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# Absolute minima & maxima (entire domain)

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

## Video transcript

so we have the function G of X is equal to x squared times the natural log of X and what I want to do in this video is see if we can figure out the absolute extrema for G of X so are there X values where G takes on an absolute maximum value or an absolute minimum value sometimes you might call them a global maximum or global minimum so the first thing I like to think about is well it's just what's the just what's the domain for which G is actually defined and we know that in the natural log of X this the the argue that the input into natural log it has to be greater than zero so the domain the domain is all real numbers greater than zero so X has to be greater than zero anything lateral log of zero is not defined there is no power that you could take e to to get to zero and natural log of negative numbers is not defined so that is the domain the domain is all real numbers such that all real numbers X such that X is greater than zero so our absolute extrema have to be within that domain so to find these let's see if we can find some local extrema and see if any of them are good candidates for absolute extrema and we could find our local extrema by looking at looking at critical points or critical values so let's take the derivative of G so G prime will just in a new color just for kicks alright so G prime of X is equal to we could use the product rule here so derivative of x squared which is 2x times the natural log of x plus x squared times the derivative of natural log of X so that is 1 over X and I can just rewrite that x squared times 1 over X and we're going to assume where that X is positive so that is going to be that is going to be adjust and actually I didn't have to make that assumption for what I'm about to do x squared divided by X is just going to be X all right and so that is G prime so now let's think about the critical critical points critical points are where the derivative are their points in the domain so they're going to have to satisfy X is great than zero such that G prime is either is either undefined or it is equal to zero so let's first think about when G prime is equal to zero so let's set it equal to zero to X natural log of X plus X is equal to zero well we can subtract we could subtract X from both sides of that and so we get 2x natural log of X is equal to negative X see if we divide both sides by X and we can do that we know X isn't going to be 0 our domain is X is greater than 0 so this is going to be actually let's divide both sides by 2x so that we get the natural log of X is equal to negative 1/2 negative X divided by 2x is negative 1/2 or we could say that X is equal to e to the negative 1/2 is equal to X remember natural log is just log base E so e to the negative 1/2 which we could also write like that e to the negative 1/2 or 1 over the square root of e so that's a point at which at which G at which at which our derivative I should say is equal to zero it is a critical point or critical value for our original function G so and that's the only place where G prime is equal to zero are there any other points where G prime is undefined and there have to be points within the domain so let's see what would make this undefined the 2x and the x that you can evaluate for any X natural log of X once again is only going to be defined for X greater than zero but that's we've already restricted ourselves to that domain so within the domain any point in the domain our derivative is actually going to be defined so given that let's see what's happening on either side of this critical point on either side of this critical point and I could draw a little number line here to to really help us visualize this so it's if this is negative one this is zero this is let's see what e to the this is going to be like one / boy this is a this is going to be a little bit less than one so let's see let me plot one here and then to here and so we have a critical point at one over the square root of e let me put it right over there one over the square root of E and we know that we're only defined from for all x is greater than zero so let's think about the interval between 0 and this critical point right over here so the open interval from 0 to 1 over the square root of e let's think about whether G prime is positive or negative there and then let's think about it for for X greater than 1 over the square root of E so that's the interval from 1 over square root of e 2 in affinity so over that yellow interval well let's just try out a value that is in there so let's just try G prime of I don't know let's try G prime of 0.1 Z prime of 0.1 is definitely going to be in this interval and so it's going to be equal to two point two times 0.1 is equal to is equal to zero point two times the natural log of 0.1 plus 0.1 and let's see this right over here this is going to be a negative value in fact it's going to be quite it's it's definitely going to be greater than negative one because e to the negative one is only gets you to let's see e to the negative one is one over e so that's 1 over two point seven so 1 over two point seven is going to be so this is going to be around around 0.3 or 0.4 so in order to get point 1 so this will be around 0.23 to 0.4 so in order to get to 0.1 you have to be even more negative so this is going to be this is going to be I could say less than negative 1 so this is less than negative 1 and I'm multiplying it times 0.2 going to get a negative value that is less than less than negative 0.2 and if I'm adding 0.1 to it well I'm still going to get a negative value so over this yellow interval G prime of X is less than zero and it would be I should have gotten a calculator or I could have gotten a calculator out I could have just evaluated a lot easier so G prime of X is less than zero in this interval and let's see in this blue interval what's going on and this will be easier we could just try out the value one so G prime of one is equal to two times the natural log of one plus one natural log of one is just zero so all of this just simplifies to one so over this blue interval light I sample the point there G prime of X is greater than zero so it looks like our function is decreasing from zero to one over square root of E and then we increase after that and we increase for for all X's f that are greater than one over the square root of e and so our function is going to hit if we're decreasing into that and then increasing after that we're hitting a global minimum point or an absolute minimum point at x equals one over the square root of e so let me write this down we hit a we hit a absolute absolute minimum minimum at x equals one over the square root of E and there is no absolute maximum as we get above 1 over square root of E we are just going to think about what's going to be happening here we're just going to 1 we know that our function just keeps on increasing and increasing and increasing forever and you can look at even this x squared is just going to get unbounded towards infinity and natural log of X it's going to grow slower than x squared but it's still going to go unbounded towards infinity so there's no global or no absolute maximum no absolute maximum point and now let's look at the graph of this to feel a little good about what we just did analytically without looking at it graphically and I looked at it ahead of time so let me copy and paste it and so this is the graph of our function so as we can see when this point right over here this is when this is one over the square root of E it's not obvious from looking at it that it's that point x equals one over the square root of e and we can see that it is indeed an absolute minimum point here and there is no absolute maximum point that it as there's arbitrarily high values that our function can take on
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