# check differentiability

• Apr 22nd 2009, 02:37 AM
check differentiability
need help...is the func. $|x|^3$ diff at x=0 or not?
• Apr 22nd 2009, 03:19 AM
mr fantastic
Quote:

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.
• Apr 22nd 2009, 03:23 AM
chisigma
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

$\chi$ $\sigma$