Absolute minima & maxima (entire domain) | AP Calculus AB | Khan Academy

So we have the function ( G(x) = x^2 \cdot \ln(x) ), and what I want to do in this video is see if we can figure out the absolute extrema for ( G(x) ). 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.

The first thing I like to think about is, well, just what’s the domain for which ( G ) is actually defined? We know that in the natural log, the argument (the input into natural log) has to be greater than zero. So the domain is all real numbers greater than zero. ( x ) has to be greater than zero; anything less than or equal to zero for natural log is not defined. There is no power that you could take ( e ) to get to zero, and the 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 > 0 ).

So our absolute extrema have to be within that domain. 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. We could find our local extrema by looking at critical points or critical values. So let’s take the derivative of ( G ).

So ( G' ) (I’m just using a new color just for kicks) is equal to, we could use the product rule here. The derivative of ( x^2 ), which is ( 2x \cdot \ln(x) + x^2 ) times the derivative of ( \ln(x) ). So that is ( \frac{1}{x} ), and I can just rewrite that as ( x^2 \cdot \frac{1}{x} ). We’re going to assume ( x ) is positive, so that’s just ( x ).

So that is ( G' ). Now let’s think about the critical points. Critical points are where the derivative is equal to zero, and they’re going to have to satisfy ( x > 0 ) such that ( G' ) is either undefined or it is equal to zero. So, let’s first think about when ( G' ) is equal to zero.

So, let’s set it equal to zero:

[ 2x \ln(x) + x = 0. ]

Well, we can subtract ( x ) from both sides of that. So we get:

[ 2x \ln(x) = -x. ]

Let’s see, if we divide both sides by ( x ) (and we can do that because we know ( x ) isn’t going to be zero since our domain is ( x > 0 )), so this is going to be:

[ \ln(x) = -\frac{1}{2}. ]

Or we could say that ( x = e^{-\frac{1}{2}} ) is equal to ( x ) (remember natural log is just log base ( e )). So ( x = e^{-\frac{1}{2}} ).

That’s a point at which ( G' ), at which our derivative, I should say, is equal to zero; it is a critical point or critical value for our original function ( G ). That’s the only place where ( G' ) is equal to zero.

Are there any other points where ( G' ) 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 > 0 ), but 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.

Given that, let’s see what’s happening on either side of this critical point. I could draw a little number line here to really help us visualize this. So if this is ( 0 ), this is ( 1 ), and let’s see... This is going to be like ( \frac{1}{\sqrt{e}} ). Let me put it right over there ( \frac{1}{\sqrt{e}} ), and we know that we’re only defined for all ( x ) 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 ( \frac{1}{\sqrt{e}} ). Let’s think about whether ( G' ) is positive or negative there and then let’s think about it for ( x > \frac{1}{\sqrt{e}} ).

That’s the interval from ( \frac{1}{\sqrt{e}} ) to infinity. So over that yellow interval, let’s just try out a value that is in there. Let’s just try ( G'(0.1) ). ( G'(0.1) ) is definitely going to be in this interval, and so it’s going to be equal to:

[ 2 \cdot 0.1 = 0.2 \cdot \ln(0.1) + 0.1. ]

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 definitely going to be greater than ( -1 ) because ( e^{-1} ) gets you to, let’s see, ( e^{1} ) is ( \frac{1}{e} ).

So ( \frac{1}{e} ) is going to be around ( 0.3679 ), so in order to get ( 0.1 ), you have to be even more negative. So this is going to be less than ( 1 ); hence, I'm multiplying it times ( 0.2 ), I’m going to get a negative value that is less than ( 0.2 ). If I’m adding ( 0.1 ) to it, well, I’m still going to get a negative value.

So over this yellow interval, ( G'(x) < 0 ). In this blue interval, what’s going on? This will be easier; we could just try out the value ( 1 ).

So ( G'(1) = 2 \cdot \ln(1) + 1 ). Natural log of ( 1 ) is just ( 0 ), so all of this just simplifies to ( 1 ). Thus, over this blue interval, I sampled a point there, and ( G'(x) > 0 ).

So it looks like our function is decreasing from ( 0 ) to ( \frac{1}{\sqrt{e}} ) and then we increase after that. We increase for all ( x ) after that, which are greater than ( \frac{1}{\sqrt{e}} ).

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 absolute minimum point at ( x = \frac{1}{\sqrt{e}} ).

So let me write this down: We hit an absolute minimum at ( x = \frac{1}{\sqrt{e}} ) and there is no absolute maximum. As we get above ( \frac{1}{\sqrt{e}} ), we are just going to think about what’s going to be happening here — we know our function just keeps on increasing and increasing forever.

You could look at, even this, ( x ) is just going to be unbounded towards infinity, and natural log of ( x ) is going to grow slower than ( x^2 ), but it’s still going to go unbounded towards infinity. So there’s no global or absolute maximum; no absolute maximum point.

Now, let’s look at the graph of this to feel good about what we just did analytically. Without looking at it graphically, I looked at it ahead of time, so let me copy and paste it.

This is the graph of our function. As we can see when this point right over here — this is when ( x = \frac{1}{\sqrt{e}} ) — it’s not obvious from looking at it that it’s that point ( x = \frac{1}{\sqrt{e}} ). We can see that it is indeed an absolute minimum point here, and there is no absolute maximum point as there are arbitrarily high values that our function can take on.