SOLUTION: Hi, looking for the Domain and the Range of Y= the Squareroot of (X-3) Also looking for the inverse of 4x+7. thank you.

Algebra ->  Functions -> SOLUTION: Hi, looking for the Domain and the Range of Y= the Squareroot of (X-3) Also looking for the inverse of 4x+7. thank you.      Log On


   

Question 148130: Hi,
looking for the Domain and the Range of Y= the Squareroot of (X-3)
Also looking for the inverse of 4x+7.
thank you.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x-3%29 Start with the given expression

Remember you cannot take the square root of a negative value. So that means the argument x-3 must be greater than or equal to zero (i.e. the argument must be positive)

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

x%3E=0%2B3Add 3 to both sides


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


So that means x must be greater than or equal to 3 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: [3,)


Now to find the range, we must graph the equation


+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+sqrt%28x-3%29%29+ Graph of y=sqrt%28x-3%29

From the graph, we can see that the lowest y-value is y=0. So the range in set-builder notation is


Also the range in interval notation is: [0,)






y=4x%2B7 Start with the given equation.


x=4y%2B7 Switch x and y.


x-7=4y Subtract 7 from both sides.


%28x-7%29%2F4=y Divide both sides by 4.


So the inverse function is