Question

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

Answer 1

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]