Question

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

$f(x) = \begin{cases} 2x+3, & \quad \text{if } x {\leq 2}\\\ 2x-3, & \quad \text{if } x \text{>2} \end{cases}$

Answer 1

$f(x) = \begin{cases} 2x+3, & \quad \text{if } x {\leq 2}\\\ 2x-3, & \quad \text{if } x \text{>2} \end{cases}$

It is evident that the given function f is defined at all the points of the real line.
Let c be a point on the real line. Then, three cases arise.
(i) c<2
(ii) c>2
(iii) c=2

Case (i): c<2
Then f(c) = 2c+3
$\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ (2x+3) = 2c+3
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous at all points x,such that x<2

Case (ii): c>2
Then f(c) = 2c-3
$\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ (2x-3) = 2c-3
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x>2

Case(iii): c=2
Then,the left hand limit of f at x = 2 is,
$\lim\limits_{x \to 2^-}$ f(x) = $\lim\limits_{x \to 2^-}$(2x+3) = 2x2+3 = 7
The right hand limit of f at x = 2 is,
$\lim\limits_{x \to 2^+}$ f(x) = $\lim\limits_{x \to 2^+}$(2x-3) = 2x2-3 = 1
It is observed that the left and right hand limit of f at x = 2 do not coincide
Therefore, f is not continuous at x = 2
Hence,x = 2 is the only point of discontinuity of f.