In this video, we are going to look at how

the idea of composition function can be extended and an extended on. We can compose several functions together. We’ll simply start in the inside and work to the outside. Start in and work out. So when we see f(g) of h(2), and we have these three functions f(x) is x plus two, g(x) is x squared minus fiv, and h(X) is square root of 3x. What we’re asked to find then is f(g) of

h(2). Make this one ah squiggly, so I can just show the difference between the parentheses, which means we’e going to do is first find starting in the inside of the h(2), and then

we will plug that result into g, and then we will get another result and we will plug

that result into f and we just keep on going as we cascade on up. So first we need to find in the very center of things, what is h of two? Well that tells us which function to go to, and to what to replace the variable with. We go to h and we replace the variable with two, so we have the square root of three times two which is the square root of six. So h(2) was the square root of six, we’re going to plug that solution into g now we’re going to find g of the answer square root of six. Well, g is x squared minus five, so it’s going to be something squared minus five, and the variable gets replace with whatever’s in the parentheses the square root of six. Well, what is the square root of six squared? Just six ’cause square and square root are inverses, minus five is one. So when g of the root six we got one, so now we’re going to plug that one into the f function. Inside of f we’re going to see the number one, so now we go back to the f. Notice how this is just compounds back to the f. X plus two, something plus two what are we plugging in, we’re plugging in one in. One plus two is three. three is our final result, which we get by starting with the two, plugging it into h plug the result to g plug the result

to f, it’s a cascade problem. Let’s try another example were we do a composition with a variable. Here we’re asked to find f, always rewriting it in the useful form, of g of h of a. a not

x sorry. Well, what this means is we first have to find what h of a is? That means we’re going to replace the variable with whatever is in parentheses. So n h we’re replace the

variable with the a. So now we have f of g of h(a). Plugging a in we get the square

root of 3a. Now we see that the square root of 3x needs to be plugged in for the variable in x. First we plug it in a then we plug it in x. Replacing that variable with whatever is inside the parentheses because whats inside the parentheses replaces the variable. So it’s x squared or something square minus five and that would be that something that would be inside the parentheses, square root of 3a, and that is still all inside the f function. Well, let’s simplify it a little bet before we more on because square and square root are inverses. We’re really finding f of 3a minus five, and so we see this 3a minus five inside of f were we replaced the variable in f with 3a minus five. So x plus two stuff plus two that stuff is 3a minus five. Simplifying, combining like terms 3a minus three becomes our final solution. Composition of composition of composition we plug one function into

the next into the next into the next. Starting from the outside working out.