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

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

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

inverse

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

\frac{x 9 + x}{3 x}, - \frac{x 9 + x}{3 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x 9 + x}{3 x}, - \frac{x 9 + x}{3 x}

5 x^{3} + 9

x^{5} + 1

x^{5} + 7

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

5 x^{3} - 8

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