Answer
$f(x)=\dfrac{6}{x^{2}-9}$
$f^{-1}(x)=\sqrt{9+\dfrac{6}{x}}$
Work Step by Step
$f(x)=\dfrac{6}{x^{2}-9}$, for $x\gt3$
Substitute $f(x)$ by $y$:
$y=\dfrac{6}{x^{2}-9}$
Solve the equation for $x$. Start by taking $x^{2}-9$ to multiply the left side:
$y(x^{2}-9)=6$
Take $y$ to divide the right side:
$x^{2}-9=\dfrac{6}{y}$
Take $9$ to the right side:
$x^{2}=9+\dfrac{6}{y}$
Finally, take the square root of both sides:
$\sqrt{x^{2}}=\sqrt{9+\dfrac{6}{y}}$
$x=\sqrt{9+\dfrac{6}{y}}$
Interchange $x$ and $y$:
$y=\sqrt{9+\dfrac{6}{x}}$
Substitute $y$ by $f^{-1}(x)$:
$f^{-1}(x)=\sqrt{9+\dfrac{6}{x}}$
The graph of both the original function and its inverse is shown in the answer section.
