让我试试：
Dear Peter:
I know you got in trouble of this question,Let me explain of it.
Frist when a is equal to one,so f(x) is equal to x²-6x+3lnx,let me set two non-know number x₁,x₂∈R(real number)&x₁≺x₂…
(没学过 鄙人，懒得写了）
Oh, my friend, I heard you were stuck on a math problem? Let me take a look.
Hmm, it’s simple.
When a = 1, the domain of the function f(x) = x^2 - 7x + 3 ln(x) is (0, ∞). Taking the derivative gives f'(x) = 2x - 7 + 3/x. Setting f'(x) = 0 gives 2x^2 - 7x + 3 = 0, which has solutions x₁ = 1/2 and x₂ = 3. Analyzing the function’s sign in intervals shows: f'(x) > 0 on (0, 1/2) and (3, ∞), so the function is increasing; f'(x) < 0 on (1/2, 3), so the function is decreasing.
I'm glad to help you solve this math problem. Whena = 1, f(x) = x^2-7x +3lnx. Its derivative f'(x)=x/(2x-1)(x-3)，So f(x) increases on (0,1/2)and (3,+∞),and decreases on(1/2,3)
For 2<x<3e and f(x)>-6Established，we can derive that a >x^2-x/-3lnx+6x-6.After calculation, a >-e/1 satisfies the condition.
If you have more questions, just ask me!
Yours, Li Hua
♥ 2↩ 1
猫猫爱吃糖_傲娇版2026-01-11
Dear Peter,
I heard you’re stuck with the math problem about the function f(x)=ax^2-(a+6)x+3\ln x, so I’m writing to help you out.
When a=1, f(x)=x^2-7x+3\ln x. Its derivative is f'(x)=\frac{(2x-1)(x-3)}{x}, so it increases on (0,\frac{1}{2}) and (3,+\infty), and decreases on (\frac{1}{2},3). For f(x)\geq-6 when 2\leq x\leq3e, a can be any number greater than or equal to \frac{1+\ln3}{e(3e-1)}.
Hope this makes it clear for you!
Yours,
Li Hua
♥ 1↩ 1
为人民服务-江理云2026-01-03
OK了 小伙子 你没说不能用数学吧[吃瓜]那你可太小瞧我大数学了 不过是不用汉字罢了
♥ 1↩ 2
闭杂入等2025-12-22
@芭比小郭
♥ 1↩ 1
爱上章北海2025-12-21
不愧是我D导，数英双强👍
♥ 1
-鲁迅周树人-2025-12-21
Because of the titile:f(x)=x²-7x+3lnx，it's easy to know that the trend of f(x) is "reduce" when x is in (-∞,7/2] and the trend is "increase" when x is in [7/2,+∞).f(x)≥-6⇒ax²-(a+6)x+3l，我不行了我为什么要做这么愚蠢的事情，作业还没写完
♥ 1↩ 1
DL除冒2025-12-21
Dear Peter
I am writing to you for your math question, I am sorry about that, because I don't know too.[doge]
Comments
让我试试： Dear Peter: I know you got in trouble of this question,Let me explain of it. Frist when a is equal to one,so f(x) is equal to x²-6x+3lnx,let me set two non-know number x₁,x₂∈R(real number)&x₁≺x₂… (没学过 鄙人，懒得写了）
♥ 129 ↩ 3
解决了数学学霸英语不好英语学霸数学不好的问题
♥ 18 ↩ 1
(f(x)=ax^2-(a+6)x+3lnx) a∈? <=> (2<=x<=3e，f(x)>=6) f(x)→df(x)/dx=2ax-a-6+3/x ddf(x)/ddx=2a-3/x，x>0→ddf(x)/ddx<2a df(x)/dx=0 →2ax^2-(a+6)x+3=0 →x∈{x1，x2}，x∈R*， ∃x1∈{x}，x1∈【2，3e】，→f(x1)>=6 ∀x1∉{x}→∀f(x1)>=6 end
♥ 14 ↩ 1
dear Peter I don't know[doge]
♥ 9 ↩ 1
Oh, my friend, I heard you were stuck on a math problem? Let me take a look. Hmm, it’s simple. When a = 1, the domain of the function f(x) = x^2 - 7x + 3 ln(x) is (0, ∞). Taking the derivative gives f'(x) = 2x - 7 + 3/x. Setting f'(x) = 0 gives 2x^2 - 7x + 3 = 0, which has solutions x₁ = 1/2 and x₂ = 3. Analyzing the function’s sign in intervals shows: f'(x) > 0 on (0, 1/2) and (3, ∞), so the function is increasing; f'(x) < 0 on (1/2, 3), so the function is decreasing.
♥ 5 ↩ 3
众所周知数学好的基本献祭英语，英语好的基本献祭了数学[笑哭]
♥ 2 ↩ 2
我很高兴能帮你解决这道数学题。当a = 1时，f(x) = x^2-7x +3lnx。它的导数f'(x)=x/(2x-1)(x-3)，所以f(x)在(0，1/2)和(3，+∞)上增加，在(1/2，3)上减少。对于2<x<3e和f(x)>-6成立，我们可以导出一个>x^2-x/-3lnx+6x-6.经过计算，a >-e/1满足条件。如果你有更多的问题，就问我！你的，李华
♥ 2
I'm glad to help you solve this math problem. Whena = 1, f(x) = x^2-7x +3lnx. Its derivative f'(x)=x/(2x-1)(x-3)，So f(x) increases on (0,1/2)and (3,+∞),and decreases on(1/2,3) For 2<x<3e and f(x)>-6Established，we can derive that a >x^2-x/-3lnx+6x-6.After calculation, a >-e/1 satisfies the condition. If you have more questions, just ask me! Yours, Li Hua
♥ 2 ↩ 1
Dear Peter, I heard you’re stuck with the math problem about the function f(x)=ax^2-(a+6)x+3\ln x, so I’m writing to help you out. When a=1, f(x)=x^2-7x+3\ln x. Its derivative is f'(x)=\frac{(2x-1)(x-3)}{x}, so it increases on (0,\frac{1}{2}) and (3,+\infty), and decreases on (\frac{1}{2},3). For f(x)\geq-6 when 2\leq x\leq3e, a can be any number greater than or equal to \frac{1+\ln3}{e(3e-1)}. Hope this makes it clear for you! Yours, Li Hua
♥ 1 ↩ 1
OK了 小伙子 你没说不能用数学吧[吃瓜]那你可太小瞧我大数学了 不过是不用汉字罢了
♥ 1 ↩ 2
@芭比小郭
♥ 1 ↩ 1
不愧是我D导，数英双强👍
♥ 1
Because of the titile:f(x)=x²-7x+3lnx，it's easy to know that the trend of f(x) is "reduce" when x is in (-∞,7/2] and the trend is "increase" when x is in [7/2,+∞).f(x)≥-6⇒ax²-(a+6)x+3l，我不行了我为什么要做这么愚蠢的事情，作业还没写完
♥ 1 ↩ 1
Dear Peter I am writing to you for your math question, I am sorry about that, because I don't know too.[doge]
♥ 1
是真的，我朋友跟我说他们学校英语和数学联姻，问数学题[笑哭]
♥ 1
如果数学学霸英语好的话也可以应付，但如果数学学霸英语本来就不好，说明他从来就没想过考好，不写也罢
♥ 1
@社会我蝶哥
♥ 1
高一表示微距
♥ 1
就我一个看up主视频的吗
♥ 1
up主互关吗
♥ 1 ↩ 1