What is the inverse of (2x)/(x^2-1) ?

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

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inverse of (2x)/(x^2-1)

inverse

\frac{2 x}{x ^{2} - 1}

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

Solution steps

y = \frac{2 x}{x ^{2} - 1}

Replace

x

with

y

x = \frac{2 y}{y ^{2} - 1}

Solve for

y, x = \frac{2 y}{y ^{2} - 1}

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

5 sin (4 x - π)

\frac{1}{( x - 4 ) ^{2}}

- 3 x^{4} + 20 x^{3} - 24 x^{2}

- \frac{9}{( x - 7 ) ^{2}}

f (x) = - 16 x^{2} + 150 x + 6

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