Answer :
To find the local minimum of the function [tex]\( f(x) \)[/tex] using the given table of values, we need to identify points where the function value is less than that of its neighboring points. A local minimum is a point in the function where the function value at that point is less than the function values at neighboring points.
Here is the table again for reference:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -4 & 105 \\ \hline -3 & 0 \\ \hline -2 & -15 \\ \hline -1 & 0 \\ \hline 0 & 9 \\ \hline 1 & 0 \\ \hline 2 & -15 \\ \hline 3 & 0 \\ \hline 4 & 105 \\ \hline 5 & 384 \\ \hline \end{tabular} \][/tex]
Consider each point and its immediate neighbors:
1. For [tex]\( x = -4 \)[/tex], we don't have neighbors on both sides, so we can't determine if it's a local minimum.
2. For [tex]\( x = -3 \)[/tex], the neighbors are [tex]\( f(-4) = 105 \)[/tex] and [tex]\( f(-2) = -15 \)[/tex]. Here, [tex]\( f(-2) = -15 < f(-3) = 0 \)[/tex], so [tex]\( x = -3 \)[/tex] is not a local minimum.
3. For [tex]\( x = -2 \)[/tex], the neighbors are [tex]\( f(-3) = 0 \)[/tex] and [tex]\( f(-1) = 0 \)[/tex]. Here, [tex]\( f(-2) = -15 \)[/tex] is less than both [tex]\( f(-3) \)[/tex] and [tex]\( f(-1) \)[/tex]. This indicates that [tex]\( x = -2 \)[/tex] is a local minimum.
4. For [tex]\( x = -1 \)[/tex], the neighbors are [tex]\( f(-2) = -15 \)[/tex] and [tex]\( f(0) = 9 \)[/tex]. Here, [tex]\( f(-2) = -15 < f(-1) = 0 \)[/tex], so [tex]\( x = -1 \)[/tex] is not a local minimum.
5. For [tex]\( x = 0 \)[/tex], the neighbors are [tex]\( f(-1) = 0 \)[/tex] and [tex]\( f(1) = 0 \)[/tex]. Here, [tex]\( f(0) = 9 \)[/tex] is greater than both [tex]\( f(-1) \)[/tex] and [tex]\( f(1) \)[/tex], so [tex]\( x = 0 \)[/tex] is not a local minimum.
6. For [tex]\( x = 1 \)[/tex], the neighbors are [tex]\( f(0) = 9 \)[/tex] and [tex]\( f(2) = -15 \)[/tex]. Here, [tex]\( f(2) = -15 < f(1) = 0 \)[/tex], so [tex]\( x = 1 \)[/tex] is not a local minimum.
7. For [tex]\( x = 2 \)[/tex], the neighbors are [tex]\( f(1) = 0 \)[/tex] and [tex]\( f(3) = 0 \)[/tex]. Here, [tex]\( f(2) = -15 \)[/tex] is less than both [tex]\( f(1) \)[/tex] and [tex]\( f(3) \)[/tex]. This indicates that [tex]\( x = 2 \)[/tex] is a local minimum.
8. For [tex]\( x = 3 \)[/tex], the neighbors are [tex]\( f(2) = -15 \)[/tex] and [tex]\( f(4) = 105 \)[/tex]. Here, [tex]\( f(2) = -15 < f(3) = 0 \)[/tex], so [tex]\( x = 3 \)[/tex] is not a local minimum.
9. For [tex]\( x = 4 \)[/tex], the neighbors are [tex]\( f(3) = 0 \)[/tex] and [tex]\( f(5) = 384 \)[/tex]. Here, [tex]\( f(3) = 0 < f(4) = 105 \)[/tex], so [tex]\( x = 4 \)[/tex] is not a local minimum.
10. For [tex]\( x = 5 \)[/tex], we don't have neighbors on both sides, so we can't determine if it's a local minimum.
Thus, we identify [tex]\( x = -2 \)[/tex] and [tex]\( x = 2 \)[/tex] as local minima. Given the choices:
- [tex]\((0, 9)\)[/tex]
- [tex]\((4, 105)\)[/tex]
- [tex]\((-1, 0)\)[/tex]
- [tex]\((2, -15)\)[/tex]
We notice that [tex]\((2, -15)\)[/tex] is among the options and is indeed one of the local minima identified.
Therefore, the ordered pair that is a local minimum of the function [tex]\( f(x) \)[/tex] is [tex]\((2, -15)\)[/tex].
Here is the table again for reference:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $f(x)$ \\ \hline -4 & 105 \\ \hline -3 & 0 \\ \hline -2 & -15 \\ \hline -1 & 0 \\ \hline 0 & 9 \\ \hline 1 & 0 \\ \hline 2 & -15 \\ \hline 3 & 0 \\ \hline 4 & 105 \\ \hline 5 & 384 \\ \hline \end{tabular} \][/tex]
Consider each point and its immediate neighbors:
1. For [tex]\( x = -4 \)[/tex], we don't have neighbors on both sides, so we can't determine if it's a local minimum.
2. For [tex]\( x = -3 \)[/tex], the neighbors are [tex]\( f(-4) = 105 \)[/tex] and [tex]\( f(-2) = -15 \)[/tex]. Here, [tex]\( f(-2) = -15 < f(-3) = 0 \)[/tex], so [tex]\( x = -3 \)[/tex] is not a local minimum.
3. For [tex]\( x = -2 \)[/tex], the neighbors are [tex]\( f(-3) = 0 \)[/tex] and [tex]\( f(-1) = 0 \)[/tex]. Here, [tex]\( f(-2) = -15 \)[/tex] is less than both [tex]\( f(-3) \)[/tex] and [tex]\( f(-1) \)[/tex]. This indicates that [tex]\( x = -2 \)[/tex] is a local minimum.
4. For [tex]\( x = -1 \)[/tex], the neighbors are [tex]\( f(-2) = -15 \)[/tex] and [tex]\( f(0) = 9 \)[/tex]. Here, [tex]\( f(-2) = -15 < f(-1) = 0 \)[/tex], so [tex]\( x = -1 \)[/tex] is not a local minimum.
5. For [tex]\( x = 0 \)[/tex], the neighbors are [tex]\( f(-1) = 0 \)[/tex] and [tex]\( f(1) = 0 \)[/tex]. Here, [tex]\( f(0) = 9 \)[/tex] is greater than both [tex]\( f(-1) \)[/tex] and [tex]\( f(1) \)[/tex], so [tex]\( x = 0 \)[/tex] is not a local minimum.
6. For [tex]\( x = 1 \)[/tex], the neighbors are [tex]\( f(0) = 9 \)[/tex] and [tex]\( f(2) = -15 \)[/tex]. Here, [tex]\( f(2) = -15 < f(1) = 0 \)[/tex], so [tex]\( x = 1 \)[/tex] is not a local minimum.
7. For [tex]\( x = 2 \)[/tex], the neighbors are [tex]\( f(1) = 0 \)[/tex] and [tex]\( f(3) = 0 \)[/tex]. Here, [tex]\( f(2) = -15 \)[/tex] is less than both [tex]\( f(1) \)[/tex] and [tex]\( f(3) \)[/tex]. This indicates that [tex]\( x = 2 \)[/tex] is a local minimum.
8. For [tex]\( x = 3 \)[/tex], the neighbors are [tex]\( f(2) = -15 \)[/tex] and [tex]\( f(4) = 105 \)[/tex]. Here, [tex]\( f(2) = -15 < f(3) = 0 \)[/tex], so [tex]\( x = 3 \)[/tex] is not a local minimum.
9. For [tex]\( x = 4 \)[/tex], the neighbors are [tex]\( f(3) = 0 \)[/tex] and [tex]\( f(5) = 384 \)[/tex]. Here, [tex]\( f(3) = 0 < f(4) = 105 \)[/tex], so [tex]\( x = 4 \)[/tex] is not a local minimum.
10. For [tex]\( x = 5 \)[/tex], we don't have neighbors on both sides, so we can't determine if it's a local minimum.
Thus, we identify [tex]\( x = -2 \)[/tex] and [tex]\( x = 2 \)[/tex] as local minima. Given the choices:
- [tex]\((0, 9)\)[/tex]
- [tex]\((4, 105)\)[/tex]
- [tex]\((-1, 0)\)[/tex]
- [tex]\((2, -15)\)[/tex]
We notice that [tex]\((2, -15)\)[/tex] is among the options and is indeed one of the local minima identified.
Therefore, the ordered pair that is a local minimum of the function [tex]\( f(x) \)[/tex] is [tex]\((2, -15)\)[/tex].
