If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Analyzing concavity (algebraic)

Sal finds the intervals where g(x)=-x⁴+6x²-2x-3 is concave down/up by finding where its second derivative, g'', is positive/negative.

Want to join the conversation?

  • leaf green style avatar for user Aehiri
    if between infinity and -1 the graph is concave down would that also mean it has a relative maximimum
    (9 votes)
    Default Khan Academy avatar avatar for user
  • starky ultimate style avatar for user LBKA
    How does concave upward relate to convexity?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Ross Henderson
      Concavity and convexity are opposite sides of the same coin. So if a segment of a function can be described as concave up, it could also be described as convex down. We find it convenient to pick a standard terminology and run with it - and in this case concave up and concave down were chosen to describe the direction of the concavity/convexity. I hope that answers your question.
      (17 votes)
  • purple pi teal style avatar for user vannirob000
    If second derivatives can be used to determine concavity, what can third or fourth derivatives determine? At in the video, the second derivative is found to be: g''(x) = -12x^2 + 12.

    For example, the function given in the video can have a third derivative g'''(x) = -24x. I realize this constitutes the slope of the tangent line of the second derivative, but does it have any applications?

    Any insight is much appreciated!
    (5 votes)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user Megachill

    When sal solves for the second derivative, he then takes it and makes it equal to zero. He then solves for x. He finds that x is equal +-1. How come it's not just 1? Why is it +- 1?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • starky ultimate style avatar for user KLaudano
      The second derivative is a quadratic equation, so it must have two solutions.

      g"(x) = -12*x^2 + 12

      -12*x^2 + 12 = 0
      -12*x^2 = -12
      x^2 = 1

      (-1)^2 = 1 and 1^2 = 1

      So, both -1 and 1 must be solutions to g"(x) = 0. (We cannot just use the principal square root of x as a solution.)
      (5 votes)
  • blobby green style avatar for user troberson3650
    How do you know that if g"(-2) is negative over the whole interval? How do I know it's not crossing through 0 or discontinuous?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user stolenunder
      That is what the checks were for. Essentially we checked to see where the function g'' becomes 0. This shows us critical points. These are locations where the function either reaches a local minimum/maximum, or has an inflection point.

      You would need to potentially check for discontinuity, however, a discontinuity would mean little for the actual concativity, unless it was potentially at an inflection point. For this simpler function though we know polynomial functions like this are continuous for all real values. And it's derivatives are polynomials as well, so they are also continuous. As for crossing through 0 this has nothing to do with concativity, but checking for where the function = 0 (for the second derivative), tells us special information. In the same way the first derivative tells us where the slope = 0 which implies there is a "hill" or "valley" and thus either concave down or up.
      (5 votes)
  • aqualine ultimate style avatar for user sarah
    Between -0.5 and 0.5 in the graph, isn't there a point where clearly the function should reach a zero before becoming positive? Why did Sal ignore it? Or doesn't it mean anything?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • stelly yellow style avatar for user 24tinat
    Why aren't we able to use the first derivative instead of the second derivative?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • male robot donald style avatar for user Venkata
      The first derivative only gives us information about the rate of change of the function. Concavity, however, is related to the rate of change of the rate of change (in other words, how the slope is changing). Hence, we need the second derivative.
      (3 votes)
  • blobby green style avatar for user Nathan Miller
    also, -1 and 1 would be points of inflection right?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Matej Šarlija
    I still don't quite get how e.g. when g is concave downwards the second derivative is negative over that whole interval. I see g' going from positive to negative over zero, but it's like I'm missing some key insight.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user ian.white
    what do you do if you encounter a function like this and you need to find its points of inflection?

    f(x) = x^6 + 3*x^5 + x^4 + 7*x^3 + x^2 + 11*x + 1

    f'(x) = 6*x^5 + 15*x^4 + 4*x^3 + 21*x^2 + 11

    f''(x) = 30*x^4 + 60*x^3 + 12*x^2 + 42*x

    How could you possibly solve these two polynomials? I am referring to the first and second derivatives of f(x). They are not as easy to solve as simply using the quadratic formula, or by using the 3 standard operations which are addition, multiplication, and exponentiation and their inverse operations.
    So can any tell me how in the world you would solve this?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user kubleeka
      Finding zeroes of higher-degree polynomials is, in general, very hard. A lot of high-caliber math has been created in trying to solve problems like these, and you very much have to play each polynomial by ear.

      There do exist formulas for the roots of 3rd and 4th-degree polynomials, much like the quadratic formula. But they're hideously complex (just writing them out would fill blackboards), and would only help us for f''(x). There aren't any such formulas for 5th-degree polynomials and higher (and it's not that we just haven't found one; they can be proven to not exist), and Wolfram|Alpha can't find any of these roots except by approximation.

      So what you do in this situation is brush up on function approximation, and give up on finding exact answers. You won't find them.
      (2 votes)

Video transcript

- [Voiceover] So I have the function g here. It's expressed as a fourth degree polynomial. And I wanna think about the intervals over which g is either concave upwards or concave downwards. Now let's just remind ourselves what these things look like. So concave, concave upwards is an interval, an interval when you're concave upwards is an interval over which the slope is increasing and it tends to look like an upward opening U like that. And you can see here that the slope over here is negative, and then as x increases, it becomes less negative, it actually approaches zero, it becomes zero, then it crosses zero and becomes slightly positive, more positive, even more positive. So you can see the slope is constantly increasing. And if you think about it in terms of derivatives, it means that you're first derivative is increasing over that interval. And in order for your first derivative to be increasing over that interval, your second derivative f prime prime of x, actually let me write it as g, because we're using g in this example. In order for your first derivative to be increasing, your, let me write this. So g, so concave upward means that your first derivative increasing, increasing, which means, which means that your second derivative is greater than zero. And concave downward is the opposite. Concave downward, downward, is an interval, or you're gonna be concave downward over an interval when your slope is decreasing. So g prime of x is decreasing or we can say that our second derivative, our second derivative is less than zero. And once again, I can draw it on this. So, when x is lower, we have a, look, or it looks like we have a positive slope, then it becomes less positive, and then it becomes less positive, as it's approaching zero, it becomes zero, then it becomes negative, then even more negative, and then even more negative. So as you can see, our slope is constantly decreasing as x increases here. So in order to think about the intervals where g is either concave upward or concave downward, what we need to do is let's find the second derivative of g, and then let's think about the points at which the second, where the second derivative can go from being positive to negative or negative to positive and those will be places where it's either undefined or where the second derivative is equal to zero, and let's see what's happening in the interval between. And then we'll know over what intervals are we concave upward or concave downward. So let's do that. So let's take the first derivative, g prime of x, just gonna apply the power rule a lot. Four times negative one is negative four x to the third power plus, okay, so you're gonna have two times six is plus 12x to the first power, you can just write it as x. And then minus two, I could say minus two x to the zeroth power, but that's just minus two. And then the derivative of negative three, of a constant is just zero. And now I can take the second derivative, g prime prime of x is going to be equal to three times negative four is negative 12x squared, decrement the exponent plus 12. And so let's see, where could this be undefined? Well, the second derivative is just a quadratic expression here which would be defined for any x. So it's not going to be undefined anywhere. So, interesting points where we could transition from going from a negative to a positive or a positive to a negative, second derivative is where this thing could be equal to zero. So let's figure that out. So let's figure out where negative 12x plus 12 could be equal to zero. See, we could subtract 12 from both sides and we get negative 12x squared is equal to negative 12. Divide both sides by negative 12, you get x squared is equal to one or x could be equal to the plus or minus, or x could be equal to the plus or minus square root of one, which is, of course, just one. So, at the second derivative at plus or minus one is equal to zero, so either between plus or minus one or on either side of them, we are going to be, we could be concave upward or concave downward. So let's think about this. And to think about this, I'm gonna make a number line. Let me find a nice, soothing color here. All right, that's a nice, soothing color. And, let's say, we should make the number line a little bit bigger. So, there we go, let's utilize the screen space. And so, if this is zero and this is negative one, this is negative two, this is positive one, this is positive two. We know that at x equals negative one and x equals one, our second derivative is equal to zero. So let's think about what's happening in between those places to see if our second derivative is positive or negative, and from that we'll be able to say where it's concave upward or concave downward. So, on this first interval right over here, so this is the interval from, this is the interval where we're going from negative infinity to negative one. Well, let's just try a value in that interval to see what whether our second derivative is positive or negative. And let's see, an easy value there could be negative two, it's in that interval, so let's take g prime prime of negative two, which is equal to negative 12 times four, because negative two squared is positive four. So it's negative 48 plus 12, so it's equal to negative 36. The important thing to realize, then, is well, if over here it's negative, then over this whole interval, because it's not crossing through zero, or it's not discontinuous at any of these points, that's why we picked this interval. That over this whole interval, g prime prime of x is less than zero, which means that over this interval we are concave downwards. So concave, concave downward, concave downward. Now let's go to the interval between negative one and one. So this is the open interval between negative one and one. And let's try a value there. Let's just try zero will be easy to compute, g prime prime of zero, well, when x is zero, this is zero, so it's just gonna be equal to 12. The important thing to realize is our second derivative here is greater than zero, so we are concave upward, concave upward on this interval between negative one and one. And then finally, let's look at, let's look at the interval where x is greater than one. So this is the interval from one to infinity, if we want to do it that way. And let's just try a value. Let's try g prime prime of two, 'cause that's in the interval. And g prime prime of two is gonna be the same thing as g prime prime of negative two, 'cause whether you have a negative two or a positive two, you square it becomes four. So you're gonna have four times negative 12, which is negative 48 plus 12, which is negative 36. So this is negative 36, and so, once again on this interval you are concave, concave downward. Now let's, I graphed this ahead of time, let's see if what we just established is actually consistent with what the graph actually looks like. We were able to come up with these, these insights about the concavity without graphing it. But now, it's kind of satisfying to take a look at a graph and actually let me see if I can match up the intervals. Actually this is pretty closely matched, all right over here. And so, this is, actually we can make it a little bit smaller, all right. And so, let me move my boundy box. So, I'm saying that I'm concave downward between negative infinity, negative infinity all the way until, all the way until negative one, all the way until this point right over here. So, all the way until that point. And that looks right. It looks like the slope is constantly decreasing all the way until we get to x equals negative one, and then the slope starts increasing. The slope starts increasing from there, from there all the way, and right at x we're transitioning, so I'm gonna leave a little, I won't color in that. And so here our slope is increasing, let me do that same color. Our slope is increasing, increasing, increasing, increasing, increasing all the way until we get to x equals one. And then our slope starts decreasing again. And we get back into concave downwards. Oops, I wanna do that in that orange color. We get back into concave downwards. So what we're able to figure out by just taking the derivatives and doing a little algebra, we can see quite clearly looking at the graph.