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Current time:0:00Total duration:4:33

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.2 (EK)

pamela was asked to find where h of x equals x to the third minus six x squared plus twelve x has a relative extrema m-- 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 a conclusion that therefore h has a relative extremum there it's Pamela's work correct if not what's her mistake so pause this video and try to work through it yourself and see if panels 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 prime of X just use the power rule multiple times it's going to be three x squared for the X to the third two times negative 6 is negative twelve or minus 12x and the derivative of 12x 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 the H prime of X is equal 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 extrema m-- at x equals two alright so she made a big conclusion here she assumed that because the derivative was 0 that we have a relative extrema so let's just see if you can even just make that conclusion in order to have a relative extrema your curve is going to look something like this and then you would have a relative extrema right over here and over here your slope goes from being positive to and then it hits the and then it goes to being negative or you can have a relative extremum like this this would be 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 aren't 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 here 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 a little bit neater so X X H prime of X right over here we know at x equals to 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 1 is 3 times 1 minus 2 squared 1 minus 2 negative 1 squared is positive 1 times 3 is still positive and then 3 well 3 minus 2 squared times 3 that's also going to be 3 so this is actually a situation where actually like I've just drawn it where our slope is positive before we hit the critical point it gets to 0 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 camels 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

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