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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.

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  • male robot hal style avatar for user 13NixonF
    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)
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  • old spice man green style avatar for user newbarker
    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)
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    • leaf green style avatar for user janismac
      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.
      (15 votes)
  • blobby green style avatar for user JAlexanderBeck
    at you said slope is increasing. Does that mean slope is becoming more positive?
    (3 votes)
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  • sneak peak green style avatar for user moose
    at why does the slope stop decreasing.
    (4 votes)
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  • old spice man green style avatar for user Jason
    Why not use convex to indicate concave downward?
    (5 votes)
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  • leafers ultimate style avatar for user Spencer Black
    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 at , 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)
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  • blobby green style avatar for user keltonthesupergenius
    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)
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    • male robot hal style avatar for user davis
      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)
  • starky ultimate style avatar for user Joshua Gammage
    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)
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  • ohnoes default style avatar for user Darth Vader
    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)
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    • hopper jumping style avatar for user Gavin
      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)
  • male robot hal style avatar for user James
    Did Sal mean to say that the derivative is negative on the point in since the derivative means the slope of the tangent line?
    (2 votes)
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    • male robot donald style avatar for user Venkata
      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.