Analyzing mistakes when finding extrema example 2 | AP Calculus AB | Khan Academy

Aaron was asked to find if f of x is equal to x squared minus 1 to the 2/3 power has a relative maximum. This is her solution, and then they give us her steps, and at the end, they say, is Aaron's work correct? If not, what's her mistake? So pause this video and see if you can figure it out yourself. Is Aaron correct, or did she make a mistake, and where was that mistake?

All right, now let's just do it together. So she says that this is the derivative. I'm just going to reevaluate it here to the right of her work. So let's see, f prime of x is just going to be the chain rule. I'm going to take the derivative of the outside with respect to the inside. So this is going to be 2/3 times (x squared minus 1) to the 2/3 minus 1, so to the negative 1/3 power times the derivative of the inside with respect to x. The derivative of (x squared minus 1) with respect to x is 2x.

There's a fire hydrant, uh, fire—fire, not a hydrant; that would be a noisy hydrant. There's a fire truck outside, but okay. I think it's past, but this looks like what she got for the derivative because if you multiply 2 times 2x, you do indeed get 4x. You have this 3 right over here in the denominator, and (x squared minus 1) to the negative 1/3—that's the same thing as (x squared minus 1) to the 1/3 in the denominator, which is the same thing as the cube root of (x squared minus 1). So all of this is looking good; that is indeed the derivative.

Step two, the critical point, is x equals 0. So let's see: a critical point is where our first derivative is either equal to 0 or it is undefined. And so it does indeed seem that f prime of 0 is going to be 4 times 0, which is going to be 0 over 3 times the cube root of 0 minus 1, which is negative 1. So this is 3 times negative 1 or 0 over negative 3. So this is indeed equal to 0. So this is true; a critical point is at x equals 0.

But the question is, is this the only critical point? Well, as we've mentioned, a critical point is where the function's derivative is either equal to 0 or undefined. This is the only one where the derivative is equal to 0. But can you find some x values where the derivative is undefined? Well, what if we make the derivative—what would make the denominator of the derivative equal to 0? Well, if x squared minus 1 is equal to 0, you take the cube root of 0, you're gonna get 0 in the denominator.

So what would make (x squared minus 1) equal to 0? Well, x is equal to plus or minus 1. These are also critical points because they make f prime of x undefined. So I'm not feeling good about step two. It is true that a critical point is x equals 0, but it is not the only critical point. So I would put that there, and the reason why it's important—you know, you might say, "Well, what's the harm in not noticing these other critical points?" She identified one; maybe this is the relative maximum point.

But as we talked about in other videos, in order to use the first derivative test, so to speak, and find this place where the first derivative is zero, in order to test whether it is a maximum or a minimum point, you have to sample values on either side of it to make sure that you have a change in sign of the derivative. But you have to make sure that when you test on either side, that you're not going beyond another critical point because critical points are places where you can change direction.

And so let's see what she does in step three right over here. Well, it is indeed in step three that she's testing—she's trying to test values on either side of the critical point, that she—the one critical point that she identified. But the problem here, the reason why this is a little shady, is this is beyond another critical point that is less than zero, and this is beyond—this is greater than another critical point that is greater than zero. This is larger than the critical point one, and this is less than the critical point negative one.

What she should have tried is x equals 0.5 and x equals negative 0.5. So this is what she should have done. She should have tried maybe negative 2, negative 1, negative 1/2, 0, 1/2, and then 1— we know is undefined—and then positive 2, because this is a candidate critical—a candidate extremum. This is a candidate extremum, and this is a candidate extremum right over here. And so you want to see in which of these situations you have a sign change of the derivative, and you just want to test in the intervals between the extremum points.

So I would say that really the main mistake she made is that step two is not identifying all of the critical points.