Q. if $f(x) = |x|^{3}$, then $f^{'}(0)$ equals
A) 0
B) 1/2
C) -1
D) -1/2
Explanation
(a)
$f(x) = |x|^{3} = \left\{ \begin{array}{r}
0,\ \ \ x = 0 \\
x^{3},\ \ \ x > 0 \\
- x^{3},\ \ \ \ x < 0
\end{array} \right.\$
Now, $Rf^{'}(0) = \lim_{h \rightarrow 0}\ \frac{h^{3} - 0}{h} = 0$
And $Lf^{'}(0) = \lim_{h \rightarrow 0}\ \frac{- h^{3} - 0}{- h} = 0$
$\because\ \ Rf^{'}(0) = Lf^{'}(0) = 0$
$\therefore\ \ f^{'}(0) = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.