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### Course: Calculus, all content (2017 edition)>Unit 3

Lesson 6: Concavity

# 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.
• The graph of f(x) can be split into two parts; the concave downwards part, and the concave upwards part. Are these parabolas?

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.
• at you said slope is increasing. Does that mean slope is becoming more positive?
• Increasing slope can mean one of two things: more positive or less negative. Whichever situation you have, increasing slope always implies concave up.
• at why does the slope stop decreasing.
• 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)
• Why not use convex to indicate concave downward?
• 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.
• 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!
• 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).
• 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!
• 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!