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Consider the function f(x) whose second derivative is f''(x) = 8x + 8sin(x). If f(0) = 3 and f'(0) = 2, what is f(x)?

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Final answer:

To find the function f(x) with the second derivative f''(x) = 8x + 8sin(x) and initial conditions f(0) = 3 and f'(0) = 2, we integrate twice and use the given conditions to find constants, resulting in f(x) = (4/3)x^3 + 8sin(x) + 10x + 3.

Step-by-step explanation:

The student is asking to find the original function f(x) when given the second derivative f''(x) = 8x + 8sin(x), and initial conditions f(0) = 3 and f'(0) = 2. To find f(x), we first need to integrate the second derivative to get the first derivative and then integrate once more to get the original function.

First integration:
f'(x) = ∫(8x + 8sin(x))dx = 4x^2 - 8cos(x) + C1
Using the initial condition f'(0) = 2, we find that C1 = 10.

Second integration:
f(x) = ∫(4x^2 - 8cos(x) + 10)dx = \frac{4}{3}x^3 + 8sin(x) + 10x + C2
Using the initial condition f(0) = 3, we find that C2 = 3.

Therefore, f(x) = \frac{4}{3}x^3 + 8sin(x) + 10x + 3.

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