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