 Hi everyone, welcome to the second half of our lesson on using the second derivative. Here we're going to talk about the second derivative test. In the last lesson we learned that the first derivative test for relative extrema can tell us where a function f is increasing or decreasing depending upon the sign, positive or negative, of the first derivative. The second derivative test is also going to be able to be used to determine the location of relative extrema. The test will involve only the critical numbers, not the intervals around them, as we saw with the first derivative test. Suppose f is a function such that f' of c equals 0 and there exists an open interval containing c on which the graph of f is concave up. Now since f' of c equals 0 that tells us we have a horizontal tangent line to the curve at c. So what can be said of the point c, f of c? Because we know the function is concave up at c, we know also that the second derivative at c is greater than 0. And observing what happens on the graph, we can obviously see a relative minimum exists at c. So notice what we're doing here. We are looking at the sign, S-I-G-N, of the second derivative at the critical number to help us determine, first of all, is the curve concave up or concave down at that point, but then also do we have a relative max or a relative min? So let's look at the other case. Suppose f is a function such that f' of c equals 0 and there exists an open interval containing c on which the graph of f is concave down. Now again, because f' of c equals 0 we still know that at the point c we have a horizontal tangent line. So what can be said about that point? Well since we know the function at that point is concave down, we know therefore that the second derivative at that point is less than 0. And we can tell from the graph that it is at c that we have a relative maximum. That brings us to the statement of the second derivative test for relative extrema. Let c be a critical number of the function f such that f' of c equals 0 and let the second derivative of f exist on the open interval from a to b containing c. If the second derivative is greater than 0 then f will have a relative minimum value at c. If the second derivative is less than 0 then f has a relative maximum value at c. If the second derivative equals 0 then the test fails and you must resort to the first derivative test instead. So think about what the second derivative information is telling us. If the second derivative is greater than 0 remember that tells us the curve is concave up and we get a minimum. If the second derivative is less than 0 that tells us the curve is concave down we get a maximum. So how do you determine then relative extrema by using the second derivative test? Well if you think back we need critical numbers so we have to find our first derivative but then we're going to need to substitute those critical numbers into the second derivative. So we are going to have to find both the first derivative and second derivative of the function. Next we will have to determine the critical numbers of f the values of x for which the first derivative equals 0 or perhaps where the first derivative does not exist. Finally we are going to determine whether there are any relative extrema at the critical numbers by finding the sign of the second derivative at each critical number. But a warning the second derivative test will not provide information as to where the function is increasing or decreasing. So if you are asked about that in a problem you won't really be able to use the second derivative test. There are times that the second derivative test will fail. However it fails if the second derivative at the critical number happens to equal 0. And as I mentioned before in those cases you will need to resort to using the first derivative test instead. So in summary the second derivative test is a little bit limited. Let's look at a couple of examples. So in this first one we are asked to use the second derivative test to find the relative extrema for the function negative 3x to the fifth plus 5x to the third. So remember the steps, the first thing we need is our first derivative. So our first derivative is simply going to be negative 15x to the fourth plus 15x squared. And we are going to need our second derivative, let me do that over here. Second derivative will be negative 60x cubed plus 30x. Remember the next thing we are going to do is find our critical number. So we are going to set our first derivative equal to 0. And it looks like we can do a greatest common factor here. So out of the negative 15x squared equaling 0 we get x equals 0. And out of x squared minus 1 equaling 0 we get positive and negative 1. So they are our critical numbers. So remember what we want to do with the second derivative test, we are going to take these critical numbers, substitute them into the second derivative and see if we get positive or negative results. So let's try 0 first. So we are substituting into the second derivative. Well that equals 0. So that is one in which the test fails. So we are going to have to come back to this and do a first derivative test on that one. We will be a good review for us, we will do that at the end. Let's try 1 and negative 1. So if we substitute 1 into that first derivative we come out with negative 30, which is obviously less than 0. So think about what that tells us. That tells us that f is concave down there, so that tells us we have a relative maximum there at x equals 1. So now we can do the same thing with negative 1. If we substitute negative 1 into the function we get a positive 30, well that is obviously greater than 0. So that tells us the original function is concave up because our second derivative came out greater than 0. And if it's concave up it tells us we have a relative minimum at x equals negative 1. So now we figured out about these two points, we need to determine about 0. And as I said we are going to have to go ahead and resort to the way we did it in the last lesson with the first derivative test. So remember we are substituting into the first derivative. The first derivative equaled 0 when x was 0. Now remember what our first derivative was, let me put it over here on the side for you. It was negative 15x to the fourth plus 15x squared, just so you have it as a reference. So if we substitute something less than 0, like a negative 1 or something, we come out with a positive answer. If we substitute something greater than 0, geez we get a positive answer there too. So we have no relative extrema there. So really all we have are the relative maximum that was at x equals 1 and the relative minimum that was at x equals negative 1. Let's take a look at a different kind of problem in which you are going to have to draw a possible graph of what a function f could look like given certain conditions. So take a look at the table of sine values and they give us information on f of x, f prime of x and the second derivative of x. And it's broken down into three parts, 0 to 1, 1 to 2 and 2 to 3. And what you have is essentially a sine chart. So the key is to take all this information and figure out about the original function, where is it going to be increasing and decreasing? Where is it going to be concave up, concave down? Where are our inflection points going to be? Where are our maximum points going to be? So remember that the first derivative tells us about increasing, decreasing nature of the function. So if the first derivative in between 0 and 1 is positive, that tells us the original function is increasing. I'm just going to write it here on the side so we know in between 0 and 1, f is increasing. So red is standing for what the original function is doing. In between 1 and 2, the first derivative is now negative. So that tells us now the original function f that we're going to draw is going to be decreasing on that interval. Then once we get to the interval from 2 to 3, notice how it keeps decreasing. So what we're trying to do is gather information about the original function. Now the second derivative tells us about the concavity of the original function. So in between 0 and 1, we are told the second derivative is negative. That tells us the original function is concave down. In between 1 and 2, it remains concave down, and actually it just keeps being concave down. So what you have now is some information. We know that in between 0 and 1, the function has to look like it's increasing but concave down. So try to envision in your head what that might look like. And then once we get to 1, it's going to switch to decreasing but it's still concave down. Now think about what might be happening at 1. We have the function changing from increasing to decreasing. So at 1, we're going to have a relative max because it's at 1 that the original function changes from increasing to decreasing. And then it looks like since it keeps decreasing from 1 to 2, then 2 to 3, there's no more extrema there. Nor are there any inflection points because it seems that on this entire interval from 0 to 3, this function's concave down. So this is the information we have to keep in mind and try to piece together. The other thing we're given is that we know whether the original function is either positive or negative if it lies above the x-axis or below the x-axis. So it looks like it's above the x-axis from 0 to 2 but once we hit 2, then it goes below. So I think we're going to have a 0 in here too at 2. So the graph is going to have to cross the x-axis, it looks like, at x equals 2. So let's try to draw this. So I have my axis set up and let's put in some of the key points. We knew this graph had to cross the x-axis at 2, so let's put a point there. We also knew we had a relative maximum at 1. So I'm going to stick it here. Now when you draw problems like this, it's very possible everybody's might come out looking a little bit different. Even when they show up on the AP exam, what we're looking for are key characteristics. So maybe you put your maximum point up a little bit higher than I did or maybe you put it a little bit lower and closer to the x-axis. That's fine. It's really the general shape we're going for. So if you go back to our chart, we knew that in between 0 and 1, the curve had to be increasing but concave down. So let's just suppose it starts here at the origin to make it easy. So increasing, concave down, it has to look kind of like that. And then in between 1 and 2, it's decreasing. It changes to decreasing, but it's still concave down. So I'm coming down like this. And once I hit that point at 2, it's still decreasing and still concave down. So it looks like it kind of keeps just going down. This looks like a little lopsided parabola. The more you draw these, the better you get at them. I promise you that. Sometimes your first attempts aren't always the greatest. But again, what we're really going for are the shape, the characteristics based upon what we gathered about the original function from the information we're given.