Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Functions with algebraic operations. If X is in the domain of two functions F&G, then F + g of X = F of X + g of X. So basically, if X is in the domain of the two functions, we can just add the two functions together. We can do the same thing for subtraction, multiplication, and division, where with the division the denominator can never equal 0. So if we look at this graphically, first of all, if we want to find the domains of F of X&G of X, let's let the blue one be F of X and we'll let the red 1 red graph be G of X. So the domain is our X values. So we want to think about how far to the left did we go? So where is this? How far left? We went 123456 to the left and we went 123 to the right. So the domain here is going to be -6 to 3:00. We're going to put brackets because it did include those dots. We're going to make them solid dots -6 and three in a minute, maybe we'll make one of the G of X's an open dot and the other one a solid. So brackets tell us that it does include the point. We're going to start from the left at -6 and we're going to end at the right at 3 for just the X values for the G of X. How far left did we go? Well, we went 12345, so we went five to the left and we went 123456. If I counted right 123456 to the right, now the point on the left was open, so it doesn't include it. So we're going to use a parenthesis. But the point on the right was solid and so it does get included. So we'll use a bracket. Now if we want F + g of X, we need all of the points where X is defined on F and on G So when we look at this, if I wanted to put in a line, say at -6, does that intersect both the F graph and the G graph? Our answer is no. What if I wanted it? Actually, I guess I was at -7. If I was at -6 there, does it intersect both the graphs? And the answer is no. Well, what if I was at -5 1/2, Would that intersect both graphs? No Where does it start intersecting both graphs? Well, we could say it -5 but that was actually an open dot. But as close as we can get to -5 it would intersect both graphs. What about going off to the right? Well, we're intersecting both. We're intersecting both, we're intersecting both. Right here would be the last time we would intersect both graphs. If I kept going. It doesn't intersect both of the graphs off to the right. So our domain for F + g of X is going to be -5 to 3:00 because I needed them both to be defined. Now if I want F + g of one, what that really is saying is what is F of 1 + g of one? Well, let's grab our line here and let's come back and say, what is the value? Whoops, let's grab only one line at a time. Come on. So let's say what is our F of one graph? When we look at this we go 123451234, let's say 4 3/4 for fun. So our F of one was 4 3/4 our Y value when X was one, our G value 123 negative 3 it looks like. So F of 1 + g of 1 is going to give us 1 3/4 when we add those two Y values together. What about F times G of 0? So that's just saying what's F of 0, IE our Y value when our X is 0 * g of 0. So if we grab this line and we move it over to the zero, we want to say what is our F of 0123 and a half? So we're going to say 3 1/2 for F of 0. What's our G of 0? Our G of 0 looks like, let's call it -2. So 3 1/2 is 7 halves times -2 that -2 we could think of as over 1. So when we multiply these, we just get negative seven F / g of -2. So F of -2 / g of -2 when I look here, negative 212 looks like my F value or my Y there is one, and it looks like my G value is also 1. So 1 / 1 would be 1. What if we wanted to solve the functions algebraically instead of geometrically? So if we have F + g of four, this is really saying F of 4 + g of four. How do we find F of four? We stick 4IN to the F equation, so we're going to have 3 * 4 - 5. That whole thing is the F of four equation. How do we find G of four? We're going to stick 4IN to the G equation, so 4 ^2 + 7. That is the G of four. Now we're just going to simplify it up. So 3 * 4 is 1212 -, 5 would be 7 + 4 ^2 is 1616 + 7 would be 23 and seven and 23 would give us 30. So our F + g of four is 30. Now gets a little more interesting. If we just use X instead of a number, we have F of X + g of X. Well, now we're literally just going to stick in F of X equation three X -, 5, and now we're going to stick in the G of X equation X ^2 + 7. And then to finish it off, we're just going to combine any like terms. So we're going to get X ^2 + 3 X +2. Now a double check or a nice way to check is this F + g of four. If this is our F + g of X equation, I ought to be able to stick 4 right into this simplified equation and get the same solution. So F + g of four, what is 4 ^2 + 3 * 4 + 216 + 12 + 2 is going to give us 30. And we did get the same answer. We did it two different ways, but it's the same equivalent. This next one we're going to let you try, and then we're going to look at domains. Well, if this is a line, the domain is what X values can we put in. And the reality is we can put in any X value. So negative Infinity to Infinity for the domain. Anytime it's a polynomial 4X plus five X ^2, X ^3 -, 4, it's going to be negative Infinity to Infinity for G of X. This case we can't divide by zero, so we're going to say three X -. 2 cannot equal 0, so 3X cannot equal 2, so X cannot equal 2/3. If we thought about looking at that on a number line, we'd say everything except 2/3 would work for the domain, so we'd have negative Infinity to 2/3. We're not going to include it, so we're going to use a parenthesis union because it's true in multiple locations. It's true over here. It's also true over here. 2/3 to Infinity. Now F / g of X. We knew that F was all real numbers, and we knew it. G of X can't have 2/3 as it. But what happens when we divide this? Well, if we have four X + 5 and we divide that by 6X over three X -, 2, when we divide, we're going to multiply by the reciprocal. So we get four X + 5, and we can think about that being over 1 / 3 X -2 / 6 X. Well, now this denominator, when we simplify the new denominator, can never ever, ever equal 0. So we get X can't equals 0 from this new denominator, 6X can't equals 0X can't equals zero. So the domain for our F / g of X is going to be. It can't be 0 from the simplified. It could be anything from the F, but it couldn't be 2/3 from the G So now we're going to have this number line where it's everything but zero and everything but 2/3. So to write this correctly, we're going to have negative Infinity to 0, union 0 to 2/3, union, 2/3 to Infinity. Thank you and have a wonderful day.