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1. Circle the functions below that are one-to-one.

a.
[tex]\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
x & -2 & -1 & 0 & 1 & 2 & 3 \\
\hline
f(x) & 1 & 4 & 5 & 4 & 1 & -3 \\
\hline
\end{array}
\][/tex]



Answer :

GinnyAnswer
To determine whether the given function is one-to-one, we need to check if every value of [tex]\( f(x) \)[/tex] maps to a unique value of [tex]\( x \)[/tex]. A function is one-to-one if no two different inputs [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex] have the same output [tex]\( f(x_1) = f(x_2) \)[/tex].

Let's examine the given data:

[tex]\( \begin{array}{|c|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline f(x) & 1 & 4 & 5 & 4 & 1 & -3 \\ \hline \end{array} \)[/tex]

We list the pairs [tex]\( (x, f(x)) \)[/tex]:
- [tex]\( (-2, 1) \)[/tex]
- [tex]\( (-1, 4) \)[/tex]
- [tex]\( (0, 5) \)[/tex]
- [tex]\( (1, 4) \)[/tex]
- [tex]\( (2, 1) \)[/tex]
- [tex]\( (3, -3) \)[/tex]

Now, let's check for repeating values in the [tex]\( f(x) \)[/tex] column:

- [tex]\( f(-2) = 1 \)[/tex] and [tex]\( f(2) = 1 \)[/tex]
- [tex]\( f(-1) = 4 \)[/tex] and [tex]\( f(1) = 4 \)[/tex]

We observe that:

1. [tex]\( f(x) = 1 \)[/tex] for both [tex]\( x = -2 \)[/tex] and [tex]\( x = 2 \)[/tex]. So, [tex]\( f(-2) = f(2) \)[/tex].
2. [tex]\( f(x) = 4 \)[/tex] for both [tex]\( x = -1 \)[/tex] and [tex]\( x = 1 \)[/tex]. So, [tex]\( f(-1) = f(1) \)[/tex].

Since there are repeating [tex]\( f(x) \)[/tex] values for different [tex]\( x \)[/tex] values, the function is not one-to-one.

Therefore, none of the functions indicated in option (a) are one-to-one.

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