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

Options

A. `(0,oo)`
B. `[0,oo]`
C. `[2,3]`
D. `(2,3)`

Explanation

To find the domain of the function \( f(x) = \sqrt{(2 - x)(x - 3)} \), we need to ensure that the expression inside the square root is non-negative, as the square root of a negative number is not defined in the real number system.

### Step-by-step Solution:

1. **Set the expression inside the square root greater than or equal to zero**:
\[ (2 - x)(x - 3) \geq 0 \]

2. **Identify the critical points**:
The critical points occur when each factor is equal to zero:
\[ 2 - x = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
Thus, the critical points are \( x = 2 \) and \( x = 3 \).

3. **Test intervals around the critical points**:
We will test the intervals determined by the critical points: \( (-\infty, 2) \), \( (2, 3) \), and \( (3, \infty) \).

- **Interval \( (-\infty, 2) \)**:
Choose a test point, say \( x = 0 \):
\[ (2 - 0)(0 - 3) = 2 \times (-3) = -6 \quad (\text{negative}) \]

- **Interval \( (2, 3) \)**:
Choose a test point, say \( x = 2.5 \):
\[ (2 - 2.5)(2.5 - 3) = (-0.5)(-0.5) = 0.25 \quad (\text{positive}) \]

- **Interval \( (3, \infty) \)**:
Choose a test point, say \( x = 4 \):
\[ (2 - 4)(4 - 3) = (-2)(1) = -2 \quad (\text{negative}) \]

4. **Determine where the expression is non-negative**:
From the tests:
- The expression is negative in \( (-\infty, 2) \) and \( (3, \infty) \).
- The expression is non-negative in \( [2, 3] \).

5. **Conclusion**:
The domain of the function \( f(x) = \sqrt{(2 - x)(x - 3)} \) is:
\[ \boxed{[2, 3]} \]