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Let k equal four so that f of x is equal to one over x squared minus four x. Determine whether f has a relative minimum, a relative maximum, or neither at x equals two. Justify your answer.

All right, well, if f of x is equal to this, then f prime of x. They gave us f prime of x in terms of k, so f prime of x is going to be equal to—in this case, k is 4—so it's going to be 4 minus 2x over x squared minus 4x squared minus 4x.

So now we know f of x, and we know f prime of x, and if we were looking for relative minima or relative maxima, we would be interested in points and critical points, especially where f prime of x is equal to zero.

And so if we said f prime of x is equal to zero, well, we could say, well, when does this numerator equal zero? You could say when does four minus two x equal zero? You add two x to both sides, you get four is equal to 2x, or x is equal to 2. x is equal to 2.

And they told us that—or I guess we've confirmed that at f prime, f prime of 2 does indeed equal zero, so this is definitely an interesting point. So let's think about whether before, when x is less than two, is f prime of x increasing or decreasing, and then when x is greater than two, is f prime of x increasing or decreasing? That'll let us know if this is a minima or a maxima.

So let's say, when x is less than 2, we could test something out. We could say f prime of—so let's just say f prime of one. So if x is less than two—and you actually don't even have to test f prime of one; I mean, you could if you want—you would have four minus two times one, so that's going to be positive, and then this down over here is always going to be non-negative because you have a squared right over here.

So when x is less than 2, f prime of x is greater than zero. And you could try this out with different x's if you like. You could say, for example, f prime of one is equal to four minus two, which is 2 over 1 minus 4, which is negative 3, but then squared is equal to 2, 9. And then we could say when x is greater than 2, f prime of x.

Well, when x is greater than 2, you can have 4 minus 2 times something larger than four, so this is going to be negative. Thus, f prime of x is going to be less than zero. Up here is going to be negative; down here is not going to be negative, and so f prime of x is going to be less than zero.

So if we are increasing as we approach something and then our slope is zero, and then we are decreasing, well then this is going to be a maximum point. So that means that we have a maximum point at x is equal to—or we could say, yes, we have f has a relative maximum at x equals 2.

Actually, let me write: x has a relative maximum. Let me just use the words they are using: relative maximum at x is equal to 2. Now, another way that you could have tried to do it is you could have tried to take the second derivative of this and then saw whether that was positive or negative and whether it's concave upwards or downwards, but taking the second derivative of this gets quite hairy.

Any time you're taking the AP test and you find yourself going down a really, really hairy path, like taking the derivative of—well, taking the second derivative of f, which would be the first derivative of f prime—anytime you see yourself going down a hairy path like that, it might work, but it's probably not the optimal path.

So an easier way is just to think, okay, well, what is f prime doing as we approach this? Or another way of thinking is, is the function increasing as we approach from below, and is it increasing or decreasing as we get beyond that point beyond x equals two?