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

The inverse of f(x)=(2x)/(sqrt(x-1)) is (x^2+xsqrt(x^2-16))/8 ,(x^2-xsqrt(x^2-16))/8

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

inverse

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

\frac{x ^{2} + x x ^{2} - 16}{8}, \frac{x ^{2} - x x ^{2} - 16}{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{x ^{2} + x x ^{2} - 16}{8}, \frac{x ^{2} - x x ^{2} - 16}{8}

f (x) = 2 x^{2}, x \geq 0

f (x) = 7 x^{3} - 12

f (x) = \frac{4}{5 x + 3}, x \neq = - \frac{3}{5}

\frac{1}{2} - x

f (x) = x^{2} + 15 x + 36

What is the inverse of f(x)=(2x)/(sqrt(x-1)) ?
The inverse of f(x)=(2x)/(sqrt(x-1)) is (x^2+xsqrt(x^2-16))/8 ,(x^2-xsqrt(x^2-16))/8