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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