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Math Help - Critical Value Problem

  1. #1
    Newbie
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    Critical Value Problem

    Hey all, I'm a little new here and am having trouble toward the end of the semester with my calculus assignments and appreciate all the help I can get.

    Find the critical values for the function.
    f(x)= 20x^3 3x^5
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  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
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    722
    To solve this you need to compute the derivative of f(x) and set it equal to 0.

    See section 3 for a quick fix.

    Derivative - Wikipedia, the free encyclopedia

    So for your function we get...

    f(x) = 20x^3 - 3x^5

    f'(x) = 60x^2 - 15x^4 = 0

    Clearly x=0 is a solution but we also have...

    (Divide through by 15)...

    4x^2 - x^4 = 0

    => x^2(4 - x^2) = 0

    Divide through by x^2...

    4 - x^2 = 0 which is the same as solving...

    x^2 - 4 = 0

    => (x+2)(x-2) = 0

    Hence x=\pm 2

    So 3 solutions are x=0, x=-2 and x=2.
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