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5000 feet north of the intersection. The runner runs due west at 17 feet per second. When will the runner’s feet get wet? 31 rider 100 feet ground level 62 ft. tower 60 feet operator 24 feet (a) Impose a coordinate system. (b) Suppose a rider is located at the point in the picture, 100 feet above the ground. If the rider drops an ice cream cone straight down, where will it land on the ground? (c) The ride operator is standing 24 feet to one side of the support tower on the level ground at the location in the picture. Determine the location(s) of a rider on the Ferris Wheel so that a dropped ice cream cone lands on the operator. (Note: There are two answers.) Problem 3.5. A crawling tractor sprinkler is located as pictured below, 100 feet South of a sidewalk. Once the water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves North along the hose at the rate of 1 2 inch/second. The hose is perpendicular to the 10 ft. wide sidewalk. Assume there is grass on both sides of the sidewalk. hose W N S tractor sprinkler E sidewalk Problem 3.4. An amusement park Ferris Wheel has a radius of 60 feet. The center of the wheel is mounted on a tower 62 feet above the ground (see picture). For these questions, the wheel is not turning. (a) Impose a coordinate system. Describe the initial coordinates of the sprinkler and ﬁnd equations of the lines forming the North and South boundaries of the sidewalk. 32 CHAPTER 3. THREE SIMPLE CURVES (b) When will the water ﬁrst strike the side- (e) How long does the ferry spend inside the walk? radar zone? (c) When will the water from the sprinkler fall completely North of the sidewalk? (d) Find the total amount of time water from the sprinkler falls on the sidewalk. (e) Sketch a picture of the situation after 33 minutes. Draw an accurate picture of the watered portion of the sidewalk. (f) Find the area of GRASS watered after one hour. Problem 3.6. Erik’s disabled sailboat is ﬂoating stationary 3 miles East and 2 miles North of Kingston. A ferry leaves Kingston heading toward Edmonds at 12 mph. Edmonds is 6 miles due east of Kingston. After 20 minutes the ferry turns heading due South.
Ballard is 8 miles South and 1 mile West of Edmonds. Impose coordinates with Ballard as the origin. sailboat Kingston Edmonds North Ballard UDub (a) Find the equations for the lines along which the ferry is moving and draw in these lines. (b) The sailboat has a radar scope that will detect any object within 3 miles of the sailboat. Looking down from above, as in the picture, the radar region looks like a circular disk. The boundary is the ”edge” or circle around this disc, the interior is the inside of the disk, and the exterior is everything outside of the disk (i.e. outside of the circle). Give a mathematical (equation) description of the boundary, interior and exterior of the radar zone. Sketch an accurate picture of the radar zone by determining where the line connecting Kingston and Edmonds would cross the radar zone. (c) When does the ferry enter the radar zone? (d) Where and when does the ferry exit the radar zone? Problem 3.7. Nora spends part of her summer driving a combine during the wheat harvest. Assume she starts at the indicated position heading east at 10 ft/sec toward a circular wheat ﬁeld of radius 200 ft. The combine cuts a swath 20 feet wide and begins when the corner of the machine labeled “a” is 60 feet north and 60 feet west of the western-most edge of the ﬁeld. N W E S 20 ft combine a swath cut center wheat field (a) When does Nora’s rig ﬁrst start cutting the wheat? (b) When does Nora’s rig ﬁrst start cutting a swath 20 feet wide? (c) Find the total amount of time wheat is being cut during this pass across the ﬁeld. (d) Estimate the area of the swath cut dur- ing this pass across the ﬁeld. Problem 3.8. (a) Solve for x: x2 − 2x + 1 x + 5 = x − 2 (b) Solve for xc) If x = −2, ﬁnd ALL solutions of the equa- tion (x + 1)2 + (y − 1)2 = 10 (d) If y = 3, ﬁnd ALL solutions of the equa- tion 5(x + 1)2 + 2(y − 1)2 = 10 Chapter 4
Linear Modeling Sometimes, we will begin a section by looking at a speciﬁc problem which highlights the topic to be studied; this section offers the ﬁrst such vista. View these problems as illustrations of precalculus in action, rather than confusing examples. Don’t panic, the essential algebraic skills will be reviewed once the motivation is in place. 4.1 The Earning Power Problem The government likes to gather all kinds of data. For example, Table 4.1 contains some data on the average annual income for full-time workers; these data were taken from the 1990 Statistical Abstract of the U.S. Given this information, a natural question would be: How can we predict the future earning power of women and men? One way to answer this (a) Women. (b) Men. YEAR MEN (dollars) YEAR WOMEN (dollars) 1970 $9,521 $5,616 1970 $18,531 $28,313 1987 1987 Table 4.1: Earning power data. question would be to use the data in the table to construct two different mathematical models that predict the future (or past) earning power for women or men. In order to do that, we would need to make some kind of initial assumption about the type of mathematical model expected. Let’s begin by drawing two identical xy-coordinate systems, where the x-axis has units of “year” and the y-axis has units of “dollars;” see Figure 4.1. In each coordinate system, the data in our table gives us two points to plot: In the case of women, the data table gives us the points P = (1970, 5,616) and Q = (1987, 18,531). Likewise, for the men, the data table gives us the points R = (1970, 9,521) and S = (1987, 28,313). To study the future earning power of men and women, we are going to make an assumption: For women, if the earning power in year x is $y, 33 34 PSfrag CHAPTER 4. LINEAR MODELING y-axis (dollars) Q = (1987,18531) P = (1970,5616) x-axis (year) 30000 25000 20000 15000 10000 5000 aca 30000 25000 20000 15000 10000 5000 y-axis (
dollars) S = (1987,28313) R = (1970,5616) x-axis (yeara) Data points for women. (b) Data points for men. Figure 4.1: Visualizing the data. then the point (x, y) lies on the line connecting P and Q. Likewise, for men, if the earning power in year x is $y, then the point (x, y) lies on the line connecting R and S. In the real world, the validity of this kind of assumption would involve a lot of statistical analysis. This kind of assumption leads us to what is called a linear model, since we are demanding that the data points predicted by the model (i.e., the points (x, y)) lie on a straight line in a coordinate system. Now that we have made this assumption, our job is to ﬁnd a way to mathematically describe when a point (x, y) lies on one of the two lines pictured in Figure 4.2. Our goal in the next subsection is to review the mathematics necessary to show that the lines in Figure 4.2 are the so-called graphs of Equations 4.1 and 4.2. 30000 25000 20000 15000 10000 y y-axis (dollars) Q = (1987,18531) (x,y) on the line: This means women earn y dollars in year x. 5000 P = (1970,5616) x-axis (year 30000 25000 20000 15000 10000 5000 y y-axis (dollars) S = (1987,28313) (x,y) on the line: This means men earn y dollars in year x. R = (1970,9521) x-axis (yeara) Linear model for women. (b) Linear model for men. Figure 4.2: Linear models of earning power. 4.2. RELATING LINES AND EQUATIONS ywomen = = ymen = = 18,531 − 5,616 1987 − 1970 12,915 17 28,313 − 9,521 1987 − 1970 18,792 17 (x − 1970) + 5,616 (x − 1970) + 5,616 (x − 1970) + 9,521 (x − 1970) + 9,521 35 (4.1) (4.2) 4.2 Relating Lines and Equations A systematic approach to studying equations and their
graphs would begin with the simple cases, gradually working toward the more complicated. Thinking visually, the simplest curves in the plane would be straight lines. As we discussed in Chapter 3, a point on the vertical line in Figure 4.3(a) will always have the same x-coordinate; we refer to this line as the graph of the equation x = h. Likewise, a point on the horizontal line in Figure 4.3(b) will always have the same y-coordinate; we refer to this line as the graph of the equation y = k. Figure 4.3(c) is different, in the sense that neither the x nor the y coordinate is constant; i.e., as you move a point along the line, both coordinates of the point are changing. It is reasonable to guess that this line is the graph of some equation involving both x and y. The question is: What is the equation? y-axis graph of x = h y-axis (x,k) is a typical point on this line y-axis (h,y) is a typical point on this line. x-axis location h on x-axis graph of y = k location k on y-axis x-axis (x,y) is a typical point on this line graph some of equation involving x and y x-axis (a) Vertical line. (b) Horizontal line. (c) Sloped line. Figure 4.3: Lines in a plane. 36 CHAPTER 4. LINEAR MODELING Here is the key geometric fact needed to model lines by mathematical equations: Important Fact 4.2.1. Two different points completely determine a straight line. This fact tells us that if you are given two different points on a line, you can reconstruct the line in a coordinate system by simply lining a ruler up with the two points. In our discussion, we will need to pay special attention to the difference between vertical and non-vertical lines. 4.3 Non-vertical Lines Assume in this section that ℓ is a non-vertical line in the plane; for exIf we are given two points P = (x1, y1) ample, the line in Figure 4.3(c). and Q = (x2, y2) on a line ℓ, then Equation (2.2) on page 18 deﬁned two quantities we can calculate: ∆x = change in x going from P
to Q = x2 − x1. ∆y = change in y going from P to Q = y2 − y1. We deﬁne the slope of the line ℓ to be the ratio of ∆y by ∆x, which is usually denoted by m: m def= = slope of ℓ = ∆y ∆x y2 − y1 x2 − x1 change in y change in x (4.3) Notice, we are using the fact that the line is non-vertical to know that this ratio is always deﬁned; i.e., we will never have ∆x = 0 (which would lead to illegal division by zero). There is some additional terminology that goes along with the deﬁnition of the slope. The term ∆y is sometimes called the rise of ℓ and ∆x is called the run of ℓ. For this reason, people often refer to the slope of a line ℓ as “the rise over the run,” meaning slope of ℓ def= rise of ℓ run of ℓ = ∆y ∆x = m. In addition, notice that the calculation of ∆y involves taking the difference of two numbers; likewise, the calculation of ∆x involves taking the difference of two numbers. For this reason, the slope of a line ℓ is sometimes called a difference quotient. 4.3. NON-VERTICAL LINES For example, suppose P = (1, 1) and Q = (4, 5) lie on a line ℓ. In this case, the rise = ∆y = 4 and the run = ∆x = 3, so m = 4 3 is the slope of ℓ. When computing ∆x, pay special attention that it is the x-coordinate of the destination point Q minus the xcoordinate of the starting point P; likewise, when computing ∆y, it is the y-coordinate of the destination point Q minus the y-coordinate of the starting point P. We can reverse this order in both calculations and get the same slope: m = ∆y ∆x = y2 − y1 x2 − x1 = −(y2 − y1) −(x2 − x1) = y1 − y2 x1 − x2 = −∆y
−∆x. 37 Q = (4,5) 5 1 P = (1,1) ∆y = 4 ∆x = 3 1 4 Figure 4.4: Computing the slope of a line. We CANNOT reverse the order in just one of the calculations and get the same slope: m = y2 − y1 x2 − x1 6 = y2 − y1 x1 − x2, and m = y2 − y1 x2 − x1 6 = y1 − y2 x2 − x1.!!! CAUTION!!! It is very important to notice that the calculation of the slope of a line does not depend on the choice of the two points P and Q. This is a real windfall, since we are then always at liberty to pick our favorite two points on the line to determine the slope. The reason for this freedom of choice is pretty easy to see by looking at a picture. If we were to choose two other points P∗ = (x∗1, y∗1) and Q∗ = (x∗2, y∗2) on ℓ, then we would get two similar right triangles: See Figure 4.5. Basic properties of similar triangles tell us ratios of lengths of common sides are equal, so that m = y2 − y1 x2 − x1 = y∗2 − y∗1 x∗2 − x∗1 ; ∗ 2,y ∗ 2 = Q ∗ x (x2,y2) = Q y2 − y1 (x1,y1) = P x2 − x1 ∗, Figure 4.5: Using similar triangles. but this just says the calculation of the slope is the same for any pair of distinct points on ℓ. For example, lets redo the slope calculation when P∗ = P = (x1, y1) and Q∗ = (x, y) represents an arbitrary point on the line. Then the two ratios of lengths of common sides give us the equation m = y − y1 x − x1, y − y1 = m(x − x1). 38 CHAPTER 4. LINEAR MODELING This can be rewritten as y = m(x − x1) + y1 or y = y2 − y1 x2 − x1 (x − x1) + y1. (4.
4) (4.5) Equation 4.4 is usually called the point slope formula for the line ℓ (since the data required to write the equation amounts to a point (x1,y1) on the line and the slope m), whereas Equation 4.5 is called the two point formula for the line ℓ (since the data required amounts to the coordinates of the points P and Q). In any event, we now see that Important Fact 4.3.1. (x,y) lies on ℓ if and only if (x,y) is a solution to y = m(x − x1) + y1. We can plot the collection of ALL solutions to the equation in Fact 4.3.1, which we refer to as the graph of the equation. As a subset of the xycoordinate system, the line Important Fact 4.3.2. ℓ = {(x, m(x − x1) + y1) \|x is any real number}. graph of y = 4 3 x − 1 3 (6, 23 3 ) (4,5) (−1, − 5 3 ) (1,1) (0, − 1 3 ) Figure 4.6: Verifying points on a line. Example 4.3.3. Consider the line ℓ, in Figure 4.6, through the two points P = (1,1) and Q = (4,5). Then the slope of ℓ is m = 4/3 and ℓ consists of all pairs of points (x,y) such that the coordinates x and y satisfy the equation y = 4 3(x − 1) + 1. Letting x = 0, 1, 6 and −1, we conclude that the following four points lie on the line ℓ: (0, 4 3(0 − 1) + 1) = (0, −1 3(6 − 1) + 1) = (6, 23 3(1 − 1) + 1) = (1,1), (6, 4 3 ) and (−1, 4 3 ). By the same reasoning, the point (0,0) does not lie on the line ℓ. As a set of points in the plane, we have 3 ), (1, 4 3(−1 − 1) + 1) = (−1, −5 ℓ = x, 4 3 (x
− 1) + 1 \|x is any real number Returning to the general situation, we can obtain a third general equation for a non-vertical line. To emphasize what is going on here, plug the speciﬁc value x = 0 into Equation 4.4 and obtain the point R = (0,b) on the line, where b = m(0 − x1) + y1 = −mx1 + y1. But, Equation (4.4) can be written y = m(x − x1) + y1 = mx − mx1 + y1 = mx + b. 4.4. GENERAL LINES 39 The point R is important; it is precisely the point where the line ℓ crosses the y-axis, usually called the y-intercept. The slope intercept equation of the line is the form y = mx + b, where the slope of the line is m and b is the y-intercept of the line. Summary 4.3.4. Non-vertical Lines: Let ℓ be a non-vertical line in the xy-plane. There are three ways to obtain an equation whose graph is ℓ, depending on the data provided for ℓ: 1. If P = (x1,y1), Q = (x2,y2) are two different points on the line, then y2 − y1 x2 − x1 (x − x1) + y1 gives an equation the two-point formula y = whose graph is ℓ. 2. If P = (x1,y1) is a point on the line and m is the slope of ℓ, then the point-slope formula y = m(x − x1) + y1 gives an equation whose graph is ℓ. 3. If the line ℓ intersects the y-axis at the point (0,b) and m is the slope of the line ℓ, then the slope-intercept formula y = mx + b gives an equation whose graph is ℓ. 4.4 General Lines Summarizing, any line in the plane is the graph of an equation involving x and y and the equation always has the form Ax + By + C = 0, for some constants A, B, C. Equations like this are called linear equaIn general, non-vertical lines
will be of the most interest to us, tions. since these are the lines that can be viewed as the graphs of functions; we will discuss this in Chapter 5. 4.5 Lines and Rate of Change If we draw a non-vertical line in the xy coordinate system, then its slope will be the rate of change of y with respect to x: slope = = ∆y ∆x change in y change in x def= rate of change of y with respect to x. 40 CHAPTER 4. LINEAR MODELING We should emphasize that this rate of change is a constant; in other words, this rate is the same no matter where we compute the slope on the line. The point-slope formula for a line can now be interpreted as follows: A line is determined by a point on the line and the rate of change of y with respect to x. An interesting thing to notice is how the units for x and y ﬁgure into the rate of change calculation. For example, suppose that we have the equation y = 10,000 x+200,000, which relates the value y of a house (in dollars) to the number of years x you own it. For example, after 5 years, x = 5 and the value of the house would be y = 10,000 (5) + 200,000 = $250,000. In this case, the equation y = 10,000 x + 200,000 is linear and already written in slope-intercept form, so the slope can be read as m = 10,000. If we carry along the units in the calculation of, then the numerator involves “dollar” units and the denominator “years” units. That means that carrying along units, the slope is actually m = 10,000 dollars/year. In other words, the value of the house is changing at a rate of 10,000 dollars/year. ∆y ∆x At the other extreme, if the units for both x and y are the same, then the units cancel out in the rate of change calculation; in other words, the slope is a unit-less quantity, simply a number. This sort of thing will come up in the mathematics you see in chemistry and physics. s b speed m reference point initial location of the object location of the object at time t Figure 4.7: Motion along a line. One important type of rate encountered is the speed of a
moving object. Suppose an object moves along a straight line at a constant speed m: See Figure 4.7. If we specify a reference point, we can let b be the starting location of the moving object, which is usually called the initial location of the object. We can write down an equation relating the initial location b, the time t, the constant speed m and the location s at time t: s = (location of object at time t) s = (initial location of object) + (distance object travels in time t) s = b + mt. where t is in the same time units used to deﬁne the rate m. Notice, both b and m would be constants given to us, so this is a linear equation involving the variables s and t. We can graph the equation in the tscoordinate system: See Figure 4.8. It is important to distinguish between this picture (the graph of s = mt+b) and the path of the object in Figure 4.7. The graph of the equation should be thought of as a visual aid attached to the equation s = mt + b. The general idea is that using this visual aid can help answer various questions involving the equation, which in turn will tell us things about the motion of the object in Figure 4.7. 4.5. LINES AND RATE OF CHANGE Two other comments related to this discussion are important. First, concerning notation, the speed m is often symbolized by v to denote constant velocity and b is writ(the subscript “0” meaning “time zero”). With ten as s ◦ these changes, the equation becomes s = s + vt, which is ◦ the form in which it would be written in a typical physics text. As a second note, if you return to Figure 4.8, you will notice we only drew in the positive t axis. This was because t represented time, which is always a non-negative quantity. 41 s-axis graph of s = mt + b ∆s ∆t rate of change = b = s-intercept t-axis ∆s ∆t Figure 4.8: The graph of s = mt + b. Example 4.5.1. Linda, Asia and Mookie are all playing frisbee. Mookie is 10 meters in front of Linda and always runs 5 m/sec. Asia is 34 meters in front of Linda and always runs 4 m/sec. Linda
yells “go!” and both Mookie and Asia start running directly away from Linda to catch a tossed frisbee. Find linear equations for the distances between Linda, Mookie and Asia after t seconds. Linda Mookie Asia Solution. Let sM be the distance between Linda and Mookie and sA the distance between Linda and Asia, after t seconds. An application of the above formula tells us sM = (initial distance between Linda and Mookie) + + (distance Mookie runs in t seconds) · · · sM = 10 + 5t. · · · Likewise, sA = (initial distance between Linda and Asia) + + (distance Asia runs in t seconds) · · · sA = 34 + 4t. · · · If sMA is the distance between Mookie and Asia after t seconds, we compute sMA = sA − sM = (34 + 4t) − (10 + 5t) = (24 − t) meters. In all cases, the distances we computed are given by linear equations of the form s = b + mt, for appropriate b and rate m. 42 CHAPTER 4. LINEAR MODELING 4.6 Back to the Earning Power Problem 40000 35000 30000 25000 20000 15000 10000 5000 y-axis (dollars) U T S y R (x,y) on the line: means men earn y dollars in year x x-axis (year We now return to the motivating problem at the start of this section. Recall the plot in Figure 4.1(b). We can model the men’s earning power using the ﬁrst and last data points, using the ideas we have discussed about linear equations. To do this, we should specify a “beginning point” and an “ending point” (recall Figure 2.9) and calculate the slope: Rbegin = (1970,9,521) and Send = (1987,28,313). Figure 4.9: Linear model of Men’s Earning Power. We ﬁnd that ∆y ∆x 28,313 − 9,521 1987 − 1970 18,792 17 m ∆y ∆x 18,792 17 If we apply the “point-slope formula” for the equation of a line, we arrive at the equation: y = 18,792 17 (x − 1970) + 9,5
21. (4.6) The graph of this line will pass through the two points R and S in Figure 4.2. We sketch the graph in Figure 4.9, indicating two new points T and U. We can use the model in (4.6) to make predictions of two different sorts: (i) predict earnings at some date, or (ii) predict when a desired value for earnings will occur. For example, let’s graphically discuss the earnings in 1995: • • Draw a vertical line x = 1995 up to the graph and label the intersection point U. Draw a horizontal line ℓ through U. The line ℓ crosses the y axis at the point 18,792 17 (1995 − 1970) + 9521 = 37,156. 4.7. WHAT’S NEEDED TO BUILD A LINEAR MODEL? 43 The coordinates of the point U = (1995, 37,156). • Conclude that $37,156 is the Men’s Earning Power in 1995. For another example, suppose we wanted to know when men’s earning power will equal $33,000? This means we seek a data point T on the men’s earning curve whose y-coordinate is 33,000. By (4.6), T has the form T = x, (x − 1970) + 9,521. 18,792 17 We want this to be a data point of the form (x, 33,000). Setting these two points equal and equating the second coordinates leads to an algebra problem: 18,792 17 (x − 1970) + 9,521 = 33,000 x = 1991.24. This means men’s earning power will be $33,000 at the end of the ﬁrst quarter of 1991. Graphically, we interpret this reasoning as follows: • • Draw a horizontal line y = 33,000 and label the intersection point T on the model. Draw a vertical line ℓ through T. The line ℓ crosses the x axis at the point 1991.24. The coordinates of the point T = (1991.24,33,000). • In the exercises, you will be asked to show that the women’s earning power model is given by the equation y = 12,915 17 (x − 1970) + 5,616. Using the two linear models for the earning power of men and women, are women gaining
on men? You will also be asked to think about this question in the exercises. 4.7 What’s Needed to Build a Linear Model? As we progress through this text, a number of different “types” of mathematical models will be discussed. We will want to think about the information needed to construct that particular kind of mathematical model. Why would we care? For example, in a laboratory context, if we knew a situation being studied was given by a linear model, this would effect the amount of data collected. In the case of linear models, we can now make this useful statement: 44 CHAPTER 4. LINEAR MODELING Important Fact 4.7.1. A linear model is completely determined by: 1. One data point and a slope (a rate of change), or 2. Two data points, or 3. An intercept and a slope (a rate of change). 4.8 Linear Application Problems Example 4.8.1. The yearly resident tuition at the University of Washington was $1827 in 1989 and $2907 in 1995. Assume that the tuition growth at the UW follows a linear model. What will be the tuition in the year 2000? When will yearly tuition at the University of Washington be $10,000? 10000 8000 6000 4000 2000 y-axis (dollars) P Q x-axis (year Solution. If we consider a coordinate system where the x-axis represents the year and the y-axis represents dollars, we are given two data points: P = (1989, 1,827) and Q = (1995, 2,907). Using the two-point formula for the equation of line through P and Q, we obtain the equation Figure 4.10: Linear tuition model. y = 180(x − 1989) + 1,827. The graph of this equation gives a line through the given points as pictured in Figure 4.10. If we let x = 2000, we get y = $3,807, which tells us the tuition in the year 2000. On the other hand, if we set the equation equal to $10,000, we can solve for x: 10,000 = 180(x − 1989) + 1,827 8,173 = 180(x − 1989) 2,034.4 = x. Conclude the tuition is $10,000 in the year 2035. 4.9 Perpendicular and Parallel Lines Here is a useful fact to keep in mind. Important
Facts 4.9.1. Two non-vertical lines in the plane are parallel precisely when they both have the same slope. Two non-vertical lines are perpendicular precisely when their slopes are negative reciprocals of one another. 4.10. INTERSECTING CURVES II 45 Example 4.9.2. Let ℓ be a line in the plane passing through the points (1, 1) and (6, −1). Find a linear equation whose graph is a line parallel to ℓ passing through 5 on the y-axis. Find a linear equation whose graph is perpendicular to ℓ and passes through (4, 6). Solution. Letting P = (1, 1) and Q = (6, −1), apply the “two point formula”: y = −2 5 2 5 (x − 1) + 1 x + 7 5. = − The graph of this equation will be ℓ. This equation is in slope intercept form and we can read off that the slope is m = −2 5. The desired line a is parallel to ℓ; it must have slope m = − 2 5 and y-intercept 5. Plugging into the “slope intercept form”: y = −2 5 x + 5. The desired line b is a line perpendicular to ℓ (so its slope is m ′ = −1 = 5 2) and passes through the point (4, 6), so we can use the “point slope formula”: −2 5 y = 5 2 (x − 4) + 6. 4.10 Intersecting Curves II We have already encountered problems that require us to investigate the intersection of two curves in the plane. Ultimately, this reduces to solving a system of two (or more) equations in the variables x and y. A useful tool when working with equations involving squared terms (i.e., x2 or y2), is the quadratic formula. Important Fact 4.10.1. Quadratic Formula: Consider the equation az2 + bz + c = 0, where a,b,c are constants. The solutions for this equation are given by the formula z = −b ± √b2 − 4ac 2a. The solutions are real numbers if and only if b2 − 4ac 0. ≥ The next example illustrates a typical application of the quadratic forIn addition, we describe a very useful
technique for ﬁnding the mula. shortest distance between a “line” and a “point.” 46 CHAPTER 4. LINEAR MODELING North ﬂight path irrigated ﬁeld West Q East s e l i m 2 crop duster P 1.5 miles South Figure 4.11: The ﬂight path of a crop duster. Example 4.10.2. A crop dusting airplane ﬂying a constant speed of 120 mph is spotted 2 miles South and 1.5 miles East of the center of a circular irrigated ﬁeld. The irrigated ﬁeld has a radius of 1 mile. Impose a coordinate system as pictured, with the center of the ﬁeld the origin (0,0). The ﬂight path of the duster is a straight line passing over the labeled points P and Q. Assume that the point Q where the plane exits the airspace above the ﬁeld is the Western-most location of the ﬁeld. Answer these questions: 1. Find a linear equation whose graph is the line along which the crop duster travels. 2. Find the location P where the crop duster enters airspace above the irrigated ﬁeld. 3. How much time does the duster spend ﬂying over the irrigated ﬁeld? 4. Find the shortest distance from the ﬂight path to the center of the irrigated ﬁeld. Solution. 1. Take Q = (−1, 0) and S = (1.5, −2) = duster spotting point. Construct a line through Q and S. The slope is −0.8 = m and the line equation becomes: y = −0.8x − 0.8. (4.7) 2. The equation of the boundary of the irrigated region is x2 + y2 = 1. We need to solve this equation AND the line equation y = −0.8x − 0.8 simultaneously. Plugging the line equation into the unit circle equation gives: x2 + (−0.8x − 0.8)2 = 1 x2 + 0.64x2 + 1.28x + 0.64 = 1 1.64x2 + 1.28x − 0.36 = 0 Apply the quadratic formula and ﬁ
nd x = −1, 0.2195. Conclude that the x coordinate of P is 0.2195. To ﬁnd the y coordinate, plug into the line equation and get y = −0.9756. Conclude that P = (0.2195, −0.9756) 4.11. UNIFORM LINEAR MOTION 47 3. Find the distance from P to Q by using the distance formula: d = (−1 − 0.2195)2 + (0 − (−0.9756))2 = 1.562 miles p Now, 1.562 miles 120 mph = 0.01302 hours = 47 seconds. 4. The idea is to construct a line perpendicular to the ﬂight path passing through the origin of the coordinate system. This line will have slope m = − 1 −0.8 = 1.25. So this perpendicular line has equation y = 1.25x. Intersecting this line with the ﬂight path gives us the point closest to the center of the ﬁeld. The x-coordinate of this point is found by setting the two line equations equal and solving: −0.8x − 0.8 = 1.25x x = −0.3902 This means that the closest point on the ﬂight path is (−0.39, −0.49). Apply the distance formula and the shortest distance to the ﬂight path is d = (−0.39)2 + (−0.49)2 = 0.6263. p 4.11 Uniform Linear Motion When an object moves along a line in the xy-plane at a constant speed, we say that the object exhibits uniform linear motion. Its location can be described using a pair of linear equations involving a variable which represents time. That is, we can ﬁnd constants a, b, c, and d such that, at any time t, the object’s location is given by (x,y), where x = a + bt and y = c + dt. Such equations are called parametric equations of motion. The motion is deﬁned in terms of the parameter t. Since there are four constants to be determined, one needs four pieces of information to determine these equations. Knowing the object’s location at two points in time is sufﬁcient. Example 4.11.1. Bob is running in
the xy-plane. He runs in a straight line from the point (2,3) to the point (5, − 4), taking 6 seconds to do so. Find his equations of motion. 48 CHAPTER 4. LINEAR MODELING Solution. We begin by setting a reference for our time parameter. Let’s let t = 0 represent the instant when Bob is at the point (2,3). In this way, t will represent the time since Bob left the point (2,3). When t = 6, we know he will be at the point (5, − 4). This is enough information to determine his equations of motion. We seek constants a, b, c, and d so that at time t, Bob’s location is given by x = a + bt and y = c + dt. When t = 0, we know Bob’s location is (2,3). That is, x = 2 and y = 3. Thus, with t = 0, we have the two equations x = 2 = a + b(0) = a and y = 3 = c + d(0) = c. and so a = 2 and c = 3. We’re half-way done. When t = 6, we know Bob’s location is (5, − 4). Thus, with t = 6, we have the two equations x = 5 = a + b(6) = 2 + 6b and y = −4 = c + d(6) = 3 + 6d which we can solve to ﬁnd b = 1 2 and d = − 7 6. So, we arrive at the equations of Bob’s motion x = 2 + 1 2 t and y = 3 − 7 6 t. Notice it is easy to check that these are correct. If we plug in t = 0, we ﬁnd x = 2, y = 3 as required. If we plug in t = 6, we ﬁnd x = 5, y = −4, as required. So we know we’ve done it right. Now that we have these equations of motion, it is very easy to calculate Bob’s location at any time. For instance, 30 seconds after leaving the point (2,3), we can ﬁnd that he is at the point (17, − 32) since x = 2 + 1 2 (30) = 17
, y = 3 − 7 6 (30) = −32. Example 4.11.2. Olga is running in the xy-plane, and the coordinate are given in meters (so, for example, the point (1,0) is one meter from the origin (0,0)). She runs in a straight line, starting at the point (3,5) and running along the line y = − 1 3 x + 6 at a speed of 7 meters per second, heading away from the y-axis. What are her parametric equations of motion? 4.11. UNIFORM LINEAR MOTION 49 Solution. This example differs in some respects from the last example. In particular, instead of knowing where the runner is at two points in time, we only know one point, and have other information given to us about the speed and path of the runner. One approach is to use this new information to ﬁnd where the runner is at some other point in time: this will then give us exactly the same sort of information as we used in the last example, and so we may solve it in an identical manner. We know that Olga starts at the point (3,5). Letting t = 0 represent the time when she starts, we then know that when t = 0, x = 3 and y = 5. To get another point (and time), we can use the fact that we know what line she travels along, and which direction she runs. We may consider any point on the line in the correct direction: any will do. For instance, the point (6,4) is on the line. We then need to ﬁnd when Olga reaches this point. To do this, we ﬁnd the distance from her starting point to the point (6,4), and divide this by her speed. The time she takes to get to (6,4) is thus (6 − 3)2 + (4 − 5)2 7 p = 0.45175395 seconds. At this point, we are now in the same situation as in the last example. We know two facts: when t = 0, x = 3 and y = 5, and when t = 0.45175395, x = 6 and y = 4. As we saw in the last example, this is enough information to ﬁnd the parametric equations of motion. We seek a, b, c, and d such that Ol
ga’s location t seconds after she starts is (x,y) where x = a + bt and y = c + dt. When t = 0, x = 3, and y = 5, so x = 3 = a + b(0) = a and y = 5 = c + d(0) = c and so a = 3 and c = 5. Also, when t = 0.45175395, x = 6 and y = 4, so x = 6 = a+b(0.45175395) = 3+b(0.45175395) and y = 4 = c+d(0.45175395) = 5+d(0.45175395). Solving these equations for b and d, we ﬁnd b = 3 0.45175395 = 6.64078311 and d = −1 0.45175395 = −2.21359436. Thus, Olga’s equations of motion are x = 3 + 6.64078311t, y = 5 − 2.21359436t. 50 CHAPTER 4. LINEAR MODELING 4.12 Summary • • • The equation of every non-vertical line can be expressed in the form y = m(x − h) + k (the point-slope form) and y = mx + b (the slope-intercept form) A vertical line has an equation of the form x = c. The shortest distance between a point, P, and a line, l, can be found by determining a line l2 which passes through P and is perpendicular to l. Then the point at which l and l2 intersect is the point on l which is closest to l. The distance from this point to P is the shortest distance between P and l. • The location of an object moving at constant speed along a line can be described using a pair of equations (parametric equations) x = a + bt, y = c + dt. 4.13. EXERCISES 4.13 Exercises 51 Problem 4.1. This exercise emphasizes the “mechanical aspects” of working with linear equations. Find the equation of a line: Problem 4.4. Complete Table 4.2 on page 52. In many cases there may be several possible correct answers. (a) Passing through the points (1, − 1) and (−2,4). (b
) Passing through the point (−1, − 2) with slope m = 40. (c) With y-intercept b = −2 and slope m = −2. (d) Passing through the point (4,11) and having slope m = 0. (e) Perpendicular to the line in (a) and pass- ing through (1,1). (f) Parallel to the line in (b) and having y- intercept b = −14. (g) Having the equation 3x + 4y = 7. (h) Crossing the x-axis at x = 1 and having slope m = 1. Problem 4.2. Sketch an accurate picture of the line having equation y = 2 − 1 2 x. Let α be an unknown constant. (a) Find the point of intersection between the line you have graphed and the line y = 1 + αx; your answer will be a point in the xy plane whose coordinates involve the unknown α. (b) Find α so that the intersection point in (a) has x-coordinate 10. (c) Find α so that the intersection point in (a) lies on the x-axis. Problem 4.3. (a) What is the area of the tri2 x + angle determined by the lines y = − 1 5, y = 6x and the y-axis? (b) If b > 0 and m < 0, then the line y = mx + b cuts off a triangle from the ﬁrst quadrant. Express the area of that triangle in terms of m and b. (c) The lines y = mx + 5, y = x and the yaxis form a triangle in the ﬁrst quadrant. Suppose this triangle has an area of 10 square units. Find m. Problem 4.5. The (average) sale price for single family property in Seattle and Port Townsend is tabulated below: YEAR SEATTLE PORT TOWNSEND $8400 1970 $168,400 1990 $38,000 $175,000 (a) Find a linear model relating the year x and the sales price y for a single family property in Seattle. (b) Find a linear model relating the year x and the sales price y for a single family property in Port Townsend. (c) Sketch the graph of both modeling equations in a common coordinate system; restrict your attention to x 1970. ≥ (d) What is the sales price
in Seattle and Port Townsend in 1983 and 1998? (e) When will the average sales price in Seattle and Port Townsend be equal and what is this price? (f) When will the average sales price in Port Townsend be $15,000 less than the Seattle sales price? What are the two sales prices at this time? (g) When will the Port Townsend sales price be $15,000 more than the Seattle sales price? What are the two sales prices at this time? (h) When will the Seattle sales price be double the Port Townsend sales price? (i) Is the Port Townsend sales price ever double the Seattle sales price? Problem 4.6. The cup on the 9th hole of a golf course is located dead center in the middle of a circular green that is 70 feet in diameter. Your ball is located as in the picture below: 52 CHAPTER 4. LINEAR MODELING Equation Slope y-intercept Point on the line Point on the line y = 2x + 1 (3, −4) (−1, 7) 1 1,000 −2 1 2 0 (0, 1) (3, 3) (5, −9) (3, −2) Table 4.2: Linear equations table for Problem 4.4. 9 green cup ball path 50 feet rough ball 40 feet The ball follows a straight line path and exits the green at the right-most edge. Assume the ball travels a constant rate of 10 ft/sec. (a) Where does the ball enter the green? (b) When does the ball enter the green? (c) How long does the ball spend inside the green? (d) Where is the ball located when it is closest to the cup and when does this occur. Problem 4.7. Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person’s ankle. The cord is 30 feet long, but can stretch up to 90 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 8 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (a) Sketch an accurate picture of the situation at time t = 7 seconds. Make sure to label the locations of Allyson and Adrian; also, compute the length of the bungee cord at t = 7 seconds. (
b) Where is Allyson when the bungee reaches its maximum length? 4.13. EXERCISES 53 20 ft Building 30 ft Allyson Adrian start Problem 4.8. Dave is going to leave academia and go into business building grain silos. A grain silo is a cylinder with a hemispherical top, used to store grain for farm animals. Here is a 3D view, a cross-section, and the top view: silo h 3D-view r cross-section y-axis blind spot line of sight #1 (12,0) x-axis Dave line of sight #2 TOP VIEW If Dave is standing next to a silo of crosssectional radius r = 8 feet at the indicated position, his vision will be partially obstructed. Find the portion of the y-axis that Dave cannot see. (Hint: Let a be the x-coordinate of the point where line of sight #1 is tangent to the silo; compute the slope of the line using two points (the tangent point and (12,0)). On the other hand, compute the slope of line of sight #1 by noting it is perpendicular to a radial line through the tangency point. Set these two calculations of the slope equal and solve for a.) Problem 4.9. While speaking on the phone to a friend in Oslo, Norway, you learned that the current temperature there was −23◦ Celsius (−23◦C). After the phone conversation, you wanted to convert this temperature to Fahrenheit degrees ◦F, but you could not ﬁnd a reference with the correct formulas. You then remembered that the relationship between ◦F and ◦C is linear. (a) Using this and the knowledge that 32◦F = 0◦C and 212◦F = 100◦C, ﬁnd an equation that computes Celsius temperature in terms of Fahrenheit temperai.e., an equation of the form C= ture; “an expression involving only the variable F.” (b) Likewise, ﬁnd an equation that computes Fahrenheit temperature in terms of Celsius temperature; i.e. an equation of the form F= “an expression involving only the variable C.” (c) How cold was it in Oslo in ◦F? Problem 4.10. Pam is taking a train from the town of Rome to the town of Florence. Rome is located 30 miles due West of the town
of Paris. Florence is 25 miles East, and 45 miles North of Rome. On her trip, how close does Pam get to Paris? Problem 4.11. Angela, Mary and Tiff are all standing near the intersection of University and 42nd streets. Mary and Tiff do not move, but Angela runs toward Tiff at 12 ft/sec along a straight line, as pictured. Assume the roads are 50 feet wide and Tiff is 60 feet north of the nearest corner. Where is Angela located when she is closest to Mary and when does she reach this spot? 54 CHAPTER 4. LINEAR MODELING tiff 42 nd St. mary angela University Way Problem 4.12. The infamous crawling tractor sprinkler is located as pictured below, 100 feet South of a 10 ft. wide sidewalk; notice the hose and sidewalk are not perpendicular. Once the water is turned on, the sprinkler waters a circular disc of radius 20 feet and moves North along the hose at the rate of 1 2 inch/second. (a) Impose a coordinate system. Describe the initial coordinates of the sprinkler and ﬁnd the equation of the line forming the southern boundary of the sidewalk. (b) After 33 minutes, sketch a picture of the wet portion of the sidewalk; ﬁnd the length of the wet portion of the Southern edge of the sidewalk. (c) Find the equation of the line forming the northern boundary of the sidewalk. (Hint: You can use the properties of right triangles.) Problem 4.13. Margot is walking in a straight line from a point 30 feet due east of a statue in a park toward a point 24 feet due north of the statue. She walks at a constant speed of 4 feet per second. (a) Write parametric equations for Margot’s position t seconds after she starts walking. (b) Write an expression for the distance from Margot’s position to the statue at time t. (c) Find the times when Margot is 28 feet from the statue. Problem 4.14. Juliet and Mercutio are moving at constant speeds in the xy-plane. They start moving at the same time. Juliet starts at the point (0, − 6) and heads in a straight line toward the point (10,5), reaching it in 10 seconds. Mercutio starts at (9, − 14) and moves in a straight line. Mercutio passes through the same point on the x axis as Juliet
, but 2 seconds after she does. How long does it take Mercutio to reach the y-axis? Problem 4.15. (a) Solve for x. hose N S E W 100 ft (b) Solve for t: 2 = (1 + t)2 + (1 − 2t)2. (c) Solve for t: 3 √5 p = (1 + t)2 + (1 − 2t)2. (d) Solve for t: 0 = p (1 + t)2 + (1 − 2t)2. p 20 ft Problem 4.16. (a) Solve for x: 100 ft. sidewalk circular watered zone x4 − 4x2 + 2 = 0 (b) Solve for y: y − 2√y = 4 Chapter 5 Functions and Graphs Pictures are certainly important in the work of an architect, but it is perhaps less evident that visual aids can be powerful tools for solving mathematical problems. If we start with an equation and attach a picture, then the mathematics can come to life. This adds a new dimension to both interpreting and solving problems. One of the real triumphs of modern mathematics is a theory connecting pictures and equations via the concept of a graph. This transition from “equation” to “picture” (called graphing) and its usefulness (called graphical analysis) are the theme of the next two sections. The importance of these ideas is HUGE and cannot be overstated. Every moment spent studying these ideas will pay back dividends in this course and in any future mathematics, science or engineering courses. 5.1 Relating Data, Plots and Equations Imagineyou are standinghigh atop an oceanside cliff and spot a seagull hovering in the air-current. Assuming the gull moves up and down along a vertical line of motion, how can we best describe its location at time t seconds? There are three different (but closely linked) ways to describe the location of the gull: • • • a table of data of the gull’s height above cliff level at various times t; a plot of the data in a “time” (seconds) vs. “height” (feet) coordinate system; an equation relating time t (seconds) and height s (feet). gull line of motion cliff level ocean Figure 5.1: Seagull’s height. To make sure we really understand how to pass back and forth between
these three descriptive modes, imagine we have tabulated (Figure 5.2) the height of the gull above cliff level at one-second time intervals 55 56 CHAPTER 5. FUNCTIONS AND GRAPHS for a 10 second time period. Here, a “negative height” means the gull is below cliff level. We can try to visualize the meaning of this data by plotting these 11 data points (t, s) in a time (sec.) vs. height (ft.) coordinate system. t (sec) 0 1 2 3 Gull Height (feet above cliff level) t (sec) 8 9 10 t (sec) 4 5 6 7 s (ft) 20 6.88 -2.5 -8.12 s (ft) -10 -8.12 -2.5 6.88 Feet 50 40 30 20 10 −10 s (ft) 20 36.88 57.5 2 4 6 8 10 Seconds (a) Symbolic data. (b) Visual data. Figure 5.2: Symbolic versus visual view of data. We can improve the quality of this description by increasing the number of data points. For example, if we tabulate the height of the gull above cliff level at 1/2 second or 1/4 second time intervals (over the same 10 second time period), we might get these two plots: Feet 50 40 30 20 10 4 Seconds Feet 50 40 30 20 10 4 Seconds −10 2 6 8 10 −10 2 6 8 10 (a) 1 2 second intervals. (b) 1 4 second intervals. Figure 5.3: Shorter time intervals mean more data points. We have focused on how to go from data to a plot, but the reverse process is just as easy: A point (t, s) in any of these three plots is interpreted to mean that the gull is s feet above cliff level at time t seconds. Furthermore, increasing the amount of data, we see how the plotted points are “ﬁlling in” a portion of a parabola. Of course, it is way too tedious to create longer and longer tables of data. What we really want is a “formula” (think of it as a prescription) that tells us how to produce a data point for the gull’s height at any given time t. If we had such a formula, then we could completely dispense with the tables of data and just use the formula to crank out data points. For example
, look at this equation involving the variables t and s: s = 15 8 (t − 4)2 − 10. 5.2. WHAT IS A FUNCTION? 57 If we plug in t = 0, 1, 2, 9, 10, then we get s = 20, 6.88, −2.5, 36.88, 57.5, respectively; this was some of our initial tabulated data. This same equation produces ALL of the data points for the other two plots, using 1/2 second and 1/4 second time intervals. (Granted, we have swept under the rug the issue of “...where the heck the equation comes from...”; that is a consequence of mathematically modeling the motion of this gull. Right now, we are focusing on how the equation relates to the data and the plot, assuming the equation is in front of us to start with.) In addition, it is very important to notice that having this equation produces an inﬁnite number of data points for our gull’s location, since we can plug in any t value between 0 and 10 and get out a corresponding height s. In other words, the equation is A LOT more powerful than a ﬁnite (usually called discrete) collection of tabulated data. 5.2 What is a Function? Our lives are chock full of examples where two changing quantities are related to one another: • • • • The cost of postage is related to the weight of the item. The value of an investment will depend upon the time elapsed. The population of cells in a growth medium will be related to the amount of time elapsed. The speed of a chemical reaction will be related to the temperature of the reaction vessel. In all such cases, it would be beneﬁcial to have a “procedure” whereby we can assign a unique output value to any acceptable input value. For example, given the time elapsed (an input value), we would like to predict Informally, a unique future value of an investment (the output value). this leads to the broadest (and hence most applicable) deﬁnition of what we will call a function: Deﬁnition 5.2.1. A function is a procedure for assigning a unique output to any allowable input. The key word here is “procedure.” Our discussion of the hovering seagull in 5.1 highlights three ways to
produce such a “procedure” using data, plots of curves and equations. y-axis P = (x,y) y x x-axis • A table of data, by its very nature, will relate two columns of data: The output and input values are listed as column entries of the table and reading across each row is the “procedure” which relates an input with a unique output. Figure 5.4: Graph of a procedure. 58 • • CHAPTER 5. FUNCTIONS AND GRAPHS Given a curve in Figure 5.4, consider the “procedure” which associates to each x on the horizontal axis the y coordinate of the pictured point P on the curve. Given an equation relating two quantities x and y, plugging in a particular x value and going through the “procedure” of algebra often produces a unique output value y. 5.2.1 The deﬁnition of a function (equation viewpoint) Now we focus on giving a precise deﬁnition of a function, in the situation when the “procedure” relating two quantities is actually given by an equation. Keep in mind, this is only one of three possible ways to describe a function; we could alternatively use tables of data or the plot of a curve. We focus on the equation viewpoint ﬁrst, since it is no doubt the most familiar. If we think of x and y as related physical quantities (e.g. time and distance), then it is sometimes possible (and often desirable) to express one of the variables in terms of the other. For example, by simple arithmetic, the equations 3x + 2y = 4 x2 − x = 1 2 y − 4 y x2 + 1 = 1, p can be rewritten as equivalent equations y = 1 2 (4 − 3x) 2x2 − 2x + 8 = y y = 1 √x2 + 1. This leads to THE MOST IMPORTANT MATH DEFINITION IN THE WORLD: Deﬁnition 5.2.2. A function is a package, consisting of three parts: An equation of the form • y = “a mathematical expression only involving the variable x,” which we usually indicate via the shorthand notation y = f(x). This equation has the very special property that each time we plug in an x value, it produces exactly one (
a unique) y value. We call the mathematical expression f(x) ”the rule”. • A set D of x-values we are allowed to plug into f(x), called the ”domain” of the function. 5.2. WHAT IS A FUNCTION? 59 • The set R of output values f(x), where x varies over the domain, called the ”range” of the function. Any time we have a function y = f(x), we refer to x as the independent variable (the “input data”) and y as the dependent variable (the “output data”). This terminology emphasizes the fact that we have freedom in the values of x we plug in, but once we specify an x value, the y value is uniquely determined by the rule f(x). Examples 5.2.3. (i) The equation y = −2x + 3 is in the form y = f(x), where the rule is f(x) = −2x+3. Once we specify a domain of x values, we have a function. For example, we could let the domain be all real numbers. y-axis Graph of y = b b x-axis (ii) Take the same rule f(x) = −2x + 3 from (i) and let the domain be all non-negative real numbers. This describes a function. However, the functions f(x) = −2x+ 3 on the domain of all non-negative real numbers and f(x) = −2x + 3 on the domain of all real numbers (from (i)) are different, even though they share the same rule; this is because their domains differ! This example illustrates the idea of what is called a restricted domain. In other words, we started with the function in (i) on the domain of all real numbers, then we “restricted” to the subset of non-negative real numbers. Figure 5.5: Constant function. (iii) The equation y = b, where b is a constant, deﬁnes a function on the domain of all real numbers, where the rule is f(x) = b; we call these the constant functions. Recall, in Chapter 3, we observed that the solutions of the equation y = b, plotted in the xy coordinate system, will give a horizontal line. For example, if b = 0, you get the
horizontal axis. (iv) Consider the equation y = 1 x, then the rule f(x) = 1 x deﬁnes a function, as long as we do not plug in x = 0. For example, take the domain to be the non-zero real numbers. (v) Consider the equation y = √1 − x2. Before we start plugging in x values, we want to know the expression under the radical symbol (square root symbol) is nonnegative; this insures the square root is a real number. This amounts to solving an inequality equation: 1. These remarks show that 0 the rule f(x) = √1 − x2 deﬁnes a function, where the domain of x values is −1 x 1 − x2; i.e., −1 1. ≤ ≤ ≤ x ≤ ≤ 60 CHAPTER 5. FUNCTIONS AND GRAPHS Typically, the domain of a function y = f(x) will either be the entire number line, an interval on the number line, or a ﬁnite union of such intervals. We summarize the notation used to represent intervals in Table 5.1. Common Intervals on the Number Line Description Symbolic Notation Picture All numbers x between a and b, x possibly equal to either a or b a b x ≤ ≤ All numbers x between a = a and x and b, x = b a < x < b All numbers x between a = b and x possiand b, x bly equal to a x < b a ≤ All numbers x between a = a and x possiand b, x bly equal to Table 5.1: Interval Notations We can interpret a function as a “prescription” that takes a given x value (in the domain) and produces a single unique y value (in the range). We need to be really careful and not fall into the trap of thinking that every equation in the world is a function. For example, if we look at this equation x + y2 = 1 and plug in x = 0, the equation becomes y2 = 1. ± This equation has two solutions, y = 1, so the conclusion is that plugging in x = 0 does NOT produce a single output value. This violates one of the conditions of our function deﬁnition, so the equation x + y2 = 1 is NOT a function in the independent variable x. Notice, if you were
to try and solve this equation for y in terms of x, you’d ﬁrst write y2 = 1 − x and then take a square root (to isolate y); but the square root introduces TWO roots, which is just another way of reﬂecting the fact there can be two y values attached to a single x value. Alternatively, you can solve the equation for x in terms of y, getting x = 1 − y2; this shows the equation does deﬁne a function x = g(y) in the independent variable y. 6 6 6 6 5.2. WHAT IS A FUNCTION? 61 5.2.2 The deﬁnition of a function (conceptual viewpoint) Conceptually, you can think of a function as a “process”: An allowable input goes into a “black box” and out pops a unique new value denoted by the symbol f(x). Compare this with the machine making “hula-hoops” in Figure 5.6. While you are problem solving, you will ﬁnd this to be a useful viewpoint when a function is described in words. tube in hoop out in x out y tubes f(tube) domain f(x) (a) A hoop machine as a “process” taking “tubes” to “hoops.” (b) A function as a “process” which takes x to y. Figure 5.6: Viewing a function as a “process.” Examples 5.2.4. Here are four examples of relationships that are functions: (i) The total amount of water used by a household since midnight on a particular day. Let y be the total number of gallons of water used by a household between 12:00am and a particular time t; we will use time units of “hours.” Given a time t, the household will have used a speciﬁc (unique) amount of water, call it S(t). Then y = S(t) deﬁnes a function in the independent variable t with dependent variable y. The domain would be 0 24 and the largest possible value of S(t) on this domain is S(24). This tells us that the range would be the set of values 0 S(24). ≤ ≤ y t ≤ ≤ (ii) The height of the
center of a basketball as you dribble, depending on time. Let s be the height of the basketball center at time t seconds after you start dribbling. Given a time t, if we freeze the action, the center of the ball has a single unique height above the ﬂoor, call it h(t). So, the height of the basketball center is given by a function s = h(t). The domain would be a given interval of time you are dribbling the ball; for example, maybe 0 2 (the ﬁrst 2 seconds). In this case, the range would be all of the possible heights attained by the center of the basketball during this 2 seconds. ≤ ≤ t (iii) The state sales tax due on a taxable item. Let T be the state tax (in dollars) due on a taxable item that sells for z dollars. Given a taxable item that costs z dollars, the state tax due is a single unique 62 CHAPTER 5. FUNCTIONS AND GRAPHS amount, call it W(z). So, T = W(z) is a function, where the independent variable is z. The domain could be taken to be 0 1,000,000, which would cover all items costing up to one-million dollars. The range of the function would be the set of all values W(z), as z ranges over the domain. ≤ ≤ z (iv) The speed of a chemical reaction depending on the temperature. Let v be the speed of a particular chemical reaction and T the temperature in Celsius ◦C. Given a particular temperature T, one could experimentally measure the speed of the reaction; there will be a unique speed, call it r(T ). So, v = r(T ) is a function, where the independent variable is T. The domain could be taken to be 0 100, which would cover the range of temperatures between the freezing and boiling points of water. The range of the function would be the set of all speeds r(T ), as T ranges over the domain. ≤ ≤ T 5.3 The Graph of a Function Let’s start with a concrete example; the function f(x) = −2x + 3 on the domain of all real numbers. We discussed this Example 5.2.3. Plug in the speciﬁc x values, where x = −1, 0, 1, 2 and tabulate the resulting y values of the function: x -1 0 1 2
... x y 5 3 1 -1... −2x + 3 point (x,y) (-1,5) (0,3) (1,1) (2,-1)... (x, − 2x + 3) y-axis Graph of y = −2x + 3. x-axis (a) Tabulated data. (b) Visual data. Figure 5.7: Symbolic versus visual view of data. · This tells us that the points (0, 3), (1, 1), (2, −1), (−1, 5) are solutions of the equation y = −2x + 3. For example, if y = −2x + 3, x = 0, y = 3, 0 + 3 (which is true), or if y = −2x + 3, x = 2, y = −1, then then 3 = −2 −1 = −2 2 + 3 (which is true), etc. In general, if we plug in x we get out −2x + 3, so the point (x, − 2x + 3) is a solution to the function equation y = f(x). We can plot all of these solutions in the xy-coordinate system. The set of points we obtain, as we vary over all x in the domain, is called the set of solutions of the equation y = −2x + 3: · Solutions = { (x, −2x + 3) \| x any real number}. Notice that plotting these points produces a line of slope m = −2 with y-intercept 3. In other words, the graph of the function f(x) = −2x + 3 is 5.4. THE VERTICAL LINE TEST 63 the same as the graph of the equation y = −2x + 3, as we discussed in Chapter 4. In general, by deﬁnition, we say that a point (x,y) is a solution to the function equation y = f(x) if plugging x and y into the equation gives a true statement. HowcanweﬁndALLthesolutionsoftheequation y = f(x)? In general, the deﬁnition of a function is “rigged” so it is easy to describe all solutions of the equation y = f(x): Each time we specify an x value (in the domain), there is only one y value
, namely f(x). This means the point P = (x, f(x)) is the ONLY solution to the equation y = f(x) with ﬁrst coordinate x. We deﬁne the graph of the function y = f(x) to be the plot of all solutions of this equation (in the xy coordinate system). It is common to refer to this as either the “graph of f(x)” or the “graph of f.” Graph = {(x,f(x)) \| x in the domain} (5.1) Important Procedure 5.3.1. Points on a graph. The description of the graph of a function gives us a procedure to produce points on the graph AND to test whether a given point is on the graph. On the one hand, if you are given u in the domain of a function y = f(x), then you immediately can plot the point (u, f(u)) on the graph. On the other hand, if someone gives you a point (u, v), it will be on the graph only if v = f(u) is true. We illustrate this in Example 5.3.2. t Example 5.3.2. The function s = h(t) = 15 8 (t − 4)2 − 10 deﬁnes a function in the independent variable t. If we restrict 10, then the discussion in Chapter 7 to the domain 0 tells us that the graph is a portion of a parabola: See Figure 5.8. Using the above procedure, you can verify that the data points discussed in the seagull example (in 5.1) all § lie on this parabola. On the other hand, the point (0,0) is NOT on the graph, since h(0) = 20 = 0. ≤ ≤ s-axis 60 50 40 30 20 10 4 t-axis −10 2 6 8 10 Figure 5.8: s = h(t). 5.4 The Vertical Line Test There is a pictorial aspect of the graph of a function that is very revealing: Since (x, f(x)) is the only point on the graph with ﬁrst coordinate equal to x, a vertical line passing through x on the x-axis (with x in the domain) crosses the graph of y = f(x) once and only once. This gives us a decisive
way to test if a curve is the graph of a function. Important Procedure 5.4.1. The vertical line test. Draw a curve in the xy-plane and specify a set D of x-values. Suppose every vertical line through a value in D intersects the curve exactly once. Then the curve is 6 64 CHAPTER 5. FUNCTIONS AND GRAPHS the graph of some function on the domain D. If we can ﬁnd a single vertical line through some value in D that intersects the curve more than once, then the curve is not the graph of a function on the domain D. For example, draw any straight line m in the plane. By the vertical line test, if the line m is not vertical, m is the graph of a function. On the other hand, if the line m is vertical, then m is not the graph of a function. These two situations are illustrated in Figure 5.9. As another example, consider the equation x2 + y2 = 1, whose graph is the unit circle and specify the domain D to be −1 1; recall Example 3.2.2. The vertical line passing through the point will intersect the unit circle twice; by the vertical line test, the unit circle is not the graph of a function on the domain −1 x ≤ 1 2, 0 1. ≤ x ≤ ≤ y-axis y-axis m m y-axis l C l crosses curve twice x-axis x-axis x-axis Figure 5.9: Applying the vertical line test. 5.4.1 Imposed Constraints In physical problems, it might be natural to constrain (meaning to “limit” or “restrict”) the domain. As an example, suppose the height s (in feet) of a ball above the ground after t seconds is given by the function s = h(t) = −16t2 + 4. s-axis Physically interesting portion of graph. t-axis Figure 5.10: Restricting the domain. We could look at the graph of the function in the tsplane and we will review in Chapter 7 that the graph looks like a parabola. The physical context of this problem makes it natural to only consider the portion of the graph in the ﬁrst quadrant; why? One way of specifying this quadrant would be to restrict the domain of possible t values to lie between 0 and 1 2; notationally, we
would 1 write this constraint as 0 2. t ≤ ≤ 5.5 Linear Functions A major goal of this course is to discuss several different kinds of functions. The work we did in Chapter 4 actually sets us up to describe one 5.6. PROFIT ANALYSIS 65 very useful type of function called a linear function. Back in Chapter 4, we discussed how lines in the plane can be described using equations in the variables x and y. One of the key conclusions was: Important Fact 5.5.1. A non-vertical line in the plane will be the graph of an equation y = mx + b, where m is the slope of the line and b is the y-intercept. Notice that any non-vertical line will satisfy the conditions of the vertical line test, which means it must be the graph of a function. What is the function? The answer is to use the equation in x and y we already obtained in Chapter 4: The rule f(x) = mx + b on some speciﬁed domain will have a line of slope m and y-intercept b as its graph. We call a function of this form a linear function. Example 5.5.2. You are driving 65 mph from the Kansas state line (mile marker 0) to Salina (mile marker 130) along I-35. Describe a linear function that calculates mile marker after t hours. Describe another linear function that will calculate your distance from Salina after t hours. Solution. Deﬁne a function d(t) to be the mile marker after t hours. Using “distance=rate time,” we conclude that 65t will be the distance traveled after t hours. Since we started at mile marker 0, d(t) = 65t is the rule for the ﬁrst function. A reasonable domain would be to take 0 2, since it takes 2 hours to reach Salina. × ≤ ≤ t For the second situation, we need to describe a different function, call it s(t), that calculates your distance from Salina after t hours. To describe the rule of s(t) we can use the previous work: s(t) = (mile marker Salina) − (your mile marker at t hrs.) = 130 − d(t) = 130 − 65t. y-axis 120 100 80 60 40 20 0.5 1 1.5 2 t-axis Figure 5.
11: Distance functions. For the rule s(t), the best domain would again be 0 2. We have graphed these two functions in the same coordinate system: See Figure 5.11 (Which function goes with which graph?). ≤ ≤ t 5.6 Proﬁt Analysis Let’s give a ﬁrst example of how to interpret the graph of a function in the context of an application. 66 CHAPTER 5. FUNCTIONS AND GRAPHS Example 5.6.1. A software company plans to bring a new product to market. The sales price per unit is $15 and the expense to produce and market x units is $100(1 + √x). What is the proﬁt potential? Two functions control the proﬁt potential of the new software. The ﬁrst tells us the gross income, in dollars, on the sale of x units. All of the costs involved in developing, supporting, distributing and marketing x units are controlled by the expense equation (again in dollars): g(x) = 15x e(x) = 100(1 + √x) (gross income function) (expense function) A proﬁt will be realized on the sale of x units whenever the gross income exceeds expenses; i.e., this occurs when g(x) > e(x). A loss occurs on the sale of x units when expenses exceed gross income; i.e., when e(x) > g(x). Whenever the sale of x units yields zero proﬁt (and zero loss), we call x a break-even point; i.e., when e(x) = g(x). The above approach is “symbolic.” Let’s see how to study proﬁt and loss visually, by studying the graphs of the two functions g(x) and e(x). To begin with, plot the graphs of the two individual functions in the xycoordinate system. We will focus on the situation when the sales ﬁgures are between 0 to 100 units; so the domain of x values is the interval 100. Given any sales ﬁgure x, we can graphically relate three 0 x ≤ ≤ y-axis (dollars) y-axis (dollars) 1500 1000 500 x, 15x) P x-axis (units sold) 20 40 60 80
100 e s n e p x e 1500 1000 500 (x, e(x)) Q x-axis (units sold) 20 40 60 80 100 (a) Gross income graph. (b) Expenses graph. Figure 5.12: Visualizing income and expenses. things: x on the horizontal axis; a point on the graph of the gross income or expense function; • • y on the vertical axis. • If x = 20 units sold, there is a unique point P = (20, g(20)) = (20, 300) on the gross income graph and a unique point Q = (20, e(20)) = (20, 547) on the expenses graph. Since the y-coordinates of P and Q are the function values at x = 20, the height of the point above the horizontal axis is controlled by the function. 5.6. PROFIT ANALYSIS 67 dollars y (x, g(x)) (x, e(x)) sold units x 80 100 Figure 5.13: Modelling proﬁt and loss. 1400 1200 1000 800 600 400 200 If we plot both graphs in the same coordinate system, we can visually study the distance between points on each graph above x on the horizontal axis. In the ﬁrst part of this plot, the expense graph is above the income graph, showing a loss is realized; the exact amount of the loss will be e(x) − g(x), which is the length of the pictured line segment. Further to the right, the two graphs cross at the point labeled “B”; this is the break-even point; i.e., expense and income agree, so there is zero proﬁt (and zero loss). Finally, to the right of B the income graph is above the expense graph, so there is a proﬁt; the exact amount of the proﬁt will be g(x) − e(x), which is the length of the right-most line segment. Our analysis will be complete once we pin down the break-even point B. This amounts to solving the equation g(x) = e(x). (x, g(x)) (x, e(x)) 20 40 B 60 15x = 100(1 + √x) 15x − 100 = 100√x 225x2 − 3000x + 10000 = 10000x 225x2 − 13000x
+ 10000 = 0. Applying the quadratic formula, we get two answers: x = 0.78 or 57. Now, we face a problem: Which of these two solutions is the answer to the original problem? We are going to argue that only the second solution x = 57 gives us the break even point. What about the other ”solution” at x = 0.78? Try plugging x = 0.78 into the original equation: = 100(1 + √0.78). What has happened? Well, when going from 15(0.78) the second to the third line, both sides of the equation were squared. Whenever we do this, we run the risk of adding extraneous solutions. What should you do? After solving any equation, look back at your steps and ask yourself whether or not you may have added (or lost) solutions. In particular, be wary when squaring or taking the square root of both sides of an equation. Always check your ﬁnal answer in the original equation. We can now compute the coordinates of the break-even point using either function: B = (57, g(57)) = (57, 855) = (57, e(57)). 6 68 CHAPTER 5. FUNCTIONS AND GRAPHS 5.7 Exercises Problem 5.1. For each of the following functions, ﬁnd the expression for padelford gould f(x + h) − f(x) h. Simplify each of your expressions far enough so that plugging in h = 0 would be allowed. (a) f(x) = x2 − 2x. (b) f(x) = 2x + 3 (c) f(x) = x2 − 3 (d) f(x) = 4 − x2 (e) f(x) = −πx2 − π2 (f) f(x) = √x − 1. (Hint: Rationalize the nu- merator) x ≤ Problem 5.2. Here are the graphs of two linear functions on the domain 0 20. Find the formula for each of the rules y = f(x) and y = g(x). Find the formula for a NEW function v(x) that calculates the vertical distance between the two lines at x. Explain in terms of the picture what v(x) is calculating. What is v(5)? What
is v(20)? What are the smallest and largest values of v(x) on the domain 0 20? ≤ x ≤ ≤ y-axis 60 40 20 (0,4) g(x) (20,60) (0,24) f(x) (20,20) 10 20 x-axis Problem 5.3. Dave leaves his ofﬁce in Padelford Hall on his way to teach in Gould Hall. Below are several different scenarios. In each case, sketch a plausible (reasonable) graph of the function s = d(t) which keeps track of Dave’s distance s from Padelford Hall at time t. Take distance units to be “feet” and time units to be “minutes.” Assume Dave’s path to Gould Hall is along a straight line which is 2400 feet long. (a) Dave leaves Padelford Hall and walks at a constant speed until he reaches Gould Hall 10 minutes later. (b) Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute. He then continues on to Gould Hall at the same constant speed he had when he originally left Padelford Hall. (c) Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Then he gets confused and stops for 1 minute to ﬁgure out where he is. Dave then continues on to Gould Hall at twice the constant speed he had when he originally left Padelford Hall. (d) Dave leaves Padelford Hall and walks at a constant speed. It takes him 6 minutes to reach the half-way point. Dave gets confused and stops for 1 minute to ﬁgure out where he is. Dave is totally lost, so he simply heads back to his ofﬁce, walking the same constant speed he had when he originally left Padelford Hall. (e) Dave leaves Padelford heading for Gould Hall at the same instant Angela leaves Gould Hall heading for Padelford Hall. Both walk at a constant speed, but Angela walks twice as fast as Dave. Indicate a plot of “distance from Padelford” vs. “time” for both Angela and Dave. (f) Suppose you want to sketch the graph of a new function s = g(t)
that keeps track of Dave’s distance s from Gould Hall at time t. How would your graphs change in (a)-(e)? Problem 5.4. At 5 AM one day, a monk began a trek from his monastery by the sea to the monastery at the top of a mountain. He reached the mountain-top monastery at 11 AM, spent the rest of the day in meditation, 5.7. EXERCISES 69 and then slept the night there. In the morning, at 5 AM, he began walking back to the seaside monastery. Though walking downhill should have been faster, he dawdled in the beautiful sunshine, and ending up getting to the seaside monastery at exactly 11 AM. (a) Was there necessarily a time during each trip when the monk was in exactly the same place on both days? Why or why not? (b) Suppose the monk walked faster on the second day, and got back at 9 AM. What is your answer to part (a) in this case? (c) Suppose the monk started later, at 10 AM, and reached the seaside monastery at 3 PM. What is your answer to part (a) in this case? Problem 5.5. Sketch a reasonable graph for each of the following functions. Specify a reasonable domain and range and state any assumptions you are making. Finally, describe the largest and smallest values of your function. (a) Height of a person depending on age. (b) Height of the top of your head as you jump on a pogo stick for 5 seconds. (c) The amount of postage you must put on a ﬁrst class letter, depending on the weight of the letter. (d) Distance of your big toe from the ground as you ride your bike for 10 seconds. (e) Your height above the water level in a swimming pool after you dive off the high board. y-axis x-axis Recall the procedure 5.3.1 on page 63. (a) Find the x and y intercepts of the graph. (b) Find the exact coordinates of all points (x,y) on the graph which have ycoordinate equal to 5. (c) Find the coordinates of all points (x,y) on the graph which have y-coordinate equal to -3. (d) Which of these points is on the graph: (1, − 2), (−1,3), (2.4,8), (√3
,7 − 3√3). (e) Find the exact coordinates of the point (x,y) on the graph with x = 1 + √2. p Problem 5.7. After winning the lottery, you decide to buy your own island. The island is located 1 km offshore from a straight portion of the mainland. There is currently no source of electricity on the island, so you want to run a cable from the mainland to the island. An electrical power sub-station is located 4 km from your island’s nearest location to the shore. It costs $50,000 per km to lay a cable in the water and $30,000 per km to lay a cable over the land. ocean power your island cable path 1 km x 4 km Problem 5.6. Here is a picture of the graph of the function f(x) = 3x2 − 3x − 2. (a) Explain why we can assume the cable follows the path indicated in the picture; 70 CHAPTER 5. FUNCTIONS AND GRAPHS explain why the path consists of i.e. two line segments, rather than a weird curved path AND why it is OK to assume the cable reaches shore to the right of the power station and the left of the island. (b) Let x be the distance downshore from the power sub-station to where the cable reaches the land. Find a function f(x) in the variable x that computes the cost to lay a cable out to your island. (c) Make a table of values of f(x), where x = 0, 1 2,4. Use these calculations to estimate the installation of minimal cost. 2,2,..., 7 2,1, 3 Problem 5.8. This problem deals with the “mechanical aspects” of working with the rule of a function. For each of the functions listed ), in (a)-(c), calculate: f(0), f(−2), f(x + 3), f( f( ♥ + ). △ ♥ (a) The function f(x) = 1 main of all real numbers. 2 (x − 3) on the do- Problem 5.9. Which of the curves in Figure 5.14 represent the graph of a function? If the curve is not the graph of a function, describe what goes wrong and how you might “ﬁx it.” When you describe how to “
ﬁx” the graph, you are allowed to cut the curve into pieces and such that each piece is the graph of a function. Many of these problems have more than one correct answer. Problem 5.10. Find an EXACT answer for each problem. (a) Solve for = 30 x2 − 4x − 21 (b) Solve for x √5x − 4 = x 2 + 2 (c) Solve for x √x + √x − 20 = 10 (b) The function f(x) = 2x2 − 6x on the do- (d) Solve for t main of all real numbers. (c) The function f(x) = 4π2. √2t − 1 + √3t + 3 = 5 5.7. EXERCISES 71 (a) (e) (i) (b) (c) (d) (f) (g) (h) (j) (k) (l) (p) (m) (n) (o) Figure 5.14: Curves to consider for Problem 5.9 72 CHAPTER 5. FUNCTIONS AND GRAPHS Chapter 6 Graphical Analysis We ended the previous section with an in-depth look at a “Proﬁt Analysis Problem.” In that discussion, we looked at the graphs of the relevant functions and used these as visual aids to help us answer the questions posed. This was a concrete illustration of what is typically called “graphical analysis of a function.” This is a fundamental technique we want to carry forward throughout the course. Let’s highlight the key ideas for future reference. 6.1 Visual Analysis of a Graph A variety of information can be visually read off of a function graph. To see this, we ask ourselves the following question: What is the most basic qualitative feature of a graph? To answer this, we need to return to the deﬁnition of the graph (see Equation (5.1) on page 63) and the surrounding discussion. The key thing about the graph of a function f(x) is that it keeps track of a particular set of points in the plane whose coordinates are related by the function rule. To be precise, a point P = (x,y) will be on the graph of the function f(x) exactly when y = f(x). 6.1.1
Visualizing the domain and range A function is a package that consists of a rule y = f(x), a domain of allowed x-values and a range of output y-values. The domain can be visualized as a subset of the x-axis and the range as a subset of the y-axis. If you are handed the domain, it is graphically easy to describe the range values obtained; here is the procedure: Important Procedure 6.1.1. Look at all points on the graph corresponding to domain values on the x-axis, then project these points to the y-axis. The collection of all values you obtain on the y-axis will be the range of the function. This idea of “projection” is illustrated in the two graphs be- 73 74 CHAPTER 6. GRAPHICAL ANALYSIS low. We use arrows “ graph, then over to the y-axis: ” to indicate going from a domain x-value, up to the → graph of f (x) = −2x + 3 domain = [3,5] range = [1,9] range = [−7, − 3] domain = [−3,1] graph of f (x) = −2x + 3 (a) Domain: [3, 5]. (b) Domain: [−3, 1]. Figure 6.1: Example projections. 6.1.2 Interpreting Points on the Graph We can visually detect where a function has positive or negative values: Important Fact 6.1.2. The function values f(x) control the height of the point P(x) = (x,f(x)) on the graph above the x-axis; if the function value f(x) is negative, the point P(x) is below the x-axis. P (x) = (x,f (x)) f (x) units above x-axis f(x3) units above the x-axis x2 P (x3) = (x3,f (x3)) x x3 x-axis \|f(x)\| units below the x-axis P(x2) = (x2,f(x2)) y-axis Figure 6.2: Interpreting points on a graph. In Figure 6.2 we can now divide the domain (in this case the whole number line) into segments where the function is above, below or crossing the axis
. Keeping track of this information on a number line is called a sign plot for the function. We include a “shadow” of the graph in Figure 6.3 6.1. VISUAL ANALYSIS OF A GRAPH 75 positive positive negative negative x-axis Figure 6.3: Sign plot. to emphasize how we arrived at our “positive” and “negative” labeling of the sign plot; in practice we would only provide a labeled number line. By moving through a sequence of x values we can investigate how the corresponding points on the graph move “up and down”; this then gives us a dynamic visual sense of how the function values are changing. For example, in Figure 6.4, suppose we let x move from 1 to 5, left to right; we have indicated how the corresponding points on the curve will move and how the function values will change. #3: f(x) values move like this y-axis f(1) P f(2) f(3) f(4) 1 f(5) #2: points on graph move from P to Q graph of y = f(x) Q 5 #1: x values move from 1 to 5 x-axis Figure 6.4: Dynamic interpretation of a graph. 76 CHAPTER 6. GRAPHICAL ANALYSIS 6.1.3 Interpreting Intercepts of a Graph y-axis y-intercept = (0,f(0)) (x3,0) (x2,0) x-axis (x1,0) x-intercepts have the form “(x,0)” Figure 6.5: Intercepts of a graph. The places where a graph crosses the axes are often signiﬁcant. We isolate each as an important feature to look for when doing graphical analysis. The graph of the function y = f(x) crosses the y-axis at the point (0,f(0)); so, the y-intercept of the graph is just f(0). The graph of the function y = f(x) crosses the x-axis at points of the form (x0,f(x0)), where f(x0) = 0. The values x0 are called roots or zeros of the function f(x). There can be at most one y-intercept, but there can be several x-intercepts or no x
-intercept: See Figure 6.5 The graph of a function y = f(x) crosses the vertical line x = h at the point (h,f(h)). To ﬁnd where the graph of a function y = f(x) crosses the horizontal line y = k, ﬁrst solve the equation k = f(x) for x. If the equation k = f(x) has solutions x1, x2, x3, x4, then the points of intersection would have the coordinates given. (h,(f(h)) y-axis graph of f(x) (x1,k) (x2,k) (x3,k) (x4,k) y = k x1 x2 x3 x4 x-axis x = h y-axis upper semicircle radius r = 2 centered at (2,1-axis (a) General curve. (b) Semicircle. Figure 6.6: Crossing horizontal and vertical lines. As another example, the graph in Figure 6.6(b) above will cross the horizontal line y = k twice if and only if 1 k < 3; the graph will cross the ≤ horizontal line y = k once if and only if k = 3. The graph will not intersect the line y = 1 2 and the graph will cross the vertical line x = h if and only if 0 4. h ≤ ≤ 6.1.4 Interpreting Increasing and Decreasing y-axis uphill x-axis = local extrema downhill Figure 6.7: Graphically interpreting increasing and decreasing. We use certain terms to describe how the function values are changing over some domain of x values. Typically, we want to study what is happening to the values f(x) as x moves from “left to right” in some interval. This can be linked graphically with the study of “uphill” and “downhill” portions of the function graph: If you were “walking to the right” along the graph, the function values are increasing if you are walking uphill. Likewise, if you were 6.2. CIRCLES AND SEMICIRCLES 77 “walking to the right” along the graph, the function values are decreasing if you are walking downhill. Once we understand where the graph is moving uphill and downhill, we can isolate the places where we change from moving uphill to downhill,
or vice versa; these “peaks” and “valleys” are called local maxima and local minima. Some folks refer to either case as a local extrema. People have invested a lot of time (centuries!) and energy (lifetimes!) into the study of how to ﬁnd local extrema for particular function graphs. We will see some basic examples in this course and others will surface in future courses once you have the tools of calculus at your disposal. Examples range from business applications that involve optimizing proﬁt to understanding the three-dimensional shape a of biological molecule. Example 6.1.3. A hang glider launches from a gliderport in La Jolla. The launch point is located at the edge of a 500 ft. high cliff over the Paciﬁc Ocean. The elevation of the pilot above the gliderport after t minutes is given by the graph in Figure 6.8: 1. When is the pilot climbing and descending? 2. When is the pilot at the glider port elevation? 3. How much time does the pilot spend ﬂying level? ft above gliderport 5 minutes 10 600 400 200 −200 −400 −600 Figure 6.8: Hanglider elevations. Solution. 1. Graphically, we need to determine the portions of the graph that are increasing or decreasing. In this example, it is increasing when 5 and 0 9 9. And, it is decreasing when 3 2 and 7 10. Graphically, this question amounts to asking when the elevation is 0, which is the same as ﬁnding when the graph crosses the horizontal axis. We can read off there are four such times: t = 0, 4, 8, 10. 3. Graphically, we need to determine the portions of the graph that are 3 made up of horizontal line segments. This happens when 2 7. So, our pilot ﬂies level for a total of 3 minutes. and 5 ≤ ≤ t t ≤ ≤ 6.2 Circles and Semicircles Back in Chapter 3, we discussed equations whose graphs were circles: We found that the graph of the equation (x − h)2 + (y − k)2 = r2 (6.1) 78 CHAPTER 6. GRAPHICAL ANALYSIS is a circle of radius r centered at the point (h, k). It is possible to manipulate this equation and become confused. We
could rewrite this as (y − k)2 = r2 − (x − h)2, then take the square root of each side. However, the resulting equivalent equation would be y = k r2 − (x − h)2 ± and the presence of that ± r2 − (x − h)2 or r2 − (x − h)2. sign is tricky; it means we have two equations: Each of these two equations deﬁnes a function: p f(x) = k + g(x) = k − p r2 − (x − h)2 or r2 − (x − h)2. (6.2) (6.3) p So, even though the Equation 6.1 is not a function, we were able to obtain two different functions f(x) and g(x) from the original equation. The relationship between the graph of the original equation and the graphs of the two functions in (6.2) and (6.3) is as follows: The upper semicircle is the graph of the function f(x) and the lower semicircle is the graph of the function g(x). y-axis upper semicircle y-axis (h,k) x-axis (h,k) lower semicircle x-axis Graph of y = f(x) Graph of y = g(x) Figure 6.9: Upper and lower semicircles. Example 6.2.1. A tunnel connecting two portions of a space station has a circular cross-section of radius 15 feet. Two walkway decks are constructed in the tunnel. Deck A is along a horizontal diameter and another parallel Deck B is 2 feet below Deck A. Because the space station is in a weightless environment, you can walk vertically upright along Deck A, or vertically upside down along Deck B. You have been assigned to paint “safety stripes” on each deck level, so that a 6 foot person can safely walk upright along either deck. Determine the width of the “safe walk zone” on each deck. 6.3. MULTIPART FUNCTIONS 79 Deck A x-axis r = 15ft Deck B (a) Cross-section of tunnel. Deck A Deck B = Safe walk zone (b) Walk zones. Figure 6.10: Space station tunnels. Solution. Impose a coordinate system so that the origin is at the center of the circular
cross section of the tunnel; by symmetry the walkway is centered about the origin. With this coordinate system, the graph of the equation x2 + y2 = 152 = 225 will be the circular cross-section of the tunnel. In the case of Deck A, we basically need to determine how close to each edge of the tunnel a 6 foot high person can stand without hitting his or her head on the tunnel; a similar remark applies to Deck B. This means we are really trying to ﬁt two six-foot-high rectangular safe walk zones into the picture: Our job is to ﬁnd the coordinates of the four points P, Q, R, and S. Let’s denote by x1, x2, x3,and x4 the xcoordinates of these four points, then P = (x1, 6), Q = (x2, 6), R = (x3, −8), and S = (x4, −8). To ﬁnd x1, x2, x3, and x4, we need to ﬁnd the intersection of the circle in Figure 6.10(b) with two horizontal lines: • • Intersecting the the upper semicircle with the horizontal line having equation y = 6 will determine x1 and x2; the upper semicircle is the graph of f(x) = √225 − x2. Intersecting the lower semicircle with the horizontal line having equation y = −8 will determine x3 and x4; the lower semicircle is the graph of g(x) = −√225 − x2. For Deck A, we simultaneously solve the system of equations y = √225 − x2 y = 6. Plugging in y = 6 into the ﬁrst equation of the system gives x2 = 225 − 62 = 189; i.e., x = 13.75. This tells us that P = (−13.75, 6) and Q = (13.75, 6). In a similar way, for Deck B, we ﬁnd R = (−12.69, −8) and S = (12.69, −8). √189 = ± ± In the case of Deck A, we would paint a safety stripe 13.75 feet to the right and left of the centerline. In the case of Deck B, we would paint a safety stripe 12.69 feet
to the right and left of the centerline. 6.3 Multipart Functions So far, in all of our examples we have been able to write f(x) as a nice compact expression in the variable x. Sometimes we have to work harder. 80 CHAPTER 6. GRAPHICAL ANALYSIS As an example of what we have in mind, consider the graph in Figure 6.11(a): y-axis 1 2 3 4 x-axis 1 −1 (a) Graphing a multipart function. −1 if 0 1 if 1 −1 if 2 1 if 1 if x = 4 f(x) =    (b) Writing a multipart function. Figure 6.11: A multipart function. x ≤ The curve we are trying to describe in this picture is made up of ﬁve pieces; four little line segments and a single point. The ﬁrst thing to 4, this curve will deﬁne the graph of notice is that on the domain 0 ≤ some function f(x). To see why this is true, imagine a vertical line moving from left to right within the domain 0 4 on the x-axis; any one of these vertical lines will intersect the curve exactly once, so by the vertical line test, the curve must be the graph of a function. Mathematicians use the shorthand notation above to describe this function. Notice how the rule for f(x) involves ﬁve cases; each of these cases corresponds to one of the ﬁve pieces that make up the curve. Finally, notice the care with the “open” and “closed” circles is really needed if we want to make sure the curve deﬁnes a function; in terms of the rule, these open and closed circles translate into strict inequalities like < or weak inequalities like. ≤ This is an example of what we call a multipart function. ≤ ≤ x The symbolic appearance of multipart functions can be somewhat frightening. The key point is that the graph (and rule) of the function will be broken up into a number of separate cases. To study the graph or rule, we simply “home in” on the appropriate case. For example, in the above illustration, suppose we wanted to compute f(3.56). First,
we would ﬁnd which of the ﬁve cases covers x = 3.56, then apply that part of the rule to compute f(3.56) = 1. Our ﬁrst multipart function example illustrated how to go from a graph in the plane to a rule for f(x); we can reverse this process and go from the rule to the graph. Example 6.3.1. Sketch the graph of the multipart function 1 1 + √1 − x2 1 g(x) =    if x ≤ if −1 if x ≥ −1 x ≤ 1 1 ≤ 6.3. MULTIPART FUNCTIONS 81 y-axis ≤ Solution. The graph of g(x) will consist of three pieces. The ﬁrst case consists of the graph of the function y = −1, this consists of all points g(x) = 1 on the domain x on the horizontal line y = 1 to the left of and including the point (−1,1). We have “lassoed” this portion of the graph in Figure 6.12. Likewise, the third case in the deﬁnition yields the graph of the function y = g(x) = 1 on the domain 1; this is just all points on the horizontal line y = 1 x to the right of and including the point (1, 1). Finally, we need to analyze the middle case, which means we need to look at the graph of 1 + √1 − x2 on the domain −1 1. This is just the upper semicircle of the circle of radius 1 centered at (0,1). If we paste these three pieces together, we arrive at the graph of g(x). ﬁrst part of graph ≤ ≤ ≥ x = graph of g(x) second part of graph x-axis third part of graph Figure 6.12: Multipart function g(x). Example 6.3.2. You are dribbling a basketball and the function s = h(t) keeps track of the height of the ball’s center above the ﬂoor after t seconds. Sketch a reasonable graph of s = h(t). seconds 2 Solution. If we take the domain to be 0 2 (the ﬁrst 2 seconds), a reasonable graph might look like Figure 6.13. This is a multipart function
. Three portions of the graph are decreasing and two portions are increasing. Why doesn’t the graph touch the t axis? ≤ ≤ t Figure 6.13: Dribbling. 82 CHAPTER 6. GRAPHICAL ANALYSIS 6.4 Exercises Problem 6.1. The absolute value function is deﬁned by the multipart rule: Problem 6.4. (a) Let f(x) = x + \|2x − 1\|. Find all solutions to the equation x if 0 x −x if x < 0 ≤ \|x\| = f(x) = 8. The graph of the absolute value function is pictured below: (b) Let g(x) = 3x − 3 + \|x + 5\|. Find all values of a which satisfy the equation y-axis g(a) = 2a + 8. y = \|x\| x-axis (c) Let h(x) = \|x\| − 3x + 4. Find all solutions to the equation h(x − 1) = x − 2. Problem 6.5. Express the area of the shaded region below as a function of x. The dimensions in the ﬁgure are centimeters. (a) Calculate: \|0\|, \|2\|, \| − 3\|. (b) Solve for x: \|x\| = 4; \|x\| = 0, \|x\| = −1. (c) Sketch the graph of y = 1 2 x + 2 and y = \|x\| in the same coordinate system. Find where the two graphs intersect, label the coordinates of these point(s), then ﬁnd the area of the region bounded by the two graphs. Problem 6.2. For each of the following functions, graph f(x) and g(x) = \|f(x)\|, and give the multipart rule for g(x). 3 (a) f(x) = −0.5x − 1 (b) f(x) = 2x − 5 (c) f(x) = x + 3 x 6 5 Problem 6.3. Solve each of the following equations for x. (a) g(x) = 17, where g(x) = \|3x + 5\| (b) f(x) = 1.5 where f(x) = (c) h(x) = −1 where 2x if x
< 3, 3. 4 − x if x ≥ h(x) = −8 − 4x if x 1 + 1 −2, 3 x if x > −2. ≤ Problem 6.6. Pizzeria Buonapetito makes a triangular-shaped pizza with base width of 30 inches and height 20 inches as shown. Alice wants only a portion of the pizza and does so by making a vertical cut through the pizza and taking the shaded portion. Letting x be the bottom length of Alice’s portion and y be the length of the cut as shown, answer the following questions: 6.4. EXERCISES 83 y x 10 x 20 y 20 20 Problem 6.8. Arthur is going for a run. From his starting point, he runs due east at 10 feet per second for 250 feet. He then turns, and runs north at 12 feet per second for 400 feet. He then turns, and runs west at 9 feet per second for 90 feet. Express the (straight-line) distance from Arthur to his starting point as a function of t, the number of seconds since he started. Problem 6.9. A baseball diamond is a square with sides of length 90 ft. Assume Edgar hits a home run and races around the bases (counterclockwise) at a speed of 18 ft/sec. Express the distance between Edgar and home plate as a function of time t. (Hint: This will be a multipart function.) Try to sketch a graph of this function. (a) Find a formula for y as a multipart function of x, for 0 30. Sketch the graph of this function and calculate the range. ≤ ≤ x (b) Find a formula for the area of Alice’s portion as a multipart function of x, for 0 30. x ≤ ≤ (c) If Alice wants her portion to have half the area of the pizza, where should she make the cut? Problem 6.7. This problem deals with cars traveling between Bellevue and Spokane, which are 280 miles apart. Let t be the time in hours, measured from 12:00 noon; for example, t = −1 is 11:00 am. (a) Joan drives from Bellevue to Spokane at a constant speed, departing from Bellevue at 11:00 am and arriving in Spokane at 3:30 pm. Find a function j(t) that computes her distance from Bellevue at time t. Sketch the graph, specify the domain
and determine the range. (b) Steve drives from Spokane to Bellevue at 70 mph, departing from Spokane at 12:00 noon. Find a function s(t) for his distance from Bellevue at time t. Sketch the graph, specify the domain and determine the range. (c) Find a function d(t) that computes the distance between Joan and Steve at time t. 90 ft Edgar d(t) home plate Problem 6.10. Pagliacci Pizza has designed a cardboard delivery box from a single piece of cardboard, as pictured. (a) Find a polynomial function v(x) that computes the volume of the box in terms of x. What is the degree of v? (b) Find a polynomial function a(x) that computes the exposed surface area of the closed box in terms of x. What is the degree of a? What are the explicit dimensions if the exposed surface area of the closed box is 600 sq. inches? 84 x x x x 50 in remove shaded squares and fold to get: CHAPTER 6. GRAPHICAL ANALYSIS (a) When will the ditch be completely full? (b) Find a multipart function that models the vertical cross-section of the ditch. 20 in (c) What is the width of the ﬁlled portion of the ditch after 1 hour and 18 minutes? (d) When will the ﬁlled portion of the ditch be 42 feet wide? 50 feet wide? 73 feet wide? Problem 6.12. The graph of a function y = g(x) on the domain −6 6 consistes of line ≤ segments and semicircles of radius 2 connecting the points (−6,0),(−4,4), (0,4), (4,4), (6,0). ≤ x y-axis x-axis Problem 6.11. The vertical cross-section of a drainage ditch is pictured below: (a) What is the range of g? (b) Where is the function increasing? Where is the function decreasing? (c) Find the multipart formula for y = g(x). (d) If we restrict the function to the smaller domain −5 x ≤ ≤ 0, what is the range? (e) If we restrict the function to the smaller domain 0 x ≤ ≤ 4, what is the range? 3D−view of ditch 20 ft 20 ft R R Problem 6.13. (a)
Simply as far as possible R R vertical cross-section Here, R indicates a circle of radius 10 feet and all of the indicated circle centers lie along the common horizontal line 10 feet above and parallel to the ditch bottom. Assume that water is ﬂowing into the ditch so that the level above the bottom is rising 2 inches per minuteb) Find a, b, c that simultaneously satisfy these three equations: a + b − c = 5 2a − 3b + 1 Chapter 7 Quadratic Modeling If you kick a ball through the air enough times, you will ﬁnd its path tends to be parabolic. Before we can answer any detailed questions about this situation, we need to get our hands on a precise mathematical model for a parabolic shaped curve. This means we seek a function y = f(x) whose graph reproduces the path of the ball. ground level Figure 7.1: Possible paths for a kicked ball are parabolic. 7.1 Parabolas and Vertex Form OK, suppose we sit down with an xy-coordinate system and draw four random parabolas; let’s label them I, II, III, and IV: See Figure 7.2. The relationship between these parabolas and the ﬁxed coordinate system can vary quite a bit: The key distinction between these four curves is that only I and IV are the graphs of functions; this follows from the vertical line test. A parabola that is the graph of a function is called a standard parabola. We can see that any standard parabola has three basic features: 85 86 CHAPTER 7. QUADRATIC MODELING y-axis II I III not graphs of functions IV x-axis Figure 7.2: Relationship between a ﬁxed coordinate system and various parabolas. • • • the parabola will either open “upward” or “downward”; the graph will have either a “highest point” or “lowest point,” called the vertex; the parabola will be symmetric about some vertical line called the axis of symmetry. Our ﬁrst task is to describe the mathematical model for any standard parabola. In other words, what kind of function equations y = f(x) give us standard parabolas as their graphs? Our approach is geometric and visual: • Begin with one speciﬁc
example, then show every other standard parabola can be obtained from it via some speciﬁc geometric maneuvers. • As we perform these geometric maneuvers, we keep track of how the function equation for the curve is changing. This discussion will amount to a concrete application of a more general set of tools developed in the following section of this chapter. y-axis 35 30 25 20 15 10 5 x-axis Using a graphing device, it is an easy matter to plot the graph of y = x2 and see we are getting the parabola pictured in Figure 7.3. The basic idea is to describe how we can manipulate this graph and obtain any standard parabola. In the end, we will see that standard parabolas are obtained as the graphs of functions having the form −6 −4 −2 2 4 6 Figure 7.3: Graph of y = x2. y = ax2 + bx + c, 7.1. PARABOLAS AND VERTEX FORM 87 = 0. A function of this type is for various constants a, b, and c, with a called a quadratic function and these play a central role throughout the course. We will divide our task into two steps. First we show every standard parabola arises as the graph of a func- tion having the form y = a(x − h)2 + k, = 0. This is called the vertex for some constants a, h, and k, with a form of a quadratic function. Notice, if we were to algebraically expand out this equation, we could rewrite it in the y = ax2 + bx + c form. For example, suppose we start with the vertex form y = 2(x − 1)2 + 3, so that a = 2, h = 1, k = 3. Then we can rewrite the equation in the form y = ax2 + bx + c as follows: 2(x − 1)2 + 3 = 2(x2 − 2x + 1) + 3 = 2x2 − 4x + 5, so a = 2, b = −4, c = 5. The second step is to show any quadratic function can be written in vertex form; the underlying algebraic technique used here is called completing the square. This is a bit more involved. For example, if you are simply handed the quadratic function y = −3x2+6x−1,
it not at all obvious why the vertex form is obtained by this equality: −3x2 + 6x − 1 = −3(x − 1)2 + 2. The reason behind this equality is the technique of completing the square. In the end, we will almost always be interested in the vertex form of a quadratic. This is because a great deal of qualitative information about the parabolic graph can simply be “read off” from this form. 7.1.1 First Maneuver: Shifting Suppose we start with the graph in Figure 7.3 and horizontally shift it h units to the right. To be speciﬁc, consider the two cases h = 2 and h = 4. To visualize this, imagine making a wire model of the graph, set in on top of the curve, then slide the wire model h units to the right. What you will obtain are the two “dashed curves” in Figure 7.4. We will call the process just described a horizontal shift. Since the “dashed curves” are no longer the original parabola in Figure 7.3, the corresponding function equations must have changed. y-axis 40 30 20 10 x-axis −6 −4 −2 2 4 6 Figure 7.4: Shift to the right. Using a graphing device, you can check that the corresponding equa- tions for the dashed graphs would be y = (x − 2)2 = x2 − 4x + 4, 6 6 88 CHAPTER 7. QUADRATIC MODELING which is the plot with lowest point (2, 0) and y = (x − 4)2 = x2 − 8x + 16, which is the plot with lowest point (4, 0). In general, if h is positive, the graph of the function y = (x − h)2 is the parabola obtained by shifting the graph of y = x2 by h units to the right. y-axis 40 30 20 10 x-axis −6 −4 −2 2 4 6 Figure 7.5: Shift to the left. Next, if h is negative, shifting h units to the right is the same as shifting \|h\| units left! On the domain −6 6, Figure 7.5 indicates this for the cases h = −2, − 4, using “dashed curves” for the shifted graphs and a solid line for the graph of y
= x2. Using a graphing device, we can check that the corresponding equations for the dashed graphs would be ≤ ≤ x y = (x − (−2))2 = (x + 2)2 = x2 + 4x + 4, which is the plot with lowest point (−2, 0) and y = (x − (−4))2 = (x + 4)2 = x2 + 8x + 16, which is the plot with lowest point (−4, 0). In general, if h is negative, the graph of the function y = (x − h)2 gives the parabola obtained by shifting the graph of y = x2 by \|h\| units to the LEFT. The conclusion thus far is this: Begin with the graph of y = x2 in Figure 7.3. Horizontally shifting this graph h units to the right gives a new (standard) parabola whose equation is y = (x − h)2. y-axis 40 30 20 10 x-axis −6 −4 −2 2 4 6 −10 Figure 7.6: Vertical shifts. We can also imagine vertically shifting the graph in Figure 7.3. This amounts to moving the graph k units vertically upward. It turns out that this vertically shifted graph corresponds to the graph of the function y = x2 + k. We can work out a few special cases and use a graphing device to illustrate what all this really means. Figure 7.6 illustrates the graphs of y = x2 + k in the cases when k = 4, 10 and k = −4, −10, leading to vertically shifted graphs. Positive values of k lead to the upper two “dashed curves” and negative values of k lead to the lower two “dashed curves”; the plot of y = x2 is again the solid line. The equations giving these graphs would be y = x2 − 10, y = x2 − 4, y = x2 + 4 and y = x2 + 10, from bottom to top dashed plot. If we combine horizontal and vertical shifting, we end up with the graphs of functions of the form y = (x − h)2 + k. Figure 7.7(a) illustrates 7.1. PARABOLAS AND VERTEX FORM 89 the four cases with corresponding equations y = (x identify which equation goes with each curve. ± 2)2 ± 4; as an exercise, 7.1.
2 Second Maneuver: Reﬂection Next, we can reﬂect any of the curves y = p(x) obtained by horizontal or vertical shifting across the x-axis. This procedure will produce a new curve which is the graph of the new function y = −p(x). For example, begin with the four dashed curves in the previous ﬁgure. Here are the reﬂected parabolas and their equations are y = −(x 4: See Figure 7.7(b). 2)2 ± ± 7.1.3 Third Maneuver: Vertical Dilation y-axis 40 30 20 10 x-axis −4 −6 −2 2 4 6 (a) Combined shifts. y-axis If a is a positive number, the graph of y = ax2 is usually called a vertical dilation of the graph of y = x2. There are two cases to distinguish here: x-axis −6 −4 −2 30 20 10 −10 −20 −30 2 4 6 If a > 1, we have a vertically expanded graph. If 0 < a < 1, we have a vertically compressed graph. • • This is illustrated for a = 2 (upper dashed plot) and a = 1/2 (lower dashed plot): See Figure 7.7(c). 7.1.4 Conclusion Starting with y = x2 in Figure 7.3, we can combine together all three of the operations: shifting, reﬂection and dilation. This will lead to the graphs of functions that have the form: y = a(x − h)2 + k, (b) Reﬂections. y-axis 30 25 20 15 10 5 x-axis −6 −4 −2 2 4 6 (c) Vertical dilations. Figure 7.7: Shifts, reﬂections, and dilations. = 0. If you think about it for awhile, for some a, h and k, a it seems pretty easy to believe that any standard parabola arises from the one in Figure 7.3 using our three geometric maneuvers. In other words, what we have shown is that any standard parabola is the graph of a quadratic equation in vertex form. Let’s summarize. Important Fact 7.1.1. A standard parabola is the graph of a function y = f(x) = a(x − h)2
+ k, for some constants a, h, and k and a = 0. The vertex of the parabola is (h, k) and the axis of symmetry is the line x = h. If a > 0, then the parabola opens upward; if a < 0, then the parabola opens downward. 6 6 90 CHAPTER 7. QUADRATIC MODELING Example 7.1.2. Describe a sequence of geometric operations leading from the graph of y = x2 to the graph of y = f(x) = −3(x − 1)2 + 2. reﬂect across x-axis horizontal shift by h = 1 −3(x − 1)2 + 2 vertical shift by k = 2 vertical dilate by 3 (a) What do the symbols of an equation mean? y-axis 20 15 10 5 x-axis −6 −4 −2 2 4 6 −5 −10 −15 (b) What does the equation look like? Figure 7.8: Interpreting an equation. Solution. To begin with, we can make some initial conclusions about the speciﬁc shifts, reﬂections and dilations involved, based on looking at the vertex form of the equation. In addition, by Fact 7.1.1, we know that the vertex of the graph of y = f(x) is (1, 2), the line x = 1 is a vertical axis of symmetry and the parabola opens downward. We need to be a little careful about the order in which we apply the four operations highlighted. We will illustrate a procedure that works. The full explanation for the success of our procedure involves function compositions and we will return to that at the end of Chapter 8. The order in which we will apply our geometric maneuvers is as follows: horizontal shift vertical dilate reﬂect vertical shift ⇒ ⇒ ⇒ Figure 7.8(b) illustrates the four curves obtained by applying these successive steps, in this order. As a reference, we include the graph of y = x2 as a “dashed curve”: • A horizontal shift by h = 1 yields the graph of y = (x − 1)2; this is the fat parabola opening upward with vertex (1, 0). • • • A dilation by a = 3 yields the graph of y = 3(x − 1)2; this is
the skinny parabola opening upward with vertex (1, 0). A reﬂection yields the graph of y = −3(x − 1)2; this is the downward opening parabola with vertex (1, 0). A vertical shift by k = 2 yields the graph of y = −3(x − 1)2 + 2; this is the downward opening parabola with vertex (1, 2). 7.2 Completing the Square By now it is pretty clear we can say a lot about the graph of a quadratic function which is in vertex form. We need a procedure for rewriting a given quadratic function in vertex form. Let’s ﬁrst look at an example. 7.2. COMPLETING THE SQUARE 91 Example 7.2.1. Find the vertex form of the quadratic function y = −3x2 + 6x − 1. Solution. Since our goal is to put the function in vertex form, we can write down what this means, then try to solve for the unknown constants. Our ﬁrst step would be to write −3x2 + 6x − 1 = a(x − h)2 + k, for some constants a, h, k. Now, expand the right hand side of this equation and factor out coefﬁcients of x and x2: −3x2 + 6x − 1 = a(x − h)2 + k −3x2 + 6x − 1 = a(x2 − 2xh + h2) + k −3x2 + 6x − 1 = ax2 − 2xah + ah2 + k (−3)x2 + (6)x + (−1) = (a)x2 + (−2ah)x + (ah2 + k). If this is an equation, then it must be the case that the coefﬁcients of like powers of x match up on the two sides of the equation in Figure 7.9. Now we have three equations and three unknowns (the a, h, k) and we Equal (−3) x2 + (6) x + (−1) = (a) x2 + (−2ah) x + (ah2 + k) z}\|{ \| {z } \| {z } \|{z} z }\| { Equal \| {z } Equal Figure 7.9: Balancing the
coefﬁcients. can proceed to solve for these: −3 = a 6 = −2ah −1 = ah2 + k The ﬁrst equation just hands us the value of a = −3. Next, we can plug this value of a into the second equation, giving us 6 = −2ah = −2(−3)h = 6h, 92 CHAPTER 7. QUADRATIC MODELING so h = 1. Finally, plug the now known values of a and h into the third equation: −1 = ah2 + k = −3(12) + k = −3 + k, so k = 2. Our conclusion is then −3x2 + 6x − 1 = −3(x − 1)2 + 2. Notice, this is the quadratic we studied in Example 7.1.2 on page 90. The procedure used in the preceding example will always work to rewrite a quadratic function in vertex form. We refer to this as completing the square. Example 7.2.2. Describe the relationship between the graphs of y = x2 and y = f(x) = −4x2 + 5x + 2. y-axis 20 10 −6 −4 −2 2 4 6 x-axis −10 −20 Figure 7.10: Maneuvering y = x2. Solution. We will go through the algebra to complete the square, then interpret what this all means in terms of graphical maneuvers. We have −4x2 + 5x + 2 = a(x − h)2 + k (−4)x2 + 5x + 2 = ax2 + (−2ah)x + (ah2 + k). This gives us three equations: −4 = a 5 = −2ah 2 = ah2 + k. We conclude that a = −4, h = 5 16 = 3.562. So, this tells us that we can obtain the graph of y = f(x) from that of y = x2 by these steps: 8 = 0.625 and k = 57 Horizontally shifting by h = 0.625 units gives y = (x − 0.625)2. Vertically dilate by the factor a = 4 gives y = 4(x − 0.625)2. Reﬂecting across the x-axis gives y = −4(x − 0.625)2. Vertically
shifting by k = 3.562 gives y = f(x) = −4(x − 0.625)2 + 3.562. • • • • Example 7.2.3. A drainage canal has a cross-section in the shape of a parabola. Suppose that the canal is 10 feet deep and 20 feet wide at the top. If the water depth in the ditch is 5 feet, how wide is the surface of the water in the ditch? 7.3. INTERPRETING THE VERTEX 93 Solution. Impose an xy-coordinate system so that the parabolic cross-section of the canal is symmetric about the y-axis and its vertex is the origin. The vertex form of any such parabola is y = f(x) = ax2, for some a > 0; this is because (h, k) = (0, 0) is the vertex and the parabola opens upward! The dimension information given tells us that the points (10, 10) and (−10, 10) are on the graph of f(x). Plugging into the expression for f, we conclude that 10 = 100a, so a = 0.1 and f(x) = (0.1)x2. Finally, if the water is 5 feet deep, we must solve the equation: 5 = (0.1)x2, leading to x = 7.07. Conclude the surface of the water is 14.14 feet wide when the water is 5 feet deep. √50 = ± ± Figure 7.11: 20 feet centerline 10 feet A drainage canal. 7.3 Interpreting the Vertex minimum value f −b 2a maximum value f −b 2a vertex −b 2a vertex −b 2a Figure 7.12: The vertex as the extremum of the quadratic function. If we begin with a quadratic function y = f(x) = ax2+bx+c, we know the graph will be a parabola. Graphically, the vertex will correspond to either If a > 0, the vertex the “highest point” or “lowest point” on the graph. is the lowest point on the graph; if a < 0, the vertex is the highest point on the graph. The maximum or minimum value of the function is the second coordinate of the vertex and the value of the variable x for which this extreme value is
achieved is the ﬁrst coordinate of the vertex. As we know, it is easy to read off the vertex coordinates when a quadratic function is written in vertex form. If instead we are given a quadratic function y = ax2 + bx + c, we can use the technique of completing the square and arrive at a formula for the coordinates of the vertex in terms of a, b, and c. We summarize this below and label the two situations (upward or downward opening parabola) in the Figure 7.12. Keep in 94 CHAPTER 7. QUADRATIC MODELING mind, it is always possible to obtain this formula by simply completing the square. Important Fact 7.3.1. In applications involving a quadratic function f(x) = ax2 + bx + c, the vertex has coordinates P =. The second coordinate of the vertex will detect the maximum or minimum value of f(x); this is often a key step in problem solving. −b 2a, f −b 2a Example 7.3.2. Discuss the graph of the quadratic function y = f(x) = −2x2 + 11x − 4. −4 −2 2 4 6 8 −20 −40 −60 −80 Figure 7.13: Sketching y = f(x). Solution. We need to place the equation y = f(x) in vertex form. We can simply compute a = −2, h = −b 4 and k = f( 11 8, using Fact 7.3.1: 2a = 11 4 ) = 89 f(x) = −2x2 + 11x − 4 = −2 x − 11 4 2 + 89 8. This means that the graph of f(x) is a parabola opening 4 ; see Figdownward with vertex ure 7.13. and axis x = 11 4, 89 11 8 7.4 Quadratic Modeling Problems The real importance of quadratic functions stems from the connection with motion problems. Imagine one of the three kicked ball scenarios in Figure 7.1 and impose a coordinate system with the kicker located at the origin. We can study the motion of the ball in two ways: • • Regard time t as the important variable and try to ﬁnd a function y(t) which describes the height of the ball t seconds after the ball is kicked; this would just be the y-
coordinate of the ball at time t. The function y(t) is a quadratic function. If we had this function in hand, we could determine when the ball hits the ground by solving the equation 0 = y(t), but we would not be able to determine where the ball hits the ground. A second approach is to forget about the time variable and simply try to ﬁnd a function y = f(x) whose graph models the exact path of the ball. In particular, we could ﬁnd where the ball hits the ground by solving 0 = f(x), but we would not be able to determine when the ball hits the ground. 7.4. QUADRATIC MODELING PROBLEMS 95 Example 7.4.1. Figure 7.14(a) shows a ball is located on the edge of a cliff. The ball is kicked and its height (in feet) above the level ground is given by the function s = y(t) = −16t2 + 48t + 50, where t represents seconds elapsed after kicking the ball. What is the maximum height of the ball and when is this height achieved? When does the ball hit the ground? How high is the cliff? Solution. The function y(t) is a quadratic function with a negative leading coefﬁcient, so its graph in the tscoordinate system will be a downward opening parabola. We use a graphing device to get the picture in Figure 7.14(b). The vertex is the highest point on the graph, which can be found by writing y(t) in standard form using Fact 7.3.1: y(t) = −16t2 + 48t + 50 = −16 t − 3 2 2 + 86. path of kicked ball cliff ground level (a) What it looks like physically. 50 −2 −1 1 2 3 4 5 −50 −100 (b) What it looks like graphically. Figure 7.14: Different views of the ball’s trajectory. The vertex of the graph of y(t) is, so the maximum height of the ball above the level ground is 86 feet, occuring at time t = 3 2. 3 2, 86 The ball hits the ground when its height above the ground is zero; using the quadratic formula: y(t) = −16t2 + 48t + 50 = 0 t = −48 ±
(48)2 − 4 (−16) (50) = 3.818 sec or − 0.818 sec p 2 16 · Conclude the ball hits the ground after 3.818 seconds. Finally, the height of the cliff is the height of the ball zero seconds after release; i.e., y(0) = 50 feet is the height of the cliff. Here are two items to consider carefully: 1. The graph of y(t) is NOT the path followed by the ball! Finding the actual path of the ball is not possible unless additional information is given. Can you see why?!!! CAUTION!!! 2. The function y(t) is deﬁned for all t; however, in the context of the problem, there is no physical meaning when t < 0. 96 CHAPTER 7. QUADRATIC MODELING The next example illustrates how we must be very careful to link the question being asked with an appropriate function. Example 7.4.2. A hot air balloon takes off from the edge of a mountain lake. Impose a coordinate system as pictured in Figure 7.15 and assume that the path of the balloon follows the graph of y = f(x) = − 2 5x. The land rises at a constant incline from the lake at the rate of 2 vertical feet for each 20 horizontal feet. What is the maximum height of the balloon above lake level? What is the maximum height of the balloon above ground level? Where does the balloon land on the ground? Where is the balloon 50 feet above the ground? 2500 x2 + 4 height above lake (ft) 200 100 lake 500 balloon balloon path Solution. In the coordinate system indicated, the origin is the takeoff point and the graph of y = f(x) is the path of the balloon. Since f(x) is a quadratic function with a negative leading coefﬁcient, its graph will be a parabola which opens downward. The difﬁculty with this problem is that at any instant during the balloon’s ﬂight, the “height of the balloon above the ground” and the “height of the balloon above the lake level” are different! The picture in Figure 7.16 highlights this difference; consequently, two different functions will be needed to study these two different quantities. 1000 (ft) Figure 7.15: Visualizing. 200 100 lake A B
500 1000 height of balloon above lake level A B height of balloon above ground level height of ground above lake level Figure 7.16: The height of the balloon y as a function of x. The function y = f(x) keeps track of the height of the balloon above lake level at a given x location on the horizontal axis. The line ℓ with slope 20 = 1 m = 2 10 passing through the origin models the ground level. This says that the function y = 1 10 x keeps track of the height of the ground above lake level at a given x location on the horizontal axis. We can determine the maximum height of the balloon above lake level by analyzing the parabolic graph of y = f(x). Putting f(x) in vertex form, via Fact 7.3.1, f(x) = − 2 2500 (x − 500)2 + 200. 7.4. QUADRATIC MODELING PROBLEMS 97 The vertex of the graph of y = f(x) is (500, 200). This just tells us that the maximum height of the balloon above lake level is 200 feet. To ﬁnd the landing point, we need to solve the system of equations 2500 x2 + 4 5x y = − 2 y = 1 10x. As usual, plugging the second equation into the ﬁrst and solving for x, we get 1 10 x = − 2 2500 x2 + 4 5 x x2 = 875x x2 − 875x = 0 x(x − 875) = 0 height above lake level (feet) vertex (high point above lake) 200 100 landing point horizontal distance from launch (feet) takeoff point 500 1000 Figure 7.17: Locating the takeoff and landing points. From the algebra, we see there are two solutions: x = 0 or x = 875; these correspond to the takeoff and landing points of the balloon, which are the two places the ﬂight path and ground coincide. (Notice, if we had divided out x from the last equation, we would only get one solution; the tricky point is that we can’t divide by zero!) The balloon lands at the position where x = 875 and to ﬁnd the y coordinate of this landing point we plug x = 875 into our function for the balloon height above lake level: y = f(875) = 87.5 feet. So, the landing point has coordinates (
875,87.5). Next, we want to study the height of the balloon above the ground. Let y = g(x) be the function which represents the height of the balloon above 98 CHAPTER 7. QUADRATIC MODELING the ground when the horizontal coordinate is x. We ﬁnd g(x) = height of the balloon above lake level with horizontal coordinate x − elevation of ground above lake level with horizontal coordinate x = f(x) − g(x) = − 2 2500 x2 + 4 5 x (balloon above lake level) − minus 1 10 x (ground above lake level) \|{z} {z } (x − 437.5)2 + 153.12, \| = − 2 2500 \| {z } Notice that g(x) itself is a NEW quadratic function with a negative leading coefﬁcient, so the graph of y = g(x) will be a downward opening parabola. The vertex of this parabola will be (437.5, 153.12), so the highest elevation of the balloon above the ground is 153.12 feet. We can now sketch the graph of g(x) and the horizontal line determined by y = 50 in a common coordinate system, as below. Finding where the balloon is 50 feet above the ground amounts to ﬁnding where these two graphs intersect. We need to now solve the system of equations y = − 2 y = 50 2500 x2 + 7 10x. Plug the second equation into the ﬁrst and apply the quadratic formula to get x = 796.54 or 78.46. This tells us the two possible x coordinates when the balloon is 50 feet above the ground. In terms of the original coordinate system imposed, the two places where the balloon is 50 feet above the ground are (78.46, 57.85) and (796.54, 129.6). 7.4.1 How many points determine a parabola? We all recall from elementary geometry that two distinct points in the plane will uniquely determine a line; in fact, we used this to derive equations for lines in the plane. We could then ask if there is a similar characterization of parabolas. Important Fact 7.4.3. Let P = (x1, y1), Q = (x2, y2) and R = (x3, y3) be three distinct non-coll
inear points in the plane such that the x-coordinates are all different. Then there exists a unique standard parabola passing through these three points. This parabola is the graph of a quadratic function y = f(x) = ax2 + bx + c and we can ﬁnd these coefﬁcients by simultaneously 7.4. QUADRATIC MODELING PROBLEMS 99 vertex (high point above ground) 100 graph of height above ground function line y = 50 500 1000 places where 50 feet above ground Figure 7.18: Finding heights above the ground. solving the system of three equations and three unknowns obtained by assuming P, Q and R are points on the graph of y = f(x): ax2 1 + bx1 + c = y1 ax2 2 + bx2 + c = y2 ax2 3 + bx3 + c = y3.   .    Example 7.4.4. Assume the value of a particular house in Seattle has increased in value according to a quadratic function y = v(x), where the units of y are in dollars and x represents the number of years the property has been owned. Suppose the house was purchased on January 1, 1970 and valued at $50,000. In 1980, the value of the house on January 1 was $80,000. Finally, on January 1, 1990 the value was $200,000. Find the value function v(x), determine the value on January 1, 1996 and ﬁnd when the house will be valued at $1,000,000. Solution. The goal is to explicitly ﬁnd the value of the function y = v(x). We are going to work in a xy-coordinate system in which the ﬁrst coordinate of any point represents time and the second coordinate represents value. We need to decide what kind of units will be used. The x-variable, which represents time, will denote the number of years the house is owned. For the y-variable, which represents value, we could use dollars. But, instead, we
will follow a typical practice in real estate and use the units of K, where K = $1,000. For example, a house valued 100 CHAPTER 7. QUADRATIC MODELING at $235,600 would be worth 235.6 K. These will be the units we use, which essentially saves us from drowning in a sea of zeros! We are given three pieces of information about the value of a particular house. This leads to three points in our coordinate system: P = (0, 50), Q = (10, 80) and R = (20, 200). If we plot these points, they do not lie on a common line, so we know there is a unique quadratic function v(x) = ax2 + bx + c whose graph (which will be a parabola) passes through these three points. In order to ﬁnd the coefﬁcients a, b, and c, we need to solve the system of equations: a02 + b0 + c = 50 a(10)2 + b(10) + c = 80 a(20)2 + b(20) + c = 200   which is equivalent to the system    , = 50 c 100a + 10b + c = 80 400a + 20b + c = 200.     Plugging c = 50 into the second two equations gives the system   100a + 10b = 30 400a + 20b = 150 Solve the ﬁrst equation for a, obtaining a = 30−10b 100 the second equation to get: (7.1), then plug this into 400 30 − 10b 100 + 20b = 150 120 − 40b + 20b = 150 3 2 b = −. 2 into the ﬁrst equation of Equation 7.1 to get 100a + Now, plug b = − 3 10 = 30; i.e., a = 9 − 3 2 y = v(x) = 9 20 x2 − 20. We conclude that 3 2 x + 50, keeping in mind the units here are K. To ﬁnd the value of the house on January 1, 1996, we simply note this is after x = 26 years of ownership. Plugging in, we get y = v(
26) = 9 20(26)2 − 3 226 + 50 = 315.2; i.e., the value of the house is $315,200. To ﬁnd when the house will be worth $1,000,000, we note that $1,000,000 = 1,000 K and need to solve the equation 1000 = v(x) = 0 = 9 20 9 20 x2 − x2 − 3 2 3 2 x + 50 x − 950. 7.5. WHAT’S NEEDED TO BUILD A QUADRATIC MODEL? 101 By the quadratic formula 20 √1712.25 0.9 = 47.64 or − 44.31. 1.5 ± = 9 20 (−950) Because x represents time, we can ignore the negative solution and so the value of the house will be $1,000,000 after approximately 47.64 years of ownership. 7.5 What’s Needed to Build a Quadratic Model? Back in Fact 4.7.1 on page 44, we highlighted the information required to determine a linear model. We now describe the quadratic model analog. Important Facts 7.5.1. A quadratic model is completely determined by: 1. Three distinct non-collinear points, or 2. The vertex and one other point on the graph. The ﬁrst approach is just Fact 7.4.3. The second approach is based on the vertex form of a quadratic function. The idea is that we know any quadratic function f(x) has the form f(x) = a(x − h)2 + k, where (h, k) is the vertex. If we are given h and k, together with another point (x0, y0) on the graph, then plugging in gives this equation: y0 = a(x0 − h)2 + k. The only unknown in this equation is a, which we can solve for using algebra. A couple of the exercises will depend upon these observations. 7.6 Summary A quadratic function is one of the form • f(x) = ax2 + bx + c. where a = 0. 6 102 CHAPTER 7. QUADRATIC MODELING • The graph of a quadratic function is a parabola which is symmetric about the vertical line through the highest (or lowest
) point on the graph. This highest (or lowest) point is known as the vertex of the graph; its location is given by (h,k) where h = − b 2a and k = f(h). If a > 0, then the vertex is the lowest (or minimum) point on the graph, and the parabola ”opens upward”. If a < 0, then the vertex is the highest (or maximum) point on the graph, and the parabola ”opens downward”. Every quadratic function can be expressed in the form • • f(x) = a(x − h)2 + k where (h,k) is the vertex of the function’s graph. 7.7. EXERCISES 7.7 Exercises Problem 7.1. Write the following quadratic functions in vertex form, ﬁnd the vertex, the axis of symmetry and sketch a rough graph. (a) f(x) = 2x2 − 16x + 41. (b) f(x) = 3x2 − 15x − 77. (c) f(x) = x2 − 3 7 x + 13. (d) f(x) = 2x2. (e) f(x) = 1 100 x2. Problem 7.2. In each case, ﬁnd a quadratic function whose graph passes through the given points: (a) (0,0), (1,1) and (3, − 1). (b) (−1,1), (1, − 2) and (3,4). (c) (2,1), (3,2) and (5,1). (d) (0,1), (1,1) and (1,3). Problem 7.3. (a) Sketch the graph of the function f(x) = x2 − 3x + 4 on the interval −3 5. What is the maximum value of f(x) on that interval? What is the minimum value of f(x) on that interval? ≤ ≤ x (b) Sketch the graph of the function f(x) = x2 − 3x + 4 on the interval 2 7. What is the maximum value of f(x) on that interval? What is the minimum value of f(x) on that interval? ≤ ≤ x (c) Sketch the graph of the
function g(x) = −(x + 3)2 + 3 on the interval 0 4. What is the maximum value of g(x) on that interval? What is the minimum value of g(x) on that interval? ≤ ≤ x Problem 7.4. If the graph of the quadratic function f(x) = x2 + dx + 3d has its vertex on the x-axis, what are the possible values of d? What if f(x) = x2 + 3dx − d2 + 1? 103 (a) Find the multipart function s(t) giving the stock price after t days. If you buy 1000 shares after 30 days, what is the cost? (b) To maximize proﬁt, when should you sell shares? How much will the proﬁt be on your 1000 shares purchased in (a)? the graph Problem 7.6. Sketch of y = x2 − 2x − 3. Label the coordinates of the x and y intercepts of the graph. In the same coordinate system, sketch the graph of y = \|x2 − 2x − 3\|, give the multipart rule and label the x and y intercepts of the graph. Problem 7.7. A hot air balloon takes off from the edge of a plateau. Impose a coordinate system as pictured below and assume that the path the balloon follows is the graph of the quadratic function y = f(x) = − 4 5 x. The land drops at a constant incline from the plateau at the rate of 1 vertical foot for each 5 horizontal feet. Answer the following questions: 2500 x2 + 4 height above plateau (feet) balloon takeoff horizontal distance from launch (feet) ground incline (a) What is the maximum height of the bal- loon above plateau level? (b) What is the maximum height of the bal- loon above ground level? (c) Where does the balloon land on the ground? (d) Where is the balloon 50 feet above the ground? Problem 7.5. The initial price of buzz.com stock is $10 per share. After 20 days the stock price is $20 per share and after 40 days the price is $25 per share. Assume that while the price of the stock is not zero it can be modeled by a quadratic function. Problem 7.8. (a) Suppose f(x) = 3x2 − 2. Does the
point (1,2) lie on the graph of y = f(x)? Why or why not? (b) If b is a constant, where does the line y = 1 + 2b intersect the graph of y = x2 + bx + b? 104 CHAPTER 7. QUADRATIC MODELING (c) If a is a constant, where does the line y = 1 − a2 intersect the graph of y = x2 − 2ax + 1? (d) Where does the graph of y = −2x2 + 3x + 10 intersect the graph of y = x2 + x − 10? Problem 7.9. Sylvia has an apple orchard. One season, her 100 trees yielded 140 apples per tree. She wants to increase her production by adding more trees to the orchard. However, she knows that for every 10 additional trees she plants, she will lose 4 apples per tree (i.e., the yield per tree will decrease by 4 apples). How many trees should she have in the orchard to maximize her production of apples? Problem 7.10. Rosalie is organizing a circus performance to raise money for a charity. She is trying to decide how much to charge for tickets. From past experience, she knows that the number of people who will attend is a linear function of the price per ticket. If she charges 5 dollars, 1200 people will attend. If she charges 7 dollars, 970 people will attend. How much should she charge per ticket to make the most money? Problem 7.11. A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular Norman window is to be 24 feet. What should its dimensions be in order to maximize the area of the window and, therefore, allow in as much light as possible? Problem 7.12. Jun has 300 meters of fencing to make a rectangular enclosure. She also wants to use some fencing to split the enclosure into two parts with a fence parallel to two of the sides. What dimensions should the enclosure have to have the maximum possible area? Problem 7.13. You have $6000 with which to build a rectangular enclosure with fencing. The fencing material costs $20 per meter. You also want to have two parititions across the width of the enclosure, so that there will be three separated spaces in the enclosure. The material for the partitions costs $15 per meter. What is the maximum area you can achieve for the enclosure? a piece of
wire 60 inches long and cuts it into two pieces. Steve takes the ﬁrst piece of wire and bends it into the shape of a perfect circle. He then proceeds to bend the second piece of wire into the shape of a perfect square. Where should Steve cut the wire so that the total area of the circle and square combined is as small as possible? What is this minimal area? What should Steve do if he wants the combined area to be as large as possible? Problem 7.15. Two particles are moving in the xy-plane. The move along straight lines at constant speed. At time t, particle A’s position is given by x = t + 2, y = 1 2 t − 3 and particle B’s position is given by x = 12 − 2t, y = 6 − 1 3 t. (a) Find the equation of the line along which particle A moves. Sketch this line, and label A’s starting point and direction of motion. (b) Find the equation of the line along which particle B moves. Sketch this line on the same axes, and label B’s starting point and direction of motion. (c) Find the time (i.e., the value of t) at which the distance between A and B is minimal. Find the locations of particles A and B at this time, and label them on your graph. Problem 7.16. Sven starts walking due south at 5 feet per second from a point 120 feet north of an intersection. At the same time Rudyard starts walking due east at 4 feet per second from a point 150 feet west of the intersection. (a) Write an expression for the distance between Sven and Rudyard t seconds after they start walking. (b) When are Sven and Rudyard closest? What is the minimum distance between them? Problem 7.14. Steve likes to entertain friends at parties with “wire tricks.” Suppose he takes Problem 7.17. After a vigorous soccer match, Tina and Michael decide to have a glass of 7.7. EXERCISES 105 their favorite refreshment. They each run in a straight line along the indicated paths at a speed of 10 ft/sec. Parametrize the motion of Tina and Michael individually. Find when and where Tina and Michael are closest to one another; also compute this minimum distance. Find values of α that make this equation true (your answer will involve x). Problem 7.19. For each of the following
equations, ﬁnd the value(s) of the constant α so that the equation has exactly one solution, and determine the solution for each value. (−50,275) beet juice (200,300) soy milk (400,50) Tina Michael Problem 7.18. Consider the equation: αx2 + 2α2x + 1 = 0. Find the values of x that make this equation true (your answer will involve α). (a) αx2 + x + 1 = 0 (b) x2 + αx + 1 = 0 (c) x2 + x + α = 0 (d) x2 + αx + 4α + 1 = 0 Problem 7.20. (a) Solve for t s = 2(t − 1)2 + 1 (b) Solve for x y = x2 + 2x + 3 106 CHAPTER 7. QUADRATIC MODELING Chapter 8 Composition A new home takes its shape from basic building materials and the skillful use of construction tools. Likewise, we can build new functions from known functions through the application of analogous mathematical tools. There are ﬁve tools we want to develop: composition, reﬂection, shifting, dilation, arithmetic. We will handle composition in this section, then discuss the others in the following two sections. To set the stage, let’s look at a simple botany experiment. Imagine a plant growing under a particular steady light source. Plants continually give off oxygen gas to the environment at some rate; common units would be liters/hour. If we leave this plant unbothered, we measure that the plant puts out 1 liter/hour; so, the oxygen output is a steady constant rate. However, if we apply a ﬂash of high intensity green light at the time t = 1 and measure the oxygen output of the plant, we obtain the plot in Figure 8.1(a). Using what we know from the previous section on quadratic functions, we can check that a reasonable model for the graph is this multipart function f(t) (on the domain 0 10): t ≤ ≤ f(t) =   1 2 3t2 − 8 1 if t 3t + 3 if 1 if.8 0.6 0.4 0.2 1 0.8 0.6 0.4 0.2 oxygen rate 1 hr hours 2 4 6 8 10 (a
) Flash at t = 1. oxygen rate 1 hr 2 4 6 hours 8 10 (b) Flash at t = 5. Figure 8.1: Light ﬂashes.  Suppose we want to model the oxygen consumption when a green light pulse occurs at time t = 5 (instead of time t = 1), what is the mathematical model? For starters, it is pretty easy to believe that the graph for this new situation will look like the new graph in Figure 8.1(b). But, can we somehow use the model f(t) in hand (the known function) to build the model we want (the new function)? We will return in Exam- 107 108 CHAPTER 8. COMPOSITION ple 8.2.4 to see the answer is yes; ﬁrst, we need to develop the tool of function composition. 8.1 The Formula for a Composition The basic idea is to start with two functions f and g and produce a new function called their composition. There are two basic steps in this process and we are going to focus on each separately. The ﬁrst step is fairly mechanical, though perhaps somewhat unnatural. It involves combining the formulas for the functions f and g together to get a new formula; we will focus on that step in this subsection. The next step is of varying complexity and involves analyzing how the domains and ranges of f and g affect those of the composition; we defer that to the next subsection once we have the mechanics down. in out in out u x y the “g” function x = g(u) the “f” function y = f(x) in out “composed” function y = f(g(u)) Figure 8.2: Visualizing a composite function y = h(u) = f(g(u)). Here is a very common occurrence: We are handed a function y = f(x), which means given an x value, the rule f(x) produces a new y value. In addition, it may happen that the variable x is itself related to a third variable u by some different function equation x = g(u). Given u, the rule g(u) will produce a value of x; from this x we can use the rule f(x) In other words, we can regard y as a function to produce a y value. depending on the new independent variable u. It is important to know the mechanics of
working with this kind of setup. Abstractly, we have just described a situation where we take two functions and build a new 8.1. THE FORMULA FOR A COMPOSITION 109 function which “composes” the original ones together; schematically the situation looks like this: Example 8.1.1. A pebble is tossed into a pond. The radius of the ﬁrst circular ripple is measured to increase at the constant rate of 2.3 ft/sec. What is the area enclosed by the leading ripple after 6 seconds have elapsed? How much time must elapse so that the area enclosed by the leading ripple is 300 square feet? Solution. We know that an object tossed into a pond will generate a series of concentric ripples, which grow steadily larger. We are asked questions that relate the area of the circular region bounded by the leading ripple and time elapsed. leading ripple after t seconds leading ripple after 2 seconds r = r(t) Let r denote the radius of the leading ripple after t seconds; units of feet. The area A of a disc bounded by a leading ripple will be A = A(r) = πr2. This exhibits A as a function in the variable r. However, the radius is changing with respect to time: leading ripple after 1 second Figure 8.3: Concentric ripples. r = r(t) = radius after t seconds = 2.3 feet sec t seconds = 2.3t feet. So, r = r(t) is a function of t. In the expression A = A(r), replace “r” by “r(t),” then A = π(2.3t)2 = 5.29πt2. The new function a(t) = 5.29πt2 gives a precise relationship between area and time. To answer our ﬁrst question, a(6) = 598.3 feet2 is the area of the region bounded by the leading ripple after 6 seconds. On the other hand, if a(t) = 300 ft2;, 300 = 5.29πt2, so t = 4.25. Since t represents time, only the positive solution t = 4.25 seconds makes sense. p We can formalize the key idea used in solving this problem, which is 300/(5.29π) = ± ± familiar from previous courses. Suppose that y = f(
x) and that additionally the independent variable x is itself a function of a different independent variable t; i.e., x = g(t). Then we can replace every occurrence of “x” in f(x) by the expression “g(t),” thereby obtaining y as a function in the independent variable t. We usually denote this new function of t: y = f(g(t)). 110 CHAPTER 8. COMPOSITION We refer to f(g(t)) as the composition of f and g or the compositefunction. The process of forming the composition of two functions is a mechanical procedure. If you are handed the actual formulas for y = f(x) and x = g(t), then Procedure 8.1.2 is what you need. Important Procedure 8.1.2. To obtain the formula for f(g(t)), replace every occurrence of “x” in f(x) by the expression “g(t).” Here are some examples of how to do this: Examples 8.1.3. Use the composition procedure in each of these cases. (i) If y = f(x) = 2 and x = g(t) = 2t, then f(g(t)) = f(2t) = 2. (ii) If y = 3x − 7 and x = g(t) = 4, then f(g(t)) = f(4) = 3 4 − 7 = 5. · (iii) If y = f(x) = x2 + 1 and x = g(t) = 2t − 1, then f(g(t)) = f(2t − 1) = (2t − 1)2 + 1 = 4t2 − 4t + 2. (iv) If y = f(x) = 2 + 1 + (x − 3)2 and x = g(t) = 2t2 − 1, then p f(g(t)) = f(2t2 − 1) = 2 + 1 + (2t2 − 1 − 3)2 = 2 + 4t4 − 16t2 + 17. p (v) If y = f(x) = x2 and x = g(t) = t + p, then ♥ ) f(g(t)) = f(t + ♥ )2 = (t + ♥ = t2
+ 2t ♥ + 2. ♥ It is natural to ask: What good is this whole business about compositions? One way to think of it is that we can use composite functions to break complicated functions into simpler parts. For example, y = h(x) = x2 + 1 p can be written as the composition f(g(x)), where y = f(z) = √z and z = g(x) = x2+1. Each of the functions f and g is “simpler” than the original h, which can help when studying h. Examples 8.1.4. Here we use composite functions to “simplify” a given function. 8.1. THE FORMULA FOR A COMPOSITION 111 (i) The function y = where y = f(z) = 1 1 (x−3)2+2 can be written as a composition y = f(g(x)), z2+2 and z = g(x) = x − 3. (ii) The upper semicircle of radius 2 centered at (1,2) is the graph of the 4 − (x − 1)2. This function can be written as a comfunction y = 2 + position y = f(g(x)), where y = f(z) = 2 + √4 − z2 and z = g(x) = x − 1. p 8.1.1 Some notational confusion In our discussion above, we have used different letters to represent the domain variables of two functions we are composing. Typically, we have been writing: If y = f(x) and x = g(t), then y = f(g(t)) is the composition. This illustrates that the three variables t, x, and y can all be of different types. For example, t might represent time, x could be speed and y could be distance. If we are given two functions that involve the same independent variable, like f(x) = x2 and g(x) = 2x + 1, then we can still form a new function f(g(x)) by following the same prescription as in Procedure 8.1.2: Important Procedure 8.1.5. To obtain the formula for f(g(x)), replace every occurrence of “x” in f(x) by the expression “g(x).” For our example, this gives
us: f(g(x)) = f(2x + 1) = (2x + 1)2. Here are three other examples: If f(x) = √x, g(x) = 2x2 + 1, then f(g(x)) = √2x2 + 1. If f(x) = 1 x, g(x) = 2x + 1, then f(g(x)) = 1 2x+1. If f(x) = x2, g(x) = △ − x, then f(g(x)) = 2 − 2x △ △ + x2. • • • 112 y-axis 10 8 6 4 2 x-axis −3 −2 −1 1 2 3 Figure 8.4: Sketching composite functions. CHAPTER 8. COMPOSITION Example 8.1.6. Let f(x) = x2, g(x) = x + 1 and h(x) = x − 1. Find the formulas for f(g(x)), g(f(x)), f(h(x)) and h(f(x)). Discuss the relationship between the graphs of these four functions. Solution. If we apply Procedure 8.1.5, we obtain the composition formulas. The four graphs are given on the domain −3 3, together with the graph of f(x) = x2. x ≤ ≤ f(g(x)) = f(x + 1) = (x + 1)2 g(f(x)) = g(x2) = x2 + 1 f(h(x)) = f(x − 1) = (x − 1)2 h(f(x)) = h(x2) = x2 − 1. We can identify each graph by looking at its vertex: f(x) has vertex (0,0) f(g(x)) has vertex (-1,0) g(f(x)) has vertex (0,1) f(h(x)) has vertex (1,0) h(f(x)) has vertex (0,-1) • • • • • Horizontal or vertical shifting of the graph of f(x) = x2 gives the other four graphs: See Figure 8.4. 8.2 Domain, Range, etc. for a Composition A function is a “package” consisting of a rule, a
domain of allowed input values, and a range of output values. When we start to compose functions, we sometimes need to worry about how the domains and ranges of the composing functions affect the composed function. First off, when you form the composition f(g(x)) of f(x) and g(x), the range values for g(x) must lie within the domain values for f(x). This may require that you modify the range values of g(x) by changing its domain. The domain values for f(g(x)) will be the domain values for g(x). 8.2. DOMAIN, RANGE, ETC. FOR A COMPOSITION 113 in out in out domain g the “g” function the “f” function range of f(g(x)) range of g domain of f Figure 8.5: What is the domain and range of a composite function? In practical terms, here is how one deals with the domain issues for a composition. This is a reﬁnement of Procedure 8.1.5 on page 111. Important Procedure 8.2.1. To obtain the formula for f(g(x)), replace every occurrence of “x” in f(x) by the expression “g(x).” In addition, if there is a condition on the domain of f that involves x, then replace every occurrence of “x” in that condition by the expression “g(x).” The next example illustrates how to use this principle. Example 8.2.2. Start with the function y = f(x) = x2 on the domain −1 1. Find the rule and domain of y = f(g(x)), where g(x) = x − 1. x ≤ ≤ Solution. We can apply the ﬁrst statement in Procedure 8.2.1 to ﬁnd the rule for y = f(g(x)): y = f(g(x)) = f(x − 1) = (x − 1)2 = x2 − 2x + 1. To ﬁnd the domain of y = f(g(x)), we apply the second statement in Procedure 8.2.1; this will require that we solve an inequality equation: −1 −1 0 ≤ ≤ ≤ g(x) x − 1 x 1 1 2 ≤ ≤ ≤ The conclusion is that y =
f(g(x)) = x2 −2x+1 on the domain 0 x ≤ ≤ 2. Example 8.2.3. Let y = f(z) = √z, z = g(x) = x + 1. What is the largest possible domain so that the composition f(g(x)) makes sense? 114 CHAPTER 8. COMPOSITION y range y = f(z) z domain z z = g(x) = x + 1 −1 desired range x required domain (a) y = √z. (b) z = x = 1. Figure 8.6: Finding the largest domain for f(g(x)). Solution. The largest possible domain for y = f(z) will consist of all nonnegative real numbers; this is also the range of the function f(z): See Figure 8.6(a). To ﬁnd the largest domain for the composition, we try to ﬁnd a domain of x-values so that the range of z = g(x) is the domain of y = f(z). So, in this case, we want the range of g(x) to be all non-negative real numbers, z. We graph z = g(x) in the xz-plane, mark the desired range denoted 0 0 z on the vertical z-axis, then determine which x-values would lead to points on the graph with second coordinates in this zone. We ﬁnd that the domain of all x-values greater or equal to −1 (denoted −1 x) leads to the desired range. In summary, the composition y = f(g(x)) = √x + 1 is deﬁned on the domain of x-values −1 x. ≤ ≤ ≤ ≤ Let’s return to the botany experiment that opened this section and see how composition of functions can be applied to the situation. Recall that plants continually give off oxygen gas to the environment at some rate; common units would be liters/hour. Example 8.2.4. A plant is growing under a particular steady light source. If we apply a ﬂash of high intensity green light at the time t = 1 and measure the oxygen output of the plant, we obtain the plot below and the mathematical model f(t). f(t) =   1 2 3t2 − 8 1 if t 3
t + 3 if 1 if 3 1 t t ≤ ≤ ≤ 3 ≤  Now, suppose instead we apply the ﬂash of high intensity green light at the time t = 5. Verify that the mathematical model for this experiment is given by f(g(t)), where g(t) = t − 4. 8.2. DOMAIN, RANGE, ETC. FOR A COMPOSITION 115 Solution. Our expectation is that the plot for this new experiment will have the “parabolic dip” shifted over to occur starting at time t = 5 instead of at time t = 1. In other words, we expect the graph in Figure 8.7(b). Our job is to verify that this graph is obtained from the function f(g(t)), where g(t) = t − 4. This is a new terrain for us, since we need to look at a composition involving a multipart function. Here is how to proceed: When we are calculating a composition involving a multipart function, we need to look at each of the parts separately, so there will be three cases to consider: First part: f(t) = 1 when t 1. To get the formula for f(g(t)), we now appeal to Procedure 8.2.1 and just replace every occurrence of t in f(t) by g(t). That gives us this NEW domain condition and function equation: ≤ 1 0.8 0.6 0.4 0.2 1 0.8 0.6 0.4 0.2 oxygen rate 1 hr hours 2 4 6 8 10 (a) Flash at t = 1. oxygen rate 1 hr hours 2 4 6 8 10 (b) Flash at t = 5. f(g(t)) = f(t − 4) = 1 when t − 4 1 ≤ = 1 when t Second part: f(t) = 2 5. ≤ 3t2 − 8 3. We now appeal to Procedure 8.2.1 and just replace every occurrence of t in this function by g(t). That gives us this NEW domain condition and function equation: 3t + 3 when 1 ≤ ≤ t Figure 8.7: Applying light at time t. f(g(t)) = f(t − 4) = (t − 4)2 − 8 3 (t − 4) + 3 when 1 t − 4 3 ≤ ≤ 2 3 Third part: f(t) = 1 when 3
t2 − 8t + = when 5 t ≤ ≤ 7. t. We now appeal to Procedure 8.2.1 and just replace every occurrence of t in this function by g(t). That gives us this NEW domain condition and function equation: ≤ 2 3 73 3 f(g(t)) = f(t − 4) = 1 when 3 = 1 when 7 t. ≤ t − 4 ≤ The multipart rule for this composition can now be written down and using a graphing device you can verify its graph is the model for our experiment. 1 3t2 − 8t + 73 2 1 3 if t if 5 if (g(t)) =    116 CHAPTER 8. COMPOSITION 8.3 Exercises Problem 8.1. For this problem, f(t) = t − 1, g(t) = −t − 1 and h(t) = \|t\|. (a) Compute the multipart rules for h(f(t)) and h(g(t)) and sketch their graphs. (b) Compute the multipart rules for f(h(t)) and g(h(t)) and sketch their graphs. (c) Compute the multipart rule for h(h(t)−1) and sketch the graph. Problem 8.2. Write each of the following functions as a composition of two simpler functions: (There is more than one correct answer.) (a) y = (x − 11)5. (b) y = 3√1 + x2. (c) y = 2(x − 3)5 − 5(x − 3)2 + 1 2 (x − 3) + 11. (d) y = 1 x2+3. (e) y = √x + 1. p (f) y = 2 − 5 − (3x − 1)2. p Problem 8.3. (a) Let f(x) be a linear function, f(x) = ax + b for constants a and b. Show that f(f(x)) is a linear function. (b) Find a function g(x) such that g(g(x)) = 6x − 8. Problem 8.4. Let f(x) = 1 2 x + 3. (a) Sketch the graphs of f(x),f(f(x)),f(f(f
(x))) on the interval −2 x 10. ≤ ≤ (b) Your graphs should all intersect at the point (6,6). The value x = 6 is called a ﬁxed point of the function f(x) since f(6) = 6; that is, 6 is ﬁxed - it doesn’t move when f is applied to it. Give an explanation for why 6 is a ﬁxed point for any function f(f(f(...f(x)...))). (c) Linear functions (with the exception of f(x) = x) can have at most one ﬁxed point. Quadratic functions can have at most two. Find the ﬁxed points of the function g(x) = x2 − 2. Problem 8.5. A car leaves Seattle heading east. The speed of the car in mph after m minutes is given by the function C(m) = 70m2 10 + m2. (a) Find a function m = f(s) that converts seconds s into minutes m. Write out the formula for the new function C(f(s)); what does this function calculate? (b) Find a function m = g(h) that converts hours h into minutes m. Write out the formula for the new function C(g(h)); what does this function calculate? (c) Find a function z = v(s) that converts mph s into ft/sec z. Write out the formula for the new function v(C(m); what does this function calculate? Problem 8.6. Compute the compositions f(g(x)), f(f(x)) and g(f(x)) in each case: (a) f(x) = x2, g(x) = x + 3. (b) f(x) = 1/x, g(x) = √x. (c) f(x) = 9x + 2, g(x) = 1 (d) f(x) = 6x2 + 5, g(x) = x − 4. (e) f(x) = 4x3 − 3, g(x) = 3√2x + 6 (f) f(x) = 2x + 1, g(x) = x3. 9 (x − 2). (g) f(x) = 3
, g(x) = 4x2 + 2x + 1. (h) f(x) = −4, g(x) = 0. Problem 8.7. Let y = f(z) = √4 − z2 and z = g(x) = 2x + 3. Compute the composition y = f(g(x)). Find the largest possible domain of x-values so that the composition y = f(g(x)) is deﬁned. Problem 8.8. Suppose you have a function y = f(x) such that the domain of f(x) is 1 6 and the range of f(x) is −3 5. y x ≤ ≤ (a) What is the domain of f(2(x − 3))? ≤ ≤ (b) What is the range of f(2(x − 3))? (c) What is the domain of 2f(x) − 3? (d) Give a quadratic function whose ﬁxed (d) What is the range of 2f(x) − 3? points are x = −2 and x = 3. (e) Can you ﬁnd constants B and C so that the domain of f(B(x − C)) is 8 x ≤ ≤ 9? 8.3. EXERCISES 117 (f) Can you ﬁnd constants A and D so that this simpliﬁed expression: the range of Af(x) + D is 0 y ≤ ≤ 1? Problem 8.9. For each of the given functions y = f(x), simplify the following expression so that h is no longer a factor in the denominator, then calculate the result of setting h = 0 in f(x + h) − f(x) h. (a) f(x) = 1 x−1. (b) f(x) = (2x + 1)2. (c) f(x) = √25 − x2. 118 CHAPTER 8. COMPOSITION Chapter 9 Inverse Functions The experimental sciences are loaded with examples of functions relating time and some measured quantity. In this case, time represents our “input” and the quantity we are measuring is the “output.” For example, maybe you have just mixed together some chemical reactants in a vessel. As time goes by, you measure the fraction of reactants
remaining, tabulate your results, then sketch a graph as indicated in Figure 9.1. 1 fraction Viewing the input value as “time” and the output value as “fraction of product,” we could ﬁnd a function y = f(t) modeling this data. Using this function, you can easily compute the fraction of reactants remaining at any time in the future. However, it is probably just as interesting to know how to predict the time when a given fraction of reactants exists. In other words, we would like a new function that allows us to input a “fraction of reactants” and get out the “time” when this occurs. This “reverses” the input/output roles in the original function. Is there a systematic way to ﬁnd the new function if we know y = f(t)? The answer is yes and depends upon the general theory of inverse functions. 0 time Figure 9.1: Fraction of reactants as a function of time. 9.1 Concept of an Inverse Function Suppose you are asked to solve the following three equations for x. How do you proceed? (x + 2) = 64 (x + 2)2 = 64 (x + 2)3 = 64. In the ﬁrst equation, you add “−2” to each side, then obtain x = 62. In the third equation, you take the cube root of both sides of the equation, giving you x + 2 = 4, then subtract 2 getting x = 2. In second equation, you take a square root of both sides, BUT you need to remember both the 119 120 CHAPTER 9. INVERSE FUNCTIONS positive and negative results when doing this. So, you are reduced down to x + 2 = 8 or that x = −10 or 6. Why is it that in two of these cases you obtain a single solution, while in the remaining case there are two different answers? We need to sort this out, since the underlying ideas will surface when we address the inverse circular functions in Chapter 20. ± In x Out f(x) The function Figure 9.2: A function as a process. Let’s recall the conceptual idea of a function: A function is a process which takes a number x and outputs a new number f(x). So far, we’ve only worked with this process from “left to right
;” i.e., given x, we simply put it into a symbolic rule and out pops a new number f(x). This is all pretty mechanical and straightforward. 9.1.1 An Example Let’s schematically interpret what happens for the speciﬁc concrete example y = f(x) = 3x − 1, when x = −1, − 1 2, 1, 2: See Figure 9.3. 2, 0, 1 −1 − 1 2 in in in out 3x − 1 −4 3x − 1 − 5 2 out out 0 3x − 1 −1 in in in 1 2 1 2 out 3x − 1 out 3x − 1 out 3x − 1 1 2 2 5 Figure 9.3: Function process y = 3x − 1. We could try to understand the function process in this example in “reverse order,” going “right to left;” namely, you might ask what x value can be run through the process so you end up with the number 11? This is somewhat like the “Jeopardy” game show: You know what the answer is, you want to ﬁnd the question. For our example, if we start out with some given y values, then we can deﬁne a “reverse process” x = 1 3(y + 1), which returns the x value required so that f(x) = y: See Figure 9.4. 9.1. CONCEPT OF AN INVERSE FUNCTION 121 eplacements −1 − 1 2 0 (y+1) 3 (y+1) 3 (y+1) 3 −4 − 5 2 1 2 1 −1 2 (y+1) 3 (y+1) 3 (y+1) 3 1 2 2 5 Figure 9.4: The reverse process x = 1 3 (y + 1). 9.1.2 A Second Example If we begin with a linear function y = f(x) = mx + b, where m = 0, then we can always ﬁnd a “reverse process” for the function. To ﬁnd it, you must solve the equation y = f(x) for x in terms of y: y = mx + b y − b = mx 1 m (y − b) = x So, if m = 3 and b =
−1, we just have the ﬁrst example above. For another example, suppose y = −0.8x + 2; then m = −0.8 and b = 2. In this case, the reverse process is −1.25(y − 2) = x. If we are given the value y = 11, we simply compute that x = −11.25; i.e., f(−11.25) = 11. 9.1.3 A Third Example The previous examples hide a subtle point that can arise when we try to understand the “reverse process” for a given function. Suppose we begin with the function y = f(x) = (x − 1)2 + 1. Figure 9.5 is a schematic of how the function works when we plug in x = −1, − 1 2, 1, 2; what is being illustrated is a “forward process”, in that each input generates a unique output. 2, 0, 1 For this example, if we start out with some given y values, then we can try to deﬁne a “reverse process” x =??? which returns an x value required so that f(x) = y. Unfortunately, there is no way to obtain a single formula for this reverse process; Figure 9.6(a) shows what happens if you are given y = 3 and you try to solve for x. 6 122 CHAPTER 9. INVERSE FUNCTIONS in in in −1 − 1 2 0 (x − 1)2 + 1 (x − 1)2 + 1 (x − 1)2 + 1 out out out 5 13 4 2 in in in 1 2 1 2 (x − 1)2 + 1 (x − 1)2 + 1 (x − 1)2 + 1 out out out 5 4 1 2 Figure 9.5: Function process y = f(x) = (x − 1)2 + 1. The conclusion is that the “reverse process” has two outputs. This violates the rules required for a function, so this is NOT a function. The solution is to create two new “reverse processes.” 1 + √2 reverse process x = − +√y − 1 3 1 − √2 (a) Reverse process but not a function. reverse process 1 + √2 x = +√y − 1 reverse process 1 − √
2 x = −√y − 1 3 3 (b) Two new reverse processes that are functions. Figure 9.6: What to do if a reverse process is not a function. Each of these “reverse processes” has a unique output; in other words, each of these “reverse processes” deﬁnes a function. So, given y = 3, there are TWO possible x values, √2, so that f(1 + √2) = 3 and f(1 − √2) = 3. namely x = 1 In other words, the reverse process is not given by a single equation; there are TWO POSSIBLE reverse processes. ± 9.2 Graphical Idea of an Inverse We have seen that ﬁnding inverses is related to solving equations. However so far, the discussion has been symbolic; we have pushed around a few equations and in the end generated some confusion. Let’s use the tools of Chapter 6 to visualize what is going on here. Suppose we are given the graph of a function f(x) as in Figure 9.7(a). What input x values result in an output value of 3? This involves ﬁnding all x such that f(x) = 3. Graphically, this means we are trying to ﬁnd points on the graph of f(x) so that their y-coordinates are 3. The easiest way to to do this is to draw the line y = 3 and ﬁnd where it intersects the graph. In Figure 9.7(b) we can see the points of intersection are (−5, 3), (−1, 3), and (9, 3). That means that x = −5, −1, 9 produce the output value 3; i.e., f(−5) = f(−1) = f(9) = 3. 9.2. GRAPHICAL IDEA OF AN INVERSE 123 y = f(x) 5 5 y = f(x) y = 3 −5 5 10 −5 5 10 −5 −5 (a) Given the function y = f(x). (b) What values of x give f(x) = 3? Figure 9.7: Using the horizontal line y = 3 to ﬁnd values on the x-axis. This leads to our ﬁrst important fact about the “reverse
process” for a function: Important Fact 9.2.1. Given a number c, the x values such that f(x) = c can be found by ﬁnding the x-coordinates of the intersection points of the graphs of y = f(x) and y = c. Example 9.2.2. Graph y = f(x) = x2 and discuss the meaning of Fact 9.2.1 when c = 3, 1, 6. Solution. We graph y = x2 and the lines y = 1, y = 3 and y = 6. Let’s use c = 6 as an example. We need to simultaneously solve the equations y = x2 and y = 6. Putting these together, we get x2 = 6 or x = 2.449; i.e., 1.732) = 3. 2.449) = 6. If c = 3, we get x = f( Finally, if c = 1, we get x = √6 1.732; i.e., f( ± 1; i.e., f( ≈ ± ± 1) = 1. ± ± ± ± y = x2 y = 6 y = 3 y = 1 The pictures so far indicate another very important piece of information. For any number c, we can tell exactly “how many” input x values lead to the same output value c, just by counting the number of times the graphs of y = f(x) and y = c intersect. Figure 9.8: Graph of y = f(x) = x2. Important Fact 9.2.3. For any function f(x) and any number c, the number of x values so that f(x) = c is the number of times the graphs of y = c and y = f(x) intersect. 124 CHAPTER 9. INVERSE FUNCTIONS Examples 9.2.4. (i) If f(x) is a linear function f(x) = mx + b, m = 0, then the graph of f(x) intersects a given horizontal line y = c EXACTLY once; i.e., the equation c = f(x) always has a unique solution. (ii) If f(x) = d is a constant function and c = d, then every input x value = d, in the domain leads to the output value c. On
the other hand, if c then no input x value will lead to the output value c. For example, if f(x) = 1 and c = 1, then every real number can be input to produce an output of 1; if c = 2, then no input value of x will lead to an output of 2. y-axis f(x) = mx + b y-axis y = c x-axis the only input which leads to an output of c f(x) = 1 x-axis any of these inputs leads to an output of 1 linear functions constant functions Figure 9.9: Does a horizontal line y = c intersect a curve once or more than once? 9.2.1 One-to-one Functions For a speciﬁed domain, one-to-onefunctions are functions with the property: Given any number c, there is at most one input x value in the domain so that f(x) = c. Among our examples thus far, linear functions (degree 1 polynomials) are always one-to-one. However, f(x) = x2 is not one-to-one; we’ve already seen that it can have two values for some of its inverses. By Fact 9.2.3, we can quickly come up with what’s called the horizontal line test. Important Fact 9.2.5 (Horizontal Line Test). On a given domain of x-values, if the graph of some function f(x) has the property that every horizontal line crosses the graph at most only once, then the function is one-to-one on this domain. 6 6 9.3. INVERSE FUNCTIONS Example 9.2.6. By the horizontal line test, it is easy to see that f(x) = x3 is one-to-one on the domain of all real numbers. Although it isn’t common, it’s quite nice when a function is one-to-one because we don’t need to worry as much about the number of input x values producing the same output y value. In effect, this is saying that we can deﬁne a “reverse process” for the function y = f(x) which will also be a function; this is the key theme of the next section. 125 y-axis x-axis horizontal lines Figure 9.10: A one-to
-one function f(x) = x3. 9.3 Inverse Functions Let’s now come face to face with the problem of ﬁnding the “reverse process” for a given function y = f(x). It is important to keep in mind that the domain and range of the function will both play an important role in this whole development. For example, Figure 9.11 shows the function f(x) = x2 with three different domains speciﬁed and the corresponding range values. range range domain domain range domain Figure 9.11: Possible domains for a given range. These comments set the stage for a third important fact. Since the domain and range of the function and its inverse rule are going to be intimately related, we want to use notation that will highlight this fact. We have been using the letters x and y for the domain (input) and range (output) variables of f(x) and the “reverse process” is going to reverse these roles. It then seems natural to simply write y (instead of c) for the input values of the “reverse process” and x for its output values. 126 CHAPTER 9. INVERSE FUNCTIONS Important Fact 9.3.1. Suppose a function f(x) is one-to-one on a domain of x values. Then deﬁne a NEW FUNCTION by the rule f−1(y) = the x value so that f(x) = y. The domain of y values for the function f−1(y) is equal to the range of the function f(x). The rule deﬁned here is the “reverse process” for the given function. It is referred to as the inverse function and we read f−1(y) as “...eff inverse of y...”. Both the “domain” of f(x) and the “rule” f(x) have equal inﬂuence on whether the inverse rule is a function. Keep in mind, you do NOT get an inverse function automatically from functions that are not one-to-one!!!! CAUTION!!! 9.3.1 Schematic Idea of an Inverse Function Suppose that f(x) is one-to-one, so that f−1(y) is a function. As a result, we can model f−1(y)
as a black box. What does it do? If we put in y in the input side, we should get out the x such that f(x) = y. in y f−1(y) out x such that f(x) = y Figure 9.12: A new function x = f−1(y). Now, let’s try to unravel something very special that is happening on a symbolic level. What would happen if we plugged f(a) into the inverse function for some number a? Then the inverse rule f−1(f(a)) tells us that we want to ﬁnd some x so that f(x) = f(a). But, we already know x = a works and since f−1(y) is a function (hence gives us unique answers), the output of f−1(f(a)) is just a. Symbolically, this means we have Fact 9.3.2. Important Fact 9.3.2. For every a value in the domain of f(x), we have f−1(f(a)) = a. (9.1) This is better shown in the black box picture of Figure 9.13. in a out in f(x) f(a) f−1(x) out a Figure 9.13: Visualizing f−1(f(a)) = a. A good way to get an idea of what an inverse function is doing is to remember that f−1(y) reverses the process of f(x). We can think of f−1(y) as a “black box” running f(x) backwards. 9.4. TRYING TO INVERT A NON ONE-TO-ONE FUNCTION 127 9.3.2 Graphing Inverse Functions How can we get the graph of an inverse function? The idea is to manipulate the graph of our original one-to-one function in some prescribed way, ending up with the graph of f−1(y). This isn’t as hard as it sounds, but some confusion with the variables enters into play. Remember that a typical point on the graph of a function y = f(x) looks like (x, f(x)). Now let’s take a look at the inverse function x = f−1(y). Given a number y in the domain of f−1(y), y = f(x
) for some x in the domain of f(x); i.e., we are using the fact that the domain of f−1 equals the range of f. The function f−1(y) takes the number f(x) and sends it to x, by Fact 9.3.2. So when f(x) is the input value, x becomes the output value. Conclude a point on the graph of f−1(y) looks like (f(x), x). It’s similar to the graph of y = f(x), only the x and y coordinates have reversed! What does that do to the graph? Essentially, you reorient the picture so that the positive x-axis and positive y-axis are interchanged. Figure 9.14 shows the process for the function y = f(x) = x3 and its inverse function x = f−1(y) = 3√y. We symbols on the graph to help keep track of what is happenplace some ing. ∗ +y−axis * +x−axis * * rotate 900 clockwise * * * +x−axis +y−axis ﬂip across horiz axis +x−axis * * * +y−axis Figure 9.14: Graphically ﬁnding x = f−1(y) = 3√y. 9.4 Trying to Invert a Non one-to-one Func- tion Suppose we blindly try to show that √y is the inverse function for y = x2, without worrying about all of this one-to-one stuff. We’ll start out with the number −7. If f−1(y) = √y, then we know that f−1(f(−7)) = f−1(49) = 7. On the other hand, the formula in Fact 9.3.2 tells us that we must have f−1(f(−7)) = −7, so we have just shown 7 = −7! So clearly f−1(y) = √y. Even if we try f−1(y) = −√y, we produce a contradiction. It seems that if you didn’t have 6 128 CHAPTER 9. INVERSE FUNCTIONS to worry about negative numbers, things would be all right. Then you could say that f−1(y) = √y. Let
’s try to see what this means graphically. Let’s set f(x) = x2, but only for non-negative x-values. That means that we want to erase the graph to the left of the y-axis (so remember - no negative x-values allowed). The graph would then look like Figure 9.15. +y y = x2 +x inverse function √y +x +y domain non-negative x domain non-negative y Figure 9.15: Restricting the domain: No negative x-values. This is a now a one-to-one function! And now, one can see that its inverse function is √y. Similarly, we could have taken f(x) = x2 but only for the non-positive x-values. In that case, f−1(y) = −√y. In effect, we have split the graph of y = x2 into two parts, each of which is the graph of a one-to-one function; Figure 9.16. +y y = x2 +x domain non-negative y +x +y domain non-positive x inverse function −√y Figure 9.16: Restricting the domain: No positive x-values. It is precisely this splitting into two cases that leads us to multiple solutions of an equation like x2 = 5. We obtain x = √5 and x = −√5; one 9.5. SUMMARY 129 solution comes from the side of the graph to the left of the y-axis, and the other from the right of the y-axis. This is because we have separate inverse functions for the left and right side of the graph of y = x2. 9.5 Summary • • • • • • Two functions f and g are inverses if f(g(x)) = x and g(f(x)) = x for all x in the domain of f and the domain of g. A function f is one-to-one if every equation f(x) = k If there is a value of k such that the has at most one solution. equation f(x) = k has more than one solution, then f is not one-toone. A function is one-to-one if every horizontal line intersects the function’s graph at most once. A function has an inverse if the function is one-to-one, and
every one-to-one function has an inverse. The domain of a function is the range of its inverse, and the range of a function is the domain of its inverse. The graph of a function and its inverse are mirror images of each other across the line y = x. 130 CHAPTER 9. INVERSE FUNCTIONS 9.6 Exercises Problem 9.1. Let f(x) = 2 domain for which the formula makes sense. 3x−4 on the largest A (a) Find the domain and range of f(x), then sketch the graph. (b) Find the domain, range and rule for the inverse function f−1, then sketch its graph. C Problem 9.2. Find the inverse function of each of the following functions. Specify the domains of the inverse functions. B D (a) f(x) = (b) h(xc) g(x) = 4√3 − x − 7 (d) j(x) = √x + √x − 1 (e) k(x) = 16 − x2, 0 p 4 x ≤ ≤ Problem 9.3. For this problem, y = f(x) = 2x2 − 3x − 1 on the domain of all real numbers. (a) Sketch the function graph and ﬁnd the coordinates of the vertex P = (a,b). (b) Explain why y = f(x) does not have an inverse function on the domain of all real numbers. (c) Restrict y = f(x) to the domain {a ≤ x} and ﬁnd the formula for the inverse function f−1(y). What are the domain and range of the inverse function? (d) Restrict y = f(x) to the domain {x ≤ a} and ﬁnd the formula for the inverse function f−1(y). What are the domain and range of the inverse function? Problem 9.5. Show that, for every value of a, the function f(x) = a + 1 x − a is its own inverse. Problem 9.6. Clovis is standing at the edge of a cliff, which slopes 4 feet downward from him for every 1 horizontal foot. He launches a small model rocket from where he is standing. With the origin of the coordinate system located where he is standing, and the x-axis extending horizontally, the path of the rocket is described by
the formula y = −2x2 + 120x. (a) Give a function h = f(x) relating the height h of the rocket above the sloping ground to its x-coordinate. (b) Find the maximum height of the rocket above the sloping ground. What is its x-coordinate when it is at its maximum height? (c) Clovis measures its height h of the rocket above the sloping ground while it is going up. Give a function x = g(h) relating the x-coordinate of the rocket to h. (d) Does this function still work when the rocket is going down? Explain. Problem 9.4. Which of the following graphs are one-to-one? If they are not one-to-one, section the graph up into parts that are one-toone. Problem 9.7. For each of the following functions: (1) sketch the function, (2) ﬁnd the inverse function, and (3) sketch the inverse function. In each case, indicate the correct domains and ranges. (4) Finally, make sure you 9.6. EXERCISES 131 test each of the functions you propose as an inverse with the following compositions: (b) After how many hours will the surface of the water have width of 6 feet? and f(f−1(x))?= x f−1(f(x))?= x. (c) Give a function t = f−1(w) relating the time to the width of the surface of the water. Make sure to specify the domain and compute the range too. (a) f(x) = 3x − 2 (b) f(x) = 1 2 x + 5 (c) f(x) = −x2 + 3, x 0 ≥ (d) f(x) = x2 + 2x + 5, x (e) f(x) = √4 − x2, 0 −1 2 ≤ x ≤ ≤ Problem 9.8. A trough has a semicircular cross section with a radius of 5 feet. Water starts ﬂowing into the trough in such a way that the depth of the water is increasing at a rate of 2 inches per hour. water 5 ft cross-section of trough (a) Give a function w = f(t) relating the width w of the surface of the water to the time t, in hours.
Make sure to specify the domain and compute the range too. Problem 9.9. A biochemical experiment involves combining together two protein extracts. Suppose a function φ(t) monitors the amount (nanograms) of extract A remaining at time t (nanoseconds). Assume you know these facts: 1. The function φ is invertible; i.e., it has an inverse function. 2. φ(0) = 6, φ(1) = 5, φ(2) = 3, φ(3) = 1, φ(4) = 0.5, φ(10) = 0. (a) At what time do you know there will be 3 nanograms of extract A remaining? (b) What is φ−1(0.5) and what does it tell you? (c) (True or False) There is exactly one time when the amount of extract A remaining is 4 nanograms. (d) Calculate φ(φ−1(1)) = (e) Calculate φ−1(φ(6)) = (f) What is the domain and range of φ? 132 CHAPTER 9. INVERSE FUNCTIONS Chapter 10 Exponential Functions If we start with a single yeast cell under favorable growth conditions, then it will divide in one hour to form two identical “daughter cells”. In turn, after another hour, each of these daughter cells will divide to produce two identical cells; we now have four identical “granddaughter cells” of the original parent cell. Under ideal conditions, we can imagine how this “doubling effect” will continue: cells Figure 10.1: Observing cell growth. The question is this: Can we ﬁnd a function of t that will predict (i.e. model) the number of yeast cells after t hours? If we tabulate some data (as at right), the conclusion is that the formula N(t) = 2t predicts the number of yeast cells after t hours. Now, let’s make a very slight change. Suppose that instead of starting with a single cell, we begin with a population of 3 If we assume 106 cells; a more realistic situation. × 133 TIME t=0 hours t=1 hours t=2 hours t=3 hours Total hours 0 1 2 3 4 5 6 Number of yeast cells 1=20 2=21 4=22

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