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 (−332,0)∪(0,332) using a method I just googled, and it is unlikely that it was a C1 method.
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)
Is that really a C1 question? Because I get the set of values to be (−332,0)∪(0,332) using a method I just googled, and it is unlikely that it was a C1 method.
The discriminant of a cubic ax3+bx2+cx+d which satisfies ax3+bx2+cx+d=0 is given by (brace yourself)
b2c2−4ac3−4b3d−27a2d2+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−k2, expanded the brackets to get x3−2x2+x−k2, calculated the discrimant and solved for the discriminant <0. Following through the working gives
The discriminant of a cubic ax3+bx2+cx+d which satisfies ax3+bx2+cx+d=0 is given by (brace yourself)
b2c2−4ac3−4b3d−27a2d2+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−k2, expanded the brackets to get x3−2x2+x−k2, calculated the discrimant and solved for the discriminant <0. Following through the working gives
x2(x−332)(x+332)<0
which gives the set of values.
Much better working with what we already know (look at my post above)
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