# Math: Calculus

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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.

• Math: Calculus -

y '' = 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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