**Question:**

Find all the points of discontinuity of $f$ defined by $f(x)=|x|-|x+1|$.

**Solution:**

Given: $f(x)=|x|-|x+1|$

The two functions, $g$ and $h$, are defined as

$g(x)=|x|$ and $h(x)=|x+1|$

Then, $f=g-h$

The continuity of $g$ and $h$ is examined first.

$g(x)=|x|$ can be written as

$g(x)= \begin{cases}-x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{cases}$

Clearly, $g$ is defined for all real numbers.

Let $c$ be a real number.

Case I:

If $c<0$, then $g(c)=-c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(-x)=-c$

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

So, $g$ is continuous at all points $x<0$.

Case II:

If $c>0$, then $g(c)=c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c} x=c$

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

So, $g$ is continuous at all points $x>0$

Case III:

If $c=0$, then $g(c)=g(0)=0$

$\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0$

$\lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0$

$\therefore \lim _{x \rightarrow 0^{0}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=g(0)$

So, $g$ is continuous at $x=0$

From the above three observations, it can be concluded that $g$ is continuous at all points.

$h(x)=|x+1|$ can be written as

$h(x)= \begin{cases}-(x+1), & \text { if, } x<-1 \\ x+1, & \text { if } x \geq-1\end{cases}$

Clearly, $h$ is defined for every real number.

Let $c$ be a real number.

Case I:

If $c<-1$, then $h(c)=-(c+1)$ and $\lim _{x \rightarrow c} h(x)=\lim _{x \rightarrow c}[-(x+1)]=-(c+1)$

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

So, $h$ is continuous at all points $x>-1$.

Case III:

If $c=-1$, then $h(c)=h(-1)=-1+1=0$

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

$\therefore \lim _{x \rightarrow-1^{-}} h(x)=\lim _{h \rightarrow-1^{+}} h(x)=h(-1)$

So, $h$ is continuous at $x=-1$

From the above three observations, it can be concluded that $h$ is continuous at all points of the real line.

So, $g$ and $h$ are continuous functions.

Thus, $f=g-h$ is also a continuous function.

Therefore, $f$ has no point of discontinuity.