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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Show that the function $\text{f(x)} = |\text{x - 1| + | x + 1|.}$ for all $\text{x} \in \Re,$ is not differentiable at the point$\text{x = -1 and x = 1}.$
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Answer

$\text{f(x)} = |\text{x - 1| +| x + 1|}$
$\text{L f'(-1) }= \lim\limits_{ x \rightarrow (-1)^{-}} \frac{-\left\{\text{(x - 1) - (x + 1) - 2}\right\}}{\text{x - (-1)}} = \lim\limits_{ x \rightarrow (-1)^{-}}\frac{-2(\text{x + 1)}}{\text{x + 1}} = -2$
$\text{R f'(-1) }= \lim\limits_{ x \rightarrow (-1)^{+}} \frac{-\left\{\text{(x - 1) + (x + 1) - 2}\right\}}{\text{x - (-1)}} = \lim\limits_{ x \rightarrow (-1)^{+}}\frac{0}{\text{x + 1}} = 0$
$-2 \neq 0 \therefore \text{f(x)}$ is not differentiable at $\text{x = -1}$
$\text{L f'(1) }= \lim\limits_{ x \rightarrow (-1)^{-}} \frac{-\left\{\text{(x - 1) + (x + 1) - 2}\right\}}{\text{x -1}} = \lim\limits_{ x \rightarrow (-1)^{-}}\frac{0}{\text{x - 1}} = 0$
$\text{R f'(1) }= \lim\limits_{ x \rightarrow 1^{+}} \frac{\big\{\text{x}-1+\text{x}+1\big\}-2}{\text{x}-1} = \lim\limits_{ x \rightarrow 1^{+}}\frac{2(\text{x - 1)}}{\text{x - 1}} = 2$
$0 \neq 2 \therefore\text{f(x)}$ is not differentiable at $\text{x = 1}$

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