# Real Analysis: Derivative of the function

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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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.