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

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

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

inverse

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

\frac{2 x + 1 + 4 x + 1}{2 x}, \frac{2 x + 1 - 4 x + 1}{2 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{2 x + 1 + 4 x + 1}{2 x}, \frac{2 x + 1 - 4 x + 1}{2 x}

f (x) = 5 e^{- x} - 2

f (x) = \frac{3 - 7 x}{4 + 5 x}

Y

y = sin (e^{2 x})

\frac{1}{4 - 1 - x}

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