To solve this problem, let's start by understanding the definitions of the sets [tex]\( F \)[/tex] and [tex]\( H \)[/tex]:
- [tex]\( F = \{ y \mid y < 3 \} \)[/tex]: This set includes all real numbers that are less than 3.
- [tex]\( H = \{ y \mid y \geq 6 \} \)[/tex]: This set includes all real numbers that are greater than or equal to 6.
### Finding the Union [tex]\( F \cup H \)[/tex]
The union of two sets includes all elements that are in either set. In this case, it means we combine all values that are in [tex]\( F \)[/tex] with all values that are in [tex]\( H \)[/tex]:
- [tex]\( F = (-\infty, 3) \)[/tex]
- [tex]\( H = [6, \infty) \)[/tex]
To write the union of these two sets in interval notation, we simply combine the intervals:
[tex]\[ F \cup H = (-\infty, 3) \cup [6, \infty) \][/tex]
This union statement means that any real number less than 3 or any real number 6 or more is included in the set.
### Finding the Intersection [tex]\( F \cap H \)[/tex]
The intersection of two sets includes only the elements that are in both sets. Here we need to find where [tex]\( y \)[/tex] belongs to both [tex]\( F \)[/tex] and [tex]\( H \)[/tex]:
- In [tex]\( F \)[/tex], [tex]\( y < 3 \)[/tex]
- In [tex]\( H \)[/tex], [tex]\( y \geq 6 \)[/tex]
There is no overlap between these conditions since a number cannot simultaneously be less than 3 and greater than or equal to 6. Thus, the intersection is empty.
In interval notation, we represent an empty set by [tex]\( \varnothing \)[/tex]:
[tex]\[ F \cap H = \varnothing \][/tex]
### Summary of the Solution
- The union [tex]\( F \cup H = (-\infty, 3) \cup [6, \infty) \)[/tex]
- The intersection [tex]\( F \cap H = \varnothing \)[/tex]
