What is the inverse of (x-3)/(x^2-3x-18) ?

The inverse of (x-3)/(x^2-3x-18) is (3x+1+sqrt(81x^2-6x+1))/(2x),(3x+1-sqrt(81x^2-6x+1))/(2x)

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inverse of (x-3)/(x^2-3x-18)

inverse

\frac{x - 3}{x ^{2} - 3 x - 18}

\frac{3 x + 1 + 81 x ^{2} - 6 x + 1}{2 x}, \frac{3 x + 1 - 81 x ^{2} - 6 x + 1}{2 x}

Solution steps

y = \frac{x - 3}{x ^{2} - 3 x - 18}

Replace

x

with

y

x = \frac{y - 3}{y ^{2} - 3 y - 18}

Solve for

y, x = \frac{y - 3}{y ^{2} - 3 y - 18}

\frac{3 x + 1 + 81 x ^{2} - 6 x + 1}{2 x}, \frac{3 x + 1 - 81 x ^{2} - 6 x + 1}{2 x}

\frac{6 + 2 x}{6 - 2 x}

y = 4 x - 3

f (x) = \frac{x + 2}{10}

f (x) = \frac{5}{x - 5} + 1

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

What is the inverse of (x-3)/(x^2-3x-18) ?
The inverse of (x-3)/(x^2-3x-18) is (3x+1+sqrt(81x^2-6x+1))/(2x),(3x+1-sqrt(81x^2-6x+1))/(2x)