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

    I've done that and the stationary points are:

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

    Now it says:

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

    Hmm, I'm not sure.
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    Sketch/visualise the graph, you want values of k for which y = k intesects y = x^3 + x^2 3 times. You can get the answer easily from your answer to the first part
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    (Original post by dnumberwang)
    Sketch/visualise the graph, you want values of k for which y = k intesects y = 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...
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    (Original post by Dededex)
    (-\frac{1}{3}, \frac{2}{27})
    are you sure?
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    (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:

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

    Then add 1/9 (3/27) to -1/27 you get 2/27?
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    (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  x^3 + x^2 = k will have 3 roots whenever a < k < b
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    nvm spotted it.
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    (Original post by Dededex)
    Hmm I was gonna ask - the answer in book give 4/27 but surely if x = -1/3 then:

    (-\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
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    Differentiate it again. You will see where you have gone wrong. Particularly the x^2
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    (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! Sorry
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    (Original post by dnumberwang)
    I just noticed that as well



    The equation  x^3 + x^2 = k will have 3 roots whenever a < 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.
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    (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 x^3 + x^2 = k. So the lines y = x^3 + x^2 and  y = k will intersect 3 times.  y = k has no variables (x's) so it'll be horizontal

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

    Sketch the graph of y = 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
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    (Original post by dnumberwang)
    3 distinct real roots means that there are 3 different solutions to the equation x^3 + x^2 = k. So the lines y = x^3 + x^2 and  y = k will intersect 3 times.  y = k has no variables (x's) so it'll be horizontal

    Sketch the graph of y = 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?
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    (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 &lt; k &lt; \dfrac {4}{27}
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    (Original post by dnumberwang)
    So k must be between 4/27 and 0

     0 &lt; k &lt; \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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