First, let's determine the sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex] by solving the given inequalities.
### Solving for [tex]\(A\)[/tex]:
We have the inequality:
[tex]\[ 3x + 4 \geq 13 \][/tex]
Subtract 4 from both sides:
[tex]\[ 3x \geq 9 \][/tex]
Divide both sides by 3:
[tex]\[ x \geq 3 \][/tex]
So, the set [tex]\(A\)[/tex] is:
[tex]\[ A = \{ x \mid x \geq 3 \} \][/tex]
### Solving for [tex]\(B\)[/tex]:
We have the inequality:
[tex]\[ \frac{1}{2}x + 3 \leq 4 \][/tex]
Subtract 3 from both sides:
[tex]\[ \frac{1}{2}x \leq 1 \][/tex]
Multiply both sides by 2:
[tex]\[ x \leq 2 \][/tex]
So, the set [tex]\(B\)[/tex] is:
[tex]\[ B = \{ x \mid x \leq 2 \} \][/tex]
### Finding [tex]\(A \cup B\)[/tex]:
The union of sets [tex]\(A\)[/tex] and [tex]\(B\)[/tex] is essentially combining all the values of [tex]\(x\)[/tex] that belong to either set [tex]\(A\)[/tex] or set [tex]\(B\)[/tex] or both.
- Set [tex]\(A\)[/tex] is all [tex]\(x\)[/tex] such that [tex]\( x \geq 3 \)[/tex].
- Set [tex]\(B\)[/tex] is all [tex]\(x\)[/tex] such that [tex]\( x \leq 2 \)[/tex].
Since [tex]\(x \geq 3\)[/tex] and [tex]\(x \leq 2\)[/tex] do not overlap, every possible value of [tex]\(x\)[/tex] either belongs to [tex]\(A\)[/tex] or [tex]\(B\)[/tex]. However, there is no single value of [tex]\(x\)[/tex] that can satisfy [tex]\( x \geq 3 \)[/tex] and [tex]\( x \leq 2 \)[/tex] simultaneously.
Hence, [tex]\( A \cup B \)[/tex] can never be empty since all real numbers will belong to either [tex]\(A\)[/tex] or [tex]\(B\)[/tex].
Therefore, there are no values of [tex]\(x\)[/tex] for which [tex]\(A \cup B = \varnothing\)[/tex].
None of the given options indicate that [tex]\( A \cup B \)[/tex] is empty.
So, we conclude that the correct interpretation leads us to recognize that it is impossible for [tex]\( A \cup B \)[/tex] to be empty.
The correct answer is:
[tex]\[ \text{There are no values of } x \text{ for which } A \cup B = \varnothing. \][/tex]
