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

Solution:

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

The given function f 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 $f(c)=-c+3$

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

 

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

Therefore, $f$ is continuous at all points $x$, such that $x<-3$

Case II:

If $c=-3$, then $f(-3)=-(-3)+3=6$

$\lim _{x \rightarrow-3} f(x)=\lim _{x \rightarrow-3}(-x+3)=-(-3)+3=6$

$\lim _{x \rightarrow-3} f(x)=\lim _{x \rightarrow-3}(-2 x)=-2 \times(-3)=6$

 

$\therefore \lim _{x \rightarrow-3} f(x)=f(-3)$

Therefore, $f$ is continuous at $x=-3$

Case III:

If $-3

$\lim _{x \rightarrow 3} f(x)=\lim _{x \rightarrow 3}(-2 x)=-2 \times 3=-6$

The right hand limit of $f$ at $x=3$ is,

$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3^{+}}(6 x+2)=6 \times 3+2=20$

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 $\mathrm{V}$ :

If $c>3$, then $f(c)=6 c+2$ and $\lim f(x)=\lim (6 x+2)=6 c+2$

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

Therefore, $f$ is continuous at all points $x$, such that $x>3$

Hence, $x=3$ is the only point of discontinuity of $f$.