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Thread: impossible stationary points

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    impossible stationary points

    Hi I would like some assistance on this question Thanks

    find any turning point on the curve y=(x-3)√(4-x) and determine their nature.

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    Junior Member OkDen's Avatar
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    Re: impossible stationary points

    Quote Originally Posted by cos16mh View Post
    Hi I would like some assistance on this question Thanks

    find any turning point on the curve y=(x-3)√(4-x) and determine their nature.
    For Turning Points, dy/dx = 0

    To find derivative of y=(x-3)√(4-x) use Product Rule.

    So, dy/dx = (-3x+11)/2√(-x+4)
    Since, dy/dx = 0, 0 =(-3x+11)/2√(-x+4)

    Which gives: x = 11/3
    Sub into original function: y=(x-3)√(4-x), giving y = (2√3)/9

    To find nature of Turning Point, sub in turning points (11/3 , (2√3)/9) to the second derivative to determine nature.

    positive = minimum
    negative = maximium

    Or you can graph it, or test coordinates.

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    Re: impossible stationary points

    Quote Originally Posted by OkDen View Post


    For Turning Points, dy/dx = 0

    To find derivative of y=(x-3)√(4-x) use Product Rule.

    So, dy/dx = (-3x+11)/2√(-x+4)
    Since, dy/dx = 0, 0 =(-3x+11)/2√(-x+4)

    Which gives: x = 11/3
    Sub into original function: y=(x-3)√(4-x), giving y = (2√3)/9

    To find nature of Turning Point, sub in turning points (11/3 , (2√3)/9) to the second derivative to determine nature.

    positive = minimum
    negative = maximium

    Or you can graph it, or test coordinates.

    Its the second derivative that I'm struggling with

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    617 pages fan96's Avatar
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    Re: impossible stationary points

    Quote Originally Posted by cos16mh View Post
    Its the second derivative that I'm struggling with
    Then test the first derivative around the turning points.

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