 Hello, and welcome to the screencast about the second derivative test. All right, so the second derivative test says that if P is a critical value of a continuous function such that F prime of P is equal to 0, I'll highlight that part, and the second derivative at P is not equal to 0. Then F has a relative max at P if and only if the second derivative at P is greater than 0. Okay, so let's kind of refresh a little bit why this makes sense. So remember from before that if our second derivative at a number is less than 0, that means your graph is concave down. Concave down kind of has this upside down parabola-like shape. So if you think about it then up here at the top of this parabola is a maximum. So that's why that makes sense. Okay, here if the second derivative is greater than 0, remember that means the graph is concave up. So it has kind of a parabola-like shape to it that's facing upwards. So then down here at the bottom this graph has a minimum. Okay, so that's just kind of a good intuitive feel for why this derivative test works. Okay, so now we're going to go ahead and use this test on a polynomial function. So that polynomial is f of x equals 12 x to the fifth plus 15 x to the fourth minus 40 x to the third. So we're going to find the extreme values for this. And if you watch the previous screencast, we actually use the same function, but we use the first derivative test. So this is kind of a place where you can compare and contrast them. Which one do you think is easier for this function and which one do you think maybe is a little bit more complicated? In certain ways, we use the first derivative test. In certain functions, it's going to be easier for one or the other. All right, just in case you didn't watch that screencast though, let's go ahead and go through and find our critical values again. This is good practice. So we need to find our first derivative. And that's going to give us 60 x to the fourth plus 60 x to the third minus 120 x squared. Okay, polynomial, you can go ahead and use your power rules. Not too bad. Now we've got to set this function equal to zero. So I think while I do that, I'm going to go ahead and do some factoring. So hopefully you notice that each of these terms has a 60 x squared in it. So if I factor that out, I'm going to be left with an x squared plus an x minus a 2. And then again, you notice we have a nice polynomial in here, a nice trinomial that can be factored again. So that would be 60 x squared and then x plus 2 and x minus 1 because 2 times negative 1 gives me my negative 2 at the end and 2 plus negative 1 gives me a positive one here in the middle. Okay, so then our critical values are going to be when each of these pieces is equal to zero. So 60 x squared is zero, x plus 2 is zero, and x minus 1 is zero. Let me scoot this down, get a little bit more room. Okay, so that's going to give us that x is zero, x is negative 2, and x is a positive 1. Okay, so these are our critical values. So now we've got to see what kind of extreme values are these. So are they minimums, are they maximums, or are they just kind of fake? Alright, so take a look at the second derivative since that's the test that we're going to use. And we take our second derivative. Let's see, that gives us from the original derivative over here, not the factored form because then we'd have to use the product rule and that gets too messy. So in here, let's see what would that be, 240 x to the third plus 180 x squared and then minus a 240 x. Okay, again, let's go ahead and set this function equal to zero. Whoops, nope, we don't need to do that, never mind, I lied. I forgot, we're not finding inflection points, we're just finding extreme values. Okay, so now what we need to do is we need to be able to use these points then to test whether or not the function is going to be concave up or concave down. And then that'll tell us whether or not their minimums are maximums. Okay, but we do need to look at the conditions here because it says f prime of p is zero and the second derivative at p is not zero. So hopefully you notice that if you plug zero into this function, you're not going to get anything. So zero must be one of those funny points where it's neither a minimum or a maximum and maybe the graph is just kind of like at a flat point there or something. But zero is not going to be a minimum or maximum. So this is going to be neither. And again, I know that because f double prime of zero is zero. So that means it's going to be an inflection point which doesn't tell us anything about whether it's a minimum or maximum. Okay, but I do want to go ahead and test negative two and one. Now you could plug them into this function how they are, but oh man, okay. Those are some big coefficients. So I think it might be easier to take and factor the second derivative. So I notice that it looks like they all have a 60x in them. Because then all we can do is just check our signs. So if we factor a 60x out of the first one, that leaves us with a 4x squared. And let's see, plus a 3x and then minus a 4. Okay, now I don't think this will factor anymore, but at least now we can kind of make these coefficients a little bit smaller. Okay, so let's check our signs. So if we look at the second derivative at negative 2. So that means I'm going to be doing 60x negative 2. And then I'm going to be doing 4x negative 2 squared plus 3x negative 2 minus 4. And remember, I don't care about the number here. I just care is this going to be a positive number or a negative number? Okay, so let me change colors real quick so you can see what I'm doing. All right, so here this is obviously a negative number. Okay, because 60 times negative 2 is negative 120. So now what's happening in these other set of parentheses? Well, let's see, this gives us, what is that, 16 plus a negative 6 and then minus a 4. So I believe that's going to give us a positive number. So now when you multiply a negative number and a positive number, that gives us something that is negative. Change my color back. So this is a negative number. So then that is less than 0. And because that's less than 0, that means we're going to have a relative max. Okay, so again, I'm not sure if doing this is easier than doing a number line. I'm kind of walking through it a little bit more slowly maybe than you would when you're actually going to do the problems. But in this case, I think the second derivative is fairly easy to do and it's fairly easy to check. Okay, so let's go ahead and do the first or the second derivative at 1 as well. And if you want to go ahead and pause the video and see if you can come up with the same answer that I do, I think that's a fantastic thing to try. So I'm plugging in a 1 everywhere I see an x. So again, let me go ahead and check my signs. So here 60 times 1 gives me a positive. And then let's see, this would be 4 plus 3 minus 4. That's going to give me another positive. Positive times positive gives me a positive. So this is a positive number. That's obviously greater than 0. So that's going to give me a relative min. Okay, practice this. Maybe try the first derivative test and try the second derivative test with the problem. See if you like it. I mean, some problems are going to be good for this. Other problems are not. But it's just kind of another good gut check that you can do. All right, thank you for watching.