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Math Help - Average value of a function.

  1. #1
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    Average value of a function.

    Find the number b such that the average value of f(x)=2+6x-3x^2 on the intervak [0, b] is equal to 3.

    so I did the integal from 0 to b of 2+6x-3x^2 which is 2x+3x^2-x^3

    now what? Do i set it equal to 3 and find out what b is?

    Thanks
    Last edited by mr fantastic; July 28th 2010 at 07:37 PM.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    f(c)=\frac{1}{b-a}\int^b_a {f(x)dx} when c\in[a,b]

    You looking for b when f(c)=3.
    Last edited by Also sprach Zarathustra; July 28th 2010 at 04:48 PM. Reason: mistake
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  3. #3
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    For F(x) = \int f(x)~dx find F(b) = 3
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  4. #4
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    right, so I set 2x+3x^2-x^3 = 3 right?

    = x(2+3x-x^2)
    =x(x -1) (x-2)

    ahhh i'm confused....
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  5. #5
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    Quote Originally Posted by softballchick View Post
    right, so I set 2x+3x^2-x^3 = 3 right?
    Yep, then

    2b+3b^2-b^3 = 3

    2b+3b^2-b^3 - 3=0

    Now use the factor thm to find the roots. You know this theorem?
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  6. #6
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    um... I don't think I do. could u elaborate more? Thanks
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  7. #7
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    You need to colve this cubic 2b+3b^2-b^3 - 3=0

    To do that you should employ the factor theorem. It's really not something that can be explained very well in a few lines but here goes.

    The Factor Theorem: For the polynomial P(x) if P(a)=0 then x-a is a factor of P(x)

    Example to follow is here. The Fundamental Theorem of Algebra
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    Mr. pickslides I convinced you aware that the roots not so "nice"...
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  9. #9
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Mr. pickslides I convinced you aware that the roots not so "nice"...
    I did not look that closely!
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