\begin{tabular}{|c|c|}
\hline
[tex]$x$[/tex] & [tex]$f(x)$[/tex] \\
\hline
-3 & -16 \\
\hline
-2 & -1 \\
\hline
-1 & 2 \\
\hline
0 & -1 \\
\hline
1 & -4 \\
\hline
2 & -1 \\
\hline
\end{tabular}

Analyze the table of values for the continuous function [tex]$f(x)$[/tex] to complete the statements:

A local maximum occurs over the interval [tex]$\square$[/tex]

A local minimum occurs over the interval [tex]$\square$[/tex]



Answer :

To analyze the table of values, we need to inspect the function [tex]\( f(x) \)[/tex] at each given point and determine where the local maximum and minimum occur.

Given table of values:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -3 & -16 \\ \hline -2 & -1 \\ \hline -1 & 2 \\ \hline 0 & -1 \\ \hline 1 & -4 \\ \hline 2 & -1 \\ \hline \end{array} \][/tex]

1. Local Maximum:
- We need to find where [tex]\( f(x) \)[/tex] changes from increasing to decreasing.
- At [tex]\( x = -2 \)[/tex]: [tex]\( f(x) \)[/tex] changes from [tex]\( f(-3) = -16 \)[/tex] (increasing) to [tex]\( f(-2) = -1 \)[/tex] (increasing).
- At [tex]\( x = -1 \)[/tex]: [tex]\( f(x) \)[/tex] changes from [tex]\( f(-2) = -1 \)[/tex] (increasing) to [tex]\( f(-1) = 2 \)[/tex] (increasing).
- At [tex]\( x = 0 \)[/tex]: [tex]\( f(x) \)[/tex] changes from [tex]\( f(-1) = 2 \)[/tex] (decreasing) to [tex]\( f(0) = -1 \)[/tex] (decreasing).
- Therefore, [tex]\( x = -1 \)[/tex] is a local maximum because the function increases to this point and then starts to decrease.
- The interval around this maximum can be identified as from [tex]\( x = -2 \)[/tex] to [tex]\( x = 0 \)[/tex].

2. Local Minimum:
- We need to find where [tex]\( f(x) \)[/tex] changes from decreasing to increasing.
- At [tex]\( x = 1 \)[/tex]: [tex]\( f(x) \)[/tex] changes from [tex]\( f(0) = -1 \)[/tex] (decreasing) to [tex]\( f(1) = -4 \)[/tex] (decreasing).
- At [tex]\( x = 2 \)[/tex]: [tex]\( f(x) \)[/tex] changes from [tex]\( f(1) = -4 \)[/tex] (increasing) to [tex]\( f(2) = -1 \)[/tex] (increasing).
- Therefore, [tex]\( x = 0 \)[/tex] is a local minimum because the function decreases to this point and then starts to increase.
- The interval around this minimum can be identified as from [tex]\( x = 0 \)[/tex] to [tex]\( x = 2 \)[/tex].

Thus, the statements can be completed as follows:

A local maximum occurs over the interval [tex]\((-2, 0)\)[/tex]

A local minimum occurs over the interval [tex]\((0, 2)\)[/tex]