answer：Okay, so I have this problem about sets A and B. Let me read it again to make sure I understand.Set A is defined as all real numbers x such that -3 is less than or equal to x, which is less than or equal to 2. So, A is the interval [-3, 2]. Set B is defined as all real numbers x such that 1 - m is less than or equal to x, which is less than or equal to 3m - 1. So, B is the interval [1 - m, 3m - 1]. There are two parts to this problem. The first part is when m equals 3, find the intersection of A and B, which is A ∩ B, and the union of A and B, which is A ∪ B. The second part is a bit trickier: if A ∩ B equals B, find the range of the real number m.Starting with part 1. When m is 3, let's substitute m into set B. So, B becomes [1 - 3, 3*3 - 1]. Calculating that, 1 - 3 is -2, and 3*3 is 9, minus 1 is 8. So, set B is [-2, 8].Now, set A is [-3, 2], and set B is [-2, 8]. To find A ∩ B, which is the intersection, I need the overlap between these two intervals. So, A starts at -3 and goes to 2, while B starts at -2 and goes to 8. The overlap would be from -2 to 2 because that's where both intervals are present. So, A ∩ B is [-2, 2].Next, A ∪ B is the union of the two intervals. Since A goes from -3 to 2 and B goes from -2 to 8, the union would start at the smallest left endpoint, which is -3, and end at the largest right endpoint, which is 8. So, A ∪ B is [-3, 8].Okay, that seems straightforward. Let me just double-check. When m is 3, B is [-2, 8]. A is [-3, 2]. The intersection is where they both overlap, which is from -2 to 2. The union is from -3 to 8. Yep, that makes sense.Moving on to part 2. This is where it gets a bit more complex. The condition is that A ∩ B equals B. So, what does that mean? If A ∩ B is equal to B, that implies that B is a subset of A. Because the intersection of A and B would only be B if every element of B is already in A. So, B must be entirely contained within A.So, for B to be a subset of A, the interval [1 - m, 3m - 1] must lie entirely within [-3, 2]. That means both endpoints of B must be within A. So, 1 - m must be greater than or equal to -3, and 3m - 1 must be less than or equal to 2. Additionally, we need to make sure that the interval B is valid, meaning that the lower bound is less than or equal to the upper bound. So, 1 - m must be less than or equal to 3m - 1.Let me write that down as inequalities:1. 1 - m ≥ -32. 3m - 1 ≤ 23. 1 - m ≤ 3m - 1Let me solve each of these inequalities one by one.Starting with the first inequality: 1 - m ≥ -3.Subtract 1 from both sides: -m ≥ -4.Multiply both sides by -1, remembering to flip the inequality sign: m ≤ 4.Second inequality: 3m - 1 ≤ 2.Add 1 to both sides: 3m ≤ 3.Divide both sides by 3: m ≤ 1.Third inequality: 1 - m ≤ 3m - 1.Let's solve this. Bring all terms to one side:1 - m - 3m + 1 ≤ 0Combine like terms:2 - 4m ≤ 0Subtract 2 from both sides:-4m ≤ -2Divide both sides by -4, flipping the inequality:m ≥ (-2)/(-4) => m ≥ 0.5So, putting it all together, from the three inequalities:1. m ≤ 42. m ≤ 13. m ≥ 0.5So, combining these, m must be greater than or equal to 0.5 and less than or equal to 1. So, m ∈ [0.5, 1].Wait, hold on. But the initial thought was that if A ∩ B = B, then B is a subset of A. So, does that mean that all three inequalities must be satisfied? Let me think.Yes, because for B to be a subset of A, both endpoints of B must lie within A, and also, the lower bound of B must be less than or equal to the upper bound of B. So, all three conditions must hold.But let me verify. If m is 0.5, then B becomes [1 - 0.5, 3*0.5 - 1] = [0.5, 1.5 - 1] = [0.5, 0.5]. So, it's just the single point 0.5, which is within A. If m is 1, then B is [1 - 1, 3*1 - 1] = [0, 2], which is entirely within A.What if m is less than 0.5? Let's say m is 0. Then B would be [1 - 0, 3*0 - 1] = [1, -1]. Wait, that's not a valid interval because the lower bound is greater than the upper bound. So, in that case, B would be an empty set. Is the empty set considered a subset of A? Yes, because the empty set is a subset of every set.Hmm, so if m is less than 0.5, B becomes an empty set, which is still a subset of A. So, does that mean that m can be less than 0.5 as well?Wait, let's go back to the condition A ∩ B = B. If B is empty, then A ∩ B is also empty, which is equal to B. So, in that case, the condition holds. So, m can be less than 0.5 as well.But in my earlier inequalities, I had m ≥ 0.5. So, that was only considering when B is a non-empty interval. But if B is empty, it's still a subset of A, so m can be less than 0.5 as well.So, perhaps I need to consider two cases:Case 1: B is non-empty. Then, the inequalities 1 - m ≤ 3m - 1, which gives m ≥ 0.5, and also 1 - m ≥ -3 and 3m - 1 ≤ 2, which gives m ≤ 1. So, in this case, m is between 0.5 and 1.Case 2: B is empty. That happens when 1 - m > 3m - 1, which is when m < 0.5. In this case, B is empty, which is a subset of A, so the condition A ∩ B = B holds.Therefore, combining both cases, m can be less than 0.5 or between 0.5 and 1. So, overall, m can be any real number less than or equal to 1.Wait, but if m is greater than 1, what happens? Let's test m = 2. Then, B would be [1 - 2, 3*2 - 1] = [-1, 5]. Now, A is [-3, 2]. So, the intersection A ∩ B would be [-1, 2], which is not equal to B, because B is [-1, 5]. So, in this case, A ∩ B ≠ B, so m cannot be greater than 1.Similarly, if m is exactly 1, B is [0, 2], which is entirely within A, so A ∩ B = B. If m is less than 1, say m = 0.75, then B is [1 - 0.75, 3*0.75 - 1] = [0.25, 1.25], which is within A. If m is 0.5, B is [0.5, 0.5], which is a single point in A. If m is less than 0.5, say m = 0.25, then B is [1 - 0.25, 3*0.25 - 1] = [0.75, -0.25], which is an empty set, which is still a subset of A.Therefore, the range of m is all real numbers less than or equal to 1.So, putting it all together, m can be any real number such that m ≤ 1.I think that's the correct answer. Let me just recap:For part 1, when m = 3, B is [-2, 8], so intersection with A is [-2, 2], and union is [-3, 8].For part 2, to have A ∩ B = B, B must be a subset of A. This happens when m ≤ 1, considering both cases where B is empty or non-empty.Yeah, that seems right.**Final Answer**1. A cap B = boxed{[-2, 2]} and A cup B = boxed{[-3, 8]}.2. The range of m is boxed{(-infty, 1]}.

answer：Okay, so I have this problem where there's a square ABCD, and they cut out an isosceles right triangle BFC from it and then translate it to the left, creating this shaded figure ABFCDE. The perimeter of the original square ABCD is 40 inches, and I need to find the perimeter of this new figure ABFCDE.First, let me try to visualize the square and the triangle. Since it's a square, all sides are equal. The perimeter is 40 inches, so each side must be 40 divided by 4, which is 10 inches. So, each side of square ABCD is 10 inches long.Now, they're cutting out an isosceles right triangle BFC. An isosceles right triangle has two legs of equal length and a hypotenuse. Since it's cut out from the square, I think the legs of the triangle must be along the sides of the square. Let me see, point B is one corner of the square, and point C is another. So, triangle BFC is probably attached at points B and C, and point F is somewhere on the square.Wait, in the Asymptote code, they mention points A, B, C, D, E, F. From the coordinates given, it looks like A is at (2,2), B at (4,2), C at (4,0), D at (2,-2), E at (0,0), and F at (2,0). So, the square is centered at (2,0), with sides of length 4 units in the drawing, but in reality, each side is 10 inches. Hmm, maybe the drawing is scaled down.But regardless, the key is that triangle BFC is an isosceles right triangle. So, the legs BF and FC must be equal. Since it's cut out from the square, BF and FC are each equal to the side length of the square, which is 10 inches. Wait, no, because if you cut out a triangle from the square, the legs can't be the full side length, otherwise, you wouldn't have anything left.Wait, maybe the legs are half the side length? Let me think. If the square has side length 10, then the diagonal is 10√2. But triangle BFC is an isosceles right triangle, so its legs are equal, and the hypotenuse would be leg * √2.But where exactly is point F? From the Asymptote code, point F is at (2,0), which is the midpoint of the bottom side of the square. So, in the original square, point C is at (4,0), and point F is at (2,0). So, the length of FC is 2 units in the drawing, but in reality, it's 10 inches. Wait, no, in the drawing, the square goes from (0,0) to (4,2), but in reality, it's 10 inches per side. Maybe the coordinates are just for illustration.Wait, perhaps I should not rely on the coordinates but just on the description. So, square ABCD has side length 10 inches. They cut out an isosceles right triangle BFC. So, points B and C are adjacent corners of the square, right? Because in a square, adjacent corners are connected by sides, and opposite corners are connected by diagonals.Wait, but if BFC is an isosceles right triangle, then angle at F must be the right angle. So, points B, F, and C form a right angle at F. So, BF and FC are the legs, and BC is the hypotenuse.But BC is a side of the square, which is 10 inches. So, if BC is the hypotenuse of the triangle BFC, then each leg BF and FC must be 10 / √2 inches, which simplifies to 5√2 inches.Wait, but if BF and FC are each 5√2 inches, then the perimeter contributions from these sides would change when we move the triangle.Wait, let me think again. The original square has four sides, each 10 inches, so the perimeter is 40 inches. When we cut out triangle BFC and move it to the left, we're removing a part of the square and adding new edges.Specifically, cutting out triangle BFC means we're removing the side BC (10 inches) and replacing it with the two legs BF and FC (each 5√2 inches). But since we're moving the triangle to the left, we're also adding new edges where the triangle was moved.Wait, actually, when you translate the triangle to the left, you're creating a new figure ABFCDE. So, the original square had sides AB, BC, CD, DA. After cutting out BFC and moving it, the new figure has sides AB, BF, FC, CD, DE, and EA.Wait, let me list the sides of the new figure:- From A to B: AB, which is 10 inches.- From B to F: BF, which is 5√2 inches.- From F to C: FC, which is 5√2 inches.- From C to D: CD, which is 10 inches.- From D to E: DE, which is the same as the side of the square, so 10 inches.- From E to A: EA, which is another side, 10 inches.Wait, but that doesn't seem right because DE and EA are not sides of the original square. Wait, maybe I need to think about how the triangle is moved.When you cut out triangle BFC and move it to the left, point F is moved to a new position, say E. So, the new figure includes the original square minus triangle BFC, plus the translated triangle.Wait, looking back at the Asymptote code, the figure is drawn as (0,0)--(2,2)--(4,2)--(2,0)--(4,0)--(2,-2)--(0,0)--cycle. So, in the drawing, it's a hexagon with points A(2,2), B(4,2), F(2,0), C(4,0), D(2,-2), E(0,0), and back to A.So, in the actual problem, each side of the square is 10 inches, but in the drawing, it's scaled down. So, the coordinates are just for illustration, but the actual lengths are 10 inches per side.So, in the new figure ABFCDE, the sides are:- AB: 10 inches- BF: which is the same as FC, which is 10 inches each? Wait, no, because triangle BFC is an isosceles right triangle, so BF = FC.Wait, if the original square has side 10, then BC is 10 inches. Since triangle BFC is an isosceles right triangle with hypotenuse BC, then legs BF and FC are each 10 / √2 = 5√2 inches.So, when we cut out triangle BFC and move it to the left, we're removing the side BC (10 inches) and adding two sides BF and FC (each 5√2 inches). But also, moving the triangle to the left creates new sides where the triangle was moved.Wait, actually, when you translate the triangle to the left, you're creating a new edge from F to E and from E to A. So, the new figure has:- AB: 10 inches- BF: 5√2 inches- FC: 5√2 inches- CD: 10 inches- DE: which is the same as the side of the square, so 10 inches- EA: which is another side, 10 inchesWait, but that would make the perimeter AB + BF + FC + CD + DE + EA = 10 + 5√2 + 5√2 + 10 + 10 + 10 = 40 + 10√2 inches.But wait, in the original square, the perimeter was 40 inches. By cutting out the triangle and moving it, we're adding two sides of the triangle (each 5√2 inches) and removing one side (10 inches). So, the change in perimeter is +10√2 -10 inches. Therefore, the new perimeter would be 40 + 10√2 -10 = 30 + 10√2 inches.Wait, that contradicts my earlier calculation. Hmm.Wait, maybe I need to think differently. The original square has four sides, each 10 inches. When we cut out triangle BFC, we're removing a part of the square, specifically the side BC, and replacing it with two sides BF and FC. So, instead of having side BC (10 inches), we now have BF and FC, each 5√2 inches. So, the perimeter contribution from that part changes from 10 inches to 10√2 inches (since 5√2 + 5√2 = 10√2).Additionally, when we move the triangle to the left, we're creating a new edge where the triangle was moved. So, the triangle is moved from position BFC to a new position, say EFC. So, we're adding two new sides: from E to F and from F to C. Wait, but in the figure, it's ABFCDE, so after moving, the figure includes the original square minus triangle BFC, plus the translated triangle.Wait, perhaps it's better to count all the sides of the new figure. The new figure ABFCDE has six sides:1. AB: 10 inches2. BF: 5√2 inches3. FC: 5√2 inches4. CD: 10 inches5. DE: 10 inches6. EA: 10 inchesSo, adding these up: 10 + 5√2 + 5√2 + 10 + 10 + 10 = 40 + 10√2 inches.But wait, that seems too much because the original perimeter was 40 inches, and we're adding 10√2 inches. But intuitively, when you cut out a corner and move it, you might be adding more perimeter.Wait, let me double-check. The original square has four sides, each 10 inches. When we cut out triangle BFC, we're removing the side BC (10 inches) and replacing it with two sides BF and FC (each 5√2 inches). So, the perimeter change is +10√2 -10 inches. Then, when we move the triangle to the left, we're creating a new edge where the triangle was moved. So, we're adding two more sides: from E to F and from F to A? Wait, no, because in the new figure, after moving, the triangle is attached at point E.Wait, maybe the new figure has the following sides:- AB: 10 inches- BF: 5√2 inches- FC: 5√2 inches- CD: 10 inches- DE: 10 inches- EA: 10 inchesSo, total perimeter is 10 + 5√2 + 5√2 + 10 + 10 + 10 = 40 + 10√2 inches.But wait, in the original square, the side BC was 10 inches, and now it's replaced by BF and FC, which are 5√2 each, so 10√2 total. So, the perimeter increases by 10√2 -10 inches. Therefore, the new perimeter is 40 + (10√2 -10) = 30 + 10√2 inches.Wait, now I'm confused because I have two different results. Let me try to clarify.When you remove triangle BFC from the square, you're taking away the side BC (10 inches) and adding two sides BF and FC (each 5√2 inches). So, the perimeter contribution from that area changes from 10 inches to 10√2 inches. So, the perimeter increases by 10√2 -10 inches.Additionally, when you move the triangle to the left, you're creating a new edge where the triangle was moved. So, you're adding two more sides: from E to F and from F to A? Wait, no, because in the new figure, after moving, the triangle is attached at point E.Wait, looking back at the Asymptote code, the figure is (0,0)--(2,2)--(4,2)--(2,0)--(4,0)--(2,-2)--(0,0)--cycle. So, in terms of the actual problem, the new figure has points A, B, F, C, D, E, and back to A.So, the sides are:- AB: 10 inches- BF: 5√2 inches- FC: 5√2 inches- CD: 10 inches- DE: 10 inches- EA: 10 inchesWait, but EA is not a side of the original square. Wait, in the original square, point E is a new point created by moving the triangle. So, EA is a new side.Wait, perhaps the perimeter is the sum of all these sides: AB + BF + FC + CD + DE + EA.But AB is 10, BF is 5√2, FC is 5√2, CD is 10, DE is 10, and EA is 10. So, total perimeter is 10 + 5√2 + 5√2 + 10 + 10 + 10 = 40 + 10√2 inches.But wait, in the original square, the perimeter was 40 inches. By cutting out the triangle, we're removing 10 inches and adding 10√2 inches, so the perimeter becomes 40 -10 +10√2 = 30 +10√2 inches. But then, when we move the triangle to the left, we're adding more sides.Wait, I think I'm overcomplicating it. Let me try to count the sides of the new figure ABFCDE.Looking at the figure, it's a hexagon with vertices A, B, F, C, D, E, and back to A.So, the sides are:1. AB: 10 inches2. BF: which is a leg of the triangle, so 5√2 inches3. FC: another leg, 5√2 inches4. CD: 10 inches5. DE: which is the same as the side of the square, so 10 inches6. EA: which is another side, 10 inchesSo, adding these up: 10 + 5√2 + 5√2 + 10 + 10 + 10 = 40 + 10√2 inches.Wait, but that seems to be the case. So, the perimeter is 40 +10√2 inches.But let me check again. The original square had a perimeter of 40 inches. When we cut out triangle BFC, we're removing side BC (10 inches) and adding two sides BF and FC (each 5√2 inches). So, the perimeter increases by 10√2 -10 inches. Then, when we move the triangle to the left, we're creating a new edge from E to F and from F to A? Wait, no, because in the new figure, after moving, the triangle is attached at point E, so we're adding sides DE and EA.Wait, maybe the perimeter is:Original perimeter: 40 inchesMinus BC: -10 inchesPlus BF + FC: +10√2 inchesPlus DE + EA: +20 inchesWait, but that would be 40 -10 +10√2 +20 = 50 +10√2 inches, which seems too much.Wait, no, because when you move the triangle, you're not adding DE and EA as new sides, but rather, those sides are already part of the figure. Wait, I'm getting confused.Let me try to think differently. The new figure ABFCDE is a hexagon. Let's list all its sides:1. AB: 10 inches2. BF: 5√2 inches3. FC: 5√2 inches4. CD: 10 inches5. DE: 10 inches6. EA: 10 inchesSo, total perimeter is 10 +5√2 +5√2 +10 +10 +10 = 40 +10√2 inches.Yes, that makes sense. So, the perimeter is 40 +10√2 inches.Wait, but earlier I thought it was 40 +20√2 inches. Hmm.Wait, maybe I made a mistake in the length of BF and FC. If triangle BFC is an isosceles right triangle with hypotenuse BC =10 inches, then legs BF and FC are each 10 /√2 =5√2 inches. So, BF + FC =10√2 inches.So, in the new figure, instead of having side BC (10 inches), we have BF + FC (10√2 inches). So, the perimeter increases by 10√2 -10 inches.Additionally, when we move the triangle to the left, we're creating a new edge from E to F and from F to A? Wait, no, because in the new figure, after moving, the triangle is attached at point E, so we're adding sides DE and EA.Wait, but DE and EA are already sides of the original square. Wait, no, because when you move the triangle, point E is a new point, so DE and EA are new sides.Wait, perhaps the perimeter is:Original perimeter: 40 inchesMinus BC: -10 inchesPlus BF + FC: +10√2 inchesPlus DE + EA: +20 inchesSo, total perimeter: 40 -10 +10√2 +20 =50 +10√2 inches.But that contradicts my earlier calculation.Wait, maybe I need to look at the figure again. The new figure ABFCDE has six sides:1. AB: 10 inches2. BF:5√2 inches3. FC:5√2 inches4. CD:10 inches5. DE:10 inches6. EA:10 inchesSo, total perimeter is 10 +5√2 +5√2 +10 +10 +10 =40 +10√2 inches.Yes, that seems correct. So, the perimeter is 40 +10√2 inches.Wait, but in the original square, the perimeter was 40 inches. By cutting out the triangle, we're removing 10 inches and adding 10√2 inches, so the perimeter becomes 40 -10 +10√2 =30 +10√2 inches. But then, when we move the triangle to the left, we're adding two more sides: DE and EA, each 10 inches, so adding 20 inches. So, total perimeter would be 30 +10√2 +20 =50 +10√2 inches.Wait, but that can't be right because the figure ABFCDE is a hexagon, and the sides are AB, BF, FC, CD, DE, EA. So, it's six sides, each of which is either 10 inches or 5√2 inches.Wait, perhaps the confusion is about whether DE and EA are new sides or not. In the original square, DE and EA were not present because the square only had four sides. When we cut out the triangle and move it, we're creating new edges DE and EA, which are each 10 inches.So, the perimeter is:Original perimeter:40 inchesMinus BC: -10 inchesPlus BF + FC: +10√2 inchesPlus DE + EA: +20 inchesTotal:40 -10 +10√2 +20=50 +10√2 inches.But in the figure, the sides are AB, BF, FC, CD, DE, EA, which are 10,5√2,5√2,10,10,10. So, total is 40 +10√2 inches.Wait, so which is correct? Is it 40 +10√2 or 50 +10√2?I think the key is to realize that when you move the triangle, you're not adding DE and EA as new sides, but rather, those sides are already part of the figure because the triangle is moved to create them.Wait, perhaps the correct way is to count all the outer edges of the new figure. So, ABFCDE has the following outer edges:- AB:10- BF:5√2- FC:5√2- CD:10- DE:10- EA:10So, total perimeter is 10 +5√2 +5√2 +10 +10 +10=40 +10√2 inches.Yes, that makes sense. So, the perimeter is 40 +10√2 inches.Wait, but in the original square, the perimeter was 40 inches. By cutting out the triangle and moving it, we're adding 10√2 inches to the perimeter. So, the new perimeter is 40 +10√2 inches.Yes, that seems correct.

answer：First, I need to determine the percentage of meters rejected as defective. I know that the inspector examined 2857.142857142857 meters and rejected 2 of them as defective.To find the percentage, I'll divide the number of defective meters by the total number of meters examined and then multiply by 100.So, the calculation is (2 / 2857.142857142857) * 100.Performing the division, 2 divided by 2857.142857142857 equals approximately 0.0007.Multiplying by 100 to convert it to a percentage gives 0.07%.Therefore, the inspector rejects approximately 0.07% of the meters as defective.

answer：Alright, so I have this problem about a geometric sequence. Let me read it again to make sure I understand.It says: Given a geometric sequence {a_n} with the first term 3/2 and common ratio -1/2, and the sum of the first n terms is S_n. Then, when n is a positive integer, we need to find the sum of the maximum and minimum values of S_n - 1/S_n.Okay, so first, let me recall what a geometric sequence is. It's a sequence where each term is multiplied by a common ratio to get the next term. So, starting with 3/2, the next term would be 3/2 * (-1/2) = -3/4, then the next term would be -3/4 * (-1/2) = 3/8, and so on.Now, the sum of the first n terms of a geometric sequence has a formula. I think it's S_n = a_1 * (1 - r^n) / (1 - r), where a_1 is the first term and r is the common ratio. Let me write that down:S_n = (3/2) * [1 - (-1/2)^n] / [1 - (-1/2)]Simplify the denominator: 1 - (-1/2) = 1 + 1/2 = 3/2.So, S_n = (3/2) * [1 - (-1/2)^n] / (3/2) = [1 - (-1/2)^n]Wait, that simplifies nicely. So, S_n = 1 - (-1/2)^n.Hmm, interesting. So, depending on whether n is odd or even, (-1/2)^n will be negative or positive.Let me think about that. If n is odd, (-1/2)^n is negative, so S_n = 1 - (negative number) = 1 + positive number. If n is even, (-1/2)^n is positive, so S_n = 1 - positive number.Therefore, for odd n, S_n is greater than 1, and for even n, S_n is less than 1.Let me write that down:- If n is odd: S_n = 1 + (1/2)^n- If n is even: S_n = 1 - (1/2)^nOkay, so now, I need to find the maximum and minimum values of S_n - 1/S_n.First, let's analyze S_n.For odd n: S_n = 1 + (1/2)^n. Since (1/2)^n decreases as n increases, S_n is decreasing for odd n.Similarly, for even n: S_n = 1 - (1/2)^n. Here, (1/2)^n also decreases as n increases, so S_n is increasing for even n.Therefore, the maximum value of S_n occurs at the smallest odd n, which is n=1: S_1 = 1 + 1/2 = 3/2.The minimum value of S_n occurs at the smallest even n, which is n=2: S_2 = 1 - 1/4 = 3/4.So, S_n ranges between 3/4 and 3/2.Now, let's define f(S_n) = S_n - 1/S_n.We need to find the maximum and minimum of f(S_n) as S_n varies between 3/4 and 3/2.Since f(t) = t - 1/t, let's analyze this function.First, compute its derivative to see if it's increasing or decreasing.f'(t) = 1 + 1/t^2.Since 1 + 1/t^2 is always positive for t > 0, the function f(t) is strictly increasing on the interval (0, ∞).Therefore, since f(t) is increasing, its maximum occurs at the maximum t, which is 3/2, and its minimum occurs at the minimum t, which is 3/4.So, compute f(3/2) and f(3/4):f(3/2) = (3/2) - 1/(3/2) = (3/2) - (2/3) = (9/6 - 4/6) = 5/6.f(3/4) = (3/4) - 1/(3/4) = (3/4) - (4/3) = (9/12 - 16/12) = -7/12.Therefore, the maximum value of f(S_n) is 5/6, and the minimum value is -7/12.Now, the problem asks for the sum of the maximum and minimum values.So, 5/6 + (-7/12) = 5/6 - 7/12.To subtract these, convert to a common denominator:5/6 = 10/12, so 10/12 - 7/12 = 3/12 = 1/4.Therefore, the sum of the maximum and minimum values is 1/4.Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, S_n:For n=1: 3/2, n=2: 3/4, n=3: 1 + 1/8 = 9/8, n=4: 1 - 1/16 = 15/16, etc.So, S_n alternates between being above 1 and below 1, approaching 1 as n increases.So, the maximum S_n is indeed 3/2, and the minimum is 3/4.Then, f(t) = t - 1/t.At t=3/2: 3/2 - 2/3 = (9 - 4)/6 = 5/6.At t=3/4: 3/4 - 4/3 = (9 - 16)/12 = -7/12.Adding them: 5/6 - 7/12 = (10 - 7)/12 = 3/12 = 1/4.Yes, that seems correct.I think my reasoning is solid here. The key was recognizing that S_n alternates around 1, and since the function f(t) is increasing, the extrema occur at the endpoints of the interval for S_n.So, the final answer is 1/4.**Final Answer**boxed{dfrac{1}{4}}

answer：To factorize the expression ( m^2 - 4 ), I recognize that it is a difference of squares. The difference of squares formula is ( a^2 - b^2 = (a + b)(a - b) ). Applying this formula to the expression, where ( a = m ) and ( b = 2 ), I can rewrite ( m^2 - 4 ) as ( m^2 - 2^2 ). Using the formula, this becomes ( (m + 2)(m - 2) ). Therefore, the factorized form of ( m^2 - 4 ) is ( (m + 2)(m - 2) ).

answer：First, I need to determine the number of ducks and cows in the group. I know there are 13 cows.Next, I'll calculate the total number of legs contributed by the cows. Since each cow has 4 legs, the cows contribute 13 multiplied by 4, which equals 52 legs.Let ( D ) represent the number of ducks. Each duck has 2 legs, so the total number of legs from the ducks is ( 2D ).The total number of legs in the group is the sum of the legs from the cows and the ducks, which is ( 52 + 2D ).The total number of heads in the group is the sum of the number of ducks and cows, which is ( D + 13 ).According to the problem, the total number of legs is 13 more than three times the number of heads. This can be expressed as:[52 + 2D = 3(D + 13) + 13]Expanding the right side of the equation:[52 + 2D = 3D + 39 + 13][52 + 2D = 3D + 52]Subtracting ( 2D ) from both sides:[52 = D + 52]Subtracting 52 from both sides:[0 = D]This implies that there are no ducks in the group. Therefore, the number of legs that are more than three times the number of heads is 13.