题目

已知函数y=sin平方x+2sinx·cosx+3cos平方 求y的取值范围

提问时间：2020-12-29

答案

y=(sinx)^2+2sinx·cosx+3(cosx)^2
= 1+ 2sinxcosx + 2(cosx)^2
= cos2x + sin2x + 2
y' = -2sin2x + 2cos2x
y' =0
=> cos2x-sin2x =0
tan2x =1
2x = π/4 or 5π/4
x = π/8 or 5π/8
y'' = -4(sin2x+cos2x)
y''(π/8) 0 ) (min)
max y = y(π/8)
= cosπ/4 + sinπ/4 + 2
= √2 +2
min y = y(5π/8)
= cos5π/4 + sin5π/4 + 2
= -√2 +2