AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 5Lesson 7: Determining concavity of intervals and finding points of inflection: algebraic
- 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
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.
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- 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:
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:
And it turns out this has been discussed elsewhere:
- can you have more than one inflection point?(3 votes)
- Certainly. sin(x) and cos(x) both have infinitely many inflection points.(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.(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)
- 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)
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)
- If the second derivative is 0 and is not an inflection point, what point that is? Referring to the example Sal did in the video.(1 vote)
- Whenever you come across situation like this just sketch the second derivative. What this will do is tell you where the graph is concave up or down. For all the points the second derivative is positive the graph is concave up and for any points the graph is negative it is concave down.
Now if you think about this visually if a graph concave down then you can say it has maximum(assuming dy/dx=0 at a certain point) as the positive grad is decreasing and will go negative as the second derivative is negative.
Similarly if the graph is concave up then there minimum as the gradient is increasng at an increasing rate assuming dy/dx =0 at certain point x.
With inflection one part of the graph is concave up and the other concave down.
You will need take a look at the graph graph of the second derivative if is equal to zero to determine if has inflection maximum or minimum(2 votes)
- [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.