What is the inverse of f(x)=(6x)/(x^2+9) ?

The inverse of f(x)=(6x)/(x^2+9) is (3(sqrt(-x^2+1)+1))/x ,(3(-sqrt(-x^2+1)+1))/x

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inverse of f(x)=(6x)/(x^2+9)

inverse

f (x) = \frac{6 x}{x ^{2} + 9}

\frac{3 ( - x ^{2} + 1 + 1 )}{x}, \frac{3 ( - - x ^{2} + 1 + 1 )}{x}

Solution steps

y = \frac{6 x}{x ^{2} + 9}

Replace

x

with

y

x = \frac{6 y}{y ^{2} + 9}

Solve for

y, x = \frac{6 y}{y ^{2} + 9}

\frac{3 ( - x ^{2} + 1 + 1 )}{x}, \frac{3 ( - - x ^{2} + 1 + 1 )}{x}

6 sin (\frac{π}{3} x) + 1

f (x) = 9 x - x^{2}

\frac{1}{x + 3}

\frac{x}{x ^{2} - 25}

- \frac{1}{2} sin (\frac{1}{4} x)

What is the inverse of f(x)=(6x)/(x^2+9) ?
The inverse of f(x)=(6x)/(x^2+9) is (3(sqrt(-x^2+1)+1))/x ,(3(-sqrt(-x^2+1)+1))/x