Challenge question to the community - - Determine the inverse function of

[checkitagain]

$f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}$$Edit: \ \ I \ sent \ a \ PM \ to \ a \ mentor.$

 

[micromass]

[STRIKE]Can you PM the solution to one of the mentors please?? This way we can verify that it's a challenge question and not a disguised homework problem.[/STRIKE]

OK, this is verified.

 

[mathman]

Finding √x is trivial. Just square it afterward.

 

[DivisionByZro]

Thanks for posting this. The inverse of f(x) is:

$$<br /> f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}<br />$$

and x>=0 to prevent singularities

If you'd like to see my work, then just ask. It's easy to show that f(f^1(x)) = x.

Edit:

But I forgot to check whether f^-1 satisfied the requirements of inverse, I'll repost later, as I made a mistake :(

 

[checkitagain]

Finding √x is trivial.
$$\text{I don't know what you mean by/from particulars that you didn't type.}$$


Just square it afterward.

$$\text{And what about any possible restrictions on a domain?}$$

 

[checkitagain]

Thanks for posting this. I didn't see the challenge here though, it's really elementary algebra. The inverse of f(x) is:

$$<br /> f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}<br />$$

If you'd like to see my work, then just ask. It's easy to show that f(f^1(x)) = x.

Hint: What you have typed is not a one-to-one function.

 

[DivisionByZro]

Hint: What you have typed is not a one-to-one function.

If I restrict x>=0, then my inverse is correct (I think...for now..). So I would say for x>=0, f^-1 is

$$f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}<br />$$

 

[checkitagain]

What you posted was not one-to-one either.
If I restrict x>=0, then my inverse is correct. So I would say for x>=0, f^-1 is

$$f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}<br />$$

Mine (meaning the original function) is one-to-one.

Recommendation:

Graph/sketch my function and see.

 

[DivisionByZro]

Mine (meaning the original function) is one-to-one.

Recommendation:

Graph/sketch my function and see.

Yeah I had made a slight typo, I looked at the graph of a completely different function. My answer stands,

$$<br /> f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}<br />$$

on x>=0 only.

 

[checkitagain]

Yeah I had made a slight typo, I looked at the graph of a completely different function. My answer stands,

$$<br /> f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}<br />$$

on x>=0 only.

No to your restriction.


Hint: The range of f is the domain of f inverse, numerically speaking
(as when given in interval notation).


I must go away from the computer for a couple of hours starting now,
but I will check back for any updates.


--------------------------------------------------------


$Edit: \ \ What \ is \ the \ range \ of \ f(x) \ ?$


$2nd \ edit: \ \ \ I \ \ am \ \ agreeing \ \ with \ \ Curious3141,$

$but \ \ I \ \ am \ \ stating \ \ it \ \ in \ \ this \ \ edit \ \ so \ \$

$that \ \ Curious3141's \ \ post \ \ will \ \ show \ \ the$

$answer \ \ more \ \ recently \ \ than \ \ my \ \ own \ \ recent \ \ post.$

 

[Curious3141]

The inverse is:


$$<br /> f^{-1}(x) = \frac{(1-x)^{2}}{(1+x)^{2}}<br />$$

with a domain of (-1,1] and a range of [0,∞).

 

[chiro]

Just for future reference for checkitagain and other interested readers, you can use identities for derivatives of inverse functions and generate a taylor series that corresponds to the inverse function.

The key thing to be aware of is if the derivative (of the original function) is zero which indicates a branch cut in the function which means that you have to treat each branch as an individual function in its own right (if it is to have an inverse)