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) = -1<0 (max)
f''(3π/2) = 1 >0 (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]

Answer 2

f′=cosx +sinx +1
所以f′=√2sin(π/4+x)+1
所以当5π/4＜x+π/4＜7π/4时，f′＜0
所以在[π,3π/2]上，f单调递减，在区间[0,π],[3π/2,2π]上单调递增.
当x=π时,f取极大值，为π+1.
当x=3π/2时，f取最小值，为1/2-3π/2.