求函数f(x)=(sinx+√2)*(cosx+√2)的最大值和最小值

因为(sinx+cosx)^2=1+2sinxcosx
sinxcosx=[(sinx+cosx)^2-1]/2
令sinx+cosx=t t=√2sin(x+π/4) -√2≤t≤√2
f(x)=(sinx+√2)(cosx+√2)
=sinxcosx+√2(sinx+cosx)+2
=(t^2-1)/2+√2t+2
=t^2/2+√2t+3/2
=1/2(t+√2)^2+1/2
对称轴t=-√2
y在[-√2,√2]上单调递增
t=-√2 最小值y=1/2
t=√2 最大值y=9/2