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

1. ## 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

2. Originally Posted by Punch
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 &#61; &#40;2x&#94;2 &#43; 6x&#41;&#47;&#40;x-1&#41; - 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.

3. could u show me the calculus approach?

4. Originally Posted by Punch
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.

5. Originally Posted by Punch
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 10–8 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.

6. Originally Posted by Opalg
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 10–8 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.