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The function [tex]\( f(x) = x^5 + (x+3)^2 \)[/tex] is used to create this table.

\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline
-2 & -31 \\
\hline
-1 & ? \\
\hline
0 & 9 \\
\hline
1 & 17 \\
\hline
\end{tabular}

Which value completes the table?

A. [tex]$-17$[/tex]

B. [tex]$-3$[/tex]

C. 1

D. 7


Sagot :

To determine the value of [tex]\( f(x) \)[/tex] when [tex]\( x = -1 \)[/tex] for the function [tex]\( f(x) = x^5 + (x + 3)^2 \)[/tex]:

First, substitute [tex]\( x = -1 \)[/tex] into the function [tex]\( f(x) \)[/tex]:

[tex]\[ f(-1) = (-1)^5 + (-1 + 3)^2 \][/tex]

Let's evaluate each term inside the function step-by-step:

1. Evaluate [tex]\( (-1)^5 \)[/tex]:
[tex]\[ (-1)^5 = -1 \][/tex]

2. Evaluate [tex]\( (-1 + 3) \)[/tex]:
[tex]\[ -1 + 3 = 2 \][/tex]

3. Then, square the result of the second evaluation:
[tex]\[ 2^2 = 4 \][/tex]

Now, combine these results to find [tex]\( f(-1) \)[/tex]:
[tex]\[ f(-1) = -1 + 4 = 3 \][/tex]

So, the value of [tex]\( f(-1) \)[/tex] is [tex]\( 3 \)[/tex].

Therefore, the completed table will be:
\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}

Hence, the value that completes the table is [tex]\( 3 \)[/tex].
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