[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.

[Zii]

(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 using a method I just googled, and it is unlikely that it was a C1 method.

[Intriguing Alias]

(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 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)

[TheJ0ker]

(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

[Intriguing Alias]

(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.

[Julii92]

(Original post by Zii)
Is that really a C1 question? Because I get the set of values to be 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?

[dongtian]

math is my enermy

[Intriguing Alias]

(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:




Very nasty C1 question!

(Sorry about the messy start, I screwed up the differentiation )

[Zii]

(Original post by Julii92)
That's the right answer - what method was that?

The discriminant of a cubic which satisfies is given by (brace yourself)



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

So I took your cubic, , expanded the brackets to get , calculated the discrimant and solved for the discriminant . Following through the working gives



which gives the set of values.

Edit: Should have written , not !

[Intriguing Alias]

(Original post by Zii)
The discriminant of a cubic which satisfies is given by (brace yourself)



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

So I took your cubic, , expanded the brackets to get , calculated the discrimant and solved for the discriminant . Following through the working gives



which gives the set of values.

Much better working with what we already know (look at my post above)

[Zii]

(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

[Intriguing Alias]

(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