Lecture notes on Algebra and Trigonometry

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College Algebra and Trigonometry a.k.a. Precalculus by Carl Stitz, Ph.D. Jeff Zeager, Ph.D. Lakeland Community College Lorain County Community College August 30, 2010Chapter 1 Relations and Functions 1.1 The Cartesian Coordinate Plane In order to visualize the pure excitement that is Algebra, we need to unite Algebra and Geometry. Simply put, we must nd a way to draw algebraic things. Let's start with possibly the greatest 1 mathematical achievement of all time: the Cartesian Coordinate Plane. Imagine two real number lines crossing at a right angle at 0 as below. y 4 3 2 1 4 3 2 1 1 2 3 4 x 1 2 3 4 The horizontal number line is usually called the x-axis while the vertical number line is usually 2 called they-axis. As with the usual number line, we imagine these axes extending o inde nitely in both directions. Having two number lines allows us to locate the position of points o of the number lines as well as points on the lines themselves. 1 So named in honor of Ren e Descartes. 2 The labels can vary depending on the context of application.2 Relations and Functions For example, consider the point P below on the left. To use the numbers on the axes to label this point, we imagine dropping a vertical line from the x-axis to P and extending a horizontal line from the y-axis to P . We then describe the point P using the ordered pair (2;4). The rst number in the ordered pair is called the abscissa or x-coordinate and the second is called the 3 ordinate or y-coordinate. Taken together, the ordered pair (2;4) comprise the Cartesian coordinates of the point P . In practice, the distinction between a point and its coordinates is blurred; for example, we often speak of `the point (2;4).' We can think of (2;4) as instructions on how to reach P from the origin by moving 2 units to the right and 4 units downwards. Notice that the order in the ordered pair is important if we wish to plot the point (4; 2), we would move to the left 4 units from the origin and then move upwards 2 units, as below on the right. y y 4 4 3 3 (4; 2) 2 2 1 1 4 3 2 1 1 2 3 4 4 3 2 1 1 2 3 4 x x 1 1 2 2 3 3 4 4 P (2;4) P  5 Example 1.1.1. Plot the following points: A(5; 8), B ; 3 , C(5:8;3), D(4:5;1), E(5; 0), 2 4 F (0; 5), G(7; 0), H(0;9), O(0; 0). Solution. To plot these points, we start at the origin and move to the right if the x-coordinate is positive; to the left if it is negative. Next, we move up if the y-coordinate is positive or down if it is negative. If the x-coordinate is 0, we start at the origin and move along the y-axis only. If the y-coordinate is 0 we move along the x-axis only. 3 Again, the names of the coordinates can vary depending on the context of the application. If, for example, the horizontal axis represented time we might choose to call it the t-axis. The rst number in the ordered pair would then be the t-coordinate. 4 The letter O is almost always reserved for the origin.1.1 The Cartesian Coordinate Plane 3 y 9 8 A(5; 8) 7 6 5 F (0; 5) 4 3  5 B ; 3 2 2 1 G(7; 0) O(0; 0) E(5; 0) 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 x 1 D(4:5;1) 2 3 C(5:8;3) 4 5 6 7 8 9 H(0;9) When we speak of the Cartesian Coordinate Plane, we mean the set of all possible ordered pairs (x;y) asx andy take values from the real numbers. Below is a summary of important facts about Cartesian coordinates. Important Facts about the Cartesian Coordinate Plane • (a;b) and (c;d) represent the same point in the plane if and only if a =c and b =d. • (x;y) lies on the x-axis if and only if y = 0. • (x;y) lies on the y-axis if and only if x = 0. • The origin is the point (0; 0). It is the only point common to both axes.4 Relations and Functions The axes divide the plane into four regions called quadrants. They are labeled with Roman numerals and proceed counterclockwise around the plane: y 4 Quadrant II Quadrant I 3 x 0, y 0 x 0, y 0 2 1 4 3 2 1 1 2 3 4 x 1 2 Quadrant III Quadrant IV 3 x 0, y 0 x 0, y 0 4 For example, (1; 2) lies in Quadrant I, (1; 2) in Quadrant II, (1;2) in Quadrant III, and (1;2) in Quadrant IV. If a point other than the origin happens to lie on the axes, we typically refer to the point as lying on the positive or negative x-axis (if y = 0) or on the positive or negative y-axis (if x = 0). For example, (0; 4) lies on the positive y-axis whereas (117; 0) lies on the negative x-axis. Such points do not belong to any of the four quadrants. 5 One of the most important concepts in all of mathematics is symmetry. There are many types of symmetry in mathematics, but three of them can be discussed easily using Cartesian Coordinates. Definition 1.1. Two points (a;b) and (c;d) in the plane are said to be • symmetric about the x-axis if a =c and b =d • symmetric about the y-axis if a =c and b =d • symmetric about the origin if a =c and b =d 5 According to Carl. Je thinks symmetry is overrated.1.1 The Cartesian Coordinate Plane 5 Schematically, y Q(x;y) P (x;y) 0 x R(x;y) S(x;y) In the above gure, P and S are symmetric about the x-axis, as are Q and R; P and Q are symmetric about the y-axis, as are R and S; and P and R are symmetric about the origin, as are Q and S. Example 1.1.2. LetP be the point (2; 3). Find the points which are symmetric to P about the: 1. x-axis 2. y-axis 3. origin Check your answer by graphing. Solution. The gure after De nition 1.1 gives us a good way to think about nding symmetric points in terms of taking the opposites of the x- and/or y-coordinates of P (2; 3). 1. To nd the point symmetric about the x-axis, we replace the y-coordinate with its opposite to get (2;3). 2. To nd the point symmetric about the y-axis, we replace the x-coordinate with its opposite to get (2; 3). 3. To nd the point symmetric about the origin, we replace the x- andy-coordinates with their opposites to get (2;3). y 3 P (2; 3) 2 (2; 3) 1 321 1 2 3 x 1 2 3 (2;3) (2;3)6 Relations and Functions One way to visualize the processes in the previous example is with the concept of re ections. If we start with our point (2; 3) and pretend the x-axis is a mirror, then the re ection of (2; 3) across thex-axis would lie at (2;3). If we pretend they-axis is a mirror, the re ection of (2; 3) across that axis would be (2; 3). If we re ect across the x-axis and then the y-axis, we would go from (2; 3) to (2;3) then to (2;3), and so we would end up at the point symmetric to (2; 3) about the origin. We summarize and generalize this process below. Re ections To re ect a point (x;y) about the: • x-axis, replace y withy. • y-axis, replace x withx. • origin, replace x withx and y withy. 1.1.1 Distance in the Plane Another important concept in geometry is the notion of length. If we are going to unite Algebra and Geometry using the Cartesian Plane, then we need to develop an algebraic understanding of what distance in the plane means. Suppose we have two points, P (x ;y ) and Q (x ;y ); in the 1 1 2 2 plane. By the distanced betweenP andQ, we mean the length of the line segment joiningP with Q. (Remember, given any two distinct points in the plane, there is a unique line containing both points.) Our goal now is to create an algebraic formula to compute the distance between these two points. Consider the generic situation below on the left. Q (x ;y ) Q (x ;y ) 2 2 2 2 d d P (x ;y ) P (x ;y ) (x ;y ) 1 1 1 1 2 1 With a little more imagination, we can envision a right triangle whose hypotenuse has length d as drawn above on the right. From the latter gure, we see that the lengths of the legs of the triangle arejx xj andjy yj so the Pythagorean Theorem gives us 2 1 2 1 2 2 2 jx xj +jy yj =d 2 1 2 1 2 2 2 (x x ) + (y y ) =d 2 1 2 1 (Do you remember why we can replace the absolute value notation with parentheses?) By extracting the square root of both sides of the second equation and using the fact that distance is never negative, we get1.1 The Cartesian Coordinate Plane 7 Equation 1.1. The Distance Formula: The distance d between the points P (x ;y ) and 1 1 Q (x ;y ) is: 2 2 q 2 2 d = (x x ) + (y y ) 2 1 2 1 It is not always the case that the points P and Q lend themselves to constructing such a triangle. If the points P and Q are arranged vertically or horizontally, or describe the exact same point, we cannot use the above geometric argument to derive the distance formula. It is left to the reader to verify Equation 1.1 for these cases. Example 1.1.3. Find and simplify the distance between P (2; 3) and Q(1;3). Solution. q 2 2 d = (x x ) + (y y ) 2 1 2 1 p 2 2 = (1 (2)) + (3 3) p = 9 + 36 p = 3 5 p So, the distance is 3 5. Example 1.1.4. Find all of the points with x-coordinate 1 which are 4 units from the point (3; 2). Solution. We shall soon see that the points we wish to nd are on the line x = 1, but for now we'll just view them as points of the form (1;y). Visually, y 3 (3; 2) 2 1 distance is 4 units x 2 3 1 (1;y) 2 3 We require that the distance from (3; 2) to (1;y) be 4. The Distance Formula, Equation 1.1, yields8 Relations and Functions q 2 2 d = (x x ) + (y y ) 2 1 2 1 p 2 2 4 = (1 3) + (y 2) p 2 4 = 4 + (y 2)   p 2 2 2 4 = 4 + (y 2) squaring both sides 2 16 = 4 + (y 2) 2 12 = (y 2) 2 (y 2) = 12 p y 2 =  12 extracting the square root p y 2 = 2 3 p y = 2 2 3 p p We obtain two answers: (1; 2 + 2 3) and (1; 2 2 3): The reader is encouraged to think about why there are two answers. Related to nding the distance between two points is the problem of nding the midpoint of the line segment connecting two points. Given two points, P (x ;y ) andQ (x ;y ), the midpoint,M, 1 1 2 2 ofP andQ is de ned to be the point on the line segment connectingP andQ whose distance from P is equal to its distance from Q. Q (x ;y ) 2 2 M P (x ;y ) 1 1 If we think of reaching M by going `halfway over' and `halfway up' we get the following formula. Equation 1.2. The Midpoint Formula: The midpoint M of the line segment connecting P (x ;y ) and Q (x ;y ) is: 1 1 2 2   x +x y +y 1 2 1 2 M = ; 2 2 If we letd denote the distance betweenP andQ, we leave it as an exercise to show that the distance between P and M is d=2 which is the same as the distance between M and Q. This suces to show that Equation 1.2 gives the coordinates of the midpoint.1.1 The Cartesian Coordinate Plane 9 Example 1.1.5. Find the midpoint of the line segment connecting P (2; 3) and Q(1;3). Solution.   x +x y +y 1 2 1 2 M = ; 2 2   (2) + 1 3 + (3) = ; 2 2   1 0 = ; 2 2   1 = ; 0 2   1 The midpoint is ; 0 . 2 We close with a more abstract application of the Midpoint Formula. We will revisit the following example in Exercise 14 in Section 2.1. Example 1.1.6. If a =6 b, prove the line y = x is a bisector of the line segment connecting the points (a;b) and (b;a). Solution. Recall from geometry that a bisector is a line which equally divides a line segment. To provey =x bisects the line segment connecting the (a;b) and (b;a), it suces to show the midpoint of this line segment lies on the line y =x. Applying Equation 1.2 yields   a +b b +a M = ; 2 2   a +b a +b = ; 2 2 Since thex andy coordinates of this point are the same, we nd that the midpoint lies on the line y =x, as required.10 Relations and Functions 1.1.2 Exercises p 1. Plot and label the points A(3;7), B(1:3;2), C(; 10), D(0; 8), E(5:5; 0), F (8; 4), G(9:2;7:8) and H(7; 5) in the Cartesian Coordinate Plane given below. y 9 8 7 6 5 4 3 2 1 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 x 1 2 3 4 5 6 7 8 9 2. For each point given in Exercise 1 above • Identify the quadrant or axis in/on which the point lies. • Find the point symmetric to the given point about the x-axis. • Find the point symmetric to the given point about the y-axis. • Find the point symmetric to the given point about the origin.1.1 The Cartesian Coordinate Plane 11 3. For each of the following pairs of points, nd the distance d between them and nd the midpoint M of the line segment connecting them.     24 6 11 19 (a) (1; 2), (3; 5) (e) ; , ; . 5 5 5 5 (b) (3;10), (1; 2)     p p  p p  1 3 (f) 2; 3 , 8; 12 (c) ; 4 , ;1 p p  p p  2 2     (g) 2 45; 12 , 20; 27 . 2 3 7 (d) ; , ; 2 (h) (0; 0), (x;y) 3 2 3 4. Find all of the points of the form (x;1) which are 4 units from the point (3; 2). 5. Find all of the points on the y-axis which are 5 units from the point (5; 3). 6. Find all of the points on the x-axis which are 2 units from the point (1; 1). 7. Find all of the points of the form (x;x) which are 1 unit from the origin. 8. Let's assume for a moment that we are standing at the origin and the positive y-axis points due North while the positive x-axis points due East. Our Sasquatch-o-meter tells us that Sasquatch is 3 miles West and 4 miles South of our current position. What are the coordinates of his position? How far away is he from us? If he runs 7 miles due East what would his new position be? 9. Verify the Distance Formula 1.1 for the cases when: (a) The points are arranged vertically. (Hint: Use P (a;y ) and Q(a;y ).) 1 2 (b) The points are arranged horizontally. (Hint: Use P (x ;b) and Q(x ;b).) 1 2 (c) The points are actually the same point. (You shouldn't need a hint for this one.) 10. Verify the Midpoint Formula by showing the distance between P (x ;y ) and M and the 1 1 distance between M and Q(x ;y ) are both half of the distance between P and Q. 2 2 11. Show that the points A, B and C below are the vertices of a right triangle. (a) A(3; 2), B(6; 4), and C(1; 8) (b) A(3; 1), B(4; 0) and C(0;3) 12. Find a pointD(x;y) such that the pointsA(3; 1), B(4; 0), C(0;3) andD are the corners of a square. Justify your answer. 6 13. The world is not at. Thus the Cartesian Plane cannot possibly be the end of the story. Discuss with your classmates how you would extend Cartesian Coordinates to represent the three dimensional world. What would the Distance and Midpoint formulas look like, assuming those concepts make sense at all? 6 There are those who disagree with this statement. Look them up on the Internet some time when you're bored.12 Relations and Functions 1.1.3 Answers p 1. The required points A(3;7), B(1:3;2), C(; 10), D(0; 8), E(5:5; 0), F (8; 4), G(9:2;7:8), and H(7; 5) are plotted in the Cartesian Coordinate Plane below. y 9 8 D(0; 8) 7 6 H(7; 5) 5 4 F (8; 4) p 3 C(; 10) 2 1 E(5:5; 0) 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 x 1 2 B(1:3;2) 3 4 5 6 7 A(3;7) 8 G(9:2;7:8) 91.1 The Cartesian Coordinate Plane 13 2. (a) The point A(3;7) is (e) The point E(5:5; 0) is • in Quadrant III • on the negative x-axis • symmetric about x-axis with (3; 7) • symmetric about x-axis with (5:5; 0) • symmetric about y-axis with (3;7) • symmetric about y-axis with (5:5; 0) • symmetric about origin with (3; 7) • symmetric about origin with (5:5; 0) (b) The point B(1:3;2) is (f) The point F (8; 4) is • in Quadrant IV • in Quadrant II • symmetric about x-axis with (1:3; 2) • symmetric about x-axis with (8;4) • symmetric abouty-axis with (1:3;2) • symmetric about y-axis with (8; 4) • symmetric about origin with (1:3; 2) • symmetric about origin with (8;4) p (c) The point C(; 10) is (g) The point G(9:2;7:8) is • in Quadrant I • in Quadrant IV p • symmetric aboutx-axis with (; 10) • symmetric about x-axis with (9:2; 7:8) p • symmetric abouty-axis with (; 10) • symmetric abouty-axis with (9:2;7:8) p • symmetric about origin with (; 10) • symmetric about origin with (9:2; 7:8) (d) The point D(0; 8) is (h) The point H(7; 5) is • on the positive y-axis • in Quadrant I • symmetric about x-axis with (0;8) • symmetric about x-axis with (7;5) • symmetric about y-axis with (0; 8) • symmetric about y-axis with (7; 5) • symmetric about origin with (0;8) • symmetric about origin with (7;5)     p 7 13 13 3. (a) d = 5, M = 1; (e) d = 74, M = ; . 2 10 10 p p p p 2 3 (b) d = 4 10, M = (1;4) (f) d = 3 5, M = ; 2 2   p 3 p (c) d = 26, M = 1; p p 5 3 2 (g) d = 83, M = 4 5; . 2 p     37 5 7 p x y 2 2 (d) d = , M = ; (h) d = x +y , M = ; 2 6 4 2 2 p p p p 4. (3 + 7;1), (3 7;1) 6. (1 + 3; 0), (1 3; 0)     p p p p 2 2 2 2 7. ; , ; 5. (0; 3) 2 2 2 2 8. (3;4), 5 miles, (4;4) p p 11. (a) The distance from A toB is 13, the distance from A toC is 52, and the distance from B to p C is 65. Since       p 2 p 2 p 2 13 + 52 = 65 ; we are guaranteed by the converse of the Pythagorean Theorem that the triangle is right.14 Relations and Functions 1.2 Relations We now turn our attention to sets of points in the plane. Definition 1.2. A relation is a set of points in the plane. Throughout this text we will see many di erent ways to describe relations. In this section we will focus our attention on describing relations graphically, by means of the list (or roster) method and algebraically. Depending on the situation, one method may be easier or more convenient to use than another. Consider the set of points below y 4 3 (4; 3) 2 (2; 1) 1 4 3 2 1 1 2 3 4 x 1 2 3 (0;3) 4 These three points constitute a relation. Let us call this relation R. Above, we have a graphical description ofR. Although it is quite pleasing to the eye, it isn't the most portable way to describe R. The list (or roster) method of describing R simply lists all of the points which belong to R. 1 Hence, we write: R =f(2; 1); (4; 3); (0;3)g: The roster method can be extended to describe in nitely many points, as the next example illustrates. Example 1.2.1. Graph the following relations. 1. A =f(0; 0); (3; 1); (4; 2); (3; 2)g 2. HLS =f(x; 3) :2x 4g 1 3. HLS =f(x; 3) :2x 4g 2 4. V =f(3;y) :y is a real numberg 1 We use `set braces'fg to indicate that the points in the list all belong to the same set, in this case, R.1.2 Relations 15 Solution. 1. To graph A, we simply plot all of the points which belong to A, as shown below on the left. 2. Don't let the notation in this part fool you. The name of this relation is HLS , just like the 1 name of the relation in part 1 wasR. The letters and numbers are just part of its name, just like the numbers and letters of the phrase `King George III' were part of George's name. The next hurdle to overcome is the description of HLS itself a variable and some seemingly 1 extraneous punctuation have found their way into our nice little roster notation The way to make sense of the constructionf(x; 3) :2 x 4g is to verbalize the set bracesfg as `the set of' and the colon : as `such that'. In words,f(x; 3) :2 x 4g is: `the set of points (x; 3) such that2 x 4.' The purpose of the variable x in this case is to describe in nitely many points. All of these points have the same y-coordinate, 3, but the x-coordinate is allowed to vary between2 and 4, inclusive. Some of the points which belong to HLS include some friendly points like: (2; 3), (1; 3), (0; 3), (1; 3), (2; 3), (3; 3), and 1  p 5 (4; 3). However, HLS also contains the points (0:829; 3), ; 3 , ( ; 3), and so on. It is 1 6 impossible to list all of these points, which is why the variable x is used. Plotting several friendly representative points should convince you that HLS describes the horizontal line 1 segment from the point (2; 3) up to and including the point (4; 3). y y 4 4 3 3 2 2 1 1 4 3 2 1 1 2 3 4 4 3 2 1 1 2 3 4 x x The graph of A The graph of HLS 1 3. HLS is hauntingly similar to HLS . In fact, the only di erence between the two is that 2 1 instead of `2x 4' we have `2x 4'. This means that we still get a horizontal line segment which includes (2; 3) and extends to (4; 3), but does not include (4; 3) because of the strict inequality x 4. How do we denote this on our graph? It is a common mistake to make the graph start at (2; 3) end at (3; 3) as pictured below on the left. The problem with this graph is that we are forgetting about the points like (3:1; 3), (3:5; 3), (3:9; 3), (3:99; 3), and so forth. There is no real number that comes `immediately before' 4, and so to describe the set of points we want, we draw the horizontal line segment starting at (2; 3) and draw an `open circle' at (4; 3) as depicted below on the right.16 Relations and Functions y y 4 4 3 3 2 2 1 1 4 3 2 1 1 2 3 4 4 3 2 1 1 2 3 4 x x This is NOT the correct graph of HLS The graph of HLS 2 2 4. Our last example, V , describes the set of points (3;y) such that y is a real number. All of these points have an x-coordinate of 3, but the y-coordinate is free to be whatever it wants to be, without restriction. Plotting a few `friendly' points of V should convince you that all the points of V lie on a vertical line which crosses the x-axis at x = 3. Since there is no restriction on the y-coordinate, we put arrows on the end of the portion of the line we draw to indicate it extends inde nitely in both directions. The graph of V is below on the left. y 4 y 3 2 4 3 2 1 1 2 3 4 x 1 1 2 1 2 3 4 x 1 3 2 4 3 The graph of y =2 4 The graph of V The relation V in the previous example leads us to our nal way to describe relations: alge- braically. We can simply describe the points in V as those points which satisfy the equation x = 3. Most likely, you have seen equations like this before. Depending on the context, `x = 3' could mean we have solved an equation for x and arrived at the solution x = 3. In this case, how- ever, `x = 3' describes a set of points in the plane whose x-coordinate is 3. Similarly, the equation y =2 in this context corresponds to all points in the plane whose y-coordinate is2. Since there are no restrictions on thex-coordinate listed, we would graph the relationy =2 as the horizontal line above on the right. In general, we have the following.1.2 Relations 17 Equations of Vertical and Horizontal Lines • The graph of the equation x =a is a vertical line through (a; 0). • The graph of the equation y =b is a horizontal line through (0;b). In the next section, and in many more after that, we shall explore the graphs of equations in great 2 detail. For now, we shall use our nal example to illustrate how relations can be used to describe entire regions in the plane. Example 1.2.2. Graph the relation: R =f(x;y) : 1y 3g Solution. The relationR consists of those points whosey-coordinate only is restricted between 1 and 3 excluding 1, but including 3. The x-coordinate is free to be whatever we like. After plotting 3 some friendly elements of R, it should become clear that R consists of the region between the horizontal linesy = 1 andy = 3. SinceR requires that they-coordinates be greater than 1, but not equal to 1, we dash the line y = 1 to indicate that those points do not belong to R. Graphically, y 4 3 2 1 4 3 2 1 1 2 3 4 x The graph of R 2 In fact, much of our time in College Algebra will be spent examining the graphs of equations. 3 The word `some' is a relative term. It may take 5, 10, or 50 points until you see the pattern.18 Relations and Functions 1.2.1 Exercises 1. Graph the following relations. (a) f(3; 9), (2; 4), (1; 1), (0; 0), (1; 1), (2; 4), (3; 9)g (b) f(2; 2), (2;1), (3; 5), (3;4)g   2 (c) n; 4n :n = 0;1;2   6 (d) ;k :k =1;2;3;4;5;6 k 2. Graph the following relations. (a) f(x;2) :x4g (e) f(x;y) :y 4g (b) f(2;y) :y 5g (f) f(x;y) :x 3; y 2g (c) f(2;y) :3y 4g (g) f(x;y) :x 0; y 4g p 2 9 (d) f(x;y) :x 3g (h) f(x;y) : 2x ; y g 3 2 3. Describe the following relations using the roster method. y y 4 4 3 3 2 2 1 1 4321 1 x 321 1 2 3 x 1 (c) The graph of relation C (a) The graph of relation A y y 3 3 2 2 1 1 321 1 2 3 x 4321 1 2 3 x 1 1 2 2 3 3 (b) The graph of relation B (d) The graph of relation D1.2 Relations 19 y y 5 2 4 1 3 x 4321 1 2 3 4 5 2 1 1 2 3 1 1 2 3 4 5 x 1 (f) The graph of relation F (e) The graph of relation E 4. Graph the following lines. (a) x =2 (b) y = 3 5. What is another name for the line x = 0? For y = 0? 6. Some relations are fairly easy to describe in words or with the roster method but are rather dicult, if not impossible, to graph. Discuss with your classmates how you might graph the following relations. Please note that in the notation below we are using the ellipsis, . . . , to denote that the list does not end, but rather, continues to follow the established pattern inde nitely. For the rst two relations, give two examples of points which belong to the relation and two points which do not belong to the relation. (a) f(x;y) :x is an odd integer, and y is an even integer.g (b) f(x; 1) :x is an irrational numberg (c) f(1; 0); (2; 1); (4; 2); (8; 3); (16; 4); (32; 5);:::g (d) f:::; (3; 9); (2; 4); (1; 1); (0; 0); (1; 1); (2; 4); (3; 9);:::g

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