The function [tex]\( f \)[/tex] is defined by the following rule:
[tex]\[ f(x) = 3x + 5 \][/tex]

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



Answer :

Sure, let's complete the function table step-by-step.

The given function is [tex]\( f(x) = 3x + 5 \)[/tex].

We will plug each value of [tex]\( x \)[/tex] into the function to find [tex]\( f(x) \)[/tex].

1. For [tex]\( x = -5 \)[/tex]:
[tex]\[ f(-5) = 3(-5) + 5 \][/tex]
[tex]\[ f(-5) = -15 + 5 \][/tex]
[tex]\[ f(-5) = -10 \][/tex]

2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 3(-1) + 5 \][/tex]
[tex]\[ f(-1) = -3 + 5 \][/tex]
[tex]\[ f(-1) = 2 \][/tex]

3. For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 3(2) + 5 \][/tex]
[tex]\[ f(2) = 6 + 5 \][/tex]
[tex]\[ f(2) = 11 \][/tex]

4. For [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 3(4) + 5 \][/tex]
[tex]\[ f(4) = 12 + 5 \][/tex]
[tex]\[ f(4) = 17 \][/tex]

5. For [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = 3(5) + 5 \][/tex]
[tex]\[ f(5) = 15 + 5 \][/tex]
[tex]\[ f(5) = 20 \][/tex]

Now, we can fill in the table with these values:

[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -5 & -10 \\ \hline -1 & 2 \\ \hline 2 & 11 \\ \hline 4 & 17 \\ \hline 5 & 20 \\ \hline \end{tabular} \][/tex]

Hence, the complete function table is:

[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -5 & -10 \\ \hline -1 & 2 \\ \hline 2 & 11 \\ \hline 4 & 17 \\ \hline 5 & 20 \\ \hline \end{tabular} \][/tex]