💡Okay, so I need to find the extreme value point of the function f(x) = x - 2 ln x. Hmm, extreme value points are where the function reaches a maximum or minimum, right? I remember from my calculus class that to find these points, I should take the derivative of the function and then find where that derivative is zero or undefined. First, let me write down the function again: f(x) = x - 2 ln x. I know that the natural logarithm function ln x is only defined for x > 0, so the domain of this function is all positive real numbers. That means I don't have to worry about negative x values or zero. Now, let's find the derivative of f(x). The derivative of x with respect to x is 1, that's straightforward. Then, the derivative of -2 ln x with respect to x. I remember that the derivative of ln x is 1/x, so multiplying by -2 gives me -2 * (1/x) = -2/x. So putting it all together, the derivative f'(x) should be 1 - 2/x. Let me write that down: f'(x) = 1 - 2/x. Next, I need to find where this derivative equals zero because that's where the function could have an extreme value. So I set f'(x) = 0:1 - 2/x = 0.Let me solve for x. I can rewrite this equation as:1 = 2/x.To solve for x, I can multiply both sides by x:1 * x = 2.So, x = 2. Wait, but I should also check where the derivative is undefined. The derivative f'(x) = 1 - 2/x is undefined when x = 0, but since x = 0 isn't in the domain of the original function f(x), I don't have to consider that point. So the only critical point is at x = 2.Now, I need to determine whether this critical point is a maximum or a minimum. I remember that I can use the first derivative test for this. The idea is to check the sign of the derivative just before and just after the critical point. If the derivative changes from positive to negative, the function has a maximum at that point. If it changes from negative to positive, it's a minimum.So, let's pick a point just less than 2, say x = 1. Plugging into f'(x):f'(1) = 1 - 2/1 = 1 - 2 = -1.Wait, that's negative. Hmm, but I thought before the critical point, the function was increasing. Maybe I made a mistake. Let me double-check.Wait, no, actually, if the derivative is negative, that means the function is decreasing. So, if at x = 1, the derivative is negative, the function is decreasing there. Then, let me pick a point just greater than 2, say x = 3.f'(3) = 1 - 2/3 = 1 - 2/3 = 1/3, which is positive. So, the derivative is positive after x = 2.Wait, so before x = 2, the derivative is negative, meaning the function is decreasing, and after x = 2, the derivative is positive, meaning the function is increasing. So, the function changes from decreasing to increasing at x = 2. That would mean that x = 2 is a minimum point, right? Because the function was going down before and starts going up after.But wait, that contradicts my initial thought. Let me think again. Maybe I confused the signs. If the derivative goes from negative to positive, that's a minimum. If it goes from positive to negative, that's a maximum. So, in this case, since the derivative goes from negative to positive at x = 2, it's a minimum.But wait, the original function is f(x) = x - 2 ln x. Let me see what the function looks like. For x approaching 0 from the right, ln x approaches negative infinity, so -2 ln x approaches positive infinity. So, f(x) approaches positive infinity as x approaches 0. As x increases, the x term grows linearly, while the -2 ln x term grows logarithmically, but negative. So, initially, the function is decreasing because the -2 ln x term is dominating, but as x increases, the x term starts to dominate, and the function starts increasing.So, at x = 2, the function changes from decreasing to increasing, which is a minimum point. Therefore, x = 2 is a minimum.Wait, but the question says "extreme value point," which can be either maximum or minimum. So, in this case, it's a minimum. So, the extreme value point is at x = 2.Let me just confirm by calculating the second derivative to check concavity. The second derivative f''(x) would be the derivative of f'(x) = 1 - 2/x. So, f''(x) = 0 + 2/x² = 2/x². Since x > 0, f''(x) is always positive, meaning the function is concave upward everywhere in its domain. Therefore, any critical point must be a minimum. So, x = 2 is indeed a minimum.So, putting it all together, the function f(x) = x - 2 ln x has an extreme value point at x = 2, and it's a minimum.**Final Answer**The extreme value point of the function is boxed{2}.

💡First, I recognize that after planting 9 short trees, the total number of short trees in the park will be 12.To find out how many short trees were initially in the park, I need to subtract the number of trees that will be planted from the final total.So, the initial number of short trees is 12 minus 9, which equals 3.

💡Alright, so I have this function here: y equals the square root of log base 2 of (4x minus 3). I need to find the domain of this function. Hmm, okay. Let me think about what the domain means. The domain is all the real numbers x for which the function is defined, right? So, I need to figure out for which values of x this expression makes sense.First, let's break down the function. There's a square root, and inside the square root, there's a logarithm. So, I need to make sure that both the logarithm and the square root are defined. Let me tackle each part step by step.Starting with the logarithm: log base 2 of (4x - 3). Logarithms are only defined for positive arguments. That means the expression inside the log, which is 4x - 3, has to be greater than zero. So, I can write that as:4x - 3 > 0Solving for x, I add 3 to both sides:4x > 3Then divide both sides by 4:x > 3/4Okay, so x has to be greater than 3/4 for the logarithm to be defined. But wait, there's also the square root part. The expression inside the square root has to be non-negative because you can't take the square root of a negative number in the set of real numbers. So, the logarithm itself has to be greater than or equal to zero.So, log base 2 of (4x - 3) ≥ 0Hmm, when is log base 2 of something greater than or equal to zero? Well, log base 2 of 1 is zero because 2^0 = 1. And since the logarithm function is increasing, log base 2 of a number greater than 1 will be positive. So, log base 2 of (4x - 3) ≥ 0 implies that 4x - 3 ≥ 1.Let me write that down:4x - 3 ≥ 1Solving for x, I add 3 to both sides:4x ≥ 4Then divide both sides by 4:x ≥ 1So, x has to be greater than or equal to 1 for the logarithm to be non-negative, which in turn makes the square root defined.Wait a second, earlier I found that x has to be greater than 3/4 for the logarithm to be defined, and now I have x has to be greater than or equal to 1 for the square root to be defined. So, which one is more restrictive? Well, x has to satisfy both conditions. So, the more restrictive condition is x ≥ 1 because if x is greater than or equal to 1, it's automatically greater than 3/4. So, the domain is x ≥ 1.Let me just double-check to make sure I didn't make a mistake. If x is 1, then 4x - 3 is 4*1 - 3 = 1. Log base 2 of 1 is 0, and the square root of 0 is 0, which is defined. If x is greater than 1, say x = 2, then 4*2 - 3 = 5. Log base 2 of 5 is a positive number, and the square root of that is also defined. What if x is less than 1 but greater than 3/4, say x = 0.8? Then 4*0.8 - 3 = 3.2 - 3 = 0.2. Log base 2 of 0.2 is negative because 0.2 is less than 1, and the square root of a negative number isn't defined in real numbers. So, that's why x has to be at least 1.Therefore, the domain of the function is all real numbers x such that x is greater than or equal to 1. In interval notation, that would be [1, infinity).I think that's it. I don't see any mistakes in my reasoning. I considered both the logarithm and the square root, found the conditions each imposes, and then took the intersection of those conditions to find the domain.**Final Answer**The domain of the function is boxed{[1,+infty)}.

💡Okay, so I have this problem here with four options, A through D, and I need to figure out which one is correct. Let me take it step by step.Starting with Option A: (3a^{2}b div b = 3a^{2}). Hmm, okay, so we're dividing (3a^{2}b) by (b). I remember that when you divide terms with the same base, you subtract the exponents. But here, (b) is just to the first power in both the numerator and the denominator. So, (b div b) should be (b^{1-1} = b^{0} = 1). That means the (b) cancels out, leaving us with (3a^{2}). So, Option A seems correct. I'll keep that in mind.Moving on to Option B: (a^{12} div a^{3} = a^{4}). Alright, using the same exponent rule as before, when dividing like bases, subtract the exponents. So, (a^{12} div a^{3} = a^{12-3} = a^{9}). But the answer given is (a^{4}), which is not the same as (a^{9}). So, Option B is incorrect.Next is Option C: ((a + b)^{2} = a^{2} + b^{2}). I recall that squaring a binomial involves the formula ((a + b)^{2} = a^{2} + 2ab + b^{2}). The given answer is missing the (2ab) term, so it's incomplete. Therefore, Option C is also incorrect.Finally, Option D: ((-2a^{2})^{3} = 8a^{6}). Let's break this down. When raising a product to a power, each factor is raised to that power. So, ((-2)^{3} = -8) and ((a^{2})^{3} = a^{6}). Multiplying these together gives (-8a^{6}). The answer provided is (8a^{6}), which is positive, but it should be negative. So, Option D is incorrect as well.After reviewing all the options, only Option A is correct. It properly simplifies the expression by canceling out the (b) terms, resulting in (3a^{2}).

💡First, I need to determine the set ( B ) by solving the inequality ( 3x^2 + x - 8 leq 0 ).To solve this quadratic inequality, I'll start by finding the roots of the equation ( 3x^2 + x - 8 = 0 ). Using the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]where ( a = 3 ), ( b = 1 ), and ( c = -8 ). Plugging in these values:[x = frac{-1 pm sqrt{1^2 - 4 times 3 times (-8)}}{2 times 3} = frac{-1 pm sqrt{1 + 96}}{6} = frac{-1 pm sqrt{97}}{6}]This gives the roots approximately as ( x approx -2 ) and ( x approx frac{4}{3} ).Since the coefficient of ( x^2 ) is positive, the parabola opens upwards. Therefore, the inequality ( 3x^2 + x - 8 leq 0 ) holds between the roots:[-2 leq x leq frac{4}{3}]So, ( B = [-2, frac{4}{3}] ).Next, I'll find the intersection ( A cap B ). Set ( A ) is ( [0, 6] ) and set ( B ) is ( [-2, frac{4}{3}] ). The intersection consists of the overlapping values:[A cap B = [0, frac{4}{3}]]

💡Okay, so I need to find the enthalpy of combustion for 10 moles of glucose, which is C6H12O6, when it reacts with oxygen to produce carbon dioxide and water. Hmm, I'm not entirely sure what enthalpy of combustion means, but I think it's the amount of heat released when a substance burns completely in oxygen. First, I guess I need to write the balanced chemical equation for the combustion of glucose. I remember that combustion reactions involve reacting a substance with oxygen to produce carbon dioxide and water. So, glucose plus oxygen should produce carbon dioxide and water. Let me try to write that out:C6H12O6 + O2 → CO2 + H2ONow, I need to balance this equation. Let's see, on the left side, I have 6 carbon atoms in glucose, and on the right side, each CO2 has one carbon atom, so I need 6 CO2 molecules to balance the carbon. For hydrogen, glucose has 12 hydrogen atoms, and each H2O has 2 hydrogen atoms, so I need 6 H2O molecules to balance the hydrogen. Now, for oxygen, it's a bit trickier. On the left side, glucose has 6 oxygen atoms, and there's also O2. On the right side, each CO2 has 2 oxygen atoms, so 6 CO2 would have 12 oxygen atoms, and each H2O has 1 oxygen atom, so 6 H2O would have 6 oxygen atoms. That makes a total of 18 oxygen atoms on the right side. On the left side, I have 6 oxygen atoms from glucose and some from O2. Let's calculate how many O2 molecules I need. Each O2 molecule has 2 oxygen atoms, so if I need 18 oxygen atoms on the right, and I already have 6 from glucose, I need 12 more from O2. That means I need 6 O2 molecules because 6 O2 molecules provide 12 oxygen atoms.So, the balanced equation should be:C6H12O6 + 6O2 → 6CO2 + 6H2OOkay, now that I have the balanced equation, I need to find the enthalpy of combustion for 10 moles of glucose. I think the enthalpy of combustion is the heat released when one mole of a substance burns completely. So, I probably need the standard enthalpy of combustion for glucose and then multiply it by 10 to get the value for 10 moles.I'm not sure what the standard enthalpy of combustion for glucose is, but I think it's a negative value because combustion releases heat. Maybe around -2800 kJ per mole? I'm not entirely sure, but I'll go with that for now.So, if one mole of glucose releases -2800 kJ, then 10 moles would release 10 times that amount. Let me calculate that:-2800 kJ/mol * 10 mol = -28000 kJWait, that seems like a lot. Is that right? I mean, glucose is a common sugar, and I know that it releases a significant amount of energy when burned, so maybe it is around that range. But I should probably double-check the standard enthalpy of combustion for glucose to make sure.I think I remember that the standard enthalpy of combustion for glucose is approximately -2800 kJ/mol, so multiplying that by 10 gives me -28000 kJ for 10 moles. That seems reasonable, but I'm still a bit unsure. Maybe I should look it up to confirm.Assuming that the standard enthalpy of combustion for glucose is indeed -2800 kJ/mol, then my calculation should be correct. So, the enthalpy of combustion for 10 moles of glucose would be -28000 kJ.