Question

求f(x)=(1+sinx)(1+cosx)的最大值和最小值.

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)