Question

What are the domain and range of the function f of x is equal to the quantity x squared plus 5x plus 6 end quantity divided by the quantity x plus 2 end quantity?

D: {x ∊ ℝ | x ≠ 2}; R: {y ∊ ℝ | y ≠ −1}
D: {x ∊ ℝ | x ≠ − 3}; R: {y ∊ ℝ | y ≠ 0}
D: {x ∊ ℝ | x ≠ −2}; R: {y ∊ ℝ | y ≠ 1}
D: {x ∊ ℝ | x ≠ −2}; R: {y ∊ ℝ | y ≠ 0}

Answer 1

To find the domain and range of the function \( f(x) = \frac{x^2 + 5x + 6}{x + 2} \), we start by simplifying the function.

1. **Factor the numerator**:
The numerator \( x^2 + 5x + 6 \) factors as:
\[
x^2 + 5x + 6 = (x + 2)(x + 3)
\]
Thus, the function simplifies to:
\[
f(x) = \frac{(x + 2)(x + 3)}{x + 2} = x + 3 \quad \text{for } x \neq -2
\]

2. **Determine the domain**:
The function \( f(x) \) is not defined where the denominator is zero, so we set the denominator equal to zero:
\[
x + 2 = 0 \implies x = -2
\]
Therefore, the domain \( D \) is:
\[
D: \{x \in \mathbb{R} \mid x \neq -2\}
\]

3. **Determine the range**:
The simplified function \( f(x) = x + 3 \) is a linear function that is defined for all \( x \) except \( x = -2 \). To find the corresponding value when \( x = -2 \):
\[
f(-2) = -2 + 3 = 1
\]
Thus, the function will approach 1 but never reach it when \( x \) approaches -2. Therefore, the range \( R \) is:
\[
R: \{y \in \mathbb{R} \mid y \neq 1\}
\]

Putting this together, we find:
- Domain: \( D: \{x \in \mathbb{R} \mid x \neq -2\} \)
- Range: \( R: \{y \in \mathbb{R} \mid y \neq 1\} \)

Thus, the correct option is:
**D: {x ∊ ℝ | x ≠ −2}; R: {y ∊ ℝ | y ≠ 1}**