Find the domain of the range of each of the following real functions:
`f(x)=sqrt((x-5)/(3-x))`

Explanation

To find the domain of the function \( f(x) = \sqrt{\frac{x-5}{3-x}} \), 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:

\[ \frac{x-5}{3-x} \geq 0 \]

### Step 2: Determine when the numerator and denominator are zero

1. **Numerator**: \( x - 5 = 0 \) gives \( x = 5 \).
2. **Denominator**: \( 3 - x = 0 \) gives \( x = 3 \).

### Step 3: Analyze the sign of the expression

We will analyze the sign of \( \frac{x-5}{3-x} \) in the intervals determined by the critical points \( x = 3 \) and \( x = 5 \). The intervals are:

- \( (-\infty, 3) \)
- \( (3, 5) \)
- \( (5, \infty) \)

### Step 4: Test each interval

1. **Interval \( (-\infty, 3) \)**: Choose \( x = 0 \):
\[ \frac{0-5}{3-0} = \frac{-5}{3} < 0 \]

2. **Interval \( (3, 5) \)**: Choose \( x = 4 \):
\[ \frac{4-5}{3-4} = \frac{-1}{-1} = 1 > 0 \]

3. **Interval \( (5, \infty) \)**: Choose \( x = 6 \):
\[ \frac{6-5}{3-6} = \frac{1}{-3} < 0 \]

### Step 5: Include the critical points

- At \( x = 3 \), the denominator becomes zero, so \( f(x) \) is undefined.
- At \( x = 5 \), the expression equals zero, which is acceptable since we can take the square root of zero.

### Conclusion: Domain of the function

From the analysis, the function is defined in the interval:

\[ (3, 5] \]