求函数y=(1/2)^(-x^2+2x+1)的定义域、值域、单调区间

定义域为Rt=-x²+2x+1=-(x-1)²+2≤2∵y=(1/2)^t是减函数∴y=(1/2)^t≥(1/2)^2=1/4∴函数值域为[1/4,+∞)x∈(-∞,1],t=-(x-1)²+2递增,y=(1/2)^t递减∴单调递减区间为(-∞,1]x∈[1,+∞),t=-(x-1)²+...