If f(x) = sgn ( x^3 ), then

MCQ

If $f(x) = {\mathop{\rm sgn}} ({x^3})$, then

A
$f$ is continuous but not derivable at $x = 0$
B
$f'({0^ + }) = 2$
C
$f'({0^ - }) = 1$
✓
$f$ is not derivable at $x = 0$

✓

Answer

Correct option: D.

$f$ is not derivable at $x = 0$

d
(d) Here, $f(x) = {\mathop{\rm sgn}} {x^3} = \left\{ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{\frac{{{x^3}}}{{|{x^3}|}},}&{{\rm{for}}}&{{x^3} \ne 0}\\{0{\rm{ }},}&{{\rm{for}}}&{{x^3} = 0}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{\frac{x}{{|x|}},}&{{\rm{for}}}&{x \ne 0}\\{0{\rm{ ,}}}&{{\rm{for}}}&{x = 0}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{ - 1,}&{x < 0}\\{0,}&{x = 0}\\{1,}&{x > 0}\end{array}} \right.\end{array} \right.$

Thus, $f(x) = {\mathop{\rm sgn}} {x^3} = {\mathop{\rm sgn}} x,$ which is neither continuous nor derivable at $0$. Note that

$f'({0^ + }) = \mathop {{\rm{lim}}}\limits_{h \to {0^ + }} \,\frac{{f(0 + h) - f(0)}}{h}$

$ = \mathop {{\rm{lim}}}\limits_{h \to {0^ + }} \,\frac{{1 - 0}}{h} \to \infty $

and $f'({0^ - }) = \mathop {{\rm{lim}}}\limits_{h \to {0^ - }} \,\frac{{f(0 - h) - f(0)}}{h}$

$ = \mathop {{\rm{lim}}}\limits_{h \to {0^ - }} \,\frac{{ - 1 - 0}}{h} \to \infty $.

$\therefore$ $f'({0^ + }) \ne f'({0^ - })$,

$\therefore$ $f$ is not derivable at $x = 0$.

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