What is the inverse of x/(sqrt(x^2-9)) ?

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

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inverse of x/(sqrt(x^2-9))

inverse

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

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

Solution steps

y = \frac{x}{x ^{2} - 9}

Replace

x

with

y

x = \frac{y}{y ^{2} - 9}

Solve for

y, x = \frac{y}{y ^{2} - 9}

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

f (x) = \frac{4 x + 2}{x - 3}

48 x - 4 x^{2}

f (x) = \frac{5 x - 4}{9 x + 8}

f (x) = 3 x^{2} - 12 x + 3

f (x) = x + \frac{10}{3}

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