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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

\frac{( s + 1 )}{s ^{2} + 2 s + 10}

f (- 1) = \frac{3 - 2 x}{2 x - 5}

f (x) = 3 \cdot + 1

f (x) = 3 x - 6 + 5

f (y) = x^{2} - 20 x + 100

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