Question

analyze the graphed function to find the local minimum and the local maximum for the given function.

which statements about the local maximums and minimums for the given function are true? choose three options.

over the interval 1, 3, the local minimum is 0
over the interval 2, 4, the local minimum is −8.
over the interval 3, 5, the local minimum is −8.
over the interval 1, 4, the local maximum is 0.
over the interval 3, 5, the local maximum is 0.

Answer:

Part 1: Find Local Minimum and Maximum of the Function

Step 1: Recall Definitions

A local minimum on an interval is the lowest point (smallest \( y \)-value) in that interval. A local maximum is the highest point (largest \( y \)-value) in that interval.

Step 2: Analyze the Graph

• The graph has roots at \( (0,0) \), \( (2,0) \), \( (4,0) \).
• The local minima (lowest points) seem to be at \( y = -8 \) (let's assume the bottom points are at \( y=-8 \), maybe around \( x = 1 \) and \( x = 3 \) or similar, but from the graph, the troughs are at \( y=-8 \)).
• The local maxima on intervals would be the peaks, but looking at the intervals, we check the \( y \)-values.

Part 2: Evaluate Each Statement

1. Over \([1, 3]\), local minimum is 0?

In \([1, 3]\), the lowest point is not 0 (the trough is at \( y=-8 \) or lower? Wait, the graph has a root at \( (2,0) \), but the minima between \( x=1 \) and \( x=3 \): the function dips down to \( y=-8 \)? Wait, no, looking at the graph, between \( x=1 \) and \( x=3 \), the function goes from a minimum (around \( x=1 \) maybe \( y=-8 \)) up to \( (2,0) \). Wait, maybe I misread. Wait the first minimum (left) is at \( y=-8 \) (below \( x=0 \) to \( x=1 \)), then up to \( (2,0) \), then down to \( y=-8 \) again (between \( x=2 \) and \( x=4 \)), then up to \( (4,0) \), then up. Wait, the graph: at \( x=0 \), \( y=0 \); then it goes down to \( y=-8 \) (maybe at \( x=1 \)), then up to \( (2,0) \), then down to \( y=-8 \) (at \( x=3 \)), then up to \( (4,0) \), then up.

Wait, let's re-express:

• Interval \([1, 3]\): The function at \( x=1 \) is at \( y=-8 \) (minimum), then rises to \( (2,0) \), then falls to \( y=-8 \) at \( x=3 \)? No, maybe the minima are at \( x=1 \) ( \( y=-8 \)) and \( x=3 \) ( \( y=-8 \)), and the maximum in \([1,3]\) is at \( (2,0) \) ( \( y=0 \)). Wait, the statement says "local minimum is 0"—no, 0 is a maximum here. Wait, no, the question for part 1 is to find local min and max, but part 2 is the true statements. Let's focus on part 2.

1. Over \([2, 4]\), local minimum is \(-8\)?

In \([2, 4]\), the function goes from \( (2,0) \) down to a minimum (at \( x=3 \), \( y=-8 \)) then up to \( (4,0) \). So the local minimum here is \(-8\). This is true.

1. Over \([3, 5]\), local minimum is \(-8\)?

In \([3, 5]\), the function goes from \( y=-8 \) (at \( x=3 \)) up to \( (4,0) \), then up. So the local minimum in \([3, 5]\) is \(-8\) (at \( x=3 \)). Wait, but after \( x=4 \), it goes up, so the minimum in \([3,5]\) is \(-8\) (at \( x=3 \))? Wait, no—between \( x=3 \) and \( x=5 \), the function at \( x=3 \) is \( y=-8 \), then rises to \( (4,0) \), then rises further. So the local minimum in \([3,5]\) is \(-8\) (at \( x=3 \)). But wait, the next statement:

1. Over \([1, 4]\), local maximum is 0?

In \([1, 4]\), the highest points are at \( (2,0) \) and \( (4,0) \), both \( y=0 \). So the local maximum in \([1,4]\) is 0. This is true.

1. Over \([2, 4]\), local minimum is \(-8\)?

In \([2, 4]\), the function goes from \( (2,0) \) down to \( y=-8 \) (at \( x=3 \)) then up to \( (4,0) \). So the local minimum is \(-8\). True.

1. Over \([3, 5]\), local maximum is 0?

In \([3, 5]\), the function at \( (4,0) \) is a local maximum (since it rises from \( x=3 \) ( \( y=-8 \)) to \( (4,0) \), then rises further). So the local maximum is 0. True? Wait, but let's check the options for the second part (choose three true statements):

Wait the options are:

• A: Over \([1, 3]\), local min is 0 (False, min is -8)
• B: Over \([2, 4]\), local min is -8 (True)
• C: Over \([3, 5]\), local min is -8 (True? Wait, in \([3,5]\), the function at \( x=3 \) is -8, then rises to \( (4,0) \), then rises. So the local minimum in \([3,5]\) is -8 (at \( x=3 \)).
• D: Over \([1, 4]\), local max is 0 (True, since the peaks are at \( (2,0) \) and \( (4,0) \), both 0)
• E: Over \([3, 5]\), local max is 0 (True, at \( (4,0) \), \( y=0 \))

Wait, but the problem says "choose three options". Let's re-express:

Wait maybe my initial graph analysis was wrong. Let's re-express the graph:

• The function has roots at \( (0,0) \), \( (2,0) \), \( (4,0) \).
• It has two minima (valleys) at \( y=-8 \): one between \( x=0 \) and \( x=2 \) (say at \( x=1 \), \( y=-8 \)), and one between \( x=2 \) and \( x=4 \) (at \( x=3 \), \( y=-8 \)).

Now evaluate each statement:

1. Over \([1, 3]\), local minimum is 0?

In \([1, 3]\), the function goes from \( x=1 \) ( \( y=-8 \)) up to \( (2,0) \), then down to \( x=3 \) ( \( y=-8 \)). So the local minimum is \(-8\), not 0. False.

1. Over \([2, 4]\), local minimum is \(-8\)?

In \([2, 4]\), the function goes from \( (2,0) \) down to \( x=3 \) ( \( y=-8 \)) then up to \( (4,0) \). So the local minimum is \(-8\). True.

1. Over \([3, 5]\), local minimum is \(-8\)?

In \([3, 5]\), the function at \( x=3 \) is \( y=-8 \), then rises to \( (4,0) \), then rises further. So the local minimum is \(-8\) (at \( x=3 \)). True.

1. Over \([1, 4]\), local maximum is 0?

In \([1, 4]\), the function has peaks at \( (2,0) \) and \( (4,0) \), both \( y=0 \). So the local maximum is 0. True.

1. Over \([3, 5]\), local maximum is 0?

In \([3, 5]\), the function at \( (4,0) \) is a local maximum (since it rises from \( x=3 \) ( \( y=-8 \)) to \( (4,0) \), then rises further). So the local maximum is 0. True.

But the problem says "choose three options". Wait, maybe I made a mistake. Let's check again:

Wait the options are:

• Option 1: Over \([1, 3]\), local min is 0 → False (min is -8)
• Option 2: Over \([2, 4]\), local min is -8 → True
• Option 3: Over \([3, 5]\), local min is -8 → True (at \( x=3 \), \( y=-8 \))
• Option 4: Over \([1, 4]\), local max is 0 → True (peaks at \( (2,0) \) and \( (4,0) \))
• Option 5: Over \([3, 5]\), local max is 0 → True (at \( (4,0) \))

But the problem says "choose three options". Maybe the correct three are:

• Over \([2, 4]\), local min is -8 (True)
• Over \([3, 5]\), local min is -8 (True)
• Over \([1, 4]\), local max is 0 (True)

Wait, but let's confirm with the graph:

• Interval \([2, 4]\): The function decreases from \( (2,0) \) to \( x=3 \) ( \( y=-8 \)) then increases to \( (4,0) \). So local minimum is -8. True.
• Interval \([3, 5]\): The function at \( x=3 \) is -8 (minimum), then increases to \( (4,0) \) (maximum in \([3,5]\) is 0), then increases. So local minimum is -8. True.
• Interval \([1, 4]\): The function has maxima at \( (2,0) \) and \( (4,0) \) (both \( y=0 \)), so local maximum is 0. True.

Final Answers

Part 1: Local Minimum and Maximum

(Assuming the graph’s minima are at \( y=-8 \) and maxima at \( y=0 \) in relevant intervals.)

• Local minimum: \( \boldsymbol{-8} \) (at \( x=1, 3 \), etc.)
• Local maximum: \( \boldsymbol{0} \) (at \( x=2, 4 \), etc.)

Part 2: True Statements (Choose Three)

The three true options are:

• Over the interval \([2, 4]\), the local minimum is \(-8\).
• Over the interval \([3, 5]\), the local minimum is \(-8\).
• Over the interval \([1, 4]\), the local maximum is \( 0 \).

(For the multiple-choice, the correct options are:

• Over the interval \([2, 4]\), the local minimum is \(-8\).
• Over the interval \([3, 5]\), the local minimum is \(-8\).
• Over the interval \([1, 4]\), the local maximum is \( 0 \).)