数学题类英语作文（导数大题）

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完成以下一道题：

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

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

\qquad 定义域为 ( 0 , + ∞ ) (0,+\infty) (0,+∞)， f ′ ( x ) = 2 x − 7 + 3 x = 2 x 2 − 7 x + 3 x = ( x − 3 ) ( 2 x − 1 ) x f'(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： x 1 = 1 2 , x 2 = 3 x_1=\dfrac 12,x_2=3 x1=21,x2=3

	( 0 , 1 2 ) ( 0,\dfrac 12) (0,21)	1 2 \dfrac 12 21	( 1 2 , 3 ) (\dfrac 12,3) (21,3)	3 3 3	( 3 , + ∞ ) (3,+\infty) (3,+∞)
f ′ ( x ) f'(x) f′(x)	+ + +	0 0 0	− - −	0 0 0	+ + +
f ( x ) f(x) f(x)	↗ \nearrow ↗	极大值	↘ \searrow ↘	极小值	↗ \nearrow ↗

\qquad 单调递增区间为 ( 0 , 1 2 ] (0,\dfrac 12] (0,21]和 [ 3 , + ∞ ) [3,+\infty) [3,+∞)，单调递减区间为 [ 1 2 , 3 ] [\dfrac 12,3] [21,3]

( 2 ) f ′ ( x ) = 2 a x − ( a + 6 ) x + 3 ln ⁡ x = ( a x − 3 ) ( 2 x − 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 ) = 2 a − 12 + 3 ln ⁡ 2 ≥ − 6 f(2)=2a-12+3\ln 2\geq-6 f(2)=2a−12+3ln2≥−6，即 a ≥ 3 − 3 ln ⁡ 2 2 a\geq 3-\dfrac{3\ln 2}{2} a≥3−23ln2

\qquad 当 a ≤ 6 a\leq6 a≤6时， 3 a ≥ 1 2 \dfrac{3}{a}\geq\dfrac 12 a3≥21， ( 3 a , + ∞ ) (\dfrac 3a,+\infty) (a3,+∞)为单调递增区间

\qquad 当 a > 6 a>6 a>6时， 3 a < 1 2 \dfrac 3a<\dfrac 12 a3<21， ( 1 2 , + ∞ ) (\dfrac 12,+\infty) (21,+∞)为单调递增区间

\qquad 所以当 a ≥ 3 − 3 ln ⁡ 2 2 a\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 ( 3 e ) > f ( 2 ) ≥ − 6 f(3e)>f(2)\geq-6 f(3e)>f(2)≥−6

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

题目解完了，接下来就是用英文写信。

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 ) = x 2 − 7 x + 3 ln ⁡ x .And f ′ ( x ) = 2 x − 7 + 3 x = ( x − 3 ) ( 2 x − 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 x 1 = 1 2 , x 2 = 3 .Therefore the monotone increasing interval is ( 0 , 1 2 ] 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 [ 1 2 , 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 ) = 2 a − 12 + 3 ln ⁡ 2 ≥ − 6 .It means that a ≥ 3 − 3 ln ⁡ 2 2 .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 x 1 = 1 2 , x 2 = 3 a .We can prove that : When a ≤ 6 , ( 3 a , + ∞ ) 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 , ( 1 2 , + ∞ ) is a monotone increasing interval.Therefore when a ≥ 3 − 3 ln ⁡ 2 2 , \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 ( 3 e ) > 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 − 3 ln ⁡ 2 2 , + ∞ ) . \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。

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数学题类英语作文（导数大题）

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