Question

Discuss the continuity of the function $f$, where $f$ is defined by
$f\left(x\right)=$ $\{\begin{array}{c}3,if0\le x\le 1\\ 4,if1<x<3\\ 5,if3\le x\le 10\end{array}$

Answer 1

The given function is f(x) = $\{\begin{array}{c}3,if0\le x\le 1\\ 4,if1<x<3\\ 5,if3\le x\le 10\end{array}$
The given function is defined at all points of the interval $[0,10]$.
Let c be a point in the interval $[0,10]$.

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

Case II :
If $c=1,$ then $f\left(3\right)=3$
The left hand limit of f at $x=-1$ is
$\underset{x\to 1}{lim}$ f(x) = $\underset{x\to 1}{lim}$ (3) = 3
The right hand limit of f at $x=1$ is,
$\underset{x\to 1}{lim}$ f(x) = $\underset{x\to 1}{lim}$ (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 :
If 1 < c < 3, then f(c) = 4 and $\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ (4) = 4
$\therefore$ $\underset{x\to c}{lim}$ $f\left(x\right)=f\left(c\right)$
Therefore, f is continuous at all points of the interval $(1,3).$

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

Case V:
If 3 < c $\le$ 10, then $f\left(c\right)=5$ and $\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ (5) = 5
$\underset{x\to c}{lim}$ 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