Posted by timmy on Wednesday, September 22, 2010 at 8:48pm.
f(x)= xsquared 2x
for y= absolute value of f(x) does the derivitive exist at x=0 and explain why.

Hard Calculus  MathMate, Wednesday, September 22, 2010 at 9:17pm
f(x)=x²2x
=x²2x (∞,0]U[2,∞)
=(x²2x) (0,2)
f'(x)
=2x2 (∞,0)U(2,∞)
=(2x2) (0,2)
f'(0)=2
f'(0+)=2
Therefore the derivative is not continuous at x=0, and therefore f'(0) does not exist.
See:
http://img683.imageshack.us/img683/654/1285202904.png