Find the domain and range of real function ,`f(x)=(x+1)/(x-2)`

Answer

To find the domain and range of the function \( f(x) = \frac{x+1}{x-2} \), we will follow these steps:

### Step 1: Determine the Domain
The domain of a function consists of all the possible input values (x-values) for which the function is defined. For rational functions like \( f(x) \), the function is undefined wherever the denominator equals zero.

1. **Identify the denominator**: In this case, the denominator is \( x - 2 \).
2. **Set the denominator to zero**:
\[ x - 2 = 0 \]
3. **Solve for x**:
\[ x = 2 \]
4. **State the domain**: The function is defined for all real numbers except \( x = 2 \). Therefore, the domain in interval notation is:
\[ \text{Domain: } (-\infty, 2) \cup (2, \infty) \]

### Step 2: Determine the Range
To find the range, we will express \( f(x) \) in terms of \( y \) and solve for \( x \).

1. **Set \( f(x) \) equal to \( y \)**:
\[ y = \frac{x+1}{x-2} \]
2. **Cross-multiply to eliminate the fraction**:
\[ y(x - 2) = x + 1 \]
3. **Rearrange the equation**:
\[ yx - 2y = x + 1 \]
4. **Bring all terms involving \( x \) to one side**:
\[ yx - x = 2y + 1 \]
5. **Factor out \( x \)**:
\[ x(y - 1) = 2y + 1 \]
6. **Solve for \( x \)**:
\[ x = \frac{2y + 1}{y - 1} \]
7. **Identify restrictions on \( y \)**: The function is undefined when the denominator equals zero:
\[ y - 1 = 0 \implies y = 1 \]
8. **State the range**: The function can take all real values except \( y = 1 \). Therefore, the range in interval notation is:
\[ \text{Range: } (-\infty, 1) \cup (1, \infty) \]

### Final Answer
- **Domain**: \( (-\infty, 2) \cup (2, \infty) \)
- **Range**: \( (-\infty, 1) \cup (1, \infty) \)