 Hi there, and welcome to this screencast where we're going to use the first derivative test to find the local extreme values of a particular function. Namely, this function f of x equals 12x to the fifth plus 15x to the fourth minus 40x cubed. So, our goal is to find the local extreme values of this function, and we're going to use the first derivative test to do so. Remember what the first derivative test actually says, it says that given a differentiable function f, the local maximum values of f will happen where f prime changes sign from positive to negative, and the local minimum values of f will happen where f prime changes sign from negative to positive. So, this is all about finding f prime and finding the places where it could change its sign from positive to negative, or vice versa. So, first of all, let's find f prime because obviously we're going to need that. This is a simple polynomial function, so it's very easy to find its derivative. f prime of x is 60x to the fourth plus 60x cubed minus 120x squared. I'm going to do a little more with that derivative here in a second, but let's just kind of look ahead and see what we need to do. We want to find the places where this derivative in the blue here, f prime, changes its sign. Now, that can only happen at critical values of my function, places where the derivative is zero, which makes sense. It's pretty natural to think that a quantity will only change sign when it crosses zero, but it could also happen where a derivative is undefined, for example, crossing a cusp. So, we're going to find the critical values of my function first, and then use those to help locate where the derivative changes sign. To do that, I'm going to do a little algebra on the derivative here. So, I'm going to factor out the greatest common factor from this expression, and that would be 60x squared, and that's going to leave me with x squared plus x minus two. I can factor that even further into 60x squared times, and then the quadratic in here factors into x plus two times x minus one. Now, what are the critical values of f? That would be the places where f prime is either zero or undefined. f prime is a polynomial, and so there are no places where f prime is undefined, so we can forget that part. So, where is f prime equal to zero? Well, if this expression here is equal to zero, then it means that either x squared is zero, x plus two is zero, or x minus one is zero. That would give me critical values of x equals zero, which makes this term equal to zero, negative two, which would make this term equal to zero, or plus one, which would make this term equal to zero. So, I have these three critical values for my function, zero, negative two, and one. Those are the only places where f prime could possibly change sign. It doesn't mean that f prime definitely does change sign at those points. f prime might change its sign from positive to negative or vice versa at those points, or it might not. We don't know, but we do know is that f prime cannot change sign anywhere else, because otherwise we would have picked up another critical number, and we didn't. So, these are the only places where f prime can possibly change sign, i.e., it's the only places where you could possibly have local extreme values. Once again, not all of these critical values could actually lead to local extreme values, so we need to use the first derivative test to see if they do. I'm going to go on to a different page here to set up the first derivative test and find out. So, I'd like now to find where f prime is positive and where f prime is negative. And I'm going to do that by actually looking at my critical numbers, literally looking at them. So, I'm going to draw a little chart here, starting with a number line. I'm going to put my critical values on that number line, negative two, zero, and one, like so. Now, these three critical values separate that number line into four distinct intervals. I'm going to mark them off with some dashed lines here just to visually help me separate those intervals out. You see, there's four intervals now. Now, on each of those four intervals, f prime can only have one sign. It cannot change from positive to negative or negative to positive within those intervals, because if it did, we would encounter another critical number, but we already have all the critical numbers. So, on each of those intervals, f prime is going to have one sign. It will not change. So, on each of those intervals, I would like to record two pieces of information, the sign of the derivative, f prime, and then the behavior of f in terms of whether f is increasing or decreasing on that interval. Of course, these two guys are directly related to each other. If f prime is positive, it means that f is increasing. If f prime is negative, it means f is decreasing. Really, all I need to know is the sign of f prime on these four intervals here. Now, f prime is right here, as you see. Now, let's go through each of these four intervals one by one and just think about what f prime's sign must be on those intervals. Notice that f prime is a factor of four things, 60, x squared, x plus two, and x minus one. Now, 60, of course, is a constant. It never changes its sign. So, what I really want to think about is what is the sign of each of those three variable quantities on the four intervals that you see here. Let's look at the interval here on the far left and just think about the numbers in here. These would include numbers such as negative three, negative ten, and so forth. Now, the first factor, this factor of f prime, is always going to be positive on that interval. It's a squared x, so x squared is always positive, period, as long as x is not equal to zero. So, on that interval, x squared is always positive, and I'm going to denote that with a plus sign. x plus two, on the other hand, the second factor, if you use a number out of this interval, such as negative three, that's going to give you a negative quantity if you evaluate it into x plus two. For example, x equals negative three belongs to that interval, negative three plus two is minus one. And so, x plus two is always negative on this interval, so I'm going to denote that with a minus sign. Lastly, this third factor here, x minus one, is also negative for every x value on this interval. Pick your favorite x value to see that, say x equals negative three again, and negative three minus one is negative. So, I'm going to denote that with a minus sign. Now, what this tells me is f prime is the product of those three things, and a plus 60. So, on this interval, I would be, if I evaluated a number into f prime, I would be taking a positive number for x squared, times a negative number for x plus two, times another negative number for x minus one. And the result there is entirely positive. A negative number times a negative number times a positive number is positive. And what that indicates is that f is increasing on this interval. Now, let's move on to the second interval and try the same idea. Inside this second interval here are numbers such as negative one. Now, the x squared factor is positive on that interval. The x plus two factor this time is also positive on that interval. For example, if I chose x equals negative one, then evaluating that in for x inside this factor right there, negative one plus two, that would be a positive number. The third factor here would be negative on this interval. For example, if I chose negative one for my x, then negative one minus one, that's a negative value. So, f prime, which is the product of those three things, would be a positive times a positive times a negative, and that would be a net negative result. So, f prime is negative on that interval. That means f, the original function, is decreasing on that interval. Let's move on to the third interval and keep constructing the sign chart. Now, these numbers in between zero and one are things like a half or point five. The first factor, x squared is positive on that interval. x squared is positive on all of these intervals, as you might notice. The second factor, x plus two, is also going to be positive because you're adding small positive numbers to two, and that gives you a positive result. The third factor, x minus one, is still going to be negative on this interval because, for example, x equals point five when evaluated into here would give you a negative result. And so, f prime, which, again, is the product of those three things, will be a negative times a negative times a positive for no matter what x value you choose in that interval. That means f prime is going to be negative throughout that interval, and so f is going to be decreasing throughout that interval. Finally, let's look at the last interval here from one to infinity. Now, the first factor, x squared, again, is going to remain positive. I should probably switch to my black here. The second factor, x plus two, is going to be positive as well. If I choose a number that's bigger than one and add two to it, the result is positive. The third factor, x plus one, is finally positive on this interval here. If I choose a number that's bigger than one and subtract one from it, it's going to be a positive result. And so, f prime, which is the product of those three factors, is going to be a net positive value, and that means, since f prime is positive, that f is increasing. Okay, so now let's put all this information together and see if we can answer the question about where the local extreme values of f are. The first derivative test tells us that f attains a local maximum whenever f prime changes from positive to negative. That would be right here, from positive to negative. That just means that f is changing from increasing to decreasing, so it looks like this, so that makes sense. So f is going to have a local maximum value. Let's write this down. It has a local maximum value at x equals negative two. The first derivative test also says that f will have a local minimum value where f prime changes sign from negative to positive, and that's right here. Of course, that just means that f itself is changing from decreasing to increasing, so it looks like this, so that's obviously a local minimum. So we have a local minimum value, local minimum, at x equals one. Now what are the values themselves? Well, the local maximum value is what I get when I put that negative two into f, to find the height value. So the local maximum value is f of negative two, and you can do the calculation on that yourself and get 176 as its local maximum value. And likewise, the local minimum value for f occurs at x equals one, and the value itself is f of one, which you can work out to be negative 13. So the use of the sign chart is just a handy way to organize the information about the sign of f prime, which you can then run through the first derivative test to find the local maximum and local minimum values of f. Thanks for watching.