I see the image contains a set of problems, but you've already completed some of them. I'll focus on the unanswered parts:

Let's proceed with solving the problems and finding the domains for the operations between $f(x) = \sqrt{1 - x}$ and $g(x) = \sqrt{16 - x^2}$.

Problem 4: What is the domain of $f - g$?

We need to find the domain where both $f(x) = \sqrt{1 - x}$ and $g(x) = \sqrt{16 - x^2}$ are defined.

1. For $f(x) = \sqrt{1 - x}$, the expression inside the square root must be non-negative: $1 - x \geq 0 \implies x \leq 1$ So, $f(x)$ is defined for $x \in (-\infty, 1]$.

2. For $g(x) = \sqrt{16 - x^2}$, the expression inside the square root must be non-negative: $16 - x^2 \geq 0 \implies -4 \leq x \leq 4$ So, $g(x)$ is defined for $x \in [-4, 4]$.

To find the domain of $f - g$, we need the intersection of the domains of $f(x)$ and $g(x)$: [ x \in (-\infty, 1] \cap [-4, 4] = [-4, 1] ] Thus, the domain of $f - g$ is $[-4, 1]$.

Problem 6: What is the domain of $f \cdot g$?

This is similar to $f - g$. The product $f(x) \cdot g(x)$ is defined where both $f(x)$ and $g(x)$ are defined. As calculated before, the intersection of the domains is $[-4, 1]$.

Thus, the domain of $f \cdot g$ is $[-4, 1]$.

Problem 8: What is the domain of $\frac{f}{g}$?

Here, we need to consider two things:

1. Both $f(x)$ and $g(x)$ must be defined, which gives the intersection of their domains, $[-4, 1]$, as previously calculated.
2. Additionally, we need $g(x) \neq 0$, since division by zero is undefined.

To find where $g(x) = 0$, solve: $\sqrt{16 - x^2} = 0 \implies 16 - x^2 = 0 \implies x = \pm 4$ Thus, $g(x) = 0$ at $x = -4$ and $x = 4$.

Therefore, the domain of $\frac{f}{g}$ is $[-4, 1]$ excluding $x = -4$, which gives: $(-4, 1]$

If you have any further questions or need additional details, feel free to ask. Here are some related questions to deepen your understanding:

1. How would the domain change if $f(x)$ were $\sqrt{x + 1}$?
2. What is the graphical interpretation of $f(x) + g(x)$ for these functions?
3. How does the domain of $f(x) \cdot g(x)$ change if you replace $g(x)$ with another quadratic function?
4. How would you handle the case if $g(x)$ had no real roots (i.e., no points where it becomes zero)?
5. What is the effect on the range of $\frac{f}{g}$ given its domain?

Tip: When finding the domain of a function involving square roots or division, always ensure the radicand is non-negative and the denominator is non-zero!