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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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# Analyzing concavity (algebraic)

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.4 (EK)

, FUN‑4.A.5 (EK)

, FUN‑4.A.6 (EK)

Sal finds the intervals where g(x)=-x⁴+6x²-2x-3 is concave down/up by finding where its second derivative, g'', is positive/negative.

## Want to join the conversation?

- if between infinity and -1 the graph is concave down would that also mean it has a relative maximimum(9 votes)
- No. Counterexample - the natural log function - or log function for any base bigger than 1. Always increasing so no relative max, yet always concave down.(17 votes)

- How does concave upward relate to convexity?(4 votes)
- Concavity and convexity are opposite sides of the same coin. So if a segment of a function can be described as
*concave up*, it could also be described as*convex down*. We find it convenient to pick a standard terminology and run with it - and in this case concave up and concave down were chosen to describe the direction of the concavity/convexity. I hope that answers your question.(16 votes)

- If second derivatives can be used to determine concavity, what can third or fourth derivatives determine? At4:20in the video, the second derivative is found to be: g''(x) = -12x^2 + 12.

For example, the function given in the video can have a third derivative g'''(x) = -24x. I realize this constitutes the slope of the tangent line of the second derivative, but does it have any applications?

Any insight is much appreciated!(5 votes)- One use in math is that if
`f"(x) = 0`

and`f"'(x)≠0`

, then you**do**have an inflection point. Unfortunately, there are cases where`f"'(x)=0`

that**are**inflection points so this isn't always useful, but if the third derivative is easy to determine (e.g. for a polynomial) then it is worth trying.

The only other use I know of is in physics, where it called the "jerk":

https://en.wikipedia.org/wiki/Jerk_(physics)

ADDENDUM: Another thing that higher order derivatives are used for is to calculate higher order terms in a Taylor series.

https://www.khanacademy.org/math/ap-calculus-bc/bc-series#bc-taylor-series(8 votes)

- Question:

When sal solves for the second derivative, he then takes it and makes it equal to zero. He then solves for x. He finds that x is equal +-1. How come it's not just 1? Why is it +- 1?(3 votes)- The second derivative is a quadratic equation, so it must have two solutions.

g"(x) = -12*x^2 + 12

-12*x^2 + 12 = 0

-12*x^2 = -12

x^2 = 1

(-1)^2 = 1 and 1^2 = 1

So, both -1 and 1 must be solutions to g"(x) = 0. (We cannot just use the principal square root of x as a solution.)(5 votes)

- How do you know that if g"(-2) is negative over the whole interval? How do I know it's not crossing through 0 or discontinuous?(2 votes)
- That is what the checks were for. Essentially we checked to see where the function g'' becomes 0. This shows us critical points. These are locations where the function either reaches a local minimum/maximum, or has an inflection point.

You would need to potentially check for discontinuity, however, a discontinuity would mean little for the actual concativity, unless it was potentially at an inflection point. For this simpler function though we know polynomial functions like this are continuous for all real values. And it's derivatives are polynomials as well, so they are also continuous. As for crossing through 0 this has nothing to do with concativity, but checking for where the function = 0 (for the second derivative), tells us special information. In the same way the first derivative tells us where the slope = 0 which implies there is a "hill" or "valley" and thus either concave down or up.(5 votes)

- Between -0.5 and 0.5 in the graph, isn't there a point where clearly the function should reach a zero before becoming positive? Why did Sal ignore it? Or doesn't it mean anything?(2 votes)
- Note that the point you seem to be talking about is where the slope is zero (
`f'(x)=0`

), not the function! That is a local minimum.

Sal didn't mention it because he is focusing on concavity rather than extrema(maxima/minima) in this video.(3 votes)

- also, -1 and 1 would be points of inflection right?(2 votes)
- Yup! g"(x) = -12*(x+1)*(x-1) = 0 for x = {-1,1}(2 votes)

- In1:30you say that if the first derivative is increasing, then the 2nd derivative is positive. Is it true the other way around? Say, my second derivative is positive, then the first derivative is increasing?(1 vote)
- It should be. The second derivative tells us about the rate of change of the first derivative. If the rate of change is positive, it means the quantity must be increasing.(3 votes)

- what do you do if you encounter a function like this and you need to find its points of inflection?

f(x) = x^6 + 3*x^5 + x^4 + 7*x^3 + x^2 + 11*x + 1

f'(x) = 6*x^5 + 15*x^4 + 4*x^3 + 21*x^2 + 11

f''(x) = 30*x^4 + 60*x^3 + 12*x^2 + 42*x

How could you possibly solve these two polynomials? I am referring to the first and second derivatives of f(x). They are not as easy to solve as simply using the quadratic formula, or by using the 3 standard operations which are addition, multiplication, and exponentiation and their inverse operations.

So can any tell me how in the world you would solve this?(1 vote)- Finding zeroes of higher-degree polynomials is, in general, very hard. A lot of high-caliber math has been created in trying to solve problems like these, and you very much have to play each polynomial by ear.

There do exist formulas for the roots of 3rd and 4th-degree polynomials, much like the quadratic formula. But they're hideously complex (just writing them out would fill blackboards), and would only help us for f''(x). There aren't any such formulas for 5th-degree polynomials and higher (and it's not that we just haven't found one; they can be proven to not exist), and Wolfram|Alpha can't find any of these roots except by approximation.

So what you do in this situation is brush up on function approximation, and give up on finding exact answers. You won't find them.(2 votes)

- at8:08how do you label the intervals of concave up and down and where it increases and decreases(1 vote)

## Video transcript

- [Voiceover] So I have
the function g here. It's expressed as a
fourth degree polynomial. And I wanna think about the
intervals over which g is either concave upwards
or concave downwards. Now let's just remind ourselves
what these things look like. So concave, concave upwards is an interval, an interval when you're concave upwards is an interval over which
the slope is increasing and it tends to look like an
upward opening U like that. And you can see here that the
slope over here is negative, and then as x increases,
it becomes less negative, it actually approaches
zero, it becomes zero, then it crosses zero and
becomes slightly positive, more positive, even more positive. So you can see the slope
is constantly increasing. And if you think about it
in terms of derivatives, it means that you're first
derivative is increasing over that interval. And in order for your first
derivative to be increasing over that interval, your second derivative f prime prime of x, actually
let me write it as g, because we're using g in this example. In order for your first
derivative to be increasing, your, let me write this. So g, so concave upward means that your first derivative increasing, increasing, which means, which means
that your second derivative is greater than zero. And concave downward is the opposite. Concave downward, downward, is an interval, or you're
gonna be concave downward over an interval when
your slope is decreasing. So g prime of x is decreasing or we can say that our second derivative, our second derivative is less than zero. And once again, I can draw it on this. So, when x is lower, we have a, look, or it looks like we have a positive slope, then it becomes less positive, and then it becomes less positive, as it's approaching zero, it becomes zero, then it becomes negative,
then even more negative, and then even more negative. So as you can see, our slope
is constantly decreasing as x increases here. So in order to think about the intervals where g is either concave
upward or concave downward, what we need to do is let's
find the second derivative of g, and then let's think about the points at which the second, where
the second derivative can go from being positive to negative
or negative to positive and those will be places
where it's either undefined or where the second
derivative is equal to zero, and let's see what's happening
in the interval between. And then we'll know over what intervals are we concave upward or concave downward. So let's do that. So let's take the first
derivative, g prime of x, just gonna apply the power rule a lot. Four times negative one is negative four x to the
third power plus, okay, so you're gonna have two times six is plus 12x to the first power,
you can just write it as x. And then minus two, I could say minus two x to the zeroth power, but that's just minus two. And then the derivative of
negative three, of a constant is just zero. And now I can take the second
derivative, g prime prime of x is going to be equal to
three times negative four is negative 12x squared, decrement the exponent plus 12. And so let's see, where
could this be undefined? Well, the second derivative
is just a quadratic expression here which would be defined for any x. So it's not going to
be undefined anywhere. So, interesting points
where we could transition from going from a negative to a positive or a positive to a
negative, second derivative is where this thing
could be equal to zero. So let's figure that out. So let's figure out where
negative 12x plus 12 could be equal to zero. See, we could subtract 12 from both sides and we get negative 12x squared
is equal to negative 12. Divide both sides by negative 12, you get x squared is equal to one or x could be equal to the plus or minus, or x could be equal to the plus
or minus square root of one, which is, of course, just one. So, at the second derivative
at plus or minus one is equal to zero, so either
between plus or minus one or on either side of
them, we are going to be, we could be concave upward
or concave downward. So let's think about this. And to think about this, I'm
gonna make a number line. Let me find a nice, soothing color here. All right, that's a nice, soothing color. And, let's say, we should
make the number line a little bit bigger. So, there we go, let's
utilize the screen space. And so, if this is zero
and this is negative one, this is negative two,
this is positive one, this is positive two. We know that at x equals negative one and x equals one, our second
derivative is equal to zero. So let's think about what's happening in between those places to
see if our second derivative is positive or negative, and
from that we'll be able to say where it's concave upward
or concave downward. So, on this first
interval right over here, so this is the interval from, this is the interval where we're going from negative infinity to negative one. Well, let's just try a
value in that interval to see what whether our second derivative is positive or negative. And let's see, an easy value
there could be negative two, it's in that interval, so let's take g prime
prime of negative two, which is equal to negative 12 times four, because negative two
squared is positive four. So it's negative 48 plus 12, so it's equal to negative 36. The important thing to realize, then, is well, if over here it's negative, then over this whole interval, because it's not crossing through zero, or it's not discontinuous
at any of these points, that's why we picked this interval. That over this whole interval, g prime prime of x is less than zero, which means that over this interval we are concave downwards. So concave, concave downward, concave downward. Now let's go to the interval
between negative one and one. So this is the open interval
between negative one and one. And let's try a value there. Let's just try zero
will be easy to compute, g prime prime of zero, well,
when x is zero, this is zero, so it's just gonna be equal to 12. The important thing to realize
is our second derivative here is greater than zero, so
we are concave upward, concave upward on this interval between negative one and one. And then finally, let's
look at, let's look at the interval where x is greater than one. So this is the interval
from one to infinity, if we want to do it that way. And let's just try a value. Let's try g prime prime of two, 'cause that's in the interval. And g prime prime of two
is gonna be the same thing as g prime prime of negative two, 'cause whether you have a
negative two or a positive two, you square it becomes four. So you're gonna have
four times negative 12, which is negative 48 plus
12, which is negative 36. So this is negative 36, and
so, once again on this interval you are concave, concave downward. Now let's, I graphed this ahead of time, let's see if what we just
established is actually consistent with what the graph actually looks like. We were able to come up
with these, these insights about the concavity without graphing it. But now, it's kind of satisfying
to take a look at a graph and actually let me see if I
can match up the intervals. Actually this is pretty closely matched, all right over here. And so, this is, actually we can make it a
little bit smaller, all right. And so, let me move my boundy box. So, I'm saying that I'm concave downward between negative infinity,
negative infinity all the way until, all the
way until negative one, all the way until this
point right over here. So, all the way until that point. And that looks right. It looks like the slope
is constantly decreasing all the way until we get
to x equals negative one, and then the slope starts increasing. The slope starts increasing from there, from there all the way, and right at x we're transitioning, so I'm gonna leave a little,
I won't color in that. And so here our slope is increasing, let me do that same color. Our slope is increasing,
increasing, increasing, increasing, increasing all the way until we get to x equals one. And then our slope
starts decreasing again. And we get back into concave downwards. Oops, I wanna do that
in that orange color. We get back into concave downwards. So what we're able to figure out by just taking the derivatives
and doing a little algebra, we can see quite clearly
looking at the graph.