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differentiation turning points help

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    a cubic curve has equation y=x^3-3x^2+1
    use calculus to find the coordinates of the turning points on this curve. Determine the nature
    of these turning points.

    so dy/dx= 3x^2-6x
    turning point dy/dx=0
    3x^2-6x=0
    so x^2-2x=0 x(x-2) so x=2

    to determine the nature what do i do?
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    x=2 or x=?


    Second differential (or alternate methods ... depends on which you have been taught)
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    (Original post by dongonaeatu)
    a cubic curve has equation y=x^3-3x^2+1
    use calculus to find the coordinates of the turning points on this curve. Determine the nature
    of these turning points.

    so dy/dx= 3x^2-6x
    turning point dy/dx=0
    3x^2-6x=0
    so x^2-2x=0 x(x-2) so x=2

    to determine the nature what do i do?
     3x^2 -6x =0 \implies x^2 - 2x=0 \implies x(x-2)=0

    You should get two x values.
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    Second derivative test?
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    (Original post by TenOfThem)
    x=2 or x=?


    Second differential (or alternate methods ... depends on which you have been taught)

    (Original post by raheem94)
     3x^2 -6x =0 \implies x^2 - 2x=0 \implies x(x-2)=0

    You should get two x values.
    x(x-2) so x=2 or x=0

    so d2y/dx2= 6x-6

    so when x=0 d2y/dx2= -6 so its a maximum?

    when x=2 d2y/dx2= 6(2)-6 = 6 so x=2 is a minmum

    is this right
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    (Original post by dongonaeatu)


    is this right
    What do you think?

    Perhaps a quick sketch of the graph will confirm (or not)
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    (Original post by dongonaeatu)
    x(x-2) so x=2 or x=0

    so d2y/dx2= 6x-6

    so when x=0 d2y/dx2= -6 so its a maximum?

    when x=2 d2y/dx2= 6(2)-6 = 6 so x=2 is a minmum

    is this right
    See the graph of the function:




    If you still can't ensure that you are right or wrong, then see the spoiler.

    Spoiler:
    Show

    You are correct!
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    (Original post by dongonaeatu)
    x(x-2) so x=2 or x=0

    so d2y/dx2= 6x-6

    so when x=0 d2y/dx2= -6 so its a maximum?

    when x=2 d2y/dx2= 6(2)-6 = 6 so x=2 is a minmum

    is this right
    Also remember that you need to find the y-coordinates of the stationary points as well.
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    (Original post by TenOfThem)
    What do you think?

    Perhaps a quick sketch of the graph will confirm (or not)

    (Original post by raheem94)
    See the graph of the function:




    If you still can't ensure that you are right or wrong, then see the spoiler.

    Spoiler:
    Show

    You are correct!
    yay! thank you guys
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    (Original post by raheem94)
    Also remember that you need to find the y-coordinates of the stationary points as well.
    ohh ok so i just substitute x into y. so

    when x=0 y=1 so (0,1)
    when x=2 y= (2)^3-3(2)^2+1 y=-3 so (2,-3)
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    (Original post by dongonaeatu)
    ohh ok so i just substitute x into y. so

    when x=0 y=1 so (0,1)
    when x=2 y= (2)^3-3(2)^2+1 y=-3 so (2,-3)
    :yep:
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    part ii (8 marks)
    Show that the tangent to the curve at the point where x=-1 has gradient 9.
    Find the coordinates of the other point, P, on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P.
    Show that the area of the triangle bounded by the normal at P and the x- and y-axes is 8 square units.

    so dy/dx is 3x^2-6x
    where x=-1 dy/dx= 3(-1)^2-6(-1)=9 so i have shown the tangent to the curve at x=-1 has gradient 9.

    now what do i do
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    (Original post by raheem94)
    :yep:
    please look up at part ii
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    Find the equation of the tangent

    Note ... you know a point and the gradient
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    (Original post by TenOfThem)
    Find the equation of the tangent

    Note ... you know a point and the gradient
    i need to find the coordinates of P on the curve at which the tangent has gradient 9

    so is it 3x^2-6x=9

    so 3x^2-6x-9=0 then i factorise?
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    (Original post by dongonaeatu)
    please look up at part ii
    To find the other point, solve,
     3x^2-6x=9
    Also find the y-coordinate.

    Then write the gradient of the normal, as -1 divided by gradient of tangent.

    Then form the equation.

    (Original post by TenOfThem)
    Find the equation of the tangent

    Note ... you know a point and the gradient
    Why find the equation of the tangent, he needs to find the equation of the normal.
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    (Original post by dongonaeatu)
    i need to find the coordinates of P on the curve at which the tangent has gradient 9

    so is it 3x^2-6x=9

    so 3x^2-6x-9=0 then i factorise?
    Its obvious that you need to factorise it and find the solutions.

    By the way, have you read the complete chapter?
    Or do you want us to help you on every step?
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    (Original post by raheem94)
    To find the other point, solve,
     3x^2-6x=9
    Also find the y-coordinate.

    Then write the gradient of the normal, as -1 divided by gradient of tangent.

    Then form the equation.



    Why find the equation of the tangent, he needs to find the equation of the normal.
    I was just going a different way

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    (Original post by raheem94)
    Its obvious that you need to factorise it and find the solutions.

    By the way, have you read the complete chapter?
    Or do you want us to help you on every step?

    (Original post by TenOfThem)
    I was just going a different way

    sorry, i just find you guys are very good at explaining and it helps me

    so 3x^2-6x-9=0 (3x+3)(x-3) so x=3 put this into y so y=1 so p is (3,1)

    equation of normal to P so gradient of normal= -1/9

    y-1=-1/9(x-3)
    9y-9=-x+3
    equation of normal to the curve at p is 9y=-x+12
    how do i show that the area of the triangle bounded by the normal at P and the x axis is 8 square units
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    (Original post by TenOfThem)
    I was just going a different way

    What was your different way?

    You said you know a point and a gradient, so the point is (-1,-3), gradient is 9.
    Equation of this tangent is: y=9x+6

    So what should we do next if we follow your approach?

    :confused:

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Updated: May 11, 2012
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