To determine the inverse of the function [tex]\( f(x) = 9x^2 - 12 \)[/tex], we follow these steps:
1. Rewrite [tex]\( f(x) \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[
y = 9x^2 - 12
\][/tex]
2. Solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[
y = 9x^2 - 12 \implies y + 12 = 9x^2
\][/tex]
[tex]\[
\frac{y + 12}{9} = x^2
\][/tex]
[tex]\[
x = \pm \sqrt{\frac{y + 12}{9}}
\][/tex]
Since the domain is restricted to [tex]\( x \geq 0 \)[/tex], we take the positive square root:
[tex]\[
x = \sqrt{\frac{y + 12}{9}}
\][/tex]
3. Express the inverse function:
The inverse function [tex]\( f^{-1}(y) \)[/tex]:
[tex]\[
f^{-1}(y) = \sqrt{\frac{y + 12}{9}}
\][/tex]
Simplifying, we get:
[tex]\[
f^{-1}(y) = \frac{\sqrt{y + 12}}{3}
\][/tex]
4. Identify the correct option:
Now we look at the given choices:
- A. [tex]\( q(x) = \frac{\sqrt{x+12}}{9} \)[/tex]
- B. [tex]\( p(x) = \frac{\sqrt{x-12}}{9} \)[/tex]
- C. [tex]\( g(x) = \frac{\sqrt{x+12}}{3} \)[/tex]
- D. [tex]\( h(x) = \frac{\sqrt{x-12}}{3} \)[/tex]
Comparing these with our inverse function [tex]\( f^{-1}(x) = \frac{\sqrt{x+12}}{3} \)[/tex], we see that choice C matches.
Therefore, the inverse function of [tex]\( f(x) = 9x^2 - 12 \)[/tex] with the domain [tex]\( x \geq 0 \)[/tex] is:
[tex]\[
\boxed{\frac{\sqrt{x+12}}{3}}
\][/tex]
Hence, the correct option is:
[tex]\[
\boxed{C}
\][/tex]
