Find inverse for f(x)= x/(1+x)

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Homework Help Overview

The discussion revolves around finding the inverse of the function f(x) = x/(1+x) and demonstrating that it is one-to-one. Participants express confusion regarding the process of solving for the inverse and the manipulation of the function's equation.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants attempt to manipulate the equation to isolate y and explore different algebraic approaches. Questions arise about the validity of switching variables and the steps taken in the manipulation process.

Discussion Status

There is ongoing exploration of various methods to find the inverse function, with some participants questioning each other's approaches and assumptions. A few participants express uncertainty about their steps, while others provide suggestions for isolating variables.

Contextual Notes

Some participants mention a perceived misunderstanding regarding the classification of the function, with references to it being labeled as "transcendental," which they dispute. The discussion includes attempts to clarify the nature of the function and its inverse.

jimjames
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Homework Statement


Show that the functions f are one-to-one and calculate the inverse function.

Homework Equations


f(x)= x/(1+x) (It is the equation I am having trouble with)

The Attempt at a Solution


I know the solution is y= x/(1-x) But no idea how to solve it.

My attempt:
x(1+y)=y or x+xy=y

y+y=x/x
 
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jimjames said:

Homework Statement


Show that the functions f are one-to-one and calculate the inverse function.

Homework Equations


f(x)= x/(1+x) (It is the equation I am having trouble with)

The Attempt at a Solution


I know the solution is y= x/(1-x) But no idea how to solve it.

My attempt:
x(1+y)=y or x+xy=y
Start with ##y = \frac x {1 + x}##
What's the first thing you need to do?
jimjames said:
y+y=x/x
I'm not sure what you did here.
 
Where did you get y = x/(1+x) ?

What i normally do is
f(x) = x/(1+x) <=> x=f^-1(y)
f(y) = y/(1+y) <=> y=f^-1(x)

x=y/(1+y)
y=x+xy

And now I'm stuck
 
jimjames said:
Where did you get y = x/(1+x) ?

What i normally do is
f(x) = x/(1+x) <=> x=f^-1(y)
f(y) = y/(1+y) <=> y=f^-1(x)
The above doesn't help you with this problem. All you're doing is manipulating symbols.
jimjames said:
x=y/(1+y)
In the line above, all you did was switch x and y.
jimjames said:
y=x+xy
How did you get the equation above?
jimjames said:
And now I'm stuck
Start with y = ##\frac x {1 + x}##

For this problem DO NOT switch the variables x and y.
DO solve for x in the equation just above. In other words, x should appear only on one side of the equation.
 
y(1+x)=x
1+x=x/y
 
Last edited:
jimjames said:
y(1+x)=x
1+x=x/y
OK.

Now divide both sides by x.
 
jimjames said:
Where did you get y = x/(1+x) ?
That's the function you're trying to find the inverse of.
Multiplying both sides of this equation by 1 + x gives you the equation just below.
jimjames said:
y(1+x)=x
1+x=x/y
SammyS said:
OK.

Now divide both sides by x.
It's probably simpler to expand y(1 + x), get all terms that involve x on one side, and then isolate x.
 
jimjames said:

Homework Statement


Show that the functions f are one-to-one and calculate the inverse function.

Homework Equations


f(x)= x/(1+x) (It is the equation I am having trouble with)

The Attempt at a Solution


I know the solution is y= x/(1-x) But no idea how to solve it.

My attempt:
x(1+y)=y or x+xy=y

y+y=x/x

Why do you call this a "transcendental function"? It is just about as far from transcendental as you can get.
 
Ray Vickson said:
Why do you call this a "transcendental function"? It is just about as far from transcendental as you can get.
I changed the title a while ago for that very reason.
 
  • #10
jimjames said:
Where did you get y = x/(1+x) ?

What i normally do is
f(x) = x/(1+x) <=> x=f^-1(y)
f(y) = y/(1+y) <=> y=f^-1(x)

x=y/(1+y)
y=x+xy

And now I'm stuck
Subtract xy from both sides: y- xy= y(1-x)= x
 
  • #11
Managed to solve this late yesterday.
Thanks for trying to help.:smile:
 

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