2015 AP Calculus BC 5b

- [Voiceover] 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 four. So it's gonna be four minus two x. Four minus two x. Over x squared minus four x squared minus four x minus four x squared. 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. And 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 would say, when does this numerator equal zero? And so you could say, when does four minus two x equal zero? You add two x to both sides, you get four is equal to two x. Or x is equal to two, x is equal to two. And they told us that. Or I guess we've confirmed that at f prime of two does indeed equal to 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? And that will let us know if this is a minima or a maxima. So let's say, so when x is less than two, 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. You know, you actually don't even have to test f prime of one. Ii mean, you could if you want. You would have four minus two. 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 two, f prime of x is greater than zero. And you could try this out with different exes if you like. You could say, for example, for example, f prime of one is equal to four minus two, which is two over one minus four. Which is negative three, but then squared. Is equal to two ninths. And then we could say when x is greater than two, f prime of x, f prime of x. Well, when x is greater than two, you're gonna have four minus two times something larger than four. So this is going to be negative. So f prime of x is going to be less than zero. Up here's gonna be negative. Down here's 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, maximum point point at x is equal at x is equal to, or we could say, yes. That we have, f has a relative maximum, relative maximum at x equals, at x equals two. Actually, let's make it right. X has a relative maximum. We're just going to use the words they are using. Relative, relative maximum at x is equal to two. 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 was concave upwards or downwards. But taking the second derivative of this, it gets quite hairy. And any time you're taking the AP test and you find yourself going down a really, really hairy path. Like taking the derivative of, or like taking the second derivative of f, which would be the first derivative of f prime. Any time 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 to just think, okay, well what is f prime doing as we approach this? Or another way is thinking, 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?