What is the inverse of f(x)= 6/((x^2-9)),x>3 ?

The inverse of f(x)= 6/((x^2-9)),x>3 is sqrt((6+9x)/x),-sqrt((6+9x)/x)

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inverse of f(x)= 6/((x^2-9)),x>3

inverse

f (x) = \frac{6}{( x ^{2} - 9 )}, x > 3

\frac{6 + 9 x}{x}, - \frac{6 + 9 x}{x}

Solution steps

y = \frac{6}{x ^{2} - 9}

Replace

x

with

y

x = \frac{6}{y ^{2} - 9}

Solve for

y, x = \frac{6}{y ^{2} - 9}

\frac{6 + 9 x}{x}, - \frac{6 + 9 x}{x}

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

3 ln (x + 2) - 3

6 x^{4} - 2 x^{2} + 5 x - 2

\frac{6 x - 8}{5 - x}

f (x) = \frac{x ^{7} - 7}{9}

What is the inverse of f(x)= 6/((x^2-9)),x>3 ?
The inverse of f(x)= 6/((x^2-9)),x>3 is sqrt((6+9x)/x),-sqrt((6+9x)/x)