The function given is:

$f(x) = \frac{x^2 - 16}{x + 4}$

Step 1: Simplify the function

First, observe that the denominator $x + 4$ cannot be zero because division by zero is undefined. So, we need to find when the denominator is zero:

$x + 4 = 0 \implies x = -4$

Thus, $x = -4$ is not in the domain of the function.

Step 2: Factorize the numerator

Now, let's factorize the numerator $x^2 - 16$:

$x^2 - 16 = (x - 4)(x + 4)$

So, the function can be rewritten as:

$f(x) = \frac{(x - 4)(x + 4)}{x + 4}$

For $x \neq -4$, the $x + 4$ terms cancel out, leaving:

$f(x) = x - 4 \quad \text{for} \quad x \neq -4$

Step 3: Identify the domain

The only value of $x$ for which the function is not defined is $x = -4$. So, the domain of the function is all real numbers except $x = -4$:

$D = \mathbb{R} \setminus \{-4\}$

Step 4: Find the set $A$

The set $A$ is defined as the set of integers that are not in the domain of $f$. Since the only point excluded from the domain is $x = -4$, the set $A$ is:

$A = \{-4\}$

Step 5: Determine the cardinality of $A$

The set $A = \{-4\}$ contains only one element, so the cardinality of $A$ is:

$\boxed{1}$

Would you like further details on any step or have more questions?

Related Questions:

1. What is the range of the function $f(x) = x - 4$ for $x \neq -4$?
2. How do you find the limit of $f(x)$ as $x$ approaches -4?
3. Can the function $f(x) = \frac{x^2 - 16}{x + 4}$ have any vertical or horizontal asymptotes?
4. How does removing $x = -4$ from the domain affect the continuity of the function?
5. What are the critical points of the simplified function $f(x) = x - 4$?

Tip:

Always check the denominator for values that make the function undefined when simplifying rational expressions.