What is the inverse of f(x)= 1/(x^2-9) ?

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

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inverse of f(x)= 1/(x^2-9)

inverse

f (x) = \frac{1}{x ^{2} - 9}

\frac{1 + 9 x}{x}, - \frac{1 + 9 x}{x}

Solution steps

y = \frac{1}{x ^{2} - 9}

Replace

x

with

y

x = \frac{1}{y ^{2} - 9}

Solve for

y, x = \frac{1}{y ^{2} - 9}

\frac{1 + 9 x}{x}, - \frac{1 + 9 x}{x}

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