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    I'm really stuck on part d).

    I understand that the stationary point will be at (x-k)^2, but apart from that I really don't know how to solve this.

    Help pls?
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    (Original post by frostyy)


    I'm really stuck on part d).

    I understand that the stationary point will be at (x-k)^2, but apart from that I really don't know how to solve this.

    Help pls?
    It's to be done in terms of k. So find when the function is 0 in terms of k, and find what happens when x is 0 in terms of k. Similarly, coordinates of stationary points can only be expressed in terms of k. ie it's the same as b, just shifted k units to the right.

    edit: are you sure about the stationary point being at (x-k)^2 ?
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    (Original post by frostyy)
    ...
    First of all, coefficient of x^3 is positive, so the graph will start in the bottom left corner and will end in the top right corner.

    Let's look at our...

    I stopped there because I've just read the question. Surely you've done part (b)? If so, just translate your sketch from part (b) k units to the right.
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    (Original post by frostyy)


    I'm really stuck on part d).

    I understand that the stationary point will be at (x-k)^2, but apart from that I really don't know how to solve this.

    Help pls?
    What is the behaviour for big +ve and -ve x? That gives you the general layout of the cubic.

    Then note that at x=k, it will be tangent to the x-axis, and will have another root at x-k+2=0 \Rightarrow x=k-2
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    (Original post by Zacken)
    I stopped there because I've just read the question. Surely you've done part (b)? If so, just translate your sketch from part (b) k units to the right.
    Well spotted - I really must learn to read the f'ing question.
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    (Original post by atsruser)
    What is the behaviour for big +ve and -ve x? That gives you the general layout of the cubic.

    Then note that at x=k, it will be tangent to the x-axis, and will have another root at x-k+2=0 \Rightarrow x=2-k
    Lol are you sure about that last line? Aren't you some undergrad mathematician..
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    (Original post by Bath~Student)
    Lol are you sure about that last line? Aren't you some undergrad mathematician..
    No, I'm not an undergrad mathematician. But yes, that's a lol, which I will correct.
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    (Original post by atsruser)
    No, I'm not an undergrad mathematician. But yes, that's a lol, which I will correct.
    ye.. so a-level student still?
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    (Original post by Bath~Student)
    Aren't you some undergrad mathematician..
    I'm not entirely sure why you think being an undergrad mathematician means that you aren't prone to algebraic slips.
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    (Original post by Bath~Student)
    ye.. so a-level student still?
    No. I have a degree. But I never did much on equations with k in them (mainly x, y for me) so I have a good excuse.
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    (Original post by Zacken)
    I'm not entirely sure why you think being an undergrad mathematician means that you aren't prone to algebraic slips.
    It's well known that the contrary is true: it's notoriously known undergraduates spend so much time on abstract crap that they lose practise of the basics. However,
    obviously that was a slip.


    Anyway, engineering > mathematics. I'll show you my paycheck one day.
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    (Original post by atsruser)
    No. I have a degree. But I never did much on equations with k in them (mainly x, y for me) so I have a good excuse.
    So what are you up to today?


    Unemployed? Having an existential crisis? Suicidal? Just listing up some of the stereotypes.
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    (Original post by Bath~Student)
    It's well known that the contrary is true: it's notoriously known undergraduates spend so much time on abstract crap that they lose practise of the basics.
    This is completely true: when I went to university, I could integrate any damn thing you threw at me - by the time I graduated, I needed to look up everything in a book of standard integrals (except for \int e^{-ax^2} dx due to a surfeit of stat. mech.)

    Anyway, engineering > mathematics. I'll show you my paycheck one day.
    Well, I've always had a foot in both camps, so I'm happy to leave you to argue it out either way.
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    (Original post by atsruser)
    This is completely true: when I went to university, I could integrate any damn thing you through at me - by the time I graduated, I needed to look up everything in a book of standard integrals (except for \int e^{-ax^2} dx due to a surfeit of stat. mech.)


    Well, I've always had a foot in both camps, so I'm happy to leave you to argue it out either way.
    haha nice..

    So, what are you doing now???
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    (Original post by Bath~Student)
    So what are you up to today?
    Well yesterday I was painting the wall of a house..

    Unemployed? Having an existential crisis? Suicidal? Just listing up some of the stereotypes.
    .. and today I'm having an existential crisis and am suicidal as I don't want to have to paint the rest of it. I'd rather post mistakes on maths forums.
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    (Original post by Zacken)
    I stopped there because I've just read the question. Surely you've done part (b)? If so, just translate your sketch from part (b) k units to the right.
    why would that work tho?
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    (Original post by atsruser)
    Well yesterday I was painting the wall of a house..



    .. and today I'm having an existential crisis and am suicidal as I don't want to have to paint the rest of it. I'd rather post mistakes on maths forums.
    I mean professionally. Why are you so secretive? I'm just curious about hthe life of a mathematics graduate.
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    (Original post by frostyy)
    why would that work tho?
    Consider the rules regarding the shifting of functions.
    If f(x) is some function, then f(x-k) is the same function, shifted k units to the right.
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    (Original post by frostyy)
    why would that work tho?
    f(x) = x^2(x+2) \Rightarrow f(x-k) = (x-k)^2(x-k+2) which is a shift to the right of k units.
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    (Original post by Zacken)
    f(x) = x^2(x+2) \Rightarrow f(x-k) = (x-k)^2(x-k+2) which is a shift to the right of k units.
    Thank you so much!
 
 
 
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