Question

Discuss the continuity of the function$f$, where $f$ is defined by
$f\left(x\right)=$ $\{\begin{array}{c}-2,ifx\le -1\\ 2x,if-1<x\le 1\\ 2,ifx>1\end{array}$

Answer 1

The given function f is f(x) = $\{\begin{array}{c}-2,ifx\le -1\\ 2x,if-1<x\le 1\\ 2,ifx>1\end{array}$
The given function is defined at all points of the real line
Let c be a point on the real line.
Case I
If c < -1, then f(c) = -2 and $\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ (-2) = -2
$\therefore$ $\underset{x\to c}{lim}$f(x) = f(c)
Therefore, f is continuous at all points x, such that x < -1
Case II
If c = -1, then f(c) = f(-1) = -2
The left hand limit of f at x = -1 is,
$\underset{x\to -1}{lim}$f(x) = $\underset{x\to -1}{lim}$ (-2) = -2
The right hand limit of f at x = -1 is,
$\underset{x\to -1}{lim}$ f(x) = $\underset{x\to -1}{lim}$(2x_ = 2 x (-1) = -2
$\therefore$ $\underset{x\to -1}{lim}$ f(x) = f(-1)
Therefore, f is continuous at x = -1
Case III :
If -1 < c < 1, then f(c) = 2c
$\underset{x\to c}{lim}$f(x) = $\underset{x\to c}{lim}$(2x) = 2c
$\therefore$ $\underset{x\to c}{lim}$ f(x) = f(c)
Therefore, f is continuous at all points of the interval (-1, 1)
Case IV :
If c = 1, then f(c) = f(1) = 2 x 1 = 2
The left hand limit of f at x = 1 is,
$\underset{x\to 1}{lim}$ f(x) = $\underset{x\to 1}{lim}$ (2x) = 2 x 1 = 2
The right hand limit of f at x = 1 is,
$\underset{x\to 1}{lim}$ f(x) = $\underset{x\to 1}{lim}$ 2 = 2
$\therefore$ $\underset{x\to 1}{lim}$ f(x) = f(c)
Therefore, f is continuous at x = 2
Case V
If c > 1, then f(c) = 2 and $\underset{x\to c}{lim}$ f(x) = $\underset{x\to c}{lim}$ (2) = 2
$\underset{x\to c}{lim}$ f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 1
Thus, from the above observations, it can be concluded that f is continuous at all points of the real line.