Question

急求！！已知函数f(X)=sinx+cosx+sinxcosx,求该函数的值域

Answer 1

解：令t=sinx+cosx=√2sin(x+π/4)
故：-√2≤t≤√2
故：t²=1+2sinxcosx
故：sinxcosx=（t²-1）/2
故：f（x）=sinx+cosx+sinxcosx
=t+（t²-1）/2
=1/2•(t²+2t-1)
=1/2•(t+1)²-1
当t=-1时，函数f（x）=sinx+cosx+sinxcosx=1/2•(t+1)²-1取最小值-1
当t=√2时，函数f（x）=sinx+cosx+sinxcosx=1/2•(t+1)²-1取最大值3/2+√2
故：函数f（x）=sinx+cosx+sinxcosx的值域为[-1，3/2+√2]

Answer 2

解：
设t=sinx+cosx
=√2[sinx*(√2/2)+cosx*(√2/2)]
=√2[sinx*cos(π/4)+cosxsin(π/4)]
=√2sin(x+π/4)∈[-√2,√2]
t²=1+2sinxcosx

y=t+(t²-1)/2
=(t²+2t-1)/2
=(t+1)²/2-1
t=-1时，y有最小值-1
t=√2时，y有最大值√2+(1/2)
所以，值域是[-1, 1/2+√2]