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{100}{1 + 20 e ^{- 0.05 x}}

- \frac{3}{4} x + 6

f (x) = \frac{- 2 x + 7}{6 x - 10}

f (x) = 10 - 6 x^{3}

f (x) = 4 - \frac{2 x}{3}

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)