Question

求函数f(+x)=sinx++cosx(0≤x≤2π)的单调区间和极值

Answer 1

问：求函数f(+x)=sinx++cosx(0≤x≤2π)的单调区间和极值

答：f(+x)=sinx++cosx(0≤x≤2π)=√2/2*sin(x+π/4)∵0≤x≤2π∴π/4≤x+π/4≤9π/4单调增区间【0，π/4】与【5π/4，2π】

答：单调减区间【π/4，5π/4】

答：当x+π/4=π/2时即x=π/4时，函数f(+x)=sinx++cosx(0≤x≤2π)极大值为1当x+π/4=3π/2时即x=5π/2时，函数f(+x)=sinx++cosx(0≤x≤2π)的极小值为-1

答：总之正确答案是f(+x)=sinx++cosx(0≤x≤2π)=√2/2*sin(x+π/4)2kπ-π/2≤x+π/4≤2kπ+π/2时函数单调增2kπ-5π/4≤x≤2kπ+π/4时函数单调增同理2kπ+π/4≤Ⅹ≤2kπ+5π/4时函数单调减∵0≤x≤2π∴π/4≤x+π/4≤9π/4∴单调增区间【0，π/4】与【5π/4，2π】单调减区间【π/4，5π/4】当x+π/4=π/2时即x=π/4时，函数f(+x)=sinx++cosx(0≤x≤2π)极大值为1当x+π/4=3π/2时即x=5π/2时，函数f(+x)=sinx++cosx(0≤x≤2π)的极小值为-1