Question

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

Answer 1

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)=\left\\{\begin{matrix} -x &if\,x<0 \\\ x&if\,x\geq 0 \end{matrix}\right.

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
∴$\lim_{x\rightarrow c}$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
∴$\lim_{x\rightarrow c}$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
∴$\lim_{x\rightarrow 0^-}$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)={-(x+1),if x<-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 $\lim_{x\rightarrow c}$ h(x)=$\lim_{x\rightarrow c}$[(-x+1))=-(c+1)
∴$\lim_{x\rightarrow c}$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
∴$\lim_{x\rightarrow c}$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
∴$\lim_{x\rightarrow 1^-}$h(x)=$\lim_{x\rightarrow 1^+}$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.