# Homework Help: Inverse of (2x^2+x+1)/(2x^2+x-1)

1. Sep 16, 2011

### aurellinlmnsc

1. The problem statement, all variables and given/known data

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

2. Relevant 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)?

3. 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.

2. Sep 16, 2011

### SammyS

Staff Emeritus
Are you sure that there is no typo here?

3. Sep 16, 2011

### 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)))

4. Sep 16, 2011

### 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].

Last edited: Sep 16, 2011
5. Sep 16, 2011

### aurellinlmnsc

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

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

6. Sep 16, 2011

### 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)

7. Sep 16, 2011

### aurellinlmnsc

8. Sep 17, 2011

### SammyS

Staff Emeritus
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 "±" .)

9. Sep 17, 2011

### 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".

10. Sep 17, 2011

### uart

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.

11. Sep 17, 2011

### 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, $(y>1) \, \cup \, (y< -\frac{7}{9})$.

Last edited: Sep 17, 2011
12. Sep 17, 2011

### dynamicsolo

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 [Broken]

Last edited by a moderator: May 5, 2017