SOLUTION: Find the domain of the following function f(x)=the square root of the quantity 2x+5

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: Find the domain of the following function f(x)=the square root of the quantity 2x+5      Log On


   

Question 134220: Find the domain of the following function f(x)=the square root of the quantity 2x+5
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282x%2B5%29 Start with the given expression

Remember you cannot take the square root of a negative value. So that means the argument 2x%2B5 must be greater than or equal to zero (i.e. the argument must be nonnegative)

2x%2B5%3E=0 Set the inner expression greater than or equal to zero

2x%3E=0-5Subtract 5 from both sides


2x%3E=-5 Combine like terms on the right side


x%3E=%28-5%29%2F%282%29 Divide both sides by 2 to isolate x



So that means x must be greater than or equal to -%285%29%2F%282%29 in order for x to be in the domain

So the domain in set-builder notation is


So here is the domain in interval notation: [-%285%29%2F%282%29,)



Notice if we graph y=sqrt%282x%2B5%29 , we get
+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+sqrt%282x%2B5%29%29+ notice how the graph never crosses the line x=-%285%29%2F%282%29. So this graphically verifies our answer.

and we can see that x must be greater than or equal to -%285%29%2F%282%29 in order to lie on the graph