Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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Finding the Domain of a Function - Problem 4

Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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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.

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