To find the inverse of the function $f(x) = -8x^2 - 3$ with the restriction $x \geq 0$:

Steps:

1. Replace $f(x)$ with $y$:
   $y = -8x^2 - 3$

2. Swap $x$ and $y$ to prepare for finding the inverse:
   $x = -8y^2 - 3$

3. Solve for $y$ in terms of $x$:
   $x + 3 = -8y^2$
   $y^2 = -\frac{x + 3}{8}$
   $y = \sqrt{-\frac{x + 3}{8}}$

   Since $x \geq 0$, only the positive square root is valid:
   $y = \sqrt{-\frac{x + 3}{8}}$

4. Rewrite $y$ as $f^{-1}(x)$:
   $f^{-1}(x) = \sqrt{-\frac{x + 3}{8}}$

Final Answer:

$f^{-1}(x) = \sqrt{-\frac{x + 3}{8}}$

Let me know if you'd like detailed clarification on any step or have additional questions!

Related Questions:

1. What are the conditions for a function to have an inverse?
2. How does the domain of $f(x)$ affect its inverse?
3. What happens if $x < 0$ in this problem?
4. Can you graph $f(x)$ and $f^{-1}(x)$ to verify symmetry?
5. How do square roots appear in inverses, and why is the positive root chosen?

Tip: Always check the domain and range of the original function to verify the inverse's validity!