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

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In summary, the given rational function y = (2x^2+x+1)/(2x^2+x-1) has an inverse function that is not well-defined due to the function not being one-to-one. However, if we restrict the domain to x > -1/4, an inverse function can be found by completing the squares and solving for x in terms of y. The domain of this inverse function is (y>1) ∪ (y< -(7/9)).
  • #1
aurellinlmnsc
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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.
 
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  • #2
Are you sure that there is no typo here?
 
  • #3
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
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].
 
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  • #5
(2x - 1)(x + 1)y - 2 = (2x - 1)(x + 1) (2x - 1)(x + 1) does NOT = (2x^2 + x +1)
 
  • #6
(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)
 
  • #8
Comparing the answer with the question ... WOW !

Here's how it appears that they solved it.

Letting [itex]\displaystyle y=\frac{2x^2+x+1}{2x^2+x-1}[/itex] gives [itex]\displaystyle y+1=2\frac{2x^2+x}{2x^2+x-1}[/itex] and [itex]\displaystyle y-1=2\frac{1}{2x^2+x-1}\,.[/itex]

Therefore, [itex]\frac{y+1}{y-1}=2x^2+x\,.[/itex]

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
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
aurellinlmnsc said:
(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.
 
  • #11
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 [itex]x = -\frac{1}{4}[/itex]. 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 [itex]x > -\frac{1}{4}[/itex]

Define,

[tex] f(x) = \frac{2 x^2 + x +1}{2 x^2 + x -1} \,\,\,\,\, : x \ge -\frac{1}{4} [/tex]

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

[tex] f(x) = \frac{2 (x + 0.25)^2 + \frac{7}{8}}{2 (x + 0.25)^2 - \frac{9}{8}} [/tex]

Put this in the form,

[itex] y = \frac{z+a}{z+b} [/itex], where [itex] z = 2 (x+.025)^2[/itex].

Solving for z gives,

[tex] z = \frac{9y + 7}{8(y-1)}[/tex]

And hence

[tex] x = \sqrt{z/2} - \frac{1}{4} [/tex]

[tex] x = \sqrt{\frac{9y + 7}{16(y-1)}} - \frac{1}{4} [/tex]

The domain of this inverse function is, [itex](y>1) \, \cup \, (y< -\frac{7}{9})[/itex].
 
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  • #12
uart said:
...

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

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]
 
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1. What is the inverse of the given function?

The inverse of a function is a function that reverses the effect of the original function. For the given function, the inverse can be found by swapping the x and y variables and solving for y. In this case, the inverse of (2x^2+x+1)/(2x^2+x-1) is (1-y^2)/(y^2+y+1).

2. How do you find the domain of the inverse function?

The domain of the inverse function is the range of the original function. To find the range, we can use the horizontal line test to determine if the function is one-to-one. If it passes the horizontal line test, the inverse function will have the same domain as the original function. In this case, the domain of the inverse function is all real numbers except -1 and 1, which are the roots of the original function.

3. Can the inverse function be written in a simpler form?

Yes, the inverse function can be simplified by using algebraic manipulations. In this case, we can use the difference of squares and difference of cubes formulas to simplify the inverse function to y = (-y^2-2y+1)/(2y^2-1).

4. How do you graph the inverse function?

To graph the inverse function, we can use the graph of the original function and its inverse symmetry. This means that the points (x,y) on the original function will have corresponding points (y,x) on the inverse function. By plotting these points and connecting them, we can graph the inverse function.

5. Is the inverse function always the reflection of the original function over the line y=x?

Yes, the inverse function and the original function are reflections of each other over the line y=x. This is because the inverse function reverses the effect of the original function, essentially swapping the x and y coordinates. So, any point (x,y) on the original function will have a corresponding point (y,x) on the inverse function, resulting in a reflection over the line y=x.

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