# Inflection Point Without Sign Change

1. Nov 2, 2013

### Qube

1. The problem statement, all variables and given/known data

Find the points of inflection of y. (See picture for y).

https://scontent-b-dfw.xx.fbcdn.net/hphotos-prn2/v/1385610_10200987969443169_488083853_n.jpg?oh=d014eb5324c27653f6f94d43a08957b2&oe=5277D415

2. Relevant equations

Inflection points occur where the concavity of the graph changes. This means the sign of f''(x) changes.

However, the sign of f''(x) does not change around -sqrt8. I confirmed this with Wolfram Alpha by having it take the second derivative and having it find the value of the second derivative at points around the points that would zero the second derivative.

Nonetheless, Wolfram Alpha insists that -sqrt8 is a point of inflection.

What's up?

WA page for reference below. Note that it says -sqrt8 and sqrt8 are inflection points.

http://www.wolframalpha.com/input/?i=points+of+inflection+of+1+-+6/x+++8/x^3

This is the computed second derivative.

http://www.wolframalpha.com/input/?i=second+derivative+1+-+6/x+++8/x^3

We can see this second derivative is positive for BOTH x = -3 and x = 1. Note that -3 < -sqrt8 < 1.

http://www.wolframalpha.com/input/?i=-(12+(-8+x^2))/x^5+at+x+=+(-3)

http://www.wolframalpha.com/input/?i=-(12+(-8+x^2))/x^5+at+x+=+1

Last edited: Nov 2, 2013
2. Nov 2, 2013

### Staff: Mentor

Actually, the sign does change at this point.

Notice that f''(x) = -12/x3 + 96/x5 = 12(8 - x2)/x5.

The are three points of interest here: x = -2√2, x = 0, and x = 2√2.
If x < -2√2, the numerator of f''(x) is negative, and so is the denominator, so f'' < 0.
If -2√2 < x < 0, the numerator is positive, and the denominator is negative, so f'' > 0. Although it's hard to see in your first WA graph below, the two inflection points are marked in red.

3. Nov 2, 2013

### Qube

Alright, so even though x = 0 is not an inflection point since it does not exist on the domain of the original function I should include it anyway in my number line analysis of f''(x) just to be safe?

4. Nov 2, 2013

### Ray Vickson

What's up is that the second derivative DOES change sign at both x = -√8 and x = +√8. I cannot figure out why you think otherwise.

5. Nov 2, 2013

### Qube

I think the interval I tested was too big;I excluded 0 which is not an inflection point.

6. Nov 3, 2013

### Staff: Mentor

No, which is the conclusion you reconsidered below.

Correct.

7. Nov 3, 2013

### Qube

What I'm saying is that I should probably include points which are definitely not inflection points just as I include points which may or may not be inflection points in my analysis because it's a slippery slope. Not all points at which the second derivative is zero odds going to be an inflection point and why exclude some instead of others in considering the sign change of the second derivative?

8. Nov 3, 2013

### Staff: Mentor

The key thing about whether a point is an inflection point isn't necessarily that the second derivative is zero - it's whether the second derivative changes sign from one side to the other.

Here's a simple example, with f(x) = x4.

f'(x) = 4x3, and f''(x) = 12x2. We can see that f'' = 0 for x = 0, so we would be tempted to say that there is an inflection point at x = 0. This would be incorrect, though, as f'' doesn't change sign around x = 0, as 12x2 ≥ 0 for all real x.

If you sketch a graph of f, you'll see that there are no inflection points. The shape of the graph is similar to that of y = x2, but isn't parabolic.

So it isn't enough just to list all of the numbers x for which f''(x) = 0. These are just candidates for inclusion in the set of inflection points. You should check each one to see whether f'' changes sign going from one side to the other. Also, as with critical points, they have to be points on the graph of the function.

9. Nov 3, 2013

### vela

Staff Emeritus
The reason you ran into a problem here is because you didn't recognize the fact that the function blows up at x=0. Because of the singularity, you can't assume the sign of f'' remains the same as you cross x=0.

The domain of f isn't ℝ; it actually consists of the two intervals (-∞,0) and (0,∞). After identifying the points where f''(x)=0, you end up with four intervals you need to test: (-∞, -√8), (-√8,0), (0,√8), and (√8, ∞).

10. Nov 3, 2013

### Ray Vickson

I'm not sure what your "slippery slope" refers to, but if you want to find all the inflection points of $f(x)$ (in an open interval) you start with the points having $f{''}(x) = 0$ then examine them further. (Note that an inflection point does not need to be stationary; for example, x = 0 is an inflection point of the function $x^3 + x$, but it is not a stationary point.)

Anyway, for any given value $p$ giving $f{''}(p) = 0$, how can you tell if it is an inflection point or not? One way is to look at nearby points $p-r$ and $p+r$ for small r > 0 and see if $f{''}$ changes sign between them. Another way that sometimes (not always) works is to look at the third derivative $f{'''}(p)$. If it is nonzero it will tell you how the second derivative behaves; if it is also zero, you might look at the 4th derivative, etc., etc.