 Hi, I'm Zor. Welcome to Unisor Education. Today we will continue solving little problems. Actually, I call them exercises because there is nothing actually new about whatever we are going to do. It's just something which we were talking about during the theoretical lectures and now we are implementing whatever we were talking about in practice. Now, this lecture, as all others, is part of the advanced course of mathematics at Unisor.com. I suggest you to watch this lecture from the website because the website has very detailed notes which can serve as a textbook. Plus, those who are signed in, there is some kind of functionality built into the website. Like, for instance, taking exams or enrolling into a specific topic and the site is free, there is no advertising, so please use the website. Now, okay, so let's just do one by one. We have six different exercises on the material which we have covered before in lectures. Alright, so number one is there is a function and at point x equals to zero we would like to draw a tangential line and we need the linear equation which represents this tangential line. Well, first of all, let me just draw the graph of this function. Well, e to the power of minus x is something like this, right? And this is one at point zero. And x is this. Now we have to multiply them, right? Now, obviously, starting from this to the left, this goes to minus infinity, this goes to plus infinity, their product would be minus infinity going through zero. So, in this case, my graph will go something like this. Not linearly, I mean it looks like a straight line, it's not definitely a straight line, it's something which goes to infinity in both cases. Now, as far as here is concerned, again, when x goes to infinity, e to the power of minus x goes to zero significantly faster, and we did cover this material, this is something which is little o of this, if you wish. So, the result would be on the infinity, it would be going to zero asymptotically. And here, so they're probably rising something and this is general behavior of the function. So, let me just put this as a dotted line, these are my initial functions. And the resulting function is the solid black line, it goes like this. Now, this is point one, this is point two, so we are looking at tangential line which looks something like this. And we need the equation of the tangential line. Well, first of all, it's a linear function, obviously it's a straight line, which means its equation is this. Now, what is a? Well, a is always a tangent of this line with the x-axis, right? At the same time, we know that the tangential line to a function at certain point has a tangent of the angle with x-axis equals to the derivative of this function at this point. So, if we will take derivative of this function at point two, we will basically have a, right? So, it's that a is equal to derivative of function f at point x zero, which is two. We can just put two here. Now, what is this? Well, let's talk about derivative of this function. It's a product of two functions, which means its derivative is derivative of the first by second plus first on derivative of second. Okay, derivative of first by second would be e to the power of minus x and plus x times derivative of a to the power of minus x, which is the compound function. First is the minus x and then it's e to the power. And its derivative is, first we do the main function and then we multiply it by derivative of the inner function, which is minus x, which is minus one times x, actually, right? So, its derivative would be minus one. So, I'll just put minus here. And I should put it at x is equal to two, which is equal to e to the power of minus two, minus two times e to the power of minus two. That's my derivative and that's the a, right? e to the power of minus two, minus two, e to the power of two. Okay. Now, let's talk about b. What we do know is that at point x is equal to two, this particular function should give exactly the same value as this particular function, right? So, that means that a times two plus b should be equal to this function, which is two times e to the power of minus two. Now, we know a, so how can we determine b? Well, b is equal to two to the power e to the minus two, minus two a, minus two a, which is two e to the power of minus two, plus four e to the power of minus two. That's b. This cancels out. So, what do I have now? I have y is equal to e to the power of minus two, minus two e to the power of minus two x, plus four e to the power of minus two. So, that's the equation. The end. Next. Okay. Next, the same function. I have to find out maximums, minimums, and inflection points. All right. So, I know how it looks graphically. It looks something like this. So, this is maximum. And this looks like an inflection point, because I have here the function has a hump, basically. So, it's concave down. And here, since the function goes asymptotically this way, the concavity goes upward. So, that's where we are changing the curvature of the function. Well, let's just do it analytically using the calculus. Well, first of all, we need the derivative. And derivative is f of x equals e to the power of minus x. That's derivative of this times this, plus this times derivative of this and the minus sign. Okay. That's a derivative. Now, my first question is, where is derivative equals to zero? That's the stationary points. So, if this is zero, then this is zero. Now, this is obviously e to the power of minus x1 minus x equals to zero. This is not equal to zero, which means x is equal to one. So, this looks like this. Okay. That's fine. We can definitely think about whether this is a minimum or maximum. I mean, we kept graphically that it looks like a maximum, but you remember that we can use the second derivative to basically make our determination more precise. So, let's talk about the second derivative. So, second derivative is derivative of this thing. And again, it's a multiplication of two functions. I will take derivative of this times that. Now, this is minus e to the power of minus x times one minus x, plus this times derivative of this, which is minus one. So, it's minus. So, let's think about where this... Well, let me just e to the power of minus x times minus one minus x minus one. Am I right? No, plus one. And plus one. Sorry. Equals two e to the power of minus x times x minus two. This is minus. Minus x, one. And now there is no minus. I changed the signs here. Yeah, that's the derivative. Now, where is this second derivative equals to zero? Well, obviously, that x is equal to two. Before, to the left of two, I have this derivative negative. So, the second derivative is negative, in particular at point one, signifies that this is a local maximum. And second derivative being greater than zero, if x is greater than two, signifies the concavity would be upward. So, this is exactly... Point x is equal to two is exactly where our inflection actually point is. And since there are no other stationary points but x is equal to one, there is no minimum. So, there is maximum, local maximum. There is an inflection pointed to two and there is no minimum. Okay, that's it. Okay. Three. f at x is equal to sin x plus cosine x. I have to determine intervals of increasing and decreasing of this function. Now, you remember, we are talking about obviously smooth function. They are differentiable and derivatives are continuous, etc. So, we can talk about monotonically increasing function having a derivative where it's increasing positive. And if it's a monotonic and a decreasing function on certain interval, that's where its derivative is supposed to be negative. So, we are talking about this monotonic behavior and derivative connection. Monotonically increasing have positive derivative, monotonically decreasing negative. Alright, so let's just talk about this particular function, find its derivative and find where it's positive and where it's negative. Right? So, derivative is cosine of x, that's derivative of this, and minus sin of x, that's derivative of that. So, let's talk about positive. So, number one, where is cosine of x minus sin of x greater or equal to zero. Or, if you wish, cosine of x greater or sin of x. So, we have to find this. Alright, let's just talk about sin and cosine as they are defined on a unit circle. I think this is the best. Because sometimes people just divide by cosine. This is not exactly the right way to do it, because the cosine can be positive or negative, and if it's negative, then this sin should be reversed. Right? So, let's not go into these details. Let's talk about the strict definition, and let's talk about sin and cosine when one is greater or less than another. Okay. Now, sin, as you remember, is this is sin, and this is cosine. Sin and cosine. Well, let's start from zero. Now, when we are at zero, sin is zero. Cosine is an entire radius, right? So, point zero definitely satisfies this. Then, as we move counterclockwise, up to five over four, four to five degrees. Sin is increasing, cosine is decreasing, right? Sin is increasing, cosine is decreasing, and they are equal to each other exactly at P over four, at four to five degrees. Right? So, the whole area of the angles, which is in this particular sector, satisfies this particular inequality. Next, let's go this way. In this way, you have sin, let's say it's here. Sin is this one, which is negative. Cosine is positive. Definitely satisfies our inequality. So, any point in this quarter also satisfies our inequality. Let's move further. Now, when we are moving here, our cosine is negative, and our sin is also negative. But while we are here, absolute value of a sin, which is this, is greater than absolute value of cosine. But since we are talking about negative, the absolute value will be, the sin was absolute value greater than the cosine, since both of them are negative, would be actually less. So, it's also our area where this inequality is satisfied. Until they are equal, both are negative and the same absolute value. And starting from this point up to this, very easily we can see that this inequality is not actually satisfied. Because the absolute value in this case of a sin is, which is this, is less than absolute value of this. But since they are both negative, sin is greater. And same thing for any point in this quarter, because in this quarter our cosine is negative and sin is positive. So, again, this is not satisfied. So, the whole half a circle is where this particular inequality is satisfied. So, let's just describe it analytically. Well, if this is p over 4, so we start from this point, this is minus p over 4, another another. So, it's minus 3 pi over 4 plus 2 pi n, x up to pi over 4 plus 2 pi n. Well, obviously I added the period, right? So, that's the solution to this inequality. And that's where our function is increasing. Now, symmetrically you can derive that it's decreasing, obviously, in this area, which is from here to here. So, it's from pi over 4 plus 2 pi n, 1, 2, 3, 1, 2, 3, 4 plus pi actually. It's 5 quarters, 5 pi over 4 plus 2 pi n. Oh, I didn't put x. So, x between these two is where the function is decreasing. So, that's basically all analysis of the behavior of this function, increasing and increasing. Next. Next, we need to find inflection points at this function. Well, the graph of this function, well, you know. So, it looks like, you see, in all these areas, we have maximums and these are minimums local. And so, the concavity is down here and upwards there. So, that must be the inflection point. Now, how can we determine it? Well, remember, the second derivative is supposed to be equal to zero, because here, second derivative is supposed to be negative, since it's a local maximum. Here, second derivative is supposed to be positive, because it's local minimum. So, in between, it's supposed to be somewhere in between zero, right? Negative here, positive there, negative zero in between. Okay, let's just take the second derivative. So, the first one will be cosine of x. That's the first derivative. And the second derivative would be minus sin of x, right? From cosine, minus sin of x. Now, it's equal to zero, minus doesn't really matter. That's x is equal to pi n, which obviously are these points. So, inflection points are actually those points where sin is equal to zero, which is pi n. Okay, next. Okay, next is our old friend. But now, I would like to have... Remember, we were talking about tangential line at x is equal to two. Now, I would like to talk about normal, which is perpendicular to a tangential line. Now, how can I find out the equation of the normal? Well, let me start from the beginning. Let's start from the tangential line. So, the first derivative is e to the power of minus x, derivative of this times this, plus this times derivative of this, which is minus x e to the power of minus x. That's my first derivative. And at point two, it's a tangent of this angle. Well, that's minus two. So, I've got tangent of this angle. Now, if two lines are perpendicular to each other, how knowing one tangent, how to find another one? Well, I have to turn this by 90 degree, in this case, counterclockwise, which means I'm reducing the angle by 90 degrees, right, by pi over two. So, if I know that tangent of pi is equal to whatever, let's go to call it A. Now, I'm looking for tangent of phi minus A. Sorry, phi minus pi over two. What is it? Well, let me just remind you that if you have a triangle, if this is angle phi, this is pi over two minus phi. It's a right triangle, right? Which means that the tangent of this angle equals cotangent of this angle. And correspondingly, tangent of phi minus pi over two equals to cotangent, cotangent, not cosine, cotangent of phi, right? Now, this is true not only these two identities are true, not only with the right triangle where all angles are from zero to 90 degree, to pi over two, but for any angle. I mean, if you will consider the unit circle, you will have exactly the same thing. So, I'm going to use this for here. I'm sorry, I made a slight mistake. This is pi over two minus phi. Now, I see this, I see the difference. So, what's the difference? The difference is in the sign. But I know the tangent is an odd function, which means if I change my argument to a negative, my tangent is changed to the negative. So, it's minus tangent of pi over two minus phi. So, I change the sign here, I change the sign of the function. And now I'm knowing this, this is a minus cotangent of phi. So, that's what basically I need. I need tangent of this angle and this is the angle, the tangent of the angle of the normal. Because it's this angle minus pi over two, you know, the clockwise is negative, counterclockwise is positive direction. That's why I subtracted pi over two. And then I change the sign to get to this, to use this identity. And that's my answer. Now, what is this? Do I know this? Well, of course, I know the tangent, right? So, cotangent is one over tangent. So, that's my formula which I would like to use. Since I know my tangent phi, I therefore determine my tangent of phi minus pi over two, which is this, minus one over A. Where A is my tangent, right? So, now I know the tangent of this normal, which is one over minus one over tangent is my derivative at point two, which is e to the power minus two minus two, e to the power minus two x. Okay. Plus B, some B. And how do I determine B? Well, if I substitute two here, I have to get corresponding value here, right? Because they are the same. At this point, they're crossing, they're intersecting. So, the value of y is supposed to be equal to the value of function, right? So, the value of function is equal to e to the power minus two minus two e to the power of minus two. And that should be equal to, if I substitute two here, minus two, divided by e to the power minus two minus two e to the power minus two, plus B. From which I can determine the B, whatever the B is. I don't want to go into the details in this. And I've got the equation. Okay. And the last one. General f of x at point x zero. I need to determine basically the equation of tangent, tangential line and the normal at point x zero to this function. Again, considering that the function is smooth, et cetera, et cetera. All right. So, whatever the function graph is. For instance, this is the point. So, this is my tangential line and this is my normal. Okay. Now, up front, the tangential line should have the tangent of the angle with x axis equal to the function's derivative at point, at given point. So, f point, f derivative at point x zero is my coefficient at x. Right? Now, how to find the three member? What kind of B should I put here? Well, if I put x zero here, I should have f of x zero, right? So, f of x zero, because they are intersecting, right? So, f of x zero is equal to f of x zero, x zero plus B. So, B is equal to f of x zero minus f of x zero times x zero. So, the whole equation would look like f of x zero, x plus f of x zero minus f of x zero, x zero. Which usually is, they are putting these two together, which is f of x zero times x minus x zero plus f of x zero. So, this is the derivative at point x zero times x minus x zero plus this. So, this is an equation of tangential line. Okay? Now, how about the normal? Well, very easy. Normal line is at angle of 90 degree to the tangential line, right? So, I immediately, so this is equation of the tangential line. So, I immediately know, remember the previous, 1 minus 1 over a. If I have a as a tangent, then tangent of phi minus p over 2 would be this. So, immediately I can see that my normal is minus 1 over f of x zero. Obviously, assume that it's not equal to zero for all these cases. By the way, if derivative at point x zero is equal to zero, which means it's some kind of a local maximum and tangential line is parallel, then what is the normal? Normal is this. And the formula for the normal would be x is equal to x zero, right? That's the vertical line, so to speak, in Cartesian coordinates. So, let's put this aside and we consider f derivative at x zero is not equal to zero. So, that would be x plus b. And again, how do we determine b? Exactly the same way, because this normal, again at this point, x zero, is supposed to have the value equals to f of x zero. So, the same thing, same technique here, right? It's supposed to satisfy this, from which b is equal to f of x zero minus 1, no plus, now plus, it's plus f prime of x zero times x zero. So, my equation would be, instead of b, I will put this value prime of x zero times x zero, or as people, again, usually do, they factor out this having x minus x zero here. This is minus, that's why this is minus, this is minus, and plus f of x zero. So, this is equation of the normal. So, whenever you have a function f of x, these are two formulas, which I never remember by the way, I always derive them. But, I mean, if you remember it, it's fine. This is the equation for tangential line at point x zero, and this is equation of the normal line for x is equal to x zero. Well, that's it for today. I do suggest you to read all the notes for this lecture on Unisor.com. And what probably would be best, if you can just do exactly these exercises by yourself, without looking at the notes, without looking at the lecture itself, and if you come up with exactly the same results, the answers, by the way, are in the notes for this lecture. That's great. Other than that, that's the end of it. Thank you very much and good luck.