 Hi, and welcome to this video where we're going to put to work where we learned about global optimization. We're going to find the absolute maximum and absolute minimum values of the function f of x equals x minus 2 cosine x on the interval from negative 3 to 3. Now, first of all, note that the extreme value theorem does actually tell us that we must have absolute maximum and absolute minimum values for this function on this interval. The function f of x is continuous because it's x minus 2 cosine x, and we're just simply adding and subtracting continuous functions from each other. And we are on a closed interval that starts at negative 3, ends at negative 3, and includes its endpoints. We don't always get absolute extreme values on any given function. However, if the function is continuous and on a closed interval, we have to. That's what the extreme value theorem says. Now, in the section, you learned a little bit of a workflow for how to go about finding these absolute extreme values of your function. First of all, we need to find the critical values of the function. So find the critical values first in the usual way by taking the first derivative of f and finding where the first derivative is either 0 or undefined. Secondly, we're going to take the critical values and evaluate them into the original function f. So we're going to evaluate the critical values, and not only the critical values, but also the endpoints of the interval, okay, the negative 3 and the plus 3 in this case. Take those numbers, both the critical values within the interval and the endpoints on the edge of the interval, and plug them into f, the original. The reason we're going to use f, the original function, is because we're concerned about where does f hit its highest point. So I need to know about f, not f prime or anything else. And then we're just going to simply compare the results of those evaluations, and we did it in number two. We're going to compare those results and pick out the largest and smallest values that we got. So let's go over to another slide and proceed with finding the critical values of f. This, of course, involves finding the first derivative of f, so let's do that first. f prime of x in this case is going to be 1 plus 2 sine of x. Now the critical values of f are going to show up where f prime is either zero or undefined. In this case, f prime is not ever undefined, because sine of x is always defined for all values of x. And so I just need to think about where is this function, this derivative, equal to zero. Well, this 1 plus 2 sine x is equal to zero whenever sine of x is equal to negative one-half. So I need to solve this equation, but I have to be careful about it this time, because sine of x equals negative one-half has infinitely many solutions. However, I'm only concerned with the interval from negative three to three. So what I'm trying to find here are the solutions to this equation that lie within this interval, and I want to just ignore all others. Normally when we solve this equation, we think on the interval from zero to two pi, and when we do that, on the interval from zero to two pi, our unit circle knowledge tells us that x would be seven pi over six and eleven pi over six. The problem with that this time is that both of those values are larger than three. Neither of them lie inside the interval that I want. So I have to start looking at other solutions. Now if I go and subtract two pi from each of these two solutions, I'm going to get two more solutions, and I think that they may lie within the interval that I want. If I go backwards, in other words, on my angle and look at negative five pi over six and negative pi over six, those are just the two green angles we found over here with two pi subtracted off of them. You'll see that those two values are within the interval from negative three to three. In fact, they are the only solutions to this equation, sine x equals negative a half, that do lie within the interval from negative three to three. So those two guys are the critical values that I want to keep. There are many, many other places where sine of x equals negative a half, but none of them lie inside the interval, those two do. So I'm going to hold on to those. So that's step one. I found the critical values of f that lie within the interval that I'm looking at. Now what I need to do is go make a table of values. So here's my table of values. Notice that it's a table where I'm going to have x's that I'm putting in and I'm putting them into the original function f, not f prime. Remember what we're trying to do is find the high and low points of f and so I'm concerned with heights, not necessarily with slopes. In fact, the only role that f prime plays in this whole process is in step one, which is finding the critical values. Once we found the critical values, you can pretty much put f prime away for good until the end of the problem. So what did we put into f here? Well, we're going to put in the two critical values that we found. So negative pi over six and negative five pi over six are going to go in here. But not just those critical values, because the absolute extreme values of a function on a closed interval could potentially happen at the end points. So I also need to check the values of negative three and positive three. Those could be extreme values as well. So I have gone off to the side with a calculator and calculated these four values here. I'll just show you what they came out to be. Negative pi over six, when evaluated into the original function f, gives an exact value of negative radical three minus pi over six. It's nice to know that exact value, but since we're going to be comparing different values, it might be helpful to approximate just for right now. So that's approximately equal to negative 2.25565. Likewise, f of negative five pi over six is exactly equal to the square root of three minus five pi over six. And that is about equal to negative 0.885943. Now, let's put in the end points now. f of negative three doesn't give a real nice exact value. So I'm just going to give the approximation. That's negative 1.02002. And f of three is 4.97998. Okay, so we've now evaluated the two critical numbers and the two end points of the interval into my original function f. And now I can make my final answer here. The largest of these values that I calculated is going to be the absolute maximum value. That will be this. Okay, so here's your absolute maximum value right there on the interval. It occurs at the extreme right edge of the interval, and it has a value of about 4.97998. Likewise, the smallest of these values is right here. This occurs somewhere within the interval at a critical number. So it's both a local minimum and an absolute minimum. But since it's the lowest of all these values, it's an absolute minimum value. It's exactly equal to negative three, negative radical three minus pi over six about equal to negative 2.25565. Now just to put the finishing touches on this, let's go to Geogebra or a graphing tool of your choice and plot this function f on the interval from negative three to three and just reality check that we are getting the correct answers here. So here in the blue is the plot of f of x equals x minus 2 cosine x. And I've put points at the four places we're interested in. Points a and f are at negative three and plus three respectively. And d is at negative pi over six and c is at negative five pi over six. And it's very easy to see here that we've gotten the right stuff here that the function f has an absolute minimum value there at negative pi over six and the absolute maximum value is wherever here pretty close to five at the extreme right edge of the interval at x equals three. Thanks for watching.