So here's my problem:
Given $\displaystyle f\prime\prime(x)=-9sin(3x)$
$\displaystyle f\prime(0)=-4$
and $\displaystyle f(0)=-1$
Find $\displaystyle f(\frac{\pi}{2})=$
Any suggestions are greatly appreciated.
$\displaystyle f'(x) = \int f''(x)\ dx = \int -9\sin{(3x)}\ dx = 3\cos{(3x)} + C$
$\displaystyle f'(0) = -4 \implies 3\cos{(3\cdot 0)} + C = -4 \implies 3 + C = -4 \implies C = -7$
Now you know $\displaystyle f'(x) = 3\cos{(3x)} - 7$
$\displaystyle f(x) = \int f\ '(x)\ dx = \int 3\cos{(3x)} - 7 \ dx = \sin{(3x)}-7x +D$
$\displaystyle f(0)=-1 \implies \sin{(3\cdot 0)}-7\cdot 0 +D = -1 \implies D = -1 $
$\displaystyle f(x) = \sin{(3x)}-7x -1$
$\displaystyle f\left(\frac{\pi}{2}\right) = \sin{\left(\frac{3\pi}{2}\right)}- \frac{7\pi}{2}-1=-1-\frac{7\pi}{2}-1 = \boxed{-\frac{7\pi}{2}-2}$
Hope that helps
Mathemagister