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

The inverse of f(x)=((x+1))/(x(x-1)) is (x+1+sqrt(x^2+6x+1))/(2x),(x+1-sqrt(x^2+6x+1))/(2x)

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

inverse

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

\frac{x + 1 + x ^{2} + 6 x + 1}{2 x}, \frac{x + 1 - x ^{2} + 6 x + 1}{2 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x + 1 + x ^{2} + 6 x + 1}{2 x}, \frac{x + 1 - x ^{2} + 6 x + 1}{2 x}

f (4) = 9 x - 13

x^{2} - x - 5

x^{2} - x - 3

f (x) = \frac{x + 3}{x + 13}

f (x) = 0.5 x + 7

What is the inverse of f(x)=((x+1))/(x(x-1)) ?
The inverse of f(x)=((x+1))/(x(x-1)) is (x+1+sqrt(x^2+6x+1))/(2x),(x+1-sqrt(x^2+6x+1))/(2x)