Domain of Function Part 1
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Domain of Function Part 1


>>This is your Math Gal Julie Harland. Please visit my website at YourMathGal.com where all of my videos are organized by topic. In this video, you’re going to learn how to find the domain of functions that are given using function notation such as the four you see right here. So here’s the first one. State the domain of F of X equals 5X2 minus 7X. Basically, the question is what are the values that you can replace X with so that you got a real number. And there’s actually no restrictions here. You could put any number in for X and plug it in and you’ll get a real number. So in this case, the solution is just all real numbers. The trick to this is to see if there are any restrictions and eliminate those possibilities. If you can’t find any restrictions, we’ll just put all real numbers. So the solution here is all real numbers. I was not very good in writing, sorry. All real numbers. You know, some people write all real numbers like this, it’s like T lines with an R. That means all real numbers. And interval rotation, you might write from negative infinity to infinity. Those are all perfectly good ways of writing the domain, they’re all real numbers. So that’s the domain of the function, OK? Usually, you say kind of script capital D to represent the domain. Here’s our next one. In G of X we have a fraction. So here’s the trick. If the denominator of a fraction is 0, the fraction is undefined, which means it would not be a real number. So in this particular function, we have to make sure that we don’t let X be any number that yields a 0 in the denominator because then that would not be a real number, when we plug in that value. So the denominator is 3X minus 5, so 3X minus 5 cannot be 0. So since 3X minus 5 can’t be 0, that’s what we write. 3X minus 5 cannot be 0. So let’s solve this to see what X can’t be. So if we add 5 to both sides, it’s not equal. 3X is not equal to 5 and then divided by 3, this gives us the answer. X cannot be 5/3. Now, this seems that it’s all real numbers except 5/3. So somebody– sometimes people write in words all real numbers except 5/3 or except X equals 5/3, different ways of writing it. And in fact, in interval notation, you might say, well, it’s all real numbers of 5/3. Here’s a way of writing that. You can go all the way up to 5/3, not included, union from 5/3 to infinity. So different ways you might see the problem written for the domain, OK? So this might be kind of the more formal way of doing it. Since it’s kind of writing it out in words, I’ll say that’s the domain and just writing X’s not equal to 5/3, usually that means all real numbers except that number. All Right, here’s our next one. State the domain of H of X equals 7 minus 2 radical X plus 4. We don’t have a fracture here, but we do have a square root. I mean you take the square root of something we have– we can’t take the square root of a negative number to get a real number. So the square root of a negative number is not real. So we have to make sure that the part under the square root cannot be negative, because to be a real number the number under the square root must be greater than or equal to 0. So we’re going to take this part into the square root, X plus 4, and make sure that it’s greater than or equal to 0. All right, so we want to make sure that we’re going to get a real number when we plug a value of X. So X plus 4 must be greater than or equal to 0. Notice it’s OK for it to equal 0 because the square root of 0 is 0. So if we plug in a number for X like negative 4, we’d have the square root of 0, back here. Let’s just do it real quickly. What happens when I plug in 0? That’s 7 minus 2 times the square root of negative 4 plus– oops, I don’t want to say I’m plugging in 0, I’m plugging in negative 4, taking in negative 4. That would give you 7 minus 2 squared to 0. Square root of 0 is 0. Seven minus 0 is 7. So, yeah, when I plug in negative 4, I do get a real number, right? So, it’s OK to be 0. So it’s underneath this square root. So, we need to solve this equation. We just have to make sure what’s under the square root is greater than or equal 0 cannot be a negative number. So, if we just subtract 4 from both sides, X must be greater than equal to 4. So, that’s our solution. All real numbers where X is greater than our equal to negative 4, or if you are going to write that in interval notation, you could write negative 4 to infinity. Now, that shows that it’s all real numbers. And you could do a little quick check what would happen if I had plugged in a number smaller than negative 4, like for instance, negative 10. What happens if you plugged in a negative 10 into here? We would have 7 minus 2 times a square root, what happens when you plug in negative 10, negative 10 plus 4 is negative 6. This is now not a real number and we are only looking for outputs that are real numbers. So there’s an example that’s not going to work. How about if I pick a number bigger than negative 4 like 0? What about when X is 0? You would get 7 minus 2 times– well, I’m putting 0 for X, that’s square root of 4. And 7 minus 2 times 2, 7 minus 4 is 3. Sure, our answer would be 3, that’s a real number. So, when you’re done it’s always kind of good to look in see if it makes sense if you just plugged in a solution or something that’s not supposed to be a solution. Here is one more. We want to find the domain of this function. Oh, we have a fraction again. No problem with the numerator. We don’t care what’s in the numerator but we cannot have a 0 in the denominator. So, again, we know the denominator cannot be equal to 0. X squared minus 3x minus 10 cannot be equal to 0. So, what values of X will make it 0? Well, we could do that by factoring. So, this would X minus 5 times X plus 2, that can’t be 0. But neither of these numbers can be 0. So, X minus 5 can’t be 0. And this is different that when you put it to equal to 0 you always write or but actually can’t be either of these numbers. X plus 2 can’t be 0 either. So, if we solve, we get X cannot equal 5 and X cannot be equal to negative 2. So, there’s two things that X can’t be. So, our answer, we could– often you’ll just see it as X cannot be 5 or negative 2. We just put both like that. That means we want all the real numbers except 5 and negative 2. You can use to write this in interval notation. It does get a little bit hairy looking. Let me make this a little bit smaller so we have space. Here we go. If you’re going to write it in interval notation, we could write, well, all the numbers up to negative 2 but not including it union from negative 2 up to 5 union, 5 to infinity. That’s how we could write that using the interval notation. All right. I’m going to just pint out of the thing here. First of all, we could plug in 5 and see what happens into this function and see that it really is kind of fine. So, if you plug in 5, so in other words, that’s F of 5. You’d put in 5 for X, notice you’ll got a 0 in the numerator. And if you put 5 in the denominator, let’s see, so you got 5 squared, what’s that, 25, minus 3 times 5, that’s 15 minus 10. So you do get 0 over 0 which is undefined. So, you could see that 5 definitely gives us a problem. And you’re going to do the same thing for negative 2. But, I just want to show you something. If you had simplified this fraction fist, you have X minus 5 in the numerator, X minus 5 times X plus 2 in the denominator, this after you cancel will give you 1 over X plus 2. Now, what happens if I put in 5 into here? Here’s the problem here, this is not the same function. If I put 5 into here, this is a new function. G of X let’s say is 1 over X plus 2 is not the same as X minus 5 over X squared minus 3x minus 10 because the denominators are different. So, it’s true that in this new function after you seemingly simplify it, you would get 1/7 by plugging in 5 for X. But what matters is that when you plug that value into the original function, it still has to be defined before it could be simplified at all. So, that’s just a little nuance to think about. All right, so we’re done with several examples finding the domain, basically you’re looking for numbers that make the function yield something that’s not a real number. Those are the problem values. So, if there are no problem values, usually it’s just going to be all real numbers. So, what you’re looking for is there, are there any restrictions. This is You Math Gal, Julie Harland. Please visit my website at YourMathGal.com where all of my videos are organized by topic.
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