What is the inverse of f(x)= x/(9+x^2) ?

The inverse of f(x)= x/(9+x^2) is (1+sqrt(1-36x^2))/(2x),(1-sqrt(1-36x^2))/(2x)

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inverse of f(x)= x/(9+x^2)

inverse

f (x) = \frac{x}{9 + x ^{2}}

\frac{1 + 1 - 36 x ^{2}}{2 x}, \frac{1 - 1 - 36 x ^{2}}{2 x}

Solution steps

y = \frac{x}{9 + x ^{2}}

Replace

x

with

y

x = \frac{y}{9 + y ^{2}}

Solve for

y, x = \frac{y}{9 + y ^{2}}

\frac{1 + 1 - 36 x ^{2}}{2 x}, \frac{1 - 1 - 36 x ^{2}}{2 x}

f (x) = 4 - 7 x

f (x) = - 6 x + 8

f (x) = 4 - 4 \frac{3 x + 2}{15}

f (x) = - x - 7

f (x) = e^{\frac{1}{3} x}

What is the inverse of f(x)= x/(9+x^2) ?
The inverse of f(x)= x/(9+x^2) is (1+sqrt(1-36x^2))/(2x),(1-sqrt(1-36x^2))/(2x)