- 已知函数$f(x) = e^x + x$，$g(x) = 3x$，且$f(m) = g(n)$，则$n - m$的最小值为
$A. 1 - \ln 2$ $B. 2(1 - \ln 2)$ $C. \frac{1}{3}(2 - \ln 2)$ $D. \frac{2}{3}(1 - \ln 2)$
解：依题意可知$e^m + m = 3n$
$\therefore n - m = \frac{e^m + m}{3} - m = \frac{e^m - 2m}{3}$
设$h(x) = e^x - 2x$，则$h'(x) = e^x - 2$
$\therefore x > \ln 2$时$h(x)$单增，$x < \ln 2$时$h(x)$单减
$\therefore h(x)_{极小值} = h(\ln 2) = 2 - 2\ln 2$
$\therefore n - m \geq 2 - 2\ln 2$