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C1 Graphs

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.
Reply 1
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 (233,0)(0,233)(-\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.
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(x1)2x(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 :tongue: 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)
(edited 11 years ago)
Reply 3
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 :biggrin:


It's not a quadratic equation so can't use that method :smile:
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 :biggrin:


That only works with quadratics, unfortunately.
Reply 5
Original post by Zii
Is that really a C1 question? Because I get the set of values to be (233,0)(0,233)(-\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?
Reply 6
math is my enermy
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:

maths 001.jpg


Very nasty C1 question!

(Sorry about the messy start, I screwed up the differentiation :facepalm: )
Reply 8
Original post by Julii92
That's the right answer - what method was that?


The discriminant of a cubic ax3+bx2+cx+dax^3+bx^2+cx+d which satisfies ax3+bx2+cx+d=0ax^3+bx^2+cx+d = 0 is given by (brace yourself)

b2c24ac34b3d27a2d2+18abcdb^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(x1)2k2x(x-1)^2-k^2, expanded the brackets to get x32x2+xk2x^3-2x^2+x-k^2, calculated the discrimant and solved for the discriminant <0< 0. Following through the working gives

k2(k233)(k+233)<0k^2(k-\frac{2}{3\sqrt{3}})(k+\frac{2}{3\sqrt{3}}) < 0

which gives the set of values.

Edit: Should have written kk, not xx!
(edited 11 years ago)
Original post by Zii
The discriminant of a cubic ax3+bx2+cx+dax^3+bx^2+cx+d which satisfies ax3+bx2+cx+d=0ax^3+bx^2+cx+d = 0 is given by (brace yourself)

b2c24ac34b3d27a2d2+18abcdb^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd

When the discrimant is strictly less than 00, this corresponds to the cubic having precisely one real root.

So I took your cubic, x(x1)2k2x(x-1)^2-k^2, expanded the brackets to get x32x2+xk2x^3-2x^2+x-k^2, calculated the discrimant and solved for the discriminant <0< 0. Following through the working gives

x2(x233)(x+233)<0x^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) :tongue:
Reply 10
Original post by hassi94
Much better working with what we already know (look at my post above) :tongue:


I agree but I like my method because it's cool
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 :tongue:

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