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)={

−x	‌ if ‌x<0
x	‌ if ‌≥0

.
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

n→c

g(x)=

lim

x→c

(−x)=−c
∴

lim

x→c

g(x)=g(c)
Therefore, g is a continuous at all points x, such that x<O Case II
If c>, then g(c)=c and

lim

x→c

g(x)=

lim

x→c

x=c
∴

lim

x→c

g(x)=g(c)
Therefore, g is continuous at all points x, such that x>0
Case III
If c=o, then g(c)=g(o)=o

lim

x→0

g(x)=

lim

x→0

(−x)=0

lim

x→0

g(x)=

lim

x→0

(x)=0
∴

lim

x→0

g(x)=

lim

x→0

(x)=g(0)
Therefore, g is continuous at x=o
From the above three observation, it can be concluded that g is continuous at all points.
h(x)=|x+1| can be written as
h(x)‌{

−(x+1)	‌ if ‌c<−1
x+1	‌ if ‌x≥−1

.

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

n→c

h(x)=

lim

x→c

[−(x+1)]=−(c+1)
∴

lim

h→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→c

h(x)=

lim

x→c

(x+1)=c+1
∴

lim

n→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→−1

h(x)=

lim

x→−1

[−(x+1)]=−(−1+1)=0

lim

x→−1

h(x)=

lim

x→−1

(x+1)=(−1+1)=0
∴

lim

x→−1

h(x)=

lim

x→−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=gh is also a continuous function.
Therefore, f has no point of discontinuity.