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Math Help - Proving that a curve cannot lie between two values algebraically

  1. #1
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    Proving that a curve cannot lie between two values algebraically

    Consider the curve y=\frac{2x(x+3)}{x-1}

    i) State the coordinates of any points of intersection with the axes

    ii) State the equation of the asymptotes

    iii) Prove, using an algebraic method, that y=\frac{2x(x+3)}{x-1} cannot lie between two values (to be determined)



    i) (0,0) and (-3,0)

    ii)x=1 and y=2x+8

    iii) i couldnt start
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  2. #2
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    Quote Originally Posted by Punch View Post
    Consider the curve y=\frac{2x(x+3)}{x-1}

    i) State the coordinates of any points of intersection with the axes

    ii) State the equation of the asymptotes

    iii) Prove, using an algebraic method, that y=\frac{2x(x+3)}{x-1} cannot lie between two values (to be determined)



    i) (0,0) and (-3,0)

    ii)x=1 and y=2x+8

    iii) i couldnt start
    The graph of the function suggests an obvious calculus approach: plot y = (2x^2 + 6x)/(x-1) - Wolfram|Alpha

    But since you posted this question in the PRE-calculus subforum, calculus cannot be used to answer it. Right?


    Edit: Thread moved to Calculus subforum, given post #3.
    Last edited by mr fantastic; June 1st 2011 at 03:28 AM.
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  3. #3
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    could u show me the calculus approach?
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    Quote Originally Posted by Punch View Post
    could u show me the calculus approach?
    The graph has two turning points, at (x1, y1) and (x2, y2). Clearly then the range of the function is all real numbres, excluding the interval (y1, y2). So use calculus to find x1 and x2 (get dy/dx, solve dy/dx = 0 etc. etc.) , hence get y1 and y2.
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    Quote Originally Posted by Punch View Post
    iii) Prove, using an algebraic method, that y=\frac{2x(x+3)}{x-1} cannot lie between two values (to be determined)
    The question specifically asks for an algebraic method, so I'm not sure why the thread has been moved to the Calculus forum.

    Having said that, it doesn't seem to be that easy to show this result by purely algebraic means. Suppose we start by looking at a simpler result: The function y = x + \frac4x cannot take values between 4 and +4. To see that, notice that y^2 = \left(x+\tfrac4x\right)^2 = x^2 + 8 + 16x^{-2} =  \left(x-\tfrac4x\right)^2 + 16. Thus y^2\geqslant 16, which means that y cannot lie between 4 and +4.

    Now let's try to use a similar method on the function y=\frac{2x(x+3)}{x-1}. At this point, I will cheat slightly, by using the helpful graph linked to in Mr F's first comment above. That makes it look as though y cannot take values between 2 and 18, or in other words between 108 and 10+8. So we should try to prove that |y-10|\geqslant8.

    The first step is to calculate

    \begin{aligned}y-10 = \frac{2x(x+3) - 10(x-1)}{x-1} &= \frac{2(x^2-2x+5)}{x-1} \\ &= \frac{2\bigl((x-1)^2+4\bigr)}{x-1} = 2{\color{red}\bigl(}(x-1) + 4(x-1)^{-1}{\color{red}\bigr)}.\end{aligned}

    Apart from having x-1 in place of x, the expression in the red parentheses looks just like x+\tfrac4x. So you should be able to complete the argument from there.
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    Quote Originally Posted by Opalg View Post
    The question specifically asks for an algebraic method, so I'm not sure why the thread has been moved to the Calculus forum.

    Mr F says: Post #3.

    Having said that, it doesn't seem to be that easy to show this result by purely algebraic means. Suppose we start by looking at a simpler result: The function y = x + \frac4x cannot take values between 4 and +4. To see that, notice that y^2 = \left(x+\tfrac4x\right)^2 = x^2 + 8 + 16x^{-2} = \left(x-\tfrac4x\right)^2 + 16. Thus y^2\geqslant 16, which means that y cannot lie between 4 and +4.

    Now let's try to use a similar method on the function y=\frac{2x(x+3)}{x-1}. At this point, I will cheat slightly, by using the helpful graph linked to in Mr F's first comment above. That makes it look as though y cannot take values between 2 and 18, or in other words between 108 and 10+8. So we should try to prove that |y-10|\geqslant8.

    The first step is to calculate

    \begin{aligned}y-10 = \frac{2x(x+3) - 10(x-1)}{x-1} &= \frac{2(x^2-2x+5)}{x-1} \\ &= \frac{2\bigl((x-1)^2+4\bigr)}{x-1} = 2{\color{red}\bigl(}(x-1) + 4(x-1)^{-1}{\color{red}\bigr)}.\end{aligned}

    Apart from having x-1 in place of x, the expression in the red parentheses looks just like x+\tfrac4x. So you should be able to complete the argument from there.
    Many misplaced posts have wasted a lot of time and effort when it has been thought that a simple approach was precluded because of the subforum the question was posted in. My instinct was that this had the potential to be one of those times, and post #3 seemed to support this. I moved the thread after post #3.
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