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Current time:0:00Total duration:4:42

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.2 (EK)

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 in 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 so the derivative of x squared minus 1 with respect to X is 2x there's a fire hydrant fi-fire fire not a hydrant that would be a noisy hydrant there's a fire truck outside but okay I think it's 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 a cube root of x squared minus 1 so all of this is looking good that is indeed the derivative step to 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 see seen that f prime of 0 is going to be 4 times 0 is going to be 0 over 3 times the cube root of 0 minus 1 of negative 1 and 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 a question is is this the only critical point well as we've mentioned a critical point is where functions derivative is either equal to 0 or it's 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 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 know you might say well what's the harm and 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 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 3 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 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 have tried is x equals zero point five and x equals negative zero point five so this is what she should have done is try maybe negative two negative one negative one half zero one half and then one we know is undefined and then positive two 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 and 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

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