Consider the function f(x) whose second derivative is f(x)=9x+6sin(x). If f(0)=4 and f(0)=3, what is f(x)? Please do not include the constant (+C) in your answer.
You wrote << f(0)=4 and f(0)=3,>> but
both cannot be true. One of the f(x) functions must be the derivative, f'(x). Which is it?
Anyway, all you have to do is integrate your f''(x) once and use the value of f'(x) at x=0 to get the first cosntant, and then integrate it again and use the value of f(0)at x=0 to get the second constant
oh im sorry it is f(0)=4 and f'(0)=3
and it's f''(x)=9x+6sin(x)
sorry
i got the answer 1.5x^3-6sinx+9x+4
To find the function f(x), we need to integrate the second derivative of f(x).
Given that the second derivative of f(x) is f''(x) = 9x + 6sin(x), we can integrate it twice to get the original function f(x).
Step 1: Integrate f''(x) to find the first derivative f'(x):
∫(9x + 6sin(x)) dx
To integrate 9x with respect to x, we use the power rule of integration, which states that ∫x^n dx = (x^(n+1))/(n+1).
∫(9x + 6sin(x)) dx = (9/2)x^2 - 6cos(x) + C1
Step 2: Integrate f'(x) to find f(x):
∫[(9/2)x^2 - 6cos(x)] dx
To integrate (9/2)x^2 with respect to x, we use the power rule of integration.
∫[(9/2)x^2 - 6cos(x)] dx = (9/6)x^3 - 6sin(x) + C2
Since f(0) = 4, we can substitute x = 0 into the expression we obtained:
4 = (9/6)(0)^3 - 6sin(0) + C2
Simplifying, we find:
4 = C2
Therefore, C2 = 4.
Finally, we can write the expression for f(x) without the constant (+C):
f(x) = (9/6)x^3 - 6sin(x) + 4