Question

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

Answer 1

Step1:

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

When $x<-1$

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

When $-1\le x<0$

$f\left(x\right)=-x-(x+1)=-x-x-1=-2x-1$

When $x\ge 0$

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

Step2:

At $x=-1$

$LHL=\begin{array}{c}lim\\ x\to -1\end{array}f\left(x\right)=\begin{array}{c}lim\\ x\to -1\end{array}\left(1\right)=1$

$RHL=\begin{array}{c}lim\\ x\to {(-1)}^{+}\end{array}f\left(x\right)=\begin{array}{c}lim\\ x\to {(-1)}^{+}\end{array}(-2x-1)=-2(-1)-1=2-1=2$

$f(-1)=-2(-1)-1=2-1=1$\\ $LHL=RHL=f(-1)$

$f$ is continuous at $x=-1$

Step3:

At $x=0$

$LHL=\begin{array}{c}lim\\ x\to {\left(0\right)}^{-}\end{array}f\left(x\right)=\begin{array}{c}lim\\ x\to {\left(0\right)}^{-}\end{array}(-2x-1)=-2\left(0\right)-1=0-1=-1$

$RHL=\begin{array}{c}lim\\ x\to {\left(0\right)}^{+}\end{array}f\left(x\right)=\begin{array}{c}lim\\ x\to {\left(0\right)}^{+}\end{array}(-1)=-1$

$LHL=RHL=f\left(0\right)$

$f$ is continuous at $x=0$

Hence f is continuous for all $x\in R$