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

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

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

inverse

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

f (x) = 0.17 - \frac{1}{0.17}

f (x) = (3 x - 4)^{\frac{1}{3}}

tan (- \frac{1}{3})

f (x) = \frac{x + 2}{x ^{2} + 2}

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