The function shown in the image is:

$f(x) = \frac{x}{\sqrt{5 - |x + 3|}}$

The task is to find the domain of this function, i.e., the values of $x$ for which the function is defined.

Conditions for the domain:

1. The expression inside the square root must be non-negative: $5 - |x + 3| \geq 0$ This ensures that the square root is real (since we can't take the square root of a negative number in the set of real numbers).

2. The denominator cannot be zero: $\sqrt{5 - |x + 3|} \neq 0$ This avoids division by zero.

Let's solve these conditions.

Step 1: Solve $5 - |x + 3| \geq 0$

This inequality means: $|x + 3| \leq 5$ We now solve this absolute value inequality.

Remove the absolute value:

$-5 \leq x + 3 \leq 5$

Solve for $x$:

Subtract 3 from all sides: $-8 \leq x \leq 2$

Thus, $x$ must lie between $-8$ and $2$ inclusive.

Step 2: Exclude the points where the denominator is zero

We need to ensure that $5 - |x + 3| \neq 0$. This happens when:

$5 - |x + 3| = 0$

$|x + 3| = 5$

Solve this for $x$:

$x + 3 = 5 \quad \text{or} \quad x + 3 = -5$

$x = 2 \quad \text{or} \quad x = -8$

These are the points where the denominator would be zero, so we must exclude $x = 2$ and $x = -8$ from the domain.

Final Answer:

The domain of the function is:

$x \in (-8, 2)$

This is the interval where the function is defined and real.

Let me know if you'd like a more detailed explanation or have any other questions!

Relative Questions:

1. How do we handle absolute value inequalities in general?
2. Why can't we have negative values inside a square root in real functions?
3. What happens if the denominator of a function is zero?
4. Can a function be defined at the endpoints of an interval?
5. How would the domain change if the square root wasn't in the denominator?

Tip:

Always check for conditions like square roots and denominators when finding a function's domain to avoid undefined operations.