If \( f: \mathbb{R} \to \mathbb{R} \) is defined by: \[ f(x) = \begin{cases} \frac{x + 2}{x^2 + 3x + 2} & \text{if } x \in \mathbb{R} - \{-1, -2\} \\ -1 & \text{if } x = -2 \\ 0 & \text{if } x = -1 \end{cases} \] Then \( f(x) \) is continuous on the set:
Step-by-step Solution:
\[ \mathop {\lim }\limits_{x \to - 2} \frac{x + 2}{x^2 + 3x + 2} = \frac{x + 2}{(x + 2)(x + 1)} = \frac{1}{x + 1} = -1 \] Also, \[ f(-2) = -1 \] So, the function is continuous at \(x = -2\). When \(x \to -1\), the limit of the function does not exist. Hence, the function is not continuous at \(x = -1\). The correct answer is \( \mathbb{R} - \{-1\} \).