The problem in the image is asking for the domain of the function $h(x)$, which is defined as a quotient of two functions:

$F(x) = \sqrt{2 - |x|} \quad \text{and} \quad G(x) = 1 - x$ $h(x) = \frac{F(x)}{G(x)}$

To determine the domain of $h(x)$, we need to consider two things:

1. The domain of $F(x)$, which involves a square root.
2. The domain of $G(x)$, which involves a potential division by zero.

1. Domain of $F(x) = \sqrt{2 - |x|}$:

• For the square root function to be defined, the expression inside the square root must be non-negative: $2 - |x| \geq 0$ Solving this inequality: $|x| \leq 2$ Which means $x \in [-2, 2]$. So, the domain of $F(x)$ is $x \in [-2, 2]$.

2. Domain of $G(x) = 1 - x$:

• The only restriction for $G(x)$ comes from the fact that it cannot be zero (since division by zero is undefined): $1 - x \neq 0 \quad \Rightarrow \quad x \neq 1$

Combining both conditions:

• The domain of $F(x)$ is $[-2, 2]$, and the function $G(x)$ excludes $x = 1$.
• Therefore, the domain of $h(x) = \frac{F(x)}{G(x)}$ is: $x \in [-2, 2] \setminus \{1\}$ Which means the domain is $x \in [-2, 1) \cup (1, 2]$.

Final Answer:

The domain of $h(x)$ is $(-2, 1) \cup (1, 2)$.

Would you like further details or have any questions about this?

Here are 5 related questions for further exploration:

1. What is the range of the function $F(x) = \sqrt{2 - |x|}$?
2. How does the absolute value function affect the domain of $F(x)$?
3. How would the domain change if $G(x) = x^2 - 1$?
4. What happens to the domain if we modify $F(x)$ to $\sqrt{3 - |x|}$?
5. Can you graph the function $h(x)$ and verify the domain visually?

Tip: Always check for restrictions like square roots or division when finding domains of functions.