Analyzing mistakes when finding extrema (example 1) | AP Calculus AB | Khan Academy

Pamela was asked to find where ( h(x) = x^3 - 6x^2 + 12x ) has a relative extremum. This is her solution.

So, step one, it looks like she tried to take the derivative. Step two, she tries to find the solution to find where the derivative is equal to zero, and she found that it happens at ( x = 2 ). So she says that's a critical point. Step three, she says she makes a conclusion that therefore ( H ) has a relative extremum. There is Pamela's work.

Is her work correct? If not, what's her mistake? So, pause this video and try to work through it yourself and see if Pamela's work is correct.

All right, well, I'm just going to try to do it again in parallel. So first, let me just take the derivative here. So ( H' (x) ), just using the power rule multiple times, is going to be ( 3x^2 ) for the ( x^3 ), ( -6 \times 2 ) is ( -12 ) or ( -2x ), and then the derivative of ( 12x ) is ( +12 ).

Let's see, you can factor out a three here, so it's ( 3(x^2 - 4x + 4) ), and this part is indeed equal to ( (x - 2)^2 ). So this is equal to ( 3(x - 2)^2 ).

Her step one looks right on target.

Okay, step two, the solution of ( H' (x) = 0 ) is equal to ( x = 2 ). Yeah, that works out. If you were to say ( 3(x - 2)^2 ) which is ( H' (x) ), the first derivative, and set that equal to ( 0 ), this is going to be true when ( x = 2 ). So, any point where your first derivative is equal to zero or it's undefined, it is indeed a critical point.

So this step looks good so far. Step three, ( H ) has a relative extremum at ( x = 2 ). All right, so she made a big conclusion here. She assumed that because the derivative was zero that we have a relative extremum.

Let's just see if you can even just make that conclusion. In order to have a relative extremum, your curve is going to look something like this, and then you would have a relative extremum right over here and over here. Your slope goes from being positive, then it hits zero, and then it goes to being negative.

Or you could have a relative extremum like this. This would be a maximum point; this would be a minimum point right over here, and then in a minimum point, your slope is zero right over there, but right before it, your slope was negative and it goes to being positive.

But you actually have cases where your derivative, your first derivative, is zero, but you don't have an extremum. So for example, you could have a point like this where right over here your slope or your derivative could be equal to zero.

So your first derivative would be equal to zero, but notice your slope is positive. It hits zero and then it goes back to being positive again. So you can't make the conclusion just because your derivative is zero that it's definitely an extremum. You could say it's a critical point.

In order to make this conclusion, you would have to test what the derivative is doing before that point and after that point and verify that it is switching sides. We could try to do that.

So let's make a little table here. Make a little table, do a little bit neater. So ( x ), ( H' (x) ) right over here. We know at ( x = 2 ), ( H' (2) = 0 ); that's our critical point.

But let's try, I don't know, let's see what happens when ( x = 1 ), and then let's see what happens when ( x = 3 ). I'm just sampling points on either side of two. Let's see.

We are going to have when ( x = 1 ), ( H' (1) = 3(1 - 2)^2 = 1 ); thus, it remains positive. And then for ( x = 3 ), well, ( 3 - 2 = 1 ), so ( H' (3) ) is also going to be positive.

So this is actually a situation where, like I just drawn it, our slope is positive before we hit the critical point, it gets to zero, but then it starts becoming positive again.

That's why you actually have to do this test in order to identify whether it's an extremum. It turns out that this is not an extremum; this is not a maximum or minimum point here.

So Pamela's work is not correct, and her mistake is in step three. In order to make this conclusion, you would have to test on either side of that critical point and test the first derivative.