f' ''(x)= 5 + cos(x) , f(0) = -1 , f(9pi/2) = 0

Find f(x).

I got f(x)= (5/2)x^2 - cos(x) + (45pi/4)x but I got the answer wrong. Not sure what I did wrong. I also found D=0

Ruublrr Apr 14, 2019

#1**+3 **

is that f''(x) (2 primes) or f'''(x) (3 primes)?

\(f''(x) = 5+\cos(x)\\ f'(x) = 5x + \sin(x)+ c_1\\ f(x) = \dfrac 5 2 x^2 - \cos(x) + c_1 x + c_2\)

\(f(0) = -1 + c_2 = -1\\ c_2=0\\ f(x) = \dfrac 5 2 x^2 - \cos(x) + c_1 x\)

\(f\left(\dfrac{9\pi}{2}\right) = 0\\ \dfrac 5 2 \cdot \dfrac{81\pi^2}{4} - 0 + c_1 \dfrac{9\pi}{2} = 0\\ \dfrac{405\pi^2}{8} + c_1 \dfrac{9\pi}{2} = 0\\ c_1 = -\dfrac{45\pi}{4}\\ f(x) = \dfrac 5 2 x^2 - \cos(x) -\dfrac{45\pi}{4}x \)

Maybe it is f'''(x)

Rom Apr 14, 2019