The domain and range (Range = f (x): x `in` domain off ) of the function `f (x)=(x^2 -3x +2)/(x^2 +x -6)`are

Answer

To find the domain and range of the function \( f(x) = \frac{x^2 - 3x + 2}{x^2 + x - 6} \), we will follow these steps:

### Step 1: Determine the Domain
The domain of a function is the set of all possible input values (x-values) for which the function is defined. For a rational function, we need to ensure that the denominator is not equal to zero.

1. **Set the denominator equal to zero**:
\[ x^2 + x - 6 = 0 \]

2. **Factor the quadratic**:
\[ (x + 3)(x - 2) = 0 \]

3. **Find the values of x that make the denominator zero**:
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

4. **State the domain**:
The function is undefined at \( x = -3 \) and \( x = 2 \). Therefore, the domain of \( f(x) \) is:
\[ \text{Domain} = \{ x \in \mathbb{R} \mid x \neq -3, x \neq 2 \} \]

### Step 2: Determine the Range
To find the range, we need to analyze the values that \( f(x) \) can take.

1. **Rewrite the function**:
We can simplify \( f(x) \):
\[ f(x) = \frac{(x - 1)(x - 2)}{(x + 3)(x - 2)} \]
Here, we notice that \( x - 2 \) cancels out, but we must remember that \( x = 2 \) is not in the domain.

2. **Simplified function**:
The function simplifies to:
\[ f(x) = \frac{x - 1}{x + 3} \quad \text{for } x \neq 2 \]

3. **Find horizontal asymptotes**:
As \( x \) approaches infinity, the function approaches:
\[ \lim_{x \to \infty} f(x) = 1 \]
This suggests that \( y = 1 \) is a horizontal asymptote.

4. **Determine values at critical points**:
We check the values of \( f(x) \) at the points where the function is undefined:
- \( f(-3) \) is undefined.
- \( f(2) \) is undefined.

5. **Check for values that cannot be reached**:
To find if there are specific values that \( f(x) \) cannot take, we set:
\[ \frac{x - 1}{x + 3} = y \]
Rearranging gives:
\[ y(x + 3) = x - 1 \quad \Rightarrow \quad x(y - 1) = -3y - 1 \]
\[ x = \frac{-3y - 1}{y - 1} \]
This expression is undefined when \( y = 1 \), indicating that \( f(x) \) cannot equal 1.

6. **Final range**:
Since \( f(x) \) approaches 1 but never reaches it, and we also have \( f(2) \) undefined, we conclude:
\[ \text{Range} = \{ y \in \mathbb{R} \mid y \neq 1 \} \]

### Final Answer:
- **Domain**: \( \{ x \in \mathbb{R} \mid x \neq -3, x \neq 2 \} \)
- **Range**: \( \{ y \in \mathbb{R} \mid y \neq 1 \} \)