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.