# Hard Calculus

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f(x)= xsquared -2x
for y= absolute value of f(x) does the derivitive exist at x=0 and explain why.

• Hard Calculus -

f(x)=|x²-2x|
=x²-2x (-∞,0]U[2,∞)
=-(x²-2x) (0,2)
f'(x)
=2x-2 (-∞,0)U(2,∞)
=-(2x-2) (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

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