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    Q: For this curve, find where the curve meets the axes, state any symmetry and the behaviour as x -> infinity. Find the coordinates of any stationary points, determine whether they are maximum or minimum points or points of inflexion, and then sketch the curve.

    The curve is y=3x^2-x^3

    I have done everything above apart from the behaviour part, as I do not understand it. I would really appreciate it if someone explains to me how to do it.
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    (Original post by TarotOfMagic)
    Q: For this curve, find where the curve meets the axes, state any symmetry and the behaviour as x -> infinity. Find the coordinates of any stationary points, determine whether they are maximum or minimum points or points of inflexion, and then sketch the curve.

    The curve is y=3x^2-x^3

    I have done everything above apart from the behaviour part, as I do not understand it. I would really appreciate it if someone explains to me how to do it.
    Look at the graph of your function or simply look at the equation, what happens when x gets really really big?
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    (Original post by Zacken)
    Look at the graph of your function or simply look at the equation, what happens when x gets really really big?
    It gets closer to infinity?....
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    (Original post by TarotOfMagic)
    It gets closer to infinity?....
    Does it really? Does it get closer to infinity or -infinity?
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    (Original post by Zacken)
    Does it really? Does it get closer to infinity or -infinity?
    Oh, it gets closer to -infinity, my mistake. But why is this? I think I know why but I can't really put it to words.
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    (Original post by TarotOfMagic)
    Oh, it gets closer to -infinity, my mistake. But why is this? I think I know why but I can't really put it to words.
    Note that y=3x^2-x^3=x^3(\frac{3}{x}-1).

    Now as x \to \infty, we have that x^3 \to \infty but what happens to the term in the bracket?
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    (Original post by TarotOfMagic)
    Oh, it gets closer to -infinity, my mistake. But why is this? I think I know why but I can't really put it to words.
    As above, or you could also note that x^3 is miles bigger than x^2, so when you do 3x^2 - x^3 you're doing big number - bigger number which gets you -infinity. :yep:
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    (Original post by atsruser)
    Note that y=3x^2-x^3=x^3(\frac{3}{x}-1).

    Now as x \to \infty, we have that x^3 \to \infty but what happens to the term in the bracket?
    It gets closer to infinity?

    Sorry, I have never done this topic.
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    (Original post by TarotOfMagic)
    It gets closer to infinity?

    Sorry, I have never done this topic.
    Nah, you have \frac{3}{x} \approx 0 for really big x, you agree with me? So when x goes off to infinity, what atsruser has written is \infty (0 -1) = \infty (-1) = -\infty, essentially. It's not really a topic, it's more of a logical thing you need to understand and be able to deduce yourself.
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    (Original post by TarotOfMagic)
    It gets closer to infinity?

    Sorry, I have never done this topic.
    No. What does \frac{3}{x} get closer to as x \to \infty? That's just asking what happens to 3/x as we replace x with larger and larger numbers (imagine replacing it with 1000, then 1 000 000, then 1 000 000 000 etc.)

    How big are the pieces of 3 cakes when you chop them into 1 000 000 pieces?
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    (Original post by Zacken)
    So when x goes off to infinity, what atsruser has written is \infty (0 -1) = \infty (-1) = -\infty, essentially.
    Not really. What I've written is \text{(a very big number)} \times -1 = -\text{(a very big number)} and that's true regardless of how big the very big number happens to be.
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    (Original post by atsruser)
    Not really. What I've written is \text{(a very big number)} \times -1 = -\text{(a very big number)} and that's true regardless of how big the very big number happens to be.
    Sorry, should have phrased it better. I meant "when you plug x = \infty into what atsruser wrote" instead of saying that that's what you've written.
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    (Original post by Zacken)
    Nah, you have \frac{3}{x} \approx 0 for really big x, you agree with me? So when x goes off to infinity, what atsruser has written is \infty (0 -1) = \infty (-1) = -\infty, essentially. It's not really a topic, it's more of a logical thing you need to understand and be able to deduce yourself.
    Thanks. I understand it slightly more now. So, the answer is just -infninity?
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    (Original post by TarotOfMagic)
    Thanks. I understand it slightly more now. So, the answer is just -infninity?
    Kind of, you're meant to say that y diverges off to -infinity, yeah.
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    (Original post by Zacken)
    Kind of, you're meant to say that y diverges off to -infinity, yeah.
    I checked the answer to this question and it was written minus/plus infinity. Is this the same thing as - infinity?
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    (Original post by TarotOfMagic)
    I checked the answer to this question and it was written minus/plus infinity. Is this the same thing as - infinity?
    No, it's definitely -infinity. But it means that the examiners are willing to accept any answer that includes infinity, whether it's plus or minus infinity regardless of whether it's correct or not.
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    (Original post by Zacken)
    I meant "when you plug x = \infty into what atsruser wrote"
    They're going to beat you with a stick at Cambridge, if you write stuff like that
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    (Original post by atsruser)
    They're going to beat you with a stick at Cambridge, if you write stuff like that
    I half considered adding a spoiler to disclaim that I know that I'm being completely unrigorous. :rofl:
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    (Original post by Zacken)
    I half considered adding a spoiler to disclaim that I know that I'm being completely unrigorous. :rofl:
    I wonder if they'll accept those in supervision assignments.

    "hand waves"

    "I know I'm being unrigorous"
 
 
 
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