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# Finding absolute extrema on a closed interval

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.A (LO)
,
FUN‑4.A.3 (EK)

## Video transcript

let's say that we've got the function f of X is equal to 8 times the natural log of X minus x squared and it's defined over the closed interval between 1 and 4 so it's a closed interval it also includes 1 and it includes 4 you can view this this is the domain of our function as we have to find it so given this give it this information this function definition what I would like you to do is come up with the absolute absolute maximum value value of f of f as defined where f is defined right over here where f is defined on this closed interval I encourage you to pause this video and think about it on your own so the extreme value theorem tells us look we got some closed interval and I'm going to speak in generalities here so let's say that's our x axis and let's say we have some function that's defined on a closed interval we have a couple of different scenarios for what that function might look like on that closed interval so we might hit a maximum point we might hit a maximum point at the beginning of the interval something like that we might hit an absolute maximum point at the end of the interval so it might look something like this so that's at the end of the interval or we might hit an absolute maximum point someplace in between and that could look something like this it could look like this and at this maximum point the slope of the tangent line is 0 so here our foot the derivative is 0 or we can have a maximum point someplace in between that looks like this and if it looks like this then here the derivative would be undefined there's a lot of different tangent lines that you could place that you could place right over there so what we need to do is let's test let's test the different endpoints let's test the function at the beginning let's test the function at the end of the interval and then let's see if there's any points where the derivative is either 0 or the derivative is is undefined and these points where the derivative is either zero or its undefined we've seen them before we call these of course critical numbers so this would be either in either case actually if we assume that that's happening the same number we would call that a critical number critical a critical number so those are the different candidates now you could have a critical number in between that where say the slope is zero say something like this but it isn't the maximum or minimum but what we can do is if we can find all the critical numbers we can then test the F at the credit the function evaluated the critical numbers and the function value at the endpoints and we could see which of those are the largest all of those are the possible candidates for where F hits a maximum value so first we could think about well actually let's just let's find the critical numbers first since we have to do it so let's take the derivative of F F prime of X is going to be equal to the derivative of natural log of X is 1 over X so it's going to be 8 over X minus 2x and let's set that equal to 0 so if we function if we focus on this part right over here we could add 2x to both sides and we get 8 over X is equal to 2x multiply both sides by X we get 8 is equal to 2x squared divide both sides by 2 you get 4 is equal to x squared and if we were just purely solving this equation we would get X is equal to plus or minus 2 now we're saying that the function is only defined over this interval so we negative 2 isn't part of its domain so we're only going to focus on X is equal to 2 this right over here is definitely a critical number now have we found all of the critical numbers critical numbers well this is the only number other than negative 2 but the only number in the interval that will make f prime of x equal to 0 what about where it's undefined well F prime of X would be undefined the only place would be undefined is if you stuck a zero right over here in the denominator but 0 is not in the interval so the only critical number in the interval is x equals 2 so now we just have to test F at the different end points and at the critical number and see which of those is the highest so we're going to test F of 1 F of 1 which is equal to 8 times the natural log of 1 minus 1 squared we'll test f of 4 which is equal to 8 times the natural log of 4 minus 4 squared which is of course 16 which is 16 and we're going to test F of 2 so these are the endpoints and this is a critical number eight times the natural log of 2 minus -2 squared now which of these is going to be the largest and it might be tempting to get a calculator out but actually let's see if we can get a little intuition here so this is the natural log of 1 is 0 8 to the EE to the 0 power is equal to 1 so 8 times 0 is 0 so this evaluates to negative 1 now let's see what does this evaluate to the natural log of 4 e is 2.7 on and on and on so this number is going to be between 1 & 2 so it's going to be between 1 & 2 and it's actually going to be well between 1 & 2 you multiply that times 8 you're going to be between 8 and 16 and then you multiply at least without using a calculator that's very rough way which of these is is large both of these are negative numbers though now what about this natural log of 2 natural log of 2 is going to be some fraction it's going to be it's going to be more than 1/2 it's going to be more than 1/2 and since it's more than 1/2 this whole thing is going to be more than 4 which means this whole thing is going to be positive so this is negative this is negative this is positive and these are only critical these are only candidates for our maximum value so I would go with this one hour maximum our maximum value happens when X is equal to 2 and that maximum value is 8 natural log of 2 minus 4 that is the absolute maximum value absolute max value over the interval or I guess we could say over the domain that this function is defined if we want to verify it with a calculator we of course could so we already figured out this 1 plus C F of 4 8 natural log of 4 - 16 is equal to negative 5 so that's that's that this is definitely not this is this one is definitely not the maximum value and then F of 2 is 8 8 natural log of 2 minus 4 which as we said is indeed a positive number so feel pretty good about what we did