Question

Find all points of discontinuity of $f$, where $f$ is defined by
$f\left(x\right)=$ $\{\begin{array}{c}2x+3,ifx\le 2\\ 2x-3,ifx>2\end{array}$

Answer 1

The given function is $f\left(x\right)=\{\begin{array}{c}2x+3,\text{if}x\le 2\\ 2x-3,\text{if}x>2\end{array}$
It is evident that the given function $f$ is defined at all the points of the real line.
Let $c$ be a point on the real line. Then, three cases arise.
(i) $c<2$
(ii) $c>2$
(iii) $c=2$

Case (i)

$c<2$
Then, $f\left(c\right)=2c+3$
$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(2x+3)=2c+3$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at all points $x$, such that $x<2$

Case (ii)

$c>2$
Then, $f\left(c\right)=2c-3$
$\underset{x\to c}{lim}f\left(x\right)=\underset{x\to c}{lim}(2x-3)=2c-3$
$\therefore \underset{x\to c}{lim}f\left(x\right)=f\left(c\right)$
Therefore, $f$ is continuous at all points $x$ such that, $x>2$

Case (iii)

$c=2$
Then, the left hand limit of $f$ at $x=2$ is,
$\underset{x\to 2}{lim}f\left(x\right)=\underset{x\to 2}{lim}(2x+3)=2\times 2+3=7$
The right hand limit of $f$ at $x=2$ is,
$\underset{x\to 2}{lim}f\left(x\right)=\underset{x\to 2}{lim}(2x-3)=2\times 2-3=1$
It is observed that the left and right hand limit of $f$ at $x=2$ do not coincide.
Therefore, $f$ is not continuous at $x=2$

Hence, $x=2$ is the only point of discontinuity of $f$.