Question 1

I don't understand why the second derivative equals zero when it's not actually an inflection point. The function is clearly defined there, so why is it that there's no inflection point?

Accepted Answer

Imagine an upward-facing parabola centered at the origin. Now imagine that instead of a smooth curve continuing through x = 4,  f(x) actually transforms into a linear function  (i.e. a straight line) at f(4), but then curves upward once again at f(5). From X = 4 to X = 5, the first derivative (i.e. the slope of f(x)) would not change, and thus the second derivative would be 0. However, f(x) never actually switched concavity. This is how we can have f''(x) = 0 without it actually being an inflection point.

Question 2

Can we apply the logic that if f''(x) = 0, which meant we had a candidate for an inflection point, then if f'''(x) > 0, our point is an inflection point with a function that was concave downward and is now approaching concave upward or if f'''(x) < 0, our point is an inflection point with a function that was concave upward and is now approaching concave downward, or if f'''(x) = 0, our point is not an inflection point. I based this logic of the second derivative test where if f'(x) = 0 and f''(x) isn't 0, then we are either at a local minimum or maximum. It seems to works for Sal's problem since f'''(x) yields 6x - 24 and plugging in 4 yields f'''(x) = 0, making it not an inflection point. However, I just want to make sure this works for all cases before using it.

Accepted Answer

You are almost correct - that was really good thinking!

I'd never heard of this but your logic seemed sound, so I poked around and found a couple of references to a third derivative test including this:
https://en.wikibooks.org/wiki/A-level_Mathematics/OCR/C1/Differentiation#Locating_And_Evaluating_Stationary_Points

Note that this only works for "stationary points" and therefore does *not* apply to all inflection points.  Specifically, if f"'(c) = 0, then c may or may not be an inflection point.

For an example:`
f(x) = x⁶ + 6x⁵
f'(x) = 6x⁴(5 + x)
f"(x) = 30x³(4 + x)
f'"(x) = 120x²(3 + x)`

At x = 0 all of these functions including the third derivative equal zero, however x=0 *is* an inflection point,  as you can see if you look at the graphs:
https://bit.ly/2FIcpkV

EDIT:
And it turns out this has been discussed elsewhere:
https://www.khanacademy.org/math/ap-calculus-ab/ab-diff-analytical-applications-new/ab-5-8/v/calculus-graphing-with-derivatives-example?qa_expand_key=kaencrypted_3e993c57b216aab2be90dbb22ea0013b_ff695775774c2d2d6a888fac5e41ccb866550c328b7a7ed94251d581a909b70c1c3404ff911953963954e7d341f643ee10ae44d3c0061222044708d0ccbf8022f09d785d64ad2f781ad8b5656011d89db4c166ed8433be8176ddccc5b02b58e36d9226a612f1ccfedabc730a9cc47583c3cd4ff56634a75a78fdb885de206054911237fe570b246b365f9c4f7705a9c1f580a54eeb4ab5d4fc502835d12bb10d

Question 3

I am confused.
I graphed this function in desmos. The graph looks like a quadratic function. Concave upward centred at 0.
Which means slope change sign after 0. left side is - right side is +. Am I right?
It is a conflict explanation which Sal mentioned. 
Can you explain, please?

Accepted Answer

Sal said that the function has no point of inflection, which is where the concavity of the function changes. As you observe, the function is concave up everywhere, i.e. the concavity doesn't change.

Question 4

can you have more than one inflection point?

Accepted Answer

Certainly. sin(x) and cos(x) both have infinitely many inflection points.

Question 5

Does anyone know whether there is a purpose to the use of a sample x value to establish the sign of a derivative for each interval? One value together with the multiplicity of the zeros should be enough. In this case I would not gave gone beyond the perfect square of x - 4. Initially I considered the didactic reasoning of explicitly confirming the sign for each interval, but given the numerous examples without any mentioning of multiplicity I am starting to wonder whether I am missing something.

Accepted Answer

The multiplicity is an interesting point.
Correct me if I'm wrong, but I believe that a factor of f''(x) with multiplicity k, or (x - c)^k, will touch but not cross the x-axis with even k, and cross the x-axis with odd k.
Thus, if (x - c)^k where k is even is a factor of f''(x), then there is no inflection point at x = c.
However, Sal simply needs to provide a general solution to deal with all types of functions, not just relatively simpler polynomial functions, so the zeros of f''(x) are found, and the intervals are tested, simply because it is easier to analyze all sorts of matters, like intervals of increase/decrease, global/local extrema, concavity, inflection points, etc. if the derivatives' signs are found.
Hope that I helped.

Question 6

Let's say that we have the second derivative of a function, and it has many roots.

I have noticed that if a root appears an even number of times (ex: double root), then the  original function (f(x)) does not change concavity at the point, but if it appears an odd number of times, then the original function does change concavity.

For example, if f''(x)=x^4*(x-3)^3*(x-5)^2*(x-1), then the original function will change concavity at x=3 and x=1, but not at x=0 or x=5.

Does this method always work, and if so, then why? Because if it does work, one does not need to analyze where the second derivative is positive and negative to find inflection points, but only needs to count how many times each root appears.

Accepted Answer

Remember that concavity relates directly to the sign (positive or negative) of the second derivative. In your example, notice what happens when we nudge the x-value through x=3 or through x=5.

If x is slightly less than 3, then (x-3) is negative, and so is (x-3)³. Then we pass through x=3 and get a second derivative of 0, and then x is slightly greater than 3, so x-3 is positive. Notice also that none of the other factors affect the sign change so long as we're in the vicinity of x=3.

On the other hand, (x-5)² is always positive or 0, so it doesn't affect the sign of the second derivative at all. This corresponds exactly to your observation.

So what you've discovered is a nice trick that works when the second derivative is a polynomial that factors completely over the real numbers (no complex roots), as in your example.

Question 7

HELP!  Below I said... Where f(x)= 1/12x^4 -1/3x^3 +1/2x^2 ==> f " (x)= x^2 -2x +x = (x-1)(x-1). No inflection point. For any polynomial in two coordinate space where the solution for x is two integers of the same value and the same sign and there are no other solutions, there is no value for x for which the graph of the function has an inflection point. That's so cool! Is this called a rule or a theorem? What's it's name?

I then graphed this function and it looks like an asymmetrical parabola with a slight but noticeable hump in the middle.  According to the graphing calculator I used, it does have inflection points where the slope changes signs.  Visually it looks like it has inflection points.

Sal are you giving us bad information?

Accepted Answer

An inflection point is where f(x) changes it's concavity, in the function f(x)= 1/12x^4 -1/3x^3 +1/2x^2 the graph of the function is continually concave upwards, so by graphical analysis only it does not have inflection points.

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 5

Inflection points (algebraic)

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