Find all points of discontinuity of f, where f is defined as following:
$f (x) = \{\begin{cases} \begin{matrix} |x| + 3, x \leq - 3 \\ - 2 x, - 3 < x < 3 \end{matrix} \\ 6 x + 2, x \geq 3 \end{cases}$

Solution

Here, f ( x ) = $\{\begin{cases} \begin{matrix} |x| + 3, x \leq 3 \\ - 2 x, - 3 < x < 3 \end{matrix} \\ 6 x + 2, x \geq 3 \end{cases}$

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

$LHL = \lim_{x \to - 3^{-}} f (x) = \lim_{h \to 0} (- (- 3 - h) + 3) = 6 RHL = \lim_{x \to - 3^{+}} f (x) = \lim_{h \to 0} (- 2 (- 3 + h)) = (- 2) x (- 3) = 6 Since, \lim_{x \to - 3^{-}} f (x) = \lim_{x \to - 3^{+}} f (x) = f (- 3)$

So, f is continuous at x = -3

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

$LHL = \lim_{x \to 3^{-}} f (x) = \lim_{h \to 0} (- 2 (3 - h)) = - 2 x 3 = - 6 RHL = \lim_{x \to 3^{+}} f (x) = \lim_{h \to 0} (6 (3 + h) + 2) = 6 x 3 = 20 Since, \lim_{x \to 3^{-}} f (x) \neq \lim_{x \to 3^{+}} f (x)$

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

Hence x = 3 is the only point of discontinuity.

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Continuity And Differentiability

Find all points of discontinuity of f, where f is defined as following:
$f (x) = \{\begin{cases} \begin{matrix} |x| + 3, x \leq - 3 \\ - 2 x, - 3 < x < 3 \end{matrix} \\ 6 x + 2, x \geq 3 \end{cases}$

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Continuity And Differentiability

Find all points of discontinuity of f, where f is defined as following:f ( x ) = x + 3 , x ≤-3 - 2x , -3 < x < 3 6x + 2 , x ≥ 3

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find all points of discontinuity of f, where f is defined as following:
$f (x) = \{\begin{cases} \begin{matrix} |x| + 3, x \leq - 3 \\ - 2 x, - 3 < x < 3 \end{matrix} \\ 6 x + 2, x \geq 3 \end{cases}$