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.
We are finding the domain of functions. Let’s do a harder example. Find the domain of k(x) equals to square root of 4 over x plus 1 minus 1 over x minus 4.
My focus here is going to be to make sure that what’s inside the radical, is greater than or equal to 0, because we can’t have a negative in the radical. So 4 over x plus 1 minus 1 over x minus 4, has to be greater than or equal to 0. When I’m solving a rational inequality like this, I need to get a common denominator. So for this first expression, 4 over x plus 1, I need to multiply the top and bottom by x minus 4. And for the second expression, 1 over x minus 4, I need to multiply the top and bottom by x plus 1. So that will give me the same denominator for both of these guys.
So I have 4x minus 16, over x plus 1, x minus 4 minus and I’ve got x plus 1 over x minus 4, x plus 1. I need to distribute this negative sign over both of these terms when I combine. So I have 4x minus 16 minus x minus 1. And then my denominator will be x plus 1 x minus 4 greater than or equal to 0.
So just simplify the numerator and I’ll take this over here. I get 3x minus 17 over x plus 1, x minus 4. This is greater than or equal to 0. This kind of inequality I can solve using sign chart. Now a sine chart is just a number line, and on this number line let’s call this giant expression. It’s basically k(x) so let’s call it that I’ll put k x over here. It’s going to undefined at -1 and 4.
And it’s going to equal 0 at 17 over 3. 17 over 3 is almost 6 so I need to put those numbers on a number line. -1 goes first for 17 over 3 it doesn’t have to be the scale and then it's undefined here. So I’ll put a u . Undefined at 4 so I’ll put a u. And 17 over 3 the whole thing is going to be 0. So I’ll put a 0 here. These 3 numbers divide the number line into 4 regions and I have to test each of these regions to see whether k(x) is positive or negative. So let me do that.
Let’s test to the left of- 1. with x equals -2. So I’m plugging -2 into this, and I get 3 times -2 – 6minus 17 is minus 23 over -2 plus 1 is -1, times -2 minus 4 -5. I don’t need to know what this comes out to be I just need to know whether it’s positive or negative. The denominator is going to work out to be positive and the numerator is negative. So this will be negative. And I have negatives to the left of -1.
Let’s try something in between -1 and 4, 0. When I plus in 0, I get –17 on top. 1 times -4 on the bottom, that’s a negative over negative which will be positive. That means it's positive here. Let’s try 5 x equals 5. We get 3 times 5, 15 minus 17, -2, over 5 plus 1, 6, times 5 minus 4, 1. A negative over positive which is negative.
So it’s negative between 4 and 17/3. 17/3 is very nearly 18/3 which is 6. I could plug in 6 to find out what happens to the right of 17/3. So I get 3 times 6, 18 minus 17 is 1. 6 plus 1, 7. 6 minus 4, 2. It’s all positive, so I have positive value over here. Let’s remind us what we are looking for.
When is this expression greater than or equal to 0? It’s greater than 0 between -1 and 4. It’s greater than 0 over here, and it equals 0 here. That’s okay too, that’s okay for our domain. But we got to stay away from -1 and 4 because the rational expressions are undefined there.
So this is going to give us our domain this interval and this interval. Our domain is from -1 to 4, and from 17/3 to infinity. And I need to include 17/3 because the function is 0 there. So 17/3 to infinity. And that’s our answer.
Unit
Introduction to Functions