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Math Help - Nature of a point

  1. #1
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    Nature of a point

    Given that \frac{dy}{dx}=6(x-1)(x-2) and the coordinates of the stationery points are (1,1) and (2,0), determine the nature of these stationery points.
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    Quote Originally Posted by Punch View Post
    Given that \frac{dy}{dx}=6(x-1)(x-2) and the coordinates of the stationery points are (1,1) and (2,0), determine the nature of these stationery points.
    The x-coordinates of these points will make the drerivative = zero.
    so what do you think about thier nature?
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    Quote Originally Posted by General View Post
    The x-coordinates of these points will make the drerivative = zero.
    so what do you think about thier nature?
    The answer is either minimum point, maximum point or a point of inflexion.
    The answer can be derived by substituting in values which are < and > the x-coordinate of the stationery point to determine if the stationery point is a maximum, minimum or a point of inflexion.

    However, the answer I have gotten does not tally with the answer from the book.
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    Quote Originally Posted by Punch View Post
    The answer is either minimum point, maximum point or a point of inflexion.
    The answer can be derived by substituting in values which are < and > the x-coordinate of the stationery point to determine if the stationery point is a maximum, minimum or a point of inflexion.

    However, the answer I have gotten does not tally with the answer from the book.
    You did not specify the choices.
    find \frac{d^2y}{dx^2} and try to see if one of two points is a point of inflection or not.
    Also, how can you know if the point has maximum/minimum value for its function?
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  5. #5
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    okay, here is an example...
    If by substituting values of x < x-coordinate into dy/dx, you get a negative gradient, this means that it is down sloping.
    If by substituting values of x > x-coordinate into dy/dx, you get a positive gradient, this means that it is up sloping.

    So we can conclude that the x-coordinate point is a minimum point as the gradient is negative before increasing after the stationery point where dy/dx=0.
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  6. #6
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    Yes, now here is your function: dy/dx= 6(x-1)(x-2).

    If x< 1, both x- 1 and x- 2 are negative. What does that tell you about their product?

    If 1< x< 2, x- 1 is positive while x- 2 is negative. What does that tell you about their product?

    If 2< x, both x- 1 and x- 2 are positive. What does that tell you about their product?
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