第2个回答 2020-01-30
f(x)=4|sin〔(1/2)x+π/3〕|+4|cos〔(1/2)x+π/3〕|的值域怎么求?
可以分段讨论,考虑一个周期
当0<=t=1/2x+π/3<π/2时,
y=4sin(x/2+π/3)+4cos(x/2+π/3)=4√2sin(t+π/4), 该区间的值域[4, 4√2]
当π/2<=t=1/2x+π/3<π时,
y=4sin(x/2+π/3)-4cos(x/2+π/3)=4√2sin(t-π/4), 该区间的值域[4, 4√2]
当π<=t=1/2x+π/3<3π/2时,
y=-4sin(x/2+π/3)-4cos(x/2+π/3)=-4√2sin(t+π/4), 该区间的值域[4, 4√2]
当3π/2<=t=1/2x+π/3<2π时,
y=-4sin(x/2+π/3)+4cos(x/2+π/3)=-4√2sin(t-π/4),该区间的值域[4, 4√2]
可求得:
值域为[4, 4√2]本回答被网友采纳