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.