Question

Find all points of discontinuity of $f$, where $f$ is defined by
$f\left(x\right)=$ $\{\begin{array}{c}|x|+3=-x+3,\text{if}x\le -3\\ -2x,\text{if}-3<x<3\\ 6x+2,\text{if}x\ge 3\end{array}$

Answer 1

The given function is $f\left(x\right)=\{\begin{array}{c}|x|+3=-x+3,\text{if}x\le -3\\ -2x,\text{if}-3<x<3\\ 6x+2,\text{if}x\ge 3\end{array}$
The given function is defined at all the points of the real line.
Let $c$ be a point on the real line.
Case I :
If $c<-3$, then $f\left(c\right)=-c+3$
$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(-x+3)=-c+3$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at all points $x$, such that $x<-3$
Case II :
If $c=-3$, then $f(-3)=-(-3)+3=6$
$\underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3}{lim}(-x+3)=-(-3)+3=6$
$\underset{x\to 3^{+}}{lim}f\left(x\right)=\underset{x\to 3}{lim}(-2x)=-2x(-3)=6$
$\therefore \underset{x\to 3}{lim}f\left(x\right)=f(-3)$
Therefore, $f$ is continuous at $x=-3$
Case III :
If $-3<c<3$, then $f\left(c\right)=-2c$ and $\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(-2x)=-2c$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous in $(-3,3)$.
Case IV :
If $c=3$, then the left hand limit of $f$ at $x=3$ is,
$\underset{x\to 3}{lim}f\left(x\right)=\underset{x\to 3}{lim}(-2x)=-2\times 3=-6$
The right hand limit of $f$ at $x=3$ is,
$\underset{x\to 3}{lim}f\left(x\right)=\underset{x\to 3}{lim}(6x+2)=6\times 3+2=20$
It is observed that the left and right hand limit of $f$ at $x=3$ do not coincide.
Therefore, $f$ is not continuous at $x=3$
Case V :
If $c>3$, then $f\left(c\right)=6c+2$ and $\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(6x+2)=6c+2$
$\therefore$ $\underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore $f$ is continuous at all points $x$, such that $x>3$
Hence, $x=3$ is the only point of discontinuity of $f$.