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Question
Discuss the continuity and differentiability of the
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Solution
\[\text { Given: } f\left( x \right) = \left| x \right| + \left| x - 1 \right|\]
\[\left| x \right| = - x \text { for } x < 0\]
\[\left| x \right| = x \text { for }x > 0\]
\[\left| x - 1 \right| = - \left( x - 1 \right) = - x + 1 \text { for } x - 1 < 0 \text { or }x < 1\]
\[\left| x - 1 \right| = x - 1 \text { for } x - 1 > 0 \text { or }x > 1\]
Now,
\[f\left( x \right) = - x - x + 1 = - 2x + 1 x \in \left( - 1, 0 \right)\]
or
\[f\left( x \right) = x - x + 1 = 1 x \in \left( 0, 1 \right)\]
or
\[f\left( x \right) = x + x - 1 = 2x - 1 x \in \left( 1, 2 \right)\]
Now,
\[\text { LHL } = \lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^-} - 2x + 1 = 0 + 1 = 1\]
\[\text { RHL } = \lim_{x \to 0^+} f\left( x \right) = \lim_{x \to 0^+} 1 = 1\]
\[\text { Hence, at x = 0, LHL = RHL}\]
Again,
\[\text { LHL } = \lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^-} 1 = 1\]
\[\text { RHL} = \lim_{x \to 1^+} f\left( x \right) = \lim_{x \to 1^+} 2x - 1 = 2 - 1 = 1\]
\[\text { Hence, at x = 1, LHL = RHL}\]
Now,
\[f\left( x \right) = - x - x + 1 = - 2x + 1 x \in \left( - 1, 0 \right)\]
\[ \Rightarrow f'\left( x \right) = - 2 x \in \left( - 1, 0 \right)\]
\[or\]
\[f\left( x \right) = x - x + 1 = 1 x \in \left( 0, 1 \right)\]
\[ \Rightarrow f'\left( x \right) = 0 x \in \left( 0, 1 \right)\]
\[or\]
\[f\left( x \right) = x + x - 1 = 2x - 1 x \in \left( 1, 2 \right)\]
\[ \Rightarrow f'\left( x \right) = 2 x \in \left( 1, 2 \right)\]
Now,
\[\text { LHL }= \lim_{x \to 0^-} f'\left( x \right) = \lim_{x \to 0^-} - 2 = - 2\]
\[\text { RHL }= \lim_{x \to 0^+} f'\left( x \right) = \lim_{x \to 0^+} 0 = 0\]
\[\text { Since, at x = 0, LHL } \neq \text { RHL}\]
\[\text { Hence,} f\left( x \right) \text { is not differentiable at } x = 0\]
Again,
\[\text { LHL }= \lim_{x \to 1^-} f'\left( x \right) = \lim_{x \to 1^-} 0 = 0\]
\[\text { RHL } = \lim_{x \to 1^+} f'\left( x \right) = \lim_{x \to 1^+} 2 = 2\]
\[\text { Since, at } x = 1, \text { LHL } \neq \text { RHL }\]
\[\text { Hence }, f\left( x \right) \text { is not differentiable at } x = 1\]
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