What is the domain and range of #y= 1/2(2)^x#?

1 Answer
Jan 24, 2016

The domain is #(-oo, oo)#. The range is #(0, oo)#.

Explanation:

#2^x# is well defined for any Real number #x#. Hence the function #f(x) = 1/2 (2)^x# is also well defined for any #x in (-oo, oo)#.

It is also continuous and strictly monotonically increasing.

As #x->-oo# we find #2^x -> 0_+#

As #x->oo# we find #2^x -> oo#

So the range is #(0, oo)#

graph{2^x/2 [-10.12, 9.88, -1.52, 8.48]}