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
)
=
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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