 So another thing that mathematicians really like to do is to try and extend the basic notions of arithmetic to anything that they come across. And functions are no exception. And so we might make the following definitions. If I have two functions, f of x and g of x, then I can try to define the function sum or difference, f plus g of x. Well, that will be defined in the obvious way. I'm going to take the function values and add them together. The difference, f minus g of x, well, that's going to be the difference of the two functions. The product of two functions, I'm going to write that f g of x. And what that really means is that's f of x times g of x. The quotient of two functions, f over g of x, no surprise, that's f of x over g of x. And function composition, this one might be a little bit surprising. This is f applied to g of x. This is f of g of x. And what that means is I'm going to apply the function to another function. We'll see how that works. So let's take a quick example here. I have a function f of x equals 2x plus 3 g of x equals x squared. I want to express the quotient f over g of x and find the domain, and then maybe evaluate f over g of 4. So by definition, this f over g of x, this is how we indicate the quotient f of x over g of x. So again, definitions are important in mathematics. If you don't have the definitions, you can't do mathematics. Here, nice simple definition, quotient is the quotient of the two functions. Well, equals says whenever I see the one, I can replace it with the other. So equals, well, f of x, I can replace it with 2x plus 3 g of x. g of x is the same thing as x squared. So wherever I see g of x, I can replace it with x squared. And so my quotient function f of x over g of x is going to be 2x plus 3 over x squared. And it's a quotient, so I need to worry about the domain. I worry that the denominator may be equal to zero. Well, that's pretty easy to figure out. As long as I make x not zero, then the denominator won't be zero. And I'll be fine with that domain. What about the value of f over g of 4? Well, again, by definition, this is the value of f at 4, the value of g at 4. And I know what f is, I know what g is, and so I can substitute in those values. So f of 4, drop that into here, 2 times 4 plus 3, g of 4, drop 4 into here, that's 4 squared. And that's going to be 11 over 16. Function composition occurs when we apply a function to another function. So here I want to find f of g of x. And again, definitions are very important. This is the same as f applied to g of x. So to find that, well, let's start off, paper is cheap. So let's go ahead and write down f of x. Well, f of x is square root of 2x plus 5. And what I want to do, again, paper is cheap. Let's drop out our variable x. Let's drop out our independent variable and leave an empty set of parentheses. So I dropped it out, and maybe I'll leave myself a little bit more space. And again, the idea is that whatever goes in any set of parentheses has to go in all of them. So originally I had an x in here, I had an x in there. Well, what I actually want to put in there is I want to put in g of x. So I'm going to drop a g of x into this empty set of parentheses, and whatever goes in the one has to go in the other. So I'll drop that g of x in there, and I have this new expression. Now, again, paper is cheap. It does you no good to save on paper if you make the problem more complicated. So let's see. Well, I do want to find a simplification for this. And again, g of x equals x plus 3. Wherever I see g of x, I can replace it with x plus 3. Well, I actually want f of g of x. I don't want to replace the g of x over here, because this is what I'm looking for. But on the right-hand side, if I replace this g of x with x plus 3, I'll have a simpler expression. So I'll make that replacement. Paper is cheap. Don't try to do this by trying to crab everything into one line. We'll go ahead and write down a new line. So here's 2 instead of g of x. I'm going to write x plus 3 still in parentheses. Parentheses, like paper, are very cheap. And here's my expression. f of g of x equals square root 2 times x plus 3 plus 5. I can do a little bit of algebraic simplification. After that, I'll expand out the 2 times x plus 3 and add 6 plus 5 to get me 11. And here's my f of g of x equals square root 2x plus 11.