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

The inverse of f(x)=((x-1))/(x+1) is (-1-x)/(x-1)

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

inverse

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

\frac{- 1 - x}{x - 1}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{- 1 - x}{x - 1}

f (x) = 4 (x + 17)

f (x) = \frac{3}{x ^{2} - 1}

ln (x + x^{2} + 1)

g (x) = x + 6

f (x) = \frac{5}{2} x - 15

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