Question

Discuss the differentiability of function f(x) = x|x|.

Please explain properly.

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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: \: f(x) = x |x|$

We know,

$\red{\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: |x| = \begin{cases} &\sf{ - x \: \: when \: x < 0} \\ \\ &\sf{ \: \: x \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}}$

So,

$\red{\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: x|x| = \begin{cases} &\sf{ - {x}^{2} \: \: when \: x < 0} \\ \\ &\sf{ \: \: {x}^{2} \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}}$

Now, to check the differentiability of function f(x), we have to check whether LHD = RHD at x = 0.

Consider, LHD

$\rm :\longmapsto\:\displaystyle\lim_{x \to {0}^{ - } } \frac{f(x) - f(0)}{x - 0}$

$\rm \:  =  \: \displaystyle\lim_{x \to {0}^{ - } } \frac{ - {x}^{2} -0}{x}$

$\rm \:  =  \: \displaystyle\lim_{x \to {0}^{ - } } \frac{ - {x}^{2}}{x}$

$\rm \:  =  \: \displaystyle\lim_{x \to {0}^{ - } } (- x)$

$\rm \:  =  \: 0$

So,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to {0}^{ - } } \frac{f(x) - f(0)}{x - 0} = 0 \: }}$

Consider, RHD

$\rm :\longmapsto\:\displaystyle\lim_{x \to {0}^{ + } } \frac{f(x) - f(0)}{x - 0}$

$\rm \:  =  \: \displaystyle\lim_{x \to {0}^{ + } } \frac{{x}^{2} -0}{x}$

$\rm \:  =  \: \displaystyle\lim_{x \to {0}^{ + } } \frac{{x}^{2}}{x}$

$\rm \:  =  \: \displaystyle\lim_{x \to {0}^{ + } } x$

$\rm \:  =  \: 0$

So,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to {0}^{ + } } \frac{f(x) - f(0)}{x - 0} = 0 \: }}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: f'(0) = 0 \: }}$

$\rm \implies\:f(x) = x |x| \: is \: differentiable \: everywhere.$

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