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Size:  514.0 KBI am stuck on Q3,4,6,7,8,9,10,11 of the questions could anyone help? I have done Q1,2 &5.
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    What exactly is it that you are unable to do?
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    Finding the coordinates mainly (found my old account hence the different username)
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    (Original post by henb99)
    Finding the coordinates mainly (found my old account hence the different username)
    (Original post by henbnyc)
    I am stuck on Q3,4,6,7,8,9,10,11 of the questions could anyone help? I have done Q1,2 &5.
    Q3 - I can't see what gradient the book shows to I'll just refer to it as m. Anyway, you have y=\frac{x}{1+x} so you need to find \frac{dy}{dx} by using an appropriate differentiation rule that you should know. Set \frac{dy}{dx}=m and solve for x to see at what x coordinates the gradient is m.

    Once you have your value(s) of x, plug them back through y=\frac{x}{1+x} to find your y coordinates.
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    (Original post by henb99)
    Finding the coordinates mainly (found my old account hence the different username)
    It may be easier to differentiate  \frac{x}{1+x} if you express it in the form  1-\frac{1}{1+x} .
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    How do I solve? ^
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    (Original post by Aklaol)


    How do I solve? ^
    Create your own thread on the Maths forum and ask this. Don't hijack other threads, thanks.
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    (Original post by RDKGames)
    Q3 - I can't see what gradient the book shows to I'll just refer to it as m. Anyway, you have y=\frac{x}{1+x} so you need to find \frac{dy}{dx} by using an appropriate differentiation rule that you should know. Set \frac{dy}{dx}=m and solve for x to see at what x coordinates the gradient is m.

    Once you have your value(s) of x, plug them back through y=\frac{x}{1+x} to find your y coordinates.
    I have tried this but only get one set of coordinates instead of two

    When I differentiate using the quotient rule I get 1/(1+x)^2
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    (Original post by henb99)
    I have tried this but only get one set of coordinates instead of two

    When I differentiate using the quotient rule I get 1/(1+x)^2
    That's correct. When it comes to square rooting, you should get 2 solutions because x^2=a \Rightarrow x=\pm \sqrt{a}
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    (Original post by RDKGames)
    That's correct. When it comes to square rooting, you should get 2 solutions because x^2=a \Rightarrow x=\pm \sqrt{a}
    So the gradient 1/9 is equal to 1+x^-2

    -8/9=x^-2
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    (Original post by RDKGames)
    Create your own thread on the Maths forum and ask this. Don't hijack other threads, thanks.
    Lmao you look like a killjoy and you sure are
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    (Original post by henb99)
    So the gradient 1/9 is equal to 1+x^-2

    -8/9=x^-2
    Nope. So then \frac{1}{9}=(1+x)^{-2} \Rightarrow 9=(1+x)^2 \Rightarrow \pm 3 = 1+x
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    (Original post by BluntBluster)
    Lmao you look like a killjoy and you sure are
    Oh definitely a lot of joy posting complicated maths problems on a simple A-Level thread.
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    (Original post by RDKGames)
    Nope. So then \frac{1}{9}=(1+x)^{-2} \Rightarrow 9=(1+x)^2 \Rightarrow \pm 3 = 1+x
    Thank you so much I know how dumb I must sound, I have worked it out correctly

    For question 4, when I differentiated I got 3x(x+1)^2 which is equal to the gradient 7

    after this I got stuck again, there should only be one point (P)
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    (Original post by henb99)
    Thank you so much I know how dumb I must sound, I have worked it out correctly

    For question 4, when I differentiated I got 3x(x+1)^2 which is equal to the gradient 7

    after this I got stuck again, there should only be one point (P)
    Please state the original function, they are hard to see. Is it y=x(x-1)^3??? If so, check your differential again.
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    (Original post by RDKGames)
    Please state the original function, they are hard to see. Is it y=x(x-1)^3??? If so, check your differential again.
    My bad I had a plus instead of a minus
    I have got it to be (2,2)
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    (Original post by henb99)
    My bad I had a plus instead of a minus
    I have got it to be (2,2)
    Correct.
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    (Original post by RDKGames)
    Correct.
    What a levels to do btec level 3 ict with
 
 
 
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