Question

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

$f\left(x\right)=\{\begin{array}{c}2x+3ifx\le 2\\ 2x-3ifx>2\end{array}$

Answer 1

We have $f\left(x\right)=\{\begin{array}{c}2x+3ifx\le 2\\ 2x-3ifx>2\end{array}$

Case$1$

At $x=2$

$f$ is continuous at $x=2$

if L.H.L$=$R.H.L$=f\left(2\right)$

${lim}_{x\to 2^{-}}f\left(x\right)={lim}_{x\to 2^{+}}f\left(x\right)=f\left(2\right)$

L.H.L$={lim}_{x\to 2^{-}}f\left(x\right)$

$={lim}_{x\to 2^{-}}(2x+3)$ as $x<2$

$=2\times 2+3=4+3=7$

R.H.L$={lim}_{x\to 2^{+}}f\left(x\right)$

$={lim}_{x\to 2^{+}}(2x-3)$ as $x>2$

$=2\times 2-3=4-3=1$

Since L.H.L$\ne$R.H.L

$\therefore f$ is not continuous at $x=2$

Case$2$

At $x=c$ where $c<2$

$f\left(x\right)=2x+3$ as $x=c,$ where $c<2$

$f$ is continuous at $x=c$

if ${lim}_{x\to c}f\left(x\right)=f\left(c\right)$

L.H.L$={lim}_{x\to c}f\left(x\right)={lim}_{x\to c}(2x+3)=2c+3$

$f\left(c\right)=2c+3$

Hence ${lim}_{x\to c}f\left(x\right)=f\left(c\right)$

$\therefore f$ is continuous at $x=c$ where $c<2$

Thus,$f$is continuous for all real number less than $2$

Case $3$

At $x=c$ where $c>2$

$\therefore f\left(x\right)=2x-3$ as $x=c,c>2$

$f$ is continuous at $x=c$

${lim}_{x\to c}f\left(x\right)=f\left(c\right)$

L.H.L$={lim}_{x\to c}f\left(x\right)={lim}_{x\to c}(2x-3)=2c-3$

$f\left(c\right)=2c-3$

Hence ${lim}_{x\to c}f\left(x\right)=f\left(c\right)$

$\Rightarrow f$ is continuous at $x=c$ where $c>2$

$\Rightarrow f$ is continuous at all real number greater than $2$

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

$\Rightarrow f$ is continuous at all real number except $2$

Thus, $f$ is continuous for $x\in R-\left\{2\right\}$