Answer :
Let's analyze the given table of the function [tex]\( f \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -28 & -9 & -2 & -1 & 0 \\ \hline \end{array} \][/tex]
To find the inverse of the function [tex]\( f \)[/tex], we need to switch the [tex]\( x \)[/tex] and [tex]\( f(x) \)[/tex] values. That is, we need to find [tex]\( f^{-1}(y) \)[/tex] such that [tex]\( f(f^{-1}(y)) = y \)[/tex].
Given the partially completed table for [tex]\( f^{-1} \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & \square & \square & \square & -1 & 0 \\ \hline f^{-1}(x) & -2 & \square & 0 & \square & \square \\ \hline \end{array} \][/tex]
We will fill in the missing values step-by-step.
1. [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = -28 \implies f^{-1}(-28) = -2 \][/tex]
So, we put [tex]\(-28\)[/tex] under [tex]\(x\)[/tex] and [tex]\(-2\)[/tex] under [tex]\(f^{-1}(x)\)[/tex].
2. [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = -9 \implies f^{-1}(-9) = -1 \][/tex]
So, we put [tex]\(-9\)[/tex] under [tex]\(x\)[/tex] and [tex]\(-1\)[/tex] under [tex]\(f^{-1}(x)\)[/tex].
3. [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -2 \implies f^{-1}(-2) = 0 \][/tex]
So, we confirm [tex]\(0\)[/tex] under [tex]\(f^{-1}(x)\)[/tex].
4. [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -1 \implies f^{-1}(-1) = 1 \][/tex]
So, we put [tex]\(-1\)[/tex] under [tex]\(x\)[/tex] and [tex]\(1\)[/tex] under [tex]\(f^{-1}(x)\)[/tex].
5. [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 0 \implies f^{-1}(0) = 2 \][/tex]
So, we confirm [tex]\(2\)[/tex] under [tex]\(x\)[/tex].
Now, we fill out the complete table:
[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]
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline f(x) & -28 & -9 & -2 & -1 & 0 \\ \hline \end{array} \][/tex]
To find the inverse of the function [tex]\( f \)[/tex], we need to switch the [tex]\( x \)[/tex] and [tex]\( f(x) \)[/tex] values. That is, we need to find [tex]\( f^{-1}(y) \)[/tex] such that [tex]\( f(f^{-1}(y)) = y \)[/tex].
Given the partially completed table for [tex]\( f^{-1} \)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline x & \square & \square & \square & -1 & 0 \\ \hline f^{-1}(x) & -2 & \square & 0 & \square & \square \\ \hline \end{array} \][/tex]
We will fill in the missing values step-by-step.
1. [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = -28 \implies f^{-1}(-28) = -2 \][/tex]
So, we put [tex]\(-28\)[/tex] under [tex]\(x\)[/tex] and [tex]\(-2\)[/tex] under [tex]\(f^{-1}(x)\)[/tex].
2. [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = -9 \implies f^{-1}(-9) = -1 \][/tex]
So, we put [tex]\(-9\)[/tex] under [tex]\(x\)[/tex] and [tex]\(-1\)[/tex] under [tex]\(f^{-1}(x)\)[/tex].
3. [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = -2 \implies f^{-1}(-2) = 0 \][/tex]
So, we confirm [tex]\(0\)[/tex] under [tex]\(f^{-1}(x)\)[/tex].
4. [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -1 \implies f^{-1}(-1) = 1 \][/tex]
So, we put [tex]\(-1\)[/tex] under [tex]\(x\)[/tex] and [tex]\(1\)[/tex] under [tex]\(f^{-1}(x)\)[/tex].
5. [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 0 \implies f^{-1}(0) = 2 \][/tex]
So, we confirm [tex]\(2\)[/tex] under [tex]\(x\)[/tex].
Now, we fill out the complete table:
[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]
