f''(x)=2+cosx, f(0)=10, f(pi/2)=1 is the last one that I am stuck on. Find f @jhannybean @freckles @FireKat97

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Let's integrate f''(x). what did you get?
\[f'(x) = \int f''(x)\]
2x+sinx

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-_- im a little off today altogether, lol
\[f'(x) = 2x+\sin(x)\]\[f(x) = \int f'(x) = ~?\]
\[x^2-\cos(x)\]
Great. Now we take care of our conditions.
\[f(x) = x^2-\cos(x) + Cx + D\] Don't forget that you add on a constant every time you integrate.
right I know I'm not sure why I left it off lol
So if \(f(0) =10\) then we're basically finding our D value. \[10=(0)^2-\cos(0)+C(0) +D\qquad \implies D=~?\]
9?
Tell me how you got 9.
no never mind its 11
Yeah
d =11
Replace D with 11*
do you mind rewriting the whole problem out again please with d replaced with 11
i believe you're using equation mode but I'm not sure if i operate it correctly lol
\[f(x)=x^2-\cos(x)+Cx+11\] And now we take care of our second condition. \(f\left(\dfrac{\pi}{2}\right) = 1\)
So \[1=\left(\frac{\pi}{2}\right)^2 - cos\left(\frac{\pi}{2}\right) +\frac{\pi}{2}C +11\]Solve for C.
I wouldn't bother simplifying anything but isolating C first. Then you can simplify allyou want.Makes thigns a lot easier
so far i got \[(-10-\pi/4)/(\pi/2)\]=C
(pi/2)^2=pi^2/4 not pi/4
oh yes that is true i forgot to add that in lo i had t on my paper but didnt type it
\[1-\left(\frac{\pi}{2}\right)^2 +\cos\left(\frac{\pi}{2}\right) -11=\frac{\pi}{2}C\]Simplify the LHS \[1-\frac{\pi^2}{4}-11 = \frac{\pi}{2}C\]\[\frac{4-\pi^2-44}{4}=\frac{\pi}{2}C\]\[\frac{-40-\pi^2}{4}\cdot \frac{2}{\pi}=C\]\[C= \frac{-40-\pi^2}{2\pi} =-\frac{20}{\pi} -\frac{\pi}{2} \]
Are you able to follow that?
so the answer should be \[x^2-\cos(x)-20/\pi-\pi/2+11\]
\[\therefore \qquad f(x) = x^2-\cos(x)+\left(-\frac{20}{\pi}-\frac{\pi}{2}\right)x +11\]
Yeah, that's right
damn i forgot the x and it said it was wrong i guess I'm getting tired lol well thank you for all your help !
np
Hmm......

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