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### Course: Differential Calculus>Unit 5

Lesson 5: Absolute (global) extrema

# Absolute minima & maxima (entire domain)

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

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• 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?
• 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).
• 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)?
• 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.
• 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?
• For a closed interval you also need to consider the endpoint(s).
• At , 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:
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, no, x = 0 is not a critical point.
• 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.
• 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.
• 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
• 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!
• At in 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)
• At Sal divided 2x from both sides, -x/(2x) is the same as -1x/(2x). The x's cancelled out to have -1/2. From he just uses properties of exponents to convert e^(-1/2) = 1/(e^(1/2)) = 1/√(e)
• 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.
• 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.