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

Question:

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

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