1 > 0 \]

**Interval 3: \( (3, \infty) \)**
- Choose a test point, e.g., \( x = 4 \):
\[ \frac{4-2}{3-4} = \frac{2}{-1} < 0 \]

### Step 4: Combine the results
From the analysis:
- The expression \( \frac{x-2}{3-x} \) is non-negative in the interval \( (2, 3) \).
- At \( x = 2 \), \( \frac{x-2}{3-x} = 0 \), which is acceptable since the square root of zero is defined.
- At \( x = 3 \), the denominator becomes zero, which is not acceptable.

### Conclusion
The domain of the function \( f(x) = \sqrt{\frac{x-2}{3-x}} \) is:
\[ [2, 3) \]