 Hi everyone, let's take a look at a couple of examples in which we are going to determine where a given function is increasing and decreasing, as well as the locations of the relative extrema. So let's start with a trig function, and this one is supposed to be on the interval from 0 to 2 pi, I forgot to include that. So if you think of the process, remember that in order to determine where a function is increasing or decreasing, it's going to be at our critical numbers. So we need to find those first, and remember that we determine critical numbers by finding the derivative. So the derivative of this function is going to be one-half minus cosine of x. Our critical numbers, remember, are where the derivative equals 0, or where the derivative does not exist. So in this case, there is no place that the derivative does not exist, setting it equal to 0, however, we have a little trig equation to solve, that cosine of x equals a half. So if you think of your unit circle and where that occurs, we have x values of pi over 3 and 5 pi over 3. Remember if you think of your unit circle, that's where cosine equals a half. Remember we were limited to the interval like one full circle, 0 to 2 pi. So these are your critical numbers. So in order to determine where the function is increasing and decreasing, remember that we want to see where the derivative is changing sign from positive to negative. So we're going to do a number line analysis for that, and we're going to do a number line analysis substituting into the derivative using pi over 3 and 5 pi over 3 on the number line. So we have pi over 3 and 5 pi over 3. One little note about creating your number lines, I know it sounds maybe silly that I even have to say it, but I say it from my own experience. Make sure when you put the x values from the critical numbers on the number line, they're in the right order, left to right. As I said, I say that from my own experience, teaching this so long, that I have seen students just put them anywhere on the number line. They have to be in order, left to right, increasing. Remember that we are taking a look at the sign of the derivative. So it is into that derivative we're substituting, and remember what the derivative was already here on the side, so just so we have it as a reference, 1 half minus cosine of x. So it's at pi over 3 and 5 pi over 3, that the derivative equals 0. So that's why I put the little 0s there. So we're going to pick a value in between 0 and pi over 3. We're just going to go through the different sections on the number line, substituting into x in our derivative. So if we pick something in between 0 and pi over 3, maybe pi over 4 for instance, and think about what kind of answer we get when we substitute into the derivative, in that case we get a negative answer. You don't really care what the exact number is you get, you just want to know if it's positive or negative. So in that case it's negative. And if we choose something in between pi over 3 and 5 pi over 3, maybe pi over 2, something like that, we do get a positive value. If we choose something in between 5 pi over 3 and 2 pi, maybe 3 pi over 2, that would work. If we substitute into x for our derivative, we get a negative answer. Alright so think about what this is telling us. Since the derivative changes from negative to positive to negative, remember that tells us the original function is going from decreasing to increasing to decreasing again. And this is how we can figure out the answers to our question. Because remember one thing we wanted to figure out was where the original function f was increasing and where it was decreasing. So we can say f is increasing, now I'm just going to abbreviate. On the interval it would be from pi over 3 to 5 pi over 3, f is decreasing on the interval from 0 to pi over 3 and 5 pi over 3 to 2 pi. And we can also determine our relative max and min points. We're going to have a relative minimum, that would be like right here, where the original function f changes from decreasing to increasing. So we have a relative minimum at x equals pi over 3 and we have a relative maximum. That's going to occur where the derivative changes from increasing to decreasing. That's going to be a 5 pi over 3. And now I think we've covered everything we needed. So let's take a look at the same kind of problem answering the same types of questions, but just a different type of function. So this is the function x squared minus 4, the quantity to the 2 thirds. So once again we need to find our derivative first. So that's going to be 2 thirds, x squared minus 4 to the negative 1 third. Definitely need chain rule here times 2x. So let me rewrite that. And I'm also going to take that quantity that's raised to the negative 1 third. And I'm going to take that down to the denominator. So I have 4x over 3 times the quantity x squared minus 4 to the 1 third. So remember we're setting this equal to 0. So obviously one of your x is where it equals 0, because you'd have 4x equals 0. So there's one critical number. But remember, critical numbers can also occur where the derivative does not exist. So because we have this x in the denominator, we need to think of, well, what x's could be in there that could cause us to have 0 then in the denominator. And we end up with two more critical numbers. x equals positive and negative 2, right? So that's where the derivative does not exist. So let's think of our number line analysis. I have three numbers to put on here. Remember I want to do them in order. So negative 2, 0, and 2. And remember we're looking at the sign of the derivative. So the derivative did not exist at these points. So I'll put a little dne, derivative equal to 0 at 0. So we're substituting into the derivative. So if we start with something less than negative 2, so negative 3, negative 4, pick nice numbers, just whatever will allow you to do it easily in your head. We get a negative in that case. If we choose something in between negative 2 and 0, such as negative 1, for instance, we get a positive there. Choose something in between 0 and 2, perhaps 1. We get a negative, and then anything larger than 2 gives us a positive. So think again what this tells us about the original function. So the original function starts out decreasing because the derivative was negative, then it switches to increasing, then decreasing, then back to increasing again. So now we can get the intervals on which our function is increasing and decreasing. So we would have that f is increasing. Of course, it does not matter which one you answer first, increasing or decreasing. So we had two intervals on which the function was increasing. That was negative 2 to 0, and then 2 to infinity. And f is decreasing on the interval from negative infinity to negative 2 and 0 to 2. In terms of our relative maximum points, if we go back to our number line analysis, it looks like we have two different minimum here at negative 2 and positive 2, and the maximum is occurring at x equals 0, it looks like. Now, if by chance you had to provide the ordered pairs for these points, for these relative extrema points, remember that obviously we have our x values here. The y values, remember, you would have to substitute into the original function. All right, so just a little note about that. Sometimes you are asked to do that. So if you needed the actual ordered pair, you would need to obtain the y value from the original function.