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# Finding the Domain of a Function - Problem 3

###### 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.

We are finding the domains of functions. This time we got a harder example let’s find the domain of h(x) equals the square root of 15 minus 2x minus x². So remember, with radical functions, we need what’s inside the radical to be greater than or equal to 0. So we have this inequality to solve; 15 minus 2x minus x², greater than or equal to 0.

When I solve polynomial inequalities usually I like to have them factored first. So let’s factor this. We'll have an x here and here. One of this will have to be minus and the other will have to be plus. I have to think about factors of 15 that add or subtract to 2. So 3 and 5 will probably work here.

So one of these will be 3 and one will be 5. If I put a 5 here and a 3 here, I’ll get 3x minus 5x is -2x. So that will work. So the way I usually solve these kinds of inequalities is I make a sign chart. And I want to analyze when this thing is positive, when it's negative and when it's 0.

So on the sign chart, a sign chart is basically a number line. I’m going to mark the numbers -5 and 3. Now these are places where the expression 5 plus x times 3 minus x is actually equal to 0. It's 0 in both these points. All we have to do is test points to the left in between and to the right of these two to see whether the expression is positive or negative.

So we test points. Let’s test to the left. Let’s test x equals - 6. 5 plus - 6 is -1, times 3 minus -6 is 9. We don’t really even have to multiply this we just need to know it's negative. Less than 0, so this is going to be negative. Then we test a point in between like 0. And we get 5 plus 0 which is 5, times 3 minus 0, 3, and that’s greater than 0. So it’s positive.

We need to know what this is. And finally for a point to the right of 3 just pick 4. 5 plus 4 is 9, times 3 minus 4, -1. That’s less than 0. So we are negative here again. What were we looking for? We wanted this quantity to the greater than or equal to 0. It’s greater than 0 between -5 and 3 and it's equal to 0 at -5 and at -3.

So this is the interval that we are looking for, that’s our domain. So the domain is the interval of -5 to 3.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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