Question

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

\[f(n) =   \begin{cases}     x^3-3,       & \quad \text{if } x {\leq 2}\\     x^2+1,  & \quad \text{if } x \text{>2}   \end{cases}\]

Answer 1

\(f(n) =   \begin{cases}     x^3-3,       & \quad \text{if } x {\leq 2}\\     x^2+1,  & \quad \text{if } x \text{>2}   \end{cases}\)
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<2, then f(c) = c³-3 and \(\lim\limits_{x \to c}\) f(x) = \(\lim\limits_{x \to c}\) f(x³-3) = c³-3
∴\(\lim\limits_{x \to c}\) f(x) = f(c)
Therefore, f is continuous at all points x, such that x<2

Case (ii):
If c = 2, then f(c) = f(2) = 2³-3 = 5
\(\lim\limits_{x \to 2^-}\)f(x) =\(\lim\limits_{x \to 2^-}\)(x³-3) = 2³-3 = 5
\(\lim\limits_{x \to 2^+}\)f(x) = \(\lim\limits_{x \to 2^+}\)(x²+1)=2²+1= 5
∴\(\lim\limits_{x \to 2}\) f(x) = f(2)
Therefore, f is continuous at x = 2

 Case(iii):
Ifc>2, then f(c) = c²+1
\(\lim\limits_{x \to c}\) f(x) = \(\lim\limits_{x \to c}\) (x²+1)=c²+1
∴\(\lim\limits_{x \to c}\) f(x) = f(c)
Therefore, f is continuous at all points x,such that x>2 
Thus, the given function f is continuous at every point on the real line. 

Hence, f has no point of discontinuity.