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April 19, 2015

April 19, 2015

Posted by **Yoona** on Tuesday, January 3, 2012 at 12:04am.

- calculus -
**Steve**, Tuesday, January 3, 2012 at 12:30amy = ax^3 + bx^2 + cx + d

y' = 3ax^2 + 2bx + c

y'' = 6ax + 2b

y''(0) = 0

so b = 0

y'(2) = 0

so c = -12a

y(0) = 0

so d = 0

y = ax^3 - 12ax

y(2) = 4

so a = -1/4

y = -1/4 x^3 + 3x

y' = -3/4 x^2 + 3

y'' = -3/2 x

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