If $f$ is one-to-one, find an equation for its inverse.
\[
f(x)=8 x^{2}-9, x \geq 0
\]
$f^{-1}(x)=\frac{8}{\sqrt{x}-9}$
B. $f^{-1}(x)=\sqrt{\frac{x+9}{8}}, x \geq-9$
C. $f^{-1}(x)=\sqrt{\frac{8}{x-9}}, x \neq-9$
D. The function is not one-to-one.

Step 1 ：Given the function \(f(x)=8x^{2}-9\), where \(x \geq 0\), we are asked to find its inverse.

Step 2 ：First, we replace \(f(x)\) with \(y\), so the equation becomes \(y=8x^{2}-9\).

Step 3 ：Next, we swap \(x\) and \(y\) to get \(x=8y^{2}-9\).

Step 4 ：Solving for \(y\), we get \(y^{2}=(x+9)/8\).

Step 5 ：Since \(x \geq 0\) in the original function, we only take the positive square root, so \(y=\sqrt{(x+9)/8}\).

Step 6 ：Finally, replacing \(y\) with \(f^{-1}(x)\), we get \(f^{-1}(x)=\sqrt{(x+9)/8}\), which is the inverse of the function \(f(x)\).

Step 7 ：So, the correct answer is \(\boxed{f^{-1}(x)=\sqrt{(x+9)/8}, x \geq-9}\).