Lecture Calculus is the study of infinite Mathematics. In essence, it is the extension of algebraic concepts to their limfinity(l). What does that even mean? Well, let's begin with the notion of functions and proceed from there. As on tb. 2, a function (f) is a rule that assigns to each element x in a set D exactly one element y (= f(x)) in set E. The set D is called the domain of f, and E is called the codomain of f. The range of f may be any set containing E. We usually call x the independent variable and y the dependent variable. Why? There are generally 4 ways to represent a function. 1. verbal description 2. tabular 3. via formula 4. Graphically Usually we can detect whether or not a relation is a function by graphically using the vertical line test (VLT). If a vertical line can pass through the Cartesian graph of the relation and intersect it at more than one point, then the relation is not a function. A function for which to every y there corresponds one and only one x is called a 1 1 funtion. Here are some examples. Ex. Consider the following relations and list their descriptives. y = x 2 + 1 This is a function, but not 1 1. The domain is R, and the codomain is [1, ). x 2 + y 2 = 1 This not a function since the graph is a circle. Both domain and codomain are [-1, 1]. x 3 3 = y This is a 1 1 function. Both domain and codomain are R. [] 1S
Lecture Some functions may be defined in a piecewise fashion. For instance the function may be defined one way on one part of the domain and another way on a second part. Here's another example. Ex. Consider the following function. f(x) = -x 2 I x (-,0) + x 2 I x [0, ). Graph this bad boy. You'll see that this is, in fact, a 1 1 function. [] Genesis 1 24Then God said, Let the earth bring forth the living creature according to its kind: cattle and creeping thing and beast of the earth, each according to its kind ; and it was so. 2S
Lecture If a function f(x) satisfies f(-x) = f(x) x D, then f is said to be an even function. If it satisfies f(-x) = -f(x) x D, then f is said to be an odd function. Ex. Are there any even or odd functions in the previous example problems? Yes, in fact, f(x) = x 2 + 1 is an even function. And x 2 + y 2 = 1 is an even relation. [] Lecture On p. 7 we see that a function is said to increasing on an interval I if f(x 1 ) < f(x 2 ) whenever x 1 < x 2 in I, and decreasing on I if f(x 1 ) > f(x 2 ) whenever x 1 < x 2 in I. Ex. Of the previous functions mentioned which are increasing on [0, 1]? f(x) = x 2 + 1 f(x) = -x 2 I x (-,0) + x 2 I x [0, ) f(x) = x 3 3 = y. [] Student problem set A (Section 1.1) 4, 6, 8, 10, 11, 13, 22, 24, 28, 36, 42, 56 Student Solution Set A A6. Yes, it's a function where D = [-2, 2] and E = [-1, 2]. A22. [(a + h) 3 a 3 ]/h = (a 3 + 3a 2 h + 3ah 2 + h 3 a 3 )/h = 3a 2 + 3ah + h 2. [] Lecture As part of the idea of the mathematics of infinity, we'll begin with the definition of a limit (tb. 31). Let f(x) be defined on an open interval containing the number a (except possibly at a itself). Then lim x-> a f(x) = L means that for every ε > 0, there exists a δ > 0 such that if x a < δ => f(x) L < ε. Similarly lim x-> a + f(x) = L means that for x > a and for every ε > 0, there exists a δ > 0 such that if x a < δ => f(x) L < ε. The expression lim x-> a - f(x) = L is defined similarly. The idea is that we want to find out what the function does as x approaches a certain number. Here's an old Eagles song to demonstrate the concept. 3S
Ex. If f(x) = 5x/4, find lim x->4 f(x). In viewing the graph to the right it is clear that the function is a line going right through the point (4, 5). Therefore we have no problem asserting that lim x->4 5x/4 = 5. [] Ex. Show that lim x->4 5x/4 = 5. Let ε > 0 and choose δ =.8ε. If x 4 < δ => So then (5/4) x 4 = 5x/4 5 < 5δ/4 = ε (def. of length). lim x->4 5x/4 = 5 (p. 3). //. [] Ex. Show lim x->3 x 2 = 9. This is actually a C-N M405 type problem, so don't get too worked up. We'll begin by choosing/letting > 0. Does there exist > 0 so that if 0 < x 3 < => x 2 9 <? 4S
Ex. (cont.) Let's choose = ( /6). Now using properties of inequalities, for x 3, x + 3 < 3 + δ, and 0 < x 2 9 < x 3 (x + 3) 2 + 3δ = ε 2 /36 + ε/2 < ε.(cauchy-schwarz Inequality). So x 2 9 < ε. Then lim x->3 x 2 = 9 (p. 3). //. [] Student problem set B (Section 1.3) 4, 6, 12, 16, 20, 24, 28, 30 Student Solution Set B B4. a) 3, c) DNE. B30. Choose δ = ε/2. [] Lecture Here are some rules for limits from tb 35. Of course, note that for the time being all limits exist, all quantities are finite and no denominators can = 0. In the common vernacular (class repeat): 1. The limit of a sum is the sum of the limits. 2. The limit of a product is the product of the limits. 3. The limit of an inverse is the inverse of the limits. 4. The limit of a quotient is the quotient of the limits. And then, stemming from these, here are some common (and useful) results. 1. lim x->a [f(x)] n = [lim x->a f(x)] n, n N. 2. lim x->a [f(x)] 1/n = [lim x->a f(x)] 1/n, n N. Further, we see that if f(x) is a polynomial, trigonometric or rational function and a D, then lim x->a f(x) = f(a). Ex. Here is a function with some interesting features. We see that lim x->a2 f(x) = L2, lim x->a1 + f(x) = L2, lim x->a1 - f(x) = L1, and L f(x) = L2. [] 5S
Lecture On tb 38, 39 we see that for two functions f(x) = g(x), x a, => lim x->a f(x) = lim x->a g(x) provided the limits exist. Further, lim x->a f(x) = L iff lim x->a- f(x) = L = lim x->a+ f(x). On tb 41 we have two theorems, the second of which is actually called the Sandwich theorem. First if f(x) g(x) in an open interval containing a (except possibly at a) => lim x->a f(x) lim x->a g(x) provided the limits exist. This leads to the Sandwich theorem. If f(x) g(x) h(x) in an open interval containing a (except possibly at a) and lim x->a f(x) = L = lim x->a h(x) => lim x->a g(x) = L. As a result of the Sandwich theorem we have the following sign of the times. lim x->0 sin(x)/x = 1 (why?). Now here are some examples. Ex. Find lim x->0 f(x) where f(x) = -xi x (-, 0) + xi x (0, ) (the original V). Well, we know that lim x->0- f(x) = lim x->0- -x = 0 = lim x->0- x = lim x->0+ f(x). So by p. 6, lim x->0 f(x) = 0. [] Ex. Find lim x->0 f(x) where f(x) = (x 2 + 6x + 9)/(2x 3). No worries. We know that lim x->0 (x 2 + 6x + 9) = 9, and that lim x->0 (2x - 3) = -3. So by p. 5, lim x->0 (x 2 + 6x + 9)/(2x 3) = 9/(-3) = -3. [] Ex. Suppose f(x) = (x 2 9)/(x + 3). Find lim x->-3 f(x). Even though we may be tempted to utilize the results from p. 5, we cannot. Why? Because we have a singularity at x = -3. What to do? Here's the graph from Geogebra. 6S
Ex. (cont.) But wait!?! Isn't there supposed to be an asymptote at x = 3??? What just happened? Actually, a little cancellation just happened (how?). As a result the new f(x) = (x 3)I x (R \ -3). So by p. 6, lim x->-3 f(x) = -6. [] Student problem set C (Section 1.4) 1, 2, 6, 18, 19, 28, 30, 40, 44, 48, 62 Student Solution Set C C18. 12. C40. L = 0. The limit must be taken from the left of c since the speed of light ( 186K mi/sec) is as fast as we can conceive of sans going back in time. [] Lecture On tb 46 we see that, simply put, a function, f, is said to be continuous at real number a if lim x->a f(x) = f(a). So what does this mean? Actually 3 things. First, the function is defined at a. Next, that the limit of f exists at a. And third, that the first two must be equal. Ex. If we revisit the graph of f(x) = (x 2 9)/(x + 3) it appears that everything is hunky dory! Of course, that is not the case. The function is discontinuous at x = -3 since it is not defined there. The limit parts are fine, but the non-existence of the function makes it discontinuous. []. Lecture Practically speaking, a function is continuous over some interval if the graph of the function can be drawn in one fell swoop, sans lifting the pen off of the paper. Recall from p 5 that the precise definition of continuity is true for all polynomial, trigonometric and rational functions. The discontinuity in the function above is called a removable discontinuity. In an earlier example we saw this graph. 7S
Lecture Here lim x->0 +f(x) is an example of an infinite discontinuity, and lim x->a1 f(x) is an example of a jump discontinuity. Further if for a real number a lim x->a +f(x) = f(a) then f is said to be right continuous at a. Left continuous is defined similarly. Now let's extend this concept. A function f is said to be continuous on an interval of R if it is continuous at every number in the interval (meaning f is right continuous at the leftmost number in the interval and vice versa). Tb 51 tells us that all polynomials are continuous on R, and all rational, root and trigonometric functions are continuous on R except at those values not in their domains. As the reader might imagine from the global result on limits on p 5, if f and g are both continuous at a real number a, then 1. The sum, f + g, is continuous. 2. The product, fg, is continuous. 3. The quotient, f/g, is continuous. Practice saying this, and let's do some examples. Ex. Consider the function f(x) given below. Note that f has an infinite discontinuity at x = -3, a jump discontinuities at x = 1, and no discontinuity at x = 0. So what the heck is going on as x??? [] Ex. Now try to draw the graph of a function f that is a) right continuous at -2, but not left continuous there, b) has a removable continuity at x = 3, c) has a jump continuity at x = 6. [] Ex. Consider f(x) = (1 x 2 )I x (-, 1) + x -1 I x [1, ). Why is this function discontinuous at a = 1? Notice that the first part is a polynomial and is defined only up to x = 1. Notwithstanding, by p 5 lim x->1- f(x) = 0. On the right of x = 1, x-1 is a rational function, so again by p 5 lim x->1+ f(x) = 1. By the definition on p 7, then, the limit DNE, therefore f is discontinuous at a = 1. [] 8S
Lecture Let's take a look now at some features of continuous functions. A function f is said to be continuous on an interval (a, b) if it is continuous at every number in (a, b). We can also say that f is continuous on [a, b] if f is right continuous at a, and left continuous at b. f is continuous on R if it is continuous on (-, ).Here a, b R ext. What does this mean? We must be definitive on the concept of. Ex. Let's revisit the beautiful graph of a function we saw earlier. Where is the function f continuous. We can assume that f continues indefinitely in both directions. So we see that lim x->- + f(x) = (maybe)? and that Also lim x-> + f(x) = 0. lim x->-3- f(x) = - (maybe) = lim x->-3+ f(x)? f also has discontinuities at x = 1, although it is right continuous at both numbers. Finally we can write that f is continuous on Ex. (-, -3) (-3, -1] and on (-1, 1] and (1, ). [] Okay what about this bad boy we saw earlier? f(x) = (1 x 2 )I x (-, 1) + x -1 I x [1, ). Since we also saw that f was discontinuous at x = 1, but is also right continuous there, then we can write that f is continuous on (-, 1) and [1, ). [] Student problem set D (Section 1.5) 4, 6, 8, 10, 16, 18, 20, 30, 32 Student Solution Set D D4. [-4, 2), (-2, 2), [2, 4), (4, 6), (6, 8). D30. lim x-> /4- sinx = (2)/2 = lim x-> /4+ cosx. => lim x-> /4 f(x) = f( /4). Since the function is continuous elsewhere (p 8), then it is continuous on R (p 7). //. [] 9S
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