What is the inverse of f(x)= 3/(sqrt(x^2-1)) ?

The inverse of f(x)= 3/(sqrt(x^2-1)) is (sqrt(9+x^2))/x ,-(sqrt(9+x^2))/x

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inverse of f(x)= 3/(sqrt(x^2-1))

inverse

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

\frac{9 + x ^{2}}{x}, - \frac{9 + x ^{2}}{x}

Solution steps

y = \frac{3}{x ^{2} - 1}

Replace

x

with

y

x = \frac{3}{y ^{2} - 1}

Solve for

y, x = \frac{3}{y ^{2} - 1}

\frac{9 + x ^{2}}{x}, - \frac{9 + x ^{2}}{x}

ln (x + x^{2} + 1)

g (x) = x + 6

f (x) = \frac{5}{2} x - 15

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

f (x) = 1 + 6 + 9 x

What is the inverse of f(x)= 3/(sqrt(x^2-1)) ?
The inverse of f(x)= 3/(sqrt(x^2-1)) is (sqrt(9+x^2))/x ,-(sqrt(9+x^2))/x