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

The inverse of y=(2x)/(sqrt(x)-1) is (-4x+x^2+xsqrt(x^2-8x))/8 ,(-4x+x^2-xsqrt(x^2-8x))/8

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inverse of y=(2x)/(sqrt(x)-1)

inverse

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

\frac{- 4 x + x ^{2} + x x ^{2} - 8 x}{8}, \frac{- 4 x + x ^{2} - x x ^{2} - 8 x}{8}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{- 4 x + x ^{2} + x x ^{2} - 8 x}{8}, \frac{- 4 x + x ^{2} - x x ^{2} - 8 x}{8}

f (x) = (3)

f (x) = \frac{8 x}{3 x - 2}

cos (\frac{18}{27})

x^{- 3}

arctan (10)

What is the inverse of y=(2x)/(sqrt(x)-1) ?
The inverse of y=(2x)/(sqrt(x)-1) is (-4x+x^2+xsqrt(x^2-8x))/8 ,(-4x+x^2-xsqrt(x^2-8x))/8