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# 1

## 已知函数 $$f(x) = (x - 2)e^x + ax^2 + bx$$ ， $$x = 1$$ 是 $$f(x)$$ 的一个极值点

## 1. 若 $$x = 1$$ 是 $$f(x)$$ 的唯一极值点，求实数 $$a$$ 的取值范围

读取信息： 1. 极值点： 当 $$x = 1$$， $$f^\prime(x) = 0$$ 2. 唯一极值点： 不知有何意义 3. $$a$$ 的取值范围： 我们要求的是一个范围

尝试：

$$
\begin{align\*}
f^\prime(x) &= 1 \cdot e^x + (x - 2)e^x + 2ax + b
\ \\
&= e^x + (x - 2)e^x + 2ax + b
\ \\
&= e^x(1 + x - 2) + 2ax + b
\ \\
&= e^x(x - 1) + 2ax + b
\ \\
f^\prime(1) &= 0 + 2a + b
\ \\
&= 2a + b
\ \ \\
\end{align\*}
$$

$$
\begin{align\*}
\&f^\prime(1) = 0
\ \\
&2a + b = 0
\ \\
& \text{Well, as long as we want } a \text{ , we should keep } a \text{ in the origin function}
\ \\
\&b = -2a
\ \\
\&f^\prime(x) = e^x(x - 1) + 2ax - 2a
\ \\
& \text{If we want to calculate the range of } a \text{ , we must make sure } a \text{ is isolated}
\ \\
\&f^\prime(x) = e^x(x - 1) + 2a(x - 1)
\ \\
&\text{Simplification is always good for the next step, maybe after that you'll find a way to go}
\ \\
\&f^\prime(x) = (x - 1)(e^x + 2a)
\ \\
&\text{Don't forget, now, the } x \text{ is equal to } 1
\ \\
\&f^\prime(x) = 0 \cdot (e^x + 2a)
\ \ \\
\end{align\*}
$$

$$
\begin{align\*}
&\text{It seems like } (e^x + 2a) \text{ could be any value!(Just an imagination)}
\ \\
&-\infty \le (e^x + 2a) \le +\infty
\ \\
& \text{But we all knows, as x goes up, } e^x \text{ only goes } +\infty \text{ , not } -\infty
\ \\
& e^x = +\infty \text{ , } e^x \not= -\infty
\ \\
& \text{And } 2a \text{ can only be any value } \in (-\infty, +\infty) \text{ in this case}
\ \\
&(e^x) + (2a) = ?
\\
& (+\infty) + (+\infty) = (+\infty)
\\
& (+\infty) + (-\infty) = (0)
\ \ \\
\end{align\*}
$$

$$
\begin{align\*}
\text{So } (e^x + 2a) &\in \[0, +\infty)
\ \\
(e^x + 2a) &\ge 0
\ \\
2a &\ge -e^x
\ \\
2a &\ge -(+ \infty)
\ \\
2a &\ge -\infty
\ \\
\text{Because } -\infty &\text{, 0, } +\infty \text{ split all number into 3 parts} \text{ , Therefore }
\ \\
2a \ge 0 \ge -\infty&
\ \\
a \ge 0&
\end{align\*}
$$

## 2. 讨论 $$f(x)$$ 的单调性

潜意识:

```
讨论函数的单调性，实际上就是讨论一阶导数的正负性
```

## 3. 若存在正数 $$x\_0$$ ，使得 $$f(x\_0) < a$$ ， 求实数 $$a$$ 的取值范围