💡First, I need to apply Horner's method to evaluate the polynomial ( f(x) = x^5 - 2x^4 + 3x^3 - 4x^2 + 5x + 6 ) at ( x = 3 ).I'll start by rewriting the polynomial in a nested form to facilitate the evaluation:[f(x) = (((x - 2)x + 3)x - 4)x + 5)x + 6]Next, I'll substitute ( x = 3 ) into the nested expression and compute step by step:[begin{align*}f(3) &= (((3 - 2) cdot 3 + 3) cdot 3 - 4) cdot 3 + 5) cdot 3 + 6 &= (((1 cdot 3 + 3) cdot 3 - 4) cdot 3 + 5) cdot 3 + 6 &= ((6 cdot 3 - 4) cdot 3 + 5) cdot 3 + 6 &= (14 cdot 3 + 5) cdot 3 + 6 &= 47 cdot 3 + 6 &= 147end{align*}]Therefore, the value of ( f(3) ) is 147.

💡Okay, so I have this problem where I need to find the absolute difference between two expressions, A and B. Let me try to understand what each expression is.Expression A is given as:[ A = 1 times 2 + 3 times 4 + 5 times 6 + cdots + 39 times 40 + 41 ]And expression B is:[ B = 1 + 2 times 3 + 4 times 5 + cdots + 38 times 39 + 40 times 41 ]Hmm, so both A and B are sums of products of consecutive numbers, but they start and end differently. A starts with 1×2 and ends with 41, while B starts with 1 and ends with 40×41. I think the first step is to figure out how many terms there are in each expression. Let me see:For A, the products are 1×2, 3×4, ..., 39×40. Each product is of two consecutive odd and even numbers. So starting from 1 and going up to 40, stepping by 2 each time. That would be (40 - 1)/2 + 1 = 20 products. But then there's an extra +41 at the end. So A has 20 products plus 41.For B, it starts with 1, then 2×3, 4×5, ..., 40×41. So the products are from 2×3 up to 40×41. Let me count how many products there are here. Starting at 2 and going up to 40, stepping by 2 each time. That would be (40 - 2)/2 + 1 = 20 products. Plus the initial 1. So B has 1 plus 20 products.So both A and B have 20 products each, but A has an extra 41 and B has an extra 1. Interesting.Now, I need to compute A - B. Let me write that out:[ A - B = (1 times 2 + 3 times 4 + 5 times 6 + cdots + 39 times 40 + 41) - (1 + 2 times 3 + 4 times 5 + cdots + 38 times 39 + 40 times 41) ]Let me try to rearrange this expression to see if I can find a pattern or simplify it.First, let's distribute the negative sign to each term in B:[ A - B = 1 times 2 + 3 times 4 + 5 times 6 + cdots + 39 times 40 + 41 - 1 - 2 times 3 - 4 times 5 - cdots - 38 times 39 - 40 times 41 ]Now, let's group similar terms together. I notice that each product in A (except for the last term 41) has a corresponding product in B, but shifted by one. For example, 1×2 in A corresponds to 2×3 in B, 3×4 in A corresponds to 4×5 in B, and so on, up to 39×40 in A corresponding to 40×41 in B.So, let's pair them up:- The first term in A is 1×2, and the first product in B is 2×3.- The second term in A is 3×4, and the second product in B is 4×5.- This pattern continues until the 20th product in A, which is 39×40, corresponding to the 20th product in B, which is 40×41.So, each pair is like (n×(n+1)) in A and ((n+1)×(n+2)) in B. Let me write this as:For each k from 1 to 20, the term in A is (2k - 1)×2k, and the corresponding term in B is (2k)×(2k + 1).So, when we subtract B from A, each pair becomes:(2k - 1)×2k - (2k)×(2k + 1)Let me compute this difference:(2k - 1)×2k = 4k² - 2k(2k)×(2k + 1) = 4k² + 2kSubtracting these:(4k² - 2k) - (4k² + 2k) = -4kSo each pair contributes -4k to the difference A - B.Now, let's consider all these pairs from k = 1 to k = 20. So the total contribution from all these pairs is:Sum from k = 1 to 20 of (-4k) = -4 * Sum from k = 1 to 20 of kThe sum of the first 20 natural numbers is (20)(21)/2 = 210So, the total contribution is -4 * 210 = -840But wait, we also have the extra terms in A and B that aren't paired. In A, there's an extra +41, and in B, there's an extra -1. So, let's include those:A - B = (-840) + 41 - 1 = -840 + 40 = -800Therefore, the absolute difference is |A - B| = | -800 | = 800Wait, let me double-check my steps to make sure I didn't make a mistake.1. Identified that A has 20 products plus 41, and B has 20 products plus 1.2. Subtracted B from A, distributing the negative sign.3. Paired each product in A with the next product in B, resulting in each pair contributing -4k.4. Summed these contributions from k=1 to 20, getting -840.5. Added the extra terms: +41 from A and -1 from B, resulting in -840 + 40 = -800.6. Took the absolute value to get 800.This seems correct. I don't see any errors in the calculations or reasoning.

💡Okay, so I have this problem about translating the graph of a cosine function and finding one of its symmetry centers. Let me try to work through it step by step.First, the original function is y = 2cos(2x). I know that the graph of a cosine function is a wave that oscillates between its maximum and minimum values. The general form of a cosine function is y = A cos(Bx + C) + D, where A is the amplitude, B affects the period, C is the phase shift, and D is the vertical shift.In this case, A is 2, so the amplitude is 2. The coefficient B is 2, which affects the period. The period of a cosine function is normally 2π, but when you have Bx inside the cosine, the period becomes 2π/B. So here, the period is 2π/2 = π. That means the function completes one full cycle every π units along the x-axis.Now, the problem says the graph is translated to the right by π/12 units. Translating a graph to the right by a certain amount is a horizontal shift. In the general form, a horizontal shift is represented by -C/B. So if we're shifting the graph to the right by π/12, that means we're replacing x with (x - π/12) in the function. So, the translated function becomes y = 2cos(2(x - π/12)). Let me simplify that. Distribute the 2 inside the cosine: y = 2cos(2x - π/6). Alright, so the new function is y = 2cos(2x - π/6). Now, I need to find one of the symmetry centers of this translated graph. Symmetry centers for cosine functions are typically the points where the function crosses the x-axis, especially the midpoints between the maximum and minimum points. These are the points where the function is symmetric about.But wait, actually, for a cosine function, the symmetry centers are the points where the function has a point of symmetry, meaning that if you rotate the graph 180 degrees around that point, the graph remains the same. For a standard cosine function, these points are at the midpoints of the peaks and troughs.In the standard cosine function y = cos(x), the symmetry centers are at (kπ, 0) where k is an integer. These are the points where the function crosses the x-axis. So, for our function, we need to find similar points.Given that our function is y = 2cos(2x - π/6), let's find where it crosses the x-axis. The function crosses the x-axis when the cosine term is zero. So, we set 2x - π/6 equal to π/2 + kπ, where k is an integer, because cos(θ) = 0 when θ = π/2 + kπ.So, let's solve for x:2x - π/6 = π/2 + kπAdding π/6 to both sides:2x = π/2 + π/6 + kπLet me compute π/2 + π/6. π/2 is 3π/6, so 3π/6 + π/6 = 4π/6 = 2π/3.So, 2x = 2π/3 + kπDivide both sides by 2:x = π/3 + (kπ)/2So, the x-coordinates where the function crosses the x-axis are x = π/3 + (kπ)/2, where k is an integer.These are the potential symmetry centers. So, let's list a few of them for different k values.When k = 0: x = π/3 + 0 = π/3When k = 1: x = π/3 + π/2 = (2π/6 + 3π/6) = 5π/6When k = 2: x = π/3 + π = (π/3 + 3π/3) = 4π/3When k = -1: x = π/3 - π/2 = (2π/6 - 3π/6) = -π/6So, the symmetry centers are at (π/3, 0), (5π/6, 0), (4π/3, 0), (-π/6, 0), etc.Looking at the answer choices, we have options A: (5π/6, 0), B: (7π/6, 0), C: (-π/3, 0), D: (π/6, 0).From our list, (5π/6, 0) is one of the symmetry centers. So, option A is correct.But wait, let me double-check. The original function is y = 2cos(2x), which has symmetry centers at x = (kπ)/2, right? Because for y = 2cos(2x), the function crosses the x-axis when 2x = π/2 + kπ, so x = π/4 + (kπ)/2. So, the symmetry centers are at (π/4 + kπ/2, 0). But after translating it to the right by π/12, the new function is y = 2cos(2x - π/6). So, the phase shift is π/12 to the right. So, the symmetry centers should also shift by π/12.Wait, so originally, the symmetry centers were at x = π/4 + kπ/2. After shifting right by π/12, the new centers should be at x = π/4 + π/12 + kπ/2.Compute π/4 + π/12: π/4 is 3π/12, so 3π/12 + π/12 = 4π/12 = π/3. So, the new centers are at x = π/3 + kπ/2, which matches what I found earlier.So, the symmetry centers are at x = π/3 + kπ/2. So, when k = 0, x = π/3; k = 1, x = 5π/6; k = 2, x = 4π/3; k = -1, x = -π/6.Looking back at the options, A is (5π/6, 0), which is one of these points. So, A is correct.But let me check why the other options aren't correct. Option B is (7π/6, 0). Let's see if that's a symmetry center. 7π/6 is greater than π, and from our list, the next center after 5π/6 is 4π/3, which is approximately 4.188, while 7π/6 is approximately 3.665. So, 7π/6 isn't in our list. So, B is not correct.Option C is (-π/3, 0). From our list, when k = -1, we get x = -π/6, not -π/3. So, C is not correct.Option D is (π/6, 0). From our list, when k = 0, x = π/3; when k = -1, x = -π/6. So, π/6 isn't in our list. So, D is not correct.Therefore, the correct answer is A: (5π/6, 0).I think that makes sense. The key was to find where the translated function crosses the x-axis, which are the symmetry centers. By solving for when the cosine term is zero, we found the x-coordinates, and then checked which one matches the options given.

💡Okay, so I have this problem here: Given that when a number n is divided by 4, the remainder is 3. I need to find out what the remainder is when 7n is divided by 4. Hmm, let's see. First, I remember that when a number leaves a remainder when divided by another number, it's related to modular arithmetic. So, if n divided by 4 leaves a remainder of 3, that means n is congruent to 3 modulo 4. In symbols, that's n ≡ 3 mod 4. Now, I need to find 7n mod 4. Since n ≡ 3 mod 4, multiplying both sides of this congruence by 7 should give me the result for 7n. So, 7n ≡ 7*3 mod 4. Let me calculate 7*3, which is 21. So now I have 7n ≡ 21 mod 4. But 21 divided by 4 is 5 with a remainder of 1, right? Because 4*5 is 20, and 21 minus 20 is 1. So, 21 mod 4 is 1. That means 7n ≡ 1 mod 4. Wait, let me double-check that. If n is 3 more than a multiple of 4, then 7n would be 7 times that. So, 7n is 21 more than a multiple of 4. Since 21 divided by 4 leaves a remainder of 1, that makes sense. Is there another way to think about this? Maybe by breaking down 7 into smaller parts. 7 is 4 + 3, so 7n is 4n + 3n. Now, 4n is obviously a multiple of 4, so when divided by 4, it leaves no remainder. That means the remainder comes from 3n. Since n ≡ 3 mod 4, 3n ≡ 3*3 mod 4, which is 9 mod 4. And 9 divided by 4 is 2 with a remainder of 1. So again, 3n ≡ 1 mod 4. Therefore, 7n ≡ 1 mod 4. Hmm, both methods give me the same result, so I think I'm confident that the remainder is 1 when 7n is divided by 4. Just to make sure, let me plug in a specific number for n. If n is 3, then 7n is 21, and 21 divided by 4 is 5 with a remainder of 1. If n is 7 (which is 3 more than 4), then 7n is 49, and 49 divided by 4 is 12 with a remainder of 1. Another example: n = 11 (which is 3 more than 8), 7n is 77, and 77 divided by 4 is 19 with a remainder of 1. All these examples confirm that the remainder is indeed 1. So, I think I've got it right.

💡Okay, so I've got this problem about a fruit market, and it's divided into two parts. Let me try to understand each part step by step.Starting with part (1): The fruit market sells a high-end fruit for 50 yuan per kilogram. After two consecutive price reductions, the price becomes 32 yuan per kilogram. The percentage of reduction is the same each time, and I need to find that percentage.Hmm, okay. So, if the price is reduced by the same percentage twice, it goes from 50 to 32. Let me think about how percentage reductions work. If something decreases by, say, x%, then the new price is the original price multiplied by (1 - x/100). Since this happens twice, it would be multiplied by (1 - x/100) each time.So, mathematically, I can write this as:50 * (1 - x/100)^2 = 32I need to solve for x. Let me rearrange this equation.First, divide both sides by 50:(1 - x/100)^2 = 32/50Simplify 32/50:32 divided by 50 is 0.64.So, (1 - x/100)^2 = 0.64Now, take the square root of both sides:1 - x/100 = sqrt(0.64)The square root of 0.64 is 0.8, because 0.8 * 0.8 = 0.64.So, 1 - x/100 = 0.8Now, solve for x/100:x/100 = 1 - 0.8x/100 = 0.2Multiply both sides by 100:x = 0.2 * 100x = 20So, the percentage reduction each time is 20%.Wait, let me double-check that. If I reduce 50 yuan by 20%, that's 50 - (0.2 * 50) = 50 - 10 = 40 yuan. Then, reducing 40 yuan by another 20% is 40 - (0.2 * 40) = 40 - 8 = 32 yuan. Yep, that matches the given final price. So, 20% seems correct.Alright, part (1) makes sense now.Moving on to part (2): The market makes a profit of 10 yuan per kilogram and sells 500 kilograms per day. They decide to increase the price, but not by more than 8 yuan per kilogram. For every 1 yuan increase, the daily sales volume decreases by 20 kilograms. They want to ensure a daily profit of 6000 yuan. I need to find how much they should increase the price per kilogram.Okay, let's parse this. Currently, the price is 50 yuan per kilogram, and the profit per kilogram is 10 yuan. So, their cost must be 50 - 10 = 40 yuan per kilogram.They are considering increasing the price. Let's denote the increase as y yuan per kilogram. So, the new price will be 50 + y yuan per kilogram.However, for every 1 yuan increase, they sell 20 kilograms less. So, if they increase the price by y yuan, their sales volume will decrease by 20y kilograms. Therefore, the new sales volume will be 500 - 20y kilograms.Their profit per kilogram will change as well. Originally, it was 10 yuan per kilogram. If they increase the price by y yuan, their new profit per kilogram will be (10 + y) yuan, assuming their cost remains the same.Wait, hold on. Is the profit per kilogram just the selling price minus the cost? If the cost is 40 yuan, and the selling price becomes 50 + y, then the new profit per kilogram is (50 + y) - 40 = 10 + y. Yes, that makes sense.So, their total profit is (profit per kilogram) multiplied by (sales volume). They want this total profit to be at least 6000 yuan.So, the equation is:(10 + y) * (500 - 20y) = 6000Let me write that down:(10 + y)(500 - 20y) = 6000I need to solve for y. Let's expand this equation.First, multiply 10 by 500: 10 * 500 = 5000Then, 10 * (-20y) = -200yThen, y * 500 = 500yThen, y * (-20y) = -20y^2So, putting it all together:5000 - 200y + 500y - 20y^2 = 6000Combine like terms:5000 + ( -200y + 500y ) - 20y^2 = 60005000 + 300y - 20y^2 = 6000Now, subtract 6000 from both sides to set the equation to zero:5000 + 300y - 20y^2 - 6000 = 0-1000 + 300y - 20y^2 = 0Let me rearrange it:-20y^2 + 300y - 1000 = 0It's a quadratic equation. Let's write it in standard form:20y^2 - 300y + 1000 = 0I can divide the entire equation by 20 to simplify:y^2 - 15y + 50 = 0Now, let's solve for y using the quadratic formula. The quadratic is:y^2 - 15y + 50 = 0The quadratic formula is:y = [15 ± sqrt( (-15)^2 - 4*1*50 )]/(2*1)y = [15 ± sqrt(225 - 200)]/2y = [15 ± sqrt(25)]/2y = [15 ± 5]/2So, two solutions:y = (15 + 5)/2 = 20/2 = 10y = (15 - 5)/2 = 10/2 = 5So, y = 10 or y = 5.But wait, the problem says the price increase is limited to no more than 8 yuan per kilogram. So, y cannot be 10 because that's more than 8. Therefore, y = 5 is the valid solution.Let me verify this. If they increase the price by 5 yuan, the new price is 55 yuan per kilogram. The sales volume decreases by 20*5 = 100 kilograms, so they sell 500 - 100 = 400 kilograms per day.The profit per kilogram is 10 + 5 = 15 yuan.Total profit is 15 * 400 = 6000 yuan, which matches the requirement.Alternatively, if y = 10, the price would be 60 yuan, sales volume would be 500 - 200 = 300 kilograms, and profit per kilogram would be 20 yuan. Total profit would be 20 * 300 = 6000 yuan as well. But since the price increase is limited to 8 yuan, y = 10 is not acceptable.Therefore, the price should be increased by 5 yuan per kilogram.I think that's it. Let me just recap:Part (1): The percentage reduction each time is 20%.Part (2): The price should be increased by 5 yuan per kilogram to achieve a daily profit of 6000 yuan.**Final Answer**(1) The percentage of reduction each time is boxed{20%}.(2) The price should be increased by boxed{5} yuan per kilogram.

💡Alright, so I've got this function ( f(x) = (ax^2 + x - 1)e^{-x} ). The problem has two parts: first, I need to discuss the monotonicity of ( f(x) ) when ( a geq 0 ), and second, I have to prove that when ( a geq 2 ), ( f(x) + sqrt{e} geq 0 ).Starting with part (I). Monotonicity refers to whether the function is increasing or decreasing over certain intervals. To determine this, I remember that I need to find the derivative of ( f(x) ) and analyze its sign. So, let's compute the derivative ( f'(x) ). The function is a product of two functions: ( u(x) = ax^2 + x - 1 ) and ( v(x) = e^{-x} ). Using the product rule, ( f'(x) = u'(x)v(x) + u(x)v'(x) ).First, find ( u'(x) ):( u'(x) = 2ax + 1 ).Next, find ( v'(x) ):( v'(x) = -e^{-x} ).Now, plug these into the product rule:( f'(x) = (2ax + 1)e^{-x} + (ax^2 + x - 1)(-e^{-x}) ).Let me factor out ( e^{-x} ) since it's common to both terms:( f'(x) = e^{-x} [ (2ax + 1) - (ax^2 + x - 1) ] ).Simplify the expression inside the brackets:( (2ax + 1) - ax^2 - x + 1 = -ax^2 + (2a - 1)x + 2 ).So, putting it all together:( f'(x) = e^{-x} (-ax^2 + (2a - 1)x + 2) ).Hmm, that's a quadratic in the numerator. Maybe I can factor it to find critical points. Let me try factoring ( -ax^2 + (2a - 1)x + 2 ).First, factor out a negative sign to make it easier:( - (ax^2 - (2a - 1)x - 2) ).Now, try to factor ( ax^2 - (2a - 1)x - 2 ). Let's see, looking for two numbers that multiply to ( a times (-2) = -2a ) and add up to ( -(2a - 1) ).Wait, maybe using the quadratic formula would be more straightforward. The roots of ( ax^2 - (2a - 1)x - 2 = 0 ) can be found using:( x = frac{(2a - 1) pm sqrt{(2a - 1)^2 + 8a}}{2a} ).Simplify the discriminant:( (2a - 1)^2 + 8a = 4a^2 - 4a + 1 + 8a = 4a^2 + 4a + 1 = (2a + 1)^2 ).Ah, that's a perfect square! So, the roots are:( x = frac{(2a - 1) pm (2a + 1)}{2a} ).Calculating both roots:1. ( x = frac{2a - 1 + 2a + 1}{2a} = frac{4a}{2a} = 2 ).2. ( x = frac{2a - 1 - (2a + 1)}{2a} = frac{-2}{2a} = -frac{1}{a} ).So, the quadratic factors as ( (x - 2)(ax + 1) ). Therefore, the derivative becomes:( f'(x) = -e^{-x} (x - 2)(ax + 1) ).Since ( e^{-x} ) is always positive, the sign of ( f'(x) ) depends on the product ( -(x - 2)(ax + 1) ).Let me analyze the critical points ( x = 2 ) and ( x = -frac{1}{a} ). These points divide the real line into intervals where the sign of ( f'(x) ) may change.Case 1: ( a = 0 ).If ( a = 0 ), the function simplifies to ( f(x) = (0 + x - 1)e^{-x} = (x - 1)e^{-x} ). The derivative becomes:( f'(x) = -e^{-x}(x - 2)(0 + 1) = -e^{-x}(x - 2) ).So, ( f'(x) = -e^{-x}(x - 2) ). Since ( e^{-x} > 0 ) always, the sign of ( f'(x) ) depends on ( -(x - 2) ).- When ( x < 2 ), ( x - 2 < 0 ), so ( -(x - 2) > 0 ). Thus, ( f'(x) > 0 ).- When ( x > 2 ), ( x - 2 > 0 ), so ( -(x - 2) < 0 ). Thus, ( f'(x) < 0 ).Therefore, when ( a = 0 ), ( f(x) ) is increasing on ( (-infty, 2) ) and decreasing on ( (2, infty) ).Case 2: ( a > 0 ).Now, with ( a > 0 ), the critical points are ( x = 2 ) and ( x = -frac{1}{a} ). Since ( a > 0 ), ( -frac{1}{a} ) is negative, so the critical points are at ( x = -frac{1}{a} ) and ( x = 2 ).To determine the intervals where ( f'(x) ) is positive or negative, let's analyze the sign of ( -(x - 2)(ax + 1) ).First, note that ( ax + 1 ) is positive when ( x > -frac{1}{a} ) and negative when ( x < -frac{1}{a} ).Similarly, ( x - 2 ) is positive when ( x > 2 ) and negative when ( x < 2 ).So, let's consider the intervals divided by ( x = -frac{1}{a} ) and ( x = 2 ):1. ( x < -frac{1}{a} ): - ( ax + 1 < 0 ) - ( x - 2 < 0 ) - Product: ( (ax + 1)(x - 2) = (-)(-) = + ) - Multiply by -1: ( -(+) = - ) - So, ( f'(x) < 0 )2. ( -frac{1}{a} < x < 2 ): - ( ax + 1 > 0 ) - ( x - 2 < 0 ) - Product: ( (+)(-) = - ) - Multiply by -1: ( -(-) = + ) - So, ( f'(x) > 0 )3. ( x > 2 ): - ( ax + 1 > 0 ) - ( x - 2 > 0 ) - Product: ( (+)(+) = + ) - Multiply by -1: ( -(+) = - ) - So, ( f'(x) < 0 )Therefore, when ( a > 0 ), ( f(x) ) is decreasing on ( (-infty, -frac{1}{a}) ), increasing on ( (-frac{1}{a}, 2) ), and decreasing again on ( (2, infty) ).So, summarizing part (I):- If ( a = 0 ), ( f(x) ) increases on ( (-infty, 2) ) and decreases on ( (2, infty) ).- If ( a > 0 ), ( f(x) ) decreases on ( (-infty, -frac{1}{a}) ), increases on ( (-frac{1}{a}, 2) ), and decreases on ( (2, infty) ).Moving on to part (II). We need to prove that when ( a geq 2 ), ( f(x) + sqrt{e} geq 0 ).So, ( f(x) + sqrt{e} = (ax^2 + x - 1)e^{-x} + sqrt{e} geq 0 ).I need to show that this expression is always non-negative for all real ( x ) when ( a geq 2 ).First, let's analyze the behavior of ( f(x) ). From part (I), when ( a geq 2 ), ( f(x) ) has critical points at ( x = -frac{1}{a} ) and ( x = 2 ). Since ( a geq 2 ), ( -frac{1}{a} ) is between ( -frac{1}{2} ) and 0.From the monotonicity, ( f(x) ) is decreasing on ( (-infty, -frac{1}{a}) ), increasing on ( (-frac{1}{a}, 2) ), and decreasing on ( (2, infty) ).Therefore, ( f(x) ) has a local minimum at ( x = -frac{1}{a} ) and a local maximum at ( x = 2 ).To find the minimum value of ( f(x) ), let's compute ( f(-frac{1}{a}) ).Compute ( f(-frac{1}{a}) ):( f(-frac{1}{a}) = [a(-frac{1}{a})^2 + (-frac{1}{a}) - 1] e^{frac{1}{a}} ).Simplify inside the brackets:( a(frac{1}{a^2}) - frac{1}{a} - 1 = frac{1}{a} - frac{1}{a} - 1 = -1 ).So, ( f(-frac{1}{a}) = (-1) e^{frac{1}{a}} = -e^{frac{1}{a}} ).Therefore, the minimum value of ( f(x) ) is ( -e^{frac{1}{a}} ).Now, we need to ensure that ( f(x) + sqrt{e} geq 0 ) for all ( x ). The minimum value of ( f(x) ) is ( -e^{frac{1}{a}} ), so the minimum of ( f(x) + sqrt{e} ) is ( -e^{frac{1}{a}} + sqrt{e} ).Thus, we need:( -e^{frac{1}{a}} + sqrt{e} geq 0 )( sqrt{e} geq e^{frac{1}{a}} )( e^{1/2} geq e^{1/a} )Since the exponential function is increasing, this inequality holds if:( frac{1}{2} geq frac{1}{a} )Which simplifies to:( a geq 2 ).Therefore, when ( a geq 2 ), ( f(x) + sqrt{e} geq 0 ) for all ( x ).Alternatively, another approach could be to consider the function ( f(x) + sqrt{e} ) and analyze its minimum. Since we found that the minimum of ( f(x) ) is ( -e^{frac{1}{a}} ), adding ( sqrt{e} ) shifts this minimum up by ( sqrt{e} ). So, the new minimum is ( -e^{frac{1}{a}} + sqrt{e} ). As shown earlier, this is non-negative when ( a geq 2 ).Another way to think about it is to consider the behavior of ( f(x) ) as ( x ) approaches infinity and negative infinity. As ( x to infty ), ( e^{-x} ) dominates, so ( f(x) to 0 ). As ( x to -infty ), ( ax^2 ) dominates, so ( f(x) ) behaves like ( ax^2 e^{-x} ), which goes to infinity since ( e^{-x} ) becomes ( e^{|x|} ), growing exponentially. However, since we're adding ( sqrt{e} ), which is a positive constant, the function ( f(x) + sqrt{e} ) will always be above zero, especially considering the minimum value analysis.Wait, actually, as ( x to -infty ), ( f(x) ) tends to infinity because ( ax^2 ) grows faster than ( e^{-x} ) decays. So, ( f(x) + sqrt{e} ) will also tend to infinity. The critical point is the local minimum at ( x = -frac{1}{a} ), which we've already determined is ( -e^{frac{1}{a}} ). Adding ( sqrt{e} ) ensures that this minimum is non-negative when ( a geq 2 ).Therefore, combining all these observations, we can confidently say that for ( a geq 2 ), ( f(x) + sqrt{e} ) is always greater than or equal to zero.**Final Answer**(I) When ( a = 0 ), ( f(x) ) is increasing on ( (-infty, 2) ) and decreasing on ( (2, +infty) ). When ( a > 0 ), ( f(x) ) is decreasing on ( (-infty, -frac{1}{a}) ), increasing on ( (-frac{1}{a}, 2) ), and decreasing on ( (2, +infty) ).(II) When ( a geq 2 ), ( f(x) + sqrt{e} geq 0 ) for all ( x ).boxed{f(x) + sqrt{e} geq 0}