Answered

At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

Which table shows a function that is decreasing over the interval [tex](-2,0)[/tex]?

[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $f(x)$ \\
\hline
-2 & 0 \\
\hline
-1 & -5 \\
\hline
0 & 0 \\
\hline
1 & 5 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $f(x)$ \\
\hline
-2 & -15 \\
\hline
0 & -5 \\
\hline
2 & -20 \\
\hline
4 & -30 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $f(x)$ \\
\hline
-3 & 2 \\
\hline
-2 & 0 \\
\hline
-1 & -10 \\
\hline
0 & -24 \\
\hline
\end{tabular}
\][/tex]

Sagot :

GinnyAnswer
To determine which table shows a function that is decreasing over the interval [tex]\( (-2,0) \)[/tex], we need to analyze the values of [tex]\( f(x) \)[/tex] at [tex]\( x = -2 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = 0 \)[/tex] for each table.

### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -2 & 0 \\ \hline -1 & -5 \\ \hline 0 & 0 \\ \hline 1 & 5 \\ \hline \end{array} \][/tex]

For [tex]\( x_1 \)[/tex] in this interval:
- At [tex]\( x = -2 \)[/tex], [tex]\( f(x) = 0 \)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( f(x) = -5 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( f(x) = 0 \)[/tex]

In this table, from [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex], [tex]\( f(x) \)[/tex] decreases from 0 to -5. However, from [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex], [tex]\( f(x) \)[/tex] increases from -5 to 0. So, this function is not consistently decreasing over the entire interval [tex]\( (-2, 0) \)[/tex].

### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -2 & -15 \\ \hline 0 & -5 \\ \hline 2 & -20 \\ \hline 4 & -30 \\ \hline \end{array} \][/tex]

For [tex]\( x_2 \)[/tex] in this interval:
- At [tex]\( x = -2 \)[/tex], [tex]\( f(x) = -15 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( f(x) = -5 \)[/tex]

Here, [tex]\( f(x) \)[/tex] increases from [tex]\( -15 \)[/tex] to [tex]\( -5 \)[/tex] over the interval [tex]\( -2 \)[/tex] to [tex]\( 0 \)[/tex]. Therefore, this function is not decreasing over the interval [tex]\( (-2, 0) \)[/tex].

### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -3 & 2 \\ \hline -2 & 0 \\ \hline -1 & -10 \\ \hline 0 & -24 \\ \hline \end{array} \][/tex]

For [tex]\( x_3 \)[/tex] in this interval:
- At [tex]\( x = -2 \)[/tex], [tex]\( f(x) = 0 \)[/tex]
- At [tex]\( x = -1 \)[/tex], [tex]\( f(x) = -10 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( f(x) = -24 \)[/tex]

In this table, from [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex], [tex]\( f(x) \)[/tex] decreases from 0 to -10. From [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex], it further decreases from -10 to -24. Hence, the function is consistently decreasing over the entire interval [tex]\( (-2, 0) \)[/tex].

Therefore, the table that shows a function that is decreasing over the interval [tex]\( (-2,0) \)[/tex] is Table 3.
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.