Question

Discuss the continuity of the function f, where f is defined by
\(f(x)=\left\{\begin{matrix} 2x, &if\,f(x)<0 \\   0,&if\,0\leq x\leq 1\\   4x,&if\,x>1  \end{matrix}\right.\)

Answer 1

The given function is 
f(x)={2x,if x<0
         0,if 0≤x≤1
         4x,if x>1
   
 The given function f is defined at all the points of the real line. 
Let c be a point on the real line.

Case I:
If c<0,then f(c)=2c 
\(\lim_{x\rightarrow c} f(x)=\lim_{x\rightarrow c}2x=2c\)
∴\(\lim_{x\rightarrow c}\)=f(c)
Therefore, f is continuous at all points x, such that x<0

Case (II)
If c=0,then f(c)=f(0)=0
The left-hand limit of f at x=0 is
limx→0- f(x)=limx→0-(2x)=2x0=0
The right hand limit of f at x=0 is,
\(\lim_{x\rightarrow 0}\)⁺ f(x)=\(\lim_{x\rightarrow 0}(0)\)=0
∴\(\lim_{x\rightarrow 0}f(x)\)=f(0)
Therefore,f is continuous at x=0

 Case(III):
If 0<c<1,then f(x)=0 and
\(\lim_{x\rightarrow c}\) f(x)=\(\lim_{x\rightarrow c}\)(0)=0
∴\(\lim_{x\rightarrow c}\) f(x)=f(c)
Therefore, f is continuous at all points of the interval (0,1).

Case(IV):
If c=1,then f(c)=f(1)=0
The left-hand limit of f at x=1 is,
\(\lim_{x\rightarrow 1^-}\) f(x)=limx→1-(0)=0
The right-hand limit of f at x=1 is,
\(\lim_{x\rightarrow 1^+}\) f(x)=limx→1+(4x)=4x1=4
It is observed that the left and right-hand limits of f at x=1 do not coincide. 
Therefore,f is not continuous at x=1

Case(V):
If c<1,then f(c)=4c and \(\lim_{x\rightarrow c}\) f(x)=\(\lim_{x\rightarrow c}\) (4x)=4c
\(\lim_{x\rightarrow c}\) f(x)=f(c)
Therefore,f is continuous at all points x, such that x>1
Hence,f is not continuous at x=1