Find all points of discontinuity of f, where f is defined by

Solution:

The given function $f$ is $f(x)=\left\{\begin{array}{l}2 x+3, \text { if } x \leq 2 \\ 2 x-3, \text { if } x>2\end{array}\right.$

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, $f(c)=2 c+3$

$\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}(2 x+3)=2 c+3$

$\therefore \lim _{x \rightarrow c} f(x)=f(c)$

Therefore, $f$ is continuous at all points $x$, such that $x<2$

Case (ii) $c>2$

Then, the left hand limit of $f$ at $x=2$ is,

$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}}(2 x+3)=2 \times 2+3=7$

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 $f$.