The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard A. Medeiros next Lesson 18 Part 2

next © 2007 Herbert I. Gross There are times when we are interested in the sign of a number; and other times when we are only interested in the magnitude (absolute value) of a signed number. Absolute Value Functions Absolute Value Functions

For example, suppose it’s 100 miles between Town A and Town B, and a driver makes a round trip between these two towns. With respect to the distance the driver is from the starting point, the 100 miles driven in one direction “cancel” the 100 miles driven in the other direction (that is, the driver returns to the starting point of the round trip). In this case, the answer is given by… © 2007 Herbert I. Gross next 100 miles miles= 0 miles next A B

© 2007 Herbert I. Gross However, the total distance traveled during the round trip is 200 miles. More specifically, if we call the distance in the direction from Town A to Town B positive, then the directed distance from Town B to Town A is negative; and the fact that = 0 tells us that the displacement (that is, the distance between the starting point and the ending point) is 0. However, the total distance traveled is given by the sum… 100 miles + | - 100|miles = 100 miles miles =200 miles next

© 2007 Herbert I. Gross By way of review, recall that the absolute value of a number is equal to the number itself if the number is non-negative, but it’s the opposite of the number if the number is negative. │n│ = n, if n ≥ 0 - n, if n < 0 next Stated more formally, if n denotes any number, then…

© 2007 Herbert I. Gross If n < 0, - n is a positive number. In other words, the absolute value of a number cannot be negative. For example, in terms of our profit and loss model a $3 profit is not the same as a $3 loss. However, the amount in both transactions is $3. That is, in this model the absolute value (magnitude) of either transaction is $3. Note

next © 2007 Herbert I. Gross Keep in mind that for a given value of x, f(x) is a number. | n | = n, if n ≥ 0 - n, if n < 0 next Thus, we may replace n by f(x) in our previous definition of |n| to obtain the result… For example, suppose f(x) = 2x. Then… if x = - 3, f( - 3) = - 6 and |f( - 3)| = | - 6| = 6 (= - ( - 6)). f(x) next

© 2007 Herbert I. Gross Two Special Cases next For instance, suppose f(x) = x 2. Since x 2 is never negative, |x 2 | = x 2. For example, |f( - 3)| = |( - 3) 2 | = ( - 3) 2 = f( - 3). 1. Suppose that for each number x in the domain of f, f(x) ≥ 0. Then |f(x)| = f(x). Since the above discussion applies to any number x, we may replace - 3 by x and see that |f( - 3 )| = f( - 3 ). x x

© 2007 Herbert I. Gross next For instance, suppose f(x) = - 3x 2. Since x 2 is either 0 or positive, and since negative times positive is negative, - 3x 2 is either 0 or negative. 2. On the other hand, suppose that for each number x in the domain of f, f(x) < 0. Then |f(x)| = - f(x). More generally… |f(x)| = │ - 3x 2 │= - ( - 3x 2 ) = - f(x). For example, f(2) = - 3(2) 2 = Therefore, |f(2)| = | - 12| = - ( - 12) = - f(2) - f(x). |f(x)|

next © 2007 Herbert I. Gross The General Case next For example, suppose f(x) = 2x – 4. Then… In general, f(x) will be positive or 0 for some values of x and negative for other values of x. │ n │ = n, if n ≥ 0 - n, if n < 0 2x – 4 2x – 4,2x – 4 - (2x – 4),2x – 4 next

© 2007 Herbert I. Gross next We may rewrite - (2x – 4) in the standard “mx + b” form by the following sequence of steps… - (2x – 4)= - 1(2x – 4) - 1(2x – 4)= - 1(2x + - 4) - 1(2x + - 4)= - 1(2x) + - 1( - 4) - 1(2x) + - 1( - 4)= - 2x + 4 So replacing - (2x – 4) by - 2x + 4, we obtain … |f(x)| = │ 2x – 4 │ = 2x – 4, if 2x – 4 ≥ 0 - (2x – 4), if 2x – 4 < 0 next - (2x – 4) - 2x x + 4,

next © 2007 Herbert I. Gross next |f(x)| = |2x – 4| = 2x – 4, if 2x – 4 ≥ 0 - 2x + 4, if 2x – 4 < 0 x ≥ 2x ≥ 2 x < 2x < 2 If we divide both sides of 2x – 4 ≥ 0 by 2, we see that x – 2 ≥ 0, and if we then add 2 to both sides of this equation, we see that x ≥ 2. In summary… In a similar way we see that 2x – 4 < 0 is equivalent to saying that x < 2. x ≥ 2 x < 2 next

© 2007 Herbert I. Gross next To check that our answer is reasonable, suppose x = 1. In this case x < 2 and hence… |f(x)| = |2x – 4| = 2x – 4, if x ≥ 2 - 2x + 4, if x < 2 next |f( x )| = - 2 x + 41(1) = = 2 This checks with the fact that since f(1) = 2(1) – 4 = - 2; |f(1)| = 2.

next © 2007 Herbert I. Gross Using the Graph of f(x) to Construct the Graph of to Construct the Graph of │f(x)│ In terms of a graph, there is a rather nice relationship between the graph of f(x) and the graph of |f(x)|. Namely the graphs are identical whenever f(x) is non-negative. However, the portion of the graph of f(x) that was below the x-axis (that is, the values of x for which f(x) < 0) is simply reflected about the x-axis to obtain the graph of |f(x)|.

next © 2007 Herbert I. Gross (a,b) (a, - b) More generally, to reflect the point (a,b) about the x-axis we simply replace b by its opposite. That is, the points (a,b) and (a, - b) are both on the line x = a, and they are the same distance from the x-axis. x = a next

© 2007 Herbert I. Gross The graph of f(x) = 2x – 4 is (2,0) (4,4) (5,6) (3,2) (1, - 2) (0, - 4) (6,8) next y = 2x – 4

next © 2007 Herbert I. Gross The graph of f(x) = - 2x + 4 is (2,0) (4, - 4) (5, - 6) (3, - 2) (1,2 ) (0,4) (6, - 8) next y = - 2x – 4

next © 2007 Herbert I. Gross The graph of g(x) = |f(x)| = |2x + 4|is (4, - 4) (5, - 6) (3, - 2) (1,2 ) next (0,4) (6, - 8) next (2,0) (4,4) (5,6) (3,2) (1, - 2) (0, - 4) (6,8) y = 2x – 4 y = - 2x + 4

next © 2007 Herbert I. Gross And we then reflected about the x-axis the portion of the line that was below the x-axis. (1,2 ) (0,4) next (2,0) (4,4) (5,6) (3,2) (6,8) More simply, we started with the line y = 2x – 4. (1, - 2) (0, - 4) ( - 1, - 6) ( - 2, - 8) ( - 1,6) ( - 2,8) next

© 2007 Herbert I. Gross Appendix 1 The Geometry of |Absolute Value| The Geometry of |Absolute Value| Enrichment As we saw in our study of signed numbers, b – c is the directed distance from c to b and c – b is the directed distance from b to c. Hence, if we disregard direction (that is, if all we are interested in is the length of the line segment that joins the points b and c), we see that this length is given by either |b – c| or |c – b| (since |c – b| = |b – c|).

next © 2007 Herbert I. Gross Appendix 1 Geometrically, b – c is represented by the arrow that starts at point c and terminates at point b. bc | On the other hand, c – b is shown by the arrow that starts at point b and terminates at point c. b – c c – b Hence, |c – b| = |b – c|. Using the number line next

© 2007 Herbert I. Gross Appendix 1 So, for example, |x – 3| represents the distance between x and 3. More generally, for any number c, |x – c| represents the distance between x and c. In this context, |x| is the special case in which c = 0. That is |x| = |x – 0|. Thus, geometrically |x| represents the distance of the point x from the origin (0). next

© 2007 Herbert I. Gross Appendix 1 In this context the inequality |x – 3| < 2 defines the set of all points x such that the distance from x to 3 is less than 2. next |x – 3| < 2 In terms of a picture, to locate the set of all such points x, we simply draw a line that begins 2 units to the left of 3 (that is, at x = 1)… and extends 2 units to the right of 3 (that is, it ends at x = 5), and we exclude the points 1 and 5 because they satisfy |x – 3| = 2. next

© 2007 Herbert I. Gross Appendix 1 Stated in the language of sets, |x – 3| < 2 = {x:1 < x < 5} next In a similar way, |x – 3| > 2 means x is more than 2 units from 3; that is x is either to the left of (less than) 1 or to the right of (greater than) 5. That is, x < 1 or x > 5. Pictorially, |x – 3| > 2 is the region represented in orange below.

© 2007 Herbert I. Gross next To abbreviate “x > 1 and (but) x < 5” we may write 1 < x < 5. However, it would be incorrect to abbreviate “x < 1 or x > 5” by writing 5 < x < 1. That is, no number (x) can be greater than 5 but less than 1. Rather the set {x: x < 1} is the interval ( - ∞,1); and {x: x > 5} is the interval (5,∞). Therefore, “x < 1 or x > 5” means that either x є ( − ∞, 1) or x є (5,∞); or in the language of sets x є ( - ∞, 1) U (5,∞) Don’t Confuse “and” with “or”

© 2007 Herbert I. Gross next Distinguishing between |f(x)| and f(|x|). next Clearly the fact that x is positive doesn’t mean that f(x) is positive (i.e. the output can be negative even if the input is positive). Appendix 2 Enrichment For example, looking again at f(x) = 2x – 4, we see that although 1 is positive… f(1) = 2(1) – 4 = 2 – 4 = f(1) is negative… next

© 2007 Herbert I. Gross Let’s illustrate the difference between the graphs of |f(x)| and f(|x|) where f(x) = 2x – 4. f( ) = 2( ) – 4 next And since |x| = x, if x ≥ 0 - x, if x < 0 2x – 4, if x ≥ 0 2( - x) – 4, if x < 0 f(|x|)= Thus, the graph of f(|x|) is y = 2x – 4 if x ≥ 0 and y = - 2x – 4 if x < 0. next |x| Since f( ) = 2( ) – 4, next

© 2007 Herbert I. Gross ( - 3,2) ( - 2,0) ( - 4,2) ( - 5,6 ) next ( - 6, 8) ( - 1, - 2) next (2,0) (4,4) (5,6) (3,2) (1, - 2) (0, - 4) (6,8) The graph of f(|x|) = - 2|x| – 4 The graph of f(x) = 2x – 4, x ≥ 0 The graph of f(x) = - 2x – 4, x ≤ 0 next

© 2007 Herbert I. Gross To reflect a curve with respect to a line, we draw its mirror image with respect to that line. More pictorially, if we were to place a mirror on the line, the reflection of the curve would be the picture we see in the mirror. Shortcut next

© 2007 Herbert I. Gross In particular, when the point (a,b) is reflected with respect to the y-axis it becomes the point ( - a,b). next ( - a,b) (a,b)

next © 2007 Herbert I. Gross To convert the graph of f(x) to the graph of f|x|, we reflect the portion of the graph that’s to the right of the y-axis with respect to the y-axis and we delete the portion of the graph of f(x) that’s to the left of the y-axis. next For example, with respect to f(x) = 2x – 4, f( - 3) = - 3(2) – 4 = Hence, the point P( - 3, - 10) is on the graph of f(x). On the other hand, f(| - 3|) = f(3) =2(3) – 4 = 2. Hence, the point Q( - 3,2) is on the graph of f(|x|); and with respect to the y-axis, Q is the reflection of the point R(3,2) (which is on the graph of f(x)).

© 2007 Herbert I. Gross y = 2x – 4 The graph of f|(x)| = 2|x|– 4 is (4,4) (5,6) (3,2) (6,8) (2,0) (1, - 2) (0, - 4) ( - 3,2) ( - 4,4) ( - 5,6 ) ( - 6,8) ( - 2,0) ( - 1, - 2) next

© 2007 Herbert I. Gross Comparing the graphs of f(|x|) and |f(x)|, f | (x) | = 2 | x | – 4 | f(x) | = | 2x – 4 | (2,0) (3,2) (4,4) (5,6) (1,2) (0,4) ( - 1,6) To obtain the curve whose equation is y = |f(x)|, reflect the portion of the curve y = f(x) that's below the x-axis about the x-axis. To obtain the curve whose equation is y = f(|x|), reflect the portion of the curve that's to the right of the y-axis about the y-axis. next ( - 5,6) (0, - 4) ( - 2,0) ( - 3,2) ( - 4,4) (2,0) (3,2) (4,4) (5,6) (1, - 2) ( - 1, - 2)

next © 2007 Herbert I. Gross Appendix 3 The Geometric Version of the Product of Two Numbers The Geometric Version of the Product of Two Numbers Enrichment From a geometric point of view, we can always visualize the product of two numbers as being an area. For example, we may think of 5 × 6 as being the area of a rectangle whose base is 5 and whose height is next

© 2007 Herbert I. Gross Appendix 3 Moreover, area often has a physical meaning. For example, suppose an object moves in a straight line (say, in the positive horizontal direction) at a constant speed of 6 miles per hour for 5 hours. If the input axis, denoted by t, represents the time in hours; and the output axis, denoted by v, represents the velocity of the object in miles per hour then the area of the -colored region in the following diagram represents the distance that the object traveled during the 5 hours.

next © 2007 Herbert I. Gross next That is, the distance the object traveled is… 6 miles per hour × 5 hours = 30 miles (speed in miles per hour) (time in hours)

© 2007 Herbert I. Gross Suppose that after traveling 6 miles per hour for 5 hours, the object reverses its direction and again travels at a constant speed of 6 miles per hour for 5 hours. Then the area of the -colored region in the following diagram represents the distance that the object traveled during the 5 hours. Notice that in this case since the velocity of the object was considered to be positive in the direction of the positive t-axis; it must be considered to be negative in the opposite direction.

next © 2007 Herbert I. Gross next That is, the distance the object traveled is… - 6 miles per hour × 5 hours = - 30 miles (miles per hour) (time) t

next © 2007 Herbert I. Gross next If we add the two areas (distances) algebraically, we obtain = 0; which tells us that the displacement is Displacement Distance 60 On the other hand, the total area (30 + |30|) tells us that the total distance traveled is 60 miles

next © 2007 Herbert I. Gross next (miles per hour) (time) In terms of absolute values, to find the total distance traveled we have to reflect the rectangular region denoted by about the horizontal axis to obtain… t

© 2007 Herbert I. Gross Velocity is speed in a given direction. Stated in terms of absolute value, speed is the absolute value of the velocity. Thus, for example, if an object is moving along a circular path at a constant speed of, say, 6 miles per hour, its velocity is always changing because the direction of the motion is always changing. In this case, the speed of the object is constant, but its velocity isn’t. Note… Speed vs. Velocity

next © 2007 Herbert I. Gross As a real-life illustration, suppose the distance between two towns A and B is 200 miles. Suppose further that you are leaving town A driving at a constant speed of 50 miles per hour. Then it will take you 4 hours to get to Town B, but only if the direction in which you are driving is in the direction from A to B. For example, if Town B is east of Town A and you were driving west from Town A You’d have to circumnavigate the Globe to get to Town B. Real Life Illustration A B

Multiplying Two Quantities In our present illustration of 6 X 5, 6 is modifying “miles per hour” and 5 is modifying “hours”. However, no matter what nouns the 6 and 5 are modifying, the adjective part of the product would still be 30. For example That is, except for the labels the graphs would all look alike. 6 kilowatts per hour × 5 hours = 30 kilowatts 6 dollar per item × 5 items = 30 dollars © 2007 Herbert I. Gross next

© 2007 Herbert I. Gross next 6 kilowatts per hour × 5 hours = 30 kilowatts (kilowatts per hour) (time in hours) Example 1

next © 2007 Herbert I. Gross next 6 dollars per item × 5 items = 30 dollars (dollars per item) (items) Example 2