What is the inverse of f(x)=(3x)/(x^2-9) ?

The inverse of f(x)=(3x)/(x^2-9) is (3+3sqrt(1+4x^2))/(2x),(3-3sqrt(1+4x^2))/(2x)

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inverse of f(x)=(3x)/(x^2-9)

inverse

f (x) = \frac{3 x}{x ^{2} - 9}

\frac{3 + 3 1 + 4 x ^{2}}{2 x}, \frac{3 - 3 1 + 4 x ^{2}}{2 x}

Solution steps

y = \frac{3 x}{x ^{2} - 9}

Replace

x

with

y

x = \frac{3 y}{y ^{2} - 9}

Solve for

y, x = \frac{3 y}{y ^{2} - 9}

\frac{3 + 3 1 + 4 x ^{2}}{2 x}, \frac{3 - 3 1 + 4 x ^{2}}{2 x}

R (q) = ln (1 + 10 q^{2})

f (x) = - x + 4 + 3

- 7 a + 5

\frac{1}{6 x}

What is the inverse of f(x)=(3x)/(x^2-9) ?
The inverse of f(x)=(3x)/(x^2-9) is (3+3sqrt(1+4x^2))/(2x),(3-3sqrt(1+4x^2))/(2x)