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}

f (x) = (x + 4)^{2}, x \geq 0

x^{2} + 6 x + 9

\frac{x ^{2} + 3 x - 10}{x ^{2} - 4}

f (x) = \frac{1}{1 + x ^{4}}

f (x) = 3.5 (2.2)^{x}

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)