最近我看到过这样一道英语作文题，这类英语作文题很少见，但也有必要讲一讲怎么写。

简化题意：帮Peter完成一下一道题：

f(x)=ax2−(a+6)x+3ln⁡xf(x)=ax^2-(a+6)x+3\ln xf(x)=ax2−(a+6)x+3lnx
（1）讨论当a=1a=1a=1时，f(x)f(x)f(x)的单调区间
（2）求得实数aaa的一个范围使得当2≤x≤3e2\leq x \leq 3e2≤x≤3e时f(x)≥−6f(x)\geq-6f(x)≥−6恒成立

解：
(1)\quad(1)(1)当a=1a=1a=1时，f(x)=x2−7x+3ln⁡xf(x)=x^2-7x+3\ln xf(x)=x2−7x+3lnx

\qquad定义域为(0,+∞)(0,+\infty)(0,+∞)，f′(x)=2x−7+3x=2x2−7x+3x=(x−3)(2x−1)xf'(x)=2x-7+\dfrac 3x=\dfrac{2x^2-7x+3}{x}=\dfrac{(x-3)(2x-1)}{x}f′(x)=2x−7+x3​=x2x2−7x+3​=x(x−3)(2x−1)​

\qquad可能的极值点：x1=12,x2=3x_1=\dfrac 12,x_2=3x1​=21​,x2​=3

\qquad单调递增区间为(0,12](0,\dfrac 12](0,21​]和[3,+∞)[3,+\infty)[3,+∞)，单调递减区间为[12,3][\dfrac 12,3][21​,3]

(2)f′(x)=2ax−(a+6)x+3ln⁡x=(ax−3)(2x−1)x\quad(2)f'(x)=2ax-(a+6)x+3\ln x=\dfrac{(ax-3)(2x-1)}{x}(2)f′(x)=2ax−(a+6)x+3lnx=x(ax−3)(2x−1)​

\qquad依题意，f(2)=2a−12+3ln⁡2≥−6f(2)=2a-12+3\ln 2\geq-6f(2)=2a−12+3ln2≥−6，即a≥3−3ln⁡22a\geq 3-\dfrac{3\ln 2}{2}a≥3−23ln2​

\qquad当a≤6a\leq6a≤6时，3a≥12\dfrac{3}{a}\geq\dfrac 12a3​≥21​，(3a,+∞)(\dfrac 3a,+\infty)(a3​,+∞)为单调递增区间

\qquad当a>6a>6a>6时，3a<12\dfrac 3a<\dfrac 12a3​<21​，(12,+∞)(\dfrac 12,+\infty)(21​,+∞)为单调递增区间

\qquad所以当a≥3−3ln⁡22a\geq3-\dfrac{3\ln 2}{2}a≥3−23ln2​时，(2,+∞)(2,+\infty)(2,+∞)为单调递增区间

\qquad所以f(3e)>f(2)≥−6f(3e)>f(2)\geq-6f(3e)>f(2)≥−6

\qquad综上所述，aaa的取值范围为(3−3ln⁡22,+∞)(3-\dfrac{3\ln 2}{2},+\infty)(3−23ln2​,+∞)

Dear Peter:\text{Dear Peter:}Dear Peter:

I’m glad to write this letter to you.And I have solved the quetion you asked me before.Now let me tell you how \qquad \text{I'm glad to write this letter to you.And I have solved the quetion you asked me before.Now let me tell you how }I’m glad to write this letter to you.And I have solved the quetion you asked me before.Now let me tell you how 
to do it.\text{to do it.}to do it.

For the first quetion,when a=1,f(x)=x2−7x+3ln⁡x.And f′(x)=2x−7+3x=(x−3)(2x−1)x.So we can know\text{For the first quetion,when a=1,}f(x)=x^2-7x+3\ln x \text{.And }f'(x)=2x-7+\dfrac 3x=\dfrac{(x-3)(2x-1)}{x}\text{.So we can know}For the first quetion,when a=1,f(x)=x2−7x+3lnx.And f′(x)=2x−7+x3​=x(x−3)(2x−1)​.So we can know
the possible extreme points are x1=12,x2=3.Therefore the monotone increasing interval is (0,12]and [3,+∞).\text{the possible extreme points are }x_1=\dfrac 12,x_2=3\text{.Therefore the monotone increasing interval is }(0,\dfrac 12]\text{ and }[3,+\infty).the possible extreme points are x1​=21​,x2​=3.Therefore the monotone increasing interval is (0,21​] and [3,+∞).
And the monotone decreasing interval is [12,3].\text{And the monotone decreasing interval is }[\dfrac 12,3]\text{.}And the monotone decreasing interval is [21​,3].

For the second question,It’s obvious that f(2)=2a−12+3ln⁡2≥−6.It means that a≥3−3ln⁡22.Therefore the\text{For the second question,It's obvious that }f(2)=2a-12+3\ln 2\geq-6\text{.It means that }a\geq 3-\dfrac{3\ln 2}{2}\text{.Therefore the}For the second question,It’s obvious that f(2)=2a−12+3ln2≥−6.It means that a≥3−23ln2​.Therefore the
possible extreme points are x1=12,x2=3a.We can prove that : When a≤6,(3a,+∞)is a monotone increasing \text{possible extreme points are }x_1=\dfrac 12,x_2=\dfrac 3a\text{.We can prove that : When }a\leq 6,(\dfrac 3a,+\infty)\text{ is a monotone increasing }possible extreme points are x1​=21​,x2​=a3​.We can prove that : When a≤6,(a3​,+∞) is a monotone increasing 
increasing interval.And when a>6,(12,+∞)is a monotone increasing interval.Therefore when a≥3−3ln⁡22,\text{increasing interval.And when }a>6,(\dfrac 12,+\infty)\text{ is a monotone increasing interval.Therefore when }a\geq 3-\dfrac{3\ln 2}{2},increasing interval.And when a>6,(21​,+∞) is a monotone increasing interval.Therefore when a≥3−23ln2​,
(2,+∞)must be a monotone increasing interval.Therefore f(3e)>f(2)≥−6.In conclusion, the real number a (2,+\infty)\text{ must be a monotone increasing interval.Therefore }f(3e)>f(2)\geq -6\text{.In conclusion, the real number a }(2,+∞) must be a monotone increasing interval.Therefore f(3e)>f(2)≥−6.In conclusion, the real number a 
should be in the range of (3−3ln⁡22,+∞).\text{should be in the range of }(3-\dfrac{3\ln 2}{2},+\infty).should be in the range of (3−23ln2​,+∞).

This is my way to solve the quetion.Look forward to your early reply.\text{This is my way to solve the quetion.Look forward to your early reply.}This is my way to solve the quetion.Look forward to your early reply.

Yours,\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{Yours,}Yours,

Li Hua\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{Li Hua}Li Hua

文章内解题过程并不严谨，但毕竟是书信，且要求100词左右，所以解题思路明确，语法无误即可。