Answer :
To solve for the points of the inverse function \( j^{-1}(t) \), we need to understand what an inverse function does. An inverse function essentially reverses the roles of \( t \) and \( j(t) \).
In other words, if \( j(t) = y \), then \( j^{-1}(y) = t \).
Given the table for \( j(t) \):
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline t & 1 & 8 & 10 & 17 \\ \hline j(t) & 1 & 2 & 10 & 250 \\ \hline \end{tabular} \][/tex]
We swap the input \( t \) and the output \( j(t) \) to find the points for the inverse function \( j^{-1}(t) \):
The new table for \( j^{-1}(t) \) would be:
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline t & 1 & 2 & 10 & 250 \\ \hline j^{-1}(t) & 1 & 8 & 10 & 17 \\ \hline \end{tabular} \][/tex]
Thus, the correct option that displays the points for the inverse function \( j^{-1}(t) \) is:
[tex]\[ \begin{tabular}{|l|l|l|l|l|} \hline t & 1 & 2 & 10 & 250 \\ \hline j^{-1}(t) & 1 & 8 & 10 & 17 \\ \hline \end{tabular} \][/tex]
In other words, if \( j(t) = y \), then \( j^{-1}(y) = t \).
Given the table for \( j(t) \):
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline t & 1 & 8 & 10 & 17 \\ \hline j(t) & 1 & 2 & 10 & 250 \\ \hline \end{tabular} \][/tex]
We swap the input \( t \) and the output \( j(t) \) to find the points for the inverse function \( j^{-1}(t) \):
The new table for \( j^{-1}(t) \) would be:
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline t & 1 & 2 & 10 & 250 \\ \hline j^{-1}(t) & 1 & 8 & 10 & 17 \\ \hline \end{tabular} \][/tex]
Thus, the correct option that displays the points for the inverse function \( j^{-1}(t) \) is:
[tex]\[ \begin{tabular}{|l|l|l|l|l|} \hline t & 1 & 2 & 10 & 250 \\ \hline j^{-1}(t) & 1 & 8 & 10 & 17 \\ \hline \end{tabular} \][/tex]