Find the domain and range of the function `f(x)=(x^(2)-9)/(x-3)`

Explanation

To find the domain and range of the function \( f(x) = \frac{x^2 - 9}{x - 3} \), we will follow these steps:

### Step 1: Identify the Domain
The function \( f(x) \) is a rational function, which is defined as long as the denominator is not equal to zero.

1. **Set the denominator to zero**:
\[ x - 3 = 0 \]
Solving this gives:
\[ x = 3 \]

2. **Determine the domain**:
The function is defined for all real numbers except \( x = 3 \). Therefore, the domain can be expressed in interval notation as:
\[ \text{Domain} = (-\infty, 3) \cup (3, \infty) \]

### Step 2: Simplify the Function
Next, we simplify the function to find its range.

1. **Factor the numerator**:
The numerator \( x^2 - 9 \) can be factored as:
\[ x^2 - 9 = (x - 3)(x + 3) \]

2. **Rewrite the function**:
Substituting the factored form into the function gives:
\[ f(x) = \frac{(x - 3)(x + 3)}{x - 3} \]
For \( x \neq 3 \), we can cancel \( x - 3 \):
\[ f(x) = x + 3 \]

### Step 3: Determine the Range
Now, we will find the range of the simplified function \( f(x) = x + 3 \).

1. **Identify the values of \( f(x) \)**:
The function \( f(x) = x + 3 \) is a linear function, which means it can take any real number value except for the value that corresponds to \( x = 3 \).

2. **Calculate the value at \( x = 3 \)**:
If \( x = 3 \):
\[ f(3) = 3 + 3 = 6 \]
Since \( x = 3 \) is not in the domain, \( f(x) \) cannot equal 6.

3. **Determine the range**:
Therefore, the range of \( f(x) \) is all real numbers except 6, which can be expressed in interval notation as:
\[ \text{Range} = (-\infty, 6) \cup (6, \infty) \]

### Final Answer
- **Domain**: \( (-\infty, 3) \cup (3, \infty) \)
- **Range**: \( (-\infty, 6) \cup (6, \infty) \)