Find all the points of discontinuity of f defined by

The given function is $f(x)=|x|-|x+1|$

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

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

Therefore, g is continuous at all points x, such that x < 0

Case II:

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

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

Therefore, $g$ is continuous at all points $x$, such that $x>0$

Case III:

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

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

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

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

Therefore, 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

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

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 $\lim _{x \rightarrow c} h(x)=\lim _{x \rightarrow c}[-(x+1)]=-(c+1)$

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

Therefore, h is continuous at all points x, such that x < −1

Case II:

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

Therefore, h is continuous at all points x, such that x > −1

Case III:

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

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

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

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

Therefore, 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.

g and h are continuous functions. Therefore, f = g − h is also a continuous function.

Therefore, f has no point of discontinuity.