💡Okay, so I have this problem here about finding the limit of a function as Δx approaches 0. The function given is f(x), and it's differentiable with f'(1) = 1. The limit expression is [f(1 - Δx) - f(1)] / (-Δx). Hmm, I need to figure out what this limit equals.First, I remember that the derivative of a function at a point is defined as the limit of [f(a + h) - f(a)] / h as h approaches 0. In this case, our a is 1, and h is Δx. But in the given limit, instead of f(1 + Δx), we have f(1 - Δx). That seems a bit different. So maybe I can manipulate the expression to make it look more like the standard derivative definition.Let me try substituting Δx with another variable to see if that helps. Let's say I let h = -Δx. Then, as Δx approaches 0, h also approaches 0. Substituting, the expression becomes [f(1 - (-h)) - f(1)] / (-(-h)) which simplifies to [f(1 + h) - f(1)] / h. Oh, that looks exactly like the definition of the derivative of f at 1!So, the limit [f(1 - Δx) - f(1)] / (-Δx) as Δx approaches 0 is the same as the derivative of f at 1, which is given as 1. Therefore, the value of the limit should be 1.Wait, let me double-check that substitution. If h = -Δx, then Δx = -h. So, replacing Δx with -h in the original limit:[f(1 - (-h)) - f(1)] / (-(-h)) = [f(1 + h) - f(1)] / h. Yep, that's correct. So as h approaches 0, this limit is just f'(1), which is 1.I think that makes sense. The negative sign in the denominator cancels out the negative sign in the argument of f, turning it into a standard derivative expression. So the limit is indeed equal to the derivative at that point, which is 1.I don't see any mistakes in this reasoning. It's a bit tricky because of the negative Δx, but by substituting variables, it becomes clearer. So I'm confident that the answer is 1.

💡Alright, so I have this function f(x) = -1/2 x² + 1, and I need to do two things: first, find the equation of the tangent line to this function that passes through the point (1,1). Second, I need to calculate the area of the closed figure formed by the graph of f(x) and the line y = -1. Hmm, okay, let's take this step by step.Starting with the first part: finding the tangent line. I remember that the equation of a tangent line at a point x₀ on the curve is given by y = f(x₀) + f’(x₀)(x - x₀), where f’(x₀) is the derivative of f at x₀. So, I need to find the derivative of f(x) first.f(x) = -1/2 x² + 1, so f’(x) would be the derivative with respect to x. The derivative of -1/2 x² is -x, and the derivative of 1 is 0. So, f’(x) = -x. That means at any point x₀, the slope of the tangent line is -x₀.So, the equation of the tangent line at x₀ is y = f(x₀) + f’(x₀)(x - x₀). Plugging in f(x₀) and f’(x₀), we get:y = (-1/2 x₀² + 1) + (-x₀)(x - x₀)Simplify that:y = (-1/2 x₀² + 1) - x₀ x + x₀²Combine like terms:y = (-1/2 x₀² + x₀²) + 1 - x₀ xWhich simplifies to:y = (1/2 x₀²) + 1 - x₀ xSo, the equation of the tangent line is y = -x₀ x + (1/2 x₀² + 1)Now, this tangent line needs to pass through the point (1,1). So, if we plug x = 1 and y = 1 into the equation, it should satisfy.So, 1 = -x₀ * 1 + (1/2 x₀² + 1)Simplify:1 = -x₀ + (1/2 x₀² + 1)Subtract 1 from both sides:0 = -x₀ + (1/2 x₀²)So, 0 = (1/2 x₀²) - x₀Multiply both sides by 2 to eliminate the fraction:0 = x₀² - 2x₀Factor:0 = x₀(x₀ - 2)So, x₀ = 0 or x₀ = 2Therefore, there are two possible points of tangency: at x₀ = 0 and x₀ = 2.Let me find the equations of these tangent lines.First, for x₀ = 0:The equation is y = -0 * x + (1/2 * 0² + 1) = 0 + 1 = 1So, the tangent line is y = 1.Second, for x₀ = 2:The equation is y = -2x + (1/2 * 4 + 1) = -2x + (2 + 1) = -2x + 3So, the tangent line is y = -2x + 3.Therefore, there are two tangent lines to the curve f(x) that pass through (1,1): y = 1 and y = -2x + 3.Wait, but the problem says "the tangent line," implying maybe just one? Hmm, but it's possible to have two tangent lines from a point outside the curve. Let me check if (1,1) is inside or outside the curve.Looking at f(x) = -1/2 x² + 1, which is a downward-opening parabola with vertex at (0,1). So, the point (1,1) is on the same horizontal level as the vertex but shifted right by 1 unit. Since the parabola opens downward, (1,1) is actually on the curve because f(1) = -1/2 * 1 + 1 = 1/2. Wait, no, f(1) is -1/2*(1)^2 + 1 = -1/2 + 1 = 1/2. So, (1,1) is above the curve at x=1. So, it's outside the curve. Therefore, it's possible to have two tangent lines from this point to the parabola.So, both y=1 and y=-2x+3 are valid tangent lines passing through (1,1). Okay, that makes sense.Now, moving on to the second part: calculating the area of the closed figure formed by the graph of f(x) and the line y = -1.First, I need to find where these two graphs intersect. So, set f(x) equal to -1:-1/2 x² + 1 = -1Subtract 1 from both sides:-1/2 x² = -2Multiply both sides by -2:x² = 4So, x = ±2Therefore, the points of intersection are at x = -2 and x = 2. So, the region is bounded between x = -2 and x = 2.Now, to find the area between the two curves from x = -2 to x = 2. The area can be found by integrating the difference between the upper function and the lower function over this interval.Looking at the functions: f(x) = -1/2 x² + 1 is a downward-opening parabola with vertex at (0,1), and y = -1 is a horizontal line below the vertex.So, between x = -2 and x = 2, f(x) is above y = -1. Therefore, the area A is the integral from -2 to 2 of [f(x) - (-1)] dx.So, A = ∫ from -2 to 2 [(-1/2 x² + 1) - (-1)] dxSimplify the integrand:(-1/2 x² + 1) + 1 = -1/2 x² + 2So, A = ∫ from -2 to 2 (-1/2 x² + 2) dxSince the function is even (symmetric about the y-axis), we can compute the integral from 0 to 2 and multiply by 2.So, A = 2 * ∫ from 0 to 2 (-1/2 x² + 2) dxCompute the integral:∫ (-1/2 x² + 2) dx = (-1/2)*(x³/3) + 2x + C = (-1/6)x³ + 2x + CEvaluate from 0 to 2:At x=2: (-1/6)*(8) + 2*(2) = (-4/3) + 4 = (-4/3) + (12/3) = 8/3At x=0: 0 + 0 = 0So, the integral from 0 to 2 is 8/3.Multiply by 2: A = 2*(8/3) = 16/3Therefore, the area is 16/3.Wait, let me double-check the integration steps.Integrand: -1/2 x² + 2Integral: (-1/2)*(x³/3) + 2x = (-1/6)x³ + 2xAt x=2: (-1/6)*(8) + 4 = (-4/3) + 4 = (-4/3 + 12/3) = 8/3At x=0: 0 + 0 = 0So, the integral from -2 to 2 is 2*(8/3) = 16/3. Yes, that seems correct.Alternatively, without using symmetry, compute the integral from -2 to 2:∫ from -2 to 2 (-1/2 x² + 2) dx = [(-1/6)x³ + 2x] from -2 to 2At x=2: (-1/6)*(8) + 4 = (-4/3) + 4 = 8/3At x=-2: (-1/6)*(-8) + (-4) = (4/3) - 4 = (4/3 - 12/3) = (-8/3)Subtracting: 8/3 - (-8/3) = 16/3Same result. So, the area is indeed 16/3.Therefore, the area of the closed figure is 16/3.**Final Answer**1. The equations of the tangent lines are boxed{y = 1} and boxed{y = -2x + 3}.2. The area of the closed figure is boxed{dfrac{16}{3}}.

💡Okay, so I have this problem where I need to find the length of diagonal DA in a regular hexagon. The hexagon is shown in the Asymptote code, and there's a label saying "10" near one of the sides, which I assume is the length of each side of the hexagon. The Asymptote code also labels points A and D, and there's a diagonal drawn between them. First, I remember that in a regular hexagon, all sides are equal, and all internal angles are 120 degrees. I also recall that a regular hexagon can be divided into six equilateral triangles by drawing lines from the center to each vertex. This might be useful because equilateral triangles have all sides equal and all angles 60 degrees, which could help in figuring out the lengths of diagonals.Looking at the Asymptote code, it seems like the hexagon is drawn with specific coordinates. The points are (1,0), (3,0), (4,1.732), (3,3.464), (1,3.464), (0,1.732), and back to (1,0). There's also a diagonal drawn from (1,0) to (1,3.464), which is labeled as DA. The label "10" is near the point (3.5,2.598), which is probably the midpoint of one of the sides, indicating that the side length is 10 units.Wait, hold on. If the side length is 10, but in the coordinates, the distance between (1,0) and (3,0) is 2 units. That seems inconsistent. Maybe the Asymptote code is just a scaled-down version for drawing purposes, and the actual problem states that the side length is 10. So, I should proceed with the side length being 10 units.In a regular hexagon, there are different types of diagonals. The shortest diagonals connect vertices that are two apart, and the longest diagonals connect vertices that are three apart. In this case, since the hexagon is labeled A to D, and looking at the coordinates, A is at (1,0) and D is at (1,3.464). So, in the drawing, DA is a vertical line. But in a regular hexagon, the diagonals can be of different lengths depending on how many vertices they skip.Wait, in a regular hexagon, the maximum distance between two vertices is twice the side length, which would be the distance between two opposite vertices. So, if the side length is 10, the maximum diagonal should be 20 units. But in the Asymptote code, the distance from (1,0) to (1,3.464) is approximately 3.464 units, which is roughly 2 times the square root of 3, because sqrt(3) is about 1.732. So, 2*sqrt(3) is about 3.464. Hmm, that suggests that in the drawing, the side length might actually be 2 units, but the problem states it's 10. So, maybe the Asymptote code is scaled down by a factor of 5, because 2*5 is 10.So, if the drawing has a side length of 2 units, but the actual problem has a side length of 10, then all the lengths in the drawing need to be multiplied by 5 to get the actual lengths. Therefore, the diagonal DA in the drawing is approximately 3.464 units, which is 2*sqrt(3). So, scaling that up by 5, the actual length would be 10*sqrt(3). But let me verify this without relying solely on the scaling factor. In a regular hexagon, the distance between two opposite vertices (the maximum diagonal) is 2 times the side length. However, in this case, DA is not the maximum diagonal. Looking at the coordinates, A is at (1,0) and D is at (1,3.464). So, in the drawing, DA is a vertical line, but in a regular hexagon, the vertical diagonal would actually be the same as the maximum diagonal if the hexagon is oriented with a flat side at the bottom. But in this case, the hexagon seems to be oriented with a vertex at the bottom.Wait, maybe I'm overcomplicating this. Let me think about the properties of a regular hexagon. In a regular hexagon, the length of the diagonal that connects two vertices with one vertex in between (i.e., skipping one vertex) is equal to the side length multiplied by sqrt(3). The longer diagonal, which connects two vertices with two vertices in between (i.e., opposite vertices), is equal to twice the side length.But in this case, DA seems to be connecting A to D, which are three vertices apart. Let me count: starting from A, the next vertex is B, then C, then D. So, A to D is three vertices apart, which would make it the maximum diagonal, right? So, in a regular hexagon, the maximum diagonal is twice the side length. So, if the side length is 10, then DA should be 20 units.But wait, in the Asymptote code, the distance from A(1,0) to D(1,3.464) is 3.464, which is approximately 2*sqrt(3). So, if that's the case, and the side length in the drawing is 2 units, then scaling up by 5 would make the actual length 10*sqrt(3). But according to the properties, the maximum diagonal should be 20 units. So, there's a discrepancy here.Maybe I'm miscounting the vertices. Let me label the hexagon based on the coordinates:- A is at (1,0)- Next vertex is (3,0)- Then (4,1.732)- Then (3,3.464)- Then (1,3.464)- Then (0,1.732)- Back to (1,0)So, starting from A(1,0), moving to (3,0) is B, then to (4,1.732) is C, then to (3,3.464) is D, then to (1,3.464) is E, then to (0,1.732) is F, and back to A.So, A to D is from (1,0) to (3,3.464). Wait, no, D is at (3,3.464). So, A is at (1,0), D is at (3,3.464). So, the distance between A and D is not vertical, but rather a diagonal. So, in the drawing, the distance is sqrt[(3-1)^2 + (3.464-0)^2] = sqrt[4 + 12] = sqrt[16] = 4 units. But in the Asymptote code, the label "10" is near (3.5,2.598), which is the midpoint of the side from (3,0) to (4,1.732). So, the length of that side is sqrt[(4-3)^2 + (1.732-0)^2] = sqrt[1 + 3] = sqrt[4] = 2 units. So, the side length in the drawing is 2 units, but the problem states it's 10 units. Therefore, the scaling factor is 5, as 2*5=10.So, the distance from A to D in the drawing is 4 units, so in reality, it's 4*5=20 units. But earlier, I thought it might be 10*sqrt(3). Hmm, which one is correct?Wait, let's calculate the distance between A(1,0) and D(3,3.464) in the drawing. The x-coordinate difference is 2, and the y-coordinate difference is 3.464. So, the distance is sqrt[(2)^2 + (3.464)^2] = sqrt[4 + 12] = sqrt[16] = 4 units. So, in the drawing, it's 4 units, which scales to 20 units in reality. But 3.464 is approximately 2*sqrt(3), so 3.464 is roughly 2*1.732. So, 3.464 is 2*sqrt(3). Therefore, the y-coordinate difference is 2*sqrt(3), and the x-coordinate difference is 2. So, the distance is sqrt[(2)^2 + (2*sqrt(3))^2] = sqrt[4 + 12] = sqrt[16] = 4. So, that's correct.But in a regular hexagon, the distance between two vertices that are three apart (i.e., opposite vertices) should be 2 times the side length. So, if the side length is 10, the distance should be 20. So, that aligns with the calculation here. So, DA is 20 units.Wait, but earlier I thought it might be 10*sqrt(3). So, why the confusion? Because in the drawing, the distance is 4 units, which is 2*sqrt(3) in y and 2 in x, but when scaled up, it's 20 units. So, 20 units is the correct answer.But let me think again. In a regular hexagon, the maximum diagonal is 2 times the side length, which is 20 in this case. However, sometimes diagonals are referred to as the shorter diagonals, which are sqrt(3) times the side length. So, maybe I need to clarify which diagonal DA is.Looking back at the coordinates, A is at (1,0), and D is at (3,3.464). So, in terms of the hexagon's structure, A is at the bottom, and D is at the top. So, in a regular hexagon, the vertical distance from the bottom to the top is 2 times the side length times sin(60 degrees), which is sqrt(3)/2. So, the vertical distance is 2*10*(sqrt(3)/2) = 10*sqrt(3). Wait, that's different from 20.Wait, maybe I'm mixing up the distance between opposite vertices and the vertical height. Let me clarify.In a regular hexagon, the distance between two opposite vertices (the maximum diagonal) is 2 times the side length. However, the vertical distance from the bottom to the top (if the hexagon is oriented with a flat side at the bottom) is 2 times the side length times sin(60 degrees), which is 2*10*(sqrt(3)/2) = 10*sqrt(3). So, that would be the vertical height.But in this case, the hexagon is oriented with a vertex at the bottom, so the vertical distance from A(1,0) to D(3,3.464) is not the same as the vertical height. Instead, it's a diagonal that spans across the hexagon.Wait, perhaps I need to calculate the distance using coordinates. In the drawing, A is at (1,0) and D is at (3,3.464). So, the distance is sqrt[(3-1)^2 + (3.464-0)^2] = sqrt[4 + 12] = sqrt[16] = 4 units. Since the side length in the drawing is 2 units, scaling up by 5 gives 20 units. So, DA is 20 units.But let me think about the properties again. In a regular hexagon, the maximum diagonal is 2 times the side length, which would be 20 in this case. So, that makes sense. Therefore, DA is 20 units.Wait, but in the Asymptote code, the label "10" is near the point (3.5,2.598), which is the midpoint of the side from (3,0) to (4,1.732). The length of that side is 2 units in the drawing, which scales to 10 units in reality. So, the scaling factor is 5. Therefore, the distance from A to D, which is 4 units in the drawing, is 4*5=20 units in reality.So, putting it all together, the length of diagonal DA is 20 units.But wait, earlier I thought it might be 10*sqrt(3). Let me check that. If the side length is 10, then 10*sqrt(3) is approximately 17.32, which is less than 20. So, which one is correct?I think the confusion arises from the orientation of the hexagon. If the hexagon is oriented with a flat side at the bottom, then the vertical distance from the bottom to the top is 10*sqrt(3). However, in this case, the hexagon is oriented with a vertex at the bottom, so the distance from A to D is indeed the maximum diagonal, which is 20 units.To confirm, let's consider the coordinates. In the drawing, A is at (1,0) and D is at (3,3.464). The distance between them is 4 units, which scales to 20 units. So, that must be the correct answer.Therefore, the length of diagonal DA is 20 units.

💡First, I need to understand the rate at which the computer's price decreases. The price decreases by one-third every two years, which means that every two years, the price is multiplied by two-thirds of its previous value.Next, I determine how many two-year periods are there in six years. Dividing six by two gives me three periods.Then, I apply the price decrease formula for each two-year period. The formula is:[ text{New Price} = text{Original Price} times left(frac{2}{3}right)^{text{number of periods}} ]Substituting the given values:[ text{New Price} = 8100 times left(frac{2}{3}right)^3 ]Calculating the exponent:[ left(frac{2}{3}right)^3 = frac{8}{27} ]Finally, I multiply the original price by the calculated factor:[ text{New Price} = 8100 times frac{8}{27} = 2400 ]Therefore, the price of the computer after six years will be 2400 yuan.

💡First, I need to determine the length of the train based on the given speed and the time it takes to cross a pole.The train is traveling at a speed of 18 kilometers per hour (km/h). To make the units consistent with the time given in seconds, I should convert the speed from km/h to meters per second (m/s).I know that:1 kilometer = 1000 meters1 hour = 3600 secondsSo, to convert 18 km/h to m/s:[ text{Speed in m/s} = frac{18 times 1000}{3600} = 5 text{ m/s} ]Now that I have the speed in meters per second, I can calculate the length of the train using the formula:[ text{Distance} = text{Speed} times text{Time} ]Given that the time taken to cross the pole is 5 seconds:[ text{Length of train} = 5 text{ m/s} times 5 text{ s} = 25 text{ meters} ]Therefore, the length of the train is 25 meters.

💡First, I need to determine the selling price at each stage of the transaction to find the retailer's profit percentage.Starting with the manufacturer, they sell the glasses to the wholesaler at an 18% profit. The manufacturer's cost price is Rs. 17. To calculate the manufacturer's selling price, I'll add 18% of Rs. 17 to the cost price.Next, the wholesaler sells the glasses to the retailer at a 20% profit. Using the manufacturer's selling price as the wholesaler's cost price, I'll calculate the wholesaler's selling price by adding 20% of that amount.Finally, the retailer sells the glasses to the customer for Rs. 30.09. To find the retailer's profit percentage, I'll subtract the wholesaler's selling price from the retailer's selling price to determine the profit amount. Then, I'll calculate the profit percentage by dividing the profit by the wholesaler's selling price and multiplying by 100.By following these steps, I can accurately determine the retailer's profit percentage.