Question

函数y=

1

2x+1

的值域是______．

Answer 1

∵y=

1

2x+1

=

1

2(x+ 1 2 )

=

1 2

x+ 1 2

又∵y =

1 2

x

的图象向左平移

1

2

可得y=

1 2

x+ 1 2

的图象，且反比例函数y=

1 2

x

≠0
∴y=

1

2x+1

=

1

2(x+ 1 2 )

=

1 2

x+ 1 2

≠0
故答案为{y|y≠0}