Main content

### Course: Calculus, all content (2017 edition) > Unit 8

Lesson 2: AP Calculus BC questions- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c
- AP Calculus BC exams: 2008 1 a
- AP Calculus BC exams: 2008 1 b&c
- AP Calculus BC exams: 2008 1 c&d
- AP Calculus BC exams: 2008 1 d
- Calculus BC 2008 2 a
- Calculus BC 2008 2 b &c
- Calculus BC 2008 2d
- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# 2015 AP Calculus BC 5b

Relative minima and maxima.

## Want to join the conversation?

- At0:50Sal mentions that we are interested in critical points, especially where f'(x) = 0. Is there a reason that Sal ignores critical points where f'(x) is undefined since I thought there can be relative extrema at either of these?(4 votes)
- I understand the first derivative part no problem. Could you please explain more on how the 2nd derivative may show rather this is a max or a min? (I know this is not relevant to this question, but might still want to know.) Thanks!

Or is it that if the f''(x) is + over x=2, that would be a min? and if the f''(x) is -, that would be max?(2 votes)- 'Generally, the second derivative is not used for examining the maxima or minima of a curve. Rather, it is used to suggest the concavity, or parabolic action, of a curve. When

f'(x) = 0, you are finding the critical points of a curve; when f"(x) = 0, you are finding the inflection points of a curve or when the concavity of the curve changes. For example, if you graph the sin(x) curve, every time it touches the x-axis there is an inflection point where f"(x) = 0. This shows the change in concavity which is evident on the sin(x) graph.'(2 votes)

## Video transcript

- [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?