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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

\frac{1}{ln ( e ^{x} + 1 )}

[\frac{z ^{2}}{( z - 1 ) ( z - 3 )}]

f (x) = (1, 5) (- 3, 2) (8, - 6)

f (x) = \frac{400}{x ^{2}}

f (x) = 2 x^{(2)} - 3 x - 1

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