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find the absolute minimum and maximum values of f(x)=x^(1/3) (3+x) on the interval [-2,1]

  • calculus -

    f' = (4x+3)/(3x^(2/3))
    This is undefined at x=0, where f(0) = 0, a local min.

    f'=0 at x = -3/4 where there is a local max. f(-3/4) = -(9/4) ∛(3/4)

    check the value of f at the end points for other max/min values.

  • calculus -

    do not get it :(

  • calculus -

    what don't you get?
    max/min occurs where f' = 0, as long as f" is not also zero

    where f' is undefined, check to see whether f is defined there. If so, it will be some kind of cusp, so a max or min. Or it will have a vertical tangent.

    As it turns out, I made a mistake at x=0; there is no local min there.
    And, the extremum at x = -3/4 is a min, not a max.

    Oddly, this error was caused by my trust in the graphs at, which showed it wrong.

    To see the real graph, visit

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