 Welcome back to our lecture series Math 1050, College Algebra for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Angela Misseldine. So lecture 50, this is the end of our series. We've made it, yay. So as you recall, for those following along with our videos here, that in our last chapter entitled Pre-Calculus, we're revisiting many topics we've seen previously in this lecture series on College Algebra, combining things that we didn't really originally combined from the first place and emphasizing why this is helpful for calculus, which many of us will be taking, perhaps in a future college class or such. So it's appropriate that the last topic, the ultimate topic of our lecture series is gonna be about difference quotients, which is one of the most important algebraic things we have to consider at the very beginning of calculus one. So to preference why we care about difference quotients, let me kind of remind you about the average rate of change which we saw previously in this lecture series. So let f of x be a function defined on some interval a to b. Then we say that the average rate of change of f of x with respect to x for a function f as x, changes from a to b is given by this. We usually denote it as delta y over delta x. And sometimes we add a superscript of a comma b to emphasize which average rate of change we're talking about because it does depend different intervals will have different average, I should say average rates of change. But the formula is just gonna be f of b minus f of a over b minus a. This is just the slope formula. So, cause one thing to remember when it comes to the average rate of change, if you have some function, here's our function f and we pick our values, here's a, here's b, then the average rate of change is just gonna be the slope of this so-called secant line that goes through the graph like so. That's all that we're trying to calculate here. And so in terms of the calculation, let's take f of x equal 3x squared on the interval one to three for which the average rate of change delta y over delta x. This is gonna equal f of three minus f of one over three minus one. So we're gonna get three times three squared which is nine minus three times one squared which is a one over three minus one. Well, I guess I noticed the tops divisible by three you can factor it out. Nine minus one over three minus one which is two. Nine minus one is an eight. So we get three times eight over two. Two goes into eight four times. So we get three times four which is 12. So the average rate of change of this function in this context would be 12, okay? For which again, it's measuring and then if we were to graph this one here y equals 3x squared, that's just your standard parabola that's been scaled by a factor of three, right? So we've seen something like the following. It just looks like a parabola, good old parabola, right? For which we wanna go from one to three that would be kinda like here's one, here's three and then connecting the dots. We get a line that looks something like this is our secant line and then we're saying the slope of that secant line m is equal to 12. That's what the average rate of change is measuring. So sometimes we desire that the average rate of change of a function f be computed on the interval a to a plus h where to interpret that, let me kinda draw us a picture. Why would we care about something like that? So let's say we have our function, here's the x-axis and here's our function, maybe it does something like this, whatever, here's our function f. And we have some value a in mind. It's like I wanna kinda know what's happening right here. Here's my a. So when you look at the, think about this, like if we wanna take us just a small step, a small step to the right, let's call that distance h. Well, if you take a small step to the right, then the address of this x-corner would be a plus h. And so if you follow this up to the top, you have this point right here. Here's what happens when x equals a. Here's a point over here when you have a plus h. So you have these points, right? So you have the point a comma f of a, that's there. But you have this point over here where you're gonna get a plus h comma a, f of a plus h, like so. We could be interested in the secant line in that situation, like so you get that secant line. And the point is that we're supposed to understand a plus h as just a little bit to the right of a, right, just a teeny bit. And so when a is really, when h is really, really, really small, right, you know, we could get closer and closer and closer and closer and closer. That's just so close, right? When h is considered this really small value, then a plus h and a will be really, really close together. Notice for example, if I take a plus h and I subtract from an a, this just gives you h, right? So when you calculate the average rate of change in this context, right, if you take delta y over delta x, this is gonna look like f of a plus h minus f of a over a plus h minus a, you're gonna see that the a's in the bottom cancel out. And this simplifies just to be, well, honestly this thing right here, right? f of a plus h minus f of a over h. And this is commonly referred to as a difference quotient. It's not a very clever name, but it's just what you see here, right? You have a difference and you have a quotient. It's like saying, oh, that, you know, I'm gonna domesticate this pet right here. Oh, it's a furry quadruped that barks, right? You know, we could call it a dog. We could come up with a name with it, but no, I'm gonna stick with furry quadruped that barks, right? That's what I'm gonna call a dog. So this is what we get for the difference quotient. This is what happens to average rate of change when you kind of replace instead of a and b, you replace b with a plus h. Again, we're emphasizing, instead of just having two points, right? We're emphasizing we have a point and then we have a second point that's close to it because it's important to ask yourself the question, what happens, what happens as h gets small, right? What happens as h gets close, you know, if a plus h gets closer, closer, closer, closer, closer, closer to a, right? What happens when h gets small or how we usually like to say it, what happens as h approaches zero, all right? Well, we can't just plug in h equals zero because if we did that, we'd end up with f of a plus zero minus f of a over zero, which would then become zero over zero. Wait, that's not a number, right? That doesn't make any sense. We can't just take zero over zero. But this actually gives us hope, right? What we've seen here is we have the numerator zero and the denominator zero at the same time. The only way that's happening is that maybe the numerator and the denominator have a common factor, which in this case would have to be h. We can potentially cancel out the h to the numerator, simplify the difference quotient and then plug in h equals zero, okay? Why would we want to do such a thing? So we talked about secant lines before, right? If I make my value get a little bit closer, right? We could draw the secant line there, like so, or we could even get it even closer, right? I'm using a different color this time to emphasize. What if we even get a little bit closer, right? We could draw our secant line. We're gonna get something like this, right? Really what our goal is though is what if we don't have a secant line? What if we have a so-called tangent line, right? What if we get a line that touches the graph only on one point? Can we get a tangent line? Well, that tangent line would be formed by taking your point, again, closer and closer and closer to some so-called point of tangency. So that's what happens when h goes to zero. So if we wanna find the so-called instantaneous rate of change, this is what happens, the instantaneous rate of change occurs when h is allowed to go to zero. But like I said, algebraically, we can't divide by zero, but we could cancel out the h on the bottom with some factor of h on the top and then evaluate h equals zero. And that would then give us the instantaneous rate of change, the rate of change at a moment. So let's consider that algebraically as an exercise for a moment. If we take the function f of x equals 3x squared, you'll notice we did that before, we don't let h be zero, right? But we wanna evaluate and simplify this difference quotient. What it would look like would be the following. We're gonna see that delta y over delta x is equal to, we're going to get 3x plus h quantity squared minus 3 times 3x squared. So we took right here, and this is all gonna be above h. So now I'll annotate this thing here, 3x squared, this is just f of x, right? But what is this right here? 3x plus h squared, this is equal to f of x plus h. That's what the difference formula requires, the difference quotient formula. But now we wanna simplify it. Our goal is to give it of the h on the bottom, right? Because h is not allowed to be zero in the current form. If we could cancel out the h, then we could be like, oh, h can't equal zero now. Let's see what happens in that situation. Now to do that, to give it of the h, we have to basically expand and combine like terms in the numerator. So I'm gonna have to foil out the x plus h, the x plus h squared. So doing that, it's gonna give me x squared plus 2xh plus h squared, and then we subtract a 3x squared right there. Next thing, distribute the three. In this case, you're gonna get 3x squared plus 6xh plus 3h squared minus 3x squared all over h. One thing you're always gonna notice when you do difference quotients is that with the difference quotient, there's this f of x part and there's the f of x plus h part. The f of x part will always cancel out entirely with some of the f of x plus h parts. Notice, for example, you have a 3x squared and 3x squared. You're subtracting them, they cancel out. And that leaves you behind a 6xh plus 3h squared all over h. Aha, notice everyone in the numerator is divisible by h. Here's an h, here's an h. We can factor those things out. And now we'll then leave behind h times 6x plus 3h all over h. For it's now this divisor of h can cancel with the divisor, so the h in the denominator can cancel with this common divisor of h in the numerator and we end up with 6x plus 3h. So this gives us a formula for the average rate of change. This is a formula for the average rate of change here. So notice now that the average rate of change is always gonna look like 6x plus 3h. For which it's like, if I allowed h to go to zero in this situation, so now if we allow h to go to zero, this would look like 6x plus 3 times zero. That is, this just becomes 6x. And so then what we can then say here is that not the average rate, so one thing we talk about is the average rate, the average rate of change we're gonna denote as this delta y over delta x. But the instantaneous rate of change, that is, this is the slope of the tangent line, the instantaneous rate of change, we actually denote that as dy over dx. What we see here is that this dy over dx always is gonna look like a 6x, all right? And so earlier we did delta y over delta x, we evaluated that from one to three right and we got the number 12, like so. So what this told us is that if we take, taking again our parabola, I'm just gonna redraw it on the screen right here. So we have our parabola, it looks something like the following. So what we saw here is if we take the point one and we take the point three, then connecting the dots, the slope of the secant line would look like m equals 12. But using our difference quotient and simplifying it, what we see here is that the instantaneous rate of change, which we call dy over dx, if we do that one, the instantaneous rate of change dy over dx, if we evaluated at x equals three, so the average rate of change requires an interval, you need a left endpoint and a right endpoint. The instantaneous rate of change only requires a single point x equals three. By the formula we computed, we should take six times three, which is equal to 18. This tells us that the slope of the tangent line, the slope of the tangent line at x equals three is gonna equal 18. Like so. And so by simplifying the difference quotient, we can not just find the average rate, we can find the instantaneous rate of change. Let's do another example. Let's take this one right here, f of x equals five x minus 17. Can we simplify the difference quotient and do what we did on the last example here? So if we look at this difference quotient, so this difference quotient, we're gonna call it delta y over delta x, right? This gives us f of x plus h. You're gonna replace the x here with an x plus h. So we get five times x plus h minus 17. Make sure you put some brackets here because we're gonna subtract from f of x, which f of x is gonna repeat the formula, five x minus 17. Like so, this is over h. Now combine like terms, right? I'm gonna distribute the five first of all. So we're gonna get a five x plus five h minus 17. And then we're gonna subtract from that five x minus 17. This all sits above h. You'll notice that the five x is canceled. Five x cancels with the five x. The 17s are gonna cancel as well because you get negative 17 minus the negative 17, they cancel out. And you're left with just a five x over h, so excuse me, five h over h, for which the h is canceled, you just get a five, like so. And so let's think about this geometrically what's going on here. If I take my x-axis, whoops, if I take my x-axis like so, and you take the function, the function is y equals five x minus 17. So that itself is a straight line with a slope of five minus up to negative 17. So if I scooch this up a little bit, this might look something like the following. Let me try that again, like so. And so what we've discovered here is that the average rate of change is always gonna be five, right? So if I take two points on the graph and I connect the dots, right? The average rate of change would should be the slope of that CK line. It's the exact same line. So that slope is still going to be five, all right? But then if we think about the instantaneous rate of change, right? The instantaneous rate of change is dy over dx. This occurs when the instantaneous rate of change is the rate of change you get when you set h equal to zero, right? Which as it doesn't depend on h, it's still five. So if we pick a point for tangency like so, what's the slope of that tangent line? It's the same line. So your slope's gonna be five again, right? The slope in all of these situations turns out to be five. So that's kind of a nice little thing you see when you have lines. The average rate of change and the instantaneous rate of change are always equal to the slope of the line. That's what we want it to be. Coming back to this picture right here, you have again in blue, this average rate of change from one to three give you a slope of 12. That's the slope of the CK line. What we're saying is on average, as you go from one to three, the functions, if it behaved like a line, it would have increased by a factor of 12. On the other hand though, if we look at the instantaneous rate of change, what we're saying is at this moment, at the moment three, right? Maybe this is like a function with respect to time. At the moment three, our function is increasing at a rate of 18. That is, if it were a line at this moment, it would continue on at a rate of three. And so this use of instantaneous rate of change is very important. Because we don't always care about like, okay, from this time period to that time period, that was the average change. Averages are important, don't get me wrong. But if you're at a specific moment in time, you might ask, where are we going right now, right? You know, like in fall 2020, when the coronavirus was at its height, we might be asking, how fast is the disease spreading at this moment in time? And that's why this instantaneous rate of change, this instantaneous spread is the most important. But this instantaneous rate of change, this DY over X, in calculus, they give it a different name. In calculus, this is known as the derivative. And so surprise everyone, we just started calculus right now. It's appropriate given that this is the last lecture in our pre-calculus unit. It seems that the baton is then passed off that, hey, we're not doing algebra anymore, we're doing calculus. You were able to do it, congratulations everyone. I knew you could do it. Some people get scared about calculus, but I want you to be aware that we've been doing calculus the whole time. It's much like when Mr. Miyagi is teaching Daniels on karate in the original karate kid, right? Daniel, he has to do all these chores for Mr. Miyagi, like sand the deck, paint the fence, wax the car and things like that. And Mr. Miyagi emphasizes all these different motions. Like, no, this is how you wax on, wax off. And so when Daniel does it kind of lazy, he corrects him or when he paints the fence, it's like, nope, you have to do it this way. And so why all these chores, why all this emphasis on muscle motions? It's because unbeknownst to Daniel, he was learning defensive karate techniques the whole time and he needed to practice those techniques by doing the muscle memory over and over and over again. That's what college algebra is all about. Yes, we learn about functions and modeling. We can approach many, many real life problems using college algebra, especially if you combine it with some statistical methods that we haven't learned about in this class. But we also should remember that college algebra could be called pre-calculus. It's preparing us for that calculus setting. We're learning the muscle memory that we need to perform in a calculus environment, which calculus, the applications of calculus are through the roof. I can't even begin to emphasize you how useful calculus is. The modern sciences would not be what they are today without the invention of calculus. And I wanna give you a pat on the back. You are now ready to be a calculus student.