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

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

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

inverse

f (x) = \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}

\frac{x - 7}{( x - 7 ) ( x - 5 )}

(y - 8) = \frac{1}{7} (x - 10)

(2, 11), (- 6, 13)

\frac{7}{\frac{7}{x}}

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

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