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(C) Given that $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \ -1, & x = -2 \ 0, & x = -1 \end{cases}$ For $x \in R - \{

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$R - \{-1\}$

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(C) Given that $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \ -1, & x = -2 \ 0, & x = -1 \end{cases}$ For $x \in R - \{-1, -2\}$, $f(x) = \frac{x + 2}{(x + 1)(x + 2)} = \frac{1}{x + 1}$. Now, we check the continuity at $x = -2$ and $x = -1$. At $x = -2$: $\lim_{x ightarrow -2} f(x) = \lim_{x ightarrow -2} \frac{1}{x + 1} = \frac{1}{-2 + 1} = -1$. Since $f(-2) = -1$, we have $\lim_{x ightarrow -2} f(x) = f(-2)$, so $f$ is continuous at $x = -2$. At $x = -1$: $\lim_{x ightarrow -1^-} f(x) = \lim_{x ightarrow -1^-} \frac{1}{x + 1} = -\infty$ and $\lim_{x ightarrow -1^+} f(x) = \lim_{x ightarrow -1^+} \frac{1}{x + 1} = \infty$. Since the limit does not exist at $x = -1$, $f$ is discontinuous at $x = -1$. Therefore, $f$ is continuous on the set $R - \{-1\}$.

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \\ -1, & x = -2 \\ 0, & x = -1 \end{cases}$ then $f$ is continuous on the set

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