Functions Question 574
Question: If $ f(x)=sgn (x^{3}) $ , then
[DCE 2001]
Options:
A) f is continuous but not differentiable at $ x=0 $
B) $ f’({0^{+}})=0 $
C) $ f’({0^{+}})=1 $
D) f is not derivable at $ x=0 $
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Answer:
Correct Answer: D
Solution:
Here, $ f(x)=sgn x^{3}= \begin{cases} & \begin{cases} \frac{x^{3}}{|x^{3}|}, & for & x^{3}\ne 0 \\ 0\text{ }, & for & x^{3}=0 \\ \end{cases} . \\ & \begin{cases} \frac{x}{|x|}, & for & x\ne 0 \\ 0\text{ ,} & for & x=0 \\ \end{cases} . \\ & \begin{cases} -1, & x<0 \\ 1, & x>0 \\ \end{cases} . \\ \end{cases} . $ Thus, $ f(x)=sgn x^{3}=sgn x, $ which is neither continuous nor derivable at 0. Note that $ {f}’({0^{+}})=\underset{h\to {0^{+}}}{\mathop{lim}}\frac{f(0+h)-f(0)}{h} $ $ =\underset{h\to {0^{+}}}{\mathop{lim}}\frac{1-0}{h}\to \infty $ and $ {f}’({0^{-}})=\underset{h\to {0^{-}}}{\mathop{lim}}\frac{f(0-h)-f(0)}{h} $ $ =\underset{h\to {0^{-}}}{\mathop{lim}}\frac{-1-0}{h}\to \infty $ . $ {f}’({0^{+}})\ne {f}’({0^{-}}) $ , f is not derivable at $ x=0 $ .
BETA
