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

The inverse of f(x)=(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 f(x)=(2x)/(sqrt(x)-1)

inverse

f (x) = \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}

4 + lo g_{3} (x)

f (x) = 9 + lo g_{6} (x - 3)

f (t) = 57 e^{0.5 t}

f (x) = 1 - 6 x

f (x) = 1 - 7 x

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