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Let $f(x) \begin{cases}=|x|+3, & \text { if } x \leq-3 \\ =-2 x, & \text { if }-3 < x < 3, \text { then } \\ =6 x+2, & \text { if } x \geq 3\end{cases}$
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Verified Answer
The correct answer is:
$f(x)$ is continuous at $x=-3$ but discontinuous at $x=3$
We have $f(x)=-x+3$, if $x \leq-3$
$$
\begin{aligned}
& =-2 x \text {, if }-3 < x < 3 \\
& =6 \mathrm{x}+2 \text {, if } \mathrm{x} \geq 3 \\
& \underset{x \rightarrow 3^{-}}{f(x)}=-(-3)+3=6 \text { and } \underset{x \rightarrow 3^{+}}{f(x)}=-2(-3)=6 \text { and } f(-3) \\
& =-(-3)+3=6 \\
&
\end{aligned}
$$
Thus $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=-3$
$$
\underset{x \rightarrow 3^{-}}{f(x)}=-2(3)=-6 \quad \text { and } \underset{x \rightarrow 3^{+}}{f(x)}=6(3)+2=20
$$
Thus $\mathrm{f}(\mathrm{x})$ is not continuous at $\mathrm{x}=3$.
$$
\begin{aligned}
& =-2 x \text {, if }-3 < x < 3 \\
& =6 \mathrm{x}+2 \text {, if } \mathrm{x} \geq 3 \\
& \underset{x \rightarrow 3^{-}}{f(x)}=-(-3)+3=6 \text { and } \underset{x \rightarrow 3^{+}}{f(x)}=-2(-3)=6 \text { and } f(-3) \\
& =-(-3)+3=6 \\
&
\end{aligned}
$$
Thus $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=-3$
$$
\underset{x \rightarrow 3^{-}}{f(x)}=-2(3)=-6 \quad \text { and } \underset{x \rightarrow 3^{+}}{f(x)}=6(3)+2=20
$$
Thus $\mathrm{f}(\mathrm{x})$ is not continuous at $\mathrm{x}=3$.
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