calculus

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Consider the function f(x)=(x^3)(e^9x),
-2 is less than or equal to x is less than or equal to 4

*The absolute maximum value is _____ and this occurs at x equals 4.

*The absolute minimum value is _____ and this occurs at x equals ______

I got the 4 but I must be doing something wrong with the other numbers because it is not working

  • calculus -

    let's take the derivative
    f'(x) = 3x^2 (e^(9x) ) + x^3 (9) e^(9x)
    = 3x^2 e^(9x) [ 1 + 3x]
    = 0 for a max/min

    x = 0 , x = -1/3

    f(0) = 0
    f(-1/3) = (-1/27) e^-3 = -1/(27e^3) = appr -.00184
    f(-2) = -8 e^-18 = -8/e^18 = appr -.0000000121

    so the absolute minimum is -8/e^18 , when x = -2

  • calculus -

    wrong is x=-4

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