函数f(x)=x-2cosx,x[0,2π)的极值点为

f'(x)=1+2sinx
令f'(x)=0
x=7π/6或x=11π/6
x0 (0,7π/6) 7π/6(7π/6,11π/6)11π/6 (11π/6,2π)
f'(x) + - 0+
f(x) ↑极大↓极小 ↑
极大 f(7π/6)=[7π+(6√3)]/6
极小 f(11π/6)=[11π-(6√3)]/6