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

Lesson 6: Concavity

# Analyzing concavity (graphical)

Sal walks through an exercise where you are asked to recognize the concavity of a function in certain regions. Created by Sal Khan.

## Want to join the conversation?

• Is there an algebraic method for determining concavity?
• Yes. However, it will often be less work (and less prone to error) to use the Calculus method. Anyway here is how to find concavity without calculus.
Step 1: Given f(x), find f(a), f(b), f(c), for x= a, b and c, where a < c < b
Where a and b are the points of interest. C is just any convenient point in between them.
Step 2: Find the equation of the line that connects the points found for a and b.
Step 3: Find that y-coordinate of the line from Step 2 at point C.
Step 4: If the value found in Step 3 is greater than f(c) of the original function, then you have a positive concavity. If less, you have a negative concavity. If equal, you have an average concavity of zero.

Example: f(x) = x⁵ - 5x³ + 8x² - 2 at x = 3 and 5
f(3) = 178 ; f(5) = 2698
For a point in between, I pick 4 (could be any point between 3 and 5, though). f(4) = 830
Step 2: The line connecting (3, 178), (5, 2698) is y = 1260x-3602
Step 3: At x=4, the line's y-coordinate is y = 1438
Step 4: 1438 > 830. Therefore, this function as a positive concavity between f(3) and f(5).
It should be obvious, though, that this method can make mistakes (particularly, if there is an inflection point between A and B).
• if the 2nd derivative means concavity, then doesn't the 3rd derivative mean the frequency? (i.e. rate of change of concavity)
• Typically in mathematics and natural sciences, we don't deal too much with the 3rd derivative. I think the easiest way to understand the 3rd derivative is through physics, in which it is the rate of change of acceleration. If you have ever driven a car and felt that your body was being sucked into the seat, then that is the feeling of acceleration, but if you can feel that you are progressively getting sucked into the seat harder or slower, then that is the jerk (3rd derivative). It's much easier to visualize in that sense, but it mathematically, the rate of change of the concavity is equivalent. Frequency typically deals with waves, so we save it for there.
• If there are two critical points where the slope is 0 -let's call them f(a) and f(b)- Can we always make the assumption that f''((a+b)/2) = 0? In other words, can we assume that the second derivative of a function at the average x value of two critical points always be 0? I'm just wondering because all graphs I've seen Sal use appear to have this property.
• Interesting question – I was sure this wasn't true, but it was harder than I expected to come up with a counter example.

Try this:
`f(x) = sin(eˣ)f'(x) = eˣ•cos(eˣ)f"(x) = eˣ•[cos(eˣ) - eˣ•sin(eˣ)]`

`f'(x) = 0: eˣ•cos(eˣ) = 0 cos(eˣ) = 0 — since eˣ is never equal to zero eˣ = π/2 + n•pi — where n can be any integer x = ln(π/2 + c•pi) — where c is any non-negative integer`

So, if we take the average of the first two of these and plug them into the equation above for f"(x):
`a = ln(π/2) b = ln(3π/2) (a + b) / 2 ~= 1.00088885 f"(1.00088885) ~= -5.50761219357`

I suspect that the graphs you are talking about are third order (cubic) polynomials, which thus have a constant third derivative. I believe that this constant rate of change in the second derivative leads to your observation.

Based on this, if you play around with fourth order polynomials I think you will find that your rule doesn't hold true for them either ...

EDIT:
Checked for `g(x) = -x⁴ + x²` – I get h"[(a+b)/2] = 0.5 ...
• What is the the general purpose of the inflection point and the second derivative (Calculus 1)?

What are the limitations of the second derivative?

One that comes to mind, though I don't fully understand, is I tried to use f'' to to find a local max or min, but it didn't work...

Thanks
• The second derivative is one of the higher order derivatives. One example is the second derivative of the position function, s(t), which is equivalent to taking the derivative of the velocity function, v(t). The result of doing so will yield the acceleration function, a(t). This especially useful in problems such as those concerning the gravitational forces of the earth and the moon, and finding the ratio.
To find the local maximum and minimum, I believe you must take the first derivative of the function f(x), and then set this derivative equal to zero. For example,
if f(x) = x^2-4x
then f'(x) = 2x-4
When you set this equal to zero,
2x-4 = 0
x=2
Substitution of this x value into f(x) will produce the corresponding y value, which will be -4.
Thus one can conclude that there is a minimum of -4, at x=2
• " f''(x)>0 means the slope of f(x) is increasing "
is the meaning of that if the slope is positive it become more positive and if the slope is negative it become more negative??
• First half is correct, but if the slope is negative and f"(x) > 0, then the slope still increases – this means it becomes less negative, i.e. becomes closer to zero.
• I have a general question, so i learned that in an inflow/ outflow problem, when it asks for the maximum amount, you graph the rate and the constant inflow and find the intersection. Can someone explain the reason behind this?
• So if the 2nd derivative is negative that doesnt necessarilly mean that the 1st derivative is also negative, but that it becomes less positive, or more negative?
(1 vote)
• Yep! That's correct. The second derivative being negative only implies that the first derivative is decreasing (becoming less positive or more negative)