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

Calculus-based justification for function increasing

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.A (LO)
,
FUN‑4.A.10 (EK)
,
FUN‑4.A.11 (EK)
A calculus-based justification is when we explain a property of a function f based on its derivative f'. See a good example (and a few wrong ones) for how to do this when explaining why a function increases.

Want to join the conversation?

  • piceratops seedling style avatar for user Nitya Sunkad
    Can someone confirm if this is correct? I am trying to learn this stuff for my calc test and want to make sure I understand it

    1) f(x) is concave up —> f'(x) is increasing —> f''(x) graph values are positive

    2) An inflection point on f(x) corresponds to a max/min on f'(x) and a x-intercept on f''(x)
    (4 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user Pat
      Let's start with your first statement:

      1.) Yes, if f(x) is assumed concave up, f'(x) must be increasing on the concaved up interval, and therefore, f''(x) must be positive on this same interval.

      -If f'(x) is increasing, it could still be negative until it would pass a critical point (f'(x) = 0) and then f'(x) would turn positive.

      -The 2nd derivative, f''(x) being positive is implying a positive rate of change of the first derivative, f'(x). So, this would imply the original function, f(x) is increasing more greatly in a positive, upward direction, thus, the concaved upward shape.

      -This circumstance you have stated would imply a local or relative minimum of the function, f(x). In other words, if f'(c) = 0 and f''(c) > 0, then there is a local minimum at the point x=c.

      2.) No, an inflection point may not necessarily mean a maximum/minimum on f'(x). For example, the max/min of f'(x) at some point x=c is stating that f''(c) = 0. This doesn't mean it's an inflection point because we do not know if f''(x) is changing signs from negative to positive or positive to negative. The derivative f'(x) could be increasing, become zero, and then continue to increase afterward. This would state that f''(x) is always positive--the slope of the first derivative is increasing on that interval. The definition of an inflection point is just a point where the concavity changes signs, so f''(x) > 0 would switch to f''(x) < 0 and vice-versa. Looking at a point, x=c, if f''(c-0.0001) < 0 and f''(c+0.0001) > 0, this would imply an inflection point at the point c.

      -As for the inflection point being an x-intercept on f''(x), this could be true, but it is not necessarily true. It's an 'x-intercept' in the sense that you're setting f''(x) = 0 and solving for the x-values; however, these are just candidate inflection points. You will have to test the interval before and after the said candidate inflection point to determine whether it is or not. If f''(x) changes signs, then yes, it is an inflection point.
      (2 votes)
  • leafers tree style avatar for user Peijie (Angela) Yu
    In the video, Sal said that an increasing/decreasing derivative doesn't mean that the function is increasing/decreasing--- Yet is there an alternative meaning for that ?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • winston baby style avatar for user Ibrahim
      There is a difference between increasing and positive. if the derivative is positive at a point, then the original function must be increasing at that point. However, the derivative can be increasing without being positive. For example, the derivative of f(x) = x^2 is 2x. if you graph f'(x) = 2x, you can see that for any negative x value, the graph is negative. However, f'(x) is still increasing; it is becoming less negative. So in this case, the derivative is increasing, but the function is decreasing.
      (3 votes)
  • blobby green style avatar for user Choon Meng Tan
    Hi, why is the answer "The derivative of h is positive when x > 0" correct? The derivative of a function may be decreasing even when it is positive where x > 0 such as the part of the curve of a function which is concaving downwards before reaching the max point.
    Thanks.
    (2 votes)
    Default Khan Academy avatar avatar for user
  • male robot donald style avatar for user khanrafialam11
    Should we expect 2020 AP Calculus AB test to have questions like this? It can't be googled easily so I suppose its a good question style to avoid cheating at home.
    (0 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] We are told the differentiable function h and its derivative h prime are graphed and you can see it here, h is in blue and then its derivative h prime is in this orange color. Four students were asked to give an appropriate calculus-based justification for the fact that h is increasing when x is greater than zero. Can you match the teacher's comments to the justifications? So before I even look at what the students wrote, you might say, hey, look, I can just look at this and see that h is increasing when x is greater than zero but just by looking at the graph of h, that by itself is not a calculus-based justification. We're not using calculus. We're just using our knowledge of what it means for a graph to be increasing. In order for it to be a calculus-based justification, we should use calculus in some way. So maybe use the derivative in some way. Now, you might recognize that you know that a function is increasing if its derivative is positive. So before I even look at what the students said, what I would say, my calculus-based justification, and I wouldn't even have to see the graph of h, I would just have to see the graph of h prime is to say, look, h prime is greater than, h prime is positive when x is greater than zero. If the derivative is positive then that means that the slope of the tangent line is positive and that means that the graph of the original function is going to be increasing. Now, let's see whether one of the students said that or what some of the other students wrote. So can you match the teacher's comments to the justifications? So one student wrote, the derivative of h is increasing when x is greater than zero. So it is indeed the case that the derivative is increasing when x is greater than zero but that's not the justification for why h is increasing. For example, the derivative could be increasing while still being negative in which case h would be decreasing. The appropriate justification is that h prime is positive, not that it's necessarily increasing 'cause you could be increasing and still not be positive. So let's see. I would say that this doesn't justify why h is increasing. When x is greater than zero, as the x-values increase, the function values also increase. Well, that is a justification for why h is increasing but that's not calculus-based. In no way are you using a derivative. So this isn't a calculus-based justification. It's above the x-axis. So this one, what is it? Are they talking about h? Are they talking about h prime? If they were saying that h prime is above the x-axis when x is greater than zero then that would be a good answer but this is just what is above the x-axis and over what interval? So I would actually, let's scroll down a little bit, this looks like a good thing for the teacher to write. Please use more precise language. This cannot be accepted as a correct justification. And then finally, this last student wrote, the derivative of h is positive when x is greater than zero and it is indeed the case. If your derivative is positive, that means that your original function is going to be increasing over that interval. So kudos, you are correct.