The function [tex]f[/tex] is defined by the following rule:
[tex]\[ f(x) = 2x - 2 \][/tex]

Complete the function table.
\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline
-5 & [tex]$\square$[/tex] \\
\hline
-4 & [tex]$\square$[/tex] \\
\hline
1 & [tex]$\square$[/tex] \\
\hline
3 & [tex]$\square$[/tex] \\
\hline
5 & [tex]$\square$[/tex] \\
\hline
\end{tabular}



Answer :

Sure! Let's complete the function table step-by-step using the given function [tex]\( f(x) = 2x - 2 \)[/tex].

1. For [tex]\( x = -5 \)[/tex]:
[tex]\[ f(-5) = 2(-5) - 2 = -10 - 2 = -12 \][/tex]
So, [tex]\( f(-5) = -12 \)[/tex].

2. For [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = 2(-4) - 2 = -8 - 2 = -10 \][/tex]
So, [tex]\( f(-4) = -10 \)[/tex].

3. For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2(1) - 2 = 2 - 2 = 0 \][/tex]
So, [tex]\( f(1) = 0 \)[/tex].

4. For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = 2(3) - 2 = 6 - 2 = 4 \][/tex]
So, [tex]\( f(3) = 4 \)[/tex].

5. For [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = 2(5) - 2 = 10 - 2 = 8 \][/tex]
So, [tex]\( f(5) = 8 \)[/tex].

Now, let's complete the function table with these values:

[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -5 & -12 \\ \hline -4 & -10 \\ \hline 1 & 0 \\ \hline 3 & 4 \\ \hline 5 & 8 \\ \hline \end{tabular} \][/tex]