Let's solve the inequalities one at a time to determine the sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex].
1. Inequality for Set [tex]\(A\)[/tex]:
[tex]\[
3x + 4 \geq 13
\][/tex]
To isolate [tex]\(x\)[/tex], we start by subtracting 4 from both sides:
[tex]\[
3x \geq 9
\][/tex]
Then, divide both sides by 3:
[tex]\[
x \geq 3
\][/tex]
So, set [tex]\(A\)[/tex] consists of all [tex]\(x\)[/tex] values such that [tex]\(x \geq 3\)[/tex].
2. Inequality for Set [tex]\(B\)[/tex]:
[tex]\[
\frac{1}{2}x + 3 \leq 4
\][/tex]
To isolate [tex]\(x\)[/tex], subtract 3 from both sides:
[tex]\[
\frac{1}{2}x \leq 1
\][/tex]
Then, multiply both sides by 2:
[tex]\[
x \leq 2
\][/tex]
So, set [tex]\(B\)[/tex] consists of all [tex]\(x\)[/tex] values such that [tex]\(x \leq 2\)[/tex].
Now, we need to determine for which values of [tex]\(x\)[/tex] the union of sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is empty, i.e. [tex]\(A \cup B = \varnothing\)[/tex].
- Set [tex]\(A\)[/tex] includes all [tex]\(x \geq 3\)[/tex].
- Set [tex]\(B\)[/tex] includes all [tex]\(x \leq 2\)[/tex].
- Notice that there are no values of [tex]\(x\)[/tex] that can simultaneously satisfy both [tex]\(x \geq 3\)[/tex] and [tex]\(x \leq 2\)[/tex].
Since there are no overlapping values between the two sets, there are no values of [tex]\(x\)[/tex] that belong to both sets at the same time. Therefore, [tex]\(A \cup B = \varnothing\)[/tex].
The appropriate values of [tex]\(x\)[/tex] for which [tex]\(A \cup B\)[/tex] is empty are:
[tex]\[ x \leq 2 \quad \text{and} \quad x \geq 3 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{x \leq 2 \text{ and } x \geq 3} \][/tex]
