Question

求函数y=sinxcosx-sinx+cosx+2的值域

Answer 1

因为sinxcosx=1/2sin2x
又 -1/2<=1/2sinzx<=1/2
因为-1<=sinx<=1
所以 -1/2<=1/2sin2x-sinx<=1/2
同理可得 -3/2<=1/2sin2x-sinx+cosx<=3/2
则 1/2<=1/2sin2x-sinx+cosx+2<=7/2
答：函数y=sinxcosx-sinx+cosx+2的值域为[1/2,7/2]

Answer 2

方法一：
利用导数求函数的单调性
方法二：
sinxcosx-1/2=-1/2(sinx-cosx)^2 ,故
y=sinxcosx-sinx+cosx+2
=-1/2(sinx-cosx)^2-(sinx-cosx)+2.5
设t=sinx-cosx=根号2[sin(x-π/4)],显然-根号2<=t<=根号2
y=-1/2t^2-t+2.5=-1/2(t+1)^2+3
显然其取值范围为：1.5-根号2<=y<=3