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    http://www.examsolutions.net/a-level...y/paper.php#Q8

    how would i go around doing part d of this?

    if i split \dfrac{8x^3 -1}{1-2x^3}

    up into \dfrac{8x^3}{1-2x^3} and\ \dfrac{1}{1-2x^3}

    can i just make the first bit into \dfrac{24x^2}{1-6x^2} or am i killing people with my "maths" here?
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    (Original post by thegreatwhale)
    http://www.examsolutions.net/a-level...y/paper.php#Q8

    how would i go around doing part d of this?

    if i split \dfrac{8x^3 -1}{1-2x^3}

    up into \dfrac{8x^3}{1-2x^3} and\ \dfrac{1}{1-2x^3}

    can i just make the first bit into \dfrac{24x^2}{1-6x^2} or am i killing people with my "maths" here?
    That would be illegal I'm afraid. You need to multiply both top and bottom by 3, giving you the result (24x^3)/(3-6x^2).

    I'd recommend differentiating gf(x) (using quotient rule; maybe the product rule) to get yourself another fraction. Put that equal to zero and solve.
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    (Original post by JLegion)
    That would be illegal I'm afraid. You need to multiply both top and bottom by 3, giving you the result (24x^3)/(3-6x^2).

    I'd recommend differentiating gf(x) (using quotient rule; maybe the product rule) to get yourself another fraction. Put that equal to zero and solve.
    couldn't i just use quotient rule as it is?(just learnt it xD)

    and say that

    \dfrac{(1-2x^3)(24x^2)-(8x^3 -1)(-6x^2)}{(1-2x^3)^2}

    \dfrac{24x^2 -48x^5 +48x^5 -6x^2}{(1-2x^3)^2}

    \dfrac{18x^2}{(1-2x^3)^2}
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    (Original post by thegreatwhale)
    couldn't i just use quotient rule as it is?(just learnt it xD)

    and say that

    \dfrac{(1-2x^3)(24x^2)-(8x^3 -1)(-6x^2)}{(1-2x^3)^2}

    \dfrac{24x^2 -48x^5 +48x^5 -6x^2}{(1-2x^3)^2}

    \dfrac{18x^2}{(1-2x^3)^2}
    Yes that works, and hence you can deduce that the stationary point is at x=0. Now just substitute into the expression for gf(x) to get the y-coordinate.
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    (Original post by HapaxOromenon3)
    Yes that works, and hence you can deduce that the stationary point is at x=0. Now just substitute into the expression for gf(x) to get the y-coordinate.
    How do i know it's a quadratic? it doesn't look like a normal one :/
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    (Original post by thegreatwhale)
    How do i know it's a quadratic? it doesn't look like a normal one :/
    At stationary points, dy/dx = 0.
    Thus 18x^2/(1-2x^3)^2 = 0.
    Multiply both sides by (1-2x^3)^2 to give 18x^2 = 0, then divide by 18 to give x^2 = 0, so x = 0.
    Notice that in general, if a fraction equals 0, we can automatically write down that the numerator must be 0, and it doesn't matter what the denominator is (as long as we don't have a division by zero).
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    (Original post by HapaxOromenon3)
    At stationary points, dy/dx = 0.
    Thus 18x^2/(1-2x^3)^2 = 0.
    Multiply both sides by (1-2x^3)^2 to give 18x^2 = 0, then divide by 18 to give x^2 = 0, so x = 0.
    Notice that in general, if a fraction equals 0, we can automatically write down that the numerator must be 0, and it doesn't matter what the denominator is (as long as we don't have a division by zero).
    i understand \dfrac{\mathrm d y}{\mathrm d x}=0 at a stationary point

    but how u know it's a quadratic? because quadratic generally have u shape or n shape and that way it has only 1 stationary point or am i wrong and does the graph of it look totally different?
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    (Original post by JLegion)
    That would be illegal I'm afraid. You need to multiply both top and bottom by 3, giving you the result (24x^3)/(3-6x^2).

    I'd recommend differentiating gf(x) (using quotient rule; maybe the product rule) to get yourself another fraction. Put that equal to zero and solve.
    Thanks for pointing out the OP's illegal move , do you have the number for the maths police ?
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    (Original post by thegreatwhale)
    couldn't i just use quotient rule as it is?(just learnt it xD)

    and say that

    \dfrac{(1-2x^3)(24x^2)-(8x^3 -1)(-6x^2)}{(1-2x^3)^2}

    \dfrac{24x^2 -48x^5 +48x^5 -6x^2}{(1-2x^3)^2}

    \dfrac{18x^2}{(1-2x^3)^2}
    Yes you can. I was just letting you know that when you split the fraction and multiplied by 3, it was done incorrectly.

    Now you know the derivative. Equate it to 0, and find the value of gf(x) for this value of x to find the coordinates of the stationary point
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    (Original post by JLegion)
    Yes you can. I was just letting you know that when you split the fraction and multiplied by 3, it was done incorrectly.

    Now you know the derivative. Equate it to 0, and find the value of gf(x) for this value of x to find the coordinates of the stationary point
    ah right oh i understand the question told me lol >.>
    thanks got the answer now
    (Original post by fefssdf)
    Thanks for pointing out the OP's illegal move , do you have the number for the maths police ?
    this is what happens when you go through the textbook up to chapter 3 and you aint done the chapter on differentiation but you still wanna do dem big mark questions an finish the rest of the question.
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    (Original post by thegreatwhale)
    ah right oh i understand the question told me lol >.>
    thanks got the answer now


    this is what happens when you go through the textbook up to chapter 3 and you aint done the chapter on differentiation but you still wanna do dem big mark questions an finish the rest of the question.
    Aha oh right Yh I know what you mean ; have you started c3 early then ? I started c3 last summer but didn't really remember much of it lol . I wish I had actually learnt it properly so I had more time for other stuff hmmm
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    (Original post by fefssdf)
    Aha oh right Yh I know what you mean ; have you started c3 early then ? I started c3 last summer but didn't really remember much of it lol . I wish I had actually learnt it properly so I had more time for other stuff hmmm
    yes, i'm still tryin to learn it now myself so i can get a head start and get things going back when i go back to school
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    (Original post by thegreatwhale)
    yes, i'm still tryin to learn it now myself so i can get a head start and get things going back when i go back to school
    Brilliant haha ; do all the exercises and keep them in a folder and mark them and then when your teacher sets homework from the book then you would've done it already aha
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    (Original post by fefssdf)
    Brilliant haha ; do all the exercises and keep them in a folder and mark them and then when your teacher sets homework from the book then you would've done it already aha
    ah thing is though my teacher doesn't do that he rushes through and tries to get us to do all the past papers so we have as much experience of papers as possible
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    (Original post by thegreatwhale)
    ah thing is though my teacher doesn't do that he rushes through and tries to get us to do all the past papers so we have as much experience of papers as possible
    Oh ok well just do all the exercises in the book then so when you come to do papers you'll be confident with the content
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    When I did the problem initially :moon: I noticed that \dfrac{8x^3-1}{1-2x^3}=3(1-2x^3)^{-1}-4 and I found the derivative using the chain rule leading to \dfrac{18x^2}{(1-2x^3)^2} as you have.
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    (Original post by fefssdf)
    Oh ok well just do all the exercises in the book then so when you come to do papers you'll be confident with the content
    (Original post by Kvothe the arcane)
    When I did the problem initially :moon: I noticed that \dfrac{8x^3-1}{1-2x^3}=3(1-2x^3)^{-1}-4 and I found the derivative using the chain rule leading to \dfrac{18x^2}{(1-2x^3)^2} as you have.
    woah where did that come from?? o..o
    i understand \dfrac{1}{1-2x^3} =(1-2x^3)^{-1}

    but i'm a little unsure of where the 3 comes from ad i'm assuming the -4 is to balance things out
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    (Original post by thegreatwhale)
    woah where did that come from?? o..o
    i understand \dfrac{1}{1-2x^3} =(1-2x^3)^{-1}

    but i'm a little unsure of where the 3 comes from ad i'm assuming the -4 is to balance things out
    -4 is the remainder I think
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    (Original post by fefssdf)
    -4 is the remainder I think
    lol still don't know where the 3 comes from and where the 8x^3 -1 went
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    (Original post by thegreatwhale)
    lol still don't know where the 3 comes from and where the 8x^3 -1 went
    I'm actually so confused right now
 
 
 
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