What is the inverse of f(x)= 1/((x^2-3x)) ?

The inverse of f(x)= 1/((x^2-3x)) is (3x+sqrt(9x^2+4x))/(2x),(3x-sqrt(9x^2+4x))/(2x)

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inverse of f(x)= 1/((x^2-3x))

inverse

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

\frac{3 x + 9 x ^{2} + 4 x}{2 x}, \frac{3 x - 9 x ^{2} + 4 x}{2 x}

Solution steps

y = \frac{1}{x ^{2} - 3 x}

Replace

x

with

y

x = \frac{1}{y ^{2} - 3 y}

Solve for

y, x = \frac{1}{y ^{2} - 3 y}

\frac{3 x + 9 x ^{2} + 4 x}{2 x}, \frac{3 x - 9 x ^{2} + 4 x}{2 x}

f (x) = \frac{6}{2 x + 3}

\frac{3 x + 2}{x - 1}

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

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

f (x) = - x^{6} + 5

What is the inverse of f(x)= 1/((x^2-3x)) ?
The inverse of f(x)= 1/((x^2-3x)) is (3x+sqrt(9x^2+4x))/(2x),(3x-sqrt(9x^2+4x))/(2x)