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Find k such that the function $f(x)=|x|^3$ is $C^{k}$ but not $C^{k+1}$

I am SO lost...any help would be appreciated

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up vote 7 down vote accepted

Well, $$ f(x)= \begin{cases} x^3 &\text{if $x \geq 0$}\\ -x^3 &\text{if $x < 0$}. \end{cases} $$ So you are gluing two function at $x=0$: the derivatives of the two pieces are $3x^2$, $6x$, $6$, on the right, and $-3x^2$, $-6x$, $-6$ on the left. Higher derivatives are identically zero. Now the question is: how manu derivatives can you glue at zero, so that the gluing is continuous? The answer is not very difficult.

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Wow...I had a major brain fart there! Thanks for the clarification, I am embarrassed. – xc92 Apr 27 '14 at 11:37

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