 So this video is going to talk about the algebra of functions, so we have a sum and difference to talk about first. So we're going to have two functions and they have overlapping domains and we're going to call them p and q. So that means that if I add the two functions I can rewrite it as f plus g of x, just one function and or if I have a subtraction of these two I can write it as f minus g of x and that means that all the elements of p and q that they have in common, all the elements that they have in common, that would be our domain. Let's try it out. So we have f of x is equal to 2x squared minus 18 and g of x is negative 3x minus 7 and we want to find the domain. Well f of x has a domain of all reels, double back r is all reels. And g of x has a domain of all reels. This is a quadratic, goes left and right forever, this is the line, goes left and right forever. So the domain of h of x then, they have everything in common, so the domain of h of x is also all reels. So given these two functions that we had, find h of 5. We can say that then h of 5 remember was f of x plus g of x. So we take 2 times 5 squared minus 18 and that's the f of x plus the g of x which is minus 3 times 5 for x minus 7. This gives us 25 times 2 or 50 minus 18. I can drop the parentheses because I'm adding. Negative 3 times 5 is negative 15 and then minus 7 and if we look at that we have, this is 50 and then all of these negatives give me minus 40 and we find out that it is 10. Now it says find a new rule for h. Well that says, okay I'm going to go back to function, notation, 2x squared minus 18 plus negative 3x minus 7. Really, I'm just going to combine light terms here. So 2x squared is the only x squared term and then my x term is over here in the g function and that's minus 3x because I'm adding and then negative 18 and a negative 7 is going to give us a minus 25. So this would be h of x. Now it says find h of 5 using the new function. So h of 5 is going to be equal to, instead of having to plug it into each one I can just say 2 times 5 squared minus 3 times 5 minus 25. This is 50 minus 15 minus 25 and I am going to have the same thing of 10. Two new functions. Determine the function of h of f of x plus g of x. If I think of this as f of x then it's x squared. So that means that it's going to go this way to negative infinity and it's also going to go this way to positive infinity. If I take this as g of x, well I see that this and what's underneath the radical has to be greater than or equal to 0. So that tells me that x has to be greater than or equal to 3 for g of x. So 1, 2, 3, so it starts here but it's greater than or equal to and it goes this way to infinity. And I want to know what these two things have in common. Where they overlap and they overlap over here. So h of x, the domain is from 3 to infinity. Same two functions, we want to find a new rule. Well remember that it was again asking us to add these two functions. And so we have 2x squared being added to the square root of x minus 3. When none of those are like terms, so I just have h of x is equal to 2x squared plus the square root of x minus 3. And then it says find h of 4 using the new function. So h of 4 is going to be equal to 2 times 4 squared plus the square root of 4 minus 3. And h of 4 is going to be 4 squared of 16 times 2 is going to be 32 plus the square root of 4 minus 3 which is 1. So it's really 32 plus 1 or 33. And we never did do a subtraction. Let's take this first two functions that we had. We started out with f of x was equal to 2x squared minus 18. This is not on your sheet. I neglected to put a subtraction in there, so I want to show you one. So I want h of x to be equal to f of x minus g of x. In fact, I don't always have to start with f of x. I could have said g of x minus f of x. It really doesn't matter which one comes first. We just want to take one function plus or minus the other function. Well, f of x is 2x squared minus 18. And minus g of x is negative 3x minus 7. And so we distribute the negative. So we have 2x squared minus 18. But then it's plus 3x and plus 7. So h of x, when we simplify, will be 2x squared plus the 3x. And then negative 18 plus 7 is going to be minus 11. So now we're going to talk about products and quotients. So we have these two functions. And again, their domains are overlapping. And this little sign right here means intersecting. Remember when we were looking at the number lines? This is really true for the addition and subtraction as well. I just didn't write this set notation for you. But we talked about when we had those two that were on the number lines, we had to take what they have in common, where they intersect, where they have the same values. It's not the combination of the two. It's just what they have in common. When we have a multiplication, we can just say it's the intersection. When we have division, we say it's all elements of p and q as long as g of x is not equal to 0. That all elements that p and q have in common, except for when g is equal to 0. So let's try it out. We have these two functions. Determine the domain of f times g of x. So the domain of f of x is equal to all reals. And the domain of g of x has got to be this x minus 7 has to be greater than or equal to 0. So x has to be greater than or equal to 7. So we would say that it's going to be from 7 to infinity. This is all real. So it goes to infinity for sure. But it also goes to negative infinity. So the 7 to infinity is what they have in common. So we could say x is an element of 7 to infinity. That's probably the better way to write it, the way that our book likes to write things. So now let's find the rule for h. Remember, we're multiplying. So negative 3x minus 5 is going to be multiplied by the square root of x minus 7. It's really just distributive. Negative 3x, that's an equal sign, times the square root of x minus 7. Because that's just a monomial over there. Minus, because it's a negative, 5 times the square root of x minus 7. So evaluate h of 8 and h of 11. h of 8, we would come in here and say negative 3 times 8 times the square root of 8 minus 7 minus 5 times the square root of 8 minus 7. Negative 3 times 8 is going to give us negative 24 times the square root of 1. So that's really just negative 24. And then minus 5 times the square root of 1, which again is just minus 5. And we are going to get negative 29. And I'll do h of 11 over here so I have enough space. And I'll probably have to write it below it. We have negative 3 times our x, which is 11, times the square root of x, which is 11 minus 7, minus 5 times that square root of 11 minus 7. So this is negative 33. And 11 minus 7 is 4 minus 5 square root 4. So it's really negative 33 times 2 minus 5 times 2. Negative 33 times 2 would be negative 66 minus 10. So we have negative 76. We have one more set of functions because now we want to do the division. And we have to determine the domain again. Well, different functions here. So f of x has an x as an element of, and it's a cubic, goes left and right forever. It doesn't have any place where it stops. So we have r reels. And g of x, x is an element of, again, it's going to be r reels. Except we have to remember that g is being divided down here. So we have to also say that x cannot be equal to 1. Because down here, we're going to have being divided by x minus 1. So x minus 1 cannot equal 0. So x cannot equal 1. That's how we got there. So we would say that our domain is going to go everywhere except for here. So h of x, x is an element of negative infinity up to the 1, but not including it. Because that would cause us to have a 0 in the denominator. Union, we want to be on the other side of 1 of 2 infinity. So there is the domain of our new function. Consider both domains. And then you also have to consider what would make the denominator 0. So now find the function for h in simplified form. So h of x, remember, was f of x, x cubed minus 1, divided by g of x, which is x minus 1. And notice it says simplified form. So if you remember when you factor an a cubed minus b cubed, it's going to be a minus b times a squared plus ab plus b squared. You might not remember that. But we don't factor a whole lot of cubes. So I'll show you what the factorization would be. Well, our a here is equal to x, and our b here is equal to 1. So we would say that this is x minus 1 times x squared plus 1 times x, or just x, plus 1 squared, or just 1, all over x minus 1. And I'm going to put it in parentheses, because I'm going to try to simplify this thing if I can. So it is just going to be x minus 1 as a quantity. And I can reduce these two and say that it is x squared plus x plus 1 as long as, don't forget, we did have a denominator down here, though. So we have to say as long as x is not equal to, and that comes because of this denominator here. All right, so let's try one more. We have this problem here where we have to divide again. But now we have a little bit more involved function. So p of x, x is an element of all reals. But remember for q of x, 3 minus x has to be greater than or equal to 0. So if I take the x to the other side, I'd have x is less than or equal to 3. So for q of x, I'm going to say x is an element of negative infinity to 3. Now that is all true. So how do I find out what r of x is? Or where do they overlap? This goes everywhere. This only goes from negative infinity up to 3. It's going to be the smaller of the two, really, since this one is infinite. So we have negative infinity up to 3. They overlap only in this interval. As long as we remember that if it is 3, we're going to have a problem. Because when you have the square root of 3 minus 3, that'll be the square root of 0, which is 0. So it's not a bracket here. It's a parentheses. So don't forget that. This is a bracket when we think about the domain for q, just q by itself. But when we put it in the denominator, it can't be that 3. So find the rule. And it was p over q. So 1 minus x, this is r of x, 1 minus x over the square root of 3 minus x. So use this new rule to evaluate r of 6 and r of negative 6. So let's start here with r of 6. And that'll be 1 minus x, which is 6, over the square root of 3 minus x, which is 6. And this, if we look at it, 6 is not in the domain of r of x. And you can see right here that it's going to cause problems, because this is going to give us the square root of negative 3, which isn't real. So let's try r of negative 6. Is that in our domain? It is a value that is 3 or smaller. So 6 was too big, but negative 6 works. So we should be able to find an answer here. So that's 1 minus a negative 6 or plus 6 over the square root of 3 minus a negative 6 or plus 6. So 7 on the top and the square root of 9. Or really, it is 7 over 3.