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

1 Answer
iceman
Sep 21, 2015

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)

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