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

If and only if  x0x0  is not a simple root.  –  did  aug 19 '12 at 13: 15      you should look at  f(x)=x3f(x)=x3.  –  chris eagle  aug 19 '12 at 13: 164  just observe that for  p(x)=x3p(x)=x3,  |p(x)||p(x)|  is also differentiable at  x=0x=0. the geometric reason that  p(x)=xp(x)=xfails 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

Maybe the answer is 20mm. all i really did was just multiply, which is obvious.

The answer would be b