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Question
Examine the differentialibilty of the function f defined by
\[f\left( x \right) = \begin{cases}2x + 3 & \text { if }- 3 \leq x \leq - 2 \\ \begin{array}xx + 1 \\ x + 2\end{array} & \begin{array} i\text { if } - 2 \leq x < 0 \\\text { if } 0 \leq x \leq 1\end{array}\end{cases}\]
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Solution
\[f\left( x \right) = \begin{cases}2x + 3 & \text { if }- 3 \leq x \leq - 2 \\ \begin{array}xx + 1 \\ x + 2\end{array} & \begin{array} i\text { if } - 2 \leq x < 0 \\\text { if } 0 \leq x \leq 1\end{array}\end{cases}\]
\[f\left( x \right) = \begin{cases}2x + 3 & \text { if }- 3 \leq x \leq - 2 \\ \begin{array}11 \\ 1\end{array} & \begin{array} i\text { if } - 2 \leq x < 0 \\\text { if } 0 \leq x \leq 1\end{array}\end{cases}\]
Now,
\[\text { LHL }= \lim_{x \to - 2^-} f'\left( x \right) = \lim_{x \to - 2^-} 2 = 2\]
\[\text { RHL } = \lim_{x \to - 2^+} f'\left( x \right) = \lim_{x \to - 2^+} 1 = 1\]
\[\text { Since, at x } = - 2, \text { LHL} \neq \text{RHL}\]
\[\text { Hence,} f\left( x \right) \text { is not differentiable at x } = - 2\]
Again,
\[\text { LHL }= \lim_{x \to 0^-} f'\left( x \right) = \lim_{x \to 0^-} 1 = 1\]
\[\text { RHL } = \lim_{x \to 0^+} f'\left( x \right) = \lim_{x \to 0^+} 1 = 1\]
\[\text { Since, at x = 0, LHL = RHL }\]
\[\text { Hence }, f\left( x \right) \text { is differentiable at x } = 0\]
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