To find the domain and range of the function \( f(x) = \frac{x^2 + 5x + 6}{x + 2} \), we start by simplifying the function.
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Factor the numerator: The numerator \( x^2 + 5x + 6 \) factors as: \[ x^2 + 5x + 6 = (x + 2)(x + 3) \] Thus, the function simplifies to: \[ f(x) = \frac{(x + 2)(x + 3)}{x + 2} = x + 3 \quad \text{for } x \neq -2 \]
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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: \[ D: {x \in \mathbb{R} \mid x \neq -2} \]
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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}
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