Question

求f(x)=(1+sinx)(1+cosx)的最大值和最小值。（要解题过程）

Answer 1

f(x)=(1+sinx)(1+cosx)＝1+(sinx+cosx)+sinxcosx
令t=sinx+cosx =√2sin(x+π/4)
则-√2≤t≤√2
f(x)=1+t+(t^2-1)/2 =1/2(t^2+2t)+1/2
=1/2(t+1)^2
最小值是0(此时t=-1),最大值是√2+3/2(此时t=√2)

Answer 2

解：f(x)=sinx+cosx+sinxcosx+1.可设t=sinx+cosx=(√2)sin[x+(π/4)].t∈[-√2,√2].sinxcosx=(t²-1)/2.∴f(x)=t+1+(t²-1)/2=(t+1)²/2.∴f(x)max=(1+√2)²/2.f(x)min=0.