已知函数f（x）=2sin（(pai\4)x+(pai\4)）；当x属于[-6,-2\3]时,求函数y=f（x）+f（x

f(x)=2sin((π/4)x+π/4),故：f(x+2)=2sin((π/4)x+3π/4)
所以：f(x)+f(x+2)=2sin((π/4)x+π/4)+2sin((π/4)x+3π/4)
=2*2sin(πx/4+π/2)cos(π/4)=2sqrt(2)cos(πx/4)
当x∈[-6,-2/3]时,πx/4∈[-3π/2,-π/6]
当πx/4=-π,即x=-4时,函数取得最小值-2sqrt(2)；
当πx/4=-π/6,即x=-2/3时,函数取得最大值sqrt(6)