Question

what ordered pair is closest to a local minimum of the function, f(x)?

x	f(x)
-1	-3
0	-2
1	4
2	1
3	3

options:

• (-1, -3)
• (0, -2)
• (1, 4)
• (2, 1)

Explanation:

Step1: Understand local minimum

A local minimum is a point where the function changes from decreasing to increasing (the function value is lower than its neighboring points around that region).

Step2: Analyze the function values

- For \( x = -2 \), \( f(x) = -8 \); \( x = -1 \), \( f(x) = -3 \) (function is increasing here: from -8 to -3).
- For \( x = -1 \), \( f(x) = -3 \); \( x = 0 \), \( f(x) = -2 \) (still increasing: from -3 to -2).
- For \( x = 0 \), \( f(x) = -2 \); \( x = 1 \), \( f(x) = 4 \) (increasing: from -2 to 4). Wait, no—wait, let's check the trend before and after each point. Wait, maybe I made a mistake. Wait, let's list the values:

\( x=-2 \): -8; \( x=-1 \): -3 (so from x=-2 to x=-1, function increases (since -3 > -8)).

\( x=-1 \): -3; \( x=0 \): -2 (increases: -2 > -3).

\( x=0 \): -2; \( x=1 \): 4 (increases: 4 > -2).

\( x=1 \): 4; \( x=2 \): 1 (decreases: 1 < 4).

\( x=2 \): 1; \( x=3 \): 3 (increases: 3 > 1).

Wait, so the function increases from x=-2 to x=1 (since f(x) goes from -8 to 4), then decreases from x=1 to x=2 (4 to 1), then increases from x=2 to x=3 (1 to 3). Wait, but a local minimum would be a point where the function changes from decreasing to increasing. Wait, but before x=0, the function is increasing (from x=-2 to x=0: -8, -3, -2—each time increasing). Then from x=0 to x=1, it's still increasing (to 4). Then from x=1 to x=2, it decreases (to 1), then from x=2 to x=3, it increases (to 3). So the point where the function changes from decreasing to increasing is between x=1 and x=3? Wait, no—wait, the local minimum in the given options. Wait, the options are (-1,-3), (0,-2), (1,4), (2,1). Wait, maybe I misread the trend. Wait, let's check the values again:

At x=-2: -8; x=-1: -3 (so f(-1) > f(-2) → increasing from x=-2 to x=-1).

x=-1: -3; x=0: -2 (f(0) > f(-1) → increasing from x=-1 to x=0).

x=0: -2; x=1: 4 (f(1) > f(0) → increasing from x=0 to x=1).

x=1: 4; x=2: 1 (f(2) < f(1) → decreasing from x=1 to x=2).

x=2: 1; x=3: 3 (f(3) > f(2) → increasing from x=2 to x=3).

So the function increases from x=-2 to x=1, then decreases from x=1 to x=2, then increases from x=2 to x=3. So the local minimum would be at the point where the function changes from decreasing to increasing, which is around x=2 (since after x=2, it increases). But wait, the value at x=2 is 1, and at x=3 is 3. But the options include (2,1). Wait, but let's check the other options. Wait, maybe I made a mistake in the trend. Wait, maybe the local minimum is before? Wait, no—wait, the function values: from x=-2 to x=1, the function is increasing (each f(x) is higher than the previous). Then from x=1 to x=2, it decreases (f(2)=1 < f(1)=4), then from x=2 to x=3, it increases (f(3)=3 > f(2)=1). So the function has a local minimum at x=2, because before x=2 (from x=1 to x=2) it was decreasing, and after x=2 (from x=2 to x=3) it's increasing. So the ordered pair (2,1) is the local minimum? Wait, but let's check the options. Wait, the options are (-1,-3), (0,-2), (1,4), (2,1). Wait, but when x increases from -2 to 1, f(x) is increasing (so those points are not local minima, since the function is still increasing). Then at x=1, it's a local maximum (since it's higher than neighbors: f(0)=-2, f(1)=4, f(2)=1 → 4 is higher than -2 and 1). Then from x=1 to x=2, it decreases, then from x=2 to x=3, it increases. So the local minimum is at x=2, because the function changes from decreasing to increasing there. So the ordered pair (2,1) is the local minimum. Wait, but let's check the values again. Wait, maybe I messed up the direction. Wait, a local minimum is a point where the function is lower than its immediate neighbors. Let's check each option:

- (-1, -3): Neighbors are x=-2 (f=-8) and x=0 (f=-2). So f(-1)=-3 is higher than f(-2)=-8 (so not a minimum) and lower than f(0)=-2? Wait, no: f(-1)=-3, f(0)=-2. So -3 < -2, so from x=-1 to x=0, the function increases. From x=-2 to x=-1, the function increases (f(-1)=-3 > f(-2)=-8). So at x=-1, the function is increasing on both sides? No, from x=-2 to x=-1: increasing (since -3 > -8), from x=-1 to x=0: increasing (since -2 > -3). So x=-1 is not a minimum.

- (0, -2): Neighbors x=-1 (f=-3) and x=1 (f=4). f(0)=-2 is higher than f(-1)=-3 (so from x=-1 to x=0, function increases) and lower than f(1)=4 (from x=0 to x=1, function increases). So x=0 is not a minimum.

- (1, 4): Neighbors x=0 (f=-2) and x=2 (f=1). f(1)=4 is higher than both, so it's a local maximum.

- (2, 1): Neighbors x=1 (f=4) and x=3 (f=3). f(2)=1 is lower than f(1)=4 (so from x=1 to x=2, function decreases) and lower than f(3)=3? Wait, no: f(3)=3 > f(2)=1, so from x=2 to x=3, function increases. So at x=2, the function changes from decreasing (x=1 to x=2) to increasing (x=2 to x=3), so it's a local minimum. And among the options, (2,1) is the local minimum.

Answer:

(2, 1)