Find the inverse function -- my problem

[lookagain]

Given $\displaystyle f(x) = \frac {1- \sqrt{x}}{1 + \sqrt{x}}.$


Find its inverse.

 

[BigGlenntheHeavy]

$\displaystyle Given: \ y \ = \ f(x) \ = \ \frac{1-\sqrt x}{1+\sqrt x}, \ x \ \ge \ 0, \ find \ its \ inverse.$

$\displaystyle To \ find \ inverse, \ alternate \ variables \ and \ solve \ for \ y.$

$\displaystyle Hence, \ x \ = \ \frac{1-\sqrt y}{1+\sqrt y}.$

$\displaystyle x+x\sqrt y \ = \ 1-\sqrt y, \ x\sqrt y+\sqrt y \ = \ 1-x, \ \sqrt y \ = \ \frac{1-x}{1+x}, \ y \ = \ \bigg(\frac{1-x}{1+x}\bigg)^2$

$\displaystyle Ergo \ f(x) \ = \ \frac{1-\sqrt x}{1+\sqrt x} \ and \ f^{-1}(x) \ = \ \bigg(\frac{1-x}{1+x}\bigg)^2, \ -1 \ < \\x \ \le \ 1, \ see \ graph.$

[attachment=0:bje8o0xr]ccc.jpg[/attachment:bje8o0xr]

 

[lookagain]

$\displaystyle Given: \ y \ = \ f(x) \ = \ \frac{1-\sqrt x}{1+\sqrt x}, \ x \ \ge \ 0, \ find \ its \ inverse.$



$\displaystyle Ergo \ f(x) \ = \ \frac{1-\sqrt x}{1+\sqrt x} \ and \ f^{-1}(x) \ = \ \bigg(\frac{1-x}{1+x}\bigg)^2, \ x \ \ge \ 0, \ see \ graph.$

The original function doesn't need $\displaystyle "x \ge 0"$ because the $\displaystyle \sqrt{x}$ already limits
it to that, and the original function is already a one-to-one function.

The domain of the inverse is the range of the original. The range of the original function is
$\displaystyle (-1, 1],$ as $\displaystyle y = -1$ is a horizontal asymptote, and the largest y-value occurs on the
extreme left-hand side. That largest function value is where $\displaystyle y = 1$ at $\displaystyle x = 0$
for the original function.

$\displaystyle f^{-1}(x) \ = \ \bigg(\frac{1-x}{1+x}\bigg)^2, \ \ -1 <\ x \ \le \ 1.$

 

[BigGlenntheHeavy]

$\displaystyle Thank \ you \ lookagain \ for \ keeping \ me \ honest; \ I \ stand \ corrected.$

$\displaystyle The \ domain \ of \ f^{-1}(x) \ is \ -1 \ \ < \ x \ \le \ 1, \ not \ x \ \ge \ 0.$

$\displaystyle Note: \ I \ edited \ my \ original \ reply \ above \ to \ avoid \ any \ future \ mathematician$

$\displaystyle from \ being \ mislead.$

 

[jengodbout]

If I have g(x) = 1-x^2. How do I solve this?

 

[BigGlenntheHeavy]

$\displaystyle If \ you \ are \ looking \ for \ the \ inverse \ of \ g(x), \ none \ exist \ as \ g(x) \ is \ not \ 1-to-1.$

 

[mmm4444bot]

g(x) = 1 - x^2

This function definition does not comprise an exercise.

How do I solve this?

There is nothing to solve, until you tell us what you are asked to do with function g.

You posted your function definition at the end of somebody else's discussion.

Please start your own thread, and read the post titled, "Read Before Posting".

Cheers ~ Mark

 

[jengodbout]

Sorry, Very new at this. Not a blogger. Just trying ot get some much needed math help.

 

[Aladdin]

Sorry, Very new at this. Not a blogger. Just trying ot get some much needed math help.

Start a new thread , and ask your question.