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}

f (x) = 2 (x - 3)^{2} - 8

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

+ lo g_{8} (x)

f (x) = \frac{1}{2 ( x - 4 ) ^{2} + 1}

(x - 1)^{3} + 5

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)