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Thread: check differentiability

  1. #1
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    Exclamation check differentiability

    need help...is the func. $\displaystyle |x|^3$ diff at x=0 or not?
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  2. #2
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    Quote Originally Posted by adhyeta View Post
    need help...is the func. $\displaystyle |x|^3$ diff at x=0 or not?
    Yes. f(x) = x^3 when x > 0 and f(x) = -x^3 when $\displaystyle x \leq 0$. Therefore f'(x) is continuus at x = 0.
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    MHF Contributor chisigma's Avatar
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    If you compute the derivative of $\displaystyle f(x)$ in $\displaystyle x=x_{0}$as...

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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