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

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

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

inverse

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

\frac{x + x ( x + 1 )}{x}, \frac{x - x ( x + 1 )}{x}

Solution steps

y = \frac{1}{x ( x - 2 )}

Replace

x

with

y

x = \frac{1}{y ( y - 2 )}

Solve for

y, x = \frac{1}{y ( y - 2 )}

\frac{x + x ( x + 1 )}{x}, \frac{x - x ( x + 1 )}{x}

f (x) = 2 x + 3 - 4

y + 1

f (x) = 5^{2} + 3

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

f (x) = x^{2} - 18 b x

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