题目

y=sinx+cosx+sinxcosx+1的最值

提问时间：2020-12-30

答案

y = sinx+cosx+sinxcosx+1
y' = cosx - sinx - (sinx)^2 + (cosx)^2
= (cosx+1/2)^2- (sinx+1/2)^2
= (cosx+sinx+1)(cosx-sinx)
y' =0
=> sinx = cosx or cosx+sinx+1=0
=> tanx = 1 or sin(π/4 +x ) = -√2/2
=> x = π/4 or x = π
max y , at x= π/4
max y = sinπ/4 + cosπ/4 + sinπ/4cosπ/4 + 1
= √2/2+√2/2+(√2/2)√2/2+1
= 3/2+ √2
miny, at x= π
miny = sinπ + cosπ+ sinπcosπ+1
= 0-1+(-1).0+1
= 0