Second derivative test | Using derivatives to analyze functions | AP Calculus AB | Khan Academy

So what I want to do in this video is familiarize ourselves with the second derivative test.

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.

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 at c is right over there. So that's the point c, f of c.

I could draw a straight or dotted line, so that is x being equal to c, and we visually see that we have a local maximum point there. We already 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.

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. Since our slope is concave, it's positive, 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. I haven't done any deep mathematical proof here, but if I have a critical point where f prime—where our 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, intuitively this makes sense that we are at a maximum value.

We could go the other way if we are at a local minimum point at x equals c, or relative minimum point. Our first derivative should still be equal to zero because our slope of the 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. So here we have a relative minimum value. 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.

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. This means that over some interval, its first and second derivatives are defined.

Let’s say there’s some point x equal to 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. Most of the functions we deal with, if it's differentiable at c, it tends to be differentiable in a neighborhood around c.

Then we also assume that the second derivative exists; it 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—it might be neither a minimum nor 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. 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. If our second derivative is zero, it’s inconclusive. There, we don't know what is actually going on at that point; we can't make any strong statement.

So 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 8 is equal to 5. I tell you that h prime of 8 is equal to 0, and I tell you that the second derivative at x = 8 is equal to 4.

Given this, can you tell me whether the point (8, 5)—so the (8, 5)—is it a relative minimum, relative 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. I think it's safe to assume that it’s, and well, for the sake of this problem, we're going to assume that the derivative exists in a neighborhood around x equals 8. So this example c is 8. The point (8, 5) is definitely on the curve. The derivative is equal to zero. So we're potentially dealing with one of these scenarios.

And our second derivative is greater than zero. The second derivative is greater than zero. So this 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 (8, 5), 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 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 8.