求极值?
求函数f(x)=sinx+cosx在[0,2π]上的极值? (用二阶导数求极值)
f'(x)=cosx-sinx, f''(x)=-(sinx+cosx) 由f'(x)=cosx-sinx=0,0≤x≤π,得驻点x1=π/4, x2=5π/4, ∵ f''(π/4)=-[sin(π/4)+cos(π/4)]=-√20, ∴ x=5π/4时,f(min)=-√2
答:方法一: y=cosx+sinx =(根2)*[(根2)/2*cosx+(根2)/2*sinx] =(根2)*[sin(π/4)cosx+cos(π/4)sin...详情>>