Inverse of (2x^2+x+1)/(2x^2+x-1)

[aurellinlmnsc]

Homework Statement

Find x as a function of y, if y = (2x^2+x+1)/(2x^2+x-1)

Homework Equations

(1/8 (-4 i x+sqrt(7)-i) (4 i x+sqrt(7)+i) ) / (x+1) (2 x-1)

for 2x^2+x+1, how do I factor that out to get 1/8 (-4 i x+sqrt(7)-i) (4 i x+sqrt(7)+i)?

quad. formula?

The Attempt at a Solution

y(x+1)(2x-1) = 1/8 (-4 i x+sqrt(7)-i) (4 i x+sqrt(7)+i)


So stuck.

 

[SammyS]

Are you sure that there is no typo here?

 

[aurellinlmnsc]

Yes, I am sure. This problem comes from the Chemistry Maths Book, by Erich Steiner.

Exercise Section 2.5 problem 44.

The answer is in the back of the book, but I don't even know how in the world they got to that answer as well. I've been running smoothly through the first 43 problems just fine.

Anyhow the answer should somehow be :

1/4(-1 +/- sqrt(1+8 ((y+1)/(y-1)))

 

[glebovg]

y = (2x^2 + x +1)/[(2x - 1)(x + 1)]

(2x - 1)(x + 1)y = 2x^2 + x + 1

(2x - 1)(x + 1)y - 2 = (2x - 1)(x + 1) Subtract 2 from both sides

y - 2/[(2x - 1)(x + 1)] = 1

y - 1 = 2/[(2x - 1)(x + 1)]

2/(y - 1) = (2x - 1)(x + 1)

2/(y - 1) = 2x^2 + x - 1 Expand

1/(y - 1) = x^2 + x/2 - 1/2

1/(y - 1) = (x + 1/4)^2 - (1/4)^2 - 1/2 Complete the square

(x + 1/4)^2 = 1/(y - 1) + 9/16

x = -1/4 ± √[1/(y - 1) + 9/16].

 

[aurellinlmnsc]

(2x - 1)(x + 1)y - 2 = (2x - 1)(x + 1) (2x - 1)(x + 1) does NOT = (2x^2 + x +1)

 

[glebovg]

(2x - 1)(x + 1)y = 2x^2 + x + 1

Subtract 2 from both side yields

(2x - 1)(x + 1)y - 2 = 2x^2 + x + 1 - 2

(2x - 1)(x + 1)y - 2 = 2x^2 + x - 1

(2x - 1)(x + 1)y - 2 = (2x - 1)(x + 1)

 

[aurellinlmnsc]

when I check in wolframalpha if the function x = -1/4 ± √[1/(y - 1) + 9/16] is the same as 1/4(-1 +/- sqrt(1+8 ((y+1)/(y-1)))

it doesn't even confirm but confused it as an entirely new function. http://www.wolframalpha.com/input/?i=1/4(-1++/-+sqrt(1+8+((y+1)/(y-1)))+=+-1/4+±+√[1/(y+-+1)+++9/16]

 

[SammyS]

Comparing the answer with the question ... WOW !

Here's how it appears that they solved it.

Letting $\displaystyle y=\frac{2x^2+x+1}{2x^2+x-1}$ gives $\displaystyle y+1=2\frac{2x^2+x}{2x^2+x-1}$ and $\displaystyle y-1=2\frac{1}{2x^2+x-1}\,.$

Therefore, $\frac{y+1}{y-1}=2x^2+x\,.$

Now, solve for x using the quadratic formula.

However, the graphs resulting from all this look inconsistent.

Another big problem with this is that the original rational function is not 1-to-1, so its inverse is not a well-defined function. (It is true that the result given but the author is not well-defined function because of the "±" .)

 

[dynamicsolo]

Taking up from where glebovg left it, some fairly meticulous algebra does indeed give Steiner's result. However, as SammyS points out, this is not a one-to-one function (it "fails the Horizontal Line Test", as you can confirm by graphing the original function). The function is symmetrical about the line x = -(1/4) , so you can only define an inverse for the half either to the left or to the right of that vertical line. That is the meaning of the "plus or minus" term following the -(1/4) in Steiner's "inverse".

 

[uart]

(2x - 1)(x + 1)y - 2 = (2x - 1)(x + 1)


(2x - 1)(x + 1) does NOT = (2x^2 + x +1)

glebovg never claimed that, (2x - 1)(x + 1) = (2x^2 + x +1).

He claimed that (2x - 1)(x + 1) = (2x^2 + x +1) - 2, which it does.

 

[uart]

This is a pretty good question from a "domain/range" perspective.

Both the numerator and denominator of the original function have an axis of symmetry at $x = -\frac{1}{4}$. So it's clear the y also has an axis of symmetry and thus is not invertible on the full domain. It is however invertible if we restrict the domain to $x > -\frac{1}{4}$

Define,

$$f(x) = \frac{2 x^2 + x +1}{2 x^2 + x -1} \,\,\,\,\, : x \ge -\frac{1}{4}$$

The best approach is to start by completing the squares top and bottom.

$$f(x) = \frac{2 (x + 0.25)^2 + \frac{7}{8}}{2 (x + 0.25)^2 - \frac{9}{8}}$$

Put this in the form,

$y = \frac{z+a}{z+b}$, where $z = 2 (x+.025)^2$.

Solving for z gives,

$$z = \frac{9y + 7}{8(y-1)}$$

And hence

$$x = \sqrt{z/2} - \frac{1}{4}$$

$$x = \sqrt{\frac{9y + 7}{16(y-1)}} - \frac{1}{4}$$

The domain of this inverse function is, $(y>1) \, \cup \, (y< -\frac{7}{9})$.

 

[dynamicsolo]

...

The domain of this inverse function is, $(y>1) \, \cup \, (y< -\frac{7}{9})$.

I thought I'd make a graph of the original function to illustrate the various points that uart and others have made. The factors (2x-1) and (x+1) that turn up in the factorization produce the vertical asymptotes (in green) at x = -1 and x = 1/2 . The symmetry about x = -1/4 (halfway between the vertical asymptotes, as would be expected) is evident. I have marked the necessary division of the function into two individually invertible halves by coloring them blue or red. We also see that the domain for either inverse is y > 1 or y < -(7/9) .

[PLAIN]http://img193.imageshack.us/img193/1471/aurellinlmnscgraph.jpg