What is the inverse of x/(x^2-4) ?

The inverse of x/(x^2-4) is (1+sqrt(1+16x^2))/(2x),(1-sqrt(1+16x^2))/(2x)

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inverse of x/(x^2-4)

inverse

\frac{x}{x ^{2} - 4}

\frac{1 + 1 + 16 x ^{2}}{2 x}, \frac{1 - 1 + 16 x ^{2}}{2 x}

Solution steps

y = \frac{x}{x ^{2} - 4}

Replace

x

with

y

x = \frac{y}{y ^{2} - 4}

Solve for

y, x = \frac{y}{y ^{2} - 4}

\frac{1 + 1 + 16 x ^{2}}{2 x}, \frac{1 - 1 + 16 x ^{2}}{2 x}

f (x) = \frac{6 x}{x ^{2} - 16}

(- 1, 3), (2, - 2)

f (x) = (x - 4)^{2} - 9

f (x) = - x^{2} + 5 x - 3

f (x) = 1 + \frac{1}{x} + \frac{1}{x ^{2}}

What is the inverse of x/(x^2-4) ?
The inverse of x/(x^2-4) is (1+sqrt(1+16x^2))/(2x),(1-sqrt(1+16x^2))/(2x)