Thank you for the clarification. The problem asks to find the inverse of the function $g(x) = x^2 - 12x$ given that $x \geq 6$.

Let's go through the steps to find the inverse function:

1. Set $g(x) = y$

$y = x^2 - 12x$

2. Swap $y$ and $x$ to solve for $y$

$x = y^2 - 12y$

3. Solve the quadratic equation for $y$

We rearrange the equation: $y^2 - 12y - x = 0$

This is a quadratic equation in the form $ay^2 + by + c = 0$, where $a = 1$, $b = -12$, and $c = -x$.

4. Use the quadratic formula to solve for $y$

The quadratic formula is: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute $a = 1$, $b = -12$, and $c = -x$: $y = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(-x)}}{2(1)}$ $y = \frac{12 \pm \sqrt{144 + 4x}}{2}$ $y = \frac{12 \pm \sqrt{4(36 + x)}}{2}$ $y = \frac{12 \pm 2\sqrt{36 + x}}{2}$ $y = 6 \pm \sqrt{36 + x}$

5. Determine the correct sign

Since $x \geq 6$ implies that the function $g(x) = x^2 - 12x$ is increasing for $x \geq 6$, we take the positive square root: $y = 6 + \sqrt{36 + x}$

Final Answer

The inverse function is: $g^{-1}(x) = 6 + \sqrt{36 + x}$

Would you like any further details on this solution? Here are some related questions that might interest you:

1. What is the domain of the inverse function $g^{-1}(x)$?
2. How do you graph the inverse function $g^{-1}(x)$ along with $g(x)$?
3. What is the significance of the condition $x \geq 6$ in the original function?
4. How would the inverse function change if $g(x)$ had a different form?
5. Can you verify the inverse function by composing $g(g^{-1}(x))$ and $g^{-1}(g(x))$?

Tip: When solving for the inverse, ensure that you consider the domain of the original function, as it will influence the correct branch of the square root to use.