Sure! Let's solve the problem step by step:
We are given a function [tex]\( f(x) = x - 1 \)[/tex]. We need to evaluate this function for the values [tex]\( x = -1 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( x = 1 \)[/tex], and then fill in the corresponding values for [tex]\( f(x) \)[/tex] in the table.
1. Evaluate [tex]\( f(x) \)[/tex] for [tex]\( x = -1 \)[/tex]:
Substitute [tex]\( x = -1 \)[/tex] into the function:
[tex]\[
f(-1) = -1 - 1 = -2
\][/tex]
So, [tex]\( f(-1) = -2 \)[/tex].
2. Evaluate [tex]\( f(x) \)[/tex] for [tex]\( x = 0 \)[/tex]:
Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[
f(0) = 0 - 1 = -1
\][/tex]
So, [tex]\( f(0) = -1 \)[/tex].
3. Evaluate [tex]\( f(x) \)[/tex] for [tex]\( x = 1 \)[/tex]:
Substitute [tex]\( x = 1 \)[/tex] into the function:
[tex]\[
f(1) = 1 - 1 = 0
\][/tex]
So, [tex]\( f(1) = 0 \)[/tex].
Now, we can fill in the table with the calculated values:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $f(x)$ \\
\hline
-1 & -2 \\
\hline
0 & -1 \\
\hline
1 & 0 \\
\hline
\end{tabular}
\][/tex]
Thus, the completed table looks like this:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $f(x)$ \\
\hline
-1 & -2 \\
\hline
0 & -1 \\
\hline
1 & 0 \\
\hline
\end{tabular}
\][/tex]
