anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Jhannybean
  • Jhannybean
Let's integrate f''(x). what did you get?
Jhannybean
  • Jhannybean
\[f'(x) = \int f''(x)\]
anonymous
  • anonymous
2x+sinx

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Jhannybean
  • Jhannybean
-_- im a little off today altogether, lol
Jhannybean
  • Jhannybean
\[f'(x) = 2x+\sin(x)\]\[f(x) = \int f'(x) = ~?\]
anonymous
  • anonymous
\[x^2-\cos(x)\]
Jhannybean
  • Jhannybean
Great. Now we take care of our conditions.
Jhannybean
  • Jhannybean
\[f(x) = x^2-\cos(x) + Cx + D\] Don't forget that you add on a constant every time you integrate.
anonymous
  • anonymous
right I know I'm not sure why I left it off lol
Jhannybean
  • Jhannybean
So if \(f(0) =10\) then we're basically finding our D value. \[10=(0)^2-\cos(0)+C(0) +D\qquad \implies D=~?\]
anonymous
  • anonymous
9?
Jhannybean
  • Jhannybean
Tell me how you got 9.
anonymous
  • anonymous
no never mind its 11
Jhannybean
  • Jhannybean
Yeah
anonymous
  • anonymous
d =11
Jhannybean
  • Jhannybean
Replace D with 11*
anonymous
  • anonymous
do you mind rewriting the whole problem out again please with d replaced with 11
anonymous
  • anonymous
i believe you're using equation mode but I'm not sure if i operate it correctly lol
Jhannybean
  • Jhannybean
\[f(x)=x^2-\cos(x)+Cx+11\] And now we take care of our second condition. \(f\left(\dfrac{\pi}{2}\right) = 1\)
Jhannybean
  • Jhannybean
So \[1=\left(\frac{\pi}{2}\right)^2 - cos\left(\frac{\pi}{2}\right) +\frac{\pi}{2}C +11\]Solve for C.
Jhannybean
  • Jhannybean
I wouldn't bother simplifying anything but isolating C first. Then you can simplify allyou want.Makes thigns a lot easier
anonymous
  • anonymous
so far i got \[(-10-\pi/4)/(\pi/2)\]=C
freckles
  • freckles
(pi/2)^2=pi^2/4 not pi/4
anonymous
  • anonymous
oh yes that is true i forgot to add that in lo i had t on my paper but didnt type it
Jhannybean
  • Jhannybean
\[1-\left(\frac{\pi}{2}\right)^2 +\cos\left(\frac{\pi}{2}\right) -11=\frac{\pi}{2}C\]Simplify the LHS \[1-\frac{\pi^2}{4}-11 = \frac{\pi}{2}C\]\[\frac{4-\pi^2-44}{4}=\frac{\pi}{2}C\]\[\frac{-40-\pi^2}{4}\cdot \frac{2}{\pi}=C\]\[C= \frac{-40-\pi^2}{2\pi} =-\frac{20}{\pi} -\frac{\pi}{2} \]
Jhannybean
  • Jhannybean
Are you able to follow that?
anonymous
  • anonymous
so the answer should be \[x^2-\cos(x)-20/\pi-\pi/2+11\]
Jhannybean
  • Jhannybean
\[\therefore \qquad f(x) = x^2-\cos(x)+\left(-\frac{20}{\pi}-\frac{\pi}{2}\right)x +11\]
Jhannybean
  • Jhannybean
Yeah, that's right
anonymous
  • anonymous
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 !
Jhannybean
  • Jhannybean
np
abb0t
  • abb0t
Hmm......

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