First slide

Differentiability

Question

$Let f (x) = x | x | . The set of points where f (x) is twice differentiable is :$

Moderate

A

$(- \infty, \infty)$
B

$(1, \infty)$
C

$(- \infty, - 1)$
D

$(- \infty, - \infty) - \{0\}$

Solution

$Since f (x) = x | x |$
$∴ f (x) = \{\begin{matrix} = - x^{2}, & i f & x < 0 \\ = 0, & i f & x = 0 \\ = x^{2}, & i f & x > 0 \end{matrix}$
The function $f$ is differentiable everywhere and its derivative is given by
$f' (x) = \{\begin{matrix} - 2 x & i f & x < 0 \\ 0 & i f & x = 0 \\ 2 x & i f & x > 0 \end{matrix}$
$f " (0^{+}) = \lim_{h \to 0} \frac{f' (0 + h) - f' (0)}{h}$
$= \lim_{h \to 0} \frac{2 h - 0}{h} = 2$
$f " (0^{-}) = \lim_{h \to 0^{-}} \frac{f (0 - h) - f' (0)}{- h}$
$= \lim_{h \to 0} \frac{- 2 (- h) - 0}{- h} = - 2$
$Hence f is not twice diff. at x = 0$
$But f^{''} (x) = - 2, if x < 0$
$= 2, if x > 0$
$Thus, the set of points. Where f is twice differentiable is (- \infty, - \infty) - {0} .$

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