Calculus

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 3x^(1/3) + 6x^(4/3). Please use an analysis of f ′(x) and f ′′(x).

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asked by Mel
  1. if y = 3x^(1/3) + 6x^(4/3)

    then
    dy/dx = x^-(2/3)+ 8 x^(1/3)

    and
    d^2y/dx^2
    =-(2/3)x^-(5/3)+(8/3)x^-(2/3)

    extrema or inflection where dy/dx = 0

    note slope undefined when x = 0


    dy/dx = x^-(2/3)+ 8 x^(1/3) = 0
    or
    1/x^(2/3) = - 8 x^(1/3)
    1 = -8 x
    x = -1/8

    so what is d^2/dy^2 when x = -1/8 ?

    d^2y/dx^2
    =-(2/3)x^-(5/3)+(8/3)x^-(2/3)

    = -(2/3)(-1/8)^-5/3 + (8/3) (-1/8)^-(2/3)

    well -1/8^(5/3) = (-1/2)^5 = -1/32
    so
    (-1/8)^-5/3 = -32
    and we have so far
    -(2/3)(-32) +(8/3) (-1/8)^-(2/3)
    64/3 + (-1/2)^-2 = 1/(1/4) = 4
    64/3 + 4
    so second derivative is positive, a relative minimum at x = -1/8
    CHECK MY ARITHMETIC !!!

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    posted by Damon
  2. You got it right bro

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    posted by John

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