💡First, I need to determine the total number of apples ordered by the cafeteria. They ordered 25 red apples and an unknown number of green apples, which I'll denote as G.Next, I know that only 10 students wanted fruit, so the number of apples taken by the students is 10.The cafeteria ended up with 32 extra apples. This means that the total number of apples ordered minus the number of apples taken by the students equals 32.I can set up the equation: 25 (red apples) + G (green apples) - 10 (apples taken by students) = 32 (extra apples).Simplifying the equation, I get G + 15 = 32.Finally, solving for G, I find that G = 17. Therefore, the cafeteria ordered 17 green apples.

💡Alright, so I've got this problem here where I need to find three numbers a, b, and c such that when I take all the pairwise sums of 1, a, b, and c, those sums form an arithmetic sequence. Also, the total sum of all these pairwise sums is 201. Hmm, okay, let's break this down step by step.First, let's list out all the pairwise sums. Since we have four numbers: 1, a, b, and c, the pairwise sums would be:1 + a, 1 + b, 1 + c, a + b, a + c, and b + c.So that's six sums in total. The problem states that these six sums form an arithmetic sequence. An arithmetic sequence is a sequence where each term after the first is obtained by adding a constant difference. So, if I can figure out what that common difference is, I might be able to find a, b, and c.Also, the total sum of all these six sums is 201. Let me write that down:(1 + a) + (1 + b) + (1 + c) + (a + b) + (a + c) + (b + c) = 201.Let me simplify this equation. If I add all these up, I get:3 + 3a + 3b + 3c = 201.Wait, because 1 appears three times (in 1+a, 1+b, 1+c), and a, b, c each appear three times as well (in the other sums). So, factoring out the 3:3(1 + a + b + c) = 201.Dividing both sides by 3:1 + a + b + c = 67.So, a + b + c = 66.Okay, that's a useful equation. Now, I need to find a, b, and c such that their pairwise sums form an arithmetic sequence.Let me think about the order of these sums. Since 1 < a < b < c, the smallest sum should be 1 + a, followed by 1 + b, then 1 + c, and then a + b, a + c, and finally b + c. So, the order of the sums from smallest to largest is:1 + a, 1 + b, 1 + c, a + b, a + c, b + c.Now, since these form an arithmetic sequence, the difference between consecutive terms should be constant. Let's denote this common difference as d.So, the sequence is:1 + a, 1 + a + d, 1 + a + 2d, 1 + a + 3d, 1 + a + 4d, 1 + a + 5d.But we also have expressions for these sums in terms of a, b, and c. So, let's set them equal:1 + a = 1 + a (obviously)1 + b = 1 + a + d1 + c = 1 + a + 2da + b = 1 + a + 3da + c = 1 + a + 4db + c = 1 + a + 5dOkay, so from the second equation:1 + b = 1 + a + d ⇒ b = a + d.From the third equation:1 + c = 1 + a + 2d ⇒ c = a + 2d.From the fourth equation:a + b = 1 + a + 3d ⇒ b = 1 + 3d.But we already have b = a + d, so:a + d = 1 + 3d ⇒ a = 1 + 2d.From the fifth equation:a + c = 1 + a + 4d ⇒ c = 1 + 4d.But we also have c = a + 2d, so:a + 2d = 1 + 4d ⇒ a = 1 + 2d.Wait, that's consistent with what we had before. So, a = 1 + 2d.From the sixth equation:b + c = 1 + a + 5d.We know b = a + d and c = a + 2d, so:b + c = (a + d) + (a + 2d) = 2a + 3d.And this should equal 1 + a + 5d.So:2a + 3d = 1 + a + 5d ⇒ a = 1 + 2d.Again, consistent.So, we have a = 1 + 2d, b = a + d = 1 + 3d, c = a + 2d = 1 + 4d.Now, recall that a + b + c = 66.Substituting the expressions in terms of d:(1 + 2d) + (1 + 3d) + (1 + 4d) = 66.Simplify:3 + 9d = 66 ⇒ 9d = 63 ⇒ d = 7.So, d = 7.Now, let's find a, b, and c:a = 1 + 2d = 1 + 14 = 15.b = 1 + 3d = 1 + 21 = 22.c = 1 + 4d = 1 + 28 = 29.Let me verify if these values satisfy the conditions.First, check if 1 < a < b < c: 1 < 15 < 22 < 29. Yes, that's true.Now, compute all pairwise sums:1 + a = 1 + 15 = 16.1 + b = 1 + 22 = 23.1 + c = 1 + 29 = 30.a + b = 15 + 22 = 37.a + c = 15 + 29 = 44.b + c = 22 + 29 = 51.Now, let's list these sums: 16, 23, 30, 37, 44, 51.Check if they form an arithmetic sequence.Compute the differences:23 - 16 = 7.30 - 23 = 7.37 - 30 = 7.44 - 37 = 7.51 - 44 = 7.Yes, the common difference is 7, so it's an arithmetic sequence.Also, the total sum of these sums is 16 + 23 + 30 + 37 + 44 + 51.Let's compute that:16 + 23 = 39.39 + 30 = 69.69 + 37 = 106.106 + 44 = 150.150 + 51 = 201.Perfect, that matches the given total sum.So, the values are a = 15, b = 22, c = 29.Wait, but earlier in my thought process, I considered another possible solution where a = 10, b = 19, c = 37. Let me check if that also works.Compute the pairwise sums:1 + 10 = 11.1 + 19 = 20.1 + 37 = 38.10 + 19 = 29.10 + 37 = 47.19 + 37 = 56.So, the sums are: 11, 20, 29, 38, 47, 56.Check if they form an arithmetic sequence.Differences:20 - 11 = 9.29 - 20 = 9.38 - 29 = 9.47 - 38 = 9.56 - 47 = 9.Yes, common difference of 9.Total sum: 11 + 20 + 29 + 38 + 47 + 56.Compute:11 + 20 = 31.31 + 29 = 60.60 + 38 = 98.98 + 47 = 145.145 + 56 = 201.That also adds up to 201.So, there are two possible solutions: one with a = 15, b = 22, c = 29, and another with a = 10, b = 19, c = 37.But wait, the problem states that 1 < a < b < c, and the sums are all distinct. Both solutions satisfy these conditions.However, in the first case, the common difference d was 7, leading to a = 15, b = 22, c = 29.In the second case, the common difference was 9, leading to a = 10, b = 19, c = 37.So, both are valid solutions.But the problem asks to find the values of a, b, c. It doesn't specify if there are multiple solutions or just one.Looking back at the problem statement, it says "the sums of every pair of these numbers are all distinct, these 6 sums form an arithmetic sequence from smallest to largest with a total sum of 201."So, both solutions satisfy this.But wait, in the first case, the order of the sums is 16, 23, 30, 37, 44, 51, which is correct.In the second case, the order is 11, 20, 29, 38, 47, 56, which is also correct.So, both are valid.But the problem might expect one solution, so perhaps I need to see if there are constraints I missed.Wait, in the first case, a = 15, b = 22, c = 29.In the second case, a = 10, b = 19, c = 37.Both satisfy 1 < a < b < c, and the sums are distinct and form an arithmetic sequence with total sum 201.So, both are correct.But the problem says "find the values of a, b, c," implying there might be a unique solution.Wait, perhaps I made a mistake in assuming both are valid.Let me double-check the initial equations.In the first case, when I set a = 1 + 2d, b = 1 + 3d, c = 1 + 4d, and found d = 7, leading to a = 15, b = 22, c = 29.In the second case, I considered a different ordering where 1 + c < a + b, leading to a different set of equations.Wait, in the first case, I assumed the order of sums as 1 + a, 1 + b, 1 + c, a + b, a + c, b + c.But in reality, depending on the values of a, b, and c, the order might change.For example, if a + b is less than 1 + c, then the order would be 1 + a, 1 + b, a + b, 1 + c, a + c, b + c.Similarly, if 1 + c is less than a + b, then the order would be 1 + a, 1 + b, 1 + c, a + b, a + c, b + c.So, there are two possible orderings depending on whether a + b is less than or greater than 1 + c.Therefore, there are two possible solutions.So, both a = 15, b = 22, c = 29 and a = 10, b = 19, c = 37 are valid.But the problem might expect both solutions.Alternatively, perhaps I need to consider that in the first case, the common difference is 7, and in the second case, it's 9.But the problem doesn't specify the common difference, just that the sums form an arithmetic sequence.So, both are valid.Therefore, the possible solutions are:Either a = 10, b = 19, c = 37 or a = 15, b = 22, c = 29.But let me check if there are more solutions.Wait, in the first case, when I set a = 1 + 2d, b = 1 + 3d, c = 1 + 4d, and found d = 7, leading to a = 15, b = 22, c = 29.In the second case, I considered a different ordering where 1 + c < a + b, leading to a different set of equations.But perhaps there are more possibilities.Wait, let me think about the ordering of the sums.The smallest sum is always 1 + a.The next could be 1 + b or a + a, but since a > 1, 1 + b is less than a + a only if b < 2a - 1.Wait, actually, since a > 1, 1 + b could be less than a + a or not.Wait, but in our case, since a, b, c are all greater than 1, and 1 < a < b < c, the next sum after 1 + a is 1 + b, then 1 + c, then a + b, then a + c, then b + c.But this depends on the relationship between a + b and 1 + c.If a + b < 1 + c, then the order would be 1 + a, 1 + b, a + b, 1 + c, a + c, b + c.If a + b > 1 + c, then the order is 1 + a, 1 + b, 1 + c, a + b, a + c, b + c.So, depending on whether a + b is less than or greater than 1 + c, we have two different orderings.Therefore, there are two cases to consider.Case 1: a + b < 1 + c.Case 2: a + b > 1 + c.So, in Case 1, the order is 1 + a, 1 + b, a + b, 1 + c, a + c, b + c.In Case 2, the order is 1 + a, 1 + b, 1 + c, a + b, a + c, b + c.So, let's handle both cases.Case 1: a + b < 1 + c.So, the sums in order are:1 + a, 1 + b, a + b, 1 + c, a + c, b + c.Let me denote the common difference as d.So,1 + a,1 + a + d = 1 + b,1 + a + 2d = a + b,1 + a + 3d = 1 + c,1 + a + 4d = a + c,1 + a + 5d = b + c.From the second equation:1 + b = 1 + a + d ⇒ b = a + d.From the third equation:a + b = 1 + a + 2d ⇒ b = 1 + 2d.But b = a + d, so:a + d = 1 + 2d ⇒ a = 1 + d.From the fourth equation:1 + c = 1 + a + 3d ⇒ c = a + 3d.From the fifth equation:a + c = 1 + a + 4d ⇒ c = 1 + 4d.But c = a + 3d, so:a + 3d = 1 + 4d ⇒ a = 1 + d.Consistent with earlier.From the sixth equation:b + c = 1 + a + 5d.We know b = a + d and c = a + 3d, so:b + c = (a + d) + (a + 3d) = 2a + 4d.This should equal 1 + a + 5d.So:2a + 4d = 1 + a + 5d ⇒ a = 1 + d.Again, consistent.Now, recall that a + b + c = 66.Substituting a = 1 + d, b = a + d = 1 + 2d, c = a + 3d = 1 + 4d.So,a + b + c = (1 + d) + (1 + 2d) + (1 + 4d) = 3 + 7d = 66 ⇒ 7d = 63 ⇒ d = 9.So, d = 9.Therefore,a = 1 + d = 10,b = 1 + 2d = 19,c = 1 + 4d = 37.So, this is the second solution I found earlier.Case 2: a + b > 1 + c.So, the order is:1 + a, 1 + b, 1 + c, a + b, a + c, b + c.Let me denote the common difference as d.So,1 + a,1 + a + d = 1 + b,1 + a + 2d = 1 + c,1 + a + 3d = a + b,1 + a + 4d = a + c,1 + a + 5d = b + c.From the second equation:1 + b = 1 + a + d ⇒ b = a + d.From the third equation:1 + c = 1 + a + 2d ⇒ c = a + 2d.From the fourth equation:a + b = 1 + a + 3d ⇒ b = 1 + 3d.But b = a + d, so:a + d = 1 + 3d ⇒ a = 1 + 2d.From the fifth equation:a + c = 1 + a + 4d ⇒ c = 1 + 4d.But c = a + 2d, so:a + 2d = 1 + 4d ⇒ a = 1 + 2d.Consistent.From the sixth equation:b + c = 1 + a + 5d.We know b = a + d and c = a + 2d, so:b + c = (a + d) + (a + 2d) = 2a + 3d.This should equal 1 + a + 5d.So:2a + 3d = 1 + a + 5d ⇒ a = 1 + 2d.Again, consistent.Now, recall that a + b + c = 66.Substituting a = 1 + 2d, b = a + d = 1 + 3d, c = a + 2d = 1 + 4d.So,a + b + c = (1 + 2d) + (1 + 3d) + (1 + 4d) = 3 + 9d = 66 ⇒ 9d = 63 ⇒ d = 7.So, d = 7.Therefore,a = 1 + 2d = 15,b = 1 + 3d = 22,c = 1 + 4d = 29.So, this is the first solution I found earlier.Therefore, there are two possible solutions:1. a = 10, b = 19, c = 37.2. a = 15, b = 22, c = 29.Both satisfy the conditions of the problem.But the problem asks to "find the values of a, b, c." It doesn't specify if there are multiple solutions or just one. So, perhaps both are acceptable.However, in the initial problem statement, it's mentioned that the sums form an arithmetic sequence from smallest to largest. So, depending on the ordering, both are valid.Therefore, the possible solutions are:Either a = 10, b = 19, c = 37 or a = 15, b = 22, c = 29.But let me check if there are any other possible cases.Wait, in the first case, when a + b < 1 + c, we found a = 10, b = 19, c = 37.In the second case, when a + b > 1 + c, we found a = 15, b = 22, c = 29.Are there any other cases? For example, what if a + b = 1 + c? But the problem states that all sums are distinct, so a + b cannot equal 1 + c.Therefore, only two cases are possible: a + b < 1 + c or a + b > 1 + c.Hence, only two solutions.So, the final answer is that there are two sets of values for a, b, c:Either a = 10, b = 19, c = 37 or a = 15, b = 22, c = 29.But the problem might expect both solutions, or perhaps only one.Wait, in the initial problem statement, it's mentioned that the sums form an arithmetic sequence from smallest to largest. So, both orderings are valid, hence both solutions are acceptable.Therefore, the values of a, b, c are either 10, 19, 37 or 15, 22, 29.But let me check if there are any other constraints.Wait, in the first solution, a = 10, b = 19, c = 37.Check if a + b = 29 < 1 + c = 38. Yes, so the order is 1 + a, 1 + b, a + b, 1 + c, a + c, b + c.In the second solution, a = 15, b = 22, c = 29.Check if a + b = 37 > 1 + c = 30. Yes, so the order is 1 + a, 1 + b, 1 + c, a + b, a + c, b + c.Therefore, both are valid.So, the answer is that there are two possible sets of values for a, b, c: (10, 19, 37) and (15, 22, 29).But the problem says "find the values of a, b, c," which might imply that there is a unique solution. However, based on the analysis, there are two solutions.Alternatively, perhaps I made a mistake in assuming both cases are valid.Wait, let me check the total sum of the arithmetic sequence.In the first case, the sums are 11, 20, 29, 38, 47, 56, which sum to 201.In the second case, the sums are 16, 23, 30, 37, 44, 51, which also sum to 201.So, both are correct.Therefore, the problem has two solutions.But perhaps the problem expects the smallest possible a, or the largest possible a.But since the problem doesn't specify, both are valid.Therefore, the values of a, b, c are either 10, 19, 37 or 15, 22, 29.But to present the answer, I think both solutions should be mentioned.However, in the initial problem statement, it's mentioned that the sums form an arithmetic sequence from smallest to largest. So, both orderings are valid, hence both solutions are acceptable.Therefore, the possible values are:Either a = 10, b = 19, c = 37 or a = 15, b = 22, c = 29.But to be thorough, let me check if there are any other possible values.Wait, in the first case, d = 9, leading to a = 10, b = 19, c = 37.In the second case, d = 7, leading to a = 15, b = 22, c = 29.These are the only two possible values of d that satisfy the conditions.Therefore, these are the only two solutions.So, the final answer is that the values of a, b, c are either 10, 19, 37 or 15, 22, 29.

💡Okay, so I've got this problem here: Let a, b, and c be positive real numbers such that their product abc equals 1. I need to find the minimum value of the expression (2a)/b + (3b)/c + (4c)/a. Hmm, okay, let's see how to approach this.First off, I remember that when dealing with expressions that have variables in both the numerator and denominator, and especially when there's a condition like abc = 1, the AM-GM inequality might be useful. AM-GM is often helpful for finding minima or maxima of such expressions. Let me recall the AM-GM inequality: for non-negative real numbers, the arithmetic mean is always greater than or equal to the geometric mean, right? And equality holds when all the numbers are equal.So, maybe I can apply AM-GM to the three terms in the expression. Let me write them down again: (2a)/b, (3b)/c, and (4c)/a. If I consider these three terms, their arithmetic mean would be [(2a)/b + (3b)/c + (4c)/a]/3, and their geometric mean would be the cube root of [(2a)/b * (3b)/c * (4c)/a]. Let me compute that product inside the geometric mean first. Multiplying the three terms together:(2a)/b * (3b)/c * (4c)/a. Let's see, the a in the numerator of the first term cancels with the a in the denominator of the third term. Similarly, the b in the denominator of the first term cancels with the b in the numerator of the second term, and the c in the denominator of the second term cancels with the c in the numerator of the third term. So, all the variables cancel out, and we're left with 2 * 3 * 4, which is 24.So, the geometric mean is the cube root of 24. Therefore, according to AM-GM, the arithmetic mean is greater than or equal to the cube root of 24. Multiplying both sides by 3, we get that the sum (2a)/b + (3b)/c + (4c)/a is greater than or equal to 3 times the cube root of 24.Wait, but the problem is asking for the minimum value. So, does that mean the minimum is 3 times the cube root of 24? Let me compute that. The cube root of 24 is approximately 2.884, so 3 times that is approximately 8.652. Hmm, but I suspect the exact value might be a nicer number. Let me see if I can express 3 times the cube root of 24 in a different way.24 can be written as 8 * 3, so cube root of 24 is cube root of 8 times cube root of 3, which is 2 times cube root of 3. So, 3 times that would be 3 * 2 * cube root of 3, which is 6 times cube root of 3. Hmm, that's still not a very clean number. Maybe I made a mistake in my approach.Wait, let me double-check the AM-GM application. AM-GM says that the arithmetic mean is greater than or equal to the geometric mean, so [(2a)/b + (3b)/c + (4c)/a]/3 >= cube root[(2a)/b * (3b)/c * (4c)/a]. That part seems correct. The product inside is indeed 24, so the geometric mean is cube root of 24. So, the inequality is correct.But maybe the minimum isn't achieved at the AM-GM equality condition? Or perhaps I need to adjust the terms so that they can be equal, given the constraint abc = 1. Because in AM-GM, equality holds when all the terms are equal. So, for equality, we need (2a)/b = (3b)/c = (4c)/a.Let me set each term equal to some constant k. So, (2a)/b = k, (3b)/c = k, and (4c)/a = k. Then, from the first equation, 2a = kb, so a = (kb)/2. From the second equation, 3b = kc, so c = (3b)/k. From the third equation, 4c = ka, so c = (ka)/4.Wait, so from the second equation, c is (3b)/k, and from the third equation, c is (ka)/4. Therefore, (3b)/k = (ka)/4. Let me write that down: (3b)/k = (ka)/4. Cross-multiplying, 12b = k^2 a. But from the first equation, a = (kb)/2. So, substituting that into the equation, 12b = k^2 * (kb)/2. Simplifying, 12b = (k^3 b)/2. Dividing both sides by b (since b is positive and non-zero), we get 12 = (k^3)/2, so k^3 = 24, which means k = cube root of 24. So, k is indeed the cube root of 24, which is approximately 2.884.So, each term equals cube root of 24, and the sum is 3 times that, which is approximately 8.652. But earlier, I thought maybe the minimum is 9. Hmm, perhaps I need to check if my assumption that equality holds is correct, considering the constraint abc = 1.Let me try to find the values of a, b, c that satisfy both the equality conditions of AM-GM and the constraint abc = 1. So, from the earlier equations:(2a)/b = (3b)/c = (4c)/a = k.We can express a, b, c in terms of k.From (2a)/b = k, we have a = (kb)/2.From (3b)/c = k, we have c = (3b)/k.From (4c)/a = k, we have c = (ka)/4.But we already have c in terms of b, so let's substitute a from the first equation into the third equation. So, c = (k*(kb)/2)/4 = (k^2 b)/8.But from the second equation, c = (3b)/k. Therefore, (k^2 b)/8 = (3b)/k. Since b is positive and non-zero, we can divide both sides by b, getting (k^2)/8 = 3/k. Multiplying both sides by 8k, we get k^3 = 24, which again gives k = cube root of 24.So, now, let's express a, b, c in terms of k. Let me choose b as a variable and express a and c in terms of b.From a = (kb)/2, and c = (3b)/k.Now, the constraint is abc = 1. Let's substitute a and c in terms of b.So, a * b * c = (kb/2) * b * (3b/k) = (kb/2) * b * (3b/k).Simplify this: the k in the numerator and denominator cancels out, as does the 1/2 and 3. So, (kb/2) * b * (3b/k) = (3/2) * b^3.Set this equal to 1: (3/2) * b^3 = 1. Therefore, b^3 = 2/3, so b = cube root of (2/3).Then, a = (k * b)/2. Since k = cube root of 24, and b = cube root of (2/3), let's compute a.a = (cube root of 24 * cube root of (2/3)) / 2.Similarly, cube root of 24 is cube root of (8*3) = 2 * cube root of 3.Cube root of (2/3) is cube root of 2 divided by cube root of 3.So, multiplying them together: 2 * cube root of 3 * cube root of 2 / cube root of 3 = 2 * cube root of 2.Therefore, a = (2 * cube root of 2) / 2 = cube root of 2.Similarly, c = (3b)/k. Let's compute that.c = (3 * cube root of (2/3)) / cube root of 24.Again, cube root of 24 is 2 * cube root of 3, so:c = (3 * cube root of (2/3)) / (2 * cube root of 3).Simplify numerator: 3 * cube root of 2 / cube root of 3.Denominator: 2 * cube root of 3.So, c = (3 * cube root of 2 / cube root of 3) / (2 * cube root of 3) = (3 / (2 * cube root of 3 * cube root of 3)) * cube root of 2.But cube root of 3 * cube root of 3 is cube root of 9, which is 3^(2/3). So, 3 / (2 * 3^(2/3)) = (3^(1 - 2/3)) / 2 = (3^(1/3)) / 2.Therefore, c = (3^(1/3) / 2) * cube root of 2 = (cube root of 3 * cube root of 2) / 2 = cube root of 6 / 2.Wait, let me check that again. Maybe I messed up the exponents.Alternatively, perhaps it's easier to compute numerically.But regardless, the key point is that with these values of a, b, c, the expression (2a)/b + (3b)/c + (4c)/a equals 3 times cube root of 24, which is approximately 8.652. However, earlier, I thought maybe the minimum is 9. Is that possible?Wait, let me think again. Maybe I made a mistake in assuming that the AM-GM gives the exact minimum here. Perhaps I need to use a different approach, like Lagrange multipliers, to find the exact minimum considering the constraint abc = 1.Alternatively, maybe I can use substitution based on the constraint. Since abc = 1, I can express one variable in terms of the others. For example, let me express c as 1/(ab). Then, substitute this into the expression.So, substituting c = 1/(ab) into the expression:(2a)/b + (3b)/(1/(ab)) + (4*(1/(ab)))/a.Simplify each term:First term: (2a)/b remains as is.Second term: (3b)/(1/(ab)) = 3b * (ab)/1 = 3a b^2.Third term: (4*(1/(ab)))/a = 4/(a^2 b).So, the expression becomes: (2a)/b + 3a b^2 + 4/(a^2 b).Hmm, that seems more complicated. Maybe this isn't the best approach. Alternatively, perhaps I can set variables in terms of exponents to make the expression homogeneous.Wait, another thought: Maybe I can use the method of weighted AM-GM. Since the coefficients 2, 3, 4 are different, perhaps I need to adjust the weights accordingly.Let me consider the expression (2a)/b + (3b)/c + (4c)/a. Let me write it as 2*(a/b) + 3*(b/c) + 4*(c/a).Now, if I think of these as three separate terms with coefficients 2, 3, 4, maybe I can apply the weighted AM-GM inequality. The weighted AM-GM states that for positive real numbers x_i and positive weights w_i, the weighted arithmetic mean is greater than or equal to the weighted geometric mean.In this case, the weights would be 2, 3, 4, and the variables would be (a/b), (b/c), (c/a). So, the weighted AM-GM would be:(2*(a/b) + 3*(b/c) + 4*(c/a)) / (2 + 3 + 4) >= ( (a/b)^2 * (b/c)^3 * (c/a)^4 )^(1/(2+3+4)).Simplify the right-hand side:First, compute the product inside the root:(a/b)^2 * (b/c)^3 * (c/a)^4.Let's break it down:(a^2 / b^2) * (b^3 / c^3) * (c^4 / a^4).Multiply the numerators and denominators:Numerator: a^2 * b^3 * c^4.Denominator: b^2 * c^3 * a^4.Simplify:a^(2 - 4) * b^(3 - 2) * c^(4 - 3) = a^(-2) * b^(1) * c^(1).So, the product is (b c) / a^2.Therefore, the weighted geometric mean is [ (b c) / a^2 ]^(1/9).But from the constraint abc = 1, we can express bc as 1/a. So, substituting that in:[ (1/a) / a^2 ]^(1/9) = [1 / a^3]^(1/9) = (a^(-3))^(1/9) = a^(-1/3).Hmm, but this seems a bit messy. Maybe this approach isn't leading me anywhere useful.Wait, perhaps I should try to make the exponents work out so that the product simplifies to a constant, given the constraint abc = 1. Let me see.Alternatively, maybe I can use substitution to reduce the number of variables. Let me set x = a/b, y = b/c, z = c/a. Then, the expression becomes 2x + 3y + 4z. But note that x * y * z = (a/b)*(b/c)*(c/a) = 1. So, we have the constraint x y z = 1.But now, the problem is to minimize 2x + 3y + 4z with x y z = 1. Hmm, this seems similar to the original problem but with different coefficients. Maybe I can apply AM-GM here as well.Using AM-GM on the three terms 2x, 3y, 4z:(2x + 3y + 4z)/3 >= cube root(2x * 3y * 4z).Compute the product inside the cube root:2x * 3y * 4z = 24 x y z.But since x y z = 1, this becomes 24 * 1 = 24.Therefore, (2x + 3y + 4z)/3 >= cube root(24), so 2x + 3y + 4z >= 3 * cube root(24), which is the same result as before.So, again, the minimum is 3 times cube root of 24, which is approximately 8.652. But earlier, I thought the minimum might be 9. Maybe I need to check if 9 is indeed the minimum or if it's actually approximately 8.652.Wait, let me compute 3 times cube root of 24 more accurately. Cube root of 24 is approximately 2.884, so 3 times that is approximately 8.652. So, it's less than 9. Therefore, the minimum value should be 3 times cube root of 24, which is approximately 8.652.But wait, the problem asks for the minimum value, and it's likely expecting an exact value, not an approximate decimal. So, 3 times cube root of 24 can be written as 3 * 24^(1/3). But 24 can be factored as 8 * 3, so 24^(1/3) is 2 * 3^(1/3). Therefore, 3 * 2 * 3^(1/3) = 6 * 3^(1/3). So, the exact minimum value is 6 times the cube root of 3.Wait, let me confirm that. 24 = 8 * 3, so cube root of 24 is cube root of 8 times cube root of 3, which is 2 * cube root of 3. Therefore, 3 times that is 3 * 2 * cube root of 3 = 6 * cube root of 3. Yes, that's correct.So, the minimum value is 6 times the cube root of 3, which is approximately 8.652. Therefore, the exact minimum value is 6 * 3^(1/3).But wait, earlier I thought maybe the minimum is 9. Let me check if that's possible. Suppose a = b = c = 1, which satisfies abc = 1. Then, the expression becomes (2*1)/1 + (3*1)/1 + (4*1)/1 = 2 + 3 + 4 = 9. So, when a = b = c = 1, the expression equals 9. But according to the AM-GM result, the minimum is approximately 8.652, which is less than 9. Therefore, the minimum is indeed 6 * 3^(1/3), which is less than 9.Wait, but how come when a = b = c = 1, the expression is 9, but the AM-GM suggests a lower value? That must mean that the equality condition of AM-GM isn't satisfied when a = b = c = 1. Because in AM-GM, equality holds when all the terms are equal. So, in our case, for equality, we need (2a)/b = (3b)/c = (4c)/a. But when a = b = c = 1, we have (2*1)/1 = 2, (3*1)/1 = 3, and (4*1)/1 = 4. These are not equal, so equality doesn't hold. Therefore, the minimum isn't achieved at a = b = c = 1, but rather at some other values of a, b, c where the terms are equal.So, the minimum value is indeed 6 * 3^(1/3), which is approximately 8.652. Therefore, the answer should be 6 times the cube root of 3.Wait, but let me double-check my earlier substitution. When I set (2a)/b = (3b)/c = (4c)/a = k, and found that k = cube root of 24, then the sum is 3k = 3 * cube root of 24 = 3 * 2 * cube root of 3 = 6 * cube root of 3. So, that's consistent.Therefore, the minimum value is 6 * cube root of 3, which is approximately 8.652. So, the exact value is 6 * 3^(1/3).But wait, let me see if I can express this in a simpler form. 6 * 3^(1/3) is already pretty simple, but maybe it can be written as 3^(4/3) * 2, but that doesn't seem simpler. Alternatively, 6 * 3^(1/3) is fine.Wait, but let me check if my earlier substitution for a, b, c actually satisfies abc = 1. Earlier, I found that a = cube root of 2, b = cube root of (2/3), and c = cube root of 6 / 2. Let me compute abc:a = 2^(1/3), b = (2/3)^(1/3), c = (6)^(1/3)/2.So, abc = 2^(1/3) * (2/3)^(1/3) * (6)^(1/3)/2.Simplify:First, multiply 2^(1/3) and (2/3)^(1/3): that's (2 * 2/3)^(1/3) = (4/3)^(1/3).Then, multiply by (6)^(1/3): (4/3)^(1/3) * (6)^(1/3) = (4/3 * 6)^(1/3) = (8)^(1/3) = 2.Then, divide by 2: 2 / 2 = 1. So, abc = 1, which satisfies the constraint. Therefore, the values I found are correct, and the minimum value is indeed 6 * 3^(1/3).Wait, but 6 * 3^(1/3) is approximately 6 * 1.442 = approximately 8.652, which is less than 9. So, the minimum value is indeed 6 * 3^(1/3), not 9.But earlier, when I set a = b = c = 1, I got 9, which is higher than the minimum. So, the minimum is achieved when a, b, c are not all equal, but rather when they satisfy the equality condition of AM-GM, i.e., when (2a)/b = (3b)/c = (4c)/a.Therefore, the minimum value is 6 times the cube root of 3.Wait, but let me see if I can write this in a different form. 6 * 3^(1/3) can also be written as 3^(4/3) * 2, but that's not necessarily simpler. Alternatively, perhaps it's better to leave it as 6 * 3^(1/3).Alternatively, since 3^(1/3) is the cube root of 3, we can write it as 6∛3.Yes, that's a concise way to write it. So, the minimum value is 6∛3.Therefore, the answer is 6 times the cube root of 3, which is approximately 8.652, but exactly 6∛3.Wait, but let me confirm once more. If I plug in the values of a, b, c that I found earlier into the original expression, do I get 6∛3?Earlier, I found a = cube root of 2, b = cube root of (2/3), c = cube root of 6 / 2.Let me compute each term:(2a)/b = 2 * cube root of 2 / cube root of (2/3).Simplify: 2 * (2^(1/3)) / ((2/3)^(1/3)) = 2 * (2^(1/3) * (3/2)^(1/3)) ) = 2 * ( (2 * 3/2)^(1/3) ) = 2 * (3^(1/3)).Similarly, (3b)/c = 3 * cube root of (2/3) / (cube root of 6 / 2).Simplify: 3 * ( (2/3)^(1/3) ) / ( (6)^(1/3) / 2 ) = 3 * (2/3)^(1/3) * 2 / 6^(1/3).Note that 6^(1/3) = (2*3)^(1/3) = 2^(1/3)*3^(1/3).So, 3 * (2/3)^(1/3) * 2 / (2^(1/3)*3^(1/3)).Simplify numerator: 3 * 2 * (2/3)^(1/3) = 6 * (2/3)^(1/3).Denominator: 2^(1/3)*3^(1/3).So, the whole term becomes 6 * (2/3)^(1/3) / (2^(1/3)*3^(1/3)).But (2/3)^(1/3) = 2^(1/3)/3^(1/3), so substituting:6 * (2^(1/3)/3^(1/3)) / (2^(1/3)*3^(1/3)) = 6 / (3^(1/3) * 3^(1/3)) = 6 / 3^(2/3).But 3^(2/3) is the same as (3^(1/3))^2, so 6 / (3^(1/3))^2.Alternatively, 6 * 3^(-2/3).But 3^(-2/3) is 1/(3^(2/3)).Wait, but earlier, I thought each term equals k = cube root of 24. Let me compute k:k = cube root of 24 = cube root of (8*3) = 2 * cube root of 3.So, each term should equal 2 * cube root of 3.Let me check the first term: (2a)/b = 2 * cube root of 2 / cube root of (2/3).As above, this simplifies to 2 * cube root of 3.Similarly, the second term: (3b)/c = 3 * cube root of (2/3) / (cube root of 6 / 2).Simplify: 3 * (2/3)^(1/3) * 2 / (6)^(1/3).As above, this becomes 6 * (2/3)^(1/3) / (6)^(1/3).But (2/3)^(1/3) / (6)^(1/3) = (2/3 / 6)^(1/3) = (2/(3*6))^(1/3) = (2/18)^(1/3) = (1/9)^(1/3) = 1/9^(1/3).But 9^(1/3) is 3^(2/3), so 1/9^(1/3) = 3^(-2/3).Therefore, the second term is 6 * 3^(-2/3).Wait, but 6 * 3^(-2/3) is equal to 6 / 3^(2/3) = 6 / (3^(1/3))^2.But earlier, I thought each term equals 2 * cube root of 3. Let me compute 2 * cube root of 3:2 * 3^(1/3) ≈ 2 * 1.442 ≈ 2.884.Now, 6 / 3^(2/3) = 6 / (3^(1/3))^2 ≈ 6 / (1.442)^2 ≈ 6 / 2.08 ≈ 2.884. So, yes, both terms equal approximately 2.884, which is cube root of 24.Therefore, each term equals cube root of 24, and the sum is 3 times that, which is 3 * 2.884 ≈ 8.652, which is 6 * 3^(1/3).Therefore, the minimum value is indeed 6 * cube root of 3, which is approximately 8.652.So, to summarize, by applying the AM-GM inequality to the three terms (2a)/b, (3b)/c, and (4c)/a, we find that their sum is at least 3 times the cube root of their product, which is 24. This gives a lower bound of 3 * cube root of 24, which simplifies to 6 * cube root of 3. We then verify that this minimum is achievable by finding specific values of a, b, c that satisfy both the equality condition of AM-GM and the constraint abc = 1. Therefore, the minimum value of the expression is 6 * cube root of 3.**Final Answer**The minimum value is boxed{6sqrt[3]{3}}.

💡Okay, so I have this problem where a bus and a cyclist start from point A at 13:00 heading towards point B, which is 4 km away. The bus arrives at B, immediately turns back, and meets the cyclist at point C at 13:10. Then, after going back to A, the bus heads to B again and catches up with the cyclist at point D, which is 2/3 km from C. I need to find the speed of the bus.Alright, let's break this down step by step.First, let's note the given information:- Distance between A and B: 4 km.- Both the bus and cyclist start at 13:00.- They meet at point C at 13:10, which is 10 minutes after departure.- Then, the bus goes back to A, starts again towards B, and meets the cyclist at D, which is 2/3 km from C.Let me denote the speed of the cyclist as v1 (km/h) and the speed of the bus as v2 (km/h). Both speeds are constant.First, let's consider the time from 13:00 to 13:10, which is 10 minutes or 1/6 of an hour.In this time, the bus travels from A to B and then back to C. The cyclist travels from A to C.So, the distance the bus covers in this time is the distance from A to B plus the distance from B to C. The cyclist covers the distance from A to C.Let me denote the distance from A to C as x km. Therefore, the distance from C to B is (4 - x) km.So, the total distance the bus covers in 10 minutes is 4 + (4 - x) = 8 - x km.Since the bus's speed is v2, the distance covered is v2 * (1/6) = (8 - x).Similarly, the cyclist covers x km in 10 minutes, so x = v1 * (1/6).So, we have two equations:1. v2 * (1/6) = 8 - x2. v1 * (1/6) = xFrom equation 2, x = v1 / 6.Plugging this into equation 1:v2 / 6 = 8 - (v1 / 6)Multiply both sides by 6:v2 = 48 - v1So, v1 + v2 = 48. That's one equation.Now, moving on to the next part. After meeting at C, the bus goes back to A, which is x km from C. Then, it immediately starts again towards B and catches up with the cyclist at D, which is 2/3 km from C.So, from point C, the bus goes back to A, which is x km, then from A to D, which is (x + 2/3) km? Wait, no. Let me think.Wait, point D is 2/3 km from C, so depending on the direction, it could be towards B or towards A. But since the bus is going from A to B again, and the cyclist is still going towards B, D must be between C and B, right? Because the bus is catching up from behind. So, D is 2/3 km beyond C towards B.So, the distance from C to D is 2/3 km.So, when the bus starts from A again, it has to cover the distance from A to D, which is x + 2/3 km.Meanwhile, the cyclist, after meeting at C, continues towards B, covering the distance from C to D, which is 2/3 km, in the same time it takes the bus to go from A to D.Wait, let me clarify the timeline.At 13:10, they meet at C. Then, the bus goes back to A, which takes some time, then immediately starts towards B again and catches up with the cyclist at D at some later time.So, the time between 13:10 and the time they meet at D is the time it takes for the bus to go from C to A and then from A to D, while the cyclist goes from C to D.Let me denote the time between 13:10 and the meeting at D as t hours.So, during this time t:- The bus travels from C to A, which is x km, at speed v2, taking x / v2 hours. Then, immediately, the bus travels from A to D, which is (x + 2/3) km, at speed v2, taking (x + 2/3) / v2 hours.Wait, but that would mean the total time t is x / v2 + (x + 2/3) / v2 = (2x + 2/3) / v2.But the cyclist, during this time t, travels from C to D, which is 2/3 km, at speed v1, so t = (2/3) / v1.Therefore, we have:(2x + 2/3) / v2 = (2/3) / v1Simplify:(2x + 2/3) / v2 = 2/(3v1)Multiply both sides by v2:2x + 2/3 = (2v2)/(3v1)But from earlier, we have v1 + v2 = 48, so v1 = 48 - v2.Plugging that into the equation:2x + 2/3 = (2v2)/(3*(48 - v2))Also, from earlier, x = v1 / 6 = (48 - v2)/6.So, x = (48 - v2)/6.Plugging x into the equation:2*(48 - v2)/6 + 2/3 = (2v2)/(3*(48 - v2))Simplify:(96 - 2v2)/6 + 2/3 = (2v2)/(3*(48 - v2))Simplify each term:(96 - 2v2)/6 = 16 - (v2)/32/3 remains as is.So, left side becomes:16 - (v2)/3 + 2/3 = 16 - (v2)/3 + 2/3Combine constants:16 + 2/3 - (v2)/3 = (48/3 + 2/3) - (v2)/3 = (50/3) - (v2)/3So, left side is (50 - v2)/3.Right side is (2v2)/(3*(48 - v2)).So, equation becomes:(50 - v2)/3 = (2v2)/(3*(48 - v2))Multiply both sides by 3 to eliminate denominators:50 - v2 = (2v2)/(48 - v2)Multiply both sides by (48 - v2):(50 - v2)(48 - v2) = 2v2Expand the left side:50*48 - 50v2 -48v2 + v2^2 = 2v2Calculate 50*48: 2400So:2400 - 98v2 + v2^2 = 2v2Bring all terms to left side:v2^2 - 98v2 - 2v2 + 2400 = 0Combine like terms:v2^2 - 100v2 + 2400 = 0Now, solve this quadratic equation:v2^2 - 100v2 + 2400 = 0Using quadratic formula:v2 = [100 ± sqrt(10000 - 9600)] / 2Calculate discriminant:sqrt(400) = 20So,v2 = [100 ± 20]/2So, two solutions:v2 = (100 + 20)/2 = 120/2 = 60v2 = (100 - 20)/2 = 80/2 = 40Now, check which one makes sense.If v2 = 60 km/h, then v1 = 48 - 60 = -12 km/h, which is impossible.So, v2 must be 40 km/h, then v1 = 48 - 40 = 8 km/h.Let me verify this.So, v2 = 40 km/h, v1 = 8 km/h.First, from 13:00 to 13:10 (10 minutes):Bus travels from A to B (4 km) and back to C.Time taken by bus to go from A to B: 4 / 40 = 0.1 hours = 6 minutes.Then, it turns back and meets the cyclist at C at 13:10, which is 10 minutes after departure.So, the bus spends 6 minutes going to B, then 4 minutes coming back to C.In 4 minutes, the bus covers distance from B to C: 40 km/h * (4/60) h = (40 * 4)/60 = 160/60 = 8/3 ≈ 2.6667 km.So, distance from C to B is 8/3 km, so distance from A to C is 4 - 8/3 = 4/3 ≈ 1.3333 km.Meanwhile, the cyclist in 10 minutes (1/6 hour) covers 8 km/h * 1/6 = 4/3 km, which matches the distance from A to C.Good.Now, after meeting at C, the bus goes back to A, which is 4/3 km away. Time taken: (4/3)/40 = (4/3)/40 = 1/30 hours = 2 minutes.So, the bus arrives at A at 13:12.Then, it immediately starts towards B again. The cyclist is moving towards B at 8 km/h.At 13:12, the cyclist is at C, which is 4/3 km from A, and continues towards B.The bus starts from A at 13:12 towards B.They meet at D, which is 2/3 km from C towards B.So, distance from C to D is 2/3 km, so distance from A to D is 4/3 + 2/3 = 6/3 = 2 km.So, the bus needs to cover 2 km from A to D, while the cyclist needs to cover 2/3 km from C to D.Time taken for the bus: 2 / 40 = 1/20 hours = 3 minutes.Time taken for the cyclist: (2/3) / 8 = (2/3)/8 = 1/12 hours ≈ 5 minutes.Wait, this doesn't match. If the bus takes 3 minutes and the cyclist takes 5 minutes, they wouldn't meet at the same time.Wait, maybe I made a mistake.Wait, the bus starts at 13:12, and the cyclist is at C at 13:12, moving towards B.The distance between the bus (at A) and the cyclist (at C) is 4/3 km.But the bus is going towards B, and the cyclist is also going towards B.Wait, actually, the distance between them is 4/3 km, but they are moving in the same direction.Wait, no, the bus is at A, and the cyclist is at C, which is 4/3 km from A. So, the distance between them is 4/3 km, but they are moving towards each other? No, the bus is going from A to B, and the cyclist is going from C to B, so they are moving in the same direction.Wait, that complicates things.Wait, actually, the bus is going from A to B, and the cyclist is going from C to B, so they are moving in the same direction. Therefore, the bus is behind the cyclist and needs to catch up.So, the distance between them is 4/3 km, and the relative speed is v2 - v1 = 40 - 8 = 32 km/h.So, time to catch up is (4/3) / 32 = (4/3)/32 = 1/24 hours ≈ 2.5 minutes.But according to the problem, they meet at D, which is 2/3 km from C. So, the cyclist travels 2/3 km from C to D in that time.So, time taken is (2/3)/8 = 1/12 hours ≈ 5 minutes.Wait, but according to relative speed, it should take 1/24 hours ≈ 2.5 minutes.This inconsistency suggests an error in my reasoning.Wait, perhaps I need to consider the total distance the bus travels from A to D and the cyclist from C to D.Wait, the bus goes from A to D, which is 2 km, while the cyclist goes from C to D, which is 2/3 km.But the bus started at 13:12, and the cyclist was at C at 13:12.So, the time taken for the bus to reach D is 2 / 40 = 1/20 hours = 3 minutes.In that time, the cyclist would have traveled 8 * (1/20) = 0.4 km, which is 2/5 km, not 2/3 km.Hmm, discrepancy here.Wait, perhaps I need to set up the equations correctly.Let me denote t as the time from 13:10 to the meeting at D.Wait, no, from 13:10, the bus goes back to A, which takes x / v2 hours, then immediately starts towards B and meets the cyclist at D, which takes (x + 2/3) / v2 hours.But the cyclist, from 13:10, travels from C to D, which is 2/3 km, taking (2/3)/v1 hours.So, the total time from 13:10 to meeting at D is t = x / v2 + (x + 2/3)/v2 = (2x + 2/3)/v2.But also, t = (2/3)/v1.So, (2x + 2/3)/v2 = (2/3)/v1.From earlier, x = v1 / 6.So, plug x into the equation:(2*(v1 / 6) + 2/3)/v2 = (2/3)/v1Simplify numerator:( (2v1)/6 + 2/3 ) = (v1/3 + 2/3 ) = (v1 + 2)/3So, equation becomes:(v1 + 2)/3 / v2 = (2/3)/v1Multiply both sides by 3:(v1 + 2)/v2 = 2 / v1Cross-multiply:v1(v1 + 2) = 2v2But from earlier, v1 + v2 = 48, so v2 = 48 - v1.Plug into equation:v1(v1 + 2) = 2(48 - v1)Expand left side:v1^2 + 2v1 = 96 - 2v1Bring all terms to left:v1^2 + 4v1 - 96 = 0Solve quadratic:v1 = [-4 ± sqrt(16 + 384)] / 2 = [-4 ± sqrt(400)] / 2 = [-4 ± 20]/2Positive solution:v1 = (16)/2 = 8 km/hSo, v1 = 8 km/h, then v2 = 48 - 8 = 40 km/h.So, the bus speed is 40 km/h.Let me verify again.From 13:00 to 13:10:Bus goes A to B (4 km) in 4/40 = 0.1 h = 6 minutes, then turns back.In the remaining 4 minutes (from 13:06 to 13:10), the bus travels back towards C.Distance covered: 40 * (4/60) = 160/60 = 8/3 ≈ 2.6667 km.So, C is 8/3 km from B, which is 4 - 8/3 = 4/3 ≈ 1.3333 km from A.Cyclist in 10 minutes covers 8 * (10/60) = 8/6 = 4/3 km, which matches.Then, after 13:10, the bus goes back to A, which is 4/3 km away.Time taken: (4/3)/40 = 1/30 h = 2 minutes, arriving at 13:12.Then, the bus starts towards B again, and the cyclist is moving towards B from C.Distance from A to D is 2 km (since D is 2/3 km from C towards B, and C is 4/3 km from A, so 4/3 + 2/3 = 2 km).So, the bus needs to cover 2 km, taking 2/40 = 1/20 h = 3 minutes, arriving at 13:15.The cyclist, from C, needs to cover 2/3 km, taking (2/3)/8 = 1/12 h ≈ 5 minutes, arriving at 13:15.Yes, that matches. So, both meet at D at 13:15.Therefore, the calculations are consistent.So, the speed of the bus is 40 km/h.

💡Okay, so I need to show that the exponential function ( x mapsto exp(x) ) is convex on the entire real line ( mathbb{R} ), and that the natural logarithm function ( x mapsto ln(x) ) is concave on the interval ( (0, +infty) ). Hmm, I remember that convexity and concavity have to do with the shape of the function's graph, but I need to recall the precise definitions and how to prove them.First, let's recall what it means for a function to be convex or concave. A function ( f ) is convex on an interval if, for any two points ( x_1 ) and ( x_2 ) in that interval and any ( lambda in [0,1] ), the following inequality holds:[f(lambda x_1 + (1 - lambda) x_2) leq lambda f(x_1) + (1 - lambda) f(x_2)]This means that the function lies below the straight line connecting any two points on its graph. Conversely, a function is concave if the inequality is reversed:[f(lambda x_1 + (1 - lambda) x_2) geq lambda f(x_1) + (1 - lambda) f(x_2)]So, the function lies above the straight line connecting any two points.But I also remember that there's a relationship between convexity/concavity and the second derivative of a function. If the second derivative of a function is non-negative over an interval, the function is convex on that interval. Similarly, if the second derivative is non-positive, the function is concave.Alright, so maybe I can use the second derivative test to determine the convexity or concavity of these functions. Let's start with the exponential function.**For ( f(x) = exp(x) ):**First, I need to compute the first and second derivatives.The first derivative of ( exp(x) ) is straightforward:[f'(x) = frac{d}{dx} exp(x) = exp(x)]Now, the second derivative is just the derivative of the first derivative:[f''(x) = frac{d}{dx} exp(x) = exp(x)]So, ( f''(x) = exp(x) ). Since the exponential function is always positive for all real numbers ( x ), ( f''(x) > 0 ) for all ( x in mathbb{R} ). According to the second derivative test, if ( f''(x) > 0 ) on an interval, the function is convex on that interval. Therefore, ( f(x) = exp(x) ) is convex on ( mathbb{R} ).That seems straightforward. Now, moving on to the natural logarithm function.**For ( g(x) = ln(x) ):**Again, I'll compute the first and second derivatives.The first derivative of ( ln(x) ) is:[g'(x) = frac{d}{dx} ln(x) = frac{1}{x}]Now, the second derivative is the derivative of ( frac{1}{x} ):[g''(x) = frac{d}{dx} left( frac{1}{x} right) = -frac{1}{x^2}]So, ( g''(x) = -frac{1}{x^2} ). Since ( x ) is in ( (0, +infty) ), ( x^2 ) is always positive, which means ( g''(x) = -frac{1}{x^2} ) is always negative on ( (0, +infty) ).According to the second derivative test, if ( g''(x) < 0 ) on an interval, the function is concave on that interval. Therefore, ( g(x) = ln(x) ) is concave on ( (0, +infty) ).Wait a minute, just to make sure I'm not making a mistake, let me think about the definitions again. Convexity is when the function lies below the chord, and concavity is when it lies above the chord. For ( exp(x) ), since it's convex, it should curve upwards, which makes sense because the exponential function grows rapidly. For ( ln(x) ), it's concave, so it should curve downwards, which also makes sense because the logarithm grows slowly.Let me also visualize both functions. The exponential function ( exp(x) ) indeed has a U-shape that curves upwards, confirming its convexity. On the other hand, the logarithm function ( ln(x) ) has a shape that curves downwards, which aligns with it being concave.Just to be thorough, maybe I can test the convexity and concavity using the definition with specific points.**Testing Convexity of ( exp(x) ):**Let's pick two points, say ( x_1 = 0 ) and ( x_2 = 1 ), and choose ( lambda = 0.5 ).Compute ( f(lambda x_1 + (1 - lambda) x_2) ):[f(0.5 cdot 0 + 0.5 cdot 1) = f(0.5) = exp(0.5) approx 1.6487]Compute ( lambda f(x_1) + (1 - lambda) f(x_2) ):[0.5 cdot exp(0) + 0.5 cdot exp(1) = 0.5 cdot 1 + 0.5 cdot 2.7183 approx 0.5 + 1.3591 = 1.8591]Since ( 1.6487 leq 1.8591 ), the inequality ( f(lambda x_1 + (1 - lambda) x_2) leq lambda f(x_1) + (1 - lambda) f(x_2) ) holds, which is consistent with ( f(x) ) being convex.**Testing Concavity of ( ln(x) ):**Again, pick two points, say ( x_1 = 1 ) and ( x_2 = e ), and choose ( lambda = 0.5 ).Compute ( g(lambda x_1 + (1 - lambda) x_2) ):[g(0.5 cdot 1 + 0.5 cdot e) = gleft( frac{1 + e}{2} right) = lnleft( frac{1 + e}{2} right) approx ln(1.718) approx 0.542]Compute ( lambda g(x_1) + (1 - lambda) g(x_2) ):[0.5 cdot ln(1) + 0.5 cdot ln(e) = 0.5 cdot 0 + 0.5 cdot 1 = 0.5]Since ( 0.542 geq 0.5 ), the inequality ( g(lambda x_1 + (1 - lambda) x_2) geq lambda g(x_1) + (1 - lambda) g(x_2) ) holds, which is consistent with ( g(x) ) being concave.Hmm, that worked out. Maybe try another set of points to be sure.**Another Test for ( exp(x) ):**Let ( x_1 = -1 ) and ( x_2 = 1 ), ( lambda = 0.5 ).Compute ( f(lambda x_1 + (1 - lambda) x_2) ):[f(0.5 cdot (-1) + 0.5 cdot 1) = f(0) = exp(0) = 1]Compute ( lambda f(x_1) + (1 - lambda) f(x_2) ):[0.5 cdot exp(-1) + 0.5 cdot exp(1) approx 0.5 cdot 0.3679 + 0.5 cdot 2.7183 approx 0.18395 + 1.35915 approx 1.5431]Again, ( 1 leq 1.5431 ), so the inequality holds.**Another Test for ( ln(x) ):**Let ( x_1 = 2 ) and ( x_2 = 4 ), ( lambda = 0.5 ).Compute ( g(lambda x_1 + (1 - lambda) x_2) ):[g(0.5 cdot 2 + 0.5 cdot 4) = g(3) = ln(3) approx 1.0986]Compute ( lambda g(x_1) + (1 - lambda) g(x_2) ):[0.5 cdot ln(2) + 0.5 cdot ln(4) approx 0.5 cdot 0.6931 + 0.5 cdot 1.3863 approx 0.34655 + 0.69315 approx 1.0397]Since ( 1.0986 geq 1.0397 ), the inequality holds.Okay, so these tests seem to confirm the convexity and concavity as per the second derivative test. I think I'm confident now that ( exp(x) ) is convex on ( mathbb{R} ) and ( ln(x) ) is concave on ( (0, +infty) ).Just to recap:- For ( exp(x) ), the second derivative is positive everywhere, so it's convex everywhere.- For ( ln(x) ), the second derivative is negative everywhere on its domain, so it's concave there.I don't see any flaws in this reasoning. Maybe I can also think about the graph of these functions. The exponential function is always increasing and its rate of increase is accelerating, which is another way to see it's convex. The logarithm function is increasing but its rate of increase is decelerating, which is characteristic of a concave function.Another thought: convex functions have the property that their chords lie above the function, and concave functions have chords lying below. So for ( exp(x) ), if I draw a straight line between any two points on its graph, the line will be above the function in between those points. For ( ln(x) ), the line will be below the function.Let me also recall that the exponential function is its own derivative, which we've used to compute the second derivative. The logarithm function's derivative is ( 1/x ), which we also used correctly.I think I've covered all the bases here. I've used the second derivative test, checked with specific examples, and thought about the graphical interpretation. I don't see any mistakes in my reasoning.**Final Answer**The function ( x mapsto exp(x) ) is convex on ( mathbb{R} ) and the function ( x mapsto ln(x) ) is concave on ( (0, +infty) ). Thus,[boxed{text{The function } x mapsto exp(x) text{ is convex on } mathbb{R} text{ and } x mapsto ln(x) text{ is concave on } (0, +infty).}]

💡Okay, so Javier is planning his visit to Universal Studios, and he wants to make sure he visits five different attractions. But he's really excited about the Harry Potter ride, so he's decided that he wants to visit that one first. The question is asking how many different orders he can visit the remaining four attractions after the Harry Potter ride.Alright, let's break this down. Javier has five attractions in total, but the first one is fixed as the Harry Potter ride. That means we don't need to worry about that one anymore; it's already decided. So, we're left with four attractions that he can visit in any order after the Harry Potter ride.Now, when we're talking about arranging things in a specific order, that's a permutation problem. Permutations are about the number of ways you can arrange a set of items where the order matters. In this case, the order in which Javier visits the attractions matters because visiting ride A first and ride B second is different from visiting ride B first and ride A second.So, for the first attraction after Harry Potter, Javier has four choices. Once he's visited one of them, he's left with three attractions for his next visit. Then, after visiting the second one, he has two attractions left, and finally, the last attraction is the only one remaining.To find the total number of possible orders, we multiply the number of choices he has at each step. That would be 4 choices for the first attraction, times 3 for the second, times 2 for the third, and times 1 for the last one. So, mathematically, that's 4 × 3 × 2 × 1, which equals 24.Wait, let me make sure I'm not missing anything here. Is there any restriction or condition that I'm overlooking? The problem says he wants to visit five different attractions, and the Harry Potter ride is fixed as the first one. The remaining four can be in any order, so I think I've accounted for all possibilities.Another way to think about it is using factorial notation. The number of ways to arrange n distinct items is n factorial, written as n!. In this case, n is 4 because we're arranging four attractions. So, 4! = 4 × 3 × 2 × 1 = 24. That confirms my earlier calculation.I guess another way to visualize this is by listing out the possibilities, but that would be tedious since there are 24 different orders. However, if I try to list a few, it might help solidify the concept. Let's say the attractions are A, B, C, D, and E, with E being the Harry Potter ride. So, Javier starts with E, and then the possible orders for A, B, C, D are:1. A, B, C, D2. A, B, D, C3. A, C, B, D4. A, C, D, B5. A, D, B, C6. A, D, C, B7. B, A, C, D8. B, A, D, C9. B, C, A, D10. B, C, D, A11. B, D, A, C12. B, D, C, A13. C, A, B, D14. C, A, D, B15. C, B, A, D16. C, B, D, A17. C, D, A, B18. C, D, B, A19. D, A, B, C20. D, A, C, B21. D, B, A, C22. D, B, C, A23. D, C, A, B24. D, C, B, AWow, that's 24 different orders, and each one starts with E (Harry Potter). So, this exhaustive list also shows that there are indeed 24 possible ways Javier can visit the remaining four attractions after the Harry Potter ride.I think I've covered all the bases here. Javier has four attractions left to visit after the first one, and the number of ways to arrange those four is 4 factorial, which is 24. There are no restrictions mentioned besides the first one being fixed, so I don't see any reason to adjust this number further.Just to recap, the key steps are:1. Identify that the first attraction is fixed (Harry Potter ride).2. Recognize that the remaining four attractions can be arranged in any order.3. Use permutations to calculate the number of possible arrangements.4. Calculate 4! = 24.This makes sense because each step reduces the number of choices by one, and multiplying them together gives the total number of permutations.I also thought about whether there's a different way to approach this problem, maybe using combinations instead of permutations, but combinations don't consider order, and since the order of visiting attractions matters here, permutations are the correct approach.In summary, Javier has 24 different orders in which he can visit the remaining four attractions after starting with the Harry Potter ride.