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## AP®︎/College Calculus AB

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

Lesson 4: Using the first derivative test to find relative (local) extrema- Introduction to minimum and maximum points
- 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 2)

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

## Want to join the conversation?

- Do we always need to find all critical points? What about functions like

x^3/(x+sin(x)) ?(8 votes)- You don't need to find all the critical points if the x value that makes the derivative 0 also makes the original function zero. For example 1/x.

Also, you don't need to find critical points of regular trig functions since they're repeating.(1 vote)

- I thought critical points cannot be 0 in the original function. If you were to plug in 1 or -1 into the original function it would be 0, so would -1 and 1 be voided critical points? Making the assumption right at the end?(4 votes)
- No, this is not true. Critical values points that make the original function undefined, not zero, are voided because they're not in the domain of the function. Clearly, critical points where the function is 0 are in the domain of the function and so would not be voided. (Think of it this way: there would be no reason to void the critical point (minimum) at 0 for the function x^2, just because this function's value happens to be 0 there.)

Note that the original function f(x) = (x^2 - 1)^(2/3), at 1 and -1, is 0 instead of undefined, since the exponent 2/3 is positive. So these critical points 1 and -1 would not be voided.(7 votes)

- Isn't the f' also undefined for x € [0, 1] since the square root can't be negative?(2 votes)
- No, because what f' actually contains is a
*cube*root which is in fact defined for all real numbers since it is an odd root.(6 votes)

- For Step 3, in the third column, first row, the equation reads "
`f'(3) = -2 < 0`

". Since this is the evalution of the test x-value, x= -3, shouldn't this read "`f'(-3) = -2 < 0`

"?(1 vote)

## Video transcript

- [Instructor] Erin was
asked to find if f of x is equal to x squared
minus one 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 Erin's work correct? If not, what's her mistake? So pause this video and see if you can figure it out yourself. Is Erin correct, or did
you 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
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 gonna take the
derivative of the outside with respect to the inside. So this is going to be 2/3
times x squared minus one to the 2/3 minus one, so
to the negative 1/3 power, times the derivative of the
inside with respect to x. So the derivative of x squared
minus one with respect to x is two x. (siren ringing) There's a fire hydrant, a fire (laughing), not a hydrant, that
would be a noisy hydrant. There's a fire truck outside, but okay, I think it's passed. But this looks like what
she got for the derivative. Because if you multiply two times two x, you do indeed get four x. You have this three right
over here in the denominator. And x squared minus one
to the negative 1/3, that's the same thing
as x squared minus one to the 1/3 in the denominator, which is the same thing as the cubed root of x squared minus one. So all of this is looking good. That is indeed the derivative. Step two, the critical
point is x equals zero. So let's see, a critical point
is where our first derivative is either equal to zero
or it is undefined. And so it does indeed
seem that f prime of zero is going to be four times zero, it's gonna be zero over
three times the cubed root of zero minus one, of negative one. And so this is three times negative one, or zero over negative three, so this is indeed equal to zero. So this is true. A critical point is at x equals zero. But a question is, is this
the only critical point? Well as we've mentioned, a critical point is where a function's derivative
is either equal to zero or it's undefined. This is the only one where the
derivative's equal to zero, 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 zero? Well if x squared minus
one is equal to zero, you take the cube root of zero, you're gonna get zero in the denominator. So what would make x squared
minus one equal to zero? Well x is equal to plus or minus one. 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 zero, but it is not the only critical point. So I would put that there. And the reason why it's important, 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 minimum point, is you have to sample
values on either side of it to make sure that you have a change, 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's she's testing, she's trying to test values on either side of the critical point that she, that 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've tried is x equals 0.5 and x equals negative 0.5. So this is what she shoulda done is try maybe negative two, negative one, negative 1/2, zero, 1/2, and then one we know is undefined, and then positive two. Because this is a candidate extremum, this is a candidate extremum, and this is a candidate
extremum right over here. And so you wanna see in
which of these situations you have a sign change of the derivative. And you just wanna test in the intervals between the extremum points. So I would say that really
the main mistake she made is in step two is not identifying all of the critical points.