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July 31, 2014

July 31, 2014

Posted by **timmy** on Wednesday, September 22, 2010 at 8:48pm.

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 9:17pmf(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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