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.

### Course: AP®︎ Calculus BC (2017 edition)>Unit 6

Lesson 8: Finding inflection points & analyzing concavity

# 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?

• if between infinity and -1 the graph is concave down would that also mean it has a relative maximimum
• No. Counterexample - the natural log function - or log function for any base bigger than 1. Always increasing so no relative max, yet always concave down.
• How does concave upward relate to convexity?
• 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.
• 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!
• One use in math is that if `f"(x) = 0` and `f"'(x)≠0`, then you do have an inflection point. Unfortunately, there are cases where `f"'(x)=0` that are inflection points so this isn't always useful, but if the third derivative is easy to determine (e.g. for a polynomial) then it is worth trying.

The only other use I know of is in physics, where it called the "jerk":
https://en.wikipedia.org/wiki/Jerk_(physics)

ADDENDUM: Another thing that higher order derivatives are used for is to calculate higher order terms in a Taylor series.
• Question:

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?
• 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.)
• 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?
• 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.
• 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?
• Note that the point you seem to be talking about is where the slope is zero (`f'(x)=0`), not the function! That is a local minimum.

Sal didn't mention it because he is focusing on concavity rather than extrema(maxima/minima) in this video.
• also, -1 and 1 would be points of inflection right?
• Yup! g"(x) = -12*(x+1)*(x-1) = 0 for x = {-1,1}
• 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)
• 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.
• Why aren't we able to use the first derivative instead of the second derivative?
(1 vote)
• 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.
(1 vote)
• at how do you label the intervals of concave up and down and where it increases and decreases
(1 vote)