The Student Room Group

Core 1 question (Stationary points etc)

Qi) Find the coordinates of the stationary points on the graph of y=x3+x2y = x^3 + x^2

I've done that and the stationary points are:

(0,0)(13,227)(0, 0) (-\frac{1}{3}, \frac{2}{27})

Now it says:

ii) Hence write down the set of values of the constant kk for which the equation x3+x2=kx^3 + x^2 = k has three distinct real roots.

Hmm, I'm not sure.
Sketch/visualise the graph, you want values of k for which y=ky = k intesects y=x3+x2y = x^3 + x^2 3 times. You can get the answer easily from your answer to the first part
Reply 2
Original post by dnumberwang
Sketch/visualise the graph, you want values of k for which y=ky = k intesects y=x3+x2y = x^3 + x^2 3 times. You can get the answer easily from your answer to the first part


So do you use 0 and 2/27? I understand that I think but in the answers they use the inequalities; I can't understand how you use them to show 3 roots...
Reply 3
Original post by Dededex
(13,227)(-\frac{1}{3}, \frac{2}{27})


are you sure?
Reply 4
Original post by Pheylan
are you sure?


Hmm I was gonna ask - the answer in book give 4/27 but surely if x = -1/3 then:

(13)3=127(-\frac{1}{3})^3 = -\frac{1}{27}

Then add 1/9 (3/27) to -1/27 you get 2/27?
Original post by Pheylan
are you sure?


I just noticed that as well

Original post by Dededex
So do you use 0 and 2/27? I understand that I think but in the answers they use the inequalities; I can't understand how you use them to show 3 roots...


The equation x3+x2=k x^3 + x^2 = k will have 3 roots whenever a<k<ba < k < b
Reply 6
nvm spotted it.
Reply 7
Original post by Dededex
Hmm I was gonna ask - the answer in book give 4/27 but surely if x = -1/3 then:

(13)3=127(-\frac{1}{3})^3 = -\frac{1}{27}

Then add 1/9 (3/27) to -1/27 you get 2/27?


so perhaps x isn't -1/3
Reply 8
Differentiate it again. You will see where you have gone wrong. Particularly the x^2
Reply 9
Original post by Pheylan
so perhaps x isn't -1/3


Hmm perhaps you are right - infact, perhaps I have indeed differentiated wrong!

..





Wait, yeah I have! :colondollar: Sorry
Reply 10
Original post by dnumberwang
I just noticed that as well



The equation x3+x2=k x^3 + x^2 = k will have 3 roots whenever a<k<ba < k < b


hmm well the answer is 0 < k < 4/27

But I can't understand what that has to do with "3 distinct real roots" or what it means - all I can see from that is where y is increasing.
Original post by Dededex
hmm well the answer is 0 < k < 4/27

But I can't understand what that has to do with "3 distinct real roots" or what it means - all I can see from that is where y is increasing.


3 distinct real roots means that there are 3 different solutions to the equation x3+x2=kx^3 + x^2 = k. So the lines y=x3+x2y = x^3 + x^2 and y=k y = k will intersect 3 times. y=k y = k has no variables (x's) so it'll be horizontal

Sketch the graph of y=x3+x2y = x^3 + x^2 and think about what horizontal lines will intersect it 3 times
(edited 13 years ago)
Reply 12
Original post by dnumberwang
3 distinct real roots means that there are 3 different solutions to the equation x3+x2=kx^3 + x^2 = k. So the lines y=x3+x2y = x^3 + x^2 and y=k y = k will intersect 3 times. y=k y = k has no variables (x's) so it'll be horizontal

Sketch the graph of y=x3+x2y = x^3 + x^2 and think about what horizontal lines will intersect it 3 times


Sorry even after I sketch it I can't understand - how can it intersect three times? What is the line y = k? eergh stupid question - don't worry I'll ask my teacher tomorrow or something I'm wasting your time on here to be honest :s-smilie:
Reply 13
Original post by dnumberwang
3 distinct real roots means that there are 3 different solutions to the equation x3+x2=kx^3 + x^2 = k. So the lines y=x3+x2y = x^3 + x^2 and y=k y = k will intersect 3 times. y=k y = k has no variables (x's) so it'll be horizontal

Sketch the graph of y=x3+x2y = x^3 + x^2 and think about what horizontal lines will intersect it 3 times


Oh wait I think I see it - so if k is less than 4/27? But then if k is also less than zero won't it intersect 3 times?
Original post by Dededex
Oh wait I think I see it - so if k is less than 4/27? But then if k is also less than zero won't it intersect 3 times?


So k must be between 4/27 and 0

0<k<427 0 < k < \dfrac {4}{27}
Reply 15
Original post by dnumberwang
So k must be between 4/27 and 0

0<k<427 0 < k < \dfrac {4}{27}


Aah yes I see it now! Finally thank you so much - I think it's also because I had to account for the fact that k is the y value not the x value which I'm used to.

Thanks mate

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