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    (Original post by RDKGames)
    It is indeed 256. Also it's the area under the curve from -5 to 2, and above it from 2 to 3 (reference to the sketch).
    But you can just work it out as -5 to 3 like ive done?
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    (Original post by kiiten)
    But you can just work it out as -5 to 3 like ive done?
    Yes that would give you the whole area enclosed by the curve and the x-axis, but your reference to the sketch is technically incorrect.
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    Ughh guys ive gone wrong again :/ ques 4.b)

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    (Original post by kiiten)
    Ughh guys ive gone wrong again :/ ques 4.b)

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    If the area of the 2 sections are equal, then the integral between 0 and k will be 0, as the net area above the x axis will be 0, as it will cancel with the part of the curve that is below the x axis.
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    (Original post by B_9710)
    If the area of the 2 sections are equal, then the integral between 0 and k will be 0, as the net area above the x axis will be 0, as it will cancel with the part of the curve that is below the x axis.
    :facepalm:thanks
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    (Original post by kiiten)
    :facepalm:thanks
    Ah crap, just realised what I've said about the previous question is wrong, sorry. Anyway, k=root(8) in case you want it validated.
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    (Original post by RDKGames)
    Ah crap, just realised what I've said about the previous question is wrong, sorry. Anyway, k=root(8) in case you want it validated.
    Ahh - i got the notif but never saw your post so its fine. Thanks
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    (Original post by kiiten)
    Ahh - i got the notif but never saw your post so its fine. Thanks
    Also your method would still work if you equate the area from 2 to k to -4 rather that positive 4, because the area is negative.
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    (Original post by RDKGames)
    Also your method would still work if you equate the area from 2 to k to -4 rather that positive 4, because the area is negative.
    Eh? where did the 4 come from??
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    (Original post by kiiten)
    Eh? where did the 4 come from??
    That's the area from 0 to 2.
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    can 4/x^-1/2 be written in a different form?
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    (Original post by kiiten)
    can 4/x^-1/2 be written in a different form?
    Of course. \frac{4}{x^{-\frac{1}{2}}}=\frac{4}{\frac{1}{  x^{\frac{1}{2}}}}=4x^{\frac{1}{2  }}=4\sqrt{x}
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    (Original post by kiiten)
    can 4/x^-1/2 be written in a different form?
    \displaystyle \frac{4}{x^{-1/2}} = 4x^{1/2} = 4\sqrt{x} = \sqrt{16x} = 4 \sqrt[4]{x} \sqrt[4]{x} = \frac{4}{\frac{1}{\sqrt{x}}} = \cdots
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    (Original post by RDKGames)
    Of course. \frac{4}{x^{-\frac{1}{2}}}=\frac{4}{\frac{1}{  x^{\frac{1}{2}}}}=4x^{\frac{1}{2  }}=4\sqrt{x}
    (Original post by Zacken)
    \displaystyle \frac{4}{x^{-1/2}} = 4x^{1/2} = 4\sqrt{x} = \sqrt{16x} = 4 \sqrt[4]{x} \sqrt[4]{x} = \frac{4}{\frac{1}{\sqrt{x}}} = \cdots

    Thanks (sorry for asking easy ques - ive forgotten a lot of maths since the exams :3 )
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    Ughhh where am i going wrong. You have to find the coordinates of the minimum. Ive completely forgotten what you have to do (x is only 1/2)

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    (Original post by kiiten)
    Ughhh where am i going wrong. You have to find the coordinates of the minimum. Ive completely forgotten what you have to do (x is only -1/2 not 2 but why?)

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    Complete the square. Or you could differentiate and find the value of x that gives f'(x)=0, then plug in to find f at that point.
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    (Original post by B_9710)
    Complete the square. Or you could differentiate and find the value of x that gives f'(x)=0, then plug in to find f at that point.
    i already found 2 values of x but they're wrong. why?
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    (Original post by kiiten)
    i already found 2 values of x but they're wrong. why?
    If f(x) = 6x^2 - 6x + 3 then f ' (x) = 12x - 6 so f ' (x) = 0 only when 12x - 6 = 0, not sure how you got two values of x from that.
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    (Original post by Zacken)
    If f(x) = 6x^2 - 6x + 3 then f ' (x) = 12x - 6 so f ' (x) = 0 only when 12x - 6 = 0, not sure how you got two values of x from that.
    yeah i forgot to differentiate. Thanks
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    (Original post by kiiten)
    yeah i forgot to differentiate. Thanks
    You even wrote "differentiation" on top!
 
 
 
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