Question

设函数y=(1/3)^(1-x)的值域为A,y=x+x^-1的值域为B,则A∪B=

Answer 1

y=(1/3)^(1-x)的值域为A A>0
y=x+x^-1的值域为B B>=2或B<=-2
A∪B=(-∞,-2]∪(0,+∞)

Answer 2

A={x|x>0}
B={x|x<=-2或x>=2}
A∪B={x|x<=-2或x>0}

Answer 3

　A∪B=（0，正无穷）∪【-2，负无穷）