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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition) > Unit 12

Lesson 2: AP Calculus AB 2015 free response- 2015 AP Calculus AB/BC 1ab
- 2015 AP Calculus AB/BC 1c
- 2015 AP Calculus AB/BC 1d
- 2015 AP Calculus AB 2a
- 2015 AP Calculus AB 2b
- 2015 AP Calculus 2c
- 2015 AP Calculus AB/BC 3a
- 2015 AP Calculus AB/BC 3b
- 2015 AP Calculus AB/BC 3cd
- 2015 AP Calculus AB/BC 4ab
- 2015 AP Calculus AB/BC 4cd
- 2015 AP Calculus AB 5a
- 2015 AP Calculus AB 5b
- 2015 AP Calculus AB 5c
- 2015 AP Calculus AB 5d
- 2015 AP Calculus AB 6a
- 2015 AP Calculus AB 6b
- 2015 AP Calculus AB 6c

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# 2015 AP Calculus AB 5b

Intervals where f is concave downwards and decreasing.

## Video transcript

- [Voiceover] Part b,
on what open intervals contained in negative three is
less than x is less than four is the graph of f both
concave down and decreasing? Give a reason for your answer. All right, so f and this should be both concave down and decreasing. So f concave down, concave down. That's going to be true if and only if the second derivative
of f is less than zero which is true if and only if f prime is decreasing, f prime is decreasing. So that's how we can tell
when f is concave down. I wanna see what do I
need to observe of f prime because that's the graph they gave us. All right, now for f to be decreasing, f is going to be decreasing if and only if, we are decreasing when
you have a negative slope if your relative change is negative. So that's true if and only if f prime of x is less than zero. So both true. Both true when f prime of x is less than zero and decreasing, and, let me write, and
f prime is decreasing. And f prime, decreasing. So let's see if we can identify that. Well, let's go back to our graph and see when is f prime
of x less than zero and decreasing. So if we look at less than zero, let me highlight this, so less than zero. I'm highlighting it in
magenta right over here. This is all the places
where we are less than zero in magenta. And now of that, where are we decreasing? Well, we are decreasing over
this interval right over here. So that would be negative
two is less than x which is less than negative one. So that's we're decreasing of the places where we are negative. I'm taking the subset
of where we are negative to think what we're decreasing. We're decreasing right there. And we are also decreasing, we are also decreasing right over there. So that is one is less than
x which is less than three. Now we are decreasing before x
equals negative two over here but we're not less than zero there so I'm not gonna include that. And so we have our intervals. So our open intervals. So we have both true for when
these two things are true. And so this happens. This happens when negative two is less than
x is less than negative one or one is less than x is less than three. So those are two open intervals. This one and that one.