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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 5

Lesson 5: Using the candidates test to find absolute (global) extrema# Absolute minima & maxima (entire domain)

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.3 (EK)

Sal analyzes the absolute minimum and maximum points of g(x)=x²ln(x) over its entire domain.

## Want to join the conversation?

- How can there be an absolute minimum if the interval is open? I thought in order for there to be an absolute min/max the interval had to be closed?(11 votes)
- If the interval is closed and bounded, any continuous function is guaranteed to have an absolute minimum. However, this does not mean that there cannot be an absolute minimum if the interval is open. If the interval is open, a continuous function could have an absolute minimum but is not guaranteed to have one.

For example, f(x) = x^2 on the open interval (-1, 1) has an absolute minimum value of 0 at x=0.

However, this same function has no absolute minimum on the open interval (0, 1).(26 votes)

- Isn't the method you would use to find the absolute minima or maxima in a closed interval the same as the one you would use to find the absolute minima or maxima in an open interval?(4 votes)
- For a closed interval you also need to consider the endpoint(s).(6 votes)

- why is sal plugging the numbers .1 and 1 into g'(x) and not g(x)? What does plugging them into the g'(x) tell us versus plugging them into g(x)?(3 votes)
- g(x) tells the value of y when x=1, but g'(x) tells the instantaneous slope when x=1. So when you determine that g'(x) is zero, you determine that there is most likely a minima or maxima.(4 votes)

- A lot of parts went unexplained here. How do we evaluate 1/sqrt(e) and ln(0.1) ? I find it very complicated to do without a calculator - we aren't allowed to use one in exams.(2 votes)
- What Sal is doing is approximating in his head.

This isn't that difficult – for example:

`1/sqrt(e) ≅ 1/sqrt(2.7)`

– simple rounding

We know that 2.25 = (1.5)² < 2.7 < 2² = 4, so:`1/1.5 = 2/3 > 1/sqrt(2.7) > 1/2`

Therefore`1/sqrt(e)`

lies between 1/2 and 2/3.

In fact, given that 2.7 is much closer to 2.25 than to 4 we can even guess that the value will be closer to 2/3 than to 1/2.(4 votes)

- is the calculator allowed in such cases. if no , how can I be expected to figure out Ln(0.1) <-1 .? also why only -1 . why not -2. or do i have to know the graph of Ln(x) which does let me know that value of Ln(x)<-1 for x<0.1. your advise please(3 votes)
- Well, we know that for all
`x : 0 < x < 1`

, we must have`ln(x) < 0`

because for`e^n < 1`

, we must have`n < 0`

. Then, we should also know that e = 2.72 (roughly). So`e^(-1) = 1/e = 1/2.72`

, which is a bit bigger than 1/3. Therefore, to get`e^n = 1/10 = 0.1`

, we would need to have`n < -1`

so that the denominator would be greater ( it'd be a higher power of`e`

). This translates to saying that`ln(0.1) < -1`

- that is, we must raise`e`

to a power`n`

that is less than -1 in order to make the expression e^n equal 0.1.

In this case, we just need to prove that ln(0.1) < -1 so that we then have`g'(0.1) < -0.2 + 0.1 = -0.1 < 0`

. So the end goal is to show that`g'(x) < 0`

, and we go with a simple calculation that will let us make that statement with no guesswork. In this case, ln(x) < -1 works. Really, ln(x) < -0.5 would have also worked. And ln(x) < -2 would still work, it's just less easy to prove because you have to square e or some other number to get a numerical denominator. But you can still prove fairly easily that because e < 3 and therefore 1/e > 1/3, and since`(1/3)^2 = 1/9 > 1/10`

,`(1/e)^2 > 1/9 > 1/10`

which implies`e^(-2) > 0.1`

, we therefore have`ln(0.1) < -2`

.

Hope that helps!(3 votes)

- At3:34, Sal says that this is the 1/sqrt(e) is the only number that will make g'(x) = 0, but isn't also x = 0 satisfies the equation. If it's true, then Sal has missed a critical point.(1 vote)
- He also says:0:21So the first thing I like to think about is well,0:23what's the domain for which g is actually defined?0:28And we know that in the natural log of x0:31the input into natural log,
**it has to be greater than zero**.

So, no, x = 0 is not a critical point.(6 votes)

- At3:00in the video, how is the value of ln(x) = -(1/2). I could not understand the calculation here and after it till x = 1/sqrt(e). Hoping someone would be kind enough to explain the calculation going on there, thank you.(1 vote)
- At3:00Sal divided 2x from both sides, -x/(2x) is the same as -1x/(2x). The x's cancelled out to have -1/2. From3:15he just uses properties of exponents to convert e^(-1/2) = 1/(e^(1/2)) = 1/√(e)(2 votes)

- Is calculating whether the function is increasing or decreasing near a critical number not just a waste of time? Is there anything wrong with computing all the critical values and endpoints (if closed interval) in the original function and comparing the results?(1 vote)
- Although the process of comparing the values of the function at the critical points and endpoints will tell you the absolute extrema, that will not work when you need to determine the relative extrema.(2 votes)

- If we bound the domain i,e 0 to 2 then how does we find an absolute maximum value in the interval

??(1 vote)- Depends on how you bound the domain. If you mean a closed interval [0,2] then in this case, the maximum occurs at 2, since the function is increasing after 1/√e. In fact, the Min-Max Theorem says that any continuous function on a closed interval will have an absolute minimum and maximum.

If you mean an open interval, (0,2), there's still no absolute maximum. If you said, for example, that the maximum occurred at x=1.9, I could find a larger value at x=1.99. So for any "largest" value you find, I could find a larger one.(2 votes)

- if the local minimum is the minimum value in the closed interval then can local minimum=absolute minimum?

and how can u tell the difference between local and absolute

why are they named as local and absolute

whats the meaning of local minimum does that mean the curve has to have sudden increments and decrements?(1 vote)- A local minimum can indeed be an absolute minimum. The difference between a local minimum and an absolute minimum is that a local minimum is just a minimum in general of a function either within a closed interval or across the entire domain. An absolute minimum means that out of every minimum of the function whether it be in an interval or the entire domain that it "goes the lowest" so to speak or has the smallest value. A local minimum just means that to the very left and right of the point there is nothing of the same value or less than that point. But only specifically right next to that point, there can be other parts of the function that are at the same height or lower and that would be another minimum.(2 votes)

## Video transcript

- [Voiceover] So we have the function g of x is equal to x
squared times the natural log of x. And what I wanna do in this video is see if you 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 a global minimum. So the first thing I like
to think about is well, what's the domain for which
g is actually defined? And we know that in the natural log of x 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 access so
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 critical
points, or critical values. So let's take the derivative of g. So g prime, using a new
color just for kicks. All right, so g prime of x is equal to, we could use the product rule here. So derivative of x squared which is two x times the natural log of x plus x squared times the derivative of natural log of x,
so that is one over x, and I can just rewrite that. X squared times one over x. And we're gonna assume that x is positive. So that is going to be, that is going to be just, and actually I didn't even
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 points. Critical points are where the derivative they're points in the domain, so they're gonna have to
satisfy x is greater than zero, such that g prime 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. Two 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
two x 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 gonna be zero, our domain is x is greater than zero. So this is going to be,
actually let's divide both sides by two x. So that we get the natural
log of x is equal to negative one half. Negative x divided by two
x is negative one half. Or we could say that x is equal to e to the negative one half is equal to x. Remember, natural log is just log base e. So e to the negative one half, which we could also write like that, e to the negative one half, or one over the square root of e. So 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. 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 they're have to be
points within the domain. So let's see, what would
make this undefined? The two x 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 really help us visualize this. So, if this is negative one, this is zero, this is let's see, e to
the, this is gonna be like one over, oh boy this is, this is going to be a
little bit less than one, so let's see, why don't we plot one here, and then two here, and so
we have a critical point at one over the square root of e, and we'll put it right over there. One over the square root of e. And we know that we're only defined from, for all x's greater than zero. So let's think about the interval between zero and this critical point. Right over here. So the open interval,
from zero to one 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 x greater than one over
the square root of e. So that's the interval from
one over square root of e to infinity. So over that yellow interval,
we 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. G 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 0.2 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 definitely going to be
greater than negative one. Cuz e to the negative
one only gets you to, let's see, e to the
negative one is one over e so that's one over 2.7, so
one over 2.7 is going to be, so this is going to be around 0.3 or 0.4. So in order to get point one, so this'll be around 0.3 to 0.4. So in order to get to 0.1 you
have to be even more negative. So this is going to be, I could
say less than negative one. So if this is less than negative one, and I'm multiplying it times 0.2, I'm gonna 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 for 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. Now let's see in this blue
interval what's going on. And this'll 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,
I sampled a 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 all
x's 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 a 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 minimum at x equals one over the square root of e, and
there is no absolute maximum. As we get above one over
the square root of e we are just going to think about what's going to be happening here. We're just going to one, we know that our function
just keeps on increasing and increasing and increasing forever. And you could look at even this, x squared is just gonna get
unbounded towards infinity, and natural log of x is gonna
grow slower than x squared, but it's still gonna 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 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 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. There's arbitrarily high values that our function can take on.