finding a function

**ascendancy523** · April 3rd 2010, 11:28 AM

So here's my problem:

Given $f\prime\prime(x)=-9sin(3x)$
$f\prime(0)=-4$
and $f(0)=-1$

Find $f(\frac{\pi}{2})=$

Any suggestions are greatly appreciated.

**mathemagister** · April 3rd 2010, 11:59 AM

Originally Posted by ascendancy523

So here's my problem:

Given $f\prime\prime(x)=-9sin(3x)$
$f\prime(0)=-4$
and $f(0)=-1$

Find $f(\frac{\pi}{2})=$

Any suggestions are greatly appreciated.

$f'(x) = \int f''(x)\ dx = \int -9\sin{(3x)}\ dx = 3\cos{(3x)} + C$

$f'(0) = -4 \implies 3\cos{(3\cdot 0)} + C = -4 \implies 3 + C = -4 \implies C = -7$

Now you know $f'(x) = 3\cos{(3x)} - 7$

$f(x) = \int f\ '(x)\ dx = \int 3\cos{(3x)} - 7 \ dx = \sin{(3x)}-7x +D$

$f(0)=-1 \implies \sin{(3\cdot 0)}-7\cdot 0 +D = -1 \implies D = -1$

$f(x) = \sin{(3x)}-7x -1$

$f\left(\frac{\pi}{2}\right) = \sin{\left(\frac{3\pi}{2}\right)}- \frac{7\pi}{2}-1=-1-\frac{7\pi}{2}-1 = \boxed{-\frac{7\pi}{2}-2}$

Hope that helps

Mathemagister

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