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Main content
Current time:0:00Total duration:9:42
AP.CALC:
FUN‑4 (EU)
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FUN‑4.A (LO)
,
FUN‑4.A.10 (EK)
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FUN‑4.A.9 (EK)

Video transcript

we've got the function f of X is equal to X to the third power minus 12x plus 2 and what I want to do in this video is think about at what points does my function f take on minimum or maximum values and to figure that out I have to first figure out what are the critical points for my function f and then which of those critical points do we achieve a minimum or maximum value and to determine the critical points we have to find the derivative of our function because our critical points are just the points at which our derivative is either equal to zero or undefined so the derivative of this thing right over here we're just going to use the power rule several times and then I guess you could call it the constant rule but the derivative of X to the third is 3x squared derivative of negative 12x is negative 12 and the derivative of a constant it doesn't change with respect to X so it's just going to be equal to zero so we're going to get a critical point when this thing right over here for some value of x is either undefined or zero well this thing is defined for all values of X so the only way places we're going to find critical points is when this thing is equal to zero so let's set it equal to zero when does three x squared minus 12 equals zero so let's add 12 to both sides you get 3x squared is equal to 12 divide both sides by 3 you get x squared is equal to 4 well this is going to happen when X is equal to 2 and X is equal to negative and X is equal to negative 2 just to be clear F of 2 is equal or let me be clear F prime of 2 you get 3 times 4 minus 12 which is equal to 0 and F prime of negative 2 is also same exact reason is also equal to 0 so we can say and I'll switch colors here that that F has critical points critical points at x equals 2 and x equals negative 2 well that's fair enough but we still don't know whether their minimum point where the function takes on a minimum values at those points maximum values at those points or neither to figure that out we have to figure out whether the relative changes signs around these points so let's actually try to graph the derivative to think about this so let's graph so I'll draw an axis right over here I'll do it down here because maybe we can use that information later on to graph f of X so let's say so let's say this is my x axis this is my this is my y axis and so we have critical points that X is equal to positive 2 so it's 1/2 and X is equal to negative 2 1/2 X is equal to negative 2 so what is this what is this derivative look like if we wanted to graph it well we have a when X is equal to 0 for the derivative we're at negative 12 so this is the point y is equal to negative 12 so this is we're graphing y is equal to F prime of X so it looks something like this it will look something something like this these are obviously the zeros of our derivative so it has to move up to cross the x axis there and over here and over here so what is the derivative doing at each of these critical points well over here our derivative is crossing from being positive we have a positive negative derivative to being a negative derivative so we're crossing from being a positive derivative to being a negative derivative that was our criteria for a critical point to be a maximum point for being a maximum point over here we're crossing from a we're crossing from a negative derivative to a positive derivative which is our criteria for a critical point - for the function to have a minimum value at a critical point so a minimum and I just want to make sure we have the derivative we have the correct intuition if our function if some function is increasing going into some point so if it's increasing going into some point and at that point we see actually we have a derivative of 0 the derivative also be undefined we have a derivative of 0 and then function begins decreasing that's why this would be a maximum point similarly if we have a situation where the function is decreasing going into a point the derivative is negative remember this is the graph of the derivative let me make this clear this is the graph of y is equal to not f of X but F prime of X so if we have a situation we're going into the point the function is the function has a negative slope we see we have a negative slope here so the function might look something like this and then right at this point the function is either undefined or has zero slope so in this case it has zero slope and then after that point let me do it right under it so going into it we have a negative slope and then right over here we have a zero slope check it draw it even better than that so if we were to imagine going into it we have a negative slope right at that point we have a zero slope and then we have a positive slope so the function begins increasing that's why we say it has we have a minimum point right over there so what I did right over here is to try to conceptualize what the function itself could look like given the derivative in this case switching from a positive derivative to a negative derivative across that critical point or going from a negative derivative to a positive derivative that's why this is a criteria for a maximum point this is a criteria for a minimum point well with that out of the way can we use this intuition that we just talked about to at least try to sketch the graph of f of X so let's try to do it let's try to do it so let's try to do it and it's just going to be a sketch it's not going to be very exact but at least will give us a sense of the shape of what f of X looks like so my best attempt so it might not be drawn completely to scale so it's my x axis this is my y axis we know we have a critical point at X is equal to positive 2 and we have a critical point at X is equal to and we have a critical point at X is equal to negative 2 we know just from inspection that the y intercept right here of the graph of y is equal to f of X when x is 0 f of X is 2 so we're going to hit right over let's say we're going to hit right / I want to give it I don't want to draw this completely to the same scale as the x-axis so let's say that this is - let's say that this is - right over here so this is where we're going to cross this is going to be our y-intercept and so we said already that we have a maximum point at X is equal to negative two so what is F of negative two F of F of negative two is equal to 8 or negative eight let me be careful it's negative eight and then we're going to have 12 times negative 2 which is negative 24 but then we're going to add it so we're subtracting negative 24 so this is plus plus 24 and then we finally have plus 2 so negative 8 plus 24 plus 2 so that's going to be negative C negative 8 plus 24 16 plus 2 is 18 so F of negative 2 is equal to 18 and I'm not drawing it completely scale but let's say that this is 18 right over here so this is the function this is the point negative 2 comma 18 and we know that it's a maximum point the derivative going into that point is negative the derivative going into that point is negative I'm sorry the derivative going into that point is positive so we are increasing we are increasing the slope is positive and then after we cross that point the slope becomes negative the slope becomes negative the derivative cross the x-axis the slope becomes negative it look might actually I want to use that same color it looks like this it looks like this and then of course the graph is going to cross the it's going to have a y-intercept something like that and then and then as we approach 2 we are approaching another critical point now what is f of 2 f of 2 is going to be equal to positive 8 minus 24 minus 24 plus 2 so this is 10 minus 24 which is equal to negative 14 so let's say this is the point negative 14 right over here which I could draw out a little bit let's say this is negative 14 so this is f of 2 right over there and we saw already that the slope is negative as we approach it so we are our function is decreasing as we approach it and then right there the slope is 0 we figured that out earlier that's how we identified it being a critical point and then the slope is increasing after that the derivative is positive the slope is increasing so this is our sketch of what func of what f of X could look like given that these are the critical points we were able to identify F of or we were able to identify two as a minimum point so this was a minimum value the function takes on a minimum value what X is equal to 2 and the function took on a maximum value when F was equal to negative 2
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