Question

函数f(x)=x-2sinx在区间[-2π/3,2π/3]上的最大值是

Answer 1

求导f'(x)=1-2cosx

令f'(x)=1-2cosx=0,cosx=1/2

得x=（+/-）π/3

f(-2π/3)=-2π/3-2*(-根号3/2)=-2π/3+根号3

f(-π/3)=-π/3-2*(-根号3/2)=-π/3+根号3

f(π/3)=π/3-2*根号3/2=π/3-根号3

f(2π/3)=2π/3-根号3

故最大值=f(-π/3)=-π/3+根号3