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Pre-Calculus

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I'm having problems solving this and showing the steps:

Find the absolute maximum and absolute minimum, if either exists, for

y = f(x)= x^3 - 12x + 12 -3 less equal to x greater than equal to 5

  • Pre-Calculus - ,

    I'll rewrite the equation.
    y = f(x) = x^3 -12x + 12
    -3 <= x >= 5

  • Pre-Calculus - ,

    dy/dx = 3x^2 -12
    function is horizontal when dy/dx = 0
    that is when
    x^2 = 4
    x = +/- 2
    when x = -2
    f(-2) = -8 + 24 + 12 = 28
    and
    d^2y/dx^2 = 6 x = -12 so a maximum
    when x = 2
    f(2) = -4
    and
    d^2y/dx^2 = 6x = +12 so a minimum
    now check the end points
    when x = -3
    f(-3) = -27+36+12 = 21
    when x=+5
    f(5) = 125 - 60 +12 = 77
    so the maximum is 77 at x = 5
    and the minimum is -4 at x = 2

  • Pre-Calculus - ,

    Thank you so much. This is a great help. I can see where I started to go wrong.

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