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Math Help - critical points

  1. #1
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    critical points

    f(x) = 18x + 15x^2 -4x^3 on the interval [-3,4]

    A) f(-3) =
    f(4) =

    B) find the critical points of f(x) state x and y show your work

    C) State the absolute maximum and absolute minimum values of the function on the interval [-3,4]


    how do i get the critical points iv found the derivative to be 18+30x-12x^2 now what do i do?
    Last edited by burton6; October 26th 2009 at 09:22 PM.
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  2. #2
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    Quote Originally Posted by burton6 View Post
    f(x) = 18x + 15x^2 -4x^3 on the interval [-3,4]

    A) f(-3) =
    f(4) =

    B) find the critical points of f(x) state x and y show your work

    C) State the absolute maximum and absolute minimum values of the function on the interval [-3,4]
    You can't just post the question.

    Where are you having trouble with it? Have you made an attempt? Where is your working?
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  3. #3
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    ow do i get the critical points iv found the derivative to be 18+30x-12x^2 now what do i do?
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  4. #4
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    Quote Originally Posted by burton6 View Post
    ow do i get the critical points iv found the derivative to be 18+30x-12x^2 now what do i do?
    Critical points are points at which the gradient of the function is zero. The gradient of a function f(x) is described by its derivative f'(x), so you're looking for the points at which f'(x) = 0.

    So set f'(x) = 0, and solve for x. Once you have the values of x that solve this equation, plug them back into the original equation, f(x), to find the corresponding y values.
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  5. #5
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    f(x) = 18x + 15x^2 - 4x^3
    f'(x) = 18 + 30x - 12x^2

    Critical numbers are numbers in an interval where f'(c) = 0

    So just set the derivative to 0 and solve for the roots. If you can't evenly factor the equation then use the quadratic formula.

    It'll make it easier on you if you factor 6 out of f'.
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  6. #6
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    tanks got it
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