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

The inverse of f(x)= x/(x^2-1) is (1+sqrt(1+4x^2))/(2x),(1-sqrt(1+4x^2))/(2x)

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

inverse

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

\frac{1 + 1 + 4 x ^{2}}{2 x}, \frac{1 - 1 + 4 x ^{2}}{2 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 + 1 + 4 x ^{2}}{2 x}, \frac{1 - 1 + 4 x ^{2}}{2 x}

x

2 x^{2} + 4 x - 10

cos (2 x + 5)

2 x - y = 1, (1, - 2)

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

What is the inverse of f(x)= x/(x^2-1) ?
The inverse of f(x)= x/(x^2-1) is (1+sqrt(1+4x^2))/(2x),(1-sqrt(1+4x^2))/(2x)