Answer:
[tex]u(2) = 3[/tex].
Step-by-step explanation:
A function [tex]f: A \to B[/tex] is a set of ordered pairs ([tex]2[/tex]-tuples in the form [tex](a,\, b)[/tex]), such that for any [tex]a \in A[/tex], there is exactly one [tex]b \in B[/tex] such that [tex](a,\, b) \in f[/tex].
In other words, for any element [tex]a[/tex] in the domain of this function, there must be exactly one ordered pair with this [tex]a\![/tex] as the first element.
If [tex]f[/tex] meets these requirements, then for any [tex]a \in A[/tex] and [tex]b \in B[/tex], write [tex]f(a) = b[/tex] if and only if [tex](a,\, b) \in f[/tex].
In other words, write [tex]f(a) = b[/tex] ([tex]f[/tex] maps [tex]a[/tex] to [tex]b[/tex]) if and only if there is an ordered pair in [tex]f\![/tex] with [tex]a\![/tex] as the first element and [tex]b\![/tex] as the second element.
The [tex]u[/tex] in this question is indeed a set of ordered pairs and meet the requirements for being a function from [tex]\lbrace 1, 2, 3\rbrace[/tex] to [tex]\lbrace 2,\, 3,\, 4 \rbrace[/tex].
The question is asking for the value of [tex]u(2)[/tex]. Thus, it would be necessary to find an ordered pair in [tex]u[/tex] with [tex]2[/tex] as the first element.
The only ordered pair in [tex]u[/tex] with [tex]2[/tex] as the first element is [tex](2,\, 3)[/tex]. Thus, [tex]u\![/tex] maps [tex]2\![/tex] to [tex]3\![/tex], such that [tex]u(2) = 3[/tex].
