Question

求函数y=(1/2)^(x^2-2x)的单调区间和值域

Answer 1

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