Consider the function f(x) whose second derivative is f''(x)=8x+4sin(x).

If
f(0)=3 and
f'(0)=2, what is f(x)?

f(x)=??

what, no ideas at all? Just straightforward integration.

f"(x) = 8x+4sin x
f'(x) = 4x^2 - 4cos x + C
So, since f'(0) = 2,
16-4+C = 2
C = -10
and so
f'(x) = 4x^2 - 4cos x - 10

Now do that all over again to get f(x)

To find the function f(x), we need to integrate the given second derivative f''(x).

∫ f''(x) dx = ∫ (8x + 4sin(x)) dx

Integration of 8x with respect to x gives:
∫ 8x dx = 4x^2 + C₁, where C₁ is the constant of integration.

Integration of 4sin(x) with respect to x gives:
∫ 4sin(x) dx = -4cos(x) + C₂, where C₂ is the constant of integration.

Adding both results together, we get:
f'(x) = 4x^2 - 4cos(x) + C

Now, we know that f'(0) = 2. So, we can substitute x = 0 into f'(x) and solve for the constant C:

f'(0) = 4(0)^2 - 4cos(0) + C
2 = 0 - 4 + C
2 = -4 + C
C = 6

Now, we have the expression for f'(x):

f'(x) = 4x^2 - 4cos(x) + 6

To find f(x), we integrate f'(x):

∫ f'(x) dx = ∫ (4x^2 - 4cos(x) + 6) dx

Integration of 4x^2 with respect to x gives:
∫ 4x^2 dx = (4/3)x^3 + C₃, where C₃ is the constant of integration.

Integration of -4cos(x) with respect to x gives:
∫ -4cos(x) dx = -4sin(x) + C₄, where C₄ is the constant of integration.

Integration of 6 with respect to x gives:
∫ 6 dx = 6x + C₅, where C₅ is the constant of integration.

Adding all the results together, we get:
f(x) = (4/3)x^3 - 4sin(x) + 6x + C

Now, we know that f(0) = 3. So, let's substitute x = 0 into f(x) and solve for the constant C:

f(0) = (4/3)(0)^3 - 4sin(0) + 6(0) + C
3 = 0 - 0 + 0 + C
C = 3

Therefore, the function f(x) is:
f(x) = (4/3)x^3 - 4sin(x) + 6x + 3

To find the function f(x), we'll integrate the second derivative f''(x) and use the initial conditions f(0) = 3 and f'(0) = 2 to determine the constant of integration.

First, let's integrate f''(x). The antiderivative of 8x with respect to x is 4x^2, and the antiderivative of 4sin(x) with respect to x is -4cos(x).

Thus, integrating f''(x) gives us f'(x) = 4x^2 - 4cos(x) + C1, where C1 is a constant of integration.

Next, since we know f'(0) = 2, we can substitute x = 0 into the equation for f'(x) to solve for C1:

f'(0) = 4(0)^2 - 4cos(0) + C1
2 = -4 + C1
C1 = 6

Now we have f'(x) = 4x^2 - 4cos(x) + 6.

Finally, let's integrate f'(x) to find f(x). The antiderivative of 4x^2 with respect to x is (4/3)x^3, the antiderivative of -4cos(x) with respect to x is -4sin(x), and the antiderivative of 6 with respect to x is 6x.

Therefore, integrating f'(x) gives us f(x) = (4/3)x^3 - 4sin(x) + 6x + C2, where C2 is another constant of integration.

Using the initial condition f(0) = 3, we can substitute x = 0 to solve for C2:

f(0) = (4/3)(0)^3 - 4sin(0) + 6(0) + C2
3 = 0 - 0 + 0 + C2
C2 = 3

Finally, we have the function f(x) = (4/3)x^3 - 4sin(x) + 6x + 3 as the solution.