Question

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

Answer 1

\left\\{\begin{matrix} -2, &if\,x\leq-1 \\\ 2x,&if\,-1<x\leq 1 \\\ 2,&if\, x>1 \end{matrix}\right.

The given function is defined at all points of the real line.
Let c be a point on the real line.

Case I:
If c<-1,then f(c)=-5
$\lim_{x\rightarrow c}$ f(x)=$\lim_{x\rightarrow c}$ f(-2)=-2
∴$\lim_{x\rightarrow c}$ f(x)=f(c)
Therefore,f is continuous at all points x, such that x<-1

Case (II)
If c=-1,then f(c)=f(-1)=-2
The left-hand limit of f at x=-1 is
$\lim_{x\rightarrow 1^-}$ f(x)=$\lim_{x\rightarrow 1^-}$(-2)=-2
The right-hand limit of f at x=-1 is,
$\lim_{x\rightarrow 1^+}$ f(x)=$\lim_{x\rightarrow 1^+}$(2x)=2(-1)=-2
∴$\lim_{x\rightarrow -1}$ f(x)=f(-1)
Therefore,f is continuous at x=-1

Case(III):
If-1<c<1,then f(c)=2c and
$\lim_{x\rightarrow c}$ f(x)=$\lim_{x\rightarrow c}$(2x)=2c
∴$\lim_{x\rightarrow c}$=f(c)
Therefore, f is continuous at all points of the interval (-1,1).

Case(IV):
If c=1,then f(c)=f(1)=2x1=2
The left-hand limit of f at x=1 is,
$\lim_{x\rightarrow 1^-}$ f(x)=$\lim_{x\rightarrow 1^-}$2x)=2x1=2
The right-hand limit of f at x=1 is,
$\lim_{x\rightarrow 1^+}$ f(x)=$\lim_{x\rightarrow 1^+}$(2)=2
∴$\lim_{x\rightarrow 1}$ f(x)=f(c)
Therefore,f is continuous at x=2

Case(V):
If c>1,then f(c)=2 and $\lim_{x\rightarrow c}$ f(x)=$\lim_{x\rightarrow c}$ (2)=2
limx→c f(x)=f(c)
Therefore, f is continuous at all points x, such that x>1
Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.