How do you find the domain and range of #Y=(2x-1)^(1/2)#?

1 Answer
Jul 23, 2017

Answer:

The domain is # x in [1/2, +oo)#
The range is #y in [0,+oo)#

Explanation:

Our function is

#y=f(x)=sqrt(2x-1)#

What's under the square root sign is #>=0#

Therefore,

#2x-1>=0#

#x>=1/2#

The domain is # x in [1/2, +oo)#

The minimum value of #y=0# and the maximum value is #+oo#

The range is #y in [0,+oo)#

graph{sqrt(2x-1) [-6.244, 6.243, -3.12, 3.123]}