Finding the inverse of a rational function

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Homework Help Overview

The discussion revolves around finding the inverse of the rational function f(x) = 2x / (x - 2). Participants explore whether the function is one-to-one and how to express the equation in terms of y.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to verify if the function is one-to-one without graphing, questioning algebraic methods for this verification. They discuss the implications of having multiple y-intercepts and how that relates to the function being one-to-one.
  • Questions arise about expressing the equation in the form where x is given in terms of y, with some participants noting the difficulty of isolating x or y on one side of the equation.
  • Some suggest factoring and rearranging terms as potential steps to solve for x.

Discussion Status

The discussion is ongoing, with participants providing various insights and methods for approaching the problem. Some guidance has been offered regarding algebraic manipulation, but no consensus has been reached on a definitive method or solution.

Contextual Notes

Participants note the challenge of working with the equation due to the presence of both x and y on both sides, which complicates the process of finding the inverse. There is also mention of homework constraints that may affect the approaches taken.

mindauggas
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Homework Statement



Find the inverse of a rational function: f(x)=\frac{2x}{x-2}

The Attempt at a Solution



1. Verify the function is one-to-one.

Question: is there a way to do that withought drawing the graph? Some algebraic method? I know that this means that the function crosess any given x=a line only once, so, e.g. the function with two y intercepts would not be 1-1 ?

2. show that f(x)=f^{-1}(x)

Q.: How to express the given equation in the form where x is given in terms of y? It is imposible since you always get either y or x on both sides.
 
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mindauggas said:

Homework Statement



Find the inverse of a rational function: f(x)=\frac{2x}{x-2}

The Attempt at a Solution



1. Verify the function is one-to-one.

Question: is there a way to do that withought drawing the graph? Some algebraic method? I know that this means that the function crosess any given x=a line only once, so, e.g. the function with two y intercepts would not be 1-1 ?

2. show that f(x)=f^{-1}(x)

Q.: How to express the given equation in the form where x is given in terms of y? It is impossible since you always get either y or x on both sides.
What is f^{-1}(f(x))\ ?
 
Q.: How to express the given equation in the form where x is given in terms of y? It is imposible since you always get either y or x on both sides.

You should be able to solve for x in terms of y. You will need to take out a common factor at some stage.
 
mindauggas said:

Homework Statement



Find the inverse of a rational function: f(x)=\frac{2x}{x-2}

The Attempt at a Solution



1. Verify the function is one-to-one.

Question: is there a way to do that withought drawing the graph? Some algebraic method? I know that this means that the function crosess any given x=a line only once, so, e.g. the function with two y intercepts would not be 1-1 ?
Technically, if there are two y intercepts, it would not be a function. That is different from saying it is not one to one. A function is not one to one if it crosses some y line more than once.

To determine whether this function is one to one, look at what happens if two values of x give the same y: suppose 2a/(a- 2)= 2b/(b- 2). Multply on both sided by (a- 2)(b- 2) to get 2a(b- 2)= 2b(a- 2). That is the same as 2ab- 2a= 2ba- 2b. We can subtract 2ba (= 2ab) from both sides to get -2a= -2b, and finally divide both sides by -2: a= b, showing that two different values of x cannot give the same y.

2. show that f(x)=f^{-1}(x)

Q.: How to express the given equation in the form where x is given in terms of y? It is imposible since you always get either y or x on both sides.
That kind of equation is the whole point of algebra! If you have, for example, 4xy= y- x, you solve for x by adding x to both sides: xy+ x= y, factoring out x: x(y+ 1)= 4, and dividing both sides by y+ 1: x= 4/(y+1).
 
CAF123 said:
You should be able to solve for x in terms of y. You will need to take out a common factor at some stage.

I do it thus:

(i) y=\frac{2x}{x-2}

(ii) y(x-2)=2x

(iii) \frac{y(x-2)}{2}=x

Is this the right path? Probably not ...
 
From (ii) y(x- 2)= 2x, multiply out the left side: xy- 2y= 2x, then get the "x"s together on one side- xy- 2x= 2y; x(y- 2)= 2y. Can you finish that?
 
HallsofIvy said:
From (ii) y(x- 2)= 2x, multiply out the left side: xy- 2y= 2x, then get the "x"s together on one side- xy- 2x= 2y; x(y- 2)= 2y. Can you finish that?

Thanks, to you and to all :)
 
Notice that for the function given,
f(f(x))=x\ .​
 

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