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    Can someone tell me any rules for this where the demoninator is a binomial (or at least has x as the denominator? I keep getting stuck on them.. I tried unsing the graphics calculator but it's like getting blood from a stone when it comes to the complicated ones such as y = (2x - 3)/(5 - x).

    Thankyou for any help!

    mike.
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    The things you need to consider are

    1) Where does the graph cross the axes.

    2) Where are there asymptotes? Note that the function undefined where the denominator is zero, so for y = (2x - 3)/(5 - x) there will be a vertical asymptote when x = 5.

    3) Where there is an asymptote, look at values of x just slightly to the left and right of this to observe the behaviour of the function. For example, for x=5 we consider x = 5.01 and see that y will be negative, for x = 4.99 y will be positive. This tells us that as we approach the asymptote from the left the graph 'shoots' upwards to infinity, while as we approach it from the right it 'shoots' downwards to minus infinity.

    4) What happens when x gets very large in the positive and negtive direction. What this usually means is that we can disgard the constants. Hence, when x is very large we see that y is roughly 2x/(-x) = -2. Hence, there is a horizontal asymptote when y = -2.

    From this you should be able to draw the graph.
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    (Original post by mik1a)
    Can someone tell me any rules for this where the demoninator is a binomial (or at least has x as the denominator? I keep getting stuck on them.. I tried unsing the graphics calculator but it's like getting blood from a stone when it comes to the complicated ones such as y = (2x - 3)/(5 - x).

    Thankyou for any help!

    mike.
    argh, P4 graph sketching...GRR
    this is how we were taught alright...
    first...find the asymptotes.
    so we got x=5 then there's another...to find it you gotta consider what happesn when x->infintiy. y->-2 therefore y=-2 is another asymptote.
    (there are more difficult examples you might have to know when the numerator is of a higher power than the denomiator)
    second...calculate your stationary points (if any)...I'll leave you to do that...
    thirdly...when x=0...what does y=
    viceversa
    then you got enough info to sketch away
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    Wow... that helps a lot, thankyou very much! I was trying to solve P4 inequalities but couldn't draw the graphs. Thanks.

    Btw how do you find stationary points with a binomial denominator? Is it the quotient rule?

    (f(x)/g(x))' = [f(x)g'(x) - f'(x)g(x)]/g^2(x) ? :confused:
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    (Original post by mik1a)
    Wow... that helps a lot, thankyou very much! I was trying to solve P4 inequalities but couldn't draw the graphs. Thanks.

    Btw how do you find stationary points with a binomial denominator? Is it the quotient rule?

    (f(x)/g(x))' = [f(x)g'(x) - f'(x)g(x)]/g^2(x) ? :confused:
    indeed it is
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    Cool, thanks!
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    If you are solving P4 inequalities I think that the method of critical regions is a lot easier to use, drawing graphs does take longer.
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    For the derivative, quotient rule, you got it the wrong way round on the top.
    d(u/v)/dx = (vdu - udv)/v^2
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    Also can you find the area under a curve such as 1/x by modelling it as a convergant series? I think my teacher said is was tricky to integrate 1/x

    Also this is an interesting thing I read about with gabriel's horn, imagine the curve y = 1/|x| (I think..), you get a horn shape. Imagine this is the cross section of the horn and the horn has a volume, and you try to fill it with paint. It has an infinite surface area, so no matter how much paint you put in you will not leave paint on the all the inside, but it has a finite area so the horn will fill up...

    Weird huh.
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    (Original post by JamesF)
    For the derivative, quotient rule, you got it the wrong way round on the top.
    d(u/v)/dx = (vdu - udv)/v^2
    I'm not sure what that means.. in the way I wrote it would it be:

    (f(x)/g(x))' = [f'(x)g(x) - f(x)g'(x)]/g^2(x) ?
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    (Original post by mik1a)
    Also can you find the area under a curve such as 1/x by modelling it as a convergant series? I think my teacher said is was tricky to integrate 1/x

    Also this is an interesting thing I read about with gabriel's horn, imagine the curve y = 1/|x| (I think..), you get a horn shape. Imagine this is the cross section of the horn and the horn has a volume, and you try to fill it with paint. It has an infinite surface area, so no matter how much paint you put in you will not leave paint on the all the inside, but it has a finite area so the horn will fill up...

    Weird huh.
    Well, the integral of 1/x is ln|x| but you are right that 1/x is a difficult function to integrate; for, I am not exactly sure about this but I think that the ln function was defined in order to deal with this particular integral (don't quote me on this). In terms of a convergent series, this is about estimating an integral in terms of spliting a region into lots of rectangles.
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    (Original post by mik1a)
    I'm not sure what that means.. in the way I wrote it would it be:

    (f(x)/g(x))' = [f'(x)g(x) - f(x)g'(x)]/g^2(x) ?
    Yea, sorry, when i wrote du, i mean du/dx. Anyway, thats correct now.
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    The graph of y = (2x - 3)/(5 - x) is not complicated. It is just a stretched and shifted version of the 1/x graph.

    y = (2x - 3)/(5 - x) = -2 + 7/(5 - x).

    Draw the graph of y = 1/(5 - x), then of y = 7/(5 - x) [stretch vertically by a factor of 7 . . . or relabel the y-axis], then finally of y = -2 + 7/(5 - x) [shift down by 2].
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    (Original post by mikesgt2)
    If you are solving P4 inequalities I think that the method of critical regions is a lot easier to use, drawing graphs does take longer.
    explain please...
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    (Original post by kikzen)
    explain please...
    The idea is the find all the values for which the function changes sign, which are at asymptotes and where the graph cuts the y axis. So, if I wanted to solve

    (2x-7)/[(3x-4)(x-1)] < 0

    I would note that the function changes sign at the points x=1, 4/3, 7/2 and then consider the regions

    x < 1
    1 < x < 4/3
    4/3 < x < 7/2
    x > 7/2

    In which we can see that the function takes the signs -, +, -, + respectively. Therefore, the solutions are x < 1 and 4/3 < x < 7/2. There is more detail on this in the first chapter of the P4 book.
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    (Original post by kikzen)
    explain please...
    I think he means points where the function 'blows up', ie. if you had f(x) = 5/(x + 3) or something, you'd find values of f(x) both sides of 3.
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    (Original post by mik1a)
    I'm not sure what that means.. in the way I wrote it would it be:

    (f(x)/g(x))' = [f'(x)g(x) - f(x)g'(x)]/g^2(x) ?
    A (possibly) slicker way to do these is to say y = u/v => lny = lnu - lnv => 1/y y' = 1/u u' - 1/v v'=> y' = (v.u' - u.v')/(v')²
 
 
 
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