What is the inverse of f(x)=(8x)/(sqrt(x^2-1)) ?

The inverse of f(x)=(8x)/(sqrt(x^2-1)) is (xsqrt(x^2-64))/(x^2-64),-(xsqrt(x^2-64))/(x^2-64)

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inverse of f(x)=(8x)/(sqrt(x^2-1))

inverse

f (x) = \frac{8 x}{x ^{2} - 1}

\frac{x x ^{2} - 64}{x ^{2} - 64}, - \frac{x x ^{2} - 64}{x ^{2} - 64}

Solution steps

y = \frac{8 x}{x ^{2} - 1}

Replace

x

with

y

x = \frac{8 y}{y ^{2} - 1}

Solve for

y, x = \frac{8 y}{y ^{2} - 1}

\frac{x x ^{2} - 64}{x ^{2} - 64}, - \frac{x x ^{2} - 64}{x ^{2} - 64}

x^{2} - 4 x + 7

f (x) = - 2^{- x} + 1

f (x) = 3 + 4^{x}

f (x) = 32 - 20 e^{- 0.4 x}

What is the inverse of f(x)=(8x)/(sqrt(x^2-1)) ?
The inverse of f(x)=(8x)/(sqrt(x^2-1)) is (xsqrt(x^2-64))/(x^2-64),-(xsqrt(x^2-64))/(x^2-64)