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Language: en
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This is sorta the theoretical part of class. It's not too theoretical but it's a little theoretical.
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So remember a function takes an element in the domain and maps it to an element of the range
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That takes an input and puts it onto an output-- a function is just an operation that does that.
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So you need to have well-defined domains and ranges. So we need to be clear what's going in and clear what you get out.
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Okay, you can't just say some number in some graph, you have to have some rule so it is obvious what's the input and output
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Another thing you can do to help visualize functions, is you can graph them.
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So you guys've all done graphing, a lot. So what we'll introduce for a minute or two, think about what a graph does.
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A graph, if you had a graph like a line, what you are doing is taking the domain and you are matching it to an element in the range.
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and you are making a mark that says "that's where those two go together."
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And then another element of domain and another element of range make a mark and you say that's where those two go, and you connect all those dots
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and that's a picture
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So when we use graphs it helps us visualize what's gong on and that's why you can use a vertical line test or a horizontal line test
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So the key for graphs is if you want to think whether something is a function of x, the vertical line test will tell you
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You need to be able to draw a vertical line anywhere on the graph and never get the graph twice
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You can not hit the graph at all (ie, miss it entirely), but no vertical line can intersect the graph in more than one place.
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So again, no vertical line can intersect the graph more than once
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Could be less than once --it could be zero-- but it can't be more than once
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Yes?
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No?
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Look at this and say wow I can draw a line and touch the graph more than once.
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Here, I can draw the lines and no matter what I do I'm only going to hit the graph once.
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I think next week we go into one-to-one functions. One-to-one functions also have to go through the horizontal line test
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and a one-to-one function has to pass both the vertical and horizontal line test.
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Now another thing to think about with graphs, which we're going to do
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May or may not get recorded on the exam, I'm not sure.
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What's called translation of a graph
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Makes you move the graph around
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okay but I don't want to talk about that to much. Today I want to talk about
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Composition of functions. So composition of functions, means one function is inside of another function.
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We will do a lot of that. There's a lot of of algebra involved in it.
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The idea is how they function f(x).
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And you can do different things if you have a second function, like g(x).
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So let's say f(x) equals x squared plus 3
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g(x) equals 1 over x.
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Nice simple function
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Okay, so what is (f+g)(x)?
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Should be pretty obvious
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*Student is answering question*
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Well it is f(x) plus g(x), right but what is that?
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So (f+g)(x) is obvious..
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There you go. Have more candy to throw up there.
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Okay all you're doing is literally adding them.
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She asked that striaght forward, we are doing an algebra of function. What we are saying is you can do rules of functions
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So (f-g)(x)
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Gets more fun.
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Okay and by the way, what happens is you'll have a domain of f(x) and a domain of g(x)
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So the domain of f(x) is all real, so you can put any number you want, square it and add three to anything you want.
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Domain of g(x) is not all reals, because you can't have a zero
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You can't have a zero in the domain
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If you put in zero, you'll get one over zero and that's undefined
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The domain of (f+g) of x will also be all reals except zero.
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So when you have two domains the more more restricted one is going to win.
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because the domain of g(x) is more restricted than the domain of f(x).
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So when we combined them the restriction doesn't go away
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So you couldn't put zero in this one so you can't put zero in this one either, same here..
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(f times g)(x)
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Is write like that
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x squared plus 3 times one over x
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Question: Can you put zero in this one?
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Why not?
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You can put zero? You can't? To me you can't put zero in that because you can't have zero in the denominator.
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And you still have a denominator
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So the problem doesn't go away just because you put two functions togeher
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So far so good?
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Okay so when you have f plus g you add them, we have f minus g you subtract them.
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f divided by g
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(f/g)(x), this one is kind of entertaining, because now you have a denominator
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So you could say to yourself, now I can flip that x. Right? That 1 over x?
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and we get x squared plus 3 times x.
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Then the problem goes away doesn't it?
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I make it this..
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And now I don't have the zero in the denominator problem anymore, right?
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Wrong.
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If you couldn't have it in the original function, you can't have it in the combined function.
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You still can't have zero here.
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Okay, you can flip that over as long as x is not zero.
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We are more likely, what's happens if f(x) is x squared plus 3, g(x) is x squared minus 1. Right now both domains are fine.
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But if you have f(g(x))
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The domain of f(g(x))
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Now x can't be one
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How we doing so far? This isn't very hard, right?
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The real reason you need to know what a domain is when you do a function because you really need to know what are you doing, what are you putting in?
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This will show up at other aspects of life, not necessary mathematically.
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Now for the fun part.
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When you put functions inside functions
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So now when you do f(g(x)).
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Which is sometimes written "f o g" of x.
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f(g(x)), what does that mean?
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Well, before we had f(x). What the x means, where you see x in the equation plug something in, yes?
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So what does g(x) mean?
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This says go to f and wherever we have x, replace it with g(x)
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Also known as "fog" of x, so the circle means composite. So f(g(x)).
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Means..take x and now replace it with g(x).
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g(x) is 1 over x.
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We genuinely don't care if you simplify, well there are a few problems you'll have to simplify. But on a test I wouldn't expect you to simplify something that defined
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And notice before domain x can't be zero so it can't be zero.
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What is g(f(x))?
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Well now, I am going the other way..
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Now I take f(x), and put it in g(x). So it'd be 1 over f(x), which is 1 over x squared plus 3
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So far so good? This is what we mean by composition of functions
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So let's say I say f(x)
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I have those two functions.
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And I want to find f(4).
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You all know how to do f(4) right?
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Just plug in 4.
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4 cubed minus 4 plus 2.
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Which is 62.
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So far so good
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Easy.
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What if I ask for f(g(4))?
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How do you do f(g(4))?
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Well..
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First, you figure out what g(4) is then you take that answer and plus it into f.
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So if f(g(4)) equals question mark
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We'll say well g(4) is 2 times 4 plus 1 equals 9.
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So this is now going to ask for f(g(4)) means find f(9)
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Which is 9 cubed minus 9 plus 2, which is 722.
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You can all do it in your head right?
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Maybe not in your head
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You guys don't know 9 cubed?
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Same as 3 to the 6?
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Okay let's do another one of these
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We got the idea, this isn't very hard, right?
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You learned this back when you did algebra the last time or the time before.
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So if asked f(a), it would just be a^2 + 2a + 1. If I asked f(b), it would be b^2 + 2b + 1
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What if i asked for f(x+h)?
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What would it be?
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but we take x and replace it with x+h.
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So we get x plus h squared, plus 2 x plus h, plus 1.
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Now heres the fun part.. this we might ask you to simplify
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So heres a couple handy things to memorize
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few good ones to memorize
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I'll put it on this board, walk over to the other side and put it on that board
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x plus h squared, we will foil that out and we get x squared plus 2xh plus h squared.
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And the other one is x plus h cubed.
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and that one is x cubed plus 3 times x squared h, plus 3xh squared, plus h cubed. So the pattern for the second one, see the power of x goes down each time?
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goes from 3 to 2 to 1to 0. And the power of h goes up each time, goes from 0 to 1 to 2 to 3.
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It also happened here, x goes from 2 to 1 to 0 and h goes from 0 to 1 to 2
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Coefficients come from Pascals Triangle, or you can find them in Pascals Triangle
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So how am I going to memorize these, because theses are going to sow up a couple of times in this class and a whole bunch of times in calculus
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If you aren't memorizing them thats fine, you just take x plus h and multiply iy by x plus h.
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and then you multiply it by x plus h again if you have to
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So here..
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x plus h squared is x squared, plus 2xh plus x squared, plus 2x plus 2h plus 1
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Cant really do anything with that so we stop
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This is something you might see over the next few days on webassign
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Lets practice one of these
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To find f(x+h) use f(x) and replace it by x+h.
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So it's 3 x plus h, squared, minus x plus h, plus 4
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Now you multiply out the x plus h
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and you get x squared plus 2xh plus h squared
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I'm going to rewrite that
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So you'll multiple out the x plus h, minus x, minus h, plus 4 and then distribute the 3
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and you get 3x squared, plus 6xh, plus 3h squared, minus x minus h plus 4
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Okay, so f(x+h) minus f(x).
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. Now you're going to take this and subtract f(x)
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So you're gonna take 3x squared, plus 6xh, plus 3h squared, minus x minus h plus 4
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Minus 3x squared minus x plus 4
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Notice what happens..
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All the terms on the left that do not contain and h are going to cancel all the terms on the right.
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Because here you have 3x squared, and here you have minus 3x squared.
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Here you have minus x, here you have minus minus x
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and here you have 4, and here you have minus 4. You are now left with 6xh plus 3 h squared minus h
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So a clue, when we give you a problem in the form of f(x+h) minus f(x) you want to look for cancelations, you'll get a bunch of those.
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-Everyone see that?
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Lets do another one
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to make sure you get the hang out it
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Okay ready for the last touch?
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So do what you were doing before, to get the answer divided it by h.
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So you are going to do this a bunch of times in calculus. So we are practicing it now to make sure, that's why this course is called introduction to calculus
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First do the hard part. Got to find f(x) plus h
Well, f(x+h), is (x plus h)cubed plus 4 x plus h minus 1.
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Okay, so if we multiply that out thats x cubed plus 3x squared h, plus 3x h squared, plus h cubed plus 4x, plus 4h minus 1
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So far so good?
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How do we know thisâ€¦
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Because that's this one, make sure you memorize that. I'll put it up later
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So f(x+h) - f(x), is going to be this minus this.
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So it's going to be x cubed plus 3 x squared h, plus 3 x h squared, plus h cubed plus 4 x plus 4 h minus 1, minus x cubed, 4x, minus 1
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Now we can cancel.. I got x cubed, minus x cubed.I got 4x and minus 4x, and minus 1 minus minus 1.
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So now i am left with 3 x squared h, plus 3x h squared, plus x cubed, plus 4h.
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Finally, I am going to take that..
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and divide it by h.
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Some of you guys saw this in precalc or whatever in high school
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How do you solve this? Factor and h out of everything
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Take an h out of each of the terms on top, they each contain an h and now you can cancel that h
And you get 3 x squared, plus 3xh,plus h squared plus 4
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You're going to see something like this on webassign, and a good chance you can see this on an exam. Can't say a hundred percent but we are practicing it for a reason
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When you did this with polynomials, all the terms on the right will cancel on the left, every time.
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Let's take x and replace it with x plus h
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So f(x + h)
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1 over x plus, plus 2
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These are composite functions, okay, get the function x plus h and then you put that inside the function f.
00:29:25.400 --> 00:29:28.120
and you get out 1 over x plus h, plus 2
00:29:28.120 --> 00:29:32.860
and now we do a little algebra on it, you take this function and subtract that function.
00:29:33.340 --> 00:29:38.780
This is going to be one over x plus h, plus 2 minus..
00:29:39.480 --> 00:29:44.060
1 over x, plus 2. The 2's cancel
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And you get 1 over x plus h, minus 1 over x.
00:29:52.160 --> 00:29:56.920
Now you might be tempted to just leave it like that, but you can actually simplify this
00:29:56.920 --> 00:30:00.740
When we have two fractions what do we have to do to combined them? Common denominator.
00:30:01.500 --> 00:30:03.880
Right, you got to do that
00:30:04.360 --> 00:30:06.620
Common denominator could be x times x plus h
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so you multiply the left by x, top and bottom
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You take the right one and multiply it by x plus h.
00:30:32.480 --> 00:30:36.320
That becomes x minus, x plus h
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over x times, x plus h
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Now distribute that minus sign and notice the x's cancel and you're left with minus h over x times, x plus h
00:31:10.780 --> 00:31:14.080
We could distribute the x on the bottom but it doesnt get you anywhere
00:31:15.860 --> 00:31:27.160
Simplifying is to make things simpler. So the problem with distributing it is here you know that zero is zero and zero is negative
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When you multiply out, it makes it less obvious.
00:31:30.860 --> 00:31:36.940
So that's my problem with the phrase simplify. I'm not sure this is really simpler than this
00:31:38.820 --> 00:31:44.020
Theres an advantage for the second form when we want to do more operations on it
00:40:29.860 --> 00:40:37.180
These are composite functions, okay, get the function x plus h and then you put that inside the function f.