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December 22, 2014

December 22, 2014

Posted by **REALLY NEED HELP!!!!** on Wednesday, November 10, 2010 at 1:05am.

I got... f'(x)=4(x^2 - cos[x]) and f(x)=(4(x^3 - 3sin[x]))/3 but it's wrong. Can anyone explain to me how to solve this.

- Math: Calculus -
**Reiny**, Wednesday, November 10, 2010 at 8:26amy '' = 8x + 4sin(x)

y ' 4x^2 - 4cosx + C

f'(0) = 2 --- > 2 = 0 - 4(cos0) + C

2 = 0-4 + C

C = 6

so y' = 4x^2 - 4cosx + 6

y = (4/3)x^3 - 4sinx + 6x + K

f(0) = 2 ---> 2 = 0 - 4sin0 + 0 + K

K = 2

then y = (4/3)x^3 - 4sinx + 6x + 2

check by differentiating and subbing in x = 0 at each level

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