Question

Find the points of discontinuity of the following function.
f(x) = \left\\{ {\begin{array}{*{20}{c}} {\left| x \right| + 3} \\\ {2x} \\\ {6x + 2} \end{array}} \right. \begin{array}{*{20}{c}} {if} \\\ {if} \\\ {if} \end{array} \begin{array}{*{20}{c}} {x \leqslant - 3} \\\ { - 3 < x < 3} \\\ {x > 3} \end{array}

Answer 1

Hint : Use the properties of continuity to check continuity in the interval and use mathematical definition of continuity to check continuity at the boundary points.

Complete step-by-step answer :
We know that, $\left| x \right|$ is continuous at every point in its domain.
Therefore, $\left| x \right|$ is continuous for all the values in $x \leqslant - 3$
We also know that polynomial functions are always continuous.
Therefore, $2x$ and $6x + 2$ are continuous in their respective intervals. i.e. $- 3 < x < 3$ and $x > 3$
But these properties are not applicable at boundary points. So we need to use the mathematical definition of continuity to check the continuity of the given function at boundary points, i.e. -3 and 3
We know that,
$f(x)$ is said to be continuous at some point $x = a$ if
$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)$
For continuity at $x = - 3$
For $x = - {3^ - }$ , $f(x) = \left| x \right| + 3$
Therefore, we get
$\mathop {\lim }\limits_{x \to - {3^ - }} f(x) = \left| { - 3} \right| + 3$
$= 3 + 3$ $\left( {\because \left| x \right| = - x,x < 0} \right)$
$\Rightarrow \mathop {\lim }\limits_{x \to - {3^ - }} f(x) = 6$
For $x = - {3^ + },$ $f(x) = 2x$
$\mathop {\lim }\limits_{x \to - {3^ + }} f(x) = 2( - 3)$
$\Rightarrow \mathop {\lim }\limits_{x \to - {3^ + }} f(x) = - 6$
Therefore, left hand limit is not equal to right hand limit.
Thus, the function is discontinuous at $x = - 3$
For continuity at $x = 3$
For $x = {3^ - },$ $f(x) = 2x$
$\Rightarrow \mathop {\lim }\limits_{x \to {3^ - }} f(x) = 2 \times 3$
$\Rightarrow \mathop {\lim }\limits_{x \to {3^ - }} f(x) = 6$
For $x = {3^ + },$ $f(x) = 6x + 2$
$\Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} f(x) = 6 \times 3 + 2$
$\Rightarrow \mathop {\lim }\limits_{x \to {3^ + }} f(x) = 20$
Therefore, the left hand limit is not equal to the right hand limit.
Thus, the function is discontinuous at $x = 3$
Therefore, $f(x)$ is discontinuous at $x = - 3$ and $x = 3$

Note : There are an infinite number of points in any interval. You cannot check the function for continuity at all those points using the mathematical definition of continuity. Therefore, it is important to know the properties of continuous functions and to know which functions are continuous on which intervals. You need to remember that the properties of continuous functions are not applicable for boundary points of the given interval. Therefore, it is important to check for continuity using mathematical definition at boundary points.