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

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

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

inverse

\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}

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