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

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

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

inverse

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

- \frac{1}{x ^{2} - 1}, - - \frac{1}{x ^{2} - 1}

Solution steps

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

Replace

x

with

y

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

Solve for

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

- \frac{1}{x ^{2} - 1}, - - \frac{1}{x ^{2} - 1}

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

f (x) = (x^{2} - 36)^{\frac{1}{3}}

f (x) = 460 x - 11040

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

3 - x^{2}

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