When is absolute value of a polynomial differentiable at a real root?

If and only if x0 x0 is NOT a simple root. - Did Aug 19 '12 at 13:15 You should look at f (x) = x3 f (x) = x3. - Chris Eagle Aug 19 '12 at 13:16 4 Just observe that for p (x) = x3 p (x) = x3, |p (x) | |p (x) | is also differentiable at x=0 x=0. The geometric reason that p (x) = x p (x) = x fails to satisfy this property is that taking absolute value flips over the negative part of the graph of y=p (x) y=p (x), resulting in a wedge-like peak. So as long as this wedge is avoided, taking absolute value does not spoil differentiability. This is the case for zeros of multiplicity ≥2 ≥2