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Keystone Problem… Keystone Problem… next Set 18 Part 2 © 2007 Herbert I. Gross.

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Presentation on theme: "Keystone Problem… Keystone Problem… next Set 18 Part 2 © 2007 Herbert I. Gross."— Presentation transcript:

1 Keystone Problem… Keystone Problem… next Set 18 Part 2 © 2007 Herbert I. Gross

2 You will soon be assigned five problems to test whether you have internalized the material in Lesson 18, Part 2 of our algebra course. The Keystone Illustration below is a prototype of the problems you’ll be doing. Work out the problem on your own. Afterwards, study the detailed solutions we’ve provided. Notice that several different ways are presented that could be used to solve each problem. Instructions for the Keystone Problems next © 2007 Herbert I. Gross

3 As a teacher/trainer, it is important for you to understand and be able to respond in different ways to the different ways individual students learn. The more ways you are ready to explain a problem, the better the chances are that the students will come to understand. next © 2007 Herbert I. Gross

4 next Problem #1 Express y as a function of x if |x| + |y| = 1 Keystone Problem for Lesson 18 Part 2 © 2007 Herbert I. Gross next

5 The fact that the absolute value of a number cannot be negative tells us that if |x| + |y| = 1 then both |x| ≤ 1 and |y| ≤ 1. © 2007 Herbert I. Gross Preamble to Our Solution Namely, if |x| > 1, then the fact that |y| ≥ 0 would mean that |x| + |y| >1. This would contradict the given information that |x| + |y| = 1. Therefore, we know that |x| ≤ 1. A similar argument applies to the assumption that |y| > 1. next

6 Do not confuse |x| ≤ 1 with x ≤ 1. © 2007 Herbert I. Gross Caution Geometrically speaking, |x| is the distance between x and 0. So to be on the x-axis but no more than 1 unit from 0, x can be no further than 1 unit to the right of 0 and no further than 1 unit to the left of 0. next Hence, x ≤ 1 and x ≥ - 1 (which we usually write as - 1 ≤ x ≤ 1). In a similar way |y| ≤ 1 means that - 1 ≤ y ≤ 1. For example, if x = - 3, then the |x| = 3. Hence, x < 1, but |x| > 1.

7 In set notation, we want to describe the set S where S = {(x,y): |x| + |y| = 1}. In terms of the graph of the function, this means we want to describe the set of all points (x,y) in the xy-plane that satisfy the equation |x| + |y| = 1. © 2007 Herbert I. Gross Geometric Interpretation Since |x| ≤ 1 and |y| ≤ 1, we see that the set S is contained within the square that is bounded on the left by the line x = - 1, on the right by the line x = 1, below by the line y = - 1, and above by the line y = 1. next

8 © 2007 Herbert I. Gross next x = - 1 In terms of a picture... x = 1 y = - 1 y = 1 next Set S

9 Caution next © 2007 Herbert I. Gross For example, ( 1 / 2, 3 / 4 ) is inside the square but | 1 / 2 | + | 3 / 4 | > 1. next What we have shown is that for any point (x,y) that lies outside the above square, |x| + |y| > 1. However, not every point (x,y) that is in or on the above square satisfies the equation |x| + |y| = 1.

10 Solution for Problem #1: The relationship between a number and its absolute value depends on the sign of the number. © 2007 Herbert I. Gross next Hence, there are four separate cases we have to consider…

11 Solution for Problem #1: © 2007 Herbert I. Gross next Case 1: (x,y) є S, and: 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 next In this case, |x| = x and |y| = y. Hence, the equation |x| + |y| = 1 becomes x + y = 1; or, equivalently, y = 1 – x. Stated in the language of functions… y = f 1 (x)… where f 1 (x) = 1 – x, dom f 1 = [0,1]

12 Geometrically, the set of points (x,y) є S for which |x| + |y| = 1, 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 is the region which is shown on the following slide. next © 2007 Herbert I. Gross In that region, the graph of f 1 is given by the equation… y = f 1 (x) = 1 – x; where dom f 1 =[0,1]. next This is the line segment whose end points are the points (1,0) and (0,1).

13 © 2007 Herbert I. Gross In terms of a picture... (0,1) (1,0) next y = f 1 (x) = 1 – x; dom f 1 = [0,1] next

14 We do not have to consider any point (x,y) on the line x + y = 1 that lies outside the square because we have already shown that for all such points, |x| + |y| > 1. © 2007 Herbert I. Gross Note

15 Solution for Problem #1: © 2007 Herbert I. Gross next Case 2: (x,y) є S, and: - 1 ≤ x ≤ 0 and 0 ≤ y ≤ 1 next In this case, |x| = - x and |y| = y. Hence, the equation |x| + |y| = 1 becomes - x + y = 1; or, equivalently, y = x + 1. Stated in the language of functions… y = f 2 (x) where f 2 (x) = x + 1, dom f 2 = [ - 1,0]

16 Geometrically, the set of points (x,y) є S for which |x| + |y| = 1, - 1 ≤ x ≤ 0 and 0 ≤ y ≤ 1 is the region which is shown on the following slide. next © 2007 Herbert I. Gross In that region, the graph of f 2 is given by the equation… y = f 2 (x) = x + 1; where dom f 2 =[0,1]. next This is the line segment whose end points are the points ( - 1,0) and (0,1).

17 © 2007 Herbert I. Gross In terms of a picture... (0,1) ( - 1,0) next y = f 2 (x) = 1 + x,dom f 2 = [ - 1,0] next (0,1) ( - 1,0) y = f 2 (x) = 1 + x dom f 2 = [ - 1,0].

18 Solution for Problem #1: © 2007 Herbert I. Gross next Case 3: (x,y) є S, and: -1 ≤ x ≤ 0 and -1 ≤ y ≤ 0. next In this case, |x| = - x and |y| = - y. So, the equation |x| + |y| = 1 becomes - x + - y = 1. Multiplying both sides of this equation by - 1, we obtain x + y = - 1 (or y = - x + - 1). Stated in the language of functions… y = f 3 (x) where f 3 (x) = - x + - 1, dom f 3 = [ - 1,0]

19 Geometrically, the set of points (x,y) є S for which |x| + |y| = 1, -1 ≤ x ≤ 0, and -1 ≤ y ≤ 0 is the region which is shown on the following slide. next © 2007 Herbert I. Gross This is the line segment whose end points are (0, - 1) and ( - 1,0). next In that region the graph of f 3 is given by the equation… y = f 3 (x) = x, dom f 3 = [ - 1,0].

20 © 2007 Herbert I. Gross In terms of a picture... (0, - 1) ( - 1, 0) next y = f 3 (x) = x, dom f 3 = [ - 1, - 1] (0, - 1) ( - 1, 0) y = f 3 (x) = x, dom f 3 = [ - 1, - 1] next

21 Solution for Problem #1: © 2007 Herbert I. Gross next Case 4: (x,y) є S, and: 0 ≤ x ≤ 1 and - 1 ≤ y ≤ 0. next In this case, |x| = x and |y| = - y. Hence, the equation |x| + |y| = 1, becomes x + - y = 1; or, equivalently, x = 1 + y or y = x – 1. Stated in the language of functions… y = f 4 (x) where f 4 (x) = x – 1, dom f 4 = [0,1]

22 Geometrically, the set of points (x,y) є S for which |x| + |y| = 1, 0 ≤ x ≤ 1 and - 1 ≤ y ≤ 0 is the region which is shown on the following slide. next © 2007 Herbert I. Gross This is the line segment whose end points are (0, - 1) and (1,0). next In that region, the graph of f 4 is given by the equation… y = f 4 (x) = x – 1; and dom f 4 = [0,1].

23 © 2007 Herbert I. Gross In terms of a picture... (1,0) (0, - 1) next y = f 4 (x) = x – 1;dom f 4 = [0,1] y = f 4 (x) = x – 1;dom f 4 = [0,1] (1,0) (0, - 1) next

24 If we now juxtaposition our four graphs, we see that the graph of |x| + |y| = 1 is the square that is shown on the next slide. © 2007 Herbert I. Gross We have enlarged the scale to make the graph easier to view. next

25 © 2007 Herbert I. Gross next (1,0) (0,1) ( - 1,0) (0, - 1) Juxtaposing the four graphs. next As shown on the next slide, this graph does not represent a function.

26 © 2007 Herbert I. Gross next (1,0) (0,1) ( - 1,0) (0, - 1) For example, the line x = 1/2 intersects the square at 2 points next

27 While the graph does not represent a function, it is the union of four graphs each of which does represent a function. © 2007 Herbert I. Gross For example, as shown in the next slide, the line x = 1/2 intersects the square twice but only once on the red line segment and once on the blue line segment. next

28 © 2007 Herbert I. Gross next (1,0) (0,1) ( - 1,0) (0, - 1) next (1/2,1/2)(1/2,1/2) ( 1 / 2, -1 / 2 )

29 © 2007 Herbert I. Gross next f 1 (x), x є [0,1] next To summarize our results in the language of functions, our graph can be represented in the form… f 2 (x), x є [ - 1,0] f 3 (x), x є [ - 1,0] f 4 (x), x є [0,1] f(x) =

30 © 2007 Herbert I. Gross next f 1 (x), x є [0,1] next And if we now recall the definitions of f 1 (x), f 2 (x), f 3 (x), and f 4 (x), we may summarize our results in the form… f 2 (x), x є [0,1] f 3 (x), x є [ - 1,0] f 4 (x), x є [ - 1,0] y = 1 – x, x є [0,1] x + 1, x є [ - 1,0] - x + - 1, x є [ - 1,0] x – 1, x є [0,1]


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