The domain of the function `f(x)= (sqrt((x+1) (x-3)))/(x-2)` is

Options

A. `[-1,2] cup [3, oo)`
B. `(-1, 2) cup [3, oo)`
C. `[-1, 2] cup [3, oo)`
D. None of these

Answer

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

### Step 1: Determine when the expression under the square root is non-negative
The expression under the square root is \( (x+1)(x-3) \). For the square root to be defined, we need:
\[ (x+1)(x-3) \geq 0 \]

### Step 2: Find the critical points
The critical points occur when \( (x+1) = 0 \) or \( (x-3) = 0 \):
1. \( x + 1 = 0 \) → \( x = -1 \)
2. \( x - 3 = 0 \) → \( x = 3 \)

### Step 3: Test intervals around the critical points
We will test the intervals determined by the critical points \( x = -1 \) and \( x = 3 \):
- Interval 1: \( (-\infty, -1) \)
- Interval 2: \( (-1, 3) \)
- Interval 3: \( (3, \infty) \)

**Testing Interval 1: \( x < -1 \) (e.g., \( x = -2 \))**
\[ (-2 + 1)(-2 - 3) = (-1)(-5) = 5 \geq 0 \quad \text{(valid)} \]

**Testing Interval 2: \( -1 < x < 3 \) (e.g., \( x = 0 \))**
\[ (0 + 1)(0 - 3) = (1)(-3) = -3 < 0 \quad \text{(invalid)} \]

**Testing Interval 3: \( x > 3 \) (e.g., \( x = 4 \))**
\[ (4 + 1)(4 - 3) = (5)(1) = 5 \geq 0 \quad \text{(valid)} \]

### Step 4: Combine the results
From the tests, the intervals where \( (x+1)(x-3) \geq 0 \) are:
- \( (-\infty, -1] \)
- \( [3, \infty) \)

### Step 5: Exclude points where the denominator is zero
The denominator \( x - 2 \) cannot be zero, so we exclude \( x = 2 \).

### Step 6: Final domain
The domain of \( f(x) \) is:
\[ (-\infty, -1] \cup (2, 3] \cup (3, \infty) \]

### Summary of the Domain
Thus, the final domain of the function \( f(x) \) is:
\[ (-\infty, -1] \cup (2, 3) \cup (3, \infty) \]