What is the inverse of f(x)=(2x)/(x^2+1) ?

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

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inverse of f(x)=(2x)/(x^2+1)

inverse

f (x) = \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}

\frac{1}{8} x^{3} - x^{2}

- 2 sin (- 2 x)

x = 2

7 x^{3} - x^{2} + 7 x - 1

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

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