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

The inverse of f(x)=((x+x^2))/((1-x)) is (-1-x+sqrt(x^2+6x+1))/2 ,(-1-x-sqrt(x^2+6x+1))/2

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inverse of f(x)=((x+x^2))/((1-x))

inverse

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

\frac{- 1 - x + x ^{2} + 6 x + 1}{2}, \frac{- 1 - x - x ^{2} + 6 x + 1}{2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{- 1 - x + x ^{2} + 6 x + 1}{2}, \frac{- 1 - x - x ^{2} + 6 x + 1}{2}

f (x) = \frac{x + 4}{2 x - 3}

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

(x - 9)^{2} + 3

f (x) = \frac{x}{6 x + 5}

What is the inverse of f(x)=((x+x^2))/((1-x)) ?
The inverse of f(x)=((x+x^2))/((1-x)) is (-1-x+sqrt(x^2+6x+1))/2 ,(-1-x-sqrt(x^2+6x+1))/2