SOLUTION: state the domain: g(x)= square root 5x+ 2 5x+2=0 x= -2/5 domain: (- indef. , -2/5] is it this correct?

Algebra ->  Radicals -> SOLUTION: state the domain: g(x)= square root 5x+ 2 5x+2=0 x= -2/5 domain: (- indef. , -2/5] is it this correct?       Log On


   



Question 405429: state the domain:
g(x)= square root 5x+ 2
5x+2=0
x= -2/5
domain: (- indef. , -2/5]
is it this correct?

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Please put parentheses around radicands (the expressions within a radical). For example:
g(x) = square root(5x+2)
What you posted could be interpreted as
g%28x%29++=+sqrt%285x%2B2%29
or
g%28x%29++=+sqrt%285x%29%2B2
which are different functions with different domains.

From your attempted solution, I guess that the correct function is:
g%28x%29++=+sqrt%285x%2B2%29
The domain of this function will be all the numbers that make the radicand non-negative (positive or zero). So you just solve the inequality:
5x%2B2+%3E=0
Subtract 2 from each side:
5x+%3E=+-2
Divide by 5:
x+%3E=+-2%2F5
This is the domain. In interval notation this is:
[-2%2F5, infinity)