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anonymous

  • one year ago

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = x3 − 9x on the interval [−1, 1]. If so, find the x-coordinates of the point(s) guaranteed to exist by the theorem

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  1. zzr0ck3r
    • one year ago
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    Is the function continuous on the interval? Is the interval closed?

  2. anonymous
    • one year ago
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    um it is continuous and i do believe it is closed

  3. zzr0ck3r
    • one year ago
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    Ok, so the theorem applies.

  4. anonymous
    • one year ago
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    oh really, okay great that was actually the part i was most unsure of

  5. zzr0ck3r
    • one year ago
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    Now we need to look at \[F(b) - F(a)=F'(c)(b-a) \] and we have \(F(x) = \int_a^xf(t) dt\) where \(a=-1\) and \(b=1\).

  6. anonymous
    • one year ago
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    okay so then \[ \int\limits_{-1}^{1}x^3 -9x\]

  7. anonymous
    • one year ago
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    so is c the x-coordinate im looking for?

  8. zzr0ck3r
    • one year ago
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    \(\int_{-1}^1(t^3-9t) dt = 2(c^3-9c)\) Correct.

  9. anonymous
    • one year ago
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    okay so im getting 2/3 fro F(b) - F(a)

  10. anonymous
    • one year ago
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    do i just solve for c now?

  11. zzr0ck3r
    • one year ago
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    hmm cubic around the origin shuold even out on that interval

  12. zzr0ck3r
    • one year ago
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    http://www.wolframalpha.com/input/?i=%5Cint_%7B-1%7D%5E1%28t%5E3-9t%29+dt

  13. zzr0ck3r
    • one year ago
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    but yes, solve for \(c\). I get \(0=2c(c^2-9)\)

  14. anonymous
    • one year ago
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    oh woops i see why i got 2/3 , i put t^2 instead of t^3

  15. zzr0ck3r
    • one year ago
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    \(-3\) is not in our interval.

  16. anonymous
    • one year ago
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    are you sure i should leave it as zero and not 8.5

  17. zzr0ck3r
    • one year ago
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    ?

  18. anonymous
    • one year ago
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    okay nvm sorry iv been solving a lot of area and volume problems like these in which area and volume cant be negative so i am told to interpret negatives as positives put i guess it doesn't apply here :)

  19. anonymous
    • one year ago
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    couldn't c = 0

  20. anonymous
    • one year ago
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    \[0 = 2c(c^2-9)\] \[0 = 2c\] \[0 = c\]

  21. zzr0ck3r
    • one year ago
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    \(0=2c(c^2-9)\\\dfrac{0}{2c}=c^2-9\\0=c^2-9\\c^2=9\)

  22. anonymous
    • one year ago
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    c = 3 , 0

  23. zzr0ck3r
    • one year ago
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    correct

  24. anonymous
    • one year ago
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    so is my answer 0 because 3 is not in the given interval?

  25. zzr0ck3r
    • one year ago
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    yes ...sorry

  26. anonymous
    • one year ago
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    okay great, thanks for the help :D

  27. zzr0ck3r
    • one year ago
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    I think at some point i said \(3\in [-1,1]\) doooh. deleted it...

  28. zzr0ck3r
    • one year ago
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    np m8

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