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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

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