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(C) Given $f(x) = \frac{x+2}{x^2+3x+2} = \frac{x+2}{(x+2)(x+1)} = \frac{1}{x+1}$ for $x 
eq -1, -2$.For $x = -2$,$f(-2) = -1$.Checking continuity at

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$R-\{-1\}$

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(C) Given $f(x) = \frac{x+2}{x^2+3x+2} = \frac{x+2}{(x+2)(x+1)} = \frac{1}{x+1}$ for $x 
eq -1, -2$.For $x = -2$,$f(-2) = -1$.Checking continuity at $x = -2$:$\lim_{x ightarrow -2} f(x) = \lim_{x ightarrow -2} \frac{1}{x+1} = \frac{1}{-2+1} = -1$.Since $\lim_{x ightarrow -2} f(x) = f(-2) = -1$,$f$ is continuous at $x = -2$.Checking continuity 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 $x = -1$.Thus,$f$ is continuous on $R-\{-1\}$.

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

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