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## AP®︎ Calculus BC (2017 edition)

### Unit 6: Lesson 4

Finding relative extrema (first derivative test)- Finding relative extrema (first derivative test)
- Worked example: finding relative extrema
- Analyzing mistakes when finding extrema (example 1)
- Analyzing mistakes when finding extrema (example 2)
- Finding relative extrema (first derivative test)
- Relative minima & maxima
- Relative minima & maxima review

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# Analyzing mistakes when finding extrema (example 1)

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.2 (EK)

Analyzing the work of someone who tried to find extrema of a function, to see whether they made mistakes.

## Want to join the conversation?

- Isn't this a trick question? Does x^3-6*x^2+12x actually have any relative extrema?(2 votes)
- No, it doesn't. It's derivative is 3x²-12x+12, which factors as 3(x-2)², which is 0 at 2 and positive elsewhere. Since extrema require the derivative to switch between positive and negative, this function has no extrema.(7 votes)

- so any point where the derivative is 0 is called a critical point right??(1 vote)
- For a continuous function, anywhere in the domain of the function where the derivative is either 0 or undefined is called a critical point.(5 votes)

- I'm a little confused. So when you go from a peak at 3, then drop back down to 0, then go back up to 3, that's not a curve? What happens with the line?Isn't the graph you showed a value of -3? Any help would be appreciated!(1 vote)
- The key is in the fact that we are working with derivatives. Even though the derivative decreases from 3 to 0 and then back to 3, it never becomes less than zero. Since taking the derivative is just the rate of change/ slope at those points, the slope is always positive over this interval, except for at zero where the tangent line is just horizontal (dy/dx = 0).

However, if we were taking the actual value of the function and got points (1,3) and (2,0), you are right, the derivate would be negative at any point where 1<x<2.(1 vote)

- To be more precise, could you just find the one sided limit on both sides of the derivative?(1 vote)
- This isn't really related to the topic, but if we had an equation like

12x^2 + 4x - 4

Can we factor out a 4x from the terms even though only two of them have it?(1 vote)- No, I recommend reviewing algebra. What you can do is factor out the 4.

12x^2 + 4x - 4 = 4(3x^2 + x + 1)(1 vote)

## Video transcript

- [Instructor] Pamela was asked to find where h of x equals x to the third minus six x squared plus 12
x 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, find where the derivative is equal to zero and she found that it
happens at x equals two so she says that's a critical point. In step three, she says,
she makes the conclusion that therefore h has a
relative extremum there. Is Pamela's 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. Alright, well, I'm just gonna try to do it again in parallel. So first, let me just
take the derivative here. So h prime of x. Just use the power rule multiple times. It's gonna be three x squared
for the x to the third. Two times negative six is
negative 12 or minus 12 x and then the derivative of 12
x is plus 12 and let's see. You can factor out a three here. So it's three times x squared
minus four x plus four and this part is indeed
equal to x minus two squared. So this is equal to three
times x minus two squared. So her step one looks right on target. Okay, step two, the
solution of h prime of x is equal to zero is equal to x equals two. Yeah, that works out. If you were to say, if you were to say three
times x minus two squared which is h prime of x,
the first derivative, and set that equal to zero, this is going to be true
when x is equal to two and 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 and then step three, h
has a relative extremum at x equals two. Alright, so she made
a big conclusion here. She assumed that because
the derivative was zero, that we have a relative extremum. So let's just see if you can
even just make that conclusion. In order to have a relative extremum, your curve is gonna
look something like this and then you would have a
relative extremum right over here and over here, your slope
goes from being positive and 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 and 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 and 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 and so step two is correct. 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 and we could try to do that. So let's make a little table here. Make a little table. Do it a little bit neater. So x, x, h prime of x right over here. We know at x equals two,
h prime of two is zero. That's our critical point. But let's try, I don't know, let's see what happens
when x is equal to one and then let's see what
happens when x equals three. I'm just sampling points
on either side of two and let's see, we are going to have, when x is equal to one, h prime of one is three
times one minus two squared. One minus two, negative
one squared is positive one times three is still
positive and then three. Well, three minus two squared times three. That's also gonna be three. So this is actually a situation where, like I've just drawn it
where our slope is positive before we hit the critical point, it gets to zero but then it
starts becoming positive again and so 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 a minimum point here. So Pamela's work is not correct and her mistake is step three. In order to make this conclusion, you would have to test on either side of that critical point. Test the first derivative.