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Sagot :
To determine whether the given relation is a function, we need to verify if every input [tex]\( x \)[/tex] has a unique corresponding output [tex]\( y \)[/tex]. In other words, for the relation to be a function, each value of [tex]\( x \)[/tex] must be paired with exactly one value of [tex]\( y \)[/tex].
Let's examine the provided table in detail:
[tex]\[ \begin{tabular}{|c|c|} \hline Input, \( x \) & Output, \( y \) \\ \hline 9 & 2 \\ \hline 6 & 4 \\ \hline 9 & 6 \\ \hline 12 & 8 \\ \hline \end{tabular} \][/tex]
- The input [tex]\( x = 9 \)[/tex] is associated with two different outputs: [tex]\( y = 2 \)[/tex] and [tex]\( y = 6 \)[/tex].
- The input [tex]\( x = 6 \)[/tex] is associated with a single output: [tex]\( y = 4 \)[/tex].
- The input [tex]\( x = 12 \)[/tex] is associated with a single output: [tex]\( y = 8 \)[/tex].
Since the input [tex]\( x = 9 \)[/tex] does not have a unique corresponding output (it is paired with both 2 and 6), the relation fails the criteria of a function. For a relation to be a function, each input must map to exactly one output, but here we can see that [tex]\( x = 9 \)[/tex] maps to more than one output.
Therefore, the relation shown in the table is not a function.
Let's examine the provided table in detail:
[tex]\[ \begin{tabular}{|c|c|} \hline Input, \( x \) & Output, \( y \) \\ \hline 9 & 2 \\ \hline 6 & 4 \\ \hline 9 & 6 \\ \hline 12 & 8 \\ \hline \end{tabular} \][/tex]
- The input [tex]\( x = 9 \)[/tex] is associated with two different outputs: [tex]\( y = 2 \)[/tex] and [tex]\( y = 6 \)[/tex].
- The input [tex]\( x = 6 \)[/tex] is associated with a single output: [tex]\( y = 4 \)[/tex].
- The input [tex]\( x = 12 \)[/tex] is associated with a single output: [tex]\( y = 8 \)[/tex].
Since the input [tex]\( x = 9 \)[/tex] does not have a unique corresponding output (it is paired with both 2 and 6), the relation fails the criteria of a function. For a relation to be a function, each input must map to exactly one output, but here we can see that [tex]\( x = 9 \)[/tex] maps to more than one output.
Therefore, the relation shown in the table is not a function.
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