Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
To unlock all 5,300 videos, start your free trial.
Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Let’s find the domain of another function. This one is g(x) equals 1 over 3 minus the square root of x plus 1. There are only two things we need to worry about when we are looking for domain. One, is division by 0 and the second is a negative in the radical. Let’s start with the second one.
Let’s make sure there isn’t a negative and a radical by saying that x plus 1 has got to be greater than or equal to 0. That means, x is greater than or equal to -1. So that’s our starting point. And then the second thing, we’d have division by 0, if 3 minus the square root of x plus 1 ever equals 0. We can’t have this.
This cannot equal 0, which means now I can add the radical to both sides. This is actually inequality when you use the not equal to symbol. You can solve it just like any other inequality. I’m going to have the radical to both sides and get 3 is not equal to the square root of x plus 1. Then I can square both sides. 9 is not equal to x plus 1. Subtract 1 and I get 8 is not equal to x. And I discover that I need x to not equal 8 and I need it to be greater and not equal to 1.
Now if I draw our little number line, we got -1 and 8. And the greater than or equal to -1, and that looks like this. But I don’t want x to be 8, so I have to avoid this point here. So I just need to translate this interval notation. So I want for my domain, the interval from -1 to 8 including -1 but not including 8. Union the interval from 8 to infinity.
And that’s it just remember two things to look for. Division by 0 can’t have that and a negative in the radical. We have to have numbers in the radical be greater than or equal to 0.
Unit
Introduction to Functions