`f(x) = sqrt([((x + 1) (x-3))/((x-2))]` is a real valued function in the domain

Options

A. `]-oo ,-1] uu [3,oo[`
B. `]-oo,-1] uu ]2,3]`
C. `[-1 ,+2 ] uu [3,oo[`
D. None of these

Answer

To find the domain of the function \( f(x) = \sqrt{\frac{(x + 1)(x - 3)}{(x - 2)}} \), we need to ensure that the expression inside the square root is non-negative, as well as that the denominator is not zero.

### Step-by-Step Solution:

1. **Set the Expression Inside the Square Root Greater Than or Equal to Zero:**
\[ \frac{(x + 1)(x - 3)}{(x - 2)} \geq 0 \]
This means that the numerator \((x + 1)(x - 3)\) must be non-negative, and the denominator \((x - 2)\) must be positive (to avoid division by zero).

2. **Find the Critical Points:**
We set the numerator and denominator to zero to find the critical points:
- \(x + 1 = 0 \Rightarrow x = -1\)
- \(x - 3 = 0 \Rightarrow x = 3\)
- \(x - 2 = 0 \Rightarrow x = 2\)

The critical points are \(x = -1\), \(x = 2\), and \(x = 3\).

3. **Create a Number Line:**
We will analyze the intervals created by these critical points:
\[ (-\infty, -1), \quad (-1, 2), \quad (2, 3), \quad (3, \infty) \]

4. **Test Each Interval:**
- **Interval \((- \infty, -1)\)**: Choose \(x = -2\)
\[ \frac{(-2 + 1)(-2 - 3)}{-2 - 2} = \frac{(-1)(-5)}{-4} = \frac{5}{-4} < 0 \quad \text{(not valid)} \]
- **Interval \((-1, 2)\)**: Choose \(x = 0\)
\[ \frac{(0 + 1)(0 - 3)}{0 - 2} = \frac{(1)(-3)}{-2} = \frac{-3}{-2} > 0 \quad \text{(valid)} \]
- **Interval \((2, 3)\)**: Choose \(x = 2.5\)
\[ \frac{(2.5 + 1)(2.5 - 3)}{2.5 - 2} = \frac{(3.5)(-0.5)}{0.5} = \frac{-1.75}{0.5} < 0 \quad \text{(not valid)} \]
- **Interval \((3, \infty)\)**: Choose \(x = 4\)
\[ \frac{(4 + 1)(4 - 3)}{4 - 2} = \frac{(5)(1)}{2} = \frac{5}{2} > 0 \quad \text{(valid)} \]

5. **Include Critical Points:**
- At \(x = -1\): \(f(-1) = \sqrt{\frac{0 \cdot (-4)}{-3}} = 0\) (valid)
- At \(x = 2\): The denominator becomes zero, so it is not included.
- At \(x = 3\): \(f(3) = \sqrt{\frac{(4)(0)}{1}} = 0\) (valid)

6. **Combine Valid Intervals:**
The valid intervals are \([-1, 2)\) and \((3, \infty)\). Therefore, the domain of \(f(x)\) is:
\[ \text{Domain} = [-1, 2) \cup [3, \infty) \]

### Final Answer:
The domain of the function \( f(x) \) is:
\[ [-1, 2) \cup [3, \infty) \]