 In this example, we're going to find the locations of the absolute extrema, if they exist, for the polynomial function f of x equals 3x to the fourth minus 4x cubed minus 12x squared plus two. Now, one thing I want to mention here is that if we just look to the in behavior of this function, if we took the limit as x approaches plus or minus infinity of f of x, because it's a polynomial function, this will be the same as taking the in behavior of its leading term 3x to the fourth as x approaches plus or minus infinity. For which case, in both cases, this will go off towards infinity. Basically, what I'm saying is if you just kind of sketch the general picture of the graph, we know it's going to point up on the right hand side, we know it's going to point up on the left hand side, and it's going to maybe squiggle in the middle. It's clearly too many squiggles for a degree 4 polynomial, but we can anticipate something like this happening, three turning points in the middle. I don't know how many there are necessary, we could find them by calculating the derivative, which will do that in just a moment. But what we can see here is because it's going off towards infinity on both sides, we can see that this graph has no absolute maximum value. So because that's the issue, this function, although it's continuous, we don't have a closed finite interval. And so the extreme value theorem doesn't apply in this context. So the absolute extreme might not exist. We do see that we see there's no absolute maximum, but there has to be some type of absolute minimum. Now I don't exactly know where these turning points are happening, but if you're positive infinity on the left and you're positive infinity on the right, then in between there's got to be some place that's finite. And the smallest finite value, because it doesn't go off towards negative infinity, guarantees there is an absolute minimum. So even when the absolute extreme value theorem doesn't apply, we can be guaranteed some absolute extreme of we investigate the graph. So let's figure out where those are. There are no boundary points because we're going off towards infinity or negative infinity. So I guess you could actually think of the boundary points themselves are infinite. So it's like negative infinity, infinity are x values. And then when we plug those into the function, we're going to get positive infinity and positive infinity. I'm not going to say infinity is the absolute maximum because that's not a number. But we see that there is no absolute maximum because we can get bigger, bigger, bigger, bigger, bigger. What about the function? Well, we need to find the critical numbers here to figure out where that minimum value is going to be. So we calculate the derivative. So f prime of x by the usual power rule, we're going to get 12x cubed minus 12x squared minus 24x. The two will disappear because, well, it's a constant. We take the derivative constant goes to zero. Our goal is to figure out where is the derivative equal to zero? Where is the derivative undefined? The derivative is a polynomial, so it's not undefined anywhere. This function is differentiable everywhere, but it could equal zero. So we need to factor it. Looking at the terms, we see there is a common, the Grayskaw divisor is going to be a 12x. We can factor it out. That leaves behind x squared minus x minus two for which that also very conveniently factors. Does it not? It'll factor. We need factors of negative two that have to be negative one. We're going to get x minus two and x plus one. So this tells us our critical numbers are going to be x equals zero from this one. We're going to get x equals two from this factor and we get negative one for this one right here. So we write those in our table. So we're going to get negative one, zero and two. Now plug these numbers into the function. So for example, if I plug in zero, zero, zero, you get a two. So f of two, so f of zero is equal to two, excuse me. If you plug in negative one, so f of negative one, you're going to get three plus four minus 12 plus two. So two and three and four, excuse me, are seven plus two is nine minus three. That's going to give us a negative three, which we add that to our table. And then lastly, we have to do f of two. So after two, you're going to get three times 16, which is 48 minus two cubed. Did I say that first one right? Two to the fourth is 16 times three is 48. Then the next one you get four times two cubed, two cubed is eight times four is going to be 32. And then the next one, you're going to get a negative 12 times four sets 48. So those conveniently cancel out and you get a plus two. So it looks like we've got a negative 30 when we're done. Just doing the calculation there. And so the absolute minimum value is going to have to occur at the smallest value in this table, which we see is negative 30. So this function does in fact have an absolute minimum value. And the value is y equals negative 30 and occurs when x equals two. So kind of like our graph suggested, although the exact picture isn't quite right. I was just a guess, right? But we knew there was going to be an absolute minimum value. It happened when x equals two, we're going to get y equals negative 30. And there is no number in the range smaller than y equals negative 30. What if we do the same problem over again, but now we require a finite interval from negative one to negative one. So remember the table we did just a moment ago. Let's keep that table for a moment. We had infinity. We had negative one, zero, two and infinity. We don't have to reinvent the wheel when we do this problem. So we had infinity, negative three, two, negative 30, and positive infinity, excuse me. So we restricted just to negative one to one. Then we have to include the endpoints. The endpoints before were positive and negative infinity. I guess that should be a negative infinity, shouldn't that be? Now the endpoints are going to be negative one and one. We want to grab every critical number that's between negative one and one, which that would be negative one and zero. Notice that two, although it's a critical number, it was where the absolute minimum was before, but it's no longer in the interval, so we're just going to get negative one, zero and one. Some of these we already did, we're going to put this into the function. Don't put this into the derivative, put this into the function. f of negative one was negative three, f of zero was equal to two. We haven't done f of one yet. So plugging that back into our function, f of one turns out to be 12, excuse me, three times one to the fourth minus four times one cubed minus 12 times one squared plus two. Because we're just taking powers of one, this will just look like a three minus four minus 12 plus two. So again, three minus four is negative one plus two is positive one minus 12 is negative 11. Like so. And so in terms of absolute maximum minimum, now that we have a closed interval, we see that the maximum value does exist. There is an absolute maximum and it occurs at x equals zero. That's because x equals zero was a local maximum. And while the total graph had no absolute maximum because it goes off towards infinity, if we shrink the picture down, right, so our picture looks something like this, this is the total picture. But if we just zero in on basically this picture right here, just look inside of that green box right here, then there is a maximum value. It happens right here at x equals zero. It'll be the maximum value be y equals two. And there is an absolute minimum, right? It doesn't coincide with two. It coincides with one, actually. This is going to be our absolute minimum value. So we see here we have an absolute maximum at y equals two. And we have an absolute minimum at y equals negative 11. So what I wanted to do in this example was demonstrate one more time how we can solve extreme value problems that is find the absolute maximum minimum. If you have a closed interval, you're guaranteed to have an absolute maximum and an absolute minimum. If your interval is not closed, like with the original one when we did the whole real line, those things might not be guaranteed. But even if we're not guaranteed an absolute maximum, we can still use calculus and a bit of geometry to help us find these absolute maximum, absolute minimum. Because sometimes if our picture looks something like this, we don't care about the absolute maximum, right? What if this, for example, is a cost function? Do we want to find maximum cost? No, no, we want to find minimum cost. And so this is the only value we care about. And this is something we're going to see later on in this chapter as we eventually approach something called optimization problems, where we want to find the best thing to do. And that'll coincide with finding an absolute extremum.