For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

## Discussion

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Maxima & Minima

Determine if the following points are boudary points using a ball B of raidus 1 at each point.

i) x = − 2

- A point →P is a boundary point of a set U if an open ball using →P as the center contains points in U and not in U.

Determine if the following points are boudary points using a ball B of raidus 1 at each point.

ii) x = 2.5

- A point →P is a boundary point of a set U if an open ball using →P as the center contains points in U and not in U.

Determine if the following points are boudary points using a ball B of raidus 1 at each point.

iii) x = 0

- A point →P is a boundary point of a set U if an open ball using →P as the center contains points in U and not in U.

Determine if the following points are boudary points using a ball B of raidus 1 at each point.

iv) x = − 3

i) { x| − 1 < x < 0}

ii) { (x,y) - − 2 ≥ x ≥ 2, − 2 ≥ y ≥ 2}

iii) { (x,y)|y ≥ √{4 − x

^{2}} ,y >− √{4 − x

^{2}} }

- A set is closed if it contains all of its boundary points, that is, the endpoints are part of the set. Note that this is a circle whose lower half does not contain its boundary points while its top half does.

_{1}= { x|0 ≥ x ≥ 1} is bounded.

- A set is bounded if we can find a number N such that for every point x ∈ D
_{1}, || x || ≥ N. - Note that the norm is equivalent to the absolute value in one dimension. Since our set D
_{2}has 1 as its maximum norm, we can choose N = 2 to show that D_{2}is bounded, as 1 < 2 = N. - ii) Show that the set D
_{2}= { (x,y)|y = [1/x] } is not bounded. - Note that the points in D
_{2}have no restriction aside x0. Hence we can not find a number N such that || (x,y) ||≥ N. - For instance (1,1) has norm √2 so N = 2 is sufficient but || ( 4,[1/4] ) ||> 2. Then N = 3 becomes sufficient but again || ( 9,[1/9] ) ||> 3, this pattern can go on infinetly.

_{2}is not bounded.

_{1}= { x| − 1 ≥ x ≥ 1} . Find the local maxima and minima of the function y = x

^{2}− 1 on S

_{1}.

- Recall that a critical point is found by solving [dy/dx] = 0 for x. In our case we obtain 0 = 2x or x = 0.
- Using the first derivative test we can see that at x = 0 we have a minimum, our point is (0, − 1).
- We must also check our endpoints x = − 1 and x = 1 for they too can be extrema.
- At x = − 0.9 ∈ S
_{1}(slightly greater than − 1) we have the value − 0.19 while the value at x = − 1 is 0. From the left the function is clearly decreasing, so x = − 1 is a maximum. - At x = 0.9 ∈ S
_{1}(slightly less than 1) we have the value − 0.19 while the value at x = 1 is 0. From x = 0 the function is now increasing, so x = 1 is a maximum.

_{2}= { x| − 1 < x < 1} . Find the local maxima and minima of the function y = x

^{2}− 1 on S

_{2}.

- Note that our set is open, our only extrema might occur at our critical points.

^{2}− y

^{2}.

- For a function of two variables, the critical points occur where all first partial derivatives equal zero.
- So [df/dx] = 4x and [df/dy] = − 2y. Now 4x = 0 yields x = 0 while − 2y = 0 yields y = 0.

^{xy}.

- For a function of two variables, the critical points occur where all first partial derivatives equal zero.
- So [df/dx] = ye
^{xy}and [df/dy] = xe^{xy}. Now ye^{xy}= 0 yields y = 0 or e^{xy}= 0 but e^{xy}0. Similarly xe^{xy}= 0 yields x = 0.

- For a function of two variables, the critical points occur where all first partial derivatives equal zero.
- So [df/dx] = sin(y) and [df/dy] = xcos(y). Now sin(y) = 0 occurs at y = np for n = 0,1,2, … while xcos(y) = 0 yields x = 0 or cos(y) = 0 which occurs at y = [((2n + 1)p)/2] for n = 0,1,2 …

^{2}− y

^{2}.

- Note that for large values of x and small values of y, f(x,y) > 0. For instance f(10,1) = 199.
- But for small values of x and large values of y, f(x,y) < 0. For instance f(1,10) = − 98.

^{xy}.

- Note that if x > 0 and y < 0, f(x,y) < 1. Similarly for x < 0 and y > 0.
- Yet if both x > 0 and y > 0 or x < 0 and y < 0, f(x,y) > 1.

- We can check two close points to (0,p,1) to determine if the critical point is an extrema.
- Now, f(0.1,3.14) ≈ 0.00548 and f( − 0.1,3.14) ≈ − 0.00548.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Maxima & Minima

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
- Maxima and Minima 0:35
- Definition 1: Critical Point
- Example 1: Find the Critical Values
- Definition 2: Local Max & Local Min
- Theorem 1
- Example 2: Local Max, Min, and Extreme
- Definition 3: Boundary Point
- Definition 4: Closed Set
- Definition 5: Bounded Set
- Theorem 2

### Multivariable Calculus

### Transcription: Maxima & Minima

*Hello and welcome back to educator.com and multi-variable calculus.*0000

*Today we are going to start out discussion of maxima and minima. Finding points in the domain where the function obtains a maximum and where it obtains a minimum, just like in single variable calculus.*0004

*So, a lot of the things you learned in single variable calculus are going to apply here.*0016

*Now, instead of one variable, x, you are going to have 2, 3, sometimes more.*0020

*So, the problems tend to become a little bit longer, a little bit more involved, but again it is still just a max, min problem based on the derivative. Let us just jump right on in.*0026

*Okay. So, let me see, let me go ahead and just write a... well, let us go ahead and start with a definition. How is that?*0039

*I think... we do not really need a preamble here, so let f be a differentiable function -- and again by differentiable we just mean well-behaved, it has a derivative, no problem, no worries, no concerns -- on an open set.*0047

*Again, you recall what an open set means. It is just a set that does not have boundary points. An open set u.*0075

*Let p be a point in u. p is called a critical point or a critical value, we will use both terms, it is called a critical point if each partial derivative evaluated at p is equal to 0.*0085

*So you see this is the exact definition as for single calculus. The derivative is equal to 0 at a critical point.*0130

*So, let us go ahead and write what this is symbolically. We will say that is d _{i}(f) = 0.*0136

*So, the first derivative, second derivative, third partial, fourth partial, whatever it is.*0147

*Or, df/dx _{i} evaluated at p is equal to 0. Let me actually add one more thing here. It is very... evaluating it at p, so we will actually put the p in there.*0153

*So, that is it. Let us just do an example. Example 1. Find the critical values of f(x,y) = x + ysin(x).*0169

*Let us go ahead and find the derivatives. The partial with respect to x, df/dx = 1 + ycos(x), right? We are holding y as a constant.*0196

*The partial derivative with respect to y is going to equal just sin(x). This one is going to be held as a constant, the x goes to 0.*0212

*Let us see what we have got here. So, here, we want each of these, of course, equal to 0.*0223

*So 1 + ycos(x) = 0, and sin(x) = 0, so let us go ahead and do this one first.*0231

*sin(x) = 0, so our possible values of x are going to be 0, and π, and 2π, and 3π, and so on.*0243

*Now, let us see... should we do... yea, okay. Let us go ahead and take a particular value. Let us just take... so the 0, the 2π, the 4π, when we take the cosine of that, so now we are going to use these to actually solve for the y variable.*0255

*We have 0, 2π, 3π... I am sorry, 0, 2π, 4π, 6π, then we have the odds. π, 3π, 5π, so we have to check both possibilities because it turns out y is going to equal 2 different numbers.*0281

*So, let us go ahead and start with, let us say 2nπ, so when I take the evens, the 0, the 2, the 4, it is going to be... all of those can actually be written as x = 2nπ, where n is just some integer.*0293

*This little, funny-looking z here, that is a symbol for the integers. You know, integers, negative, 0, 1, 2, 3, 4, 5, that is it. Natural numbers including the negative versions of those negative numbers and 0.*0313

*This is a shorthand notation for writing the even number ones.*0328

*So, when we do that, we get 1 + y × cos(2nπ) = 0.*0333

*Well, cosine of any 2nπ, so 0, 2π, 4π, is the cosine of... it is going to be 1.*0342

*So you get 1 + y = 0, you are going to get y = -1.*0351

*Now we will go ahead and do the version for the odds. Well, the odds, the π, the 3π, the 5π and so on, that is going to be (2n+1)π.*0357

*That is the shorthand notation for all of the odds. Basically you just run n. So when n = 0, you have got π, when n = 1, you have got 3 π, when n = 2, you have got 5π. That is what this means.*0370

*So, when we do this one, we get... let me write it down here, that is okay... so we get 1 + y × cos(2n+1)π, and the cos((2n+1)π) for all n is going to be -1.*0384

*So what you end up with is 1 - y = 0, so y = 1.*0406

*So our final solution is going to look like this. Let us go ahead and write... yeah, we will write it on the next page.*0415

*So, the critical values are going to be all of the points 2nπ - 1, and 2n + 1π, 1, in other words the x value is going to be 2n + 1π, and the y value is going to be 1.*0423

*This is for n in z, where n is any integer at all, so we have 2 different possibilities.*0448

*That is it. It is that simple, you just take the partial derivatives, you set them equal to 0, and now instead of just having the one equation that you solve, you are going to have 2, 3, 4, however many equations and you just have to find all of the different values.*0456

*You are just going to have a lot of values, so you might have one critical value.*0469

*It just depends. You have to use all of the resources at your disposal, all of the things you know about a function, whether it is its graph, whether it is its particular behavior, whether it is a certain limit approaching something to come up with what these critical values are.*0475

*So, let us go ahead and speak about the critical values a little bit. The critical values for a function need not be local max's or min's.*0492

*You remember from first variable calculus, just because you find a critical value, some place where the derivative equals 0, that does not necessarily mean that it is a local max or a local min.*0522

*There are situations where it is neither, where it is an inflection point. So, the same thing can happen here, you know in several variables. It does not have to be a local max or min, it just means that it might be.*0532

*We have to subject the function to further analysis, to see if it is a local max or min, and later on if we are dealing with a closed domain if it is an absolute max or min. That is what is happening.*0542

*So, recall, we can have f'(x) = 0 and have a point of inflection, which is neither max nor min.*0556

*It looks something like this. So the graph comes like this, but you get a derivative of 0, so this point right here, I mean yeah, its derivative is 0, it is a horizontal slope, but the horizontal slope... but it is not a max or min, it is a point of inflection. That is a possibility too.*0589

*So, let us start with another definition here. Let f be a function on an open set u, a point p... and again these definitions, you know what they are... but again, we want to be reasonably formal in mathematics.*0605

*A point p is a local max for the function, if... oops, let us erase that... if there is an open ball -- and I will draw a picture in a minute -- centered at p such that f(x) < or = to f(p) for all x in that ball.*0638

*Let us go ahead and think about what this means. So in this particular... this is a general definition, it works in any number of dimensions.*0696

*So, the example I am going to use to demonstrate it is going to be in R2, in the plane, because that is the one that we are the most familiar with.*0704

*So, in R2, it looks like this. It looks like... so we have R2, and we have our point p... now an open ball around point p, centered at p, in this case is just an open disc.*0710

*If you are dealing in 3 dimensions, then you are talking about a point and there is an open ball. An open ball means, again, the boundary points are not included.*0730

*It is just some region around it, that you are not specifying a boundary on it.*0737

*Basically it says for any x-value, in here, for any x-value that I pick around that point, if a function, if I evaluate the function at any of the x's, at any point that I pick in that open ball, if all of them are less than or equal to the value of the function at p, then that is a local max.*0742

*So, the graph of this, of course, if you... so if your domain is 2 dimensions, so here we are talking about a function from R2 to R.*0766

*So, this is our domain, we are picking points from here, we are sticking it in a function, we are doing something to it, we are spitting out a number. That number is used as the z coordinate.*0776

*So, a function from R2 to R, actually gives you a surface in 3-space. So if this is the domain, there is some surface on top. It is just saying that what you have is something like that.*0786

*That here is the point, and then every point around p forms some sort of a, like a, cone or something. It is a local maximum, not an absolute maximum because we are talking about an open domain.*0795

*In a minute we will actually define what we mean by absolute max and closed domains.*0808

*So, that is it. Again, exactly the same as first variable calculus, or single variable calculus.*0812

*So, let us just go ahead and formalize one last thing. So, a local min is defined the same way.*0820

*It is defined the same way except of course all of the functional values are > or = f(p).*0830

*So, you have something that looks like this where p is down here, and all of the values around it are bigger than that.*0843

*Let us go ahead and write down a theorem. Now, let f be a function defined, and differentiable on an open set u, and let p be a local extremum.*0851

*An extremum is just a single word that takes care of both max and min at the same time. So extremum, plural being extrema. So it is max or min.*0894

*Let p be a local extremum for f on u. Then p is a critical point.*0911

*This is really, really important here. This is saying that if p happens to be a local max or min, that means that it is a critical point.*0924

*It does not mean... this theorem does not say that if p is a critical point, then it is a local max or min, because we know that that is not true.*0934

*The implication only works in one direction, not the other. Let me write this out.*0944

*So, note, a local extremum implies a critical point -- this double arrow in mathematics means implies -- not, definitely not. It does not mean it goes the other way. It does not mean if it is a critical point, it is a local extremum.*0949

*This is a very important point of logic in mathematics. Most implications -- not most, many implications only work in one direction, and we have to be very, very clear that we are not just automatically assuming that it works in the other direction.*0973

*Critical point does not imply a local extremum. This is not true.*0990

*However, what is true is something called a contrapositive of this. A local extremum implies that it is a critical point.*1001

*What is also true is that if it is not a critical point, then it is not a local extremum.*1011

*So, a implies b is the same as not b implying not a.*1018

*The way theorems are actually written in mathematics, it is often the case that we do not use them as written. We often use them in another form.*1029

*This is one of the ones where we often use in the other direction, because when we are given a function, we actually do not know whether it is a local max or min, to decide whether it is a critical value. We are working in the other direction.*1038

*We are actually looking at the function and finding the derivative and the partial derivatives, and setting them equal to 0. We are finding points that are critical values and then we are checking to see if they are local max's or min's.*1050

*What we can say, what this theorem does allow us to say, is that if I find, if I take the derivatives of functions, set them equal to 0 and realize there is no value in the domain, x or y, there is no point x or y where the derivative = 0, then there is no local max or min.*1061

*That is what is going on here. In this case, the theorem is written as if -- you know -- p is a local max or min, then it is a critical point.*1080

*Often, how we will be using it is if it is not a critical point, then it is not a local max or min. We will be using the contra-positive of that.*1088

*So, if a implies b, that is the same as not b implying not a. That is very, very important.*1097

*Let us go ahead and do an example here. Example 2.*1105

*Let us check to see if the critical values of Example 1 are local extremum, or local max's or min's.*1118

*So, again, just because they are critical values, that is not a guarantee, we have to see if they are local max's or min's.*1145

*So we had the fact that f(x,y) = x + y × sin(x).*1151

*We had critical values -- what is with my h's here? -- We had critical values of (2nπ,-1) and (2n + 1π,1).*1169

*Defined for all n in z integers, so here we have a bunch of critical values. We have (0,-1), we have (2π,-1), (4π,-1), and then we have (π,1), (3π,1), so we have any infinite number of critical values.*1192

*We need to see if any of those are local max's or min's. Well, pick a value, just sort of pick on at random and work with that. Subject that one to analysis, and so that is what we are going to do.*1218

*So, just pick a critical point and check the value of the function -- that is what you are doing, that is how you check max's and min's -- Check the value of the function in a small region, in a small ball, a small ball or disc, around that point.*1229

*That is how we are going to check, right? The definition of a local max is if I have a point p and there is a certain value for the function f(p), there.*1267

*If I sort of move to the left or the right, this way or that way, then the values that I get should either be all less than f(p) or all greater than f(p).*1275

*That will tell me if it is a local max or min, and that is what I am going to do. So, now, in subsequent lessons, in the next couple of lessons we are actually going to devise ways of handling this local max/min business more systematically.*1286

*But, oftentimes, in this case it just means we want to get used to analyzing the function, just sort of taking a look at it, get familiar with it, before we develop the systematic tools for deciding whether something is a local max or a local min.*1300

*You remember from single variable calculus, we took the second derivative to find out if it is concave down or concave up. There are analogous things like that, and we will develop them.*1313

*But right now, we just want to do it by observation, which is really, really important to do.*1322

*You know when everything is said and done, it is about how familiar you are with the function and being able to analyze it using the other tools at your disposal, all of the other things that you know.*1329

*So, now let us just choose 2, π, 1 with n = 0. So I am just going to pick the easiest point, so (0,-1), that is going to be our critical point p.*1338

*Well, let us go ahead and find what the value of the function is at (0,-1).*1356

*The function is x + y sin(x), so it is 0 + -1 × sin(0), which is 0.*1361

*So the value of the function there is 0.*1371

*The first thing I am going to do, since we are dealing with a 2 dimensional domain, you know we are dealing with R2... I am over here, so what I am going to do is I am actually going to... well we have x and y... well I need to check both.*1375

*So, what I am going to do is I am actually going to hold one of the variables fixed, and then I am just going to move the other one.*1396

*In other words, I am going to move a little bit in this direction, and I am going to move a little bit in that direction to see what the function does.*1401

*If I get my answer, great! I can stop there. If not, I am going to hold the other variable, the x fixed, and I am going to move a little bit in the y direction that way to check to see what happens to the function.*1406

*Hopefully, those movements will tell me what is going on. So, that is what I am going to do.*1420

*I am going to decide to hold... so let us hold y fixed first, and we will vary x.*1425

*So what I am going to do is I am going to hold y fixed, and I am just going to move a little bit this way, and a little bit that way around x.*1441

*Let us vary x about 0, because 0 was the point where we are, so I am going to move a little bit to the right of 0, a little bit to the left of 0, to see how the function behaves.*1449

*Well, let us just take... let us take x = 0.1, so let me move 0.1 in that direction. Then, f(0.1) - 1, because we are holding y fixed, it is going to end up giving me some number which is positive, 1.7 × 10 ^{-4}.*1461

*Okay, so it is a positive number. Well, now I am going to move in the negative direction. Let us take x = -0.1.*1485

*Then, f(-0.1,-1), when I do that one, evaluate that, I am going to get -1.7 × 10 ^{-4}.*1493

*Basically what has happened is this. This is the value of my function at (0,-1), okay? Looking at it from just one... I am holding y fixed, so now what I am looking at is just the x axis, this way.*1506

*If I move in the positive direction from 0, my function actually went up. So, the value of my function is 0, at (0,-1), so the value of my function actually became positive.*1522

*When I moved to the left, the value of my function became negative. So, in this particular case, this tells me everything that I need to know.*1536

*I do not actually need to do the next check. I do not need to hold x fixed and work on y. This right here tells me what I have is some kind of a point of inflection.*1546

*In order for it to be a local max or min, it would need to be like this for a local min, or like that for a local max.*1556

*That is not what is happening. So as I move around the point, (0,-1), in the x direction, holding -1 fixed, I end up with a point of inflection.*1565

*Essentially, I have turned it into a single variable problem, because I have the case where it is neither a max nor a min, and that is what this tells me. So this is neither max... I should not say neither max nor min. There is no local max or min.*1576

*In this particular case, there is no local max or min at these critical values. The derivative might equal 0 there, but it does not actually achieve a maximum or minimum there, because with this simple analysis, we have discovered a point of inflection, some sort of point of inflection.*1599

*So, that is all that is happening here. Okay. So now let us go ahead and move onto closed domains.*1615

*We have been talking about open sets, and now we are going to talk about closed sets, and absolute max's and min's. So, definition.*1624

*So we are going to throw out a couple of definitions beforehand. So a boundary point, a boundary point, b, of a set, u, is a point such that an open ball around b contains a point in u and a point not in u.*1634

*I know this may actually sound confusing, but mathematical definitions have to be very, very, precise and formal. Now let us draw you a picture and tell you what it actually means. It is actually very simple.*1686

*If I have some open region, like this, some set, and if there is some... so a boundary point is a set u... the point, such that an open ball around b -- this is b here -- well I take some open disc around b, right?*1695

*It contains a point in u, there is my point in u, and of course I am not in u. There is my point not in u. That is all that it means. It means that it is the point on the boundary, that is it.*1719

*You already know what this means intuitively, but this is a formal definition of what a boundary point is.*1731

*If I take any little region around it, I am going to find some points that are in u, some points that are not in u. That is the definition of a boundary.*1736

*Okay, very, very important to note. A boundary point does not have to belong to the set. In fact, that is the very definition of an open set.*1745

*An open set is a set that does not contain its boundary points. In other words, there are boundary points there, but they are not part of the set.*1755

*So, a boundary point does not have to belong to the set. So, a boundary point of a set does not have to belong to the set. Very, very important to know that.*1765

*In fact, that leads us to the next definition.*1792

*A set is called closed -- again, you already know all of this, a closed interval, an open interval, now instead of intervals we are talking about regions, or regions in space, or regions in 5-space, that is it.*1799

*So, a set is called closed if it contains all its boundary points.*1814

*So, let us go ahead and draw some... let me make this a little bit better so let me see... we will draw an open set, and then we will draw the closed version.*1833

*So, that is some open set. That is the closed version. The open set does not contain its boundary points, the closed set is everything inside including its boundary points, and of course there is... I can also think of one other set from this.*1845

*I can take all of the points from on the boundary, but not on the inside or the outside.*1864

*So if I just talk about the points on the boundary that does not include this or this, that is another thing. We actually have 3 sets of points here.*1869

*We have the open set of a region, we have the closed set on that region, and then we have the boundary of that region.*1876

*If the set actually contains its boundary points, it is called closed. In other words, the boundary points belong to the set, that is all that is happening here.*1882

*Now, we are ready for one more definition, and then we can finish off with the final theorem.*1895

*Okay, definition. A set is called bounded, in other words the set itself is called bounded if there exists some number n, and it could be any number, it does not have to be an integer.*1906

*Such that, -- but we can always choose an integer -- such that for every point, x in the set, the norm of x is < or = to n.*1931

*So, if I have a particular set, in other words, let us say I have some set like that.*1957

*It consists of all of the points inside here. In this particular case I just happened to pick a closed set. So every single, these are vectors of course right? in 2-space and of course there are 2-vectors.*1964

*So, we can find the norm of these vectors, the norm of these points, in other words the distance they are from the origin.*1979

*If the norm of these, of all the points in that set, happened to all be < or = to some number, that means the set itself is bounded. That is all that it means.*1985

*This is a very, very important definition because often times we will be talking about sets that are not bounded. This is a very important definition. It just means that the norm of all of the points in that set are all < or = to some number. That is it.*1995

*There is an upper limit on the value of the points themselves.*2010

*Okay, now we can state our theorem.*2015

*Let s be a closed and bounded set, very important. Those are the hypotheses of the theorem, let s be closed and bounded.*2023

*Let f be a continuous function defined on s. Then, f has both a max and a min on s.*2042

*Okay. So let us talk about what this theorem says. So let s be a closed and bounded set, let f be a continuous function defined on f, then s has both a maximum and a minimum on s.*2081

*So if we happen to be dealing with a set that is both closed and bounded, then we know that somewhere on that set, either in the interior of the set or on the boundaries of the set... that that function, there is some point that the function obtains a maximum value.*2093

*A highest value and a lowest value. Now this theorem does not tell me where it happens. That comes from further analysis on the particular function that we happen to be dealing with in the problem.*2109

*But it does tell me that an absolute max and an absolute min exist. Notice, I did not use the words absolute maximum and absolute minimum.*2120

*I try to keep vocabulary to a minimum. I would much rather have you understand the ideas, what is going on, but here we are talking about something that is an absolute max and an absolute min.*2129

*It can happen on the interior, or it can happen on the boundary. Now, if you end up with a function where there are no critical values, let us say you have a closed and bounded set -- we are going to do some examples in a minute -- and the critical values, there is no x or y where the derivative = 0.*2142

*Well we know automatically, but if it is closed and bounded, the theorem has to achieve a maximum and min somewhere.*2160

*If there is no critical value, then there is certainly no local max or min, that automatically tells me that the maximum and minimum value of the function happens on the boundary.*2166

*Now, I no longer have to concern myself with inside, I just have to concern myself with what is on the boundary. That is the power of these things.*2176

*Okay, we are going to close off this lesson with a statement of this theorem. The next lesson what we are going to do is just a series of problems developing this theorem, working on this theorem, practicing the theorem, finding local max's and min's, finding absolute max's and min's.*2184

*So thank you for joining us here at educator.com, we will see you next time. Bye-bye.*2196

1 answer

Last reply by: Professor Hovasapian

Fri Aug 16, 2013 2:39 AM

Post by Frank Sui on August 15, 2013

Example 2, first one should be "Choose (2nÏ€,-1) w/ n=o So (0,-1)" instead of "(2nÏ€,1)", right?