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

Differentiability at a point: graphical

Let's dive into examples of functions and their graphs, focusing on finding points where the function isn't differentiable. By examining various cases such as vertical tangents, discontinuities, and sharp turns, we gain a deeper understanding of the conditions that make a function non-differentiable. We then determine the x-values where the derivative doesn't exist, further solidifying our knowledge of differentiability.

Want to join the conversation?

  • leaf yellow style avatar for user Sunil Subramanian
    I understand that the derivative at a sharp point of a function doesn't exist because the slopes of the points around them don't reach the same value. But something is happening at that point rather than nothing, right? I mean, if we throw a ball up in the air, it makes a curve (Parabola) and comes back to the ground and the derivative of the highest point on the curve of the ball is 0. Now, if we throw a ball at a wall at a sharp angle, the ball reflects back making a sharp angle right at the moment it hit the wall w.r.t to the ground, right and so according to the derivative definition we have, the slope at the point ball hits the ball doesn't exist, but if we want to make sense of that, how do we move further ?
    (25 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user kubleeka
      If we take data points very, very finely, we can actually trace the ball's velocity function through the moment that it hits the wall. It reality, its change in velocity was not instantaneous. The ball touched the wall, compressed, thereby slowing to a stop, then decompressed and began moving in the other direction. This translates to an extremely high acceleration for that moment, and a standard acceleration due to gravity elsewhere.
      (39 votes)
  • male robot donald style avatar for user Andrew Seif
    i get the point of the "sharp" point or turn, since the slopes are not equal.
    but why would it matter if the slope is equal or not to the Differentiability?
    (15 votes)
    Default Khan Academy avatar avatar for user
    • old spice man green style avatar for user Jelle van der Beek
      Each point in the derivative of a function represents the slope of the function at that point. The slope of a point in the graph that is "sharp" is undefined: we could view it as the slope as we approach it from the left side, or as we approach it from the right side. In case of a sharp point, the slopes differ from both sides.

      Or, more mathetical: if you look at how we find the derivative, it's about finding the limit of the change in y over the change in x, as the delta approaches zero:
      lim h->0 (f(x+h) - f(x)) / h

      In the case of a sharp point, the limit from the positive side differs from the limit from the negative side, so there is no limit. The derivative at that point does not exist.
      (15 votes)
  • leaf green style avatar for user SJNewton123
    At in the video, how does one tell if a turn is too sharp?
    (9 votes)
    Default Khan Academy avatar avatar for user
    • orange juice squid orange style avatar for user Rohan Kapur
      Formally, if taking the limit of the derivative up to a certain value from both the right and left side results in different values, then the turn is too sharp. The turn not being too sharp simply means that the rate of change from both sides of a certain point should converge at the same value, i.e. for some input value a:

        lim      f(h)-f(a)        lim      f(h)-f(a)
      h -> a^- ----------- = h -> a^+ -----------
      h - a h - a
      (19 votes)
  • male robot hal style avatar for user Monty
    at why is f not differentiable on 1 ?
    (6 votes)
    Default Khan Academy avatar avatar for user
    • duskpin sapling style avatar for user Vu
      Starting at Sal said the situation where it is not differentiable.
      - Vertical tangent (which isn't present in this example)
      - Not continuous (discontinuity) which happens at x=-3, and x=1
      - Sharp point, which happens at x=3

      So because at x=1, it is not continuous, it's not differentiable.
      (15 votes)
  • leafers tree style avatar for user tham.tomas
    , isn't the vertical asymptote at -3 a vertical tangent too? I thought that it is a vertical tangent because it is approaching infinity, which would mean it is not differentiable!
    (7 votes)
    Default Khan Academy avatar avatar for user
  • starky tree style avatar for user Joel Lallier
    I might be nit picking, but for clarification, if there is a vertical tangent at x=3 wouldn't that make the said function f technically not a function?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • mr pink green style avatar for user Tamara Soltys
    Why can't we find the derivative if the function is not continuous?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user pragya.natarajan
    In a parabola the slopes also don't reach the same value right? I'm not understanding why sharp turns and parabolas are different when it comes to this.
    (5 votes)
    Default Khan Academy avatar avatar for user
    • aqualine ultimate style avatar for user famousguy786
      Well, the derivative of a function at a point, as you know, is nothing but the slope of the function at that point. In a parabola or other functions having gentle turns, the slope changes gradually. So, it does not matter whether we approach a point on a parabola from the left or the right, the slope we find will be equal ,or in other words, the left hand derivative equals the right hand derivative. On the other hand, imagine a sharp turn . If you approach the point from the left the slope will seem something, and if you approach it from the right the slope seems something else. That is why LHD won't equal the RHD. The point will have 2 slopes at the point of the sharp turn ,which is absurd. Hence, it is non-differentiable at that point. I hope that helps.
      (5 votes)
  • ohnoes default style avatar for user Neo
    why sharp edges matter for example in parabola the point in the very top has a positive slope from the right and negative slope from the left and i don't understand why one of them is differentiable while the other one is not ?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user kubleeka
      f isn't even defined at x=-3, so it can't be continuous there. And the function makes a jump at x=1, i.e. it has a jump discontnuity.

      A parabola is differentiable at its vertex because, while it has negative slope to the left and positive slope to the right, the slope from both directions shrinks to 0 as you approach the vertex. But in, say, the absolute value function, the slopes are -1 to the left and 1 to the right, constantly. There's no smooth way to jump from -1 to 1 in a single point. It's that sudden change in slope that we see as a sharp corner.
      (5 votes)
  • blobby green style avatar for user Tobey
    I don't understand Sal's explanation of the vertical tangent as having infinite dy. I know that a vertical line has an undefined slope because there's no change in x so dy is being divided by zero, but what does infinity have to do with that?
    Also, how does a small change in x give an undefined slope when it's still a defined number that returns another real number when divided by a defined change in y?
    (4 votes)
    Default Khan Academy avatar avatar for user
    • leaf yellow style avatar for user Howard Bradley
      But we don't have a "defined change in y" for a vertical line. The change could be anything. If we have a vertical line, then y is not a function of x.
      Would you be happier if we said dy was undefined for a vertical line, rather than "infinite"?
      To be honest I'm not sure if you think a vertical line is differentiable with respect to x or something else is troubling you.
      (5 votes)

Video transcript

- [Voiceover] The graph of function f is given below. It has a vertical tangent at the point three comma zero. Three comma zero has a vertical tangent, let me draw that. It has a vertical tangent right over there, and a horizontal tangent at the point zero comma negative three. Zero comma negative three, so it has a horizontal tangent right over there, and also has a horizontal tangent at six comma three. Six comma three, let me draw the horizontal tangent, just like that. Select all the x-values for which f is not differentiable. Select all that apply. F prime, f prime, I'll write it in short hand. We say no f prime under it's going to happen under three conditions. The first condition you could say well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative? Well, remember, our derivative is we're really trying to find our rate of change of y with respect to x, but when you have a vertical tangent, you change your x a very small amount, you have an infinite change in y, either in the positive or the negative direction. That's one situation where you have no derivative. They tell us where we have a vertical tangent in here, where x is equal to three. We have no ... F is not differentiable at x equals three because of the vertical tangent. You might say what about horizontal tangents? No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero. F prime of six is equal to zero. F prime of zero is equal to zero. What are other scenarios? Well another scenario where you're not gonna have a defined derivative is where the graph is not continuous. Not continuous. We see right over here at x equals negative three, our graph is not continuous. X equals negative three it's not continuous. Those are thee only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right. These there I guess would be interesting cases. They haven't given us those choices here. We already said, at x equals 0, the derivative is zero. It's defined. It's differentiable there. At x equals six, the derivative is zero. We have a flat tangent. Once again it's defined there as well. Let's do another one of these. Actually, I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn. This isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that, or like, well no, that doesn't look too sharp, or like this. The reason why where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that. The reason why we're not differentiable there is as we approach this point, as we approach this point from either side, we have different slopes. Notice our slope is positive right over here, as x increases, y is increasing, While the slope is negative here. As you're trying to find the limit of our slope as we approach this point, it's not going to exist because it's different on the left hand side and the right hand side. That's why the sharp turns, I don't see any sharp turns here so it doesn't apply to this example. Let's do one more examples. Actually this one does have some sharp turns. This could be interesting. The graph of function f is given to the left right here. It has a vertical asymptote at x equals negative three, we see that, and horizontal asymptotes at y equals zero. This end of the curve as x approaches negative infinity it looks like y is approaching zero. It has another horizontal asymptote at y equals four. As x approaches infinity, it looks like our graph is trending down to y is equal to four. Select the x values for which f is not differentiable. First of all, we could think about vertical tangents. Doesn't seem to have any vertical tangents. Then we could think about where we are not continuous. Well, we're definitely not continuous where we have this vertical asymptote right over here. We're not continuous at x equals negative three. We're also not continuous at x is equal to one. Then the last situation where we are not going to be differentiable is where we have a sharp turn, or you could kind of view it as a sharp point, on our graph. I see a sharp point right over there. Notice as we approach from the left hand side, the slope looks like a constant, I don't know, it's like a positive three halves, while as we go to the right side of that it looks like our slope turns negative. If you were to try to find the limit of the slope as we approach from either side, which is essentially what you're trying to do when you try to find the derivative, well it's not going to be defined because it's different from either side. F is also not differentiable at the x value that gives us that little sharp point right over there. If you were to graph the derivative, which we will do in future videos, you will see that the derivative is not continuous at that point. Let me mark that off. Then we can check x equals zero. X equals zero's completely cool. We're at a point that our tangent line is definitely not vertical. We're definitely continuous there. We definitely do not have a sharp point or edge. We're completely cool at x equals zero.