 All right, let's look at one last example. Let's do a little bit of trigonometry to it. That'll be fun. So what we should we know about here is that the domain, right? The domain of cosine is all real numbers. The domain of sine is all real numbers. So the only issue is the denominator. If you look at the denominator two plus sine of x, does that ever equals zero? Well, subtract two from both sides, you get sine of x equals negative two. And I'm gonna stop you right there. Sine has to sit between negative one and one, positive one, it can never equal negative two. So what we actually get is that its domain is gonna be all real numbers. But I like the other graphs we've seen. Sine, this function is periodic by the nature of the trigonometric functions here, right? And so it's gonna be, in fact, periodic on two pi. Because cosine repeats itself every two pi, sine repeats itself every two pi. If we can just graph it on the interval pi, we're gonna, on the interval, excuse me, two pi from zero to two pi, we could be just fine. Absolutely just fine with that. So we can do something like that. Let's see, what else do we need to talk about? So the next thing to do, we can think about intercepts. Think of intercepts, what would the wider set be? F of zero equals cosine of zero over two plus sine of zero. Cosine of zero is a one, sine of zero is a zero. So we get one half as our y-intercept. In terms of x-intercepts, you get cosine of x that sits above two plus sine of x. When does that equal zero? Well, to find an x-intercept, it only matters what happens in the numerator. Cosine of x equals zero. And think about our standard trigonometry. Cosine will equal zero exactly at, let's see, at pi halves, right? Cosine's equal to zero at pi halves, but also at three pi halves, five pi halves, and it'll repeat itself every, again, it's periodic in the nature there. We're gonna get x-intercepts over and over and over again. In terms of discontinuities, cosine, sine are all continuous. The only way there could have been a discontinuities if we divided by zero. Like we said earlier, that's not a possibility. It was periodic where we talked about the symmetry there. So let's talk about in behavior. Well, for periodic functions, there really is no such thing as in behavior because it just repeats itself over and over and over and over and wait for it over again. So asking the limit as x goes to infinity versus negative infinity, that's not gonna give us any information there whatsoever. So we can actually jump directly to our derivatives. Let's calculate the first derivative by the quotient rule. As you can see, the quotient rules are best friend in this section. By the quotient rule, we get two plus sine of x times the derivative of cosine, which is a negative sine. Do make sure you put parentheses there so you make sure that you're taking the correct multiplication there and you don't accidentally subtract it. Then minus cosine, oh, I should have planned ahead here, times the derivative of two plus sine, which is a cosine, and this all sits above two plus sine x squared. Don't multiply out the denominator. Let me move down to give myself some more room. And so if we distribute the negative sine onto these two pieces right here, we're gonna end up with a negative two sine of x. We're gonna have a negative sine squared. And then we also have a negative cosine squared. And this sits above two plus sine x squared. Now, since everything in the numerator was divisible by negative one, I'm gonna factor out that negative one. And so we get two sine x. And now my trigonometry sense is tingling, right? You look at things like sine squared and a cosine squared. We know this comes together to give us one. This is the Pythagorean identity. So we're gonna get two sine of x plus one. And this sits above the two plus sine of x. Now, despite the fact you see a two on the top and the bottom and a sine on top and bottom, there's no cancellation that goes on there. This is our simplified derivative. To find the critical numbers, we look at what makes this thing equal to zero. And which case we set two sine of x plus one equal to zero, this gives us sine of x equals negative one half. And so think about those unit circle diagrams that we've known from the past, if you have to. When does sine equal one half? This will happen in the first quadrant and in the second quadrant. The first one will be 30 degrees, which you can repent of that and switch it to radians if you need to, right? Which case we should then get, I'm sorry, it's gonna be a negative one half, isn't it? Because I forgot something, didn't I? Let's just pause for one moment because we have a negative sign right there. I shouldn't make any bit of a difference. We've backed out the negative out of everything. I'm trying to make sure I didn't make a calculation error anywhere. Oh, okay, I did. I did make a calculation error anywhere. Oh, okay, I did make a mistake somewhere. Oh, no, it was right, I'm sorry. We see right here, we have sine is negative one half. So let me actually try this unit circle argument one more time. We're not interested in when it's equal to one half. We're interested in when it's equal to negative one half. So that's actually gonna be below down here. So we're talking about an angle in the third quadrant and in the fourth quadrant. So this will reference to 30 degrees. We're actually looking at x is seven pi sixth and also 11 pi sixth. Now it'll repeat itself, but this will give us the, this will give us the critical numbers in a single period. And so let's then take the second derivative here. If we take the second derivative, again, there is a calculation going on here. For the sake of time, I think I'm gonna skip over it. The details can be found in the lecture notes here, but if you take the second derivative by the quotient rule again, you get two times cosine of x times sine of x minus one. And this sits above a two plus sine of x and then the denominator is cubed, all right? So the PPI's here are gonna be the things that make the numerator go to zero because there's nothing in this denominator that there's no choice of x that makes the denominator go to zero. So we have to worry about when does cosine go to zero? All right, when does cosine go to zero? And we mentioned that before, those are our x intercepts. Those happen at pi halves and three pi halves. But then we also have to worry about when does sine of x minus one equals zero. That would happen when sine equals one, which would happen at pi halves like so. And again, this repeats itself, but this gives us all the critical numbers for a single period. Did I forget anything there? No, no, I think we're good. And so now let's build a sign chart using these critical numbers and these PPI's. I'm gonna put it below our picture right here. So there was a lot of stuff. Let's keep track of it here. And so what do we have to worry about? We're gonna just do one period. So from zero, so we had a PPI at pi halves. We had a critical number at seven pi six. We had a PPI at three pi halves. And then finally, we had another critical number at 11 pi six, all right? And so, and then this then stops itself at two pi. It will repeat itself, but that's all that we have to do. We're gonna get tons and tons of solutions otherwise. And so if we come to the second derivative, let's take a look at it again real quick. Our second derivative, this might be a useful situation to use test points where you plug in specific numbers in these intervals right here. You can always see the factorizations of these things, but however you wanna do it, there's ways to do it. You're gonna end up with, you get a negative between zero and pi halves. You get a positive between pi halves and seven pi sixth. You're gonna get a positive between seven pi six and three pi halves. You get a negative between three pi halves and 11 pi sixth. And you get a negative again from 11 pi sixth to two pi. And so what this tells us right here is that we have a point of inflection at pi halves, a point of inflection at three pi halves. We have a local minimum at seven pi halves and we have a local maximum at 11 pi sixth. And so let's try to put all this information together. So we had intercepts at pi halves and at three pi halves. We had a inflection point at seven pi sixth, which is gonna be right here on the ground. And then we had another inflection point at 11 pi sixth, which is almost two pi here. So here's our two pi, here's our zero. And so if we connect the dots, our whiner step before was one half, wasn't it? So if we connect the dots, we're gonna get this one right here. So let me kind of explain what's happening here. You get an inflection when you are between zero and pi halves, it should be concave up. When you're between pi halves and three pi halves, it should be not three pi halves. No, that's right, three pi halves. You should be concave upward again. I'm sorry, I said concave up before when I meant down. Let me fix that. This is concave down right here, concave up. And then concave down again. We have a maximum right here. Sorry, this is a minimum. We have a maximum right here. And this is what the picture's gonna do. And it'll replicate itself over and over again. And this, because the blue one just gives us one period. And it just kind of repeats itself over and over and over and over again. Like so, and this will be repeat. And so if we look at a computer-generated image, we see exactly this. The one in yellow is the period we first started off with and the one in green then gives us what it looks like over and over and over again. And so the examples in this one were much more challenging than the last lecture. And I do understand that, but if you're patient with yourself, you can get some really accurate drawings of these curve sketches. And that actually concludes our lecture for today. If you have any questions, feel free to post them down in the comments below. There is a link to the lecture notes for this lecture in the description below. Feel free to use those in your studies. See you next time. Bye.