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5.4 abs extrema.pdf

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Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Tues 10/28/08 5.4 - Absolute max/min It’s what you think it is: an absolute max is the biggest point on the graph, while the absolute min is the smallest point. While there can be many di▯erent relative max’s and min’s, there can be only one absolute max and min (think about it: there can only be one highest point (unless another point reaches the same height, but who cares?)) In general, there is no guarantee that there will be an absolute max or min on a given interval Example: f(x) = x ▯ 2 has an absolute min of ▯2, but no absolute max. f(x) = ▯(x ▯ 3) + 4 has an absolute max of 4 but no absolute min. f(x) = x ▯ 2x ▯ 5x + 3 has neither an absolute max nor min (why? because it shoots o▯ to ▯1 and +1). f(x) = cos(x); [0;2▯] has an absolute max at 1 and an absolute min at -1. f(x) = e ; [0;4] has an absolute max at e and an absolute min at 1. In fact, I think we can now say something about absolute max’s and min’s. So, here’s another theorem of which you must know the name (there are only a few more in this class! Hang in there!): Extreme Value Theorem if f(x) is continuous over a closed interval [a;b], then it has both an absolute max AND min on [a;b]. Why do we need continuity and a closed interval? continuity: f(x) = 1, [▯1;1]. This guy shoots up to 1 and down to ▯1. Certainly, no absolute x max or min here! closed interval: f(x) = 2x, (▯1;1). Why doesn’t this have an absolute max? Well, 2 would be the max, but x = 1 is not in the interval we’re dealing with! if you throw me something of what you think is the max (where x has to be in the interval (▯1;1)), i can give you something that’s of higher value! Same thing with the min. Here’s something else that’s neat: if you already have an absolute max or min and it’s over an open inteval (a;b), then this abs extremum occurs at a critical point! Okay, so this Extreme Value Theorem guarantees an absolute max and min of a cont function over [a;b]. How do we go about calculating them? It’s really not that bad! Here’s what you do: (1) ▯nd all the critical points. Throw out any critical points that are outside of the interval. 1 (2) calculate where the critical points are AND calculate the end poin0s (ie if x is a critical/end point, calculate 0(x )). (3) you’re done! just see which is the biggest and that’s your abs max, and which is the smallest and that’s your abs min! Example: Find the absolute min and max of f(x) = x ▯ 3x + 1 over [▯ ;4]. 2 Solution: First, ▯nd all the critical points. 0 2 f (x) = 3x ▯ 6x = 3x(x ▯ 2) So x = 0, x = 2. Now ▯nd all the values of the critical points and end points! f(0) = 1 f(2) = 8 ▯ 12 + 1 = ▯3 ▯ ▯ 1 1 3 7 1 f ▯ = ▯ ▯ + 1 = ▯ + 1 = 2 8 4 8 8 f(4) = 64 ▯ 48 + 1 = 17 So, which is the biggest? 17. And which is the smallest? -3. So, the absolute max is at (4;17) while the absolute min is at (2;▯3). Always make sure you check those endpoints! It is very much possible that an absolute extremum occurs at an endpoint rather than a critical point! Example: Find the absolute min and max of f(x) = x ▯ 2sin(x) over [▯ ; ]. 4 2 0 f (x) = 1 ▯ 2cos(x) 0 1 ▯ so, setting f (x) = 0, we get cos2x) = , which mean3 (there’s also ▯▯=3, but that’s less than ▯▯=4, so it’s not in the interval we’re working with). Now we check this critical point and the end points! ▯ ▯ p ▯ ▯ ▯ ▯ 3 ▯ p f = ▯ 2sin = ▯ 2 = ▯ 3 ▯ ▯:68 3 3 3 3 2 3 ▯ ▯ p ▯ ▯ ▯▯ ▯ 2 ▯ p f ▯ 4 = ▯ 4 ▯ 2sin 4 = ▯ 4 + 2 2 = ▯ 4 + 2 ▯ :63 ▯ ▯ ▯ ▯ ▯ ▯ f = ▯ 2sin = ▯ 2(1) = ▯ 2 ▯ ▯:43 2 2 2 2 2 ▯ p ▯ ▯ p ▯ So the abs max occurs at▯▯4;▯4+ 2 and abs min at 3;3 ▯ 3 . Well, the world isn’t always so nice to us, so sometimes if we’re not given a closed interval, we have to deal with it anyway. A lot of times this means working with (▯1;1). So although we’re no longer guaranteed abs extrema (that was the extreme value theorem, which worked on closed intervals), that doesn’t automatically mean that there are none. 2 2 Example: ▯nd absolute extrema, if any, of f(x) = x over (▯1;1). If we graph this guy, we can see that there’s an absolute min at x = 0, but how do we go about showing that? By taking the limit. 2 lim x = +1 x!1 2
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