求函数y=log1/2(x2-x-2)的值域.单调区间

问题描述:

求函数y=log1/2(x2-x-2)的值域.单调区间

x2-x-2>0,x<-1或x>2,∴y=log1/2(x2-x-2)的值域为R
x∈(-无穷,-1)时x2-x-2为减
x∈(2,正无穷)时x2-x-2为增
且y=log1/2(x)为减函数
x用x2-x-2来换.∴y=log1/2(x2-x-2)在
(-无穷,-1)为增,(2,正无穷)为减