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Lecture Comments (21)

1 answer

Last reply by: Professor Selhorst-Jones
Fri Mar 13, 2015 9:02 PM

Post by Mohammed Fayaz Amanulla on March 13, 2015

Example 2 question 2, why do i see a 12x. isnt it just 2x^2+3^2-7 which equals to 4x^2+9-7?

1 answer

Last reply by: Professor Selhorst-Jones
Sat Feb 7, 2015 3:35 PM

Post by Randall Tew on February 7, 2015

Hello Professor Selhourst-Jones. I have a question about the 4th step in green pencil. Specifically, what rule enables the cancellation of the ^3 power in 3^3sqrt with the denominator 3 in 4/3?

1 answer

Last reply by: Professor Selhorst-Jones
Tue Jan 6, 2015 12:45 PM

Post by Andrew Demidenko on January 4, 2015

Could you please give in answer another word problem / thanks / AD

1 answer

Last reply by: Hari Karan
Wed Mar 26, 2014 2:38 AM

Post by Hari Karan on March 26, 2014

sir, could i ask you, for example 2 question 2, why do i see a 12x. isnt it just 2x^2+3^2-7 which equals to 4x^2+9-7? 0.o

1 answer

Last reply by: Professor Selhorst-Jones
Thu Oct 17, 2013 8:27 AM

Post by Constance Kang on October 17, 2013

i have a question about example 3. you said taht the domain is x greater than -1 because the root will break if its not a positive number. however, i rationalized the bottom and got (x+1+root(x+1))over 2x+2 at the bottom, which makes the domain any real number except -1

1 answer

Last reply by: Professor Selhorst-Jones
Sun Jul 28, 2013 9:39 PM

Post by Chudamuni Dahal on July 28, 2013

So prof, Can we rewrite f/g(h(x)) as
[f(h(x))]/[g(h(x))] when asked for domain and other time?

0 answers

Post by Chudamuni Dahal on July 28, 2013

Prof, I have a question. Do we have to divide f(x)/g(x) first then only compose? Can't we do like this? f/g(h(x))so that the Domain is going to be all real number such that X is greater than or equal to -1. I am confused over why are we divide f/g first?I am confused Because you said earlier, composition to the nearest first. Other than that, thank you so much. So really have amazing ability to explain the concept in a such simple way

1 answer

Last reply by: Professor Selhorst-Jones
Thu Jul 11, 2013 1:09 PM

Post by Sarawut Chaiyadech on July 1, 2013

Thx:)

5 answers

Last reply by: Professor Selhorst-Jones
Thu Aug 8, 2013 3:32 PM

Post by Jorge Sardinas on May 5, 2013

isn't supposed to be 30 root 30 instead of 3 root 30 mr.jones????!!!!!!

Composite Functions

  • Two (or more) functions can interact with each other through good old arithmetic: addition, subtraction, multiplication, and division. These sorts of interactions are called arithmetic combinations. Here are the four types:
    • Sum: (f+g)(x) = f(x) + g(x)
    • Difference: (f−g)(x) = f(x) − g(x)
    • Product: (fg)(x) = f(x) ·g(x)
    • Quotient: (f/g)(x) = f(x)/g(x)        [and it must be that g(x) ≠ 0]
  • A much more interesting idea is to compose two functions. Instead of giving both functions the same input, we give the input to just one function. Then we take the first function's output, and plug that in to the second function. The second function is acting on the first function. [If this idea is confusing, make sure to watch the video where we see some analogies.]
  • For function composition, we can use the notation of f °g. [Read as "f composed with g."] If f °g acts on x, we have (f°g ) (x). This means g acts on x first, then f acts on whatever results. [Notice how the functions act in order of closeness to the original input.]
  • Another (much easier) way to see (f °g ) (x) is in the function notation format we're already used to:

    f °g
    (x) = f
    g(x)
    .
    (Recommendation: If you see the ° notation [such as f °g], rewrite it in the "normal" format [such as f ( g(x) )]. This normally makes it easier to understand and solve problems.)
  • Working with composite functions might seem intimidating at first, but it's really just about plugging in appropriately. Each function has its own "rule", so composing multiple functions just means using the rules in succession.
  • This idea of plugging in is shown beautifully by the notation f ( g(x) ). The function g acts on x, then f acts on the resulting g(x). Since we almost certainly know what g(x) looks like from the problem, we just use that as input for f.

Composite Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:04
  • Arithmetic Combinations 0:40
    • Basic Operations
    • Definition of the Four Arithmetic Combinations
  • Composite Functions 2:53
  • The Function as a Machine 3:32
  • Function Compositions as Multiple Machines 3:59
  • Notation for Composite Functions 4:46
    • Two Formats
  • Another Visual Interpretation 7:17
  • How to Use Composite Functions 8:21
    • Example of on Function acting on Another
  • Example 1 11:03
  • Example 2 15:27
  • Example 3 21:11
  • Example 4 27:06

Transcription: Composite Functions

Hi--welcome back to Educator.com.0000

Today, we are going to talk about composite functions.0002

We are often going to have 2 or even more functions that interact with each other.0005

This lesson will explore the fundamental ways that functions can interact with each other.0009

First, we will look at how functions can interact through just arithmetic: addition, subtraction, multiplication, and division.0014

These sorts of interactions are called arithmetic combinations, because they are just using arithmetic.0019

Second, we will move on to a more complex idea, using one function's output as another function's input.0024

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.0031

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.0039

Now, it is easy to imagine creating a new function that just adds f and g together; we would call it f + g.0049

That is not very imaginative, but it makes sense.0055

It would give us the sum of the two functions: the new function, f + g (x), would be equal to x2 + x.0057

We are just adding the two functions together; so we know each function is x2 and x, so we just add them together.0066

That is a simple, basic idea; we are using a basic arithmetic operation, and we are just putting them together through that.0072

We have arithmetic; let's use it.0079

We could expand this idea to the other three basic operations.0081

We could do this with subtraction: f - g would become x2 - x; fg(x) would be x2 times x (fg being f times g,0084

just like when we say 3x, we mean 3 times x); and f/g would be x2/x--as simple as that.0093

Given two functions f and g, along with an x that is in the domain of both, we defined these four different arithmetic combinations.0101

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,0108

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--0115

all of these work really well; sum is f + g(x)--just break it down into adding the two together.0121

Difference is just subtracting one, just like normal subtraction.0127

Product--when we have fg, we read it as times: f times g; and quotient is f(x)/g(x).0132

And it also has to be that g(x) does not equal 0, because otherwise we could accidentally end up0141

blowing the world up when we divide by 0, since we are not allowed to divide by 0,0146

because it is nonsense and doesn't mean anything.0149

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 0.0151

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;0157

that input is in the domain, and it is defined as an output; then we can just put the two together.0162

We just put what f(x) is together and put what g(x) is together with any arithmetic combination that we want to; great.0167

That is a nice, direct idea; an arithmetic combination makes sense.0174

We put in the same input to the two functions, and then we combine their outputs with some pre-decided arithmetic operation.0177

If it is sum, we do it through addition; if it is product we do it through multiplication; things like that.0182

But we can do something more interesting: we can compose one function with another.0188

Instead of giving both functions the same input, we give the input to one function.0193

Then we take the first function's output, and we plug that into the second function.0198

Input goes into one, and then an output comes out of that; and that immediately goes into the second function.0203

And then finally, we get an output of that; the second function is acting on the first function.0207

Many lessons back, we first introduced the idea of a function; and we talked about how we can view it as a machine.0213

It takes in inputs, and the function produces outputs: x goes into the machine, the function f;0219

and then it gets put out after having been acted on.0225

The function is some process; it does some transformation on x, so we get f(x), f having acted upon x.0228

All right, so that is the idea of it as a machine.0237

We can expand this idea into function composition.0240

Function composition is just linking multiple machines together in series; we just put multiple of them together.0243

The output of the first function goes directly into the second as its input.0248

Our first input goes into, say, g; and so, it is now g(x); and then we plug all of g(x) into f.0252

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.0264

And then, we just jam that right into the second machine.0272

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.0274

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.0280

We note the previous slide's composition, when it went into g first, and then went into f, as f composed with g.0287

This is just a little circle between them: f circle g...we read that as f composed with g.0295

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.0301

We have created a new function out of putting the two together.0308

By linking those two machines together, it is effectively one larger machine that is doing a new way of working.0311

This means that in this machine, f composed with g, g will act on x first; and then f will act on whatever results.0318

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.0327

Now, notice: the functions act in order of closeness.0336

g goes on first; and then, f goes on second, because it is farther away.0340

We hit it with the things that are closest to the input that we are putting in.0348

So, g goes onto the x, and then f goes onto what results; and if we had even more stacked up,0352

whatever was even further to the left would act after that.0356

The functions act in order of closeness to the original input.0360

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--0363

the thing that we have been using for quite a while now.0370

f composed with g of x is just f(g(x)); so f composed with g of x...remember how we broke it up:0372

this one went first; and then f acted second--well, that is what we have right here.0380

g is acting first, and then f is acting second--it makes a lot of sense.0387

I would personally recommend, any time you see this circle notation--this f composed with g stuff--rewrite it0392

in this normal format, the format that we are used to at this point, the f(g(x)).0399

This normally will make it easier to understand and solve problems; there are very few downsides to breaking it into this thing.0404

So, I would really recommend: any time you come up against a problem, and you are not quite sure what to do,0411

and it is this sort of thing: break it into f(g(x)), f acting on g(x)--0415

something acting on something else acting on the input that you are putting in.0420

This method, the second form of notation--this is really great as a way to look at things.0424

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.0429

Another way we can visually interpret it--this is a little hard to see what is going on here, but try to follow me0438

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.0444

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).0451

But we can also think of it as some new function that has been created, f composed with g.0460

We have created a new machine that can just go directly from our original x to the end result of f(g(x)).0466

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.0476

We can look at it as stair-stepping across, or we can look at it as one new-built, giant leap,0483

where it does both of these actions in one jump.0489

All right, you can take steps across the pond, or you can take one giant leap.0492

But ultimately, they do the same thing; the leap has to be informed by the steps, though.0495

So, how do we actually use these composite functions?0502

We understand the idea behind them now; and it turns out that using them actually isn't that hard.0504

It is just important to understand the idea.0509

Each function has its own rule, like f(x) = x3 means to cube your input.0511

So, composing multiple functions just means using these rules in succession.0517

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)).0521

The function says that the function g acts on x; then, f acts on the resulting g(x).0528

So, g acts on x; that is what this says right here; and then f acts on the resulting g(x).0535

f acts on what we just had there; great.0543

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.0546

We use the rules that we were given earlier, and we just apply them to these things.0552

Let's see an example: for example, if we had f(x) = x2 + 3 and g(x) = 2x - 2,0556

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.0564

Remember, it is just a placeholder: f(x) is just a way of saying f(whatever is in here), whatever f is acting on.0572

The thing that it is acting on will get squared; plus 3.0580

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 - 2.0583

So then, we plug that in for f(x); f(x) becomes x2 + 3; so if what is inside of the box is 2x - 2,0592

it is going to be (2x - 2)2; so the box has the same process happen--it is just a new thing going on.0601

Instead of x going into it, it is just 2x - 2 going into it.0612

The same processes: it is taking the input, squaring it, and adding 3.0617

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 3.0621

So, if we wanted to, at this point we could expand (2x - 2)2 + 3; but this is really the key idea--0630

getting to this point of thinking of it as boxes; we are plugging in, based on boxes.0634

And we will see a bunch of examples using this idea later on.0639

But you want to think of it as we are just swapping out; we are using x as a placeholder.0642

It is not x that we are really attached to; f(x) is saying f of box, and then what happens to box;0647

f of placeholder, and then what happens to placeholder; f of input, and then what happens to input.0653

That is the way you want to think about it; and that makes it really easy to do composite functions.0659

All right, it is time for some examples.0663

So, f + g of 3; if we have f(x) = 2x + 3, and g(x) = x2 - 7, what would f + g of 3 be?0665

We do this: f(x) is 2x + 3; g(x) is x2 - 7; we have 2x + 3 + x2 - 7.0675

So, that becomes something; we could simplify it; but at this point, let's plug in x = 3.0686

We have that x = 3 is going to get plugged in, so we have 2(3) + 3 + 32 - 7.0692

6 + 3 + 9 - 7; 9 + 9 is 18; 18 - 7 is 11; so we have 11 as the answer here.0702

All right, the next one we will do with the color blue: g - f(1); what is g?0713

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;0718

not just the 2x, but minus (2x + 3); it is a whole quantity that we have to be subtracting.0728

Now, we will plug in; what happens when we plug in x = 1?0733

Well, we have 12 - 7 - (2(1) + 3); so that is 1 - 7 - 2 + 3, which is equal to -6 - 5, equals -11.0738

Great; the next one--I will do this one in green: fg(-2).0761

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 - 7.0766

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),0776

times (-2)2, minus (oops, I accidentally made a plus sign) 7; great.0786

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 3.0795

Great; and finally, let's go back to red for our very last one, g/f(8).0813

So, what is g? g is x2 - 7, over all of f; so it is 2x + 3.0819

So, we plug in x = 8; 82 - 7, over 2(8) + 3; 82 is 64, minus 7, over 2 times 8 (is 16), plus 3.0827

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 57.0844

So, 57/19...we get 3 once again, by chance; great.0854

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.0859

So notice: f(3)...for example, we used the f + g(3) just to make a point here.0866

So, f(3) is equal to 2(3) + 3, which would be equal to 6 + 3, or 9.0876

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), 0884

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.0903

So, we can either put the functions together, and then plug in the variable;0914

or we can plug in the variable into each function and then add them together.0916

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.0920

But it is important to notice that we can do it the other way.0924

The second example: we have the same functions that we just ended up using, f(x) = 2x + 3 and g(x) = x2 - 7.0927

What is f composed with g, first?--so we will start with the green for this one.0935

f composed with g of x; what was my recommendation? It was f(g(x)); that is the exact same thing.0939

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.0948

f(g(x)): f(x) = 2x + 3, but we don't need that information yet; we need to plug in g(x) first.0957

f(x2 - 7); now, what is the x? Here is our x here, and x goes right here.0967

That means that f of box is equal to 2 times box plus 3.0979

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 3.0983

And if we wanted to, we could expand that out.1000

We expand that out pretty easily; and we would get 2x2 - 14 + 3, which is 2x2 - 11; great.1003

The next one: blue for the next one: g composed with f(x) is much easier to write as g(f(x)).1015

What is f(x)? f(x) is 2x + 3, so that is what is going into g right now: 2x + 3.1024

And now, if we plug into g, g of box equals box squared minus 7.1032

So, g(2x + 3) is equal to (2x + 3)...that is the thing in the box...squared, minus 7.1040

We work that out; we get 4x2 (2x times 2x is 4x2) + 2x + 3, and (3 + 2)x is 6x, plus 6x is 12x,1048

plus 3 times 3 is 9, minus 7, which equals 4x2 + 12x + 2.1059

Great; OK, the next one--let's use red for this one: f composed with f--f composed with itself.1069

We can also write this as f(f(x)); what is f(x)? f(x) is 2x + 3, so f(2x + 3).1080

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) + 3.1089

Great; we just work this out: 2(2x + 3) is 4x + 6 + 3 = 4x + 9.1101

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.1111

g(x) is x2; so now it is g acting on x2 - 7; remember, g of box is equal to box squared, minus 7.1121

So, if we are plugging in x2 - 7, it is going to be (x2 - 7)2 (remember, box squared minus 7), and then also - 7.1130

Great; so we square x2 - 7; x2 times x2 is x4;1139

x2 times -7 is 7x2, minus 7x2...another -7x2;1144

we have -7x2 + -7x2; that is minus 14x2;1149

and -7 times -7 is positive 49; and finally, minus 7; so it is x4 - 14x2 + 42; there we are.1154

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,1173

and remembering, in terms of the substitution: it is not about the letter x; it is about if we just had a box here.1177

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.1184

If we put in an input, what happens to the input?1190

We just happen to use x, because it is a convenient thing; we are used to using it as a placeholder.1192

But x isn't inherently important; it is just the idea of what happens to an input.1197

So, if we plug in something like 2x + 3, different things will happen than if we had just plugged in x.1202

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)).1207

For the most part, flipping the order that we do our function composition in gives us very, very different results.1226

Sometimes, it will end up being the same result; but mostly, if we take f(g) or g(f), it will be totally different.1234

So, mostly, f composed with g is a totally different function than g composed with f.1241

And this is just something to keep in mind--that composition order matters very much.1248

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.1253

So, the order that you put it in--the order that one function goes into another--the order of composition--has massive importance.1264

Great; the next example: let f(x) = x + 1, g(x) = 2x, h(x) = √(x + 1).1271

What is the domain of f, divided by g, composed with h, and then also of h composed with g composed with f?1278

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.1286

I am sorry, not f(g); what is f divided by g?1293

Well, f/g(x) is just equal to f(x)/g(x), except for when it is undefined, when g(x) is equal to 0.1296

So, what does this mean? Well, f(x) is x + 1, so (x + 1)/2x--there we are.1306

f/g(x) is equal to (x + 1)/2x; great, now we can just go back to what we are used to doing.1314

So, we use blue for this one; f/g(h(x))...it is much easier to see it written in that format.1322

f/g...what is h(x)?...f/g(√(x + 1)) = √(x + 1) + 1, over 2(√(x + 1)).1332

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.1349

f divided by g composed with h comes out to be √(x + 1) + 1, divided by 2(√(x + 1)).1359

So, the domain is going to be everywhere where it doesn't break.1365

So, what things in here can break? Well, first, square root--any time we see square root, that breaks when a negative is inside.1373

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.1391

It breaks--it is not defined, more formally--when x + 1 < 0, which is to say when x < -1.1400

So, that is one important point of information: x < -1 means failure.1409

Another failure point is if this bottom part is equal to 0; we have another break if 2√(x + 1) = 0.1414

And that is going to end up being x + 1 = 0, which means x = -1.1431

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 ≤ -1.1436

So, its actual domain, the domain of this function, is x > -1.1450

It is all of the places where the function does not fail, where the function does not break.1459

The domain is everything that can go in; we know everything that breaks it, x < -1 and x = -1.1464

So, the domain is everything that does not break it, x > -1.1470

All right, what if we composed h with g with f?1475

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))).1479

So, first, what is f(x)? It is h of g of...what was f(x)?1490

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).1495

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...1515

we plug that in here; remember, h of box is the square root of box + 1;1527

so h(2x + 2) is √(2x + 2 + 1), which is equal to √(2x + 3).1534

Great; so once again, this is not our answer; but it is going to help us figure out our answer.1549

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.1554

It is this thing right here; it is equal to the square root of (2x + 3).1565

So, when does √(2x + 3) break? Well, once again, it breaks--it fails--when there is a negative inside.1568

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.1578

It is going to be the reverse of that: everything that doesn't cause failure is the domain.1599

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.1603

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.1613

Great; the final example: The volume of a spherical balloon is given by volume = 4/3πr3.1625

The balloon starts being inflated at time t = 0, in seconds, and its radius, in centimeters, is given by r = 3√t.1633

OK, what does that mean? Let's try to figure it out really quickly.1641

We have a spherical balloon; well, a sphere is just a ball, so that is basically what we expect when we think of balloons.1644

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.1650

So, it starts at t = 0; and what is its radius at t = 0? r = 3√t, so 3√0, so its radius is 0.1661

So, it starts completely small; it is completely uninflated--it is just a dot at 0.1670

And then, from there, it inflates; it grows out from that point; it grows out from that moment in time.1675

Give the volume of the balloon as a function in time.1682

We blow into the balloon, and the radius expands, and the radius expands, and the radius expands.1685

And as the radius expands, there is now volume inside of the balloon.1689

What is the volume at 30 seconds?1693

All right, the first thing we need to do is give the volume of the balloon as a function of time.1696

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.1702

But we could easily turn them into functions: volume is really just a function of radius,1710

because the only thing that can vary in there is the radius.1715

It is 4/3πr3, a simple function; and then what about radius?1718

Well, radius is a function based off of time, because the only thing that can vary in it is time, 3√t.1725

The volume of the balloon is a function of time; well, the volume of the balloon doesn't have time inside of it.1731

But we do know that volume has radius inside of it; and radius has time inside of it.1737

So, we can just put these together; we can compose them; and volume of radius of time will be equal to a function,1742

based off of time, that will give the volume of the balloon.1752

Let's see what that is: volume of 3√t...now we are just plugging in the radius at any given time.1755

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.1763

So, it is going to be 4/3π times (3√t)3.1773

We simplify this out a bit; we get 4/3π times 33 times (√t)3.1780

Notice: 33 can cancel down to a squared and take this one out.1794

We have 4π times 32; well, 4π32...what is 32? 32 is 9.1801

4 times 9 is 36, so we have 36π.1810

What about √t3? Well, remember: √t2 (let's put it in a different color,1814

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.1820

So, that gives us times t√t; and there we are.1830

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.1834

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.1843

We have the first part of this question done.1852

The next part--volume at 30 seconds: well, we have two options for how to do this.1856

We could plug in, into the function that we just built, volume at 30 equals 36π times 30 times √30.1860

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)3.1870

And it ends up being the case that these two things actually end up equaling the exact same thing.1891

Let's just fold them together: 36π times 30 times √30...that ends up being 1080π√30.1895

And if want to get this as an approximate value, something that we could actually know as a number,1907

as opposed to just having symbols that are precise and accurate, and exactly correct and right,1910

but hard to actually grasp as a single number and know what we are talking about,1915

we could get a pretty close thing, and we could round this to 8584 using a calculator.1919

What are the units that it comes in? It is centimeters cubed, because if radius is in centimeters,1926

and volume is centimeters cubed (and it makes sense, because we are talking about volume),1932

and length is centimeters, area is centimeters squared, and volume is centimeters cubed, at least if we are using centimeters.1937

If we are using meters, it is meters, meters squared, meters cubed.1944

If we are using inches or feet, it is square inches, square feet, and cubic inches and cubic feet.1947

All right, great; that completes it for composite functions.1952

I hope you have a much better understanding of what is going on.1954

Remember, when you see that circle, it means "composed with," but it is much easier1956

to break it into f of g of x, or g of f of x, depending on the order it goes in.1959

And remember, it is just going to be based off of the order that they are hitting the x in.1965

Whichever is closer goes first; so this becomes f of...g gets to act first, because it is closer to the x.1970

That is what that means; whichever is closer goes first, so it is whatever the order is with the circles.1979

But now, f(g(x)); f, circle, g(x) becomes f(g(x)); a, circle, b, circle, c(x) becomes a(b(c(x))).1984

Great; all right, I am glad to have taught you the composite functions.1995

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.1999

And we will see you at Educator.com later--goodbye!2003