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

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

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

inverse

f (x) = \frac{x}{1 + x ^{2}}

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

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

f (x) = 5 lo g_{e} (x - 10)

\frac{2 + ( x - 3 ) ^{2}}{( x - 3 ) ^{2}}

\frac{x - 12}{4}

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

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