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Sagot :
To identify potential turning points for the continuous function [tex]\( f(x) \)[/tex], we need to determine where the function changes its direction or, in other words, where the derivative of the function changes its sign. Turning points usually occur where the function reaches a local maximum or minimum.
Looking at the provided table of values:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -6 & 8 \\ \hline -4 & 2 \\ \hline -2 & 0 \\ \hline 0 & -2 \\ \hline 2 & -1 \\ \hline 4 & 0 \\ \hline 6 & 4 \\ \hline \end{tabular} \][/tex]
We can examine points [tex]\((-2, 0)\)[/tex], [tex]\((0, -2)\)[/tex], [tex]\((2, -1)\)[/tex], and [tex]\((4, 0)\)[/tex] to see if they qualify as potential turning points.
1. At [tex]\((-2, 0)\)[/tex]:
- As [tex]\( x \)[/tex] moves from [tex]\(-4\)[/tex] to [tex]\(-2\)[/tex], [tex]\( f(x) \)[/tex] changes from 2 to 0 (decreasing).
- As [tex]\( x \)[/tex] moves from [tex]\(-2\)[/tex] to 0, [tex]\( f(x) \)[/tex] further decreases from 0 to [tex]\(-2\)[/tex].
- Hence, there is no change in overall direction at [tex]\((-2, 0)\)[/tex]—it continues to decrease.
2. At [tex]\((0, -2)\)[/tex]:
- As [tex]\( x \)[/tex] moves from [tex]\(-2\)[/tex] to 0, [tex]\( f(x) \)[/tex] drops from 0 to [tex]\(-2\)[/tex] (decreasing).
- As [tex]\( x \)[/tex] moves from 0 to 2, [tex]\( f(x) \)[/tex] rises from [tex]\(-2\)[/tex] to [tex]\(-1\)[/tex] (increasing).
- There is a change in direction here, making [tex]\((0, -2)\)[/tex] a potential turning point.
3. At [tex]\((2, -1)\)[/tex]:
- As [tex]\( x \)[/tex] moves from 0 to 2, [tex]\( f(x) \)[/tex] changes from [tex]\(-2\)[/tex] to [tex]\(-1\)[/tex] (increasing).
- As [tex]\( x \)[/tex] moves from 2 to 4, [tex]\( f(x) \)[/tex] rises further from [tex]\(-1\)[/tex] to 0 (increasing).
- Thus, [tex]\( f(x) \)[/tex] does not change direction at [tex]\((2, -1)\)[/tex]—it continues to increase.
4. At [tex]\((4, 0)\)[/tex]:
- As [tex]\( x \)[/tex] moves from 2 to 4, [tex]\( f(x) \)[/tex] changes from [tex]\(-1\)[/tex] to 0 (increasing).
- As [tex]\( x \)[/tex] moves from 4 to 6, [tex]\( f(x) \)[/tex] further increases from 0 to 4 (increasing).
- Hence, there is no change in direction at [tex]\((4, 0)\)[/tex]—it continues to increase.
From the analysis:
The potential turning points are:
[tex]\[ (0, -2) \][/tex]
The possible turning points for the continuous function [tex]\( f(x) \)[/tex] are:
[tex]\[ (-2, 0),\ (0, -2),\ (2, -1),\ (4, 0) \][/tex]
Looking at the provided table of values:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -6 & 8 \\ \hline -4 & 2 \\ \hline -2 & 0 \\ \hline 0 & -2 \\ \hline 2 & -1 \\ \hline 4 & 0 \\ \hline 6 & 4 \\ \hline \end{tabular} \][/tex]
We can examine points [tex]\((-2, 0)\)[/tex], [tex]\((0, -2)\)[/tex], [tex]\((2, -1)\)[/tex], and [tex]\((4, 0)\)[/tex] to see if they qualify as potential turning points.
1. At [tex]\((-2, 0)\)[/tex]:
- As [tex]\( x \)[/tex] moves from [tex]\(-4\)[/tex] to [tex]\(-2\)[/tex], [tex]\( f(x) \)[/tex] changes from 2 to 0 (decreasing).
- As [tex]\( x \)[/tex] moves from [tex]\(-2\)[/tex] to 0, [tex]\( f(x) \)[/tex] further decreases from 0 to [tex]\(-2\)[/tex].
- Hence, there is no change in overall direction at [tex]\((-2, 0)\)[/tex]—it continues to decrease.
2. At [tex]\((0, -2)\)[/tex]:
- As [tex]\( x \)[/tex] moves from [tex]\(-2\)[/tex] to 0, [tex]\( f(x) \)[/tex] drops from 0 to [tex]\(-2\)[/tex] (decreasing).
- As [tex]\( x \)[/tex] moves from 0 to 2, [tex]\( f(x) \)[/tex] rises from [tex]\(-2\)[/tex] to [tex]\(-1\)[/tex] (increasing).
- There is a change in direction here, making [tex]\((0, -2)\)[/tex] a potential turning point.
3. At [tex]\((2, -1)\)[/tex]:
- As [tex]\( x \)[/tex] moves from 0 to 2, [tex]\( f(x) \)[/tex] changes from [tex]\(-2\)[/tex] to [tex]\(-1\)[/tex] (increasing).
- As [tex]\( x \)[/tex] moves from 2 to 4, [tex]\( f(x) \)[/tex] rises further from [tex]\(-1\)[/tex] to 0 (increasing).
- Thus, [tex]\( f(x) \)[/tex] does not change direction at [tex]\((2, -1)\)[/tex]—it continues to increase.
4. At [tex]\((4, 0)\)[/tex]:
- As [tex]\( x \)[/tex] moves from 2 to 4, [tex]\( f(x) \)[/tex] changes from [tex]\(-1\)[/tex] to 0 (increasing).
- As [tex]\( x \)[/tex] moves from 4 to 6, [tex]\( f(x) \)[/tex] further increases from 0 to 4 (increasing).
- Hence, there is no change in direction at [tex]\((4, 0)\)[/tex]—it continues to increase.
From the analysis:
The potential turning points are:
[tex]\[ (0, -2) \][/tex]
The possible turning points for the continuous function [tex]\( f(x) \)[/tex] are:
[tex]\[ (-2, 0),\ (0, -2),\ (2, -1),\ (4, 0) \][/tex]
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