Find the domain of the following functions:
`sqrt((x-2)(4-x))`

Answer

To find the domain of the function \( f(x) = \sqrt{(x-2)(4-x)} \), 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 set of real numbers.

### Step-by-Step Solution:

1. **Set the expression inside the square root greater than or equal to zero:**
\[ (x-2)(4-x) \geq 0 \]

2. **Identify the critical points:**
The critical points occur when either factor is zero:
- \( x - 2 = 0 \) gives \( x = 2 \)
- \( 4 - x = 0 \) gives \( x = 4 \)

3. **Determine the intervals:**
The critical points divide the number line into intervals:
- \( (-\infty, 2) \)
- \( [2, 4] \)
- \( (4, \infty) \)

4. **Test each interval:**
We will test a point from each interval to see if the product \( (x-2)(4-x) \) is non-negative.

- For the interval \( (-\infty, 2) \), choose \( x = 0 \):
\[ (0-2)(4-0) = (-2)(4) = -8 \quad (\text{negative}) \]

- For the interval \( [2, 4] \), choose \( x = 3 \):
\[ (3-2)(4-3) = (1)(1) = 1 \quad (\text{positive}) \]

- For the interval \( (4, \infty) \), choose \( x = 5 \):
\[ (5-2)(4-5) = (3)(-1) = -3 \quad (\text{negative}) \]

5. **Combine the results:**
The product \( (x-2)(4-x) \) is non-negative in the interval \( [2, 4] \).

6. **Include the endpoints:**
Since the square root is defined at the endpoints (where the product is zero), we include \( x = 2 \) and \( x = 4 \).

### Conclusion:
The domain of the function \( f(x) = \sqrt{(x-2)(4-x)} \) is:
\[ \text{Domain} = [2, 4] \]