Question

If f$:$R$\to$R is defined by $f\left(x\right)=\{\begin{array}{ccc}\frac{x+2}{x^2+3x+2}& if& x\in R-\{-1,-2\}\\ -1& if& x=-2\\ 0& if& x=-1\end{array}$ then f is continuous on the set.

• A. R
• B. $R-\{-2\}$
• C. $R-\{-1\}$
• D. $R-\{-1,-2\}$

Answer 1

Answer: (C)

$R-\{-1\}$

$f\left(x\right)=\{\begin{array}{ccc}\frac{x+2}{x^2+3x+2}& if& R-\{-1,-2\}\\ -1& & \\ 0& & \end{array}$

If $x\ne -2$

$f\left(x\right)=\frac{x+2}{(x+1)(x+2)}=\frac{1}{x+1}=\frac{1}{-2+1}=-1$ (at $2^{-}$ or $2^{+}$)

which satisfies $f\left(2\right)$ so it is continuous

But $f(-1)=0$ and $f(-1^{-})$ is not defined and so is discontinuous.

$R-\{-1\}$.