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}

f (x) = - 1 + (x + 1)^{5}

f (x) = 5 + 2 x - 2

f (x) = \frac{4 x + 8}{x - 7}

f (x) = \frac{x + 5}{x + 1}

f (x) = x^{4} + 2 x^{2} + 1

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))