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### Course: AP®︎ Calculus BC (2017 edition) > Unit 6

Lesson 6: Concavity & inflection points intro# Concavity introduction

Sal introduces the concept of concavity, what it means for a graph to be "concave up" or "concave down," and how this relates to the second derivative of a function. Created by Sal Khan.

## Want to join the conversation?

- When finding maxima and minima in 3 dimensions, do you take the derivative of your function (in z=f(x,y) form) with respect to x, then with respect to y, set both the derivatives equal to zero , and then solve a system of equations? If so, how do you solve systems of equations with trig functions in them. For example: system: 2cos(x)+3sin(y) + (xy^3)-(3xy-1)=0 2cos(y)+30sin(x)+ 2(xy^2)-(xy-5) =0 or similar. As far as I know you can't use substitution. It is possible that my example has infinite solutions or no solutions , but I hope you get the idea.(17 votes)
- The graph of f(x) can be split into two parts; the concave downwards part, and the concave upwards part. Are these parabolas?(3 votes)
- Short answer: no.

Since the function f is not defined by some formula, only by the graph sal draw, you cant say wether or not these are parabolas.

That being said, let's assume f(x) = x^3 since the graph look very similar to a x^3 function.

f(x) is certainly not a parabola since a parabola has to be a 2nd order polynomial (x^2).

What happens when we split f(x) at the inflection point as you asked? Nothing. You reduced the domain, but that doesn't change the nature of the function. It is still a 3rd order polynomial so it cannot be a parabola.(16 votes)

- at7:36you said slope is increasing. Does that mean slope is becoming more positive?(3 votes)
- Increasing slope can mean one of two things:
*more*positive or*less*negative. Whichever situation you have, increasing slope always implies concave up.(11 votes)

- at3:55why does the slope stop decreasing.(4 votes)
- From that point on, the slope goes from being negative to becoming zero. Hence, it stops decreasing (in other words, it increases till it becomes zero)(2 votes)

- Why not use convex to indicate concave downward?(5 votes)
- Thank you for another great video! I appreciate all you do to teach so many people about quantitative literacy, which is vital to our ability to understand the universe.

My question is, "Can you understand why I'm not comfortable when at5:33, you say the slope of the first derivative is not changing at it's minimum point?"

This is even something that many people have a hard time understanding when they try to do the thought experiments of weightless flight in the "vomit comets" that follow a parabolic trajectory to allow people to experience "weightlessness". The airplane can be said to follow a sinusoidal (sin x) looking function, but where each of the crests have been replaced with a parabola. (The bottoms are probably also similarly replaced.)

What bothered me when thinking about this was that I had just heard highly revered science communicator Neil deGrasse Tyson make silly statements that I'm sure he could refine, given more time and less stress. His was that you are weightless as soon as the pilot starts to point the nose of the plane to the ground. This sounds like the plane is flying along at a constant altitude, then noses down, which is not the case.

The thing is, that when the plane is flying the "weightless" parabolic path, there is no change in feeling experienced at the maximum altitude of the plane's flight - because the change in velocity of the vertical component of the flight path is constant. It is changing by the same amount at each point in time.

I wish you had said that the slope of the line of the first derivative was changing by the same amount that it had been, but that at that moment in time the slope of the derivative was zero, and therefore transitioning to being positive at the next instant.

By the same token, we wouldn't say that the slope of the second derivative at the x-intercept was zero.(3 votes) - Hi! Is it possible for two critical points (only critical points where f'=0, ignoring critical points where f' it doesn't exist) to occur during one section that is concave down? For example, if some random function is concave down when x < 2, is it possible for there to be more than one x value < 0 where f' = 0?

Thanks!(3 votes)- In short, it structurally won't happen. If f has the same concavity on [a,b] then it can have no more than one local maximum (or minimum).

Some explanation:

On a given interval that is concave, then there is only one maximum/minimum. It is this way because of the structure of the conditions for a critical points. A the first derivative must change its slope (second derivative) in order to double back and cross 0 again. If second derivative does this, then it meets the conditions for an inflection point, meaning we are now dealing with 2 different concavities.(1 vote)

- Sal defined concave downwards as
`f''(x) < 0`

and concave upwards as`f''(x) > 0`

. I was wondering if there's anything special about the point where`f''(x) = 0`

(a point that is neither concave upwards nor concave downwards according to the definition).(2 votes)- A point where both f''(x) = 0 and f''(x) changes sign (i.e. f(x) changes concavity) is called a point of inflection of f(x). Visually, the graph of f(x) has a "wiggle" at a point of inflection of f(x).

Have a blessed, wonderful day!(3 votes)

- When f'(x) increases or decreases, the rate at which f(x) increases/decreases changes. How will f'(x) and f(x) change if f''(x) increases or decreases? Is there anything special to notice?(1 vote)
- When f'(x) is positive, f(x) increases

When f'(x) is negative, f(x) decreases

When f'(x) is zero, it indicates a**possible**local max or min (use the first derivative test to find the critical points)

When f''(x) is positive, f(x) is concave up

When f''(x) is negative, f(x) is concave down

When f''(x) is zero, that indicates a**possible**inflection point (use 2nd derivative test)

Finally, since f''(x) is just the derivative of f'(x), when f'(x) increases, the slopes are increasing, so f''(x) is positive (and vice versa)

Hope this helps!(5 votes)

- Did Sal mean to say that the derivative is negative on the point in1:32since the derivative means the slope of the tangent line?(2 votes)
- Observe that he was looking at the graph of f'(x). This graph is the derivative of the first graph. So, in the first graph, at that point, the derivative (slope) is positive, which is shown in the graph for f'(x) as well as the graph is above the x axis.

Now, the slope of f'(x) is indeed negative, which is shown in the graph of f"(x) as it is below the x axis around that point.(2 votes)

## Video transcript

What I have here in yellow is
the graph of y equals f of x. Then here in this
mauve color I've graphed y is equal to
the derivative of f is f prime of x. And then here in
blue, I've graphed y is equal to the second
derivative of our function. So this is the
derivative of this, of the first derivative
right over there. And we've already seen
examples of how can we identify minimum
and maximum points. Obviously if we have
the graph in front of us it's not hard for a
human brain to identify this as a local maximum point. The function might take
on higher values later on. And to identify this as
a local minimum point. The function might take on
the lower values later on. But we saw, even if we don't
have the graph in front of us, if we were able to take the
derivative of the function we might-- or even if
we're not able to take the derivative of
the function-- we might be able to identify these
points as minimum or maximum. The way that we did
it, we said, OK, what are the critical
points for this function? Well, critical points are
where the function's derivative is either undefined or 0. This is the
function's derivative. It is 0 here and here. So we would call
those critical points. And I don't see
any points at which the derivative is
undefined just yet. So we would call here
and here critical points. So these are candidate points
at which our function might take on a minimum
or a maximum value. And the way that we
figured out whether it was a minimum or
a maximum value is to look at the behavior of the
derivative around that point. And over here we saw the
derivative is positive as we approach that point. And then it becomes negative. It goes from being
positive to negative as we cross that point. Which means that the
function was increasing. If the derivative
is positive, that means that the function was
increasing as we approached that point, and then decreasing
as we leave that point, which is a pretty good way
to think about this being a maximum point. If we're increasing
as we approach it and decreasing as we
leave it, then this is definitely going
to be a maximum point. Similarly, right
over here we see that the derivative is negative
as we approach the point, which means that the
function is decreasing. And we see that the
derivative is positive as we exit that point. We go from having a
negative derivative to a positive
derivative, which means the function goes from
decreasing to increasing right around that point, which
is a pretty good indication, or that is the indication,
that this critical point is a point at which the function
takes on a minimum value. What I want to do
now is extend things by using the idea of concavity. And I know I'm
mispronouncing it. Maybe it's concavity. But thinking about
concavity, we could start to look at the second
derivative rather than kind of seeing just this transition
to think about whether this is a minimum or a maximum point. So let's think about
what's happening in this first region,
this part of the curve up here where it looks like
a arc where it's opening downward, where it
looks kind of like an A without the cross beam
or an upside down U. And then we'll
think about what's happening in this kind of upward
opening U part of the curve. So over this first
interval, right over here, if we start
over here the slope is very-- actually let me do
it in a-- actually I'll do it in that same color,
because that's the same color I used for
the actual derivative. The slope is very positive. Then it becomes less positive. Then it becomes
even less positive. It eventually gets to 0. Then it keeps decreasing. Now it becomes slightly
negative, slightly negative, then it becomes
even more negative, then it becomes
even more negative. And then it looks like it stops
decreasing right around there. So the slope stops decreasing
right around there. And you see that
in the derivative. The slope is decreasing,
decreasing, decreasing, decreasing until that point,
and then it starts to increase. So this entire section
right over here, the slope is decreasing. And you see it right over here
when we take the derivative. The derivative right over
here, over this entire interval is decreasing. And we also see that when we
take the second derivative. If the derivative
is decreasing, that means that the second
derivative, the derivative of the derivative, is negative. And we see that that
is indeed the case. Over this entire interval,
the second derivative is indeed negative. Now what happens as
we start to transition to this upward opening
U part of the curve? Well, here the derivative
is reasonably negative. It's reasonably
negative right there. But then it's still
negative, but it becomes less negative and less
negative and less negative, less negative and less
negative, and less negative. Then it becomes 0. It becomes 0 right over here. And then it becomes more
and more and more positive. And you see that
right over here. So over this entire interval,
the slope or the derivative is increasing. So the slope is increasing. And you see this over here. Over here the slope is 0. The slope of the
derivative is 0. The derivative itself isn't
changing right at this moment. And then you see that
the slope is increasing. And once again, we
can visualize that on the second derivative, the
derivative of the derivative. If the derivative
is increasing, that means the derivative of
that must be positive. And it is indeed the case that
the derivative is positive. And we have a word for
this downward opening U and this upward opening U.
We call this concave downwards. Let me make this clear. Concave downwards. And we call this
concave upwards. So let's review
how we can identify concave downward intervals
and concave upwards intervals. So if we're talking
about concave downwards, we see several things. We see that the
slope is decreasing. Which is another
way of saying that f prime of x is decreasing. Which is another way of saying
that the second derivative must be negative. If the first derivative
is decreasing, the second derivative
must be negative, which is another way of saying
that the second derivative over that interval
must be negative. So if you have a negative
second derivative, then you are in a concave
downward interval. Similarly-- I have
trouble saying that word-- let's think
about concave upwards, where you have an upward
opening U. Concave upwards. In these intervals, the
slope is increasing. We have a negative slope, less
negative, less negative, 0, positive, more positive, more
positive, even more positive. So slope is increasing. Which means that the derivative
of the function is increasing. And you see that
right over here. This derivative is
increasing in value, which means that the second
derivative over an interval where we are concave upwards
must be greater than 0. If the second derivative
is greater than 0, that means that the
first derivative is increasing, which means
that the slope is increasing. We are in a concave
upward interval. Now given all of these
definitions that we've just given for concave downwards
and concave upwards, can we come up with
another way of identifying whether a critical
point is a minimum point or a maximum point? Well, if you have
a maximum point, if you have a critical
point where the function is concave downwards, then you're
going to be at a maximum point. Concave downwards, let's
just be clear here, means that it's
opening down like this. And when we're talking
about a critical point, if we're assuming it's
concave downwards over here, we're assuming differentiability
over this interval. And so the critical
point is going to be one where the slope is 0. So it's going to be that
point right over there. So if you're concave
downwards and you have a point where f prime of,
let's say, a is equal to 0, then we have a
maximum point at a. And similarly, if
we're concave upwards, that means that our function
looks something like this. And if we found a point,
obviously a critical point could also be where the
function is not defined, but if we're assuming
that our first derivative and second derivative
is defined here, then the critical point
is going to be one where the first derivative
is going to be 0. So f prime of a is equal to 0. And if f prime of
a is equal to 0 and if we're concave
upwards in the interval around a, so if the second
derivative is greater than 0, then it's pretty
clear, you see here, that we are dealing with
a minimum point at a.