611.

Find all points of discontinuity of f, where f is defined as following:

$\mathrm{f} ( \mathrm{x} ) = \left\{\begin{array}{l}\begin{array}{c}\left| \mathrm{x} \right| + 3 , \mathrm{x} \le -3 \\ - 2\mathrm{x} , -3 < \mathrm{x} < 3 \end{array}\\ 6\mathrm{x} + 2 , \mathrm{x} \ge 3\end{array}\right.$

Here, f ( x ) = $\left\{\begin{array}{l}\begin{array}{c}\left|\mathrm{x}\right| + 3 , \mathrm{x}\le 3\\ -2 \mathrm{x} ,-3<\mathrm{x}<3\end{array}\\ 6\mathrm{x} + 2 , \mathrm{x} \ge 3\end{array}\right.$

The function is defined on all the points and hence continuous

possible points of discontinuity are  3 and  -3 . We need to check the 

continuity of the function at two points  x = 3  and  x = - 3 .

Case 1:  For   x = - 3,  f ( - 3 ) = - ( - 3 ) + 3 = 6

$$\begin{gathered} \mathrm{LHL} = \underset{\mathrm{x} \to -3^{-}}{\lim } \mathrm{f} ( \mathrm{x} ) = \underset{\mathrm{h} \to 0}{\lim } \left( - \left(- 3 - \mathrm{h} \right) + 3 \right) = 6 \\ \mathrm{RHL} = \underset{\mathrm{x} \to -3^{+}}{\lim } \mathrm{f} ( \mathrm{x} ) = \underset{\mathrm{h} \to 0}{\lim } \left( - 2 \left(- 3 + \mathrm{h} \right) \right) = \left( - 2 \right) \mathrm{x} \left( - 3 \right) =6 \\ \mathrm{Since}, \underset{\mathrm{x} \to -3^{-}}{\lim } \mathrm{f} ( \mathrm{x} ) = \underset{\mathrm{x} \to -3^{+}}{\lim } \mathrm{f} ( \mathrm{x} ) = \mathrm{f} ( - 3 ) \end{gathered}$$

So, f is continuous at  x = -3

Case 2:  For  x =  3,  f ( 3 ) = 6 ( 3 ) + 2 = 20

$$\begin{gathered} \mathrm{LHL} = \underset{\mathrm{x} \to 3^{-}}{\lim } \mathrm{f} ( \mathrm{x} ) = \underset{\mathrm{h} \to 0}{\lim } \left( - 2 \left( 3 - \mathrm{h} \right)\right) = - 2 \mathrm{x} 3 = - 6 \\ \mathrm{RHL} = \underset{\mathrm{x} \to 3^{+}}{\lim } \mathrm{f} ( \mathrm{x} ) = \underset{\mathrm{h} \to 0}{\lim } \left( 6 \left( 3 + \mathrm{h} \right) + 2 \right) = 6 \mathrm{x} 3 = 20 \\ \mathrm{Since}, \underset{\mathrm{x} \to 3^{-}}{\lim } \mathrm{f} ( \mathrm{x} ) \ne \underset{\mathrm{x} \to 3^{+}}{\lim } \mathrm{f} ( \mathrm{x} ) \end{gathered}$$

Therefore, function  f  is not continuous at point  x = 3

Hence  x = 3 is the only point of discontinuity.