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  • - [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.

- [Instructor] Erin was asked to find if f of x

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