Proving that a curve cannot lie between two values algebraically

Dec 2009
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Consider the curve \(\displaystyle y=\frac{2x(x+3)}{x-1}\)

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 \(\displaystyle y=\frac{2x(x+3)}{x-1}\) 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
 

mr fantastic

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

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 \(\displaystyle y=\frac{2x(x+3)}{x-1}\) 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:
Dec 2009
755
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could u show me the calculus approach?
 

mr fantastic

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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.
 

Opalg

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iii) Prove, using an algebraic method, that \(\displaystyle y=\frac{2x(x+3)}{x-1}\) 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. :p

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 \(\displaystyle y = x + \frac4x\) cannot take values between –4 and +4. To see that, notice that \(\displaystyle y^2 = \left(x+\tfrac4x\right)^2 = x^2 + 8 + 16x^{-2} = \left(x-\tfrac4x\right)^2 + 16.\) Thus \(\displaystyle y^2\geqslant 16\), which means that y cannot lie between –4 and +4.

Now let's try to use a similar method on the function \(\displaystyle y=\frac{2x(x+3)}{x-1}\). 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 \(\displaystyle |y-10|\geqslant8.\)

The first step is to calculate

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

Apart from having \(\displaystyle x-1\) in place of x, the expression in the red parentheses looks just like \(\displaystyle x+\tfrac4x.\) So you should be able to complete the argument from there.
 

mr fantastic

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The question specifically asks for an algebraic method, so I'm not sure why the thread has been moved to the Calculus forum. :p

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 \(\displaystyle y = x + \frac4x\) cannot take values between –4 and +4. To see that, notice that \(\displaystyle y^2 = \left(x+\tfrac4x\right)^2 = x^2 + 8 + 16x^{-2} = \left(x-\tfrac4x\right)^2 + 16.\) Thus \(\displaystyle y^2\geqslant 16\), which means that y cannot lie between –4 and +4.

Now let's try to use a similar method on the function \(\displaystyle y=\frac{2x(x+3)}{x-1}\). 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 \(\displaystyle |y-10|\geqslant8.\)

The first step is to calculate

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

Apart from having \(\displaystyle x-1\) in place of x, the expression in the red parentheses looks just like \(\displaystyle x+\tfrac4x.\) 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.
 
Oct 2019
2
1
Singapore
Apologies for the ‘necro’ but I think this will be helpful even in the future.
The question explicitly asked for an algebraic method, calculus is not to be used in this situation.

Consider y= 2x(x+3) / x-1 and the line y=k
K= 2x(x+3)/x-1
kx - k = 2x^2 + 6x
2x^2 + (6-k)x + k = 0
For the equation and the line to not intersect,
Discriminant must be less than 0
(6-k)^2 - 4(2)(k) < 0
36 - 12k + k^2 - 8k < 0
k^2 - 20k + 36 < 0
(k-18)(k-2) < 0
Therefore 2<k<18 and curve cannot lie between 2 and 18.
 
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Dec 2016
290
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Earth
Consider y= 2x(x+3) / x-1 and the line y=k
K= 2x(x+3)/x-1
kx - k = 2x^2 + 6x
2x^2 + (6-k)x + k = 0
For the equation and the line to not intersect,
Discriminant must be less than 0
(6-k)^2 - 4(2)(k) < 0
36 - 12k + k^2 - 8k < 0
k^2 - 20k + 36 < 0
(k-18)(k-2) < 0
Therefore 2<k<18 and curve cannot lie between 2 and 18.
Xarmcus, your denominator must be in grouping symbols. Also, you have to stay with either the lower case or
the upper case of a letter to represent a variable. In this matter, we should just use "k" throughout. And, for
readability, you should put more spaces in between your characters when written horizontally. Vertical spacing
helps more too.

Suggestion:

Consider y = 2x(x + 3)/(x - 1) and the line y = k.

k = 2x(x + 3)/(x - 1)

kx - k = 2x^2 + 6x

2x^2 + (6 - k)x + k = 0

For the curve and the line to not intersect, the discriminant must be less than 0.

(6 - k)^2 - 4(2)(k) < 0

36 - 12k + k^2 - 8k < 0

k^2 - 20k + 36 < 0

(k - 18)(k - 2) < 0

Therefore, 2 < k < 18, and the curve cannot lie between x = 2 and x = 18.
 
Oct 2019
2
1
Singapore
Xarmcus, your denominator must be in grouping symbols. Also, you have to stay with either the lower case or
the upper case of a letter to represent a variable. In this matter, we should just use "k" throughout. And, for
readability, you should put more spaces in between your characters when written horizontally. Vertical spacing
helps more too.

Suggestion:

Consider y = 2x(x + 3)/(x - 1) and the line y = k.

k = 2x(x + 3)/(x - 1)

kx - k = 2x^2 + 6x

2x^2 + (6 - k)x + k = 0

For the curve and the line to not intersect, the discriminant must be less than 0.

(6 - k)^2 - 4(2)(k) < 0

36 - 12k + k^2 - 8k < 0

k^2 - 20k + 36 < 0

(k - 18)(k - 2) < 0

Therefore, 2 < k < 18, and the curve cannot lie between x = 2 and x = 18.
Pardon my poor formatting, first time poster and was on mobile as well as rushing it out in about 3 minutes. Thanks for the suggestion/edit.