What is the inverse of f(x)=(4x)/(x^2+81) ?

The inverse of f(x)=(4x)/(x^2+81) is (2+sqrt(-81x^2+4))/x ,(2-sqrt(-81x^2+4))/x

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inverse of f(x)=(4x)/(x^2+81)

inverse

f (x) = \frac{4 x}{x ^{2} + 81}

\frac{2 + - 81 x ^{2} + 4}{x}, \frac{2 - - 81 x ^{2} + 4}{x}

Solution steps

y = \frac{4 x}{x ^{2} + 81}

Replace

x

with

y

x = \frac{4 y}{y ^{2} + 81}

Solve for

y, x = \frac{4 y}{y ^{2} + 81}

\frac{2 + - 81 x ^{2} + 4}{x}, \frac{2 - - 81 x ^{2} + 4}{x}

(18, 1), (2, - 1)

f (x) = ln (7 x), x > 0

y = - ln (- x), - 1 < x < 0

\frac{x ^{2} - x - 6}{2 x + 4}

f (x) = 2 x^{2} + 9 x - 3

What is the inverse of f(x)=(4x)/(x^2+81) ?
The inverse of f(x)=(4x)/(x^2+81) is (2+sqrt(-81x^2+4))/x ,(2-sqrt(-81x^2+4))/x