# Inflection Point Without Sign Change

• Qube
In summary, the points of inflection of y can be found by determining where the concavity of the graph changes, specifically where the sign of the second derivative changes. While Wolfram Alpha may list -sqrt8 as an inflection point, it is important to also include points where the second derivative is zero and check if the sign changes. It is not enough to just list all points where the second derivative is zero.
Qube
Gold Member

## Homework Statement

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

## Homework 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:
Qube said:

## Homework Statement

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

## Homework 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.
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.
Qube said:
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

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?

Qube said:

## Homework Statement

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

## Homework 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

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.

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

Qube said:
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?
No, which is the conclusion you reconsidered below.

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

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?

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.

Qube said:
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?
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, ∞).

Qube said:
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?

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.

## 1) What is an inflection point without sign change?

An inflection point without sign change is a point on a curve where the curvature changes from convex to concave or vice versa, but the direction of the curve does not change. This means that the slope of the curve at the inflection point is zero, but the curve does not cross the x-axis.

## 2) How is an inflection point without sign change different from a regular inflection point?

An inflection point without sign change differs from a regular inflection point in that the curve does not change direction at the inflection point. In a regular inflection point, the curve changes from increasing to decreasing or vice versa.

## 3) What causes an inflection point without sign change?

An inflection point without sign change is caused by a change in the concavity of the curve without a corresponding change in the direction of the curve. This can happen when the second derivative of the function is equal to zero at that point.

## 4) How do you find an inflection point without sign change?

To find an inflection point without sign change, you can take the second derivative of the function and set it equal to zero. Solve for the variable to find the x-coordinate of the inflection point. Then, plug this value into the original function to find the y-coordinate.

## 5) What is the significance of an inflection point without sign change?

An inflection point without sign change can provide useful information about the shape and behavior of a curve. It can indicate a change in the rate of change of the function or a point of transition between different behaviors. It can also be used to find the maximum or minimum value of a function.

• Calculus and Beyond Homework Help
Replies
1
Views
604
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Precalculus Mathematics Homework Help
Replies
34
Views
2K
• Calculus and Beyond Homework Help
Replies
18
Views
2K
• Calculus
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
11
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
2
Views
1K