Question

Find all points of discontinuity of f,where f is defined by 

\[f(x) =   \begin{cases}     |x|+3,      & \quad \text{if } x\  {\leq-3}\\     -2,  & \quad \text{if } -3<x<3 \text{} \\ 6x+2, & \quad \text{if } x\  {\geq 3}  \end{cases}\]

Answer 1

\(f(x) =   \begin{cases}     |x|+3,      & \quad \text{if } x\  {\leq-3}\\     -2,  & \quad \text{if } -3<x<3 \text{} \\ 6x+2, & \quad \text{if } x\  {\geq 3}  \end{cases}\)

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<-3,then f(c) = -c+3
\(\lim\limits_{x \to c}\) f(x) = \(\lim\limits_{x \to c}\) (-x+3) = -c+3
∴\(\lim\limits_{x \to c}\) f(x) = f(c)
Therefore, f is continuous at all points x, such that x<−3

Case (ii):
If c=-3
Then f(-3) = -(-3)+3 = 6
\(\lim\limits_{x \to 3^-}\) f(x) = \(\lim\limits_{x \to 3^-}\)(-x+3) = -(-3)+3 = 6
\(\lim\limits_{x \to 3^+}\) f(x) = \(\lim\limits_{x \to 3^+}\)(-2x) = -2x(-3) = 6
∴\(\lim\limits_{x \to -3}\) f(x) = f(-3)
Therefore,f is continuous at x = −3

Case(iii):
If -3<c<3,then f(c) = -2c and \(\lim\limits_{x \to c}\) f(x) = \(\lim\limits_{x \to c}\) (-2x) = -2c
∴\(\lim\limits_{x \to c}\) f(x) = f(c)
Therefore,f is continuous in (−3,3).

Case IV: 
If c = 3, then the left hand limit of f at x = 3 is,
\(\lim\limits_{x \to 3^-}\) f(x) = \(\lim\limits_{x \to 3^-}\)(-2x) = -2x3 = -6
The right hand limit of f at x = 3 is,
\(\lim\limits_{x \to 3^+}\)f(x) = \(\lim\limits_{x \to 3^+}\)(6x+2) = 6x3+2 = 20

It is observed that the left and right hand limit of f at x = 3 do not coincide.
Therefore,f is not continuous at x = 3

Case V:
If c>3,then f(c) = 6c+2
∴\(\lim\limits_{x \to c}\) f(x) = f(c)
Therefore,f is continuous at all points x,such that x>3 

Hence, x = 3 is the only point of discontinuity of f.