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

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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

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
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

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
f(x) =sinx-cosx+x+1
f'(x) = cosx +sinx +1 =0
√2(sin(x+π/4)) = -1
x+π/4 = 5π/4 or 7π/4
x=π or 3π/2
f''(x) = -sinx + cosx
f''(π) = -10 (min)
max f(x) = f(π) = π+2
min f(x)=f(3π/2) = 3π/2
单调区间
增加 (0,π] or [3π/2,2π)
减小 [π,3π/2]