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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = \frac{7}{10} x

\frac{x + 11}{15}

f (x) = \frac{3 x - 5}{x ^{2} - 2 x - 8}

\frac{( x ^{2} + 3 x - 4 )}{x + 4}

f (x) = x - 2 - 2.6

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