题目

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

提问时间：2020-08-04

答案

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]