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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

\frac{x}{x ^{2} - 6 x + 8}

f (x) = 2 x 3 x^{2} + 3

7 x - 2 y = 4

f (x) = \frac{x ^{2} + 6 x + 11}{2 x ^{2} + 12 x + 18}

f (x) = - x^{2} - 3 x - 2

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