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}

f (x) = 4 x^{2} - 16

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

f (x) = e^{- x}

x^{2} - 6 x

f (x) = \frac{2}{x + 4}

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)