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The set of all points where the function \mathrm{f(x)=2 x|x|} is differentiable is

Option: 1

(-\infty, \infty)


Option: 2

(-\infty, 0) \cup(0, \infty)


Option: 3

(0, \infty)


Option: 4

[0, \infty)


Answers (1)

best_answer

\mathrm{f(x)= \begin{cases}2 x^{2}, & x \geq 0 \\ -2 x^{2}, & x<0\end{cases}}

Since \mathrm{x^{2}} and \mathrm{-x^{2}} are differentiable functions, \mathrm{f(x)} is differentiable except possible at \mathrm{x=0}

Now \mathrm{f^{\prime}(0+)=\lim _{h \rightarrow 0+} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0+} \frac{f(h)}{h}}
                       \mathrm{=\lim _{h \rightarrow 0+} \frac{2 h^{2}}{h}=0}
\mathrm{and \: f^{\prime}(0-)=\lim _{h \rightarrow 0-} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0-} \frac{f(h)}{h}}
                      \mathrm{=\lim _{h \rightarrow 0-} \frac{-2 h^{2}}{h}=0}.

Hence \mathrm{f} is differentiable everywhere.

Posted by

vishal kumar

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