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### 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

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# 2015 AP Calculus BC 5c

Engineering expression for critical point.

## Want to join the conversation?

- It is never possible to have a critical point as undefined. It only works by f'(x) = 0(0 votes)
- It is possible, but he explains that in this particular problem, x=-5 must be in the domain. A critical point where the derivative is undefined is an asymptote and those are not in the domain of a function.(2 votes)

## Video transcript

- [Voiceover] Find the value of k for which f has a critical
point at x equals negative five. All right, so let's just remind ourselves what f of x and f prime of x were. They gave it to us at the top, f of x is equal to one
over x squared minus kx, and then f prime of x is
equal to all of this business. Let me rewrite it, let me rewrite it down here, so f, f of x is equal to one over x squared minus kx and f prime of x is equal to k minus two x over x squared minus kx, whole quantity squared. And we wanna find the value of k for which f has a critical
point at x equals negative five, so f has a critical point
at x equals negative five. This means that x, that means that x equals
negative five is in domain, so this is in domain, which means our function itself is defined at x equals negative five, and it means that f prime of negative five is equal to zero or undefined, or undefined. A critical point is a member
of the functions domain where the derivative is equal to zero, or it's undefined. So let's evaluate f
prime of negative five. F prime of negative five in terms of k is going to be equal to k
minus two times negative five, minus two times negative five, all of that over negative five squared minus k times negative five, and then we want to square all of this, and so this is going to
be equal to k plus 10, k plus 10 over, we have 25, what is this? 25 plus five k, plus five k squared. So what value of k makes f prime of negative five equal to zero? Well, f prime of negative five is equal to zero if k is equal to negative ten. So that's a value of k for which f has a critical point at negative five. Now, you might be saying, well, what values of k? That's the value of k that makes the function equal zero, it makes the numerator equal zero, and therefore makes the
whole function equal zero. But why can't I pick a value
of k that makes the derivative, or it makes the numerator
of the derivative zero, and so therefore makes
the derivative equal zero, I think I said function, not the derivative of the function. But you might be saying, well, why can't I pick a value of k that makes the derivative undefined? And so, you can think of what that is, what would make undefined, well, if 25 plus five k is equal to zero, well, then this is going to be, you're gonna have zero squared, and divide it by zero, it's going to be undefined, so you could say f prime of negative five undefined, undefined if k is equal to what? Negative five, right? Five times negative five plus 25 is zero, but this, if k is equal to negative five, then this can't be a
critical point anymore, x equals negative five
can't be a critical point because it won't be in
it's domain anymore. If k is equal to negative five, then f of x would be equal to one over x squared plus five x, plus five x, and then f will not be
defined at negative five, and so the negative five
couldn't be a critical point because it's not even in the domain. So the important thing is, in order to be a critical point, it has to be in the domain, and the derivative at that point has to be equal to zero undefined. I can get the derivative to
be undefined at negative five, if we said k equals negative five, but with we said k equals negative five, then x equals negative five
will no longer be in the domain, so we wanna go with the k
that just makes the numerator of our derivative equal to zero, or sets our entire
derivative equal to zero.