已知函数 f(x)=e2x−2ex+2xf\left( x\right) = {\mathrm{e}}^{2x} - 2{\mathrm{e}}^{x} + {2x}f(x)=e2x−2ex+2x 在点 P(x0,f(x0))P\left( {{x}_{0},f\left( {x}_{0}\right) }\right)P(x0,f(x0)) 处的切线方程为 l:y=g(x)l : y = g\left( x\right)l:y=g(x) ,若对任意 x∈Rx \in \mathrm{R}x∈R ,都有 (x−x0)(f(x)−g(x))≥0\left( {x - {x}_{0}}\right) \left( {f\left( x\right) - g\left( x\right) }\right) \geq 0(x−x0)(f(x)−g(x))≥0 成立,则 x0={x}_{0} =x0= ___.