R rushin25 Oct 2009 8 0 May 14, 2010 #1 f ''(x) = 8 + cos(x) f(0) = -1 here is what i got f '(x)= 8x + sinx +C f(x) = 4x^2 - cos x + Cx +D since f(0)= -1 f(0)= 0-1+0+D= -1 D = 0 Now the problem is..how do i substitute the value in f(x) ???

f ''(x) = 8 + cos(x) f(0) = -1 here is what i got f '(x)= 8x + sinx +C f(x) = 4x^2 - cos x + Cx +D since f(0)= -1 f(0)= 0-1+0+D= -1 D = 0 Now the problem is..how do i substitute the value in f(x) ???

Prove It MHF Helper Aug 2008 12,883 4,999 May 14, 2010 #2 rushin25 said: f ''(x) = 8 + cos(x) f(0) = -1 here is what i got f '(x)= 8x + sinx +C f(x) = 4x^2 - cos x + Cx +D since f(0)= -1 f(0)= 0-1+0+D= -1 D = 0 Now the problem is..how do i substitute the value in f(x) ??? Click to expand... If \(\displaystyle D = 0\) then \(\displaystyle f(x) = 4x^2 - \cos{x} + Cx\). So now \(\displaystyle 0 = 4\left(\frac{\pi}{2}\right)^2 - \cos{\left(\frac{\pi}{2}\right)} + \frac{\pi}{2}C\). Simplify and solve for \(\displaystyle C\).

rushin25 said: f ''(x) = 8 + cos(x) f(0) = -1 here is what i got f '(x)= 8x + sinx +C f(x) = 4x^2 - cos x + Cx +D since f(0)= -1 f(0)= 0-1+0+D= -1 D = 0 Now the problem is..how do i substitute the value in f(x) ??? Click to expand... If \(\displaystyle D = 0\) then \(\displaystyle f(x) = 4x^2 - \cos{x} + Cx\). So now \(\displaystyle 0 = 4\left(\frac{\pi}{2}\right)^2 - \cos{\left(\frac{\pi}{2}\right)} + \frac{\pi}{2}C\). Simplify and solve for \(\displaystyle C\).

Prove It MHF Helper Aug 2008 12,883 4,999 May 14, 2010 #4 rushin25 said: but what will be the value for 4(pi/2)^2 ? Click to expand... Surely \(\displaystyle 4\left(\frac{\pi}{2}\right)^2 = 4\left(\frac{\pi^2}{4}\right) = \pi^2\)...

rushin25 said: but what will be the value for 4(pi/2)^2 ? Click to expand... Surely \(\displaystyle 4\left(\frac{\pi}{2}\right)^2 = 4\left(\frac{\pi^2}{4}\right) = \pi^2\)...

R rushin25 Oct 2009 8 0 May 14, 2010 #5 lol...my bad...didn't see that... so should i use the pi value in radian?

Prove It MHF Helper Aug 2008 12,883 4,999 May 14, 2010 #6 rushin25 said: lol...my bad...didn't see that... so should i use the pi value in radian? Click to expand... Yes, if you are given values of \(\displaystyle \pi\) then it is safe to assume you are measuring in radians.

rushin25 said: lol...my bad...didn't see that... so should i use the pi value in radian? Click to expand... Yes, if you are given values of \(\displaystyle \pi\) then it is safe to assume you are measuring in radians.