Question

Discuss the continuity of the function f, where f is defined by

$f(n) = \begin{cases} 3, & \quad \text{if } {0\leq x\leq 1}\\\ 4, & \quad \text{if } {1<x<3} \\\ 5, & \quad \text{if } {3\leq x \leq 10} \end{cases}$

Answer 1

The given function is
$f(n) = \begin{cases} 3, & \quad \text{if } {0\leq x\leq 1}\\\ 4, & \quad \text{if } {1<x<3} \\\ 5, & \quad \text{if } {3\leq x \leq 10} \end{cases}$
The given function f is defined at all the points of the interval [0,10]
Let c be a point on the interval [0,10]

Case (I):
If 0≤c<1, then f(c) = 3 and $\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ f(3) = 3
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous in the interval [0,1).

Case (II):
If c=1, then f(3)=3
The left hand limit of f at x=1 is
$\lim\limits_{x \to 1^-}$ f(x) =$\lim\limits_{x \to 1^-}$(3) = 3
The right hand limit of f at x=1 is,
$\lim\limits_{x \to 1^+}$ f(x) =$\lim\limits_{x \to 1^+}$(4) = 4
It is observed that the left and right hand limit of f at x=1 do not coincide.
Therefore, f is not continuous at x=1

Case(III):
If1<c<3, then f(c)=4 and
$\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$(4) = 4
∴$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore, f is continuous at all points of the interval (1,3).

Case(IV):
If c=3, then f(c) = 5
The left hand limit of f at x=3 is,
$\lim\limits_{x \to 3^-}$ f(x) = $\lim\limits_{x \to 3^-}$(4) = 4
The right hand limit of f at x=3 is,
$\lim\limits_{x \to 3^+}$ f(x) =$\lim\limits_{x \to 3^+}$(5) = 5
It is observed that the left and right hand limits of f at x=3 do not coincide.
Therefore, f is not continuous at x=3

Case(V):
If 3<c≤10, then f(c)=5 and $\lim\limits_{x \to c}$ f(x) = $\lim\limits_{x \to c}$ (5) = 5
$\lim\limits_{x \to c}$ f(x) = f(c)
Therefore,f is continuous at all points of the interval (3,10].

Hence,f is not continuous at x=1 and x=3