# Find m for which $f(x)=\sqrt[3]{2x^2-mx+1}$ is differentiable on $\mathbb{R}$

We have this function: $$f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\sqrt[3]{2x^2-mx+1}$$

and the following set: $M=\{m\in\mathbb{Z}|\mbox{f is differentiable on } \mathbb{R}\}$

In these conditions I have to calculate the following sum: $S = \displaystyle\sum\limits_{m \in M} |m|$

Maybe I am wrong, but isn't this function always differentiable, no matter who is m?

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The quadratic function may have a root, in which case it's not differentiable at the root. –  Ishan Banerjee Mar 17 '13 at 13:30

If the quadratic polynomial has a real root, then near that root it will behave as $x^{1/3}$ (if it is a single root) or $x^{2/3}$ (in case of double root) but it will certainly not be differentiable. Therefore, we require that $m^2 < 8$. You can compute the sum yourself.