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.