设u=x^2-2x,y=(1/2)^(x^2-2x)=(1/2)^u是减函数.
u=(x-1)^2-1,在[1,+无穷)上是单调递增,在(-无穷,1]上是单调递减.
所以,函数y=(1/2)^(x^2-2x)的单调增区间是(-无穷,1],减区间是[1,+无穷)
u>=-1,则(1/2)^u