Suppose f′′(x)=−36sin(6x) and
f′(0)=−5, and f(0)=−3
f(π/3)=?
To find the value of f(π/3), we need to integrate the given second derivative function, f''(x), twice, subject to the initial conditions of f'(0) = -5 and f(0) = -3.
First, find the antiderivative of -36sin(6x):
∫-36sin(6x) dx = 6cos(6x) + C₁, where C₁ is the constant of integration.
Since f'(0) = -5, we can substitute the initial condition into the derivative function to solve for C₁:
f'(0) = 6cos(6(0)) + C₁
-5 = 6cos(0) + C₁
-5 = 6(1) + C₁
-5 = 6 + C₁
C₁ = -11
Now we have the indefinite integral of the second derivative function:
f'(x) = 6cos(6x) - 11
To find the definite integral, we integrate f'(x) and use the initial condition f(0) = -3:
∫(6cos(6x) - 11) dx = 6sin(6x) - 11x + C₂, where C₂ is the constant of integration.
By substituting the initial condition f(0) = -3, we can solve for C₂:
f(0) = 6sin(6(0)) - 11(0) + C₂
-3 = 6sin(0) + C₂
-3 = 0 + C₂
C₂ = -3
Now we have the indefinite integral of the first derivative function:
f(x) = 6sin(6x) - 11x - 3
Finally, we can find the value of f(π/3) by substituting x = π/3 into the function:
f(π/3) = 6sin(6(π/3)) - 11(π/3) - 3
This can be simplified to get the final answer for f(π/3).