Question

Define a continuity of a function at a point. Find all the points of discontinuity of f(x) defined by $f\left(x\right)=\mid x\mid -\mid x+1\mid .$

Answer 1

Consider the given function:

$f\left(x\right)=\left[x\right]-[x+1].$

$whenx>0$

$f\left(x\right)=x-(x-1)where\left[x\right]=\left\{{}_{-x,x=0}^{x,x\ge 0}\right\}$

$-x-x+1=x-x-1$

$x-x+1=x-x-1$

$x=-1$

$whenx<-1where[x+1]=\left\{{}_{-(x+1),x+1<0}^{(x+1),x+\ge 0}\right\}$

$f\left(x\right)=-x-[-(x+1\left)\right]\left\{{}_{-(x+1),x<1}^{(x+1),x\ge -1}\right\}$

$=-x+x+1$

$=1$

$when-1\le x<0$

$f\left(x\right)=-x-(x+1)$

$=-x-x+1$

$=-2x+1$

$Nowf\left(x\right)=\left\{{}_{-2x-1if-1\le x\le 0}^{1ifx\le -1}\right\}$

$Checkingcontinuityatx=0$

$Afunctioniscontinuousatx=0$

$ifL.H.L=R.H.L=f\left(0\right)$

${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0^{+}}f\left(x\right)=f\left(0\right)$

$L.H.L={lim}_{x\to 0^{-}}f\left(x\right)$

${lim}_{x\to o^{-}}(-2x-1)$

$putx=0$

$=-2\times 0-1=-1$

$R.H.L={lim}_{x\to 0^{+}}f\left(x\right)$

${lim}_{x\to 0^{+}}(-1)=-1$

$\&f\left(x\right)=-1$

$sof\left(0\right)=-1$

$ThusL.H.L=R.H.L=f\left(0\right)$

$Hencef\left(x\right)iscontinuousatx=0$