- 已知函数$f(x) = e^{2x} - 2e^{x} + 2x$在点$P(x_{0},\ f(x_{0}))$处的切线方程为$l$：$y = g(x)$，若对任意$x \in R$，都有$\left( x - x_{0} \right)\left( f(x) - g(x) \right) \geq 0$成立，则$x_{0} =$__．
解：$\because f'(x) = 2e^{2x} - 2e^{x} + 2$
$\therefore f'(x_{0}) = 2e^{2x_{0}} - 2e^{x_{0}} + 2$
$\therefore g(x) = \left( 2e^{2x_{0}} - 2e^{x_{0}} + 2 \right)\left( x - x_{0} \right) + \left( e^{2x_{0}} - 2e^{x_{0}} + 2x_{0} \right)$
$\therefore t(x) = f(x) - g(x) = e^{2x} - 2e^{x} + 2x - \left( 2e^{2x_{0}} - 2e^{x_{0}} + 2 \right)\left( x - x_{0} \right) - \left( e^{2x_{0}} - 2e^{x_{0}} + 2x_{0} \right)$
$\therefore t(x_{0}) = f(x_{0}) - g(x_{0}) = 0$
$t'(x) = 2e^{2x} - 2e^{x} + 2 - \left( 2e^{2x_{0}} - 2e^{x_{0}} + 2 \right) = 2\left( e^{2x} - e^{2x_{0}} \right) - 2\left( e^{x} - e^{x_{0}} \right)$
$= 2\left( e^{x} - e^{x_{0}} \right)\left( \left( e^{x} + e^{x_{0}} \right) - 1 \right)$
$\therefore x_{0} < \ln\frac{1}{2}$时，在$\left( x_{0}, + \infty \right)$上$t'(x) > 0$，则$t(x) > t(x_{0}) = 0$
$x_{0} > \ln\frac{1}{2}$时，在$\left( x_{0}, + \infty \right)$上$t'(x) > 0$，则$t(x) > t(x_{0}) = 0$
$\therefore x_{0} = \ln\frac{1}{2}$