Question

函数f(x)=sinx(1+cosx)在区间［0,2π］上的最大值是.......最小值是........

Answer 1

解：
f(x)=sinx(1+cosx) x∈0,2π］
f’(x)=cosx+cos²x-sin²x=cosx+2cos²x-1=(cosx+1)(2cosx-1)得
令f’(x)＞0得
cosx<-1或cosx>1/2
即0<x<π/3或5π/3<x<2π
∴f(x)在[0,π/3]，(5π/3,2π)递增
令f’(x)＜0得
-1<cosx<1/2
即π/3<x<5π/3
∴ f(x)在[π/3,5π/3]递减
∴函数f(x)在x=π/3处取最大值，f(x)在x=5π/3处取最小值
f(π/3)=sinπ/3(1+cosπ/3)=3√3/4
f(5π/3)=sin5π/3(1+cos5π/3)=-3√3/4
∴函数f(x)的最大值为3√3/4，最小值为3√3/4。