What is the domain and range for #f(x)=sqrt(2x-8)#?

1 Answer
Sep 21, 2015

Answer:

The domain is: #[4,oo)#
The range is: #(0,oo)#

Explanation:

#f(x)=sqrt(2x-8)=># square root of negative number is not real:
hence the domain is calculated as follow:
#2x-8>=0#
#2x>=8#
#x>=4#
In interval form:
#[4 , oo)#
Now to find the range, consider the possible minimum and the maximum values of y for the valid domain, so in this case the range is all positive real numbers:
#y>0#
#(0,oo)#