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anonymous

  • one year ago

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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  1. Jhannybean
    • one year ago
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    Let's integrate f''(x). what did you get?

  2. Jhannybean
    • one year ago
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    \[f'(x) = \int f''(x)\]

  3. anonymous
    • one year ago
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    2x+sinx

  4. Jhannybean
    • one year ago
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    -_- im a little off today altogether, lol

  5. Jhannybean
    • one year ago
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    \[f'(x) = 2x+\sin(x)\]\[f(x) = \int f'(x) = ~?\]

  6. anonymous
    • one year ago
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    \[x^2-\cos(x)\]

  7. Jhannybean
    • one year ago
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    Great. Now we take care of our conditions.

  8. Jhannybean
    • one year ago
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    \[f(x) = x^2-\cos(x) + Cx + D\] Don't forget that you add on a constant every time you integrate.

  9. anonymous
    • one year ago
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    right I know I'm not sure why I left it off lol

  10. Jhannybean
    • one year ago
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    So if \(f(0) =10\) then we're basically finding our D value. \[10=(0)^2-\cos(0)+C(0) +D\qquad \implies D=~?\]

  11. anonymous
    • one year ago
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    9?

  12. Jhannybean
    • one year ago
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    Tell me how you got 9.

  13. anonymous
    • one year ago
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    no never mind its 11

  14. Jhannybean
    • one year ago
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    Yeah

  15. anonymous
    • one year ago
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    d =11

  16. Jhannybean
    • one year ago
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    Replace D with 11*

  17. anonymous
    • one year ago
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    do you mind rewriting the whole problem out again please with d replaced with 11

  18. anonymous
    • one year ago
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    i believe you're using equation mode but I'm not sure if i operate it correctly lol

  19. Jhannybean
    • one year ago
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    \[f(x)=x^2-\cos(x)+Cx+11\] And now we take care of our second condition. \(f\left(\dfrac{\pi}{2}\right) = 1\)

  20. Jhannybean
    • one year ago
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    So \[1=\left(\frac{\pi}{2}\right)^2 - cos\left(\frac{\pi}{2}\right) +\frac{\pi}{2}C +11\]Solve for C.

  21. Jhannybean
    • one year ago
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    I wouldn't bother simplifying anything but isolating C first. Then you can simplify allyou want.Makes thigns a lot easier

  22. anonymous
    • one year ago
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    so far i got \[(-10-\pi/4)/(\pi/2)\]=C

  23. freckles
    • one year ago
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    (pi/2)^2=pi^2/4 not pi/4

  24. anonymous
    • one year ago
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    oh yes that is true i forgot to add that in lo i had t on my paper but didnt type it

  25. Jhannybean
    • one year ago
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    \[1-\left(\frac{\pi}{2}\right)^2 +\cos\left(\frac{\pi}{2}\right) -11=\frac{\pi}{2}C\]Simplify the LHS \[1-\frac{\pi^2}{4}-11 = \frac{\pi}{2}C\]\[\frac{4-\pi^2-44}{4}=\frac{\pi}{2}C\]\[\frac{-40-\pi^2}{4}\cdot \frac{2}{\pi}=C\]\[C= \frac{-40-\pi^2}{2\pi} =-\frac{20}{\pi} -\frac{\pi}{2} \]

  26. Jhannybean
    • one year ago
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    Are you able to follow that?

  27. anonymous
    • one year ago
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    so the answer should be \[x^2-\cos(x)-20/\pi-\pi/2+11\]

  28. Jhannybean
    • one year ago
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    \[\therefore \qquad f(x) = x^2-\cos(x)+\left(-\frac{20}{\pi}-\frac{\pi}{2}\right)x +11\]

  29. Jhannybean
    • one year ago
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    Yeah, that's right

  30. anonymous
    • one year ago
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    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 !

  31. Jhannybean
    • one year ago
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    np

  32. abb0t
    • one year ago
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    Hmm......

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