Find the domain of the following functions:
`y=sqrt((x-2)/(x+1))`

Answer

To find the domain of the function \( y = \sqrt{\frac{x-2}{x+1}} \), we need to ensure that the expression inside the square root is non-negative and that the denominator is not zero. Let's break this down step by step.

### Step 1: Identify the conditions for the square root
The expression inside the square root, \( \frac{x-2}{x+1} \), must be greater than or equal to zero:
\[ \frac{x-2}{x+1} \geq 0 \]

### Step 2: Find the critical points
To find where the expression is zero or undefined, we set the numerator and denominator to zero:
1. **Numerator:** \( x - 2 = 0 \) gives \( x = 2 \).
2. **Denominator:** \( x + 1 = 0 \) gives \( x = -1 \).

These critical points will help us determine the intervals to test.

### Step 3: Test intervals around the critical points
We will test the sign of \( \frac{x-2}{x+1} \) in the intervals defined by the critical points:
- Interval 1: \( (-\infty, -1) \)
- Interval 2: \( (-1, 2) \)
- Interval 3: \( (2, \infty) \)

**Interval 1: \( (-\infty, -1) \)**
Choose \( x = -2 \):
\[ \frac{-2-2}{-2+1} = \frac{-4}{-1} = 4 \quad (\text{positive}) \]

**Interval 2: \( (-1, 2) \)**
Choose \( x = 0 \):
\[ \frac{0-2}{0+1} = \frac{-2}{1} = -2 \quad (\text{negative}) \]

**Interval 3: \( (2, \infty) \)**
Choose \( x = 3 \):
\[ \frac{3-2}{3+1} = \frac{1}{4} \quad (\text{positive}) \]

### Step 4: Determine the intervals where the expression is non-negative
From our tests:
- In \( (-\infty, -1) \), the expression is positive.
- In \( (-1, 2) \), the expression is negative.
- In \( (2, \infty) \), the expression is positive.

### Step 5: Include the critical points
- At \( x = 2 \), \( \frac{2-2}{2+1} = 0 \) (included since we need \( \geq 0 \)).
- At \( x = -1 \), the expression is undefined (not included).

### Step 6: Write the domain
Thus, the domain of the function is:
\[ (-\infty, -1) \cup [2, \infty) \]

### Summary of the Domain
The domain of the function \( y = \sqrt{\frac{x-2}{x+1}} \) is:
\[ \text{Domain: } (-\infty, -1) \cup [2, \infty) \]