What is the inverse of 8/(x^2-x-6) ?

The inverse of 8/(x^2-x-6) is (x+sqrt(25x^2+32x))/(2x),(x-sqrt(25x^2+32x))/(2x)

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inverse of 8/(x^2-x-6)

inverse

\frac{8}{x ^{2} - x - 6}

\frac{x + 25 x ^{2} + 32 x}{2 x}, \frac{x - 25 x ^{2} + 32 x}{2 x}

Solution steps

y = \frac{8}{x ^{2} - x - 6}

Replace

x

with

y

x = \frac{8}{y ^{2} - y - 6}

Solve for

y, x = \frac{8}{y ^{2} - y - 6}

\frac{x + 25 x ^{2} + 32 x}{2 x}, \frac{x - 25 x ^{2} + 32 x}{2 x}

f (x) = 4 x + 8 x^{2}

f (x) = - 2^{(x - 2)} + 5

f (5) = 2 x + 1

f (x) = \frac{- 3 x + 9}{x + 7}

f (x) = - 5 x^{5} + 3

What is the inverse of 8/(x^2-x-6) ?
The inverse of 8/(x^2-x-6) is (x+sqrt(25x^2+32x))/(2x),(x-sqrt(25x^2+32x))/(2x)