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

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

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inverse of f(x)=(2x+1)/(x-1),x\ne 1

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = \frac{1}{7} x - 7

\frac{( 2 S - 3 )}{S ^{2} + S - 2}

y = x + 11 + 5

81 - \frac{2}{x}

\frac{2}{s} - \frac{2}{s + 1}

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