question_answer

If\[f(x)=|x{{|}^{3}}\], then\[f'(0)\]equals:

A) \[0\]                                     

B) \[1/2\]

C) \[-1\]                                    

D) \[-1/2\]

Correct Answer: A

Solution :

Let\[f(x)=|x{{|}^{3}}=\left\{ \begin{matrix}    0, & x=0  \\    {{x}^{3}}, & x>0  \\    -{{x}^{3}}, & x<0  \\ \end{matrix} \right.\] Now,\[R\,\,f'(0)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(h)-f(0)}{h}\]                          \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{3}}-0}{h}=0\] \[\therefore \]       \[Rf'(0)=Lf'(0)=0\] \[\therefore \]        \[f'(0)=0\] Alternative Solution: Let       \[f(x)=|x{{|}^{3}}\] It is clear from the figure that derivative (tangent) at \[x=0\] is\[=0\].