check differentiability

**adhyeta** · April 22nd 2009, 01:37 AM

need help...is the func. $|x|^3$ diff at x=0 or not?

**mr fantastic** · April 22nd 2009, 02:19 AM

Originally Posted by adhyeta

need help...is the func. $|x|^3$ diff at x=0 or not?

Yes. f(x) = x^3 when x > 0 and f(x) = -x^3 when $x \leq 0$ . Therefore f'(x) is continuus at x = 0.

**chisigma** · April 22nd 2009, 02:23 AM

If you compute the derivative of $f(x)$ in $x=x_{0}$ as...

$\lim_{\delta \rightarrow 0} \frac{f(x_{0} + \delta)-f(x_{0})}{\delta}$

... for $f(x)= |x|^{3}$ and $x_{0}=0$ is...

$\lim_{\delta \rightarrow 0} \frac{|\delta|^{3}}{\delta} = \lim_{\delta \rightarrow 0} \delta\cdot |\delta|=0$

... no matter which is the sign of $\delta$ . So $|x|^{3}$ is differentiable in $x=x_{0}$ ...

Kind regards

$\chi$ $\sigma$

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