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

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

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inverse of y=sqrt((x+1)/(x-1))

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = \frac{8 x}{3 x - 4}

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

f (x) = \frac{5 x + 3}{x - 2}

y = 1 + \frac{2 - x}{3}

\frac{x + 3}{x + 4}

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