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C1 Graphs

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    Find the co-ordinates of the stationary points on the curve with equation y=x(x-1)2.

    Find the set of real values of k such that the equation x(x-1)2=k2 has exactly one real root.

    I really don't know how to do the second part of the question.
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    (Original post by Julii92)
    Find the co-ordinates of the stationary points on the curve with equation y=x(x-1)2.

    Find the set of real values of k such that the equation x(x-1)2=k2 has exactly one real root.

    I really don't know how to do the second part of the question.
    Is that really a C1 question? Because I get the set of values to be (-\frac{2}{3\sqrt{3}},0) \cup (0,\frac{2}{3\sqrt{3}}) using a method I just googled, and it is unlikely that it was a C1 method.
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    (Original post by Julii92)
    Find the co-ordinates of the stationary points on the curve with equation y=x(x-1)2.

    Find the set of real values of k such that the equation x(x-1)2=k2 has exactly one real root.

    I really don't know how to do the second part of the question.
    Well since x(x-1)^2 must be a square value:

    x-1 = x(x-1) OR (x-1)^2 = x

    Then solve for x on both (that's the method I can see at first glance).

    In fact, scrap that - it doesn't really incorporate k in any way Doesn't make sense to me - are you sure you've written the question correctly?



    EDIT: Scrap all of this. Think about the shape of a cubic graph, and where your stationary points are (y-coordinate wise). You need to make sure both are above y = 0 (which you do by adjusting the value of k)
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    (Original post by atti.08)
    This might be wrong but don't you bring the k over and then use the b^2-4ac formula, the discriminant?

    Wait don't bother, it'is most likely wrong
    It's not a quadratic equation so can't use that method
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    (Original post by atti.08)
    This might be wrong but don't you bring the k over and then use the b^2-4ac formula, the discriminant?

    Wait don't bother, it'is most likely wrong
    That only works with quadratics, unfortunately.
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    (Original post by Zii)
    Is that really a C1 question? Because I get the set of values to be (-\frac{2}{3\sqrt{3}},0) \cup (0,\frac{2}{3\sqrt{3}}) using a method I just googled, and it is unlikely that it was a C1 method.
    That's the right answer - what method was that?
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    math is my enermy
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    (Original post by Julii92)
    That's the right answer - what method was that?
    Read my post. You need to consider the shape of a cubic graph.

    Look at this working here:

    Click image for larger version. 

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    Very nasty C1 question!

    (Sorry about the messy start, I screwed up the differentiation :facepalm: )
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    (Original post by Julii92)
    That's the right answer - what method was that?
    The discriminant of a cubic ax^3+bx^2+cx+d which satisfies ax^3+bx^2+cx+d = 0 is given by (brace yourself)

    b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd

    When the discrimant is strictly less than 0, this corresponds to the cubic having precisely one real root.

    So I took your cubic, x(x-1)^2-k^2, expanded the brackets to get x^3-2x^2+x-k^2, calculated the discrimant and solved for the discriminant < 0. Following through the working gives

    k^2(k-\frac{2}{3\sqrt{3}})(k+\frac{2}{  3\sqrt{3}}) < 0

    which gives the set of values.

    Edit: Should have written k, not x!
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    (Original post by Zii)
    The discriminant of a cubic ax^3+bx^2+cx+d which satisfies ax^3+bx^2+cx+d = 0 is given by (brace yourself)

    b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd

    When the discrimant is strictly less than , this corresponds to the cubic having precisely one real root.

    So I took your cubic, x(x-1)^2-k^2, expanded the brackets to get x^3-2x^2+x-k^2, calculated the discrimant and solved for the discriminant < 0. Following through the working gives

    x^2(x-\frac{2}{3\sqrt{3}})(x+\frac{2}{  3\sqrt{3}}) < 0

    which gives the set of values.
    Much better working with what we already know (look at my post above)
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    (Original post by hassi94)
    Much better working with what we already know (look at my post above)
    I agree but I like my method because it's cool
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    (Original post by Zii)
    I agree but I like my method because it's cool
    Meh, personally I think it's 'cooler' to use existing knowledge in a clever way rather than getting a complicated formula to solve it but to each their own

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Updated: May 31, 2012
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