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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = (x + 10)^{5}

f (x) = 5 + 2 x

- 5 x + 4 + 3

f (x) = \frac{1}{5} ln (\frac{1}{2} x)

e^{l n (x^{2})}

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