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August 22, 2014

August 22, 2014

Posted by **scarlette. . .need help asap** on Tuesday, January 10, 2012 at 12:07pm.

- calculus -
**Steve**, Tuesday, January 10, 2012 at 12:40pmh(r) = (1 - r^2)/(2 - r^3)

h'(r) = [(-2r)(2 - r^3) - (1 - r^2)(-3r^2)]/(2 - r^3)^2

h'(1) = [(-2)(1) - (0)(-3)]/(2-1)^2 = -2

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