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Use the tables below to find [tex]\((p+q)(2)\)[/tex].

[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $p(x)$ \\
\hline
4 & -1 \\
\hline
2 & 3 \\
\hline
-3 & 2 \\
\hline
\end{tabular}
\][/tex]
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $q(x)$ \\
\hline
4 & 1 \\
\hline
2 & -2 \\
\hline
-3 & 5 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
(p+q)(2) = \square
\][/tex]


Sagot :

To find [tex]\((p+q)(2)\)[/tex], we need to evaluate the functions [tex]\( p \)[/tex] and [tex]\( q \)[/tex] at [tex]\( x = 2 \)[/tex] and then add the results together. Here’s a step-by-step solution:

1. Determine [tex]\( p(2) \)[/tex] from the given table for [tex]\( p(x) \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & p(x) \\ \hline 4 & -1 \\ \hline 2 & 3 \\ \hline -3 & 2 \\ \hline \end{array} \][/tex]
From the table, we can see that [tex]\( p(2) = 3 \)[/tex].

2. Determine [tex]\( q(2) \)[/tex] from the given table for [tex]\( q(x) \)[/tex]:
[tex]\[ \begin{array}{|c|c|} \hline x & q(x) \\ \hline 4 & 1 \\ \hline 2 & -2 \\ \hline -3 & 5 \\ \hline \end{array} \][/tex]
From the table, we can see that [tex]\( q(2) = -2 \)[/tex].

3. Add the values obtained:
[tex]\[ (p+q)(2) = p(2) + q(2) \][/tex]
Substituting the values we found:
[tex]\[ (p+q)(2) = 3 + (-2) = 1 \][/tex]

Therefore, [tex]\((p+q)(2) = 1\)[/tex].
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