设函数f(x)=sinx-cosx+x+1,0<x<2π,求函数f(x)的单调区间与极值

问题描述:

设函数f(x)=sinx-cosx+x+1,0<x<2π,求函数f(x)的单调区间与极值

解析:f(x)=sinx-cosx+x+1则f'(x)=cosx+sinx+1=√2(√2/2cosx+√2/2sinx)+1=√2(sinπ/4cosx+cosπ/4sinx)+1=√2sin(x+π/4)+1.令0<x+π/4<π/2且3π/2<x<2π得-π/4<x<π/4且5π/4...