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Math Help - Please help with this question

  1. #1
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    Please help with this question

    if L(x)=32000/x^2+6

    and L'(x)=-64000/x^3+6

    Solve L'(x)=0 (to two decimal places)
    Last edited by PapaJones; May 30th 2012 at 12:01 AM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Please help with this question

    The derivative L'(x) is not correct.
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  3. #3
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    Re: Please help with this question

    Derivative of a constant is 0.
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  4. #4
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    Re: Please help with this question

    anyhelp please?
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Please help with this question

    You have to solve L'(x)=0 \Leftrightarrow \frac{-64000}{x^3}=0, this is impossible
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  6. #6
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    Re: Please help with this question

    It's impossible to solve Or Was the question:

    L(x) = \frac{32000}{x^2+6}
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  7. #7
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    Re: Please help with this question

    Quote Originally Posted by Siron View Post
    You have to solve L'(x)=0 \Leftrightarrow \frac{-64000}{x^3}=0, this is impossible
    L'(x) = 0 = -64000/x^3 +6

    Forgot the 6
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  8. #8
    Member Goku's Avatar
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    Re: Please help with this question

    Quote Originally Posted by PapaJones View Post
    L'(x) = 0 = -64000/x^3 +6

    Forgot the 6
    The derivative of 6 is 0.
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  9. #9
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    Re: Please help with this question

    Quote Originally Posted by PapaJones View Post
    if L(x)=32000/x^2+6

    and L'(x)=-64000/x^3+6

    Solve L'(x)=0 (to two decimal places)
    is the original function L(x) = \frac{32000}{x^2} + 6 or L(x) = \frac{32000}{x^2+6} ?

    ... in either case, your derivative is incorrect.
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  10. #10
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    Re: Please help with this question

    If L(x) =32000/(x^2+6) the derivation is:
    L'(x) = - 64000/[(x^2+6)^2]*2x and L'(x) = 0 <=> x = 0
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  11. #11
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    Re: Please help with this question

    Quote Originally Posted by PapaJones View Post
    anyhelp please?
    What kind of "help" do you want? You have simply ignored every post so far.
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