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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 5

Lesson 8: Using the second derivative test to find extrema# Second derivative test

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.7 (EK)

Sal justifies the second derivative test, which is a way of determining relative minima & maxima, and gives an example.

## Want to join the conversation?

- The second derivative is is negative at c, so we conclude that the slope is decreasing around c?

How can we conclude that when we only know the the second derivative is only negative at c and only c and not at points around c?(8 votes)- I'm not sure how to prove this to you, but I think I can give you an intuition for why this is true.

From earlier videos we know that a derivative at c, means that the function is continuous at c. This means that there must be some values for x very close to c where`f(x)`

is defined and continuous - I'm going to call this "window"`c ± ∆x`

.

Now imagine that we make a very tiny (infinitesimal) step in the negative direction to`c-∆x`

. Will`f(c-∆x)`

be less than, greater than or equal to`f(c)`

?

I hope you can see that if`∆x`

gets small enough, then`f(c-∆x) < f(c)`

because otherwise you would break the line and then the function wouldn't be continuous at x=c, which would be a contradiction!

The same argument holds for`f(c+∆x)`

.

Consequently, the output for the function`f(x)`

must drop at least a tiny amount as you move either direction from`x=c`

, and thus by definition you have a local maximum.(13 votes)

- What about if the second derivative is equal to zero? Is it always inconclusive?(6 votes)
- That's just simply what people found. For example, if you take f(x) = x^3, where f''(0) = 0, there is no relative extremum at x = 0. However, if you take f(x) = x^4, where f''(0) also = 0, there is a relative minimum at x = 0. Thus, it is inconclusive.(1 vote)

- What happens if f prime does not exist around an x value c? Why would the test not apply?(4 votes)
- If f'(x) doesn't exist then f"(x) will also not exist, so the second derivative test is impossible to carry out. However, this does
**not**mean that there is not an Inflection point!

An inflection point requires:

1) that the concavity changes and

2) that the function is defined at the point.

You can think of potential inflection points as critical points for the first derivative — i.e. they may occur if`f"(x) = 0`

**OR**if`f"(x)`

is undefined. An example of the latter situation is`f(x) = x^(1/3)`

at`x=0`

. (Note: f'(x) is also undefined.)

Relevant links:

http://depts.gpc.edu/~mcse/CourseDocs/calculus/concavity-inflectionOct12.pdf

https://math.stackexchange.com/questions/402459/an-inflection-point-where-the-second-derivative-doesnt-exist(2 votes)

- Why f"(c)=0 inconclusive?(2 votes)
- That's just simply what people found. For example, if you take f(x) = x^3, where f''(0) = 0, there is no relative extremum at x = 0. However, if you take f(x) = x^4, where f''(0) also = 0, there is a relative minimum at x = 0. Thus, it is inconclusive.(5 votes)

- What does f''(x)=0 look like on a graph? I've gotten asked this on my homework and have no idea how to interpret it.(3 votes)
- at3:44, how do we know that the second derivative is less than zero? and how do we even know if this function can have a second derivative? what is needed to, what defines a second derivative? thank you!(2 votes)
- The second derivative is the derivative of the first derivative.

e.g.

f(x) = x³ - x²

f'(x) = 3x² - 2x

f"(x) = 6x - 2

So, to know the value of the second derivative at a point (`x=c`

,`y=f(c)`

) you:

1) determine the first and then second derivatives

2) solve for f"(c)

e.g. for the equation I gave above f'(x) = 0 at x = 0, so this is a critical point.

f"(0) = 6•0 - 2 = -2

Therefore, f(x) is concave downward at x=0 and this critical point is a local maximum.

Can you do the same for the other critical point?(3 votes)

- Is Second derivarive for the maximum and minimum of the curve , right ?(1 vote)
- No. To find the maximum and minimum points, you use the first derivative. To get a max or min, the points you want to consider are where the function stops increasing and begins to decrease, or stops decreasing and begins to increase. If the function is continuous, there will be a point there where the function's rate of change (derivative) is zero. At this point, the increasing function stops increasing, creating a potential maximum point.(5 votes)

- Would love to see visual examples of how h'(x)=0 and h''(x)=0 could be a max, min or nothing.(2 votes)
- The functions x^3, -x^4, and x^4 all have the property that their first and second derivatives are zero at x = 0. Yet the first function has no min or max at all, the second function has a local (and also global) max at x = 0, and the third function has a local (and also global) min at x = 0.

If you want visual examples, you can graph these three functions manually or on a graphing calculator.

Have a blessed, wonderful day!(2 votes)

- when can the 2nd derivative test not be used?(2 votes)
- If it is not a function, or an implicit function like the equation of a circle (x-h)^2 + (y-k)^2 = r^2 Also if the first derivative or second drivative = 0 the second derivative test wouldn't tell you anything.

Let me know if it doesn't make sense why any of that wouldn't let the second derivative test work.(2 votes)

- If f(x)=x^3 for closed function of [-2,2], first derivative is 3x^2, so critical points are -2, 0, 2. Now, second derivative = 6x. Therefore, f"(-2)=-12 and f"(2)=12 So, -2 must be the max value and 2 must be the min value. But f(-2)=-8 which is lesser than f(2)=8. Can someone explain where I went wrong?(1 vote)
- Because you're not looking at the entire domain of the function (this function's domain is across all real numbers), you need to be careful on what critical points actually are.

So, you've correctly gotten that the critical points are -2, 0, and 2. This is because you're always interested what happens at the end of the interval (giving you -2 and 2) and you're also interested in what happens if the deriviative is 0 (giving you 0).

Now, the second derivate test only applies if the derivative is 0. This means, the second derivative test applies only for x=0. At that point, the second derivative is 0, meaning that the test is inconclusive. So you fall back onto your first derivative. It is positive before, and positive after x=0. Therefore, x=0 is an inflection point.

Now, we go back to x=-2 and x=2. We can't use derivatives here because they lie at the ends of the interval we are looking at. So, we simply evaluate the function. And we see that x=-2 is a minimum of that function and x=2 is a maximum of that function on that interval.(3 votes)

## Video transcript

- [Voiceover] So what I
want to do in this video is familiarize ourselves with
the second derivative test and before I even get into
the nitty-gritty of it, I really just want to
get an intuitive feel for what the second
derivative test is telling us. So let me just draw some axes here. So let's say that's my y-axis,
let's say this is my x-axis and let's say I have a function that has a relative maximum value at X equals C. So let's say we have a situation that looks something like that and X equals C is right over, so that's the point C, F of C. So if I can draw a straighter dotted line. So that is X being equal to C and we visually see that we
have a local maximum point there and we can use our calculus tools to think about what's going on there. Well one thing that we know, we know that the slope
of the tangent line, at least the way I've
drawn it right over here, is equal to zero. So we could say F prime
of C is equal to zero and the other thing we can see is that we are concave downward in the neighborhood around X equals C. So notice our slope is
constantly decreasing and since your slope,
notice it's positive, it's less positive, even less positive, it goes to zero, then it becomes negative, more negative and even more negative. So we know that F prime prime, we know that F prime prime of C is less than zero and so I haven't done any
deep mathematical proof here, but if I have a critical
point at X equals C, so F prime of C is equal to zero, and we also see that the
second derivative there is less than zero. Well intuitively this makes sense that we are at a maximum value and we could go the other way if we are at a local
minimum point at X equals C or relative minimum point. So our first derivative
should still be equal to zero 'cause our slope of a
tangent line right over there is still zero. So F prime of C is equal to zero. But in this second situation,
we are concave upwards. The slope is constantly increasing. We have an upward opening bowl and so here we have a
relative minimum value or we could say our second
derivative is greater than zero. Visually we see it's a
relative minimum value and we can tell just
looking at our derivatives, at least the way I've drawn it, first derivative is equal to zero and we are concave upwards. Second derivative is greater than zero. And so this intuition that
we hopefully just built up is what the second
derivative test tells us. So it says hey look, if we're
dealing with some function F, let's say it's a twice
differentiable function. So that means that over some interval. So that means that you
could find its first and second derivatives are defined and so let's say there's
some point, X equals C, where its first derivative
is equal to zero, so the slope of the tangent
line is equal to zero, and the derivative exists
in a neighborhood around C and most of the functions we deal with, if it's differentiable at C, it tends to be differentiable
in the neighborhood around C and then we also assume that
second derivative exists is twice differentiable. Well then we might be
dealing with a maximum point, we might be dealing with a minimum point, or we might not know
what we're dealing with and it might be neither a
minimum or a maximum point. But using the second derivative test, if we take the second
derivative and if we see that the second derivative is
indeed less than zero, then we have a relative maximum point. Then so this is a situation that we started with right up there. If our second derivative
is greater than zero, then we are in this situation right here, we're concave upwards. Where the slope is zero,
that's the bottom of the bowl. We have a relative minimum point and if our second derivative
is zero, it's inconclusive. We don't know what is actually
going on at that point. We can't make any strong statement. So with that out of the way,
let's just do a quick example just to see if this has gelled. Let's say that I have some
twice differentiable function H and let's say that I tell you that H of eight is equal to five, I tell you that H prime
of eight is equal to zero, and I tell you that the second derivative at X equals eight is
equal to negative four. So given this, can you tell me whether the point eight comma five,
so the point eight comma five, is it a relative minimum, relative minimum, maximum point or not enough info? Not enough info or inconclusive? And like always, pause the video and see if you can figure it out. Well we're assuming it's
twice differentiable and I think it's safe to assume that and for the sake of our problem we're gonna assume that
the derivative exists in a neighborhood around X equals eight. So in this example, C is eight. So point eight five is
definitely on the curve. The derivative is equal to zero. So we're dealing potentially
with one of these scenarios and our second derivative
is less than zero. Second derivative is less than zero. So this threw us. So the fact that the second derivative, so H prime prime of
eight is less than zero, tells us that we fall into
this situation right over here. So just with the information
they've given us, we can say that at the
point eight comma five we have a relative maximum value or that this is a relative
maximum point for this. If somehow they told us the
second derivative was zero, then we would say it's inconclusive. If they told us and
that's all they told us and if they told us the second derivative is greater than zero,
then we would be dealing with a relative minimum
value at X equals eight.