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) = 1 0^{- x}

7 - x + 9

\frac{- 2 s + 6}{s ^{2} + 4}

lo g_{10} (x + 3) + 5

y = 1 + ln (x + 4)

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)