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
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 $f$.