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5.2 relative extrema and graphing.pdf

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Dan Dolderman

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Calc 1 Tues 10/21/08 5.2 - Relative Extrema, Graphing Polynomials Relative Maximum: The point (x ,0(x )0 is considered a relative max if there is an open interval containing x on 0 which f(x ) is the largest value. In other words, graphically, it’s the highest point in a general 0 area. It doesn’t have to be the strictly highest point in total on the graph (although, that is also a relative max), but just a highest point in some surrounding area. Think mountains. You got a whole bunch of mountain peaks. Yes, there is probably only one tallest peak, but there are certainly a whole bunch more of smaller peaks. The relative max of a function would be all those peaks in the mountains. Relative Minimum Just like how there’s a concept called relative max, there is also the relative min. Well, it’s the same thing as max, only it’s the smallest point on the graph in a surrounding area. More formally, x0is a relative min if there is an open interval containint x 0n which f(x ) i0 the smallest value. Functions can certainly have both relative maxes and mins, perhaps just one but not the other, or neither. For both, think of f(x) = sin(x). For just a min, think f(x) = x . For just a max, think 2 f(x) = −x . And for neither maxes nor mins occuring anywhere, think f(x) = x. Relative maxes and mins are under the giant category of Relative Extrema, by the way. So, when I talk about relative extrema, I just mean either a relative max or min, whichever. Critical Point: ′ A critical point is a point in the domain of f at which there is a horizontal tangent (ie, f (x) = 0) or the derivative doesn’t exist. Stationary Point: ′ A stationary point is a point at which f (x) = 0 only. In other words, every stationary point is a critical point, but not every critical point is a stationary point. Relative extrema only occur at critical points. So, in other words, if you want to find the relative extrema of a function, you have to find the critical points! How do you do that? Well, just set the derivative equal to zero to find the stationary points, and then look for where the derivative is undefined for the rest of ’em! 1 ′ 3 Example: Find the critical points of f(x) where f (x) = x − 4x. Solution: Well, our derivative is a polynomial, so it’s nice and defined everywhere. This means we only need to worry about when f (x) = 0 (ie, just the stationary points). Therefore, ′ 3 2 f (x) = x − 4x = x(x − 4) = x(x − 2)(x + 2) and when we set this equal to zero, x = −2,0,2. The First Derivative Test: Ah, so now we can actually find out if something is a relative max or min! In fact, we can have critical points that are not relative extrema! How do we determine this? Why, with the first ′ derivative test of course! So, here’s what the test says: if f changes sign at a critical point, then there’s a relative extremum there. Moreover, as f (x) goes over a critical point: - if f (x) goes from positive to negative, then there’s a relative max. - if f (x) goes from negative to positive, then there’s a relative min. - if f (x) has the same sign (positive to positive or negative to negative), there is no relative extremum at that critical point. I highly highly suggest that you take a look at figure 5.2.6 on page 281 in your book to get an idea of different types of critical points and which give relative extrema. The different types of critical points that you may run into (at least the ones that give you relative extrema): - horizontal tangents (ok, yeah you know this, but I had to put it in here anyway!) - corners - cusps Gee, it feels like we went over this stuff already when we first started studying derivatives! Wait a minute! We did basically study this stuff already! Remember, the deal with corners and cusps are that the function is, in fact, defined and continuous at these points, but you get problems with the derivative. A corner comes from when you get a jump discontinuity in the derivative. A cusp comes about when you get a VA in the derivative (but everything was fine in the original function!). 2 Example: Find the relative extrema of f(x) where f (x) = x − 4x. 3 Solution: All right, let’s bring this guy back and determine where those maxes and mins are! We found before that f (x) = x(x + 2)(x − 2), so x = −2,0,2 when f (x) = 0. So, we need to do some checking of signs! This is going to work just like that concavity stuff we did on monday: −−−−−−− +++++++ −−−−−−− +++++++ −2 0 2 −3 −1 1 3 So, around x = −2, we’re going from negative to positive. f is decreasing and then increasing. This corresponds to a relative min! As for x = 0, we’re going from positive to negative, which is a max. Similarly, x = 2 is also a min. I’m sorry I can’t do more examples, but there’s a lot of material to get through today. I’ll do more examples if there is time in class. The Second Derivative Test: The name says it all. Take the second derivative of a function f in order to determine relative extrema. So, if f (x )0= 0 (ie, we have a stationary point x ), the0 - if f (x ) > 0, you have a relative min. 0 ′′ - if f (x0) < 0, you have a relative max. ′′ - if f (x0) = 0, you just wasted your time. x could0be a rel max, or perhaps it could be a relative min...or maybe neither of them! ′ 3 Example: Why not bring back that same function f where f (x) = x − 4x! Find relative extrema using the second derivative test.
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