- 已知函数$f(x) = \begin{cases} \frac{x}{e^{x+e}} - e^2, & x \leq 0 \\ -\sqrt{1 + x^2}, & x > 0 \end{cases}$，点$M$、$N$是函数$y = f(x)$图象上不同的两个点，设$O$为坐标原点，则$\tan\angle MON$的取值范围是__.
解：$f(x)$图像如右图所示
$x \leq 0$时，$f(x) = \frac{x}{e^{x+e}} - e^2$，$f'(x) = \frac{1 - x}{e^{x+e}}$
设直线$OM$与$f(x)$图像切点为$(x_0, \frac{x_0}{e^{x_0+e}} - e^2)$
则切线方程为$y - \frac{x_0}{e^{x_0+e}} + e^2 = \frac{1 - x_0}{e^{x_0+e}}(x - x_0)$
代入$(0, 0)$得$-\frac{x_0}{e^{x_0+e}} + e^2 = \frac{1 - x_0}{e^{x_0+e}}(-x_0)$
整理得到$\ln x_0^2 = x_0 + e + 2$
$\therefore 2\ln(-x_0) - x_0 = e + 2$（$x_0 < 0$）
设$g(x) = 2\ln(-x) - x$（$x < 0$），则$g'(x) = \frac{2}{x} - 1 = \frac{2 - x}{x}$
$\therefore x < 0$时$g(x)$单增 且$g(-e) = 2 + e$
$\therefore x_0 = -e$
$\therefore$切线方程为$y = (e + 1)x$
$x > 0$时$y = -\sqrt{1 + x^2}$，可化为$y^2 - x^2 = 1$
其渐近线方程为$y = -x$
$\therefore \tan\angle MON < \frac{-1 - (e + 1)}{1 + (-1)(e + 1)} = \frac{e + 2}{2}$
$\therefore \tan\angle MON \in \left( 0, \frac{e + 2}{2} \right)$