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Consider the table modeling function [tex]\( f \)[/tex].
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & -2 & -1 & 0 & 1 & 2 \\
\hline
$f(x)$ & -28 & -9 & -2 & -1 & 0 \\
\hline
\end{tabular}
\][/tex]

Complete the table modeling the inverse of function [tex]\( f \)[/tex].
[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & -28 & -9 & -2 & -1 & 0 \\
\hline
$f^{-1}(x)$ & -2 & -1 & 0 & 1 & 2 \\
\hline
\end{tabular}
\][/tex]

Sagot :

To solve this question, we need to create the inverse table for the function [tex]\( f \)[/tex]. In an inverse function, the roles of [tex]\( x \)[/tex] and [tex]\( f(x) \)[/tex] are reversed. Here’s a step-by-step solution:

1. Original Table:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -28 & -9 & -2 & -1 & 0 \\ \hline \end{tabular} \][/tex]

2. Establish the Relationship:
Each [tex]\( (x, f(x)) \)[/tex] pair in the original function table helps in forming the [tex]\( (f(x), x) \)[/tex] pair for the inverse table.

3. Construct the Inverse Table:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -28 & -9 & -2 & -1 & 0 \\ \hline f^{-1}(x) & -2 & -1 & 0 & 1 & 2 \\ \hline \end{tabular} \][/tex]

4. Mapping to the Given Format:
The table given in the problem is partially filled, and we need to match our inverse table to this format. According to the inverse table:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & - & - & - & -1 & 0 \\ \hline f^{-1}(x) & -2 & - & 0 & - & - \\ \hline \end{tabular} \][/tex]

5. Fill the Gaps:
Based on the above inverse table:
[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -28 & -9 & -2 & -1 & 0 \\ \hline f^{-1}(x) & -2 & -1 & 0 & 1 & 2 \\ \hline \end{tabular} \][/tex]

Therefore, the correct answer filled in the given boxes will be:

[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -28 & -9 & -2 & -1 & 0 \\ \hline f^{-1}(x) & -2 & -1 & 0 & 1 & 2 \\ \hline \end{array} \][/tex]

So the final table modeling the inverse function is:

[tex]\[ \begin{tabular}{|c|c|c|c|c|c|} \hline x & -28 & -9 & -2 & -1 & 0 \\ \hline f^{-1}(x) & -2 & -1 & 0 & 1 & 2 \\ \hline \end{tabular} \][/tex]