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

Lesson 7: Points of inflection

# Inflection points introduction

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?
• 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.
• 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 ?
• 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.
• could you say that an inflection point is given when the second derivative = 0?
• 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).
• Is there a way to determine inflection points from first derivative? I think there should be some connection...
• 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.
• Is it correct to say that an inflection point is the point where the second derivative is zero?
• 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.
• 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! :)
• 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.
• 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?
• 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!
• 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?
• 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.
• So is a point of inflection always at f''(x)=0?
• 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.