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Sagot :
To determine the intervals where the function [tex]\( f(x) \)[/tex] is increasing using the given table values, follow these steps:
1. Examine the values in the table:
\begin{tabular}{|c|c|}
\hline
[tex]\( x \)[/tex] & [tex]\( f(x) \)[/tex] \\
\hline
-3 & 18 \\
\hline
-2 & 3 \\
\hline
-1 & 0 \\
\hline
0 & 3 \\
\hline
1 & 6 \\
\hline
2 & 3 \\
\hline
\end{tabular}
2. Calculate the differences [tex]\( \Delta f(x) \)[/tex] between consecutive values of [tex]\( f(x) \)[/tex]:
- From [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]:
[tex]\( \Delta f(x) = f(-2) - f(-3) = 3 - 18 = -15 \)[/tex]
- From [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex]:
[tex]\( \Delta f(x) = f(-1) - f(-2) = 0 - 3 = -3 \)[/tex]
- From [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex]:
[tex]\( \Delta f(x) = f(0) - f(-1) = 3 - 0 = 3 \)[/tex]
- From [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]:
[tex]\( \Delta f(x) = f(1) - f(0) = 6 - 3 = 3 \)[/tex]
- From [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]:
[tex]\( \Delta f(x) = f(2) - f(1) = 3 - 6 = -3 \)[/tex]
3. Identify the intervals where the function is increasing:
The function is increasing where the differences [tex]\( \Delta f(x) \)[/tex] are positive.
- From [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex] : [tex]\( \Delta f(x) = -15 \)[/tex] (decreasing)
- From [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex] : [tex]\( \Delta f(x) = -3 \)[/tex] (decreasing)
- From [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex] : [tex]\( \Delta f(x) = 3 \)[/tex] (increasing)
- From [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex] : [tex]\( \Delta f(x) = 3 \)[/tex] (increasing)
- From [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex] : [tex]\( \Delta f(x) = -3 \)[/tex] (decreasing)
4. Combine the increasing intervals to find the largest interval:
According to the values:
- The function is increasing from [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex]
- The function is increasing from [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]
Combining these, the largest continuous interval where the function is increasing is from [tex]\( x = -1 \)[/tex] to [tex]\( x = 1 \)[/tex].
Therefore, the largest interval of [tex]\( x \)[/tex] values where the function is increasing is [tex]\((-1, 1)\)[/tex].
1. Examine the values in the table:
\begin{tabular}{|c|c|}
\hline
[tex]\( x \)[/tex] & [tex]\( f(x) \)[/tex] \\
\hline
-3 & 18 \\
\hline
-2 & 3 \\
\hline
-1 & 0 \\
\hline
0 & 3 \\
\hline
1 & 6 \\
\hline
2 & 3 \\
\hline
\end{tabular}
2. Calculate the differences [tex]\( \Delta f(x) \)[/tex] between consecutive values of [tex]\( f(x) \)[/tex]:
- From [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex]:
[tex]\( \Delta f(x) = f(-2) - f(-3) = 3 - 18 = -15 \)[/tex]
- From [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex]:
[tex]\( \Delta f(x) = f(-1) - f(-2) = 0 - 3 = -3 \)[/tex]
- From [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex]:
[tex]\( \Delta f(x) = f(0) - f(-1) = 3 - 0 = 3 \)[/tex]
- From [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]:
[tex]\( \Delta f(x) = f(1) - f(0) = 6 - 3 = 3 \)[/tex]
- From [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex]:
[tex]\( \Delta f(x) = f(2) - f(1) = 3 - 6 = -3 \)[/tex]
3. Identify the intervals where the function is increasing:
The function is increasing where the differences [tex]\( \Delta f(x) \)[/tex] are positive.
- From [tex]\( x = -3 \)[/tex] to [tex]\( x = -2 \)[/tex] : [tex]\( \Delta f(x) = -15 \)[/tex] (decreasing)
- From [tex]\( x = -2 \)[/tex] to [tex]\( x = -1 \)[/tex] : [tex]\( \Delta f(x) = -3 \)[/tex] (decreasing)
- From [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex] : [tex]\( \Delta f(x) = 3 \)[/tex] (increasing)
- From [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex] : [tex]\( \Delta f(x) = 3 \)[/tex] (increasing)
- From [tex]\( x = 1 \)[/tex] to [tex]\( x = 2 \)[/tex] : [tex]\( \Delta f(x) = -3 \)[/tex] (decreasing)
4. Combine the increasing intervals to find the largest interval:
According to the values:
- The function is increasing from [tex]\( x = -1 \)[/tex] to [tex]\( x = 0 \)[/tex]
- The function is increasing from [tex]\( x = 0 \)[/tex] to [tex]\( x = 1 \)[/tex]
Combining these, the largest continuous interval where the function is increasing is from [tex]\( x = -1 \)[/tex] to [tex]\( x = 1 \)[/tex].
Therefore, the largest interval of [tex]\( x \)[/tex] values where the function is increasing is [tex]\((-1, 1)\)[/tex].
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