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}

f (x) = \frac{3 x}{x - 3}

f (x) = - 4 x - 8 + 10

f (x) = 100 \cdot x

\frac{x - 2}{2 x + 5}

\frac{x}{x - 3}

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)