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

[Fractal]

I am just getting stumped by this. I always end up canceling out to zero. I remember something about multiplying the denominater and numerator on the left side by (1+x^(1/2)) when your in this kind of scenario but I still get stumped. Anyone got an idea? here is the equation again:
y=(1-x^(1/2))/(1+x^(1/2))

 

[tkhunny]

How about a new plan?

What if you first zubstitute \(\displaystyle \L\,z = \sqrt{x}\)?

Watch the applicable Domain on this one.

 

[Fractal]

Sorry, I don't follow. Could you elaborate?

 

[soroban]

Hello, Fractal!

Just do the usual "Find the inverse" routine:
. . "Switch" the $\displaystyle x$'s and $\displaystyle y$'s . . . and solve for $\displaystyle y.$

Find the inverse of: $\displaystyle \,y\;=\;\frac{1\,-\,\sqrt{x}}{1\,+\,\sqrt{x}}$

We have: $\displaystyle \,x\;=\;\frac{1\,-\,\sqrt{y}}{1\,+\,\sqrt{y}}\;\;\Rightarrow\;\;x(1\,+\,\sqrt{y}) \;= \;1\,-\,\sqrt{y}$

Then: $\displaystyle \,x\,+\,x\sqrt{y} \;= \;1\,-\,\sqrt{y}\;\;\Rightarrow\;\;x\sqrt{y}\,+\,\sqrt{y} \;=\;1\,-\,x$

Factor: \(\displaystyle \,\sqrt{y}(x\,+\,1)\;=\;1\,-\,x \;\;\Rightarrow\;\;\sqrt{y}\;=\;\frac{1\.-\,x}{1\,+\,x}\)

Square: $\displaystyle \,y\;=\;\left(\frac{1\,-\,x}{1\,+\,x}\right)^2$ . . . There!