\(\displaystyle Given: \ y \ = \ f(x) \ = \ \frac{1-\sqrt x}{1+\sqrt x}, \ x \ \ge \ 0, \ find \ its \ inverse.\)
\(\displaystyle To \ find \ inverse, \ alternate \ variables \ and \ solve \ for \ y.\)
\(\displaystyle Hence, \ x \ = \ \frac{1-\sqrt y}{1+\sqrt y}.\)
\(\displaystyle x+x\sqrt y \ = \ 1-\sqrt y, \ x\sqrt y+\sqrt y \ = \ 1-x, \ \sqrt y \ = \ \frac{1-x}{1+x}, \ y \ = \ \bigg(\frac{1-x}{1+x}\bigg)^2\)
\(\displaystyle Ergo \ f(x) \ = \ \frac{1-\sqrt x}{1+\sqrt x} \ and \ f^{-1}(x) \ = \ \bigg(\frac{1-x}{1+x}\bigg)^2, \ -1 \ < \\x \ \le \ 1, \ see \ graph.\)
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