To find the inverse of the given function $f(x) = \sqrt{x} - 9$, let's proceed step by step.

1. Set $f(x) = y$:

$y = \sqrt{x} - 9$

2. Switch $x$ and $y$:

$x = \sqrt{y} - 9$

3. Solve for $y$:

Add 9 to both sides: $x + 9 = \sqrt{y}$

Square both sides to eliminate the square root: $y = (x + 9)^2$

Thus, the inverse function is: $f^{-1}(x) = (x + 9)^2$

4. Determine the domain of the inverse:

The original function $f(x) = \sqrt{x} - 9$ has a domain $x \geq 0$ because the square root is only defined for non-negative values. The range of $f(x)$ is $y \geq -9$. Therefore, the domain of the inverse $f^{-1}(x)$ is: $x \geq -9$

Final Answer:

$f^{-1}(x) = (x + 9)^2, \quad x \geq -9$

Would you like a detailed explanation of any step? Here are related questions you might explore:

1. How do you verify that two functions are inverses of each other?
2. Why is squaring required when solving for $y$ in the inverse process?
3. How does the domain and range of a function affect its inverse?
4. Can you graph $f(x) = \sqrt{x} - 9$ and its inverse $f^{-1}(x) = (x+9)^2$?
5. What happens if the square root function includes a domain restriction other than $x \geq 0$?

Tip: Always check that the inverse function satisfies $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$.