WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about composite functions.
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We are often going to have 2 or even more functions that interact with each other.
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This lesson will explore the fundamental ways that functions can interact with each other.
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First, we will look at how functions can interact through just arithmetic: addition, subtraction, multiplication, and division.
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These sorts of interactions are called arithmetic combinations, because they are just using arithmetic.
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Second, we will move on to a more complex idea, using one function's output as another function's input.
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We call this idea composition of functions; if we want to talk about a specific example, we call that a composite function, when we put multiple functions together.
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All right, let's go--let's say we have two functions, f and g, and f(x) = x², and g(x) = x--nice, basic functions.
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Now, it is easy to imagine creating a new function that just adds f and g together; we would call it f + g.
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That is not very imaginative, but it makes sense.
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It would give us the sum of the two functions: the new function, f + g (x), would be equal to x² + x.
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We are just adding the two functions together; so we know each function is x² and x, so we just add them together.
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That is a simple, basic idea; we are using a basic arithmetic operation, and we are just putting them together through that.
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We have arithmetic; let's use it.
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We could expand this idea to the other three basic operations.
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We could do this with subtraction: f - g would become x² - x; fg(x) would be x² times x (fg being f times g,
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just like when we say 3x, we mean 3 times x); and f/g would be x²/x--as simple as that.
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Given two functions f and g, along with an x that is in the domain of both, we defined these four different arithmetic combinations.
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If x means something for f(x), and x means something for g(x)--it doesn't fail, like if we had √x be one of them,
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we couldn't plug in -3; but as long it is in the domain of both--it is a number that both of them can accept and work on--
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all of these work really well; sum is f + g(x)--just break it down into adding the two together.
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Difference is just subtracting one, just like normal subtraction.
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Product--when we have fg, we read it as times: f times g; and quotient is f(x)/g(x).
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And it also has to be that g(x) does not equal 0, because otherwise we could accidentally end up
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blowing the world up when we divide by 0, since we are not allowed to divide by 0,
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because it is nonsense and doesn't mean anything.
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So, since you can't divide by 0, it is not going to be defined when g(x) is 0, since you would have to divide by 0.
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But other than that, we are pretty good to go; if it means something, if it comes out as a normal output, it is defined;
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that input is in the domain, and it is defined as an output; then we can just put the two together.
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We just put what f(x) is together and put what g(x) is together with any arithmetic combination that we want to; great.
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That is a nice, direct idea; an arithmetic combination makes sense.
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We put in the same input to the two functions, and then we combine their outputs with some pre-decided arithmetic operation.
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If it is sum, we do it through addition; if it is product we do it through multiplication; things like that.
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But we can do something more interesting: we can compose one function with another.
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Instead of giving both functions the same input, we give the input to one function.
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Then we take the first function's output, and we plug that into the second function.
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Input goes into one, and then an output comes out of that; and that immediately goes into the second function.
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And then finally, we get an output of that; the second function is acting on the first function.
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Many lessons back, we first introduced the idea of a function; and we talked about how we can view it as a machine.
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It takes in inputs, and the function produces outputs: x goes into the machine, the function f;
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and then it gets put out after having been acted on.
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The function is some process; it does some transformation on x, so we get f(x), f having acted upon x.
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All right, so that is the idea of it as a machine.
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We can expand this idea into function composition.
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Function composition is just linking multiple machines together in series; we just put multiple of them together.
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The output of the first function goes directly into the second as its input.
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Our first input goes into, say, g; and so, it is now g(x); and then we plug all of g(x) into f.
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And so, we have all of g(x) now being acted upon by f; input into the first machine, and output comes out of that machine.
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And then, we just jam that right into the second machine.
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And if we wanted, we could string this up...3, 4, 5, 6, 7...we could string up as many of these machines in order as we wanted.
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We could compose as many functions as we wanted to; but it is easy to start by thinking about it in terms of two functions being composed together.
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We note the previous slide's composition, when it went into g first, and then went into f, as f composed with g.
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This is just a little circle between them: f circle g...we read that as f composed with g.
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If f composed with g acts on x, acts on some input x, we have f composed with g of x, just like we would normally.
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We have created a new function out of putting the two together.
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By linking those two machines together, it is effectively one larger machine that is doing a new way of working.
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This means that in this machine, f composed with g, g will act on x first; and then f will act on whatever results.
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So, we have f composed with g; and we can break it down into g goes first; then f goes on what results, the thing that comes out of that.
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Now, notice: the functions act in order of closeness.
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g goes on first; and then, f goes on second, because it is farther away.
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We hit it with the things that are closest to the input that we are putting in.
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So, g goes onto the x, and then f goes onto what results; and if we had even more stacked up,
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whatever was even further to the left would act after that.
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The functions act in order of closeness to the original input.
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There is another, much easier, way to see f composed with g of x in the function notation format that we are already used to--
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the thing that we have been using for quite a while now.
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f composed with g of x is just f(g(x)); so f composed with g of x...remember how we broke it up:
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this one went first; and then f acted second--well, that is what we have right here.
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g is acting first, and then f is acting second--it makes a lot of sense.
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I would personally recommend, any time you see this circle notation--this f composed with g stuff--rewrite it
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in this normal format, the format that we are used to at this point, the f(g(x)).
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This normally will make it easier to understand and solve problems; there are very few downsides to breaking it into this thing.
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So, I would really recommend: any time you come up against a problem, and you are not quite sure what to do,
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and it is this sort of thing: break it into f(g(x)), f acting on g(x)--
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something acting on something else acting on the input that you are putting in.
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This method, the second form of notation--this is really great as a way to look at things.
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I really recommend that, when you see this, you break it into this thing right here; it will really help you understand what is going on.
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Another way we can visually interpret it--this is a little hard to see what is going on here, but try to follow me
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on what I am saying here--what we do is start with some x; we start with x, and then we apply g to that x.
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g goes along and takes it to g(x); then, f comes along, and it hits this g(x), and it turns into f acting on g(x).
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But we can also think of it as some new function that has been created, f composed with g.
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We have created a new machine that can just go directly from our original x to the end result of f(g(x)).
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It does both of these actions, both of these processes, in one thing; it is a machine that is built out of both of the machines inside of it.
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We can look at it as stair-stepping across, or we can look at it as one new-built, giant leap,
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where it does both of these actions in one jump.
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All right, you can take steps across the pond, or you can take one giant leap.
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But ultimately, they do the same thing; the leap has to be informed by the steps, though.
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So, how do we actually use these composite functions?
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We understand the idea behind them now; and it turns out that using them actually isn't that hard.
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It is just important to understand the idea.
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Each function has its own rule, like f(x) = x³ means to cube your input.
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So, composing multiple functions just means using these rules in succession.
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This idea is shown beautifully in that notation--that notation that I was talking about being the one I really recommend earlier, f(g(x)).
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The function says that the function g acts on x; then, f acts on the resulting g(x).
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So, g acts on x; that is what this says right here; and then f acts on the resulting g(x).
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f acts on what we just had there; great.
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Since we almost certainly know what g(x) is, and what f(x) is, from the problem, we just use that as an input for f.
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We use the rules that we were given earlier, and we just apply them to these things.
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Let's see an example: for example, if we had f(x) = x² + 3 and g(x) = 2x - 2,
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then f(g(x)) is equal to...well, what we see here--don't get tricked by the fact that we have x showing up multiple times.
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Remember, it is just a placeholder: f(x) is just a way of saying f(whatever is in here), whatever f is acting on.
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The thing that it is acting on will get squared; plus 3.
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So, if it is acting on g(x), then what is g(x)? Well, g(x) is 2x - 2; so we are plugging in 2x - 2.
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So then, we plug that in for f(x); f(x) becomes x² + 3; so if what is inside of the box is 2x - 2,
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it is going to be (2x - 2)²; so the box has the same process happen--it is just a new thing going on.
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Instead of x going into it, it is just 2x - 2 going into it.
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The same processes: it is taking the input, squaring it, and adding 3.
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So, instead of taking an x, squaring it, and adding 3, we are taking in 2x - 2; we are squaring 2x - 2; and then we are adding 3.
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So, if we wanted to, at this point we could expand (2x - 2)² + 3; but this is really the key idea--
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getting to this point of thinking of it as boxes; we are plugging in, based on boxes.
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And we will see a bunch of examples using this idea later on.
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But you want to think of it as we are just swapping out; we are using x as a placeholder.
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It is not x that we are really attached to; f(x) is saying f of box, and then what happens to box;
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f of placeholder, and then what happens to placeholder; f of input, and then what happens to input.
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That is the way you want to think about it; and that makes it really easy to do composite functions.
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All right, it is time for some examples.
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So, f + g of 3; if we have f(x) = 2x + 3, and g(x) = x² - 7, what would f + g of 3 be?
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We do this: f(x) is 2x + 3; g(x) is x² - 7; we have 2x + 3 + x² - 7.
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So, that becomes something; we could simplify it; but at this point, let's plug in x = 3.
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We have that x = 3 is going to get plugged in, so we have 2(3) + 3 + 3² - 7.
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6 + 3 + 9 - 7; 9 + 9 is 18; 18 - 7 is 11; so we have 11 as the answer here.
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All right, the next one we will do with the color blue: g - f(1); what is g?
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g is x² - 7; what is f? it is -...and here is the key thing; it is not minus 2x; it is minus all of f;
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not just the 2x, but minus (2x + 3); it is a whole quantity that we have to be subtracting.
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Now, we will plug in; what happens when we plug in x = 1?
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Well, we have 1² - 7 - (2(1) + 3); so that is 1 - 7 - 2 + 3, which is equal to -6 - 5, equals -11.
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Great; the next one--I will do this one in green: fg(-2).
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What is f? f is 2x + 3; and then, we are multiplying that by g; so it is that whole f, times the whole of g, x² - 7.
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So, (2x + 3)(x² - 7); now let's plug in -2; when we plug in -2, we get 2(-2 + 3) (it has to be in that whole quantity),
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times (-2)², minus (oops, I accidentally made a plus sign) 7; great.
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So, that is equal to 2 times -2, which is -4, plus 3; (-2)² is 4 - 7; -4 + 3 gets us -1; 4 - 7 gets us -3; and we get positive 3.
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Great; and finally, let's go back to red for our very last one, g/f(8).
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So, what is g? g is x² - 7, over all of f; so it is 2x + 3.
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So, we plug in x = 8; 8² - 7, over 2(8) + 3; 8² is 64, minus 7, over 2 times 8 (is 16), plus 3.
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64 - 7...we get 57, over 19; and it turns out that 57/19...19 times 2 will get up to 38; 19 times 3...we get up to 57.
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So, 57/19...we get 3 once again, by chance; great.
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Also, I want to point out: if we wanted to, we could have done this by figuring out what f was and what g was, separately.
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So notice: f(3)...for example, we used the f + g(3) just to make a point here.
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So, f(3) is equal to 2(3) + 3, which would be equal to 6 + 3, or 9.
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g(3) is equal to 3² - 7, which equals 9 - 7, or 2; so f + g(3) is equal to, by the way we defined it, f(3) + g(3),
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which is equal to 9, f(3), plus g(3) is 2, which equals 11; that is the exact same thing we got when we started by adding.
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So, we can either put the functions together, and then plug in the variable;
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or we can plug in the variable into each function and then add them together.
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It depends on the way we want to approach it; sometimes it will be more useful to do it one way, sometimes more useful to do it the other way.
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But it is important to notice that we can do it the other way.
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The second example: we have the same functions that we just ended up using, f(x) = 2x + 3 and g(x) = x² - 7.
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What is f composed with g, first?--so we will start with the green for this one.
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f composed with g of x; what was my recommendation? It was f(g(x)); that is the exact same thing.
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It is another way to say this exact same thing, but it my opinion, makes it much easier to understand what it is going on.
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f(g(x)): f(x) = 2x + 3, but we don't need that information yet; we need to plug in g(x) first.
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f(x² - 7); now, what is the x? Here is our x here, and x goes right here.
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That means that f of box is equal to 2 times box plus 3.
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The thing in the box now is x² - 7; so that gives us f(x² - 7) is 2 times the thing in the box, x² - 7, plus 3.
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And if we wanted to, we could expand that out.
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We expand that out pretty easily; and we would get 2x² - 14 + 3, which is 2x² - 11; great.
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The next one: blue for the next one: g composed with f(x) is much easier to write as g(f(x)).
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What is f(x)? f(x) is 2x + 3, so that is what is going into g right now: 2x + 3.
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And now, if we plug into g, g of box equals box squared minus 7.
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So, g(2x + 3) is equal to (2x + 3)...that is the thing in the box...squared, minus 7.
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We work that out; we get 4x² (2x times 2x is 4x²) + 2x + 3, and (3 + 2)x is 6x, plus 6x is 12x,
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plus 3 times 3 is 9, minus 7, which equals 4x² + 12x + 2.
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Great; OK, the next one--let's use red for this one: f composed with f--f composed with itself.
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We can also write this as f(f(x)); what is f(x)? f(x) is 2x + 3, so f(2x + 3).
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And now, the thing in our box is 2x + 3, so it is f of box equals 2 box + 3; so f(2x + 3) is 2(2x + 3) + 3.
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Great; we just work this out: 2(2x + 3) is 4x + 6 + 3 = 4x + 9.
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Here we are on the very last one; use green again--g composed with g(x); g(g(x))--it is much easier to see what is going on that way.
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g(x) is x²; so now it is g acting on x² - 7; remember, g of box is equal to box squared, minus 7.
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So, if we are plugging in x² - 7, it is going to be (x² - 7)² (remember, box squared minus 7), and then also - 7.
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Great; so we square x² - 7; x² times x² is x⁴;
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x² times -7 is 7x², minus 7x²...another -7x²;
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we have -7x² + -7x²; that is minus 14x²;
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and -7 times -7 is positive 49; and finally, minus 7; so it is x⁴ - 14x² + 42; there we are.
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There are a bunch of different function compositions, but it is not that hard, as long as we are plugging one thing into the other,
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and remembering, in terms of the substitution: it is not about the letter x; it is about if we just had a box here.
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If we just had a placeholder here--if we just had this thing to hold a space, then we saw what happened to that space when it was held open.
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If we put in an input, what happens to the input?
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We just happen to use x, because it is a convenient thing; we are used to using it as a placeholder.
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But x isn't inherently important; it is just the idea of what happens to an input.
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So, if we plug in something like 2x + 3, different things will happen than if we had just plugged in x.
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Also, one other thing I want to point out: notice that, in general, f(g(x)) is not equal to (they are normally very different) g(f(x)).
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For the most part, flipping the order that we do our function composition in gives us very, very different results.
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Sometimes, it will end up being the same result; but mostly, if we take f(g) or g(f), it will be totally different.
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So, mostly, f composed with g is a totally different function than g composed with f.
00:20:48.600 --> 00:20:53.600
And this is just something to keep in mind--that composition order matters very much.
00:20:53.600 --> 00:21:04.400
We can see it in what we got here: f(g(x)) got us 2x² - 11, but g(f(x)) got us 4x² + 12x + 2--totally different results.
00:21:04.400 --> 00:21:10.800
So, the order that you put it in--the order that one function goes into another--the order of composition--has massive importance.
00:21:10.800 --> 00:21:18.500
Great; the next example: let f(x) = x + 1, g(x) = 2x, h(x) = √(x + 1).
00:21:18.500 --> 00:21:26.000
What is the domain of f, divided by g, composed with h, and then also of h composed with g composed with f?
00:21:26.000 --> 00:21:32.800
Well, we will start with f(g) composed with h; now, the very first thing we want to do is figure out what f(g) is.
00:21:32.800 --> 00:21:36.400
I am sorry, not f(g); what is f divided by g?
00:21:36.400 --> 00:21:46.300
Well, f/g(x) is just equal to f(x)/g(x), except for when it is undefined, when g(x) is equal to 0.
00:21:46.300 --> 00:21:54.100
So, what does this mean? Well, f(x) is x + 1, so (x + 1)/2x--there we are.
00:21:54.100 --> 00:22:01.900
f/g(x) is equal to (x + 1)/2x; great, now we can just go back to what we are used to doing.
00:22:01.900 --> 00:22:11.800
So, we use blue for this one; f/g(h(x))...it is much easier to see it written in that format.
00:22:11.800 --> 00:22:28.800
f/g...what is h(x)?...f/g(√(x + 1)) = √(x + 1) + 1, over 2(√(x + 1)).
00:22:28.800 --> 00:22:38.800
So, how do we figure out the domain? This right here is not our answer, but it is the function that comes out from that composition.
00:22:38.800 --> 00:22:45.300
f divided by g composed with h comes out to be √(x + 1) + 1, divided by 2(√(x + 1)).
00:22:45.300 --> 00:22:53.300
So, the domain is going to be everywhere where it doesn't break.
00:22:53.300 --> 00:23:11.500
So, what things in here can break? Well, first, square root--any time we see square root, that breaks when a negative is inside.
00:23:11.500 --> 00:23:20.100
If we have x + 1 going in in both cases, if x + 1 is less than 0 (that is to say, x + 1 is a negative value), then we have breaking.
00:23:20.100 --> 00:23:29.100
It breaks--it is not defined, more formally--when x + 1 < 0, which is to say when x < -1.
00:23:29.100 --> 00:23:34.200
So, that is one important point of information: x < -1 means failure.
00:23:34.200 --> 00:23:50.800
Another failure point is if this bottom part is equal to 0; we have another break if 2√(x + 1) = 0.
00:23:50.800 --> 00:23:55.700
And that is going to end up being x + 1 = 0, which means x = -1.
00:23:55.700 --> 00:24:09.700
So, it fails if either of these conditions happens--if x is equal to -1, or x is less than -1, which is to say x ≤ -1.
00:24:09.700 --> 00:24:19.400
So, its actual domain, the domain of this function, is x > -1.
00:24:19.400 --> 00:24:24.500
It is all of the places where the function does not fail, where the function does not break.
00:24:24.500 --> 00:24:30.600
The domain is everything that can go in; we know everything that breaks it, x < -1 and x = -1.
00:24:30.600 --> 00:24:35.000
So, the domain is everything that does not break it, x > -1.
00:24:35.000 --> 00:24:39.100
All right, what if we composed h with g with f?
00:24:39.100 --> 00:24:49.800
All right, h composed with g composed with f might seem scary at first; but remember, we can break it into a much more pleasant, easy-to-work-with, h(g(f(x))).
00:24:49.800 --> 00:24:55.100
So, first, what is f(x)? It is h of g of...what was f(x)?
00:24:55.100 --> 00:25:15.300
f(x) is x + 1; so what is g(x)?...g of box is 2 times box, so g(x + 1)...it is still "h of," but g(x + 1) is going to be 2(x + 1).
00:25:15.300 --> 00:25:26.700
So, let's simplify that inside just a little bit; it is h of 2...distribute; we get 2x + 2, so we have h(2x + 2), equals...
00:25:26.700 --> 00:25:34.400
we plug that in here; remember, h of box is the square root of box + 1;
00:25:34.400 --> 00:25:48.900
so h(2x + 2) is √(2x + 2 + 1), which is equal to √(2x + 3).
00:25:48.900 --> 00:25:53.900
Great; so once again, this is not our answer; but it is going to help us figure out our answer.
00:25:53.900 --> 00:26:04.700
The square root of (2x + 3) is what the function ends up being; that is what h composed with g composed with f of x is.
00:26:04.700 --> 00:26:08.200
It is this thing right here; it is equal to the square root of (2x + 3).
00:26:08.200 --> 00:26:18.500
So, when does √(2x + 3) break? Well, once again, it breaks--it fails--when there is a negative inside.
00:26:18.500 --> 00:26:39.200
So, if 2x + 3 is negative, it breaks down; 2x + 3 < 0, so 2x < -3, which is when x is less than -3/2; we have failure.
00:26:39.200 --> 00:26:43.600
It is going to be the reverse of that: everything that doesn't cause failure is the domain.
00:26:43.600 --> 00:26:53.300
So, the domain is going to be everything that isn't x < -3/2, which is going to be everything greater than -3/2 or equal to it.
00:26:53.300 --> 00:27:05.600
So, x is greater than or equal to -3/2, when it is big enough to not cause a negative to show up inside of that square root.
00:27:05.600 --> 00:27:12.700
Great; the final example: The volume of a spherical balloon is given by volume = 4/3πr³.
00:27:12.700 --> 00:27:21.300
The balloon starts being inflated at time t = 0, in seconds, and its radius, in centimeters, is given by r = 3√t.
00:27:21.300 --> 00:27:24.200
OK, what does that mean? Let's try to figure it out really quickly.
00:27:24.200 --> 00:27:30.400
We have a spherical balloon; well, a sphere is just a ball, so that is basically what we expect when we think of balloons.
00:27:30.400 --> 00:27:40.700
This is making sense; and it is being inflated--it is being blown up; and at time t = 0 (that is just when we start), its radius is given by r = 3√t.
00:27:40.700 --> 00:27:49.900
So, it starts at t = 0; and what is its radius at t = 0? r = 3√t, so 3√0, so its radius is 0.
00:27:49.900 --> 00:27:55.200
So, it starts completely small; it is completely uninflated--it is just a dot at 0.
00:27:55.200 --> 00:28:01.800
And then, from there, it inflates; it grows out from that point; it grows out from that moment in time.
00:28:01.800 --> 00:28:04.700
Give the volume of the balloon as a function in time.
00:28:04.700 --> 00:28:09.400
We blow into the balloon, and the radius expands, and the radius expands, and the radius expands.
00:28:09.400 --> 00:28:12.900
And as the radius expands, there is now volume inside of the balloon.
00:28:12.900 --> 00:28:16.100
What is the volume at 30 seconds?
00:28:16.100 --> 00:28:21.800
All right, the first thing we need to do is give the volume of the balloon as a function of time.
00:28:21.800 --> 00:28:29.700
Well, first, we might want to see these as functions, because right now, V = 4/3πr³, r = 3√t...they are not actually functions right now.
00:28:29.700 --> 00:28:34.800
But we could easily turn them into functions: volume is really just a function of radius,
00:28:34.800 --> 00:28:37.900
because the only thing that can vary in there is the radius.
00:28:37.900 --> 00:28:44.900
It is 4/3πr³, a simple function; and then what about radius?
00:28:44.900 --> 00:28:51.300
Well, radius is a function based off of time, because the only thing that can vary in it is time, 3√t.
00:28:51.300 --> 00:28:56.900
The volume of the balloon is a function of time; well, the volume of the balloon doesn't have time inside of it.
00:28:56.900 --> 00:29:02.100
But we do know that volume has radius inside of it; and radius has time inside of it.
00:29:02.100 --> 00:29:11.800
So, we can just put these together; we can compose them; and volume of radius of time will be equal to a function,
00:29:11.800 --> 00:29:15.000
based off of time, that will give the volume of the balloon.
00:29:15.000 --> 00:29:22.900
Let's see what that is: volume of 3√t...now we are just plugging in the radius at any given time.
00:29:22.900 --> 00:29:33.100
And that is going to be the volume of 3√t; so if we plugged in our box for r, that gives us box cubed, times the other things.
00:29:33.100 --> 00:29:40.200
So, it is going to be 4/3π times (3√t)³.
00:29:40.200 --> 00:29:53.800
We simplify this out a bit; we get 4/3π times 3³ times (√t)³.
00:29:53.800 --> 00:30:01.000
Notice: 3³ can cancel down to a squared and take this one out.
00:30:01.000 --> 00:30:09.700
We have 4π times 3²; well, 4π3²...what is 3²? 3² is 9.
00:30:09.700 --> 00:30:14.100
4 times 9 is 36, so we have 36π.
00:30:14.100 --> 00:30:20.400
What about √t³? Well, remember: √t² (let's put it in a different color,
00:30:20.400 --> 00:30:30.000
so we don't get it confused) would just be equal to t on its own; so √t³ is just one extra √t left over.
00:30:30.000 --> 00:30:34.600
So, that gives us times t√t; and there we are.
00:30:34.600 --> 00:30:43.000
This is the volume of this balloon, volume of radius of time; but it is also a way of seeing volume that is purely in terms of t.
00:30:43.000 --> 00:30:52.200
t shows up; t shows up; but π is a constant; 36 is a constant; so what we have now is volume based purely off of time.
00:30:52.200 --> 00:30:56.100
We have the first part of this question done.
00:30:56.100 --> 00:31:00.200
The next part--volume at 30 seconds: well, we have two options for how to do this.
00:31:00.200 --> 00:31:10.400
We could plug in, into the function that we just built, volume at 30 equals 36π times 30 times √30.
00:31:10.400 --> 00:31:30.900
Or we could plug in volume of radius at time t, which would be volume of 3√30, which would be equal to 4/3π(3√30)³.
00:31:30.900 --> 00:31:35.100
And it ends up being the case that these two things actually end up equaling the exact same thing.
00:31:35.100 --> 00:31:46.700
Let's just fold them together: 36π times 30 times √30...that ends up being 1080π√30.
00:31:46.700 --> 00:31:50.500
And if want to get this as an approximate value, something that we could actually know as a number,
00:31:50.500 --> 00:31:55.300
as opposed to just having symbols that are precise and accurate, and exactly correct and right,
00:31:55.300 --> 00:31:59.200
but hard to actually grasp as a single number and know what we are talking about,
00:31:59.200 --> 00:32:06.400
we could get a pretty close thing, and we could round this to 8584 using a calculator.
00:32:06.400 --> 00:32:12.500
What are the units that it comes in? It is centimeters cubed, because if radius is in centimeters,
00:32:12.500 --> 00:32:16.800
and volume is centimeters cubed (and it makes sense, because we are talking about volume),
00:32:16.800 --> 00:32:24.100
and length is centimeters, area is centimeters squared, and volume is centimeters cubed, at least if we are using centimeters.
00:32:24.100 --> 00:32:27.000
If we are using meters, it is meters, meters squared, meters cubed.
00:32:27.000 --> 00:32:32.000
If we are using inches or feet, it is square inches, square feet, and cubic inches and cubic feet.
00:32:32.000 --> 00:32:34.200
All right, great; that completes it for composite functions.
00:32:34.200 --> 00:32:36.000
I hope you have a much better understanding of what is going on.
00:32:36.000 --> 00:32:39.300
Remember, when you see that circle, it means "composed with," but it is much easier
00:32:39.300 --> 00:32:45.400
to break it into f of g of x, or g of f of x, depending on the order it goes in.
00:32:45.400 --> 00:32:50.500
And remember, it is just going to be based off of the order that they are hitting the x in.
00:32:50.500 --> 00:32:59.400
Whichever is closer goes first; so this becomes f of...g gets to act first, because it is closer to the x.
00:32:59.400 --> 00:33:04.300
That is what that means; whichever is closer goes first, so it is whatever the order is with the circles.
00:33:04.300 --> 00:33:15.400
But now, f(g(x)); f, circle, g(x) becomes f(g(x)); a, circle, b, circle, c(x) becomes a(b(c(x))).
00:33:15.400 --> 00:33:18.800
Great; all right, I am glad to have taught you the composite functions.
00:33:18.800 --> 00:33:23.100
I hope you can use it in a bunch of places; it will show up in a variety of things--it is really useful stuff here.
00:33:23.100 --> 00:33:24.000
And we will see you at Educator.com later--goodbye!