 Hi everyone, let's take a look at two examples of problems in which we are going to determine the intervals on which the function is increasing and decreasing, and then determine the locations of the relative extremum. The first one we have here is a rational function. Now some things to keep in mind when you come across different types of functions. For the rational function, this one in particular, there is some place that it's not existing. So we know x cannot equal zero in this case. So that might be something we want to keep in mind. Another thing we could do is maybe rewrite this function. Remember we can of course put each term in the numerator over the denominator and rewrite it. So let's rewrite it as x squared plus x raised to the negative two. This might just make finding the derivative a little bit easier. So as we go ahead and find the derivative, we get 2x and then minus 2x to the negative third, so I'm going to rewrite that as 2 over x cubed. And remember we want to find where that derivative is equal to zero so that we can determine the location of our critical numbers. Of course critical numbers are also going to exist where the derivative does not exist, which in this case would be zero because of that x cubed in the denominator. But remember that from the original function, zero was not in the domain in the first place. So it really cannot be a critical number if it doesn't exist in terms of the domain to begin with. Alright, so hopefully that makes sense. So let's go ahead and solve this. I'm going to add the two over x cubed to the other side, do a little cross-multiplying. I get that 2x to the fourth is equal to two. So divide both sides by two. I get x to the fourth equals one, so x has to equal positive or negative one. So those are my critical numbers. So from here we're going to do our number line analysis. I am going to include the fact that x is not allowed to equal zero only because if you think of the original function we know there's going to be an asymptote there and we all know weird things happen at asymptotes. So it might be beneficial to include zero on our number line just in case. So let's go ahead to the next page and do that. So I had one and negative one as my critical numbers and I am going to put zero in there okay and there's positive one. So remember we are doing a number line analysis on the derivative. It's at negative one and positive one that my derivative equals zero. I'm going to put an x here at zero because remember zero was not in the domain to begin with. So we want to substitute into our derivative, let me write that down for you really quick just as a reference. So if we start by picking something less than negative one like negative two or negative three substitute into the derivative we get a negative answer. If we choose an x value in between negative one and zero substitute into the derivative we get a positive and definitely try these on your own to make sure you're understanding why it's either positive or negative please. Substitute something in between zero and one perhaps a half. We get a negative there and then finally something larger than one we get a positive. So remember that what we had talked about with the test for increasing or decreasing functions that the sign, the S-I-G-N sign of the derivative tells us what's going on with the original function. So we know the original function is decreasing first because we know the derivative is negative then it switches to increasing then decreasing then increasing. Alright so let's lay out the intervals on which this function f is going to be increasing and decreasing first. So it's increasing on the interval from negative one to zero and one to infinity and it's decreasing on negative infinity to negative one and zero to one. Depending upon the textbook you use a little side note about these intervals of where it's increasing or decreasing being open intervals or closed intervals. Textbooks seem to disagree on that. I personally always do open really because it's easier. We know that concavity when you get to that will always have to be open intervals and since there's really no agreement on the increasing-decreasing part you might as well do open because then just do open for everything. So just a little note depending on the textbook you're using you might see them using closed intervals for where a function is increasing and decreasing and that's okay. Even on the AP exam we do not get hung up on whether intervals of increasing or decreasing are closed or open. One thing though because remember this had an asymptote at zero right? That has to be open if it's an asymptote. All the more reason to just always do open because then you won't mess it up. Finally let's lay out where we have relative maximum points. Well this one's kind of interesting. We do have minimum points. We got one at negative one and we got one at positive one but there is really no maximum. A lot of people would look at this and say well there's a maximum at zero. Not really because remember that was an asymptote so it can't have the maximum on it if it was an asymptote. So this one only has two locations at which we have relative minima. The next example we're going to look at is a piece function but we're still going to approach it the same way. We're going to start by finding our derivative. So derivative of the first piece is 2x and of the second piece is negative one. So remember we would then set our derivative equal to zero. So if we set 2x equal to zero obviously we get zero. Now another thing you want to make sure of, make sure it's in this interval because if by chance it was not we cannot use that as a critical number but it is so that is a critical number. Obviously with the second piece we cannot set that equal to zero and solve so that's out. Now another thing we'll need to consider though is is there perhaps a critical number where x equals 3 because we need to see if maybe it happens that the derivative doesn't exist there. Remember we need to check to see if the derivative from the left at 3 equal the derivative from the right because if there's going to be any place that the derivative does not exist it's going to be at that split point. So the derivative from the left at 3 remember that would be using the 2x we get a positive 6 there. The derivative from the right at 3 is negative 1 obviously they are not equal so we do have a critical number therefore at x equals 3 because that's where the derivative does not exist. So that's one reason I wanted to provide you with an example with a piece function. We're going to do our number line analysis just like we typically would I'll do that on the next slide so on our number line we have 0 and 3 and remember we're doing a number line analysis on the derivative so it was at 0 that the derivative equaled 0 it was at 3 that the derivative did not exist so we are substituting into the derivative let me write those down for you really quick it's pretty short so if we choose something less than 0 in that case the first piece applies and we obtain a negative answer if we substitute in for x if we choose something in between 0 and 3 we get a positive answer we're still substituting into 2x the first piece choose anything larger than 0 now you're talking about the second piece and it's pretty much always equal to negative 1. So in terms of where this function is increasing and decreasing we'll have that it's increasing on the interval in between 0 and 3 it's decreasing negative infinity to 0 and 3 to infinity we will have a relative minimum at x equals 0 because that's where the derivative changes from negative to positive therefore the original functions going to change from decreasing to increasing we have a relative maximum at x equals 3 now that can be a maximum because the point did exist on the original piece functions probably just a cusp point or something but it can be a maximum even if it's a cusp point so that's where our relative max is and the reason for that is since the derivative is changing positive to negative that tells us the original function is changing increasing to decreasing therefore creating the maximum.