Main content

## AP®︎ Calculus BC (2017 edition)

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

Lesson 6: Concavity & inflection points intro# Inflection points introduction

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.4 (EK)

, FUN‑4.A.5 (EK)

Inflection points are points where the function changes concavity, i.e. from being "concave up" to being "concave down" or vice versa. They can be found by considering where the second derivative changes signs. In similar to critical points in the first derivative, inflection points will occur when the second derivative is either zero or undefined. Created by Sal Khan.

## Want to join the conversation?

- Is it ever possible for an inflection point to be a maximum or minimum point?(28 votes)
- If you were asking only about inflection points then the above answer is right i.e, by Bryce.

But given that there is an interesting case,but before that it seems you may be under the impression that all points where the second derivative is 0 are inflection points. That's not true. For example, if f(x)=x^(4), then f′′(0)=0, but that's not an inflection point because f′′ does not change signs there: f′′ is positive on both sides of 0. And notice that that is an absolute minimum point. So if your question is whether a maximum or minimum point can occur where f′′ is 0, the answer is "yes". But that doesn't mean there's an inflection point there.(12 votes)

- So the slope over an inflection point must be
**undefined**, as slope = tan(Theta) and theta over here is 90* ??

But then why in the second and the third graph , it is not shown to be undefined ?(11 votes)- The slope of inflection point is not undefined, it can be any value, but its second derivative must be zero. The inflection point in tan(theta) occurs at theta = 0. At theta = 90 degrees, it is undefined as you say and so we can't stay anything about the slope at that point.(32 votes)

- could you say that an inflection point is given when the second derivative = 0?(12 votes)
- The second derivative must be 0 for an inflection point, yes; however, that is not by itself proof that you have an inflection point. You can have a second derivative that is 0 but not have an inflection point.

So, you need to establish that the second derivative is 0 at the point in question, AND the second derivative changes signs at that point (*you prove that by showing that the sign of the second derivative just before that point is different from the sign of the second derivative just after that point*).(22 votes)

- Is there a way to determine inflection points from first derivative? I think there should be some connection...(6 votes)
- Yes, but the method only works on some kinds of inflection points, so it is not reliable. Specifically, if the first derivative is 0 at some point, but that point is not a local max or a local min, then it is an inflection point.

Personally, I wouldn't recommend using this test because it does not always work. The reliable method for finding an inflection point is:

IF f''(c) = 0

AND

f''(c+ε) has a different sign than f''(c−ε). Where ε is an arbitrarily small constant.

Then f(x) has an inflection point at x=c.(11 votes)

- Is it correct to say that an inflection point is the point where the second derivative is zero?(3 votes)
- That is incorrect. It is a necessary, but not sufficient, condition that the second derivative be zero at an inflection point. The second derivative can be zero and yet you don't have an inflection point. For example, the second derivative of all straight lines is 0 at all points. However, there are no inflection points in a straight line.

It must also be the case that the second derivative just before the inflection point has a different sign than the second derivative just after the inflection point. If this condition is not met, even though you have a zero in the second derivative at that point, you do not have an inflection point.(7 votes)

- I teach AP Calculus. Need an answer to a question.

By definition, I teach my students that an inflection point exists when f"=0 and changes sign (that f ' must be differentiable). BUT the CollegeBoard has questions in which f" is DNE but changes sign and they call this an inflection point. I also see online in SOME definitions that f"=0 or DNE and changes sign refers to a point of inflection. Can you clarify whether f" DNE can actually produce an inflection point. And if so, what is a "real life example" of such a scenario. THANKS! :)(3 votes)- I think the answer depends on how rigorously "inflection point" is defined.

For a continuous piecewise function, it is possible, that, at the point where you change from one interval to the next that you have an abrupt change from positive to negative concavity and yet f'' does not exist.

I would regard such a point, if it is not an extremum, to be a point of inflection. But, I don't know whether professional mathematicians regard it as such.

f(x) = { x²-e² for x<e and ln (x) - 1 for x≥e

At x=e, the function is continuous, but not differentiable.

However, just before x=e, the function has a positive concavity, just after e it has a negative concavity, while f''(e) DNE. f(e) is not an extremum since the function is increasing, albeit at different rates, on both sides of f(e).

I would consider f(e) to be an inflection point. But, again, I am not a professional mathematician, so I am not sure they would concur with me or whether they require f''(c) = 0.(5 votes)

- I read some comments below and figured out that somehow
`f''(a) = 0`

but a is still**NOT**the inflection point of`f(x)`

. I can't come up with any example to demonstrate this statement. Can someone give me insight?(3 votes)- Here's an example: consider
`f(x)=x^4`

at`x=0`

. We can use the Power Rule to find`f"(x)=12x^2`

. Clearly`f"(0)=0`

, but from the graph of f(x) we see that there is not an inflection point at x = 0 (indeed, it's a local minimum). We can also see this by thinking about the second derivative, where we realize that`f"(x)>0`

for`x<0`

and`x>0`

. Therefore f(x) is concave up on either side of x = 0, and so it is not a point of inflection. Hope that helps!(5 votes)

- Can you have a point of inflection if the second derivative is 0 but the first derivative is not 0? If so, would this point be a critical point?(2 votes)
- The second derivative's zeros mark places where it is possible that the concavity changes sign (that is, where it is possible that the concavity switches from up to down or from down to up). They are not necessarily critical points, though they can be.(4 votes)

- So is a point of inflection always at f''(x)=0?(3 votes)
- No they can also occur at points where
`f"(x)`

is undefined.

An inflection point requires:

1) that the concavity changes and

2) that the function is defined at the point.

You can think of potential inflection points as critical points for the first derivative — i.e. they may occur if`f"(x) = 0`

**OR**if`f"(x)`

is undefined. An example of the latter situation is`f(x) = x^(1/3)`

at`x=0`

.

Relevant links:

http://depts.gpc.edu/~mcse/CourseDocs/calculus/concavity-inflectionOct12.pdf

https://math.stackexchange.com/questions/402459/an-inflection-point-where-the-second-derivative-doesnt-exist(2 votes)

- Is the critical point the same as the stationary point? because im from the UK(1 vote)
- A stationary point is one type, but not the only type, of critical point.

A critical point is a point where the function is defined and where the first derivative is EITHER equal to 0 OR fails to exist.

A stationary point is a critical point where the first derivative is equal to 0.

Thus, the type of critical point where the first derivative fails to exist is not a stationary point.(5 votes)

## Video transcript

If you were paying close
attention in the last video, an interesting question might
have popped up in your brain. We have talked about
the intervals over which the function is
concave downwards. And then we talked
about the interval over which the function
is concave upwards. But we see here that there's
a point at which we transition from being concave downwards
to concave upwards. Before that point, the
slope was decreasing, and then the slope
starts increasing. The slope was
decreasing and then the slope started increasing. So that's one way to look at it. Right here in our
function we go from being concave downwards
to concave upwards. When you look at our
derivative at that point, our derivative went from
decreasing to increasing. And when you look at our second
derivative at that point, it went from being
negative to positive. So this must have some type of
a special name you're probably thinking. And you'd be thinking correctly. This point at which we
transition from being concave downwards to concave upwards,
or the point at which our derivative has
a extrema point, or the point at which our
second derivative switches signs like this, we call it
an inflection point. And the most typical
way that people think about how could you
test for an inflection point, it's a point, well,
conceptually, it's where you go from being
a downward opening u to an upward opening u. Or when you go
from being concave downwards to concave upwards. But the easiest
test is it's a point at which your second
derivative switches signs. So in this case, we went
from negative to positive. But we could have also
switched from being positive to negative. So inflection point, your second
derivative f prime prime of x switches signs. Goes from being
positive to negative or negative to positive. Switches signs. So this is a case where we
went from concave downwards to concave upwards. If we went from concave
upwards to concave downwards, like that, then this inflection
point up until that point, the slope was increasing. So the second derivative
would to be positive. And then the slope
is decreasing, so your second derivative
would be negative. So here your second
derivative is going from positive to negative. Here your second
derivative is going from negative to positive. In either case, you are talking
about an inflection point.