Identify discontinuity for the following function as either a jum…

Question

Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:
$
\begin{aligned}
f(x) & =x^2+x-3 & & \text {, for } x \in[-5,-2) \\
& =x^2-5 & & \text {, for } x \in(-2,5]
\end{aligned}
$

✓

Answer

$f$ is continuous in $[-5,-2)$ and in $(-2,5]$ since it is a polynomial function.
Continuity at $x=-2$
$f(x)=x^2+x-3$, for $x \in[-5,-2)$
$
\therefore \lim _{x \rightarrow-2} f(x)=\lim _{x \rightarrow-2}\left(x^2+x-3\right)=4-2-3=-1
$
Also, $f(x)=x^2+5$, for $x \in[-2,5)$
$
\begin{aligned}
& \therefore \lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow-2}\left(x^2-5\right)=4-5=-1 \\
& \therefore \lim _{x \rightarrow-2^{-}} f(x)=\lim _{x \rightarrow-2^{+}} f(x)=-1 \\
& \therefore \lim _{x \rightarrow 2} f(x)=-1
\end{aligned}
$
But $\mathrm{f}(-2)$ is not defined.
$\therefore \mathrm{f}$ is discontinuous at $\mathrm{x}=-2$
This discontinuity is removable and can be removed by redefining the function as follows:
$
\begin{aligned}
& =x^2+x-3 & & \text { for } x \in[-5,-2) \\
\mathrm{f}(\mathrm{x}) & =x^2-5 & & \text { for } x \in(-2,5] \\
& =-1 & & \text { for } x=-2
\end{aligned}
$

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