函數y=(13)−2x2−8x+1(-3≤x≤1)的值域是______.

函數y=(13)−2x2−8x+1(-3≤x≤1)的值域是______.

設t=-2x2-8x+1=-2(x+2)2+9,∵-3≤x≤1,∴當x=-2時,t有最大值是9;當x=1時,t有最小值是-9,∴-9≤t≤9,由函數y=(13)x在定義域上是减函數,∴原函數的值域是[3-9,39].故答案為:[3-9,39].