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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

\frac{( x ^{2} + 7 x - 13 )}{x + 4}

\frac{1}{t ^{2} + 1}

h (x) = \frac{2 x + 1}{3}

f (x) = 39.793 \cdot x + 1637.3

f (x) = (- 10 x + 1)^{2}

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