# B I have a basic question about inflection points...

1. Jul 12, 2016

### leright

My question involves the discussion in this thread, which I contributed to back in 2009.

My question involves the function f(x) = x^(2n) for n greater than 1. Take f(x) = x^4 for example. You differentiate twice and you get f''(x) = 12x^2. Plug in x= 0 into the second derivative and you get f''(0) = 0. This implies an inflection point at x = 0 as I was taught in my introductory calculus course. However, we all know that f(x) = x^(2n) is a parabola, which does not change concavity, including at x = 0. Is this function an exception to the rule that I was never taught in school?

Thanks for any insight you can provide. God I feel so silly for asking this...this topic came up recently and I recalled the comments I made in the above thread...I guess I never really looked into it.

2. Jul 12, 2016

### FactChecker

The definition of an inflection point is where the function changes from concave to convex or vise versa. This means that f'' changes sign. So naturally you will look for them where f''=0. But that doesn't mean they are all inflection points. In fact, the example you are looking at has f'' ≥ 0. So it is not an inflection point.

3. Jul 13, 2016

### leright

Thanks fact checker. It is my understanding that a second derivative of 0 implies a change in concavity (for obvious reasons) or constant first derivative, as in a straight line. Those are really the only two possibilities I can think lf. What exactly is happening at x=0 for the function x^4 then? Clearly its first derivative is not constant and it is not an inflection point.

Thanks.

4. Jul 13, 2016

### leright

It just seems like there is something deeper going on here. I actually plotted the function in maple and zoomed in and it appears strangely flat in the region around x=0.

5. Jul 13, 2016

### FactChecker

No. Where f''>0 it is concave upwards and where f''<0 it is concave downwards. To find inflection points, you look for points where f''=0 so that you can check whether f'' changes sign there. The function you are considering is always concave upwards.

6. Jul 13, 2016

### leright

Thanks. I am fully aware of that but my question relates to the behavior of the function at x = 0. If it isnt an inflection point, or constant slope at that point then why is the second derivative 0? That means that the first derivative isnt changing, right? However that clearly is not the case here.

7. Jul 13, 2016

### leright

I am guessing that the second derivative of x^4 evaluated at x=0 is actually non-zero. I think the math might just break down. Correct me if Im wrong. Thanks.

8. Jul 13, 2016

### FactChecker

It just means that the first derivative is not changing at that very point. It is changing on both sides of that point. You should look at a graph of it.

9. Jul 13, 2016

### Staff: Mentor

No. Why would you think that?
If f(x) = x4
f'(x) = 4x3, and
f''(x) = 12x2.
f''(0) = ?

10. Jul 13, 2016

### pasmith

The actual statement is: "A twice differentiable function of one variable is concave upwards on an interval if and only if its second derivative is non-negative there." (Wikipedia, property #6.)

A non-negative quantity can be zero.

Similarly a twice differentiable function of one variable is concave downwards on an interval if and only if its second derivative is non-positive there.

A non-positive quantity can be zero.

Thus, in practical application of the test, the second derivative being non-zero is conclusive: the function is concave upwards or downwards depending on the sign. But the second derivative being zero is inconclusive and further investigation is required: the function could be concave upwards or concave downwards or there could be a point of inflection.

Last edited: Jul 13, 2016
11. Jul 13, 2016

### FactChecker

You really should learn to plot these functions, f, f', f'' to see what is going on. Either plot or calculate some values around the point of interest.

12. Jul 13, 2016

### leright

Thanks guys. I appreciate your patience.

I realize that the mathematical result is f''(0) = 0, which implies zero rate of change of the first derivative @ x = 0. However, clearly the function's first derivative is changing at x = 0.

I can easily visualize the plots of x^4, 4x^3, and 12x^2 and I understand the interpretation of the "rules" and why the rules are the way they are. That's not my issue. A second derivative of 0 at a point implies the following: the point is an inflection point [(point where function transitions from being concave up (positive second derivative) to concave down (negative second derivative)] OR in a region where the curve is perfectly straight OR the function is a straight line.

Now, I am a conceptual person. When I think second derivative I think the rate at which the first derivative is changing. When the rate at which the first derivative is changing is ZERO at a given point it is implied that the slope is constant for a brief period. This is not really the case for x^4. The first derivative is 4x^3. Going a infinitesimally small step in either direction of x = 0 for 4x^3 gives a non-zero first derivative. That means that at x = 0 the first derivative is indeed CHANGING. It has a non-zero rate of change at x = 0. Why then is the second derivative zero when evaluated at x = 0 for this function?

I hope I don't come across as being frustrated, because I am not. I really appreciate your help. I'm just trying to clear up some subtle confusion I have.

13. Jul 13, 2016

### leright

Thanks. I understand that a second derivative being zero when evaluated at a point is considered both non-negative and non-positive. There is clear ambiguity here. I am just interpreting this as a rate of change of the slope (first derivative). If a second derivative evaluated at a point is zero and it is bot an inflection point then the function must be "flat" about this point. However, it is clear the x^4 does not exhibit this behavior.

14. Jul 13, 2016

### leright

Right. But the first derivative is f'(x) = 4x^3. the first derivative at x = 0 is 0 but go an infinitesimally small step to the left or right of x = 0 on the first derivative function and you get a different value of the first derivative. Clearly, then, the first derivative IS changing at x = 0. Do you see my point?

15. Jul 13, 2016

### FactChecker

If I through a ball in the air, it is going up; at some point it stops; then it is going down. At the point where it stops, it is not changing. That is even though an infinitesimally small step before or after you get a different value of height. A similar thing is happening to the values of f' at x=0.

16. Jul 13, 2016

### leright

If you plot f'(x) = 4x^3 it is clear that its slope is zero at x = 0. But I am just having trouble seeing that characteristic in the plot of f(x) = x^4. What is it about the function x^4 that makes its second derivative zero there? Would that imply that there are two points infinitesimally close together that have the same slope? Is that accurate? I am 100% convinced that you guys are right...I am just having trouble with this small detail.

Last edited: Jul 13, 2016
17. Jul 13, 2016

### leright

https://math.dartmouth.edu/opencalc2/cole/lecture8.pdf

These notes describe this question at the end. I understand. What I don't understand is the interpretation of a second derivative evaluated to be zero at a point when the function does not have an inflection point at that point. What exactly is happening then at this point? The fact that f''(0) = 0 must describe some property of the function if it doesn't mean it is an inflection point in this case...

18. Jul 14, 2016

### Staff: Mentor

No, there is no property. Look at the graph of y = x4. Its shape is similar to that of y = x2.

Given y = f(x) = x4, then f'(x) = 4x3 and f''(x) = 12x2. The second derivative is positive at every x except x = 0. This means that if x < 0, f''(x) > 0, and if x > 0, f''(x) > 0, as well. This also means that the graph of f is concave upwards on both intervals, a fact that is obvious if you look at the graph of y = x4.

Since the concavity doesn't change, there is no inflection point. In this case, f''(0) = 0 has no significance.

19. Jul 14, 2016

### FactChecker

A plot of f(x) = x4, f', and f'' for x between -1 and 1 should make this clear. The black line is f, the red is f' and the green is f''. Clearly f becomes very flat at x=0 and curves up at both ends. That is confirmed by the green line, f'', being 0 at x=0 and positive at both ends. It is also clear that the red line, f', has no slope at x=0. f does not have an inflection point at x=0.

20. Jul 14, 2016

### leright

I have plotted these functions. I figured it was due to the flatness of x^4 near x=0. Is there a name for a point that has zero f" evaluated at that point but isn't an inflection point? Its significance is the flatness of x^4 compared to x^2 near x=0