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 determine the values of \( x \) for which the expression inside the square root is non-negative, since 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**:
\[ (x-2)(4-x) \geq 0 \]

2. **Identify the critical points**:
The critical points occur when the expression equals zero:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ 4 - x = 0 \quad \Rightarrow \quad x = 4 \]
So, the critical points are \( x = 2 \) and \( x = 4 \).

3. **Test intervals around the critical points**:
We will test the intervals determined by the critical points:
- \( (-\infty, 2) \)
- \( (2, 4) \)
- \( (4, \infty) \)

- **Interval \( (-\infty, 2) \)**: Choose \( x = 0 \)
\[ (0-2)(4-0) = (-2)(4) = -8 \quad (\text{negative}) \]
- **Interval \( (2, 4) \)**: Choose \( x = 3 \)
\[ (3-2)(4-3) = (1)(1) = 1 \quad (\text{positive}) \]
- **Interval \( (4, \infty) \)**: Choose \( x = 5 \)
\[ (5-2)(4-5) = (3)(-1) = -3 \quad (\text{negative}) \]

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

5. **Include the endpoints**:
Since the expression equals zero at \( x = 2 \) and \( x = 4 \), we include these points in the domain.

6. **Write the domain**:
Therefore, the domain of the function is:
\[ \text{Domain} = [2, 4] \]