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Math Help - Graphing the first and second derivative of a 3rd degree polynomial function

  1. #1
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    Graphing the first and second derivative of a 3rd degree polynomial function

    Given the graph of h(x) (Attached Below), sketch the graphs of h'(x) and h''(x).

    Please correct me if I am mistaken but I believe given the graph of h(x), the graph of the first derivative would be a parabola (opening up) with a minimum point at the coordinates (0,2) because there is a point of inflection on h(x) at x=2 and it is a continuously increasing function in the presented domain.

    But is there anything other than that we can deduce about the graph of the first derivative given h(x)? Like the y-intercept for example? How accurate can my sketch be?

    Thanks in advance.

    Sincerely,

    Raymond MacNeil
    Attached Thumbnails Attached Thumbnails Graphing the first and second derivative of a 3rd degree polynomial function-hx.jpg   Graphing the first and second derivative of a 3rd degree polynomial function-h-x-.jpg  
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  2. #2
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    Re: Graphing the first and second derivative of a 3rd degree polynomial function

    If it's the case that y = h(x) is a cubic, passes through the origin and the point with
    co-ordinates (3,3), and has an inflection at x = 2, (as appears to be from the graph), then you can calculate its equation exactly, (in which case you can then calculate its derivatives exactly). If the co-ordinates of any of those points is in doubt, then you can't.
    A good starting point is to assume that y' =k(x-2)^2=k(x^2-4x+4), which gives you an inflection at x=2.
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  3. #3
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    Re: Graphing the first and second derivative of a 3rd degree polynomial function

    Quote Originally Posted by BobP View Post
    If it's the case that y = h(x) is a cubic, passes through the origin and the point with
    co-ordinates (3,3), and has an inflection at x = 2, (as appears to be from the graph), then you can calculate its equation exactly, (in which case you can then calculate its derivatives exactly). If the co-ordinates of any of those points is in doubt, then you can't.
    A good starting point is to assume that y' =k(x-2)^2=k(x^2-4x+4), which gives you an inflection at x=2.
    Yes, those points are correct. None are in doubt. The graph I made (which is attached) was a makeshift I created in MS Paint because the PDF File I'm working through is secured. (Couldn't copy and paste.) Thanks for this info. I will follow up on this later in more detail as I have to go now. Thanks for your reply. Perhaps later, I may ask for additional advice should I have any complications determining the equation of the cubic function.

    All the best,

    Raymond
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