Core 3 differentiation stationary points

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    Find the co-ordinates of the stationary points on the curve y = (x+3)^2 (2-x)^3

    Is this correct??
    y' = u'v +v'u

    y' = (x +3)^2 - 3(2 - x )^2 + (2-x)^3 2(x+3)

    0 = (2-x)^2 (x+3) [-5x + 7]

    so stationary points are (0, 72) and (7/5, 13068/3125)
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    (Original post by Custardcream000)
    Find the co-ordinates of the stationary points on the curve y = (x+3)^2 (2-x)^3

    Is this correct??
    y' = u'v +v'u

    y' = (x +3)^2 - 3(2x-1)^2 + (2-x)^3 2(x+3)

    0 = (2-x)^2 (x+3) [-5x + 7]

    so stationary points are (0, 72) and (7/5, 13068/3125)
    Check the bold bit. That is not right.
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    (Original post by RDKGames)
    Check the bold bit. That is not right.
    i've typed it out wrong, i've changed the main post
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    (Original post by Custardcream000)
    i've typed it out wrong, i've changed the main post
    In that case check the factorisation. There should not be a 7.

    Also, where did you get (0,72) as a stationary point when x=0 isn't even a solution to the differential??? Plus there are 3 roots so there should be 3 stationary points, not 2.
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    (Original post by RDKGames)
    In that case check the factorisation. There should not be a 7.

    Also, where did you get (0,72) as a stationary point when x=0 isn't even a solution to the differential??? Plus there are 3 roots so there should be 3 stationary points, not 2.
    so you get x=2
    x=-3
    x=7/5
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    (Original post by Custardcream000)
    so you get x=2
    x=-3
    x=7/5
    Nope. First two are correct but not he last one. Check your factorisation.
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    (Original post by RDKGames)
    Nope. First two are correct but not he last one. Check your factorisation.
    x= -1
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    (Original post by Custardcream000)
    x= -1
    thanks for your help
 
 
 
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