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

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

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inverse of x/(x^2+1)

inverse

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

\frac{1 + 1 - 4 x ^{2}}{2 x}, \frac{1 - 1 - 4 x ^{2}}{2 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 + 1 - 4 x ^{2}}{2 x}, \frac{1 - 1 - 4 x ^{2}}{2 x}

f (x) = t^{4}

y = \frac{3 x ^{6} - 7 x + 10}{8 x ^{5} + 9 x + 10}

(1, 9), (- 4, - 3)

y = 3 x + 2

f (x) = 3 2 x

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