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The derivative of   \mathrm{f(x)=|x|^3~at~ x=0}  is

Option: 1

-1
 


Option: 2

0


Option: 3

does not exist


Option: 4

none of these.


Answers (1)

best_answer

We have,   \mathrm{f(x)=|x|^3=\left\{\begin{aligned} x^3, x \geq 0 \\ -x^3, x<0\end{aligned}\right.}
Now,   \mathrm{LHD ~at ~x=0)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0}-\frac{x^3}{x}=0}
and,    \mathrm{(RHD ~at ~x=0 ) =\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0} \frac{x^3}{x}=0.}
Hence, f(x) is differentiable at x=0 and its derivative at x=0 is 0 .

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Ritika Jonwal

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