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.
The function \( f(x) \) is defined as follows: \[ f(x) = \begin{cases} 2+x, & \text{if } x \geq 0 \\ 2-x, & \text{if } x \leq 0 \end{cases} \] Then function \( f(x) \) at \( x=0 \) is: