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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = \frac{( x + 7 )}{x}

3 lo g_{10} (x - 1)

2 (3.1)

f (x) = \frac{3 x ^{2} - 4 x + 5}{x - 3}

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