What is the inverse of y=((1-x)/(1+x))^2,-1<x<= 1 ?

The inverse of y=((1-x)/(1+x))^2,-1<x<= 1 is (-x-1+2sqrt(x))/(x-1),-(x+2sqrt(x)+1)/(x-1)

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inverse of y=((1-x)/(1+x))^2,-1<x<= 1

inverse

y = (\frac{1 - x}{1 + x})^{2}, - 1 < x \leq 1

\frac{- x - 1 + 2 x}{x - 1}, - \frac{x + 2 x + 1}{x - 1}

Solution steps

y = (\frac{1 - x}{1 + x})^{2}

Replace

x

with

y

x = (\frac{1 - y}{1 + y})^{2}

Solve for

y, x = (\frac{1 - y}{1 + y})^{2}

\frac{- x - 1 + 2 x}{x - 1}, - \frac{x + 2 x + 1}{x - 1}

f (x) = 24 - 2 x

f (x) = - 6 (x - 2)^{2}

g (x) = 7 x - 5

f (x) = 3 x + 3 + 4

What is the inverse of y=((1-x)/(1+x))^2,-1<x<= 1 ?
The inverse of y=((1-x)/(1+x))^2,-1<x<= 1 is (-x-1+2sqrt(x))/(x-1),-(x+2sqrt(x)+1)/(x-1)