Math: Calculus
posted by REALLY NEED HELP!!!! on .
Consider the function f(x) whose second derivative is f''(x) = 8x + 4sin(x). If f(0) = 2 and f'(0) = 2, what is f(x)?
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.

y '' = 8x + 4sin(x)
y ' 4x^2  4cosx + C
f'(0) = 2  > 2 = 0  4(cos0) + C
2 = 04 + 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