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    Ok so as embarrassing as this sounds I'm stuck on Example 1 from the Numerical Methods chapter of the Edexcel C3 Modular Maths Textbook...

    Name:  C3. CH4. Exa1 - Numerical (Graphical) Method.png
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    How am I meant to draw the cubic graph if I can't even get it to factorise in order to work out the points where it intercepts the x-axis??

    I find that the book hasn't really described this topic very well. Am I missing something?


    Also I think I will be posting other problems that I get stuck on in this chapter to this same thread instead of spamming TSR with new threads like I usually do :P
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    (Original post by Philip-flop)
    Ok so as embarrassing as this sounds I'm stuck on Example 1 from the Numerical Methods chapter of the Edexcel C3 Modular Maths Textbook...

    Name:  C3. CH4. Exa1 - Numerical (Graphical) Method.png
Views: 157
Size:  12.6 KB

    How am I meant to draw the cubic graph if I can't even get it to factorise in order to work out the points where it intercepts the x-axis??

    I find that the book hasn't really described this topic very well. Am I missing something?


    Also I think I will be posting other problems that I get stuck on in this chapter to this same thread instead of spamming TSR with new threads like I usually do :P
    Not embarrassing at all, I think this is one of the few places where the Edexcel textbook fails.

    I don't think you're expected to draw that - it most likely sufficies to find f(2) and f(3) and show that there
    s a change of sign.
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    (Original post by SeanFM)
    Not embarrassing at all, I think this is one of the few places where the Edexcel textbook fails.

    I don't think you're expected to draw that - it most likely sufficies to find f(2) and f(3) and show that there
    s a change of sign.
    Oh right I see. I was wondering that. I've actually wasted like the last hour thinking whether I'm just being stupid and can't work it out
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    (Original post by Philip-flop)
    Oh right I see. I was wondering that. I've actually wasted like the last hour thinking whether I'm just being stupid and can't work it out
    Do you understand why showing the signs of f(2) and f(3) are different is enough to show that the function crosses the x-axis somewhere between there?
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    (Original post by Zacken)
    Do you understand why showing the signs of f(2) and f(3) are different is enough to show that the function crosses the x-axis somewhere between there?
    Yes I think so. It's because it shows that f(2) and f(3) are leaning close to y=0 as f(2) is slightly less than y=0 and that f(3) is slightly more than y=0.

    Am I right?
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    (Original post by Philip-flop)
    Yes I think so. It's because it shows that f(2) and f(3) are leaning close to y=0 as f(2) is slightly less than y=0 and that f(3) is slightly more than y=0.

    Am I right?
    Yeah, the signs being different means that the curve is under the x-axis at f(2) (say f(2) is negative) and above the x-axis at f(3), so it must have crossed the x-axis somewhere in between there.
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    (Original post by Zacken)
    Yeah, the signs being different means that the curve is under the x-axis at f(2) (say f(2) is negative) and above the x-axis at f(3), so it must have crossed the x-axis somewhere in between there.
    Yeah that's it! Brilliant explanation again Zacken!! Thank you so much
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    Is it just me or is iteration a very "trial and error" based method? It seems very monotonous to me. :/
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    (Original post by Philip-flop)
    Is it just me or is iteration a very "trial and error" based method? It seems very monotonous to me. :/
    It's more for guessing an answer before proving it (to help you move in the right direction) or practical uses in the real world
    Spoiler:
    Show
    which is only a shadow of the far more superior mathematical world
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    (Original post by Philip-flop)
    Is it just me or is iteration a very "trial and error" based method? It seems very monotonous to me. :/
    Trial and error is much more random. Iteration always leads you straight to a very good approximation of the answer if you iterate enough times depending on your starting point.

    Interval bisection method to find a more accurate interval of something, such as a root to an equation, is much more trial and error based than iteration.
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    (Original post by RDKGames)
    Trial and error is much more random. Iteration always leads you straight to a very good approximation of the answer if you iterate enough times depending on your starting point.

    Interval bisection method to find a more accurate interval of something, such as a root to an equation, is much more trial and error based than iteration.
    Yeah that's true. It's almost like constructive guessing where you get closer and closer to the real answer.
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    (Original post by Zacken)
    Yeah, the signs being different means that the curve is under the x-axis at f(2) (say f(2) is negative) and above the x-axis at f(3), so it must have crossed the x-axis somewhere in between there.
    Strictly speaking that only works if the function is continuous, which it is in this case as all polynomials are continuous.
 
 
 
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