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

The inverse of 2/(x^2-2x-3) is (2x+sqrt(16x^2+8x))/(2x),(2x-sqrt(16x^2+8x))/(2x)

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

inverse

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

\frac{2 x + 16 x ^{2} + 8 x}{2 x}, \frac{2 x - 16 x ^{2} + 8 x}{2 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{2 x + 16 x ^{2} + 8 x}{2 x}, \frac{2 x - 16 x ^{2} + 8 x}{2 x}

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

\frac{2 ^{4 x - 1} \cdot 3 ^{x - 2}}{1 2 ^{x - 2}}

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

\frac{23 - 3 x}{6 x}

f (x) = 3 ln (x) - 2

What is the inverse of 2/(x^2-2x-3) ?
The inverse of 2/(x^2-2x-3) is (2x+sqrt(16x^2+8x))/(2x),(2x-sqrt(16x^2+8x))/(2x)