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avatar+322 

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

 Apr 14, 2019
 #1
avatar+6196 
+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)

 Apr 14, 2019
 #2
avatar+322 
+1

My answer was right except the last sign.  It was supposed to be a (-) but I had a (+).

Ruublrr  Apr 15, 2019

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