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## AP®︎ Calculus BC (2017 edition)

### Course: AP®︎ Calculus BC (2017 edition) > Unit 6

Lesson 8: Finding inflection points & analyzing concavity- Analyzing concavity (algebraic)
- Inflection points (algebraic)
- Mistakes when finding inflection points: second derivative undefined
- Mistakes when finding inflection points: not checking candidates
- Analyzing the second derivative to find inflection points
- Analyze concavity
- Find inflection points
- Concavity review
- Inflection points review

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# Inflection points (algebraic)

Sal analyzes the points of inflection of g(x)=¼x⁴-4x³+24x² by looking for values where the second derivative g'' changes signs.

## Want to join the conversation?

- 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?(13 votes)
- 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.(29 votes)

- 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.(6 votes)
- 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(5 votes)

- 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?(3 votes)- 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.(6 votes)

- can you have more than one inflection point?(3 votes)
- Certainly. sin(x) and cos(x) both have infinitely many inflection points.(5 votes)

- 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.(3 votes)
- 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.(4 votes)

- 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.(2 votes)- 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.(4 votes)

- 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?(2 votes)- 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.(3 votes)

- What happened when we check point of inflection

with 3rd or 4 th derivative?(3 votes) - Here is a nice
**shortcut**for finding out the inflection points. After you set f''(x) = 0, find the roots in the form (x - 1)(x + 2)². Now, look look at the**multiplicity**(exponent) of each bracket/root. If the multiplicity is even, then it's not an inflection point. If it's odd, then it must be an inflection point.

For proof, let's take the function: f''(x) = x³ + x² - 5x + 3.**Step 1**: Make f''(x) = 0

x³ + x² - 5x + 3 = 0

(x + 3)(x - 1)² = 0*Using my tip, we expect x = -3 to be the only inflection point, since it's the only one with an odd multiplicity.***Step 2**: Find where f''(x) changes signs

(-∞, -3) < 0

(-3, 1) > 0

(1, ∞) > 0**Step 3**: State the inflection points

Since f''(x) only changes signs at x = -3, x = -3 is an inflection point.*Try it yourself in your next exercise :)**I hope this helps!*(2 votes) - It would have been simpler to find the sign of the g(x)'' by plugging into (x-4)^2 instead of 3x^2 -24x + 48(2 votes)
- EDIT

Never mind, no difference in which is plugged into so probably better to use the simpler one.

You're not wrong, just usually good practice to use the original g''(x)(1 vote)

## Video transcript

- [Voiceover] Let G of X equal
one fourth X to the fourth, minus four X to the third
power, plus 24 X squared. For what values of X does the graph of G have an inflection point? Or have a point of inflection? So let's just remind ourselves what a point of inflection is. A point of inflection is
where we go from being con, where we change our concavity. Or you can say where our second derivative G prime of X switches signs. Switches, switches signs. So let's study our second derivative. In order to study or
secondary, let's find it. So we know that G of X
is equal to one fourth X to the fourth, minus
four X to the third power, plus 24 X squared. So given that, let's
now find G prime of X. G prime of X is going to be equal to, I'm just going to apply the
power rule multiple times. Four times one fourth is just one. I'm not going to write that one down, it's going to be one times X
to the four minus one power. So four to the third power, minus three times four is 12, X to three minus one power, X to the second power,
plus two times 24, 48. X to the two minus one,
or X to the first power, I can just write that as X. So there you have it. I have our first derivative, now we want to find our second derivative. G prime prime of X is just the derivative of the first derivative
of the first derivative with respect to X, and some
more of the power rule, three X squared minus 24 X to the first, or just 24 X, plus 48. So let's think about where
this switch is signed. And this is, this is
a continuous function, is going to be defined for all Xs, so the only potential candidates of where it could switch signs are when this thing equals zero. So let's see where it equals zero. So let's set that equal to zero. Three X squared minus 24 X plus 48 is equal to zero, let's see, everything is divisible by three so let's divide everything by three. So you get X squared minus eight X, plus 16, plus 16 is equal to zero. And let's see, can I factor this? Yeah, this would be X minus
four X times X minus 4. Or just you do this as
X minus four squared is equal to zero, or X
minus four is equal to zero. So, or where X equals four. So G prime prime of four is equal to zero. So let's see what's happening
on either side of that. Let's see if G, if we're actually, if we're actually switching signs or not. So let me draw a number line here. And so this is, so this is two, three, four, five, and I could keep going. And so we know that something interesting is happening right over here, G prime prime of four is equal to zero, G prime prime of four is equal to zero. So let's think about what
the second derivative is when we are less than four. And so, actually let me just
try G prime prime of zero, since that will be easy to evaluate. G prime prime of zero, well it's just going to be equal to 48. So when we are less than four, our second derivative, G
prime, the second derivative, is greater than zero. So we're actually going
to be concave upwards over this interval to the left of four. Now let's think about
to the right of four. Two, use a different color. So what about to the right of four? And so let me just evaluate, what would be the easy thing to evaluate? Well I could evaluate G prime, why not do, or of the second derivative, G prime prime, I should say, of 10? So, G, I'll do it right over here, Let me do it, well, I'm running a little bit of
space, so I'll just scroll down. So G prime prime of 10 is going
to be equal to three times 10 squared, so it's 300 minus 24 times 10. So minus 240, plus 48. So let's see, this is 60. This is, so 300 minus 240 is 60 plus 48, so this is equal to 108. So it still positive. So on either side of
four, G prime prime of X is greater than zero. So even though even though
the second derivative at X equals four is equal to zero, on either side we are concave upwards. On either side, the second
derivative is positive. And so, and that was the
only potential candidate. So there are no, there are no values of X for which G has a point of inflection. X equals four would have been a value of X at which G had a point of inflection, if we switch, if the secondary derivative switched signs here, if it
went from positive to negative, or negative to positive, but it's just staying
from positive to positive. So this second derivative was positive, it just touches zero right here, and then it goes positive again. So we're going back to the question, for what X values does the graph of G have a point of inflection? No X values. I'll put an exclamation
mark there just more drama.