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Math Help - Maximum and minimum of function in the interval

  1. #1
    Newbie Critter314's Avatar
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    Maximum and minimum of function in the interval

    Maximum and minimum of f(x) = 3x^4 - 8x^3 -48x^2 + 5 in the interval -3 \leq x \leq 5

    ---

    f'(x) = 12x^3 - 24x^2 - 96x

    Now what do I do?
    Last edited by mr fantastic; June 9th 2010 at 06:47 PM. Reason: Copied post title into main body of post.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Critter314 View Post
    f(x) = 3x^4 - 8x^3 -48x^2 + 5

    Interval

    -3 \leq x \leq 5

    ---

    f'(x) = 12x^3 - 24x^2 - 96x

    Now what do I do?
    set f'(x) = 0 to determine critical values since extrema occur at critical values.

    don't forget to check the value of f(x) at the endpoints.
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    note that [-3,5] is a compact set and f is continuous, thus by the Extreme Value Theorem f is f(-3)\le f(x)\le f(5).
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