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Math Help - Calculating critical points from a non factorable derivative

  1. #1
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    Calculating critical points from a non factorable derivative

    f(x) = x3 - 2x2 - 5x + 6 find critical points of f on the interval [4,8]
    I know I need to take the derivative,which is f1 = 3x2 - 4x -5. my problem is finding the critical points. I'm not sure what to do with this, because i cant factor it. well at least i don't think i can. Since the derivative is a quadratic function, can I use quadratic formula? Thanks in advance
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    Re: Calculating critical points from a non factorable derivative

    A critical point is defined to be any value in the domain were a function f is not differentiable or has the value 0. You have
    f(x) = x^3-2x^2-5x+6 => f'(x) = 3x^2-4x-5

    The function f(x) is differentiable everywhere. Thus we only have to search for the value in the domain were the derivative is 0, in the domain [4,8]
    f'(x) = 0 <=> 3x^2-4x-5 = 0 <=> x = 2/3 +/- sqrt(19)/3

    None of these values for x lies in the domain, hence the function doesn't have any critical points in the specified interval.
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    Re: Calculating critical points from a non factorable derivative

    I think some people also include the endpoints as critical points. Either way, if you're looking for the minimum or maximum of the function, you have to look at the endpoints.

    I attached a graph of your function.

    - Hollywood
    Attached Thumbnails Attached Thumbnails Calculating critical points from a non factorable derivative-temp.png  
    Last edited by hollywood; November 13th 2012 at 08:34 PM. Reason: change graph
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    Re: Calculating critical points from a non factorable derivative

    thanks for the help
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    Re: Calculating critical points from a non factorable derivative

    Maybe some people include the endpoints as critical points. However, the endpoints should not be included as endpoints.
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    Re: Calculating critical points from a non factorable derivative

    Quote Originally Posted by matthewporter1965 View Post
    f(x) = x3 - 2x2 - 5x + 6 find critical points of f on the interval [4,8]
    I know I need to take the derivative,which is f1 = 3x2 - 4x -5. my problem is finding the critical points. I'm not sure what to do with this, because i cant factor it. well at least i don't think i can. Since the derivative is a quadratic function, can I use quadratic formula? Thanks in advance
    Yes, you are correct. The original post, which is quoted above, defined the function over all real numbers. When the domain is not given, it is assumed to be the largest possible subset of real numbers, and of course a polynomial can be calculated for any real number.

    Given this function defined for all real numbers, he is asked for the critical points which are on the interval [4,8]. And since both critical points lie outside [4,8], the correct answer is that there are none.

    However, for various reasons, it is often the case that the endpoints can not be ignored.

    - Hollywood
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