Answer :
To complete the table for the function [tex]\( f(x) = x^5 + (x+3)^2 \)[/tex] at [tex]\( x = -1 \)[/tex], we need to calculate [tex]\( f(-1) \)[/tex].
Let's go through the calculation step by step:
1. Plug in [tex]\( x = -1 \)[/tex] into the function:
[tex]\[ f(-1) = (-1)^5 + (-1 + 3)^2 \][/tex]
2. Calculate each term:
[tex]\[ (-1)^5 = -1 \][/tex]
[tex]\[ (-1 + 3)^2 = 2^2 = 4 \][/tex]
3. Add the results:
[tex]\[ f(-1) = -1 + 4 = 3 \][/tex]
So, [tex]\( f(-1) = 3 \)[/tex].
Therefore, the value that completes the table is [tex]\( 3 \)[/tex].
The completed table looks like this:
\begin{tabular}{|c|c|}
\hline [tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline -2 & -31 \\
\hline -1 & 3 \\
\hline 0 & 9 \\
\hline 1 & 17 \\
\hline
\end{tabular}
Among the given options, the correct value is [tex]\( 3 \)[/tex].
Let's go through the calculation step by step:
1. Plug in [tex]\( x = -1 \)[/tex] into the function:
[tex]\[ f(-1) = (-1)^5 + (-1 + 3)^2 \][/tex]
2. Calculate each term:
[tex]\[ (-1)^5 = -1 \][/tex]
[tex]\[ (-1 + 3)^2 = 2^2 = 4 \][/tex]
3. Add the results:
[tex]\[ f(-1) = -1 + 4 = 3 \][/tex]
So, [tex]\( f(-1) = 3 \)[/tex].
Therefore, the value that completes the table is [tex]\( 3 \)[/tex].
The completed table looks like this:
\begin{tabular}{|c|c|}
\hline [tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline -2 & -31 \\
\hline -1 & 3 \\
\hline 0 & 9 \\
\hline 1 & 17 \\
\hline
\end{tabular}
Among the given options, the correct value is [tex]\( 3 \)[/tex].
