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top row contains ten bricks. How many bricks are there in the monument? 43. A ladder with nine rungs is to be built, with the bottom rung 24 inches wide and the top rung 18 inches wide. If the lengths of the rungs decrease uniformly from bottom to top, how long should each of the seven intermediate rungs be? second, 10 in the third, and so on, through row 12. Rows 12 through 20 (the last row) all have the same number of seats. Find the number of seats in the lecture hall. 44. Find the first eight numbers in an arithmetic sequence in which the sum of the first and seventh terms is 40 and the product of the first and fourth terms is 160. 1.4 Lines Objectives • Find the slopes of lines, including parallel and perpendicular lines • Describe the connection between arithmetic sequences and lines • Graph lines • Write the equations of lines, including horizontal and vertical lines A graph is a set of points in a plane. Some graphs are based on data points, such as those shown in Section 1.3 where arithmetic sequences were graphed as scatter plots. Other graphs arise from equations. A solution of an equation in two variables, say x and y, is a pair of numbers such that the substitution of the first number for x and the second for y produces a true statement. The graph of an equation in two variables is the set of points in a plane whose coordinates are solutions of the equation. Thus, the graph is a geometric picture of the solutions. Recall that an arithmetic sequence is a sequence in which the difference between each term and the preceding term is constant. For example, n 2 Position 1 un2 Term 1 3, 8, 13, 18, p 5 1 2 3 is an arithmetic sequence. p n 5 4 3 8 13 18 23 p 3 6 n 1 1 2 5 The graph of the sequence above has an infinite number of discrete points because the value of the sequence depends upon the term of the sequence, which must be a counting number. See Figure 1.4-1a. The graph of y 3 is a continuous line that contains the discrete points of the arithmetic sequence, as shown in Figures 1.4-1a and 1.4-1b on page 31. x 1 5 2 1 NOTE If needed, review graphing functions on a graphing calculator in the Technology Appendix. 25 25 Section 1.4 Lines 31 0 0 Figure 1.4-1a 6 0 0 Figure 1.4- |
1b 6 As the positions within a sequence increase by one, the value of the terms increases by 5. Notice on the graph that as the x-values move to the right a distance of one, the y-values move up by a distance of 5. This common difference represents one of the most identifiable characteristics of a line, its slope. Slope When you move from a point P to a point Q on a line, two numbers are involved, as illustrated in Figure 1.4-2. • The vertical distance you move is called the change in y, which is sometimes denoted and read “delta y.” ¢y • The horizontal distance you move is called the change in x, which is sometimes denoted and read “delta x.” ¢x Q Q Q Change in y = 4 Change in y = 4 P Change in x = 4 = 1 = 4 4 Change in y Change in x b. Figure 1.4-2 Change in x = 6 Change in y Change in x a. = 4 6 = 2 3 Change in y = 4 P Change in x = 1 Change in y Change in x c. = 4 1 = 4 y The fraction change in y change in x ¢y ¢x (x1, y1) pose P has coordinates Figure 1.4-3. measures the steepness of the line. Sup- and Q has coordinates (x2, y2), as shown in P y2 y1 • The change in y is the difference of the y-coordinates of P and Q. ¢y y2 y1 x • The change in x is the difference of the x-coordinates of P and Q. ¢x x2 x1 (x2, y2) Q y2 − y1 (x1, y1) P x2 − x1 x2 x1 Figure 1.4-3 32 Chapter 1 Number Patterns Consequently, slope is defined as follows. Slope of a Line y 1 1 2 3 4 −1 Figure 1.4-4 CAUTION When finding slopes, you must subtract the y-coordinates and the x-coordinates in the same order. With the points (3, 4) and (1, 8), for instance, if you use 8 4 in the numerator, you must write 1 – 3 in the denominator, not 3 1. and (x2, y2) (x1, y1) are |
points with If the line through these points is the ratio y2 x2 change in y change in x ¢y ¢x x1 y1 x1 x2, then the slope of Example 1 Finding Slope Given Two Points Find the slope of the line that passes through ure 1.4-4. 1 0, 1 2 and (4, 1). See Fig- x Solution Apply the formula in the previous box with x2 4, y2 1. x1 0, y1 1 and Slope y2 x2 y1 x1 1 (1) 4 0 2 4 1 2 The order of the points makes no difference; if you use (4, 1) for and in Example 1, the result is the same. (0, 1) for (x2, y2) ■ x1, y12 1 Example 2 Finding Slope From a Graph Find the slope of each line shown in Figure 1.4-5. The lines shown are determined by the following points: and 1, 1 3, 5 1 L4: L2: 2 and 2 0, 2 1 3, 1 2 1 1 and 1 2 L5: 2, 4 L3: and 6, 2 1 2, 2 2 1, 0 1 2 1 2 and 3, 2 2 1 2 L1: 1 0, 2 2 Solution The slopes are as follows. 3 0 1 3 1 L1: 2 1 0 1 L2 L4: L3: L5: 2 2 1 2 y L1 L2 5 4 3 2 1 L3 x −6 −5 −4 −3 −2 −1 −1 −2 3 L4 1 2 L5 Figure 1.4-5 ■ Section 1.4 Lines 33 Example 2, page 32, illustrates how the slope measures the steepness of the line, summarized as follows. The lines referenced refer to Figure 1.4-5. Properties of Slope The slope of a nonvertical line is a number m that measures how steeply the line rises or falls. • If m 77 0, the line rises from left to right; the larger m is, the more steeply the line rises. [Lines and L2 ], the line is horizontal. [Line L1 ] L3 • If • If m 0 m 66 0, the line falls from left to right; the larger is, m 0 0 the more steeply the line falls. [Lines L4 and ]L5 Sl |
ope-Intercept Form A nonvertical line intersects the y-axis at a point with coordinates (0, b), because every point on the y-axis has first coordinate 0. The number b is called the y-intercept of the line. For example, the line in Figure 1.4-4 has y-intercept because it crosses the y-axis at (0, 1). 1 y (x, y) y − b (0, b) x − 0 Let L be a nonvertical line with slope m and y-intercept b. Therefore, (0, b) is a point on L. Let (x, y) be any other point on L, as shown in Figure 1.4-6. Use the points (0, b) and (x, y) to compute the slope of L Multiply both sides of the equation by x, and solve for y. mx y b y mx b Figure 1.4-6 Thus, the coordinates of any point on L satisfy the equation which leads to the following. y mx b, Slope-Intercept Form The line with slope m and y-intercept b is the graph of the equation y mx b. Did you notice, as you recall your work with arithmetic sequences, that the explicit form of the sequence looks very similar to y mx b? Example 3 Graphs of Arithmetic Sequences and Lines and 8. Use the The first three terms of an arithmetic sequence are explicit form of the sequence to express the nth term, and compare it to 2, 3, 34 Chapter 1 Number Patterns the slope-intercept form of the equation of a line that passes through the points on the graph of the sequence. Graph both the sequence and the corresponding line on the same set of axes. Solution 25 The common difference for the sequence is 5 and explicit form is u1 2. Therefore, the 3 5 10 Figure 1.4-7 Connection between Arithmetic Sequences and Lines un n 1 u1 2 ˛d 1 n 1 2 2 ˛5 2 5n 5 5n 7 1 The last equation, un has the form y mx b, where m d 5 b u0 d 7 and x corresponds to n. 5n 7, u1 Figure 1.4-7 shows graphs of the sequence and the line. ■ Below is an important summary connecting the explicit form of an arithmetic sequence to the slope-intercept form of a |
line. The connection between the explicit form of an arithmetic sequence, a line, (n 1)˛d, is as follows. u1 un y mx b, and the slope-intercept form of • The slope of the line corresponds to the common difference of the sequence, m d. • The y-intercept represents the value of the first term of the sequence minus the difference, b u1 d. A linear equation expressed in slope-intercept form defines a relation for all the ordered pairs (x, y) on the line. The equation represents the rule and each x represents an input. For every input x there is one and only one output y, so y is a function of x. The graph of a linear function f is the graph of mx b. x y f˛ 1 2 Most graphing calculators are called function graphers because they graph a relation only if it can be expressed as a function Thus, slopeintercept form is useful for graphing a line both on paper and with a graphing calculator. y f ˛1 x. 2 Example 4 Graphing a Line Sketch the graph of ing calculator. 2y 5x 2, and confirm your sketch with a graph- y 6 Solution Begin by solving the equation for y. 2y 5x 2 2y 5x 2 y f 1 5 2 x 2 Section 1.4 Lines 35 x 1 2 Figure 1.4-8a 8 x Therefore, the graph is a line with slope 5 2, the coefficient of x, and y-intercept 1, the constant term. Because the y-intercept is 1, the point (0, 1) is on the line. When x 2, then, so (2, 6) is also on the line. Plotting and connecting the points (0, 1) and (2, 6) produces the line in Figure 1.4-8a. Figure 1.4-8b displays the same graph produced by a graphing calculator. ■ 3 5 Example 5 Linear Depreciation 1 Figure 1.4-8b An office buys a new computer system for $7000. Five years later its value is $800. Assume that the system depreciates linearly. a. Write the equation that represents value as a function of years. b. Find its value two years after it was purchased, that is, the y-value when x 2. c. Graph the equation. d. Find how many years before |
the system is worthless, that is, the x-value that corresponds to a y-value of 0. Solution a. Linear depreciation means that the equation that gives the value y of the computer system in year x has the form for some constants m and b. Because the system is worth $7000 new (when x 0 ) the y-intercept is 7000 and the equation can be written as y mx b y mx 7000 y mx b y 800 when Because the system is worth $800 after 5 years (i.e., x 5, 2 y mx 7000 800 m 5 7000 6200 5m m 6200 5 1240 The depreciation equation is the value of the system after x years. y 1240x 7000, where y represents 36 Chapter 1 Number Patterns 10,000 0 0 10 Figure 1.4-9 b. The value of the system after two years 1 x 2 7000 is 2 y 1240 $4520. 2 1 2 c. The graph of y 1240x 7000 1240, is shown in Figure 1.4-9. Notice represents the depreciation of the that the slope of the line, system per year. That is, the value of the system decreases $1240 each year. d. The trace feature or the zero feature of a graphing calculator shows that the x-value corresponding to Figure 1.4-9. That is, the system will be worthless in 5 years and 0.6 12 7.2 as shown in months. is y 0 x 5.6, ■ The Point-Slope Form Suppose the line L passes through the point and has slope m. Let x1, y12 (x, y) be any other point on L. Use the fixed point x1, y12 and the variable 1 to compute the slope m of L, and it will generate another useful point x x1, form of a line called the point-slope form. For x, y 1 2 1 y y1 x x1 y y1 m slope of L x x12 m 1 Thus, the coordinates of every point on L satisfy the equation y y1 m. x x12 1 Point-Slope Form The line with slope m through the point of the equation (x1, y1) is the graph y y1 m (x x1). There are two interesting observations about point-slope form. 1. Although slope-intercept form and point-slope form can be used to write the equation of |
a line, the point-slope form is easier to use, unless you know the y-intercept. 2. The point-slope form can also be used to graph a line because any point of the line can be used as the initial point and the remaining points can be found by using the equation’s slope. The slope determines how to find a second point from the initial point by moving vertically an amount equal to the numerator of the slope, which represents and then moving horizontally an amount equal to the denominator of the slope, which represents ¢x. ¢y, y 0 1 −1 −2 −3 −4 −5 −6 −7 x 1 765432 2 1 Figure 1.4-10 Section 1.4 Lines 37 Example 6 Point-Slope Form of a Line Sketch the graph and find the equation of the line that passes through the point with slope 2. Write the equation in slope-intercept form. 1, 6 1 2 Solution and identify another point on To graph the line, start at the point 1 the line by moving 2 units vertically and 1 unit horizontally. The point 2, 4 is also on the line. Because a unique line is determined by two 1 points, connecting the points produces the line, as and shown in Figure 1.4-10. 2, 4 1, 6 2 2 1 1 2 2 1, 6 To find the equation of the line, substitute 2 for m and in the point-slope equation. m˛1 2 1 1 y 6 2x 2 y 2x 8 x x12 x 1 2 y y1 6 Point-slope form y Slope-intercept form 2 1, 6 1 2 for x1, y12 1 ■ Vertical and Horizontal Lines When a line has 0 slope, it is called a horizontal line, and it can be written as y 0x b b. Example 7 Equation of a Horizontal Line Describe and sketch the graph of the equation y 3. x Solution y 3 can be written as its graph is a line with slope Because 0 and y-intercept 3. This is sufficient information to obtain the graph shown in Figure 1.4-11. y 0x 3, ■ Vertical Lines The preceding discussion does not apply to vertical lines, whose equations have a different form than those examined earlier because a vertical line is not a function. Example 8 Equation of a Vertical Line x y 1 −1 −3 2 Figure 1.4-11 |
y 3 2 1 0 −1 1 3 4 Find the equation of the vertical line shown in Figure 1.4-12. −2 Solution Figure 1.4-12 Every point on the vertical line in Figure 1.4-12 has first coordinate 2. and the line is the graph Thus, every point on the line satisfies x 0y 2, 38 Chapter 1 Number Patterns of the equation x 2. (2, 1) and (2, 4), you obtain If you try to compute the slope of the line, say using 4 1 2 2 3 0, which is not defined. ■ Parallel and Perpendicular Lines The slope of a line measures how steeply it rises or falls. Because parallel lines rise or fall equally steeply, their slopes are the same. Two lines that meet in a right angle, that is, a angle, are said to be perpendicular. There is a close relationship between the slopes of two perpendicular lines. 90° Parallel and Perpendicular Lines Two nonvertical lines are parallel when they have exactly the same slope. Two nonvertical lines are perpendicular when the product of their slopes is 1. Example 9 Parallel and Perpendicular Lines Given the line M whose equation is 2, 1 of the lines through the point 1 2 3x 2y 6 0, find the equation a. parallel to M. b. perpendicular to M. Solution First find the slope of M by rewriting its equation in slope-intercept form. 3x 2y 6 0 2y 3x 6 y 3 2 x 3 Therefore, M has slope 3 2. a. The line parallel to M must have the same slope, and because is on the parallel line, use the point-slope form to find its 2, 1 1 equation. 2 y y y1 x12 x 2 2 1 x 3 x 4 Section 1.4 Lines 39 b. The line perpendicular to M must have slope 2 3 1 3 2. Use point- slope form again to find the equation of the line perpendicular to M through 2 ■ The standard form of a line is where A, B, and C are integers, both 0., and A and B are not Ax By C, A 0 equation shown in Example 9, for the perpendicular line, Any line, including vertical lines, can be written in this form. The last x 1 3 can be written in standard form by multiplying both sides by 3, the least common denominator of all the terms, and then adding 2x to both sides. The resulting |
equation is 2x 3y 1. y 2 3, Standard Form of a Line NOTE The standard form of a line is sometimes called the general form of a line and may also be written as Ax By C 0. The following box summarizes the different forms of the equation of a line and when each form is best used. Forms of Linear Equations The forms of the equation of a line are Ax By C standard form y mx b slope-intercept form Graphing y y1 m(x x1) point-slope form Write equations A horizontal line has slope 0 and an equation of the form y b. A vertical line has undefined slope and an equation of the form x c. 40 Chapter 1 Number Patterns Exercises 1.4 1. For which of the line segments in the figure is the slope a. largest? b. smallest? c. largest in absolute value? d. closest to zero? y A In Exercises 7–10, find the slope of the line through the given points. 7. 9. 1, 2 ; 1 2 1 3. 10. 1 A 1, 2 ; 2 1 2, 1 2 22, 1 ; 1 B 2, 9 2 In Exercises 11–14, find a number t such that the line 2. passing through the two given points has slope B C D E x 11. 13. 1 1 0, t ; 9, 4 1 2 t 1, 5 2 ; 1 2 6, 3t 7 2 12. 14. 1 1 1, t ; 2 1 3, 5 2 t, t ; 1 2 5, 9 2 2. The doorsill of a campus building is 5 ft above ground level. To allow wheelchair access, the steps in front of the door are to be replaced by a straight ramp with constant slope 1 12, as shown in the figure. How long must the ramp be? [The answer is not 60 ft.] 15. Let L be a nonvertical straight line through the origin. L intersects the vertical line through (1, 0) at a point P. Show that the second coordinate of P is the slope of L. 16. On one graph, sketch five line segments, not all meeting at a single point, whose slopes are five different positive numbers. In Exercises 17–20, find the equation of the line with slope m that passes through the given point. 17. 19. 3, 5 2 m 1; 1 m 1; 6, 2 2 1 |
18. m 2; 20. m 0; 2, 1 2 4, 5 1 1 2 In Exercises 21–24, find the equation of the line through the given points. 21. 0, 5 1 and 1 2 3, 2 2 22. 4, 3 2 1 and 2, 1 1 2 23. 4 3 a, 2 3b and 1 3 a, 3 b 24. (6, 7) and (6, 15) In Exercises 25–28, determine whether the line through P and Q is parallel or perpendicular to the line through R and S, or neither. 25. P, Q 2, 5 2 1 1 1, 1 2 and R, S 4, 2 1 2 6, 1 1 2 26. P 3 2b 0, a, Q 1, 1 1 2 and R, S 2, 7 1 2 3, 9 1 2 R a m p 5 In Exercises 3–6, find the slope and y-intercept of the line whose equation is given. 3. 2x y 5 0 4. 3x 4y 7 5. 3 6 27. P S P R 28., Q 1, 1 1 2 and R 2, 0, 2 1 1 3, 3b a 4, 2 a 3b 3, 3 2 2, 2, Q 1, S 3, 1 4, 5 2 1 2 2 1 1 and Section 1.4 Lines 41 In Exercises 29–31, determine whether the lines whose equations are given are parallel, perpendicular, or neither. 1 10 2 7 3 4 4 1 5 2 29. 2x y 2 0 and 4x 2y 18 0 30. 3x y 3 0 and 6x 2y 17 0 31. y 2x 4 and 0.5x y 3 47. The first three terms of an arithmetic sequence are 5. 7, 1, and Write the sequence’s equation in slope-intercept form. 32. Use slopes to show that the points, 1, 12 7, 0 and 2 1 2 4, 6, 2 1 straight line. 1 all lie on the same 33. Use slopes to determine if (9, 6),, are the vertices of a right triangle. 1, 3 1 2 1, 2 and 1 2 34. Use slopes to show that the points 5, 2 1 1 of a parallelogram. 3, 1, 2 1 2 3, 0, and 5, 3 1 2 2 are |
the vertices 48. For a given arithmetic sequence, the common u1 difference is y-intercept of the graph of this sequence. Find the slope and 6. and 3 49. For a given arithmetic sequence, the common u1 difference is 8 and y-intercept of the graph of this sequence. Find the slope and 2. 50. Let L be a line that is neither vertical nor horizontal and which does not pass through the origin. Show that L is the graph of x a y b 1, In Exercises 35–42, find an equation for the line satisfying the given conditions. where a is the x-intercept and b is the y-intercept of L. 35. through 2, 1 1 2 with slope 3 36. y-intercept 7 and slope 1 37. through 38. through 1 1 2, 3 2 1, 2 and parallel to 3x 2y 5 and perpendicular to y 2x 3 2 39. x-intercept 5 and y-intercept 5 40. through 5, 2 2 (1, 2) and (4, 3) 1 and parallel to the line through 41. through 1, 3 through (0, 1) and (2, 3) 1 2 and perpendicular to the line 42. y-intercept 3 and perpendicular to 2x y 6 0 43. Find a real number k such that kx 2y 7 0. line 1 3, 2 2 is on the 44. Find a real number k such that the line 3x ky 2 0 has y-intercept 3. 45. Write the equation for the given arithmetic sequence in slope-intercept form. 1 2 2 2 3 6 4 10 5 14 46. Write the equation for the given arithmetic sequence in slope-intercept form. 51. Let A, B, C, and D be nonzero real numbers. Show that the lines Ax By D 0 are parallel. Ax By C 0 and 52. Sales of a software company increased linearly from $120,000 in 1996 to $180,000 in 1999 a. Find an equation that expresses the sales y in year x (where x 0 corresponds to 1996). b. Estimate the sales in 2001. 53. The poverty level income for a family of four was $9287 in 1981. Due to inflation and other factors, the poverty level income rose to approximately $18,267 in 2001. (Source: U.S. Census Bureau) a. Find a linear equation that approximates the |
x 0 poverty level income y in year x (with corresponding to 1981). b. Use the equation of part a to estimate the poverty level income in 1990 and 2005. 54. At sea level, water boils at 1100 ft, water boils at between boiling point and height is linear. a. Find an equation that gives the boiling point The relationship At a height of 210° F. 212° F. y of water at a height of x feet. Find the boiling point of water in each of the following cities (whose altitudes are given). b. Cincinnati, OH (550 ft) c. Springfield, MO (1300 ft) d. Billings, MT (3120 ft) e. Flagstaff, AZ (6900 ft) 42 Chapter 1 Number Patterns 55. A small plane costs $600,000 new. Ten years later, 61. A hat company has fixed costs of $50,000 and it is valued at $150,000. Assuming linear depreciation, find the value of the plane when it is 5 years old and when it is 12 years old. 56. In 1950, the age-adjusted death rate from heart disease was about 307.2 per 100,000 people. In 1998, the rate had decreased to 126.6 per 100,000. a. Assuming the rate decreased linearly, find an equation that gives the number y of deaths per x 0 100,000 from heart disease in year x, with corresponding to 1950. Round the slope of the line to one decimal place. b. Use the equation in part a to estimate the death rate in 1995 and in 2005. 57. According to the Center of Science in the Public Interest, the maximum healthy weight for a person who is 5 ft 5 in. tall is 150 pounds and for someone 6 ft 3 in. tall is 200 pounds. The relationship between weight and height here is linear. a. Find a linear equation that gives the maximum healthy weight y for a person whose height is x x 0 inches over 5 ft 5 in. ( corresponds to 5 ft to 5 ft 7 in., etc.). 5 in., b. Use the equation of part a to estimate the x 2 maximum healthy weight for a person whose height is 5 ft and for a person whose height is 6 ft. 58. The profit p (in thousands of dollars) on x thousand units of a specialty item is p 0.6x 14.5. The cost c of manufacturing x c 0.8x 14.5. items is given by a |
. Find an equation that gives the revenue r from selling x items. b. How many items must be sold for the company to break even (i.e., for revenue to equal cost)? variable costs of $8.50 per hat. a. Find an equation that gives the total cost y of producing x hats. b. What is the average cost per hat when 20,000 are made? 50,000? 100,000? Use the graph and the following information for Exercises 62–64. Rocky is an “independent” ticket dealer who markets choice tickets for Los Angeles Lakers home games (California currently has no laws against scalping). Each graph shows how many tickets will be demanded by buyers at a particular price. For instance, when the Lakers play the Chicago Bulls, the graph shows that at a price of $160, no tickets are demanded. As the price (y-coordinate) gets lower, the number of tickets demanded (x-coordinate) increases. e c i r P 160 140 120 100 80 60 40 20 0 Bulls Suns Mavericks 10 20 30 40 Quantity 62. Write a linear equation expressing the quantity x of tickets demanded at price y when the Lakers play the indicated team. a. Dallas Mavericks b. Phoenix Suns c. Chicago Bulls Hint: In each case, use the points where the graph crosses the two axes to determine its slope. 59. A publisher has fixed costs of $110,000 for a 63. Use the equations from Exercise 62 to find the mathematics text. The variable costs are $50 per book. The book sells for $72. Find equations that give the required information. a. the cost c of making x books b. the revenue r from selling x books c. the profit p from selling x books d. the publisher’s break-even point (see Exercise 58b) number of tickets Rocky would sell at a price of $40 for a game against the indicated team. a. Mavericks b. Bulls 64. Suppose Rocky has 20 tickets to sell. At what price could he sell them all when the Lakers play the indicated team. a. Mavericks b. Suns 60. If the fixed costs of a manufacturer are $1000 and it costs $2000 to produce 40 items, find a linear equation that gives the total cost of making x items. Section 1.5 Linear Models 43 1.5 Linear Models Objectives • Algebraically fit a linear model People working in business, medicine, agriculture, and other fields frequently want to know the |
relationship between two quantities. For instance, • Calculate finite differences How does money spent on advertising affect sales? and use residuals to determine the model of best fit • Use a calculator to determine a linear model • Find and interpret the correlation coefficient for a model • Create and interpret a residual plot for a linear model What effect does a fertilizer have on crop yield? How much do large doses of certain vitamins lengthen life expectancy? In many such situations there is sufficient data available to construct a mathematical model, such as an equation or graph, which demonstrates the desired relationship or predicts the likely outcome in cases not included in the data. In this section applications are considered in which the data can be modeled by a linear equation. More complicated models will be considered in later sections. When you are given a set of data points, you should first determine whether a straight line would be a good model for the data. This can be done graphically by making a scatter plot of the data, as shown in Figures 1.5-1a and 1.5-1b. Visual inspection suggests that the data points in Figure 1.5-1a are approximately linear but that those in Figure 1.5-1b are not. So a line would be a good model for the data points in Figure 1.5-1a, but for those in Figure 1.5-1b a line is not a good model. y y x x Figure 1.5-1a Figure 1.5-1b You can also determine whether a line is a good model for a given set of data points, without graphing, by using finite differences. To understand y 3x 1, the idea, consider the equation whose graph is known to be a line. Consider the table of values shown on the next page and look at the difference between each y-entry and the preceding one. The differences are the same; all of them are equal to the slope of the line y 3x 1. This fact suggests that if the successive differences of the y-coordinates of the data points are approximately equal, then a line should be a good model for the data. 44 Chapter 1 Number Patterns x 1 2 3 4 5 y 3x 1 2 5 8 11 14 Difference 5 2 3 8 5 3 11 8 3 14 11 3 Example 1 Linear Data Estimated cash flows from a company over the five-year period 1988– 1992 are shown in the table. Year 1988 1989 1990 1991 1992 Cash flow per share ($) 2.38 2.79 3 |
.23 3.64 4.06 Determine whether a line would be a good model for this data. Use two different methods. a. Calculate the finite differences for the data points. b. Draw a scatter plot of the data. Solution Year Cash Flow Differences a. Subtract each cash flow from the preceding one and record the difference, as shown in Figure 1.5-2a. 0.41 0.44 0.41 0.42 Because the differences are approximately equal, a line is a good model for this data. b. Let x 0 correspond to 1988. The scatter plot for the data points is shown in Figure 1.5-2b, where the points appear to be linear. Therefore, a line is a reasonable model. ■ 2.38 2.79 3.23 3.64 4.06 Figure 1.5-2a 1988 1989 1990 1991 1992 5 Once it has been determined that a line would be a good model for a set of data points, there are several ways to determine an appropriate model. The simplest way is to choose two of the data points and find the equation of the line that includes the points. This may require some experimenting to see which two points appear to produce a line that fits the data well. The number of data points above the line should balance with the number of data points below the line. Of course, there are many choices of two points and many possible lines that model the data. So there must be some way of determining which line fits the data best. Modeling Terminology Suppose (x, r) is a data point and that the corresponding point on the is called a residual. Residuals model is (x, y). Then the difference r y 0 0 Figure 1.5-2b 5 Section 1.5 Linear Models 45 are a measure of the error between the actual value of the data, r, and the value y given by the model. Graphically, the residual is the vertical distance between the data point (x, r) and the model point (x, y), as shown in Figure 1.5-3. y (x, r) Data point Residual r − y (x, y) Model point x Figure 1.5-3 The residual represents a directed distance that is positive when the data point is above the model point and negative when the data point is below the model point. When the sum of the residuals is 0, which indicates that the positive and negative errors cancel out |
each other, the model is probably a reasonable one. However, this is not always enough to determine which of several models is best because their residuals may all have the same sum. Consequently, to find which model among several fits the data best, use the sum of the squares of the residuals because this sum has no negative terms and no canceling. Using the sum of the squares as a measure of accuracy has the effect of emphasizing large errors, those with absolute value greater than 1, because the square is greater than the residual. It minimizes small errors, those with absolute value less than 1, because the square is less than the residual. Example 2 Modeling Data The data below shows the weekly amount spent on advertising and the weekly sales revenue of a small store over a seven-week period. Advertising Expenditure x (in hundreds of dollars) Sales revenue (in thousands of dollars Find two models for the data, each determined by a pair of data points. Then use residuals to see which model best fits the data. 46 Chapter 1 Number Patterns Solution Let the two models be denoted as A and B. For Model A, use the points (1, 2) and (3, 3). The slope of the line through these points is m 3 2 3 1 1 2 0.5 The equation of the line through (1, 2) and (3, 3) is y 2 0.5 x 1 2 1 y 0.5x 1.5 Model A is shown in Figure 1.5-4 and its residuals are shown in the table below. Notice that the sum of the squared residuals is 2. Data point (x, r) 0, 1 1, 2 2, 2 3, 3 4, 3 5, 5 6 Model point (x, y) 0, 1.5 1, 2 2 1 2, 2.5 3, 3 2 1 4, 3.5 5, 4 2 1 6, 4.5 1 1 1 1 2 2 2 2 Residual r y Squared residual (r y)2 0.5 0 0.5 0 0.5 1 0.5 Sums 0 0.25 0.25 0.25 0.25 0.25 1.25 0.25 2.00 For Model B, use the point (1, 2) and (6, 5). The slope of the line through these points is 5 2 6 1 3 5 y 2 0.6˛ 0.6 and its equation is x |
1 1 2 or y 0.6 x 1.4 Figure 1.5-5 shows the graph of Model B and the table below shows its residuals. The sum of the squared residuals is 1.56. Data point (x, r) 0, 1 1, 2 2, 2 3, 3 4, 3 5, 5 6 Model point (x, y) Residual r y Squared residual (r y)2 1 1 1 1 1 0, 1.4 1, 2 2 1 2, 2.6 3, 3.2 4, 3.8 5, 4.4 6.4 0 0.6 0.2 0.8 0.6 0 1.4 Sums 0.16 0.16 0.36.04 0.64 0.36 0.16 1.56 Because this sum is smaller that the sum for Model A, conclude that Model B is the better of the two models Advertising 6 Figure 1.5-4 2 4 Advertising 6 Figure 1.5-5 NOTE If needed, review how to compute linear regression equations in the Technology Appendix. Section 1.5 Linear Models 47 Least–Squares Regression Lines It can be proved that for any set of data there is one and only one line for which the sum of the squares of the residuals is as small as possible. Such a line is called the least–squares regression line, and the computational process for finding it is called linear regression. Most graphing calculators have the linear regression process built-in. For example, Figure 1.5-6 shows the approximate least–squares regression line for the data in Example 2. y 0.679x 0.964 The sum of the squared residuals for this model is approximately 1.107, slightly less than the corresponding sum for Model B in Example 2. The number r in Figure 1.5-6, which is called the correlation coefficient, is a statistical measure of how well the least–squares regression line fits 1 the data points. The value of r is always between and 1, the closer the r absolute value of r is to 1, the better the fit. When the fit is perfect: all the data points are on the regression line. Conversely, a correlation coefficient near 0 indicates a poor fit. 1, 0 0 r2 is called the coefficient of determination. It is the proportion of variation in y that can be attributed to a linear relationship between x and y in the data. Example 3 |
Modeling Data A circle can be circumscribed around any regular polygon. The lengths of the radii of the circumscribed circles around regular polygons whose sides have length of one unit are given as follows. Number of sides Radius 3 4 5 6 7 8 9 0.577 0.707 0.851 1.00 1.152 1.306 1.462 a. Draw a scatterplot. b. Calculate the finite differences for the data points. c. Find the model that best fits the data using the regression feature on a calculator. d. What does the correlation coefficient indicate about the data? Figure 1.5-6 Technology Tip Many calculators have a List command that ¢ can be entered into the label cell of a list. This will automatically calculate the differences between the items of the list specified. For example, List(L2) produces a list of differences between the items of list L2. ¢ 48 Chapter 1 Number Patterns Solution a. A scatter plot of the data is shown in Figure 1.5-7a. b. Subtract each radius from the preceding one and record the differences in a list. Notice that the differences are approximately equal, as shown in Figure 1.5-7b. 1.5 0 0 10 Figure 1.5-7a Figure 1.5-7b c. Use the linear regression feature to obtain Figure 1.5-7c, which shows that the least–squares regression line is approximately y 0.15x 0.12. d. The correlation coefficient, r 0.9996308197, is very close to 1, which indicates that this linear model is a very good fit for the data. ■ Although only linear models are constructed in this section, you should always allow for the possibility that a linear model may not be the best choice for certain data. The least–squares regression line may give a reasonable model, which is the line that fits the data best, but there may be a nonlinear equation that is an even better model for the data. Polynomial models, for example, are presented in Chapter 4. Figure 1.5-7c In addition to finding finite differences and a scatter plot of the data, another way to check that a linear model is appropriate is to construct a x, r y, scatter plot of the residuals. In other words, plot the points 2 x, y is the corresponding point on the where model. The general rule is two fold: is a data point |
and x, r 1 1 2 1 2 Use a linear model when the scatter plot of the residuals shows no obvious pattern, as shown in Figure 1.5-8. Use a nonlinear model when the scatter plot of the residuals has a pattern, as shown in Figure 1.5-9. r y x x Figure 1.5-8 Figure 1.5-9 Technology Tip The regression equation is stored in a variable that is usually called RegEQ each time the regression coefficients are calculated. The residuals are stored in a variable called RESID each time a regression is performed. See the Technology Appendix for specific instructions for graphing a regression line and residuals. Section 1.5 Linear Models 49 Example 4 Linear Regression and Residuals A local resident owns an espresso cart and has asked you to provide an analysis based on last summer’s data. To simplify things, only data for Mondays is provided. The data includes the amount the workers were paid each day, the number of cups sold, the cost of materials, and the total revenue for the day. The owner also must spend $40 each operating day on rent for her location and payment toward a business loan. Sales taxes have been removed from the data, so you need not consider them, and amounts have been rounded to the nearest dollar. Date Salaries ($) Cups sold Material cost ($) Total revenue ($) June 02 June 09 June 16 June 23 June 30 July 07 July 14 July 21 July 28 August 04 August 11 August 18 August 25 68 60 66 63 63 59 57 61 64 58 65 57 64 112 88 81 112 87 105 116 122 100 80 96 108 93 55 42 33 49 38 45 49 52 48 36 42 52 47 202 119 125 188 147 159 165 178 193 112 158 162 166 a. Find a linear regression model for the daily revenue as a function of the number of cups sold. b. Use a scatter plot of the residuals to determine if the linear model is a good fit for revenue. c. Find a linear regression model for the daily cost as a function of the number of cups sold. Be sure to include the pay for the workers, the fixed daily cost, and the cost of the material. d. Draw a scatter plot of the residuals to determine if the proposed model is a good fit for cost. e. Find the break-even point, that is, when revenue is equal to cost. 50 Chapter 1 Number Patterns Solution a. Enter two lists in your calculator, using the column labeled “C |
ups Sold” and “Total Revenue” in the chart. That is, the data points are (112, 202), (88, 119), and so on. Next, use the linear regression function on these lists to approximate the least–squares regression line. y 1.586x 0.895 Store its equation as 1.5-10a and 1.5-10b. y1 in the equation memory, as shown in Figure Figure 1.5-10a Figure 1.5-10b b. To obtain a scatter plot of the residuals for the least squares regression line for revenue, plot the points whose first coordinates are given by the CUPS list and whose second coordinates are given by the RESID, the variable that holds the residuals each time a regression is performed. 35 75 −35 125 Figure 1.5-11a Figure 1.5-11b c. First, create a new list that shows the total cost each day. For June 2, Salaries Material Cost Rent/Loan Costs Daily Cost $68 $55 $40 $163 Compute the total daily cost for each date, as shown in Figure 1.5-12a, where the salaries list is called PAY, the materials cost list is called COST, and TOTCO is the total daily cost list. To find a model for the total daily cost as a function of cups sold, use the data points given by the lists CUPS and TOTCO with the regression feature. Find the closest approximate least–squares regression line. y 0.395x 107.659 Technology Tip Placing RESID into Ylist will use the last computed regression’s residual values in a scatter plot. Figure 1.5-12a Technology Tip A formula can be placed in the upper cell of a list to perform the operation on all the elements of the list. Section 1.5 Linear Models 51 Store this equation as 1.5-12b. y2 in the equation memory, as shown in Figure Figure 1.5-12b d. Use the same procedure as in part b to obtain the scatter plot of the residuals for cost shown in Figure 1.5-12c. It shows no obvious pattern, which again indicates that a linear model is a good choice for this data. 30 80 30 Figure 1.5-12c 125 320 60 40 e. The break-even point occurs when revenue is equal to cost. Plot the revenue equation found in part a and |
the cost equation found in part c on the same screen, and find the x-coordinate of their intersection (shown in Figure 1.5-13). Since 89.6 cups cannot be sold, 90 cups a day must be sold to break even. ■ 180 Figure 1.5-13 Example 5 Prediction from a Model The total number of farm workers (in millions) in selected years is shown in the following table. Year Workers Year Workers Year Workers 1900 1920 1930 1940 29.030 42.206 48.686 51.742 1950 1960 1970 1980 59.230 67.990 79.802 105.060 1985 1990 1994 106.210 117.490 120.380 a. Use linear regression to find an equation that models the data. Use the equation to estimate the number of farm workers in 1975 and in 2000. 52 Chapter 1 Number Patterns b. According to the model, when will the number of workers be 150 million? c. Is a line the best model for the data? Solution x 0 Let correspond to 1900 and enter the data into two lists. Perform linear regression on the data, and display the scatter plot of the data together with the graph of the least–squares regression line. 100 y 1.0116x 18.3315 Figure 1.5-14 As suggested by Figure 1.5-14, the regression line provides a reasonable model for approximating the number of farm workers in a given year. a. If If x 75, x 100, then then y 1.0116 1 y 1.0116 1 18.3315 94.202 75. 2 18.3315 119.492. 100 2 Therefore, there were approximately 94,202,000 farm workers in 1975 and 119,492,000 in 2000. 130 0 0 15 −5 −15 Figure 1.5-15 b. To determine when the number of workers will be 150 million, solve the regression equation when y 150. 1.0116x 18.3315 150 100 1.0116x 131.6685 x 131.6685 1.0116 130.159 There will be 150 million farm workers in approximately 2030. c. As shown in Figure 1.5-15, there seems to be a pattern in the residuals, so there is a better, nonlinear model for the data. Nonlinear models are discussed in Section 4.3.A. ■ Correlation and Slope The correlation coefficient, r, always has the same sign as the slope of the least squares regression |
line. So when r is negative, the regression line slants downward from left to right. In other words, as x increases, y decreases. In such cases, we say that the data has a negative correlation. When r is positive, the regression line slopes upward from left to right, and the data is said to have a positive correlation. As x increases, y also increases. When r is close to 0 (regardless of sign), there is no correlation between the quantities. Exercises 1.5 1. a. In Example 2, find the equation of the line through the data points (1, 2) and (5, 5). b. Compute the sum of the squares of the errors for this line. Is it a better model than any of the models in the example? Why? 2. The linear model in Example 5 is the least squares regression line with coefficients rounded. Find the correlation coefficient for this model. 3. a. In Example 5, find the slope of the line through the data points for 1920 and 1994. b. Find the equation of the line through these two data points. c. Which model predicts the higher number of farm workers in 2010: the line in part b or the regression line found in Example 5? In Exercises 4–7, determine whether the given scatter plot of the data indicates that there is a positive correlation, negative correlation, or very little correlation. 4. y 5. y 6. y 7. y x x x x Section 1.5 Linear Models 53 8. The U.S. gross domestic product (GDP) is the total value of all goods and services produced in the United States. The table shows the GDP in billions x 0 of 1996 dollars. Let (Source: U.S. Bureau of Economic Analysis) a. Use a scatter plot to determine if the data correspond to 1990. appears to be linear. b. If so, is there a positive or negative correlation? Year 1990 1992 1994 1996 1998 2000 GDP $6707.9 $6880.0 $7347.7 $7813.2 $8495.7 $9318.5 In Exercises 9–13, construct a scatter plot for the data and answer these questions: a. What are the finite differences for the data? b. Do the finite differences confirm that the data is linear? If so, is there a positive or negative correlation? 9. The table shows the monthly premium (in dollars) for a term life |
insurance policy for a female nonsmoker. Let x represent age and y the amount of the premium. Age Premium 25 30 35 40 45 50 55 $11.57 $11.66 $11.83 $13.05 $16.18 $21.32 $29.58 10. The table shows the percent of persons in the United States below the U.S. poverty level in x 0 correspond to 1960. selected years. Let 54 Chapter 1 Number Patterns Year 1960 1965 1970 1975 1980 1985 1990 1992 1994 1996 1997 1998 1999 Percent below poverty level 22.2 17.3 12.6 12.3 13.0 14.0 13.5 14.8 14.5 13.7 13.3 12.7 11.8 11. The vapor pressure y of water depends on the temperature x, as given in the table. Temperature (C) Pressure (mm Hg) 0 10 20 30 40 50 60 70 80 90 100 4.6 9.2 17.5 31.8 55.3 92.5 149.4 233.7 355.1 525.8 760.0 12. The table shows the U.S. Census Bureau’s population data for St. Louis, Missouri in selected years. Let correspond to 1950. x 0 Year Population 1950 1970 1980 1990 2000 856,796 622,236 452,801 396,685 348,189 13. The table shows the U.S. disposable income (personal income less personal taxes) in billions of dollars. (Source: Bureau of Economic Analysis, U.S. Dept. of Commerce). Let 1990. correspond to x 0 Year 1990 1992 1994 1996 1998 1999 2000 Disposable personal income 4166.8 4613.7 5018.9 5534.7 6320.0 6618.0 7031.0 14. The table gives the annual U.S. consumption of beef and poultry, in million of pounds. (Source: U.S. Dept. of Agriculture) Year Beef Poultry 1990 24,031 22,151 1991 24,113 23,270 1992 24,261 24,394 1993 24,006 25,099 1994 25,125 25,754 1995 25,533 25,940 1996 25,875 26,614 Section 1.5 Linear Models 55 a. Make scatter plots for both beef and poultry consumption, using the actual years (1990, 1991, etc.) as x in each case. b. Without graphing, use your knowledge of slopes to determine |
which of the following equations models beef consumption and which one models poultry consumption. Confirm your answer by graphing. 717.46x 1,405,160 329.86x 632,699 y1 y2 15. The table at the bottom of the page gives the a. Find a linear model for this data, using x 0 to correspond to 1950. b. In the unlikely event that the linear model in part a remains valid far into the future, will there be a time when death from heart disease has been completely eliminated? If so, when would this occur? 17. The table shows the share of total U.S. household income earned by the poorest 20% of households and the share received by the wealthiest 5% of households. (Source: U.S. Census Bureau) median weekly earnings of full-time workers 25 years and older by their amount of education. (Source: U.S. Bureau of Labor Statistics) a. Make four scatter plots, one for each educational group, using to 1990. x 0 to correspond b. Four linear models are given below. Match each model with the appropriate data set. y1 y3 20.74x 392 y2 34.86x 543 y4 12.31x 238 15.17x 354 In Exercises 16–22, use the linear regression feature of your calculator to find the required model. Year Lowest 20% Top 5% 1985 1990 1995 1996 4 3.9 3.7 3.7 17.0 18.6 21.0 21.4 a. Find a linear model for the income share of the poorest 20% of households. b. Find a linear model for the income share of the wealthiest 5% of households. 16. The table shows the number of deaths per 100,000 c. What do the slopes of the two models suggest people from heart disease. of each? Year 1950 1960 1970 1980 1990 1999 Deaths 510.8 521.8 496.0 436.4 368.3 265.9 d. Assuming that these models remain accurate, will the income gap between the wealthy and the poor grow, stay about the same, or decline in the year 2000? Median Weekly Earnings By Amount of Education No High School Diploma High School Graduate Some College College Graduate $317 $321 $337 $346 $360 $378 $443 $461 $479 $490 $506 $520 $518 $535 $558 $580 $598 $621 $758 $779 $821 $ |
860 $896 $924 Year 1996 1997 1998 1999 2000 2001 56 Chapter 1 Number Patterns 18. The table shows what percent of federal aid is given in the form of loans to students at a particular college in selected years. Year (in which school year begins) Loans (%) 1975 1978 1984 1987 1990 18 30 54 66 78 a. Find a linear model for this data, with x 0 corresponding to 1975. b. Interpret the meaning of the slope and the y-intercept. c. If the model remains accurate, what percentage of federal student aid were loans in 2000? 19. The table shows the percent of federal aid given in the form of grants or work-study programs to students at the college of Exercise 18. Year (in which school year begins) Grants and work-study (%) 1975 1978 1984 1987 1990 82 70 46 34 22 a. Find a linear model for this data, with x 0 corresponding to 1975. b. Graph the model from part a and the model from Exercise 18 on the same axes. What appears to be the trend in the federal share of financial aid to college students? c. In what year is the percent of federal aid the same for loans as for grants and work-study? b. Find a linear model for the data. c. According to the model, what was the average number of take-out meals purchased per person in 1993? in 2000? Average number of annual take-out meals per person 43 48 53 55 57 61 65 Year 1984 1986 1988 1990 1992 1994 1996 21. The table shows the median time, in months, for the Food and Drug Administration to approve a new drug after the application has been made. (Source: U.S. Food and Drug Administration) Year 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 Median time for approval 24.3 22.1 22.6 23.0 17.5 15.9 14.3 13.4 12.0 11.6 15.6 a. Make a scatter plot of the data, with x 0 20. The table gives the average number of takeout corresponding to 1990. meals per person purchased at restaurants in selected years. (Source: NPD Group’s Crest Service) a. Make a scatter plot of the data, with x 0 corresponding to 1980. b. Find a linear model for the data. c. What are the limitations of this model? Hint: What does it say about approval time in the year 2009? 22. The ordered pairs below give production ( |
x) and consumption ( y) of primary energy in quadrillion BTUs for a sample of countries in 1995. Australia (7.29, 4.43) Brazil (4.55, 6.76) Canada (16.81, 11.72) China (35.49, 35.67) France (4.92, 9.43) Germany (5.42, 13.71) India (8.33, 10.50) Indonesia (6.65, 3.06) Iran (9.35, 3.90) Japan (3.98, 21.42) Mexico (8.15, 5.59) Poland (3.74, 3.75) Russia (39.1, 26.75) Saudi Arabia (20.34, 3.72) South Africa (6.08, 5.51) United States (69.1, 88.28) United Kingdom (10.57, 9.85) Venezuela (8.22, 2.53) a. Make a scatter plot of the data. b. Find a linear model for the data. Graph the model with the scatter plot. c. In 1995, what three countries were the world’s leading producers and consumers of energy? d. As a general trend, what does it mean if a country’s coordinates lie above the linear model? e. As a general trend, what does it mean if a f. country’s coordinates lie below the linear model? Identify any countries whose coordinates appear to differ dramatically from most of the others. 23. The table shows the winning times, in minutes, for men’s 1500-meter freestyle swimming at the Olympics in selected years. Year 1912 1924 1936 1948 1960 1972 1984 1996 Time 22.00 20.11 19.23 19.31 17.33 15.88 15.09 14.94 a. Find a linear model for this data, with corresponding to 1900. x 0 Section 1.5 Linear Models 57 e. Make a residual plot of the model. Is a linear model appropriate for the data? 24. The following table shows, for selected states, the percent of high school students in the class of 2001 who took the SAT and the average SAT math score. State Connecticut Delaware Georgia Idaho Indiana Iowa Montana Nevada New Jersey New Mexico North Dakota Ohio Pennsylvania South Carolina Washington Students who took SAT (%) Average math score 82 67 63 17 60 5 23 33 81 13 4 26 71 57 53 510 499 489 5 |
42 501 603 539 515 513 542 599 539 499 488 527 a. Make a scatter plot of the percent of students who took the SAT (x) versus the average SAT math score (y). b. Find a linear model for the data. c. What is the slope of your linear model? What does this mean in the context of the problem? d. Below is the data on four additional states. How well does the model match the actual figures for these states? State Students taking SAT (%) Average math score b. Kieren Perkins of Australia set the Olympic Oklahoma record of 14.72 minutes in 1992. How accurately did your model estimate his time? c. How long is this model likely to remain accurate? Why? d. Find the correlation coefficient for the model. Arizona Alaska Hawaii 8 34 51 52 561 525 510 515 58 Chapter 1 Number Patterns 1.6 Geometric Sequences Objectives • Recognize a geometric sequence • Find a common ratio • Graph a geometric sequence • Write a geometric sequence recursively and explicitly • Find partial sums of a geometric sequence Recall that in an arithmetic sequence, each term is obtained from the preceding term by adding a constant, d. A geometric sequence, which is sometimes called a geometric progression, is a sequence in which terms are found by multiplying a preceding term by a nonzero constant. Like an arithmetic sequence, where the difference between consecutive terms is the constant d, the quotient of consecutive terms in a geometric sequence is the constant r. The constant r is called the common ratio of the geometric sequence. Example 1 Recognizing a Geometric Sequence Are the following sequences geometric? If so, what is the common ratio? Write each sequence as a recursive function. 3, 9, 27, 81, p a. 5 6 Solution a. The sequence 5 u2 u1 3, 9, 27, 81, p 9 3 3 6 u3 u2 b 16, e p f is geometric with a common ratio of 3. 27 9 3 u4 u3 81 27 3 Because each term is obtained by multiplying the previous term by 3, the sequence may be denoted as a recursive function. u1 3 and un 3un1 for n 2 b. The sequence 16, e p f is geometric with a common ratio of 1 2. u2 u1 5 4 5 2 1 2 u3 u2 5 8 5 4 1 2 u4 u3 5 16 5 8 1 2 Each term is obtained by multiplying the |
previous term by which, 1 2 gives the following recursive function. u1 5 2 ˛ and un 1 2 ˛un1 for n 2 is a geometric sequence with common ratio r, then for each un6 If the term preceding 5 un is un1 and un un1 r, or equivalently, un run1. ■ n 2 Section 1.6 Geometric Sequences 59 Recursive Form of a Geometric Sequence In a geometric sequence { un }, run1 and some nonzero constant r and all n 2. un for some u1 y 600 500 400 300 200 100 Example 2 Graph of a Geometric Sequence 8. Find the common ratio of the geometric sequence with List the first five terms of the sequence, write the sequence as a recursive function, and graph the function. 2 and u1 u2 Solution Because the sequence is geometric, the common ratio is Therefore, the sequence begins with sive function is given below. 5 x 2, 8, 32, 128, 512, p 2 4 6 Figure 1.6-1 un 4un1, with u1 2 The graph of the function is shown in Figure 1.6-1. u2 u1 8 2 4. and the recur- 6 ■ Notice that the graph of the geometric sequence in Example 2 does not appear to be linear. If the points were connected, the graph would be an exponential function, which is discussed in Chapter 5. Explicit Form of a Geometric Sequence Geometric sequences can also be expressed in a form where the value of the sequence can be determined by the position of the term. Example 3 Writing a Geometric Sequence in Explicit Form Confirm that the sequence defined by expressed as form. 7 n1 un 2 1 2 can also be by listing the first seven terms produced by each with un u1 2un1 7 Solution Using the recursive function, the sequence is 7 u1 u2 u3 u4 u5 u6 2 7 2 14 2 7 2 2 7 22 2 7 23 2 7 24 2 7 25 u1 u2 u3 u4 u5 u6 u7 28 3 56 4 112 5 224 6 448 60 Chapter 1 Number Patterns is u3 7 22, Notice that by the common ratio twice, and that of the sequence multiplied by the common ratio three times. product of general, which is the first term of the sequence multiplied which is the first term un is the power. In and the common ratio, r, raised to the 7 23, n 1 u1 u |
4 is 1 2 Figure 1.6-2a The table in Figure 1.6-2b confirms the apparent equality of the two functions. ■ 7 2n1. un Figure 1.6-2b Explicit Form of a Geometric Sequence implies that The recursive formula for u2 u3 u4 u5 n 2, 3, 4, p u1r u2r u3r u4r u1r 2 u1r 2 u1r 3 1 1 1 r u1r2 r u1r3 2 r u1r 4 2 } is a geometric sequence with common ratio r, then for If { all un n 1, un u1r n1 Example 4 Explicit Form of a Geometric Sequence Write the explicit form of a geometric sequence where the first two terms, and find the first five terms of the sequence. are 2 and 2 5 Solution The common ratio is. Using the explicit form, the geometric sequence can be written as The sequence begins un ˛u1rn1 n1 2 1 2a 1 5b 2, 2 5, 2 52, 2 53, 2 54, p ■ Example 5 Explicit Form of a Geometric Sequence The fourth and ninth terms of a geometric sequence are 20 and the explicit form of the sequence. 640. Find Section 1.6 Geometric Sequences 61 Solution The fourth term can be written as The ninth term can be written as u9 u4 u1rn1, u1rn1, or 20 u1r3. 640 u1r8. or The ratio of the ninth term to the fourth term can be used to find r. u1r8 u1r3 640 20 r5 32 r 2 Substitute 2 for r into the equation defining the fourth term. u11 2 3 20 20 8 2 u1 Thus, un u1 r n1 5 2 ˛ 1 2 2 n1. Partial Sums 5 2 ■ If the common ratio r of a geometric sequence is the number 1, then 1n1u1 u1, u2, u3, p. Therefore, the sequence is just the constant sequence positive integer k, the kth partial sum of this constant sequence is for every n 1. un For any u1 u1 p u1 ku1 ⎫⎪⎪⎪⎬⎪⎪⎪⎭ k terms In other words, the kth partial sum of a constant sequence |
is just k times the constant. If a geometric sequence is not constant (that is, then its partial sums are given by the following formula. r 1, 2 Partial Sums of a Geometric Sequence The kth partial sum of the geometric sequence { mon ratio r 1 is un } with com- k a n1 un u1a 1 r k 1 r b Proof If S denotes the kth partial sum, then using the formula for the nth term of a geometric sequence derives the expression for S. p uk S u1 u2 u1 S rS, u1r u1r2 p u1r k1 as shown on the next page. Use this equation to compute 62 Chapter 1 Number Patterns u1r u1r2 p u1r k1 S u1 rS u1r u1r2 p u1r k1 u1r k u1r k 1 r k S rS u1 1 r S u11 2 2 both sides of the last equation can be divided by 1 r 1, Because complete the proof. S u11 1 r k 1 r 2 u1a 1 r k 1 r b Example 6 Partial Sum Find the sum 3 2 3 4 3 8 3 16 3 32 3 64 3 128 3 256 3 512. 1 r to ■ Solution This is the ninth partial sum of the geometric sequence e 3 1 2b n1, f 2 a where the common ratio is r 1 2. The formula in the box shows that n1 9 a n1 3 2 a 1 2b u1˛a 1 r9 1 r b 3 2b ˛ a 1 a 1 9 1 2b 1 2b a 3 2b a 1 9 1 2b a 3 2 3 2b a 2 ˛a 3b 1 1 ˛a 29b D 1 1 29 T 1 1 § 512 513 512 ¥ ■ Example 7 calculates the distance traveled by the ball discussed in Section 1.2 Example 4 when it hits the ground for the seventh time by using a partial sum of a geometric sequence. Example 7 Application of Partial Sum A ball is dropped from a height of 9 feet. It hits the ground and bounces to a height of 6 feet. It continues to bounce up and down. On each bounce it rises to 2 3 of the height of the previous bounce. How far has the ball traveled (both up and down) when it hits the ground for the seventh time? Section 1.6 Geometric |
Sequences 63 Solution First consider how far the ball travels on each bounce. On the first bounce, it rises 6 feet and falls 6 feets for a total of 12 feet. On the second bounce it rises and falls of the previous height, i.e., it travels of 12 feet. The 2 3 2 3 distance traveled is a geometric sequence with u1 12 and r 2 3. If un denotes the distance traveled on the nth bounce, then un 12 n1 ˛ 2 3b a So un6 5 is a geometric sequence with common ratio r 2 3b a. When the ball hits the ground for the seventh time, it has completed six bounces. Therefore, the total distance it has traveled is the distance it was originally dropped, 9 feet, plus the distance traveled in six bounces. 9 u1 u2 u3 u4 u5 u6 6 9 a n1 un 9 u1˛a 9 12 6 1 r6 1 r b 2 a 3b 1 2 3 1 § ¥ 41.84 feet ■ Exercises 1.6 In Exercises 1–8, determine whether the sequence is arithmetic, geometric, or neither. 1. 2, 7, 12, 17, 22, p 2. 2, 6, 18, 54, 162, p In Exercises 9–14, the first term, and the common ratio, r, of a geometric sequence are given. Find the sixth term and the recursive and explicit formulas for the nth term. u1, 3. 13, 13 2, 13 4, 13 8, p 4. 1, 1 2, 0,, p 1 2 5. 50, 48, 46, 44, p 6. 2, 3, 9 2, 27 4, 81 8, p 7. 3 16, p 8. 6, 3.7, 1.4, 9, 3.2, p 9. u1 5, r 2 10. u1 1, r 2 11. u1 4, r 1 4 12. u1 6, r 2 3 13. u1 10, r 1 2 14. u1 p, r 1 5 In Exercises 15–18, find the kth partial sum of the geometric sequence with common ratio r. un6 5 5, r 1 2 15. k 6, u1 16. k 8, u1 9, r 1 3 64 Chapter 1 Number Patterns 17. k 7, u2 6, r 2 18. k 9 |
, u2 6, r 1 4 In Exercises 19–22, show that the given sequence is geometric and find the common ratio. 19. n 1 2b ea 21. 5n2 5 6 f 20. 22. 23n 6 n 2 3 6 5 5 In Exercises 23–28, use the given information about and recursive the geometric sequence un6 5. and explicit formulas for un to find u5 23. u1 256, u2 64 24. u1 1 6, u2 1 18 25. u2 4, u5 1 16 27. u4 4 5, r 2 5 26. u3 4, u6 32 28. u2 6, u7 192 In Exercises 29–34, find the sum. 29. 31. 33. 7 a n1 2n n 9 a n1 a 1 3b j1 6 a j1 3 2b 4˛a 30. 32. 34. 6 a k1 k 1 2b 3˛a 5 a n1 8 a t1 5 3n1 t1 0.9 6˛1 2 35. For 1987–1998, the annual revenue per share in year n of a company’s stock are approximated by un 1.191 a. Show that the sequence represents 1987. is a geometric 1.71 where n, 2 1 n 7 un6 5 sequence. b. Approximate the total revenues per share for the period 1987–1998. 36. The annual dividends per share of a company’s stock from 1989 through 1998 are approximated, where by the sequence 1989 and 1.1999 a. Show that the sequence bn6 5 0.0228 is a geometric corresponds to bn 1 n 9 n. 2 bn6 5 sequence. b. Approximate the total dividends per share for the period 1989–1998. 37. A ball is dropped from a height of 8 feet. On each bounce it rises to half its previous height. When the ball hits the ground for the seventh time, how far has it traveled? 38. A ball is dropped from a height of 10 feet. On each bounce it rises to 45% of its previous height. When it hits the ground for the tenth time, how far has it traveled? 39. If you are paid a salary of 1¢ on the second day, and your salary March, continues to double each day, how much will you earn in the month of March? on the |
first day of 2¢ 40. Starting with your parents, how many ancestors do you have for the preceding ten generations? 41. A car that sold for $8000 depreciates in value 25% each year. What is it worth after five years? 42. A vacuum pump removes 60% of the air in a container at each stroke. What percentage of the original amount of air remains after six strokes? 43. Critical Thinking Suppose un6 sequence with common ratio un arithmetic sequence with common difference log r. is a geometric r 7 0 and each is an Show that the sequence log un6 7 0. 5 5 44. Critical Thinking Suppose is an arithmetic un6 5 sequence with common difference d. Let C be any positive number. Show that the sequence is a 5 Cd. geometric sequence with common ratio Cun 6 45. Critical Thinking In the geometric sequence 1, 2, 4, 8, 16, p sum of all preceding terms. show that each term is 1 plus the 46. Critical Thinking In the geometric sequence 2, 6, 18, 54, p sum of 1 and all preceding terms. show that each term is twice the of the outstanding balance. If the balance is 47. Critical Thinking The minimum monthly payment for a certain bank credit card is the larger of $5 or 1 25 less than $5, then the entire balance is due. If you make only the minimum payment each month, how long will it take to pay off a balance of $200 (excluding any interest that might accrue)? Important Concepts Section 1.1 Section 1.2 Section 1.3 Section 1.4 Section 1.5 Integers, whole, natural, and real numbers...... 3 Rational and irrational numbers............... 4 Real number line............................ 4 Coordinate plane, x-axis, y-axis, quadrants...... 5 Scatter plots................................ 5 Relation................................... 6 Domain and range.............. |
............. 6 Function................................... 7 Function notation........................... 9 Definition of a sequence..................... 13 Sequence notation.......................... 14 Recursively defined sequence................ 15 Recursive form of an arithmetic sequence...... 22 Explicit form of an arithmetic sequence........ 23 Summation notation........................ 25 Partial sums of an arithmetic sequence......... 27 Graph.................................... 30 Solution of an equation...................... 30 Slope of a line.............................. 32 Properties of slope.......................... 33 Slope–intercept form........................ 33 Connection between arithmetic sequences and lines....................... 34 Point-slope form........................... 36 Vertical and horizontal lines.................. 37 Parallel and perpendicular lines.............. 38 Standard form.......... |
................... 39 Mathematical model........................ 43 Finite differences........................... 43 Residual.................................. 44 Least–squares regression line................. 47 Correlation coefficient....................... 47 Correlation and slope....................... 52 Section 1.6 Recursive form of a geometric sequence........ 59 Explicit form of a geometric sequence.......... 60 Partial sums............................... 61 65 66 Chapter Review Important Facts and Formulas A sequence is an ordered list of numbers. A sequence is defined recursively if the first term is given and there is a method of determining the nth term by using the terms that precede it. A sequence is defined explicitly if terms are determined by their position. An arithmetic sequence is a sequence in which the difference between each term and the preceding term is a constant d. Facts about an arithmetic sequence un6 5 with common difference d: • the recursive form of the sequence is un1 • the explicit form of the sequence is n 2. d for un un • the kth partial sum is k a n1 un k 2 1 u1 n 1 1 d. 2 u1 uk2 and x2, y22 1 is given by The slope of a line that passes through y2 x2 m ¢y ¢x x1, y12 1 y1 x1. The slope-intercept form of the equation of a line is where m is the slope and b is the y-intercept. y mx b, The point-slope form of the equation of a line is where m is the slope and |
y y1 m x x12 1, x1, y12 1 is a given point on the line. Ax By C, where The standard form of the equation of a line is A, B, and C are integers. The equation of a vertical line has the form x h. The equation of a horizontal line has the form y k. Parallel lines have equal slopes. The product of the slopes of perpendicular lines is 1. The difference between an actual data value and a predicted data value is called a residual. The correlation coefficient always has the same sign as the slope of the least squares regression line. A geometric sequence is a sequence in which terms are found by multiplying a preceding term by a nonzero constant r. Facts about the geometric sequence with common ratio r 0: 5 un6 run1 un rn1u1 • the recursive form is for • the explicit form is un or k the kth partial sum is a n1 un un • if r 1, n 2. u1r n1 u1 a 1 r k 1 r b Chapter Review 67 Review Exercises In Exercises 1–10, identify the smallest subset of the real numbers—natural numbers, whole numbers, integers, rational numbers, or irrational numbers— that contains the given number. Section 1.1 1. 23 2. 0.255 6. 3 7. 2121 3. e 8. 4 9 4. 11 9. 5 5. 0 10. 0.255 11. List two real numbers that are not rational numbers. In Exercises 12–15, which sets of points represent a function? Why? 12. 13. 14. 15. 51 51 51 51, 2, 3 2 2, 3, 2, 3 2 2, 3 3, 4, 2 2, 4, 1 3, 3, 2 3, 5 1 2 4, 5 4, 5, 5, 6 6, 7, 2 1 2, 6, 2 1, 2 1 4, 3, 5, 3 6, 3, 2 1 5, 6, 2 1, 2 1 26 2, 7 26 6, 7 26 26 16. Let f be the function given by the rule f following table. 7 2x. Complete the 17. What is the domain of the function g given by g 2t 2 t 3? t 2 1 18. If f x 1 2 0 3 x 0 2x 3 7, then f 7 1 2 f 4 2 1. 19. What is the domain of |
the function given by g r 2r 4 2r 2? 20. What is the domain of the function f x 1 2 2 1 2x 2? 21. The radius of an oil spill (in meters) is 50 times the square root of the time t (in hours). a. Write the rule of a function f that gives the radius of the spill at time t. b. Write the rule of a function g that gives the area of the spill at time t. c. What are the radius and area of the spill after 9 hours? d. When will the spill have an area of 100,000 square meters? 22. The function whose graph is shown below gives the amount of money (in millions of dollars) spent on tickets for major concerts in selected years. (Source: Pollstar) s n o i l l i M 1500 1200 900 600 0 1990 1991 1992 1993 1994 1995 1996 68 Chapter Review a. What is the domain of the function? b. What is the approximate range of the function? c. Over what one-year interval is the rate of change the largest? Use the graph of the function f in the figure below to answer Exercises 23–26. y 1 1 f x 23. What is the domain of f? 24. What is the range of f? 25. Find all numbers x such that f x 1. 2 1 x 1 26. Find a number x such that f possible.) 1 6 f x 1 2 2. (Many correct answers are Use the graph of the function f in the figure to answer Exercises 27–33. y 3 2 1 f −2 −1−1 −2 −3 −5 −4 −3 x 1 2 3 4 5 6 27. What is the domain of f? 28. 3 f 1 2 31. True or false: 2f 32. True or false: 3f 2 2 2 2 1 1 29. 2 2 f 1 2 30 33. True or false: f 3 x 1 2 for exactly one number x. Section 1.2 34. The population of Gallatin is growing at the rate of 2.75% per year. The present population is 20,000. Find a recursive sequence that represents 5910ac01_1-75 9/21/05 2:26 PM Page 69 Chapter Review 69 Gallatin’s population each year. Represent the nth term of the sequence both explicitly and recursively. Find the first seven terms of the sequence. 35 |
. Roberta had $1525 in a savings account 2 years ago. What will be the value of her account 1 year from now, assuming that no deposits or withdrawals are made and the account earns 6.9% interest compounded annually? Find the solution using both a recursive and an explicit formula. 36. Suppose that $3,000 is invested at 6.5% annual interest, compounded monthly. a. What is the balance after 6 years? b. Suppose $150 is added to the account every month. What is the balance after 6 years? 37. The “biological” specimen Geomeuricus sequencius is 5 centimeters long when born. On the second day it grows 3 centimeters. The third day it grows 1.8 centimeters, and on each following day it grows 60% of the previous day’s growth. What is its length after two weeks? What is the maximum length that it could grow? In Exercises 38–39, let 1 25 A un 38. a. Show that 1 b. Show that the first ten terms of Exercises 30–32 Section 1.2) 1. and u1 u2 1 25 n B A 2n25 n B un6 5 are Fibonacci numbers. (See 39. a. For the ninth term, compute the ratio un un1 b. As n gets large, what number does the ratio approach? This number is also referred to as the “golden ratio.” This ratio is believed to have been used in the construction of the Great Pyramid in Egypt, where the ratio equals the sum of the areas of the four face triangles divided by the total surface area. 40. For the sequence un 3un1 1 with u1 1.5, write the first four terms. 41. For the sequence un 3un1 2 with u1 4, write the first five terms. 42. For the sequence un 3un1 with u1 1 9, write the first four terms. Section 1.3 In Exercises 43–46, find a formula for metic. un ; assume that the sequence is arith- 43. u1 3 44. u2 4 and the common difference is –6. and the common difference is 3. 45. u1 5 and u3 7. 46. u3 2 and u7 1. 70 Chapter Review 47. Find the 12th partial sum of the arithmetic sequence with u12 16. u1 3 and 48. Find numbers b, |
c, and d such that 8, b, c, d, 23 are the first five terms of an arithmetic sequence. Section 1.4 49. The national unemployment rates for 1990–1996 were as follows. (Source: U.S. Department of Labor, Bureau of Labor Statistics) Year 1990 1991 1992 1993 1994 1995 1996 Rate (%) 5.6 6.8 7.5 6.9 6.1 5.6 5.4 Sketch a scatter plot and a line graph for the data, letting to 1990. x 0 correspond 50. The table shows the average speed (mph) of the winning car in the Indianapolis 500 race in selected years. Year 1980 1982 1984 1986 1988 1990 1992 1994 1996 Speed (mph) 143 162 164 171 145 186 134 161 148 Sketch a scatter plot and a line graph for these data, letting correspond to 1980. x 0 51. a. What is the y-intercept of the graph of the line defined by y x x 2 5 3 5? b. What is the slope of the line? 52. Find the equation of the line passing through (1, 3) and (2, 5). 53. Find the equation of the line passing through 2, 1 1 2 with slope 3. 54. Find the equation of the line that crosses the y-axis at perpendicular to the line 2y x 5. y 1 and is 55. a. Find the y-intercept of the line defined by 2x 3y 4 0. b. Find the equation of the line through (1, 3) that has the same y-intercept as the line in part a. 56. Sketch the graph of the line defined by 3x y 1 0. 57. Find the equation of the line through ( 4, 5 ) that is parallel to the line through (1, 3) and ( 4, 2 ). 58. As a balloon is launched from the ground, the wind blows it due east. The conditions are such that the balloon is ascending along a straight line with slope 1 5. After 1 hour the balloon is 5000 ft directly above the ground. How far east has the balloon blown? 59. The point (u, v) lies on the line passing through (u, v) and the point y 5x 10 0, 10 1? 2. What is the slope of the line Chapter Review 71 In Exercises 60–66, determine whether the statement is true or false. 60. The graph of x 5y 6 has y-intercept |
6. 61. The graph of 2y 8 3x has y-intercept 4. 62. The lines 3x 4y 12 and 4x 3y 12 are perpendicular. 63. Slope is not defined for horizontal lines. 64. The line in the figure at right has positive slope. 65. The line in the figure does not pass through Quadrant III. 66. The y-intercept of the line in the figure is negative. y 1 1 −1−1 x 67. Which of the following lines rises most steeply from left to right? 68. Which of the following lines is not perpendicular to the line y x 5? a. c. e. a. c. e. y 4x 10 20x 2y 20 0 4x 1 y y 4 x 4 2x 2y 0 y x 1 5 a. c. e. y x y 2x 5 y 2x 5 b. d. y 3x 4 4x y 1 b. d. y x 5 x 1 y b. d. y 4x 7 y 4x 7 69. Which of the following lines does not pass through Quadrant III? 70. Let a and b be fixed real numbers. Where do the lines x a and y b intersect? a. Only at (b, a). c. These lines are parallel, so they don’t intersect. d. If a b, b. Only at (a, b). then these are the same line, so they have infinitely many points of intersection. e. Since these equations are not of the form y mx b, the graphs are not lines. 71. What is the y-intercept of the line 2x 3y 5 0? 72. For what values of k will the graphs of be perpendicular lines? 3y kx 2 0 2y x 3 0 and 73. The average life expectancy increased linearly from 62.9 years for a person born in 1940 to 75.4 years for a person born in 1990. a. Find an equation that gives the average life expectancy y of a person born in year x, with x 0 corresponding to 1940. b. Use the equation in part a to estimate the average life expectancy of a person born in 1980. 74. The population of San Diego grew in an approximately linear fashion from 334,413 in 1950 to 1,151,977 in 1994. 72 Chapter Review a. Find an equation that gives the population y of San Diego in year x, with |
x 0 corresponding to 1950. b. Use the equation in part a to estimate the population of San Diego in 1975 and 2000. In Exercises 75 –78, match the given information with the graph, and determine the slope of each line. y y 300 200 100 1000 800 600 400 200 3 6 a. 9 12 x x y y 300 200 100 600 400 200 1 2 b. 3 4 x x 2 4 6 c. 8 10 12 2 4 d. 6 8 75. A salesman is paid $300 per week plus $75 for each unit sold. 76. A person is paying $25 per week to repay a $300 loan. 77. A gold coin that was purchased for $300 appreciates $20 per year. 78. A CD player that was purchased for $300 depreciates $80 per year. Section 1.5 79. The table shows the monthly premium (in dollars) for a term life insurance policy for women who smoke. Age (years) 25 30 35 40 45 50 55 60 Premium 19.58 20.10 20.79 25.23 34.89 48.55 69.17 98.92 a. Make a scatter plot of the data, using x for age and y for amount of premium. b. Does the data appear to be linear? c. Calculate the finite differences. Do they confirm that the data is linear? Chapter Review 73 80. For which of the following scatter plots would a linear model be reasonable? Which sets of data show positive correlation, and which show negative correlation? y x c. y y a. d. y y x x x x b. e. Exercises 81–82 refer to the following table, which shows the percentage of jobs that are classified as managerial and the percentage of male and female employees who are managers. Year (since 1990) Managerial jobs (%) Female managers (%) Male managers (%) 8 5 2 0 1 3 5 12.32 12.31 12.00 11.83 11.79 11.43 11.09 6.28 6.85 7.21 7.45 7.53 7.65 7.73 16.81 16.67 16.09 15.64 15.52 14.79 14.10 81. a. Make scatter plots of each data set (managerial jobs, female managers, male managers). b. Match the following linear models with the correct data set. Explain your choices. y1 0.11x 7.34 y2 0.09x 11. |
74 y3 0.21x 15.48 82. a. According to the models in Exercise 81, is the percentage of female or male managers increasing at the greater rate? 74 Chapter Review b. Use the models to predict the percentage of female managers and the percentage of male managers in the year 2000. c. What year do the models indicate that the percentage of female managers will surpass the percentage of male managers? 83. The table shows the average hourly earnings of production workers in manufacturing. (Source: U.S. Bureau of Labor Statistics) Year 1991 1993 1995 1997 1999 2001 Hourly earnings ($) 11.18 11.74 12.37 13.17 13.90 14.84 a. Find a linear model for this data, with b. Use the model to estimate the average hourly wage in 1993 and in 2000. The actual average in 1993 was $11.74 and in 2000 it was $14.38. How far off is the model? corresponding to 1990. x 0 c. Estimate the average hourly earnings in 2004. 84. The table shows the total amount of charitable giving (in billions of dollars) in the United States during recent years. (Source: Statistical Abstract of the U.S.: 2001) Year Total charitable giving 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 101.4 107.2 110.4 116.5 119.2 124.0 138.6 153.8 172.1 190.8 203.5 a. Find a linear model for this data, with corresponding to 1990. b. Use your model to estimate the approximate total giving in 2002 and x 0 2005. c. Find the correlation coefficient for the model. d. Make a residual plot of the model. Is a linear model appropriate for the data? Section 1.6 un. In Exercises 85–88, find a formula for Assume that the sequence is geometric. Chapter Review 75 85. u1 2 and the common ratio is 3. 86. u1 5 and the common ratio is 1 2. 87. u2 192 and u7 6. 88. u3 9 2 and u6 243 16. 89. Find the 11th partial sum of the arithmetic sequence with common difference 2. u1 5 and 90. Find the fifth partial sum of the geometric sequence with u1 1 4 and common ratio 3. 91. Find the sixth partial sum of the geometric sequence with u1 5 and common ratio 1 2. 92. Find numbers c and d such that 8, c |
, d, 27 are the first four terms of a geometric sequence. 93. Is it better to be paid $5 per day for 100 days or to be paid the first day, 10 the second day, 20 the third day, and have your salary increase in this fashion every day for 100 days? 5¢ ¢ ¢ 94. Tuition at a university is now $3000 per year and will increase $150 per year in subsequent years. If a student starts school now, spends four years as an undergraduate, three years in law school, and five years earning a Ph.D., how much tuition will she have paid 12 Infinite Geometric Series NOTE Although TI, Sharp, and HP calculators use the letter “u” to denote terms of a sequence, the letter “a” is traditionally used. ctions Calculus is a branch of mathematics that deals with changing quantities. It is based on the concept of quantities that can be approached more and more closely. There are two related branches of calculus: differential calculus and integral calculus. Differential calculus is used to calculate the change in one variable produced by change in a related variable, and integral calculus is used to calculate quantities like the total change of a quantity given its rate of change, area, and volume. The Can Do Calculus features found here and at the end of each chapter are short adventures into the world of calculus. This first Can Do Calculus explores infinite series, which is closely related to infinite sequences. It is an example of the limit process, the fundamental building block of both differential and integral calculus. Infinite Geometric Series n 0.6 2 2 1 1.2 2 2 2 Consider the sequence and let Sk denote its kth partial sum. S1 S2 S3 S4 2 2 2 2 0.6 0.6 0.6 0. themselves form a sequence. This The partial sums sequence is a function whose domain is the set of natural numbers. The sequence can be described by the following function: 1 1 1 S1, S2, S3, S4,... 2 1.92 2 2 0.6 2 2 0.6 3 2.352 3 2 0.6 4 2.6112 0.6 0.6 0.6 2 2 2 2 2 1 2 20 Figure 1.C-1 un 0.6 0. un1 By using the trace feature to find large values of n, the graph in Figure 1C-1 suggests that the terms of the sequence |
of partial sums are getting closer and closer to 3. Consequently, 2 0.6 2 0.6 2 2 0.6 3 2 0.6 2 where 3 is said to be sum, or limit, of the infinite series, In the general case, an infinite series, or simply series, is defined to be an expression of the form a2 p an a5 a3 a4 p a1 in which each an is a real number. This series is also denoted by the sym- q bol a n1 an. NOTE a1, a2, a3, p If is a geometric sequence, then an expression of a3 a1 the form (sometimes written as q a2 p ) is called a geometric an a n1 series. 76 5 0 The partial sums of the series a2 a3 a4 p are a2 a2 a2 a3 a3 a4 a1 a1 a1 a1 a1, a2 S1 S2 S3 S4 k 1 a1 and in general, for any Sk a3 a4 S1, S2, S3, S4,... p ak. of the sequence of partial sums If it happens that the terms get closer and closer to a particular real number S in such a way that the is arbitrarily close to S when k is large enough, then the partial sum series converges and has a limit. Additionally, S is called the sum of the 2 convergent series. The series con2 verges, and its sum is 3. This series is an infinite geometric series because it has a common ratio of 0.6. 4 p 3 2 2 2 0.6 0.6 0.6 0.6 Sk 2 1 2 1 2 1 1 2 Definition of Infinite Geometric Series If {an} is a geometric sequence with common ratio r, then the corresponding infinite series a3 a2 a4 a1 is called an infinite geometric series. By using the formula for the nth term of a geometric sequence, the corresponding geometric series can be expressed in the form ra1 p r r 3a1 2a1 a1 Under certain circumstances, an infinite geometric series is convergent and has a sum. Sum of an Infinite Geometric Series 66 1, If r 00 00 then the infinite geometric series a1 ra1 r 2a1 r 3a1 converges, and its sum is a1 1 r. NOTE A series that is Example 1 Sum of an Infinite Geometric Series |
the 1, not convergent is said to be divergent. If 0 r 0 series is divergent. Therefore, a geometric series is only convergent 6 1. when 0 r 0 Determine whether the infinite geometric series converges. q n1 a. a n1 6 2 1 2 q b. a n1 8 n 5 77 Solution a. The first term is 6 and the common ratio is 2. The sum of the first k terms is Sk a1 Sk The graph of as shown in Figure 1.C-2, does not approach a single value. In fact, the sums get larger for each subsequent term. So this series with a common ratio of 2 does not converge. It diverges. 1 2 b. is an infinite geometric series with a n1 partial sum of this series is the same as the kth partial sum of the The kth a1. 8 5 and r 1 5 q 8 n 5 sequence 8. n f 5 e 10 Figure 1.C-2 Sk a1 5b kb Sk is shown in Figure 1.C-3. If you use the trace feature and The graph of move beyond approximately the calculator will probably tell you that every partial sum is 2. Actually, the partial sums are slightly smaller than 2 but are rounded to 2 by the calculator. The graph gets very close to 2 as x gets larger, but it never reaches 2. According to the formula, the sum of the infinite series is n 10, 30 S a1. ■ Figure 1.C-3 Example 1b is typical of the general case, as can be seen algebraically. a3 with common ratio r such Consider the geometric series 6 1. is the same as the kth partial sum of that geometric sequence a2 Sk and hence The kth partial sum an6, p 0 r 0 a1 5 1 r k 1 r b Sk a1 a r As k gets very large, the number gets very close to 0 because 1 0 Consequently, when k is very large, 1 r k a1 1 r b is very close to 1 0 1 r b is very close to a1 a 1 r a1 a 1 r Sk. k k. 6 1. 0 r 0 so that Infinite geometric series provide another way of writing an infinite repeating decimal as a rational number. Example 2 Repeating Decimal as a Rational Number Express 6.8573573573 p as a rational number. 1000 0 0 3 0 –1 78 Solution First write 0.057357 |
3573 the number as p as an infinite series: 6.8 0.0573573573 p. Then consider 0.0573 0.0000573 0.0000000573 0.0000000000573 p, which is the same as 0.0573 0.001 1 21 0.0573 2 0.001 1 2 1 2 This is a convergent geometric series with sum is 0.0573 0.001 1 0.0573 2 3 1 and 0.0573 p 2 r 0.001. 2 a1 a1 1 r 0.0573 1 0.001 0.0573 0.999 573 9990. Therefore, 6.8573573573 p 6.8 0.0573 0.0000573 p 3 4 6.8 573 9990 573 9990 4567 666 68 10 68505 9990 Its ■ Exercises In Exercises 1–9, find the sum, or limit, of the infinite series, if it converges. In Exercises 10–15, express the repeating decimal as a rational number. 1. 4. 5. 6. 7. 8. q 1 2n q 2n 3 2. 3. a a n1 n1 1 0.5 0.25 0.125 0.0625 p 500 200 80 32 p q 3 4b a n1 a 9 313 3 13 1 1 13 p 2 12 1 1 12 1 2 p 4 216 6 316 9 916 2 p q 9. a n1 a 1 2n 1 nb 3 n 10. 0.22222 p 12. 5.4272727 p 11. 0.37373737 p 13. 85.131313 p 14. 2.1425425425 p 15. 3.7165165165 p 16. If 5 an6 difference infinite series convergent. is an arithmetic sequence with common ai a4 and each a3 a2 d 7 0 a1 p is not explain why the 7 0 17. Use the graphical approach illustrated in Example 1 to find the sum of the series in q a n1 a 1 2 b n. Does the graph get very close to the horizontal line through 1 3? Describe the behavior of the series. 79 C H A P T E R 2 Equations and Inequalities And the rockets’ red glare... Many Fourth of July firework displays are timed to coincide with patriotic music. To accomplish correct timing, each rocket must |
be detonated at precisely the correct height at the right moment. The time needed for a rocket to reach a specific height is the solution of an equation representing the height of the rocket as a function of time. See Exercise 24 of Section 2.3. 80 Solving Equations Graphically Interdependence of Sections Chapter Outline 2.1 2.2 Solving Quadratic Equations Algebraically 2.3 Applications of Equations 2.4 Other Types of Equations 2.5 Inequalities 2.5.A Excursion: Absolute-Value Inequalities Chapter Review can do calculus Maximum Area 2.1 2.2 > 2.3 > 2.4 > 2.5 > 2.5.A G raphing technology is useful for solving equations, but don’t be mis- led into thinking that technology is always the best tool. For example, graphing technology is useless if you do not know enough alge- bra to understand the information displayed on the screen. When exact answers are required, algebraic techniques are usually needed. This chap- ter and the next two chapters develop both algebraic and graphical techniques for solving equations in one variable. 2.1 Solving Equations Graphically Objectives • Solve equations using the intersect method • Solve equations using the x-intercept method A solution of an equation is a number that, when substituted for the vari3x 2 17 able, produces a true statement. For example, 5 is a solution of 2 17 is a true statement. To solve an equation means to 5 because 2 find all of its solutions. 3 1 Two equations are said to be equivalent if they have the same solutions. x 2 3 For example, are equivalent because 5 is the and only solution of each equation. 3x 2 17 Algebraic techniques provide exact solutions to linear and quadratic equations; however, there are no formulas that provide solutions to many other types of equations. For such equations, graphical approximation methods are practical alternatives. NOTE If necessary, review the material on graphing in the Technology Appendix. Knowledge of a graphing calculator is assumed throughout the remainder of this book. 81 82 Chapter 2 Equations and Inequalities Technology Tip Absolute value (ABS) is in the NUM submenu of the MATH menu of TI and in the NUM submenu of the OPTN menu of RUN mode of CASIO. The graphical intersection finder is labeled INTERSECT, in the CALC menu of TI and ISCT in the G-S |
OLVE menu of Casio. Complete Graphs A viewing window is said to display a complete graph if it shows all the important features of the graph—including all peaks, valleys, and points where it touches an axis—and suggests the general shape of portions of the graph that are not in the window. Many different windows may show a complete graph, but it usually is best to use a window small enough to show as much detail as possible. Later chapters develop algebraic facts that will enable you to know when graphs are complete. Until then try several different windows to see which, if any, appear to display a complete graph. The Intersection Method The following example illustrates a graphical method of approximating solutions of equations where both sides of an equation are algebraic expressions. Each side of the equation can be viewed as the output of a function, and the solutions of the equation represent inputs that produce equal outputs. Example 1 Solving an Equation Using the Intersect Method Solve 0 x2 4x 3 0 x3 x 6. Solution 0 0 y1 y2 and x2 4x 3 x3 x 6. Set Graph both equations on the same screen and find the x-coordinate of the point where the two graphs intersect. This coordinate can be approximated by zooming in and using the trace feature or by using a graphical intersection finder. As shown in is an approximate solution. Figure 2.1-1, x3 x 6 x2 4x 3 x 2.207 0 0 10 Letting x 2.207, 10 10 0 1 left side 2 4 2 2.207 3 2.207 0 1 4.870849 8.828 3 0 6.957151 2 0 right side 3 2.207 6 2.207 1 10.74996374 2.207 6 6.956963743 2 10 Figure 2.1-1 The difference between the value of the left side and the value of the right x 2.207. side is small. Therefore, the solution to the original equation is ■ The Intersection Method To solve an equation of the form intersection method, follow two steps. f(x) g(x) by using the 1. Graph 2. Find the x-coordinate of each point of intersection. on the same screen. and y1 y2 g(x) f(x) Section 2.1 Solving Equations Graphically 83 CAUTION Check several viewing windows to ensure that a complete graph is shown for each side of the |
equation. If the graphs do not intersect, then they have no common output value. Therefore, there are no real solutions to the equation. The x-Intercept Method A zero of a function f is an input that produces an output of 0. For exam x3 8 23 8 0. ple, 2 is a zero of the function 2 2 Note that 2 is also a solution of the equation In other words, the zeros of the function f are the solutions, or roots, of the equation f The zeros of a function also have a graphical interpretation. f because 1 x3 8 0. 0. x x f 1 2 1 2 Graphing Exploration 1. Graph y x4 2x2 3x 2 using a decimal window. (See Technology Tip.) Find the points where the graph crosses the x-axis. 2. Verify that the x-coordinates found in Step 1 are zeros of the function f. That is, the x-coordinates are solutions of x4 2x2 3x 2 0. y f x intersects the x-axis is of the form A point where the graph of 2 (a, 0) because every point on the x-axis has y-coordinate 0. The number a is called an x-intercept of the graph of f. In the preceding Exploration, the x-intercepts of and x 2. The x-intercepts of the graph are the zeros of the function f. x4 2x2 3x 2 were found to be x 1 x f 1 2 1 Let f be a function. If r is a real number that satisfies any of the following statements, then r satisfies all the statements. • r is a zero of the function f • r is an x-intercept of the graph of f • x r is a solution, or root, of the equation f(x) 0 Technology Tip A decimal window produces one-decimal- place values of the x-coordinates when using the trace feature. For a decimal window select ZDECIMAL or ZOOMDEC in the TI ZOOM menu and INIT in the Casio V-WINDOW menu. If needed, review the material on decimal windows in the Technology Appendix. Zeros, x-Intercepts, and Solutions x Because the x-intercepts of the graph of 2 zeros of f are solutions of the related equation x can be used to solve f 0. 1 y f 1 2 are the |
zeros of f, and the f the x-intercepts 0, x 1 2 84 Chapter 2 Equations and Inequalities The x-Intercept Method Follow three steps to solve an equation by the x-intercept method. 1. Write the equation in the equivalent form f(x) 0. 2. Graph y f(x). 3. Find the x-intercepts of the graph. The x-intercepts of the graph are the real solutions of the equation. Technology Tip A standard viewing window displays 10 x 10 and 10 y 10. The ZOOM menu on most calculators contains a standard viewing window option. An advantage of using the x-intercept method is that solutions appear on the x-axis, and prior information about the range of the function is not needed. Example 2 Solving an Equation by Using the x-Intercept Method Solve the equation x5 x2 x3 5. Solution 10 Rewrite the equation so that one side is zero. 10 10 Graph y x5 x3 x2 5 in the standard viewing window. x5 x3 x2 5 0 10 Figure 2.1-2 Technology Tip Use the trace feature to find that the zero is between 1.3 and 1.5, then use zoom-in and trace features repeatedly, or use the graphical zero finder, to get a better approximation of 1.4242577. (See the Technology Tip for the location of the graphical zero finder.) Verify that 1.42 is an approximate solution by substituting into the original equation. x 1.42 ■ Technological Quirks The graphical zero finder is labeled ZERO in the TI CALC menu and ISCT in the Casio G-SOLVE menu. A graphical zero finder may fail to find some solutions of an equation, particularly when the graph of the equation touches, but does not cross, the x-axis. If the calculator does not show any x-intercepts on a graph or if its zero finder gives an error message, an alternative approach may be necessary, as illustrated in the next two examples. Example 3 Solving 2f(x) 0 by Solving f (x) 0 Solve 2x4 x2 2x 1 0. Solution y 2x4 x2 2x 1. The trace feature may display no y-value Graph for some points and the graphical zero finder may display an error message. See Figure 2.1-3 on the |
next page. Section 2.1 Solving Equations Graphically 85 2 This difficulty can be eliminated by using the fact that the only number whose square root is zero is zero itself. 3 3 2 Figure 2.1-3 NOTE Solving radical and rational equations is presented in Section 2.4, and radical and rational functions are presented in Chapter 4. That is, the solutions of tions of x4 x2 2x 1 0. 2x4 x2 2x 1 0 are the same as the solu- As the graphs below display, the solutions of x 0.4046978 and x 1.1841347, x4 x2 2x 1 0 are which are also approximate solutions of 2x4 x2 2x 1 0. 3 3 3 3 3 3 3 Figure 2.1-4a 3 Figure 2.1-4b The solutions can be verified by substitution. ■ 5 Example 4 Solving f (x) g(x) 0 5 5 Solve 2x2 x 1 9x2 9x 2 0. 5 Figure 2.1-5 5 Solution The graph of y 2x2 x 1 9x2 9x 2 in Figure 2.1-5 is impossible to read. Using the zoom feature will display a better graph, but it may be easier to use the fact that a fraction is zero only when its numerator is zero and its denominator is nonzero. The values that make the numerator zero can easily be found by finding y 2x2 x 1. the zeros of Discard any value that makes the denominator of the original equation zero because 5 5 an input that gives an undefined output is not in the domain. 5 Figure 2.1-6 Figure 2.1-6 shows that one x-intercept of and the other is (not identified on the graph). Neither value makes the denominator zero, so they are the solutions to the given equation, which can be verified by substitution. x 1 is y 2x2 x 1 x 0.5 ■ 86 Chapter 2 Equations and Inequalities Summary of Solving Equations Graphically To solve h(x) g(x) use one of the following. • The Intersection Method y1 and 1. Graph h(x) y2 g(x). 2. Find the x-coordinate of each point of intersection. • The x-Intercept Method 1. Rewrite the equation as f(x) 0, where f(x) |
h(x) g (x). y f(x). 2. Graph 3. Find the x-intercepts of the graph of y f(x) x-intercepts of the graph of of the equation. f(x). The are the solutions The x-Intercept Method has the advantage of needing no information about the range of the functions. Applications Graphical solution methods can be helpful in dealing with applied problems because approximate solutions are adequate in most real-world contexts. Example 5 Equal Populations According to data from the U.S. Bureau of the Census, the approximate population y (in millions) of Chicago and Los Angeles between 1950 and 2000 are given by Chicago Los Angeles y 0.0000304x3 0.0023x2 0.02024x 3.62 y 0.0000113x3 0.000922x2 0.0538x 1.97 where 0 corresponds to 1950. In what year did the two cities have the same population? Solution 50 Graph both functions on the same screen, and find the x-value of their point(s) of intersection. As shown in Figure 2.1-7, the populations were the same when which represents September of 1978. x 28.75, ■ Figure 2.1-7 4 0 −2 Section 2.1 Solving Equations Graphically 87 Exercises 2.1 In Exercises 1–6, determine graphically the number of solutions of the equation, but don’t solve the equation. You may need a viewing window other than the standard one to find all of the x-intercepts. 1. x5 5 3x4 x 2. x3 5 3x2 24x 25. 10x5 3x2 x 6 0 26. 1 4 x4 x 4 0 27. 2x 1 2 x2 1 12 x4 0 28. 1 4 x4 1 3 x2 3x 1 0 3. x7 10x5 15x 10 0 4. x5 36x 25 13x3 5. x4 500x2 8000x 16x3 32,000 6. 6x5 80x3 45x2 30 45x4 86x In Exercises 7–34, use a graphical method to find all real solutions of the equation, approximating when necessary. 29. 31. 32. 33. 5x x2 1 x2 4 0 0 2x 3 0 30. 2x x 5 1 3x2 2 |
x 1 x3 2 5 x x2 0 0 2x2 3 2x 2 5 34. 2x3 2 2x 5 4 7. x3 4x2 10x 15 0 8. x3 9 3x2 6x 9. x4 x 3 0 In Exercises 35–40, find an exact solution of the equation in the interval shown to the right of each equation. For example, if the graphical approximation of a solu- 10. x5 5 3x4 x 11. 2x4 x3 x 3 0 tion begins.3333, check to see if 1 3 is the exact solu- 12. 28x4 14x3 9x2 11x 1 0 13. 2 5 A x5 x2 2x 0 14. 2x4 x2 3x 1 0 15. x2 2x 5 16. 2x2 1 2x 9 0 17. 18. 19. 2x5 10x 5 x3 x2 12x 0 3x5 15x 5 x7 8x5 2x2 5 0 x3 4x 1 x2 x 6 0 20. 4 x 2 3 x 1 0 3 Use parentheses correctly. 4 21. 2x3 4x2 x 3 0 22. 6x3 5x2 3x 2 0 23. x5 6x 6 0 24. x3 3x2 x 1 0 tion. Similarly, if your approximation begins 1.414, 12 1.414. check to see if is a solution because 12 35. 3x3 2x2 3x 2 0 36. 4x3 3x2 3x 37. 12x4 x3 12x2 25x 2 0 0 6 x 6 1 38. 8x5 7x4 x3 16x 2 0 0 6 x 6 1 39. 4x4 13x2 3 0 40. x3 x2 2x 41. According to data from the U.S. Department of Education, the average cost y of tuition and fees at public four-year institutions in year x is approximated by the equation y 0.024x4 0.87x3 9.6x2 97.2x 2196 x 0 where corresponds to 1990. If this model continues to be accurate, during what year will tuition and fees reach $4000? 42. Use the equation in Example 5 to determine the year in which the population of Los Angeles reached 2.6 million. 88 Chapter 2 Equations and Inequalities |
43. According to data from the U.S. Department of Health and Human Services, the cumulative number y of AIDS cases (in thousands) as of year x is approximated by y 0.062x4 1.54x3 9.21x2 57.54x 199.36 0 x 11 2 1 x 0 corresponds to 1990. During what where year did the cumulative number of cases reach 750,000? 44. a. How many real solutions does the equation 0.2x5 2x3 1.8x k 0 have when k 0? b. How many real solutions does it have when k 1? c. Is there a value of k for which the equation has just one real solution? d. Is there a value of k for which the equation has no real solution? 2.2 Solving Quadratic Equations Algebraically Objectives • Solve equations by: factoring square root of both sides completing the square quadratic formula • Solve equations in quadratic form The basic strategy for solving equations is to use the basic properties of equality. • Add or subtract the same quantity from both sides of the equation. • Multiply or divide both sides of the equation by the same nonzero quantity. The properties of equality apply to all equations. They, together with other techniques that are presented in this chapter, can be used to transform a given equation into one whose solutions are easily found. This section considers quadratic equations and techniques used to find their solutions. Definition of a Quadratic Equation A quadratic, or second degree, equation is one that can be written in the form ax2 bx c 0 for real constants a, b, and c, with a 0. NOTE This chapter considers only real solutions, that is, solutions that are real numbers. Techniques Used to Solve Quadratic Equations There are four techniques normally used to algebraically find exact solutions of quadratic equations. Techniques that can be used to solve some quadratic equations include • factoring • taking the square root of both sides of an equation Techniques that can be used to solve all quadratic equations include • completing the square • using the quadratic formula Section 2.2 Solving Quadratic Equations Algebraically 89 Solving Quadratic Equations by Factoring The factoring method of solving quadratic equations is based on the Zero Product Property of real numbers. The Zero Product Property If a product of real numbers is |
zero, then at least one of the factors is zero. In other words, If ab 0, then a 0 or b 0 (or both). NOTE If needed, review factoring in the Algebra Appendix. Example 1 Solving a Quadratic Equation by Factoring Solve 3x2 x 10 by factoring. Solution Rearrange the terms so that one side is 0, and then factor. y 8 4 (−, 0) 5 3 3x2 x 10 0 x 2 0 3x 5 2 1 2 1 (2, 0) x Using the Zero Product Property, Subtract 10 from each side Factor the left side 3x 5 x 2 or must be 0. −8 −4 0 4 8 −4 −8 3x 5 0 or 3x 5 x 5 3 x 2 0 x 2 Figure 2.2-1 Therefore, the solutions are 5 3 and 2. See Figure 2.2-1. ■ CAUTION To guard against mistakes, always check solutions by substituting each solution into the original equation to make sure it really is a solution. Solving x 2 k x2 5 has no real solutions because the square of a numThe equation x 0. ber is never negative. The equation x2 7 because these are The equation the only numbers whose square is 7. Similar facts are true for equations of the form has only one solution, and where k is a real number. has two solutions, x2 0 27 x2 k, 27, Solutions of x 2 k For a real number k, k 66 0 k 0 k 77 0 Number of Solutions 0 1 2 Solutions 0 2k and 2k 90 Chapter 2 Equations and Inequalities CAUTION When taking the square root of both sides of an equation, ± remember to write on one side of the equation. When k is positive, the two solutions of x ± 2k, are often written as which is read “x equals plus or minus the square root of k.’’ x2 k Taking the Square Root of Both Sides of an Equation Example 2 Solving ax 2 b Solve 3x2 9. Solution 3x2 9 x2 3 x ± 23 ±1.732 Divide by 3 Take the square root Substitute both solutions into the original equation to check. ■ The method of taking the square root of both sides of an equation can be a used to solve equations of the form 2 k. x h 1 2 Example 3 Solving a(x h)2 |
k Solve 2 x 4 1 2 2 6. Solution The equation is in the form the procedure outlined above can be applied. au2 k, where u represents x 4. Therefore ± 23 x 4 23 x 2.27 x 4 ± 23 or x 4 23 x 5.73 Divide by 2 Take square roots Subtract 4 Exact solutions Approximate solutions ■ Figure 2.2-2 y = 2(x + 4)2 y y = 6 −8 −4 −5.73 −2.27 8 4 0 −4 −8 Completing the Square A variation of the method of taking the square root of both sides can be used to solve any quadratic equation. It is based on the fact that an exprescan be changed into a perfect square by adding sion of the form x2 6x a suitable constant. For example, adding 9 to the expression changes it into a perfect square. x2 bx x2 6x 9 x 3 2 2 1 The number added is 9, which is of x in the original expression, completing the square, works in every case. 32, x2 6x. and 3 is one-half of 6, the coefficient This technique, which is called Section 2.2 Solving Quadratic Equations Algebraically 91 Completing the Square To complete the square of the expression square of one-half the coefficient of x, namely x2 bx, add the b 2 b The a. 2 NOTE The procedure of completing the square is used in other areas of mathematics, and knowledge of this procedure is important in later chapters. addition produces a perfect square trinomial. x2 bx Solving a Quadratic Equation by Completing the Square To solve a quadratic equation by completing the square, follow the procedure below. 1. Write the equation in the form x2 bx c. CAUTION 2. Add 2 b 2 b a to both sides so that the left side is a perfect square and the Completing the square only works when the coefficient of an equation such as 2x2 6x 1 0 is 1. In x2 right side is a constant. 3. Take the square root of both sides. 4. Simplify. first divide both sides by 2 and then complete the square. Example 4 Solving a Quadratic Equation by Completing the Square Solve 2x2 6x 1 0 by completing the square. Solution y 8 4 −8 −4 0 4 8 x − |
4 −8 Figure 2.2-3 2x2 6x 1 0 2x2 6x 1 x2 3x 1 2 1 2 x2 3x Subtract 1 Divide by 2 Add 2 3 2 b a 9 4 Rewrite as perfect square and simplify 7 4 A Take square root ± A 7 4 2.823 Add 3 2 or x 3 2 7 4 A 0.177 There are two real solutions. See Figure 2.2-3. ■ The technique of completing the square can be used to solve any quadratic equation. 92 Chapter 2 Equations and Inequalities Solving ax 2 bx c 0 by Completing the Square Solve ax2 bx c 0 by completing the square as follows: 1. Subtract c from both sides. ax2 bx c 2. Divide both sides by a, the leading coefficient. x2 b 3. Add the square of half of that is, 2 b 2a b a, to both sides. a x c a b a, x2 b a x 2 b 2ab 2 b 2ab c a a 4. Write the left side of the equation as a perfect square. a x b a 2ab 2 2 b 2ab a c a 5. Take the square root of both sides. x b 2a ±± 2 b 2ab c a B a 6. Subtract b 2a from both sides. x b 2a ±± 2 b 2ab c a B a The equation in step 6 can be simplified. 2 b 2ab a b2 4a2 b2 4a2 c a b2 4a2 4ac 4a2 ± ± A b2 c A 4a2 a b2 4ac 4a2 x b 2a b 2a b 2a b ± 2b2 4ac 2a ± 2b2 4ac 2a The final expression is known as the quadratic formula. The Quadratic Formula The solutions of the quadratic equation b ±± 2b2 4ac 2a x ax2 bx c 0 are Because the quadratic formula can be used to solve any quadratic equation, it should be memorized. Section 2.2 Solving Quadratic Equations Algebraically 93 Example 5 Solving a Quadratic Equation by Using the Quadratic Formula Solve x2 3 8x by using the quadratic formula. Solution Rewrite the equation as a 1, b 8, |
and with x2 8x 3 0, c 3. and apply the quadratic formula x 8 ± 282 4 2 1 1 1 2 1 8 252 2 3 2 21 8 ± 264 12 2 8 ± 252 2 0.4 or x 8 252 2 7.6. Therefore, x y 4 0 x 4 −12 −8 −4 −4 −8 −12 Figure 2.2-4 The equation has two distinct real solutions, as confirmed in Figure 2.2-4. ■ The Discriminant b2 4ac in the quadratic formula, called the discriminant, The expression can be used to determine the number of real solutions of the equation ax2 bx c 0. Real Solutions of a Quadratic Equation Discriminant Value b2 4ac 77 0 b2 4ac 0 b2 4ac 66 0 Number of Real Solutions of ax2 bx c 0 2 distinct real solutions 1 distinct real solution 0 real solutions The discriminant can be used to determine if an equation has no real solutions without completing all computations. Example 6 Determining the Number of Solutions by Using the Discriminant Solve 2x2 x 3. Solution First, write the equation in general form. 2x2 x 3 0 2 3 b2 4ac 1 4 1 2 1 2 23 6 0 94 Chapter 2 Equations and Inequalities y 10 8 6 4 2 −8 −6 −4 −2 0 2 4 6 8 x Figure 2.2-5 Definition of Polynomial Equation NOTE Polynomials are discussed in Chapter 4. The discriminant of has no real solutions. 2x2 x 3 0 is negative. Therefore, 2x2 x 3 0 does not You can confirm this fact because the graph of touch the x-axis, as shown in Figure 2.2-5. Since the graph has no x-intercepts, the equation has no real solutions. y 2x2 x 3 ■ NOTE A quadratic equation with no x-intercepts has no real solution. Polynomial Equations A polynomial equation of degree n is an equation that can be written in the form where an, p anxn an1xn1 p a1x a0, a0 are real numbers. 0, 4x6 3x5 x4 7x3 8x2 4x 9 0 For instance, equation of degree 6. Similarly, sion of degree 3. Notice that polynomials have the following |
traits. is a polynomial is a polynomial expres- 4x3 3x2 4x 5 • no variables in denominators • no variables under radical signs As a general rule, polynomial equations of degree 3 and above are best solved by the graphical methods presented in Section 2.1. However, some equations are quadratic in form and can be solved algebraically. Polynomial Equations in Quadratic Form Example 7 Solving an Equation in Quadratic Form Solve 4x4 13x2 3 0. Solution To write 4x4 13x2 3 0 in quadratic form, substitute u for x2. 4x4 13x2 3 0 2 13x2 3 0 x2 2 4u2 13u 3 0 4 1 Write x4 as Substitute u for x2 x2 1 2 2 Then solve the resulting quadratic equation. u 3 4u 1 1 u 3 0 2 1 or u 3 2 0 4u 1 0 u 1 4 Factor Zero-Product Property Section 2.2 Solving Quadratic Equations Algebraically 95 Because u x2, x2 3 or x ± 23 x2 1 4 x ± 1 2 Therefore, the original equation has four solutions, x ± 1 2 as shown in Figure 2.2-6., x ±23 and ■ y 8 4 −4 −2 0 2 4 x −4 −8 Figure 2.2-6 Exercises 2.2 In Exercises 1–12, solve each equation by factoring. 1. x2 8x 15 0 2. x2 5x 6 0 3. x2 5x 14 4. x2 x 20 5. 2y2 5y 3 0 6. 3t2 t 2 0 7. 4t2 9t 2 0 8. 9t2 2 11t 9. 3u2 u 4 10. 5x2 26x 5 11. 12x2 13x 4 12. 18x2 23x 6 In Exercises 13–24, solve the equation by taking the square root of both sides. Give exact solutions and approximate solutions, if appropriate. 13. x2 9 15. x2 40 17. 3x2 12 14. x2 12 16. x2 10 18. 1 2 v2 10 19. 5s2 30 20. 3x2 11 21. 25x2 4 0 22. 4x2 28 0 23. 3w2 8 20 |
24. 2t2 11 5 In Exercises 25–28, solve the equation by completing the square. 25. x2 2x 12 26. x2 4x 30 0 27. w2 w 1 0 28. t2 3t 2 0 In Exercises 29–40, use the quadratic formula to solve the equation. 29. x2 4x 1 0 30. x2 2x 1 0 31. x2 6x 7 0 32. x2 4x 3 0 33. x2 6 2x 34. x2 11 6x 35. 4x2 4x 7 36. 4x2 4x 11 37. 4x2 8x 1 0 38. 2t2 4t 1 0 39. 5u2 8u 2 40. 4x2 3x 5 In Exercises 41–46, find the number of real solutions of the equation by computing the discriminant. 41. x2 4x 1 0 42. 4x2 4x 3 0 43. 9x2 12x 1 44. 9t2 15 30t 45. 25t2 49 70t 46. 49t2 5 42t In Exercises 47–56, solve the equation by any method. 47. x2 9x 18 0 48. 3t2 11t 20 0 49. 4x 1 x 1 2 1 51. 2x2 7x 15 50. 25y2 20y 1 52. 2x2 6x 3 53. t2 4t 13 0 54. 5x2 2x 2 55. 7x2 3 2x 3 1 56. 25x 4 x 20 96 Chapter 2 Equations and Inequalities In Exercises 57–60, use a calculator to find approximate solutions of the equation. In Exercises 69–72, find a number k such that the given equation has exactly one real solution. 57. 4.42x2 10.14x 3.79 0 58. 8.06x2 25.8726x 25.047256 0 59. 3x2 82.74x 570.4923 0 60. 7.63x2 2.79x 5.32 69. x2 kx 25 0 70. x2 kx 49 0 71. kx2 8x 1 0 72. kx2 24x 16 0 In Exercises 61–68, find all exact real solutions of the equation. 61. y4 7y2 6 0 62 |
. x4 2x2 1 0 63. x4 2x2 35 0 64. x4 2x2 24 0 65. 2y4 9y2 4 0 66. 6z4 7z2 2 0 67. 10x4 3x2 1 68. 6x4 7x2 3 73. Find a number k such that 4 and 1 are the solutions of x2 5x k 0. 74. Suppose a, b, and c are fixed real numbers such Let r and s be the solutions of b2 4ac 0. that ax2 bx c 0. a. Use the quadratic formula to show that a and rs c a. b. Use part a to verify that r s b ax2 bx c a c. Use part b to factor x r x s. x2 2x 1 21 1 2 and 5x2 8x 2. Section 2.3 Applications of Equations 97 2.3 Applications of Equations Objectives • Solve application problems Real-life situations are usually described verbally, but they must be interpreted and expressed as equivalent mathematical statements. The following guideline may be helpful. Applied Problems Guideline 1. Read the problem carefully, and determine what is asked for. 2. Label the unknown quantities with variables. 3. Draw a picture of the situation, if appropriate. 4. Translate the verbal statements in the problem and the relationships between the known and unknown quantities into mathematical language. 5. Consolidate the mathematical information into an equation in one variable that can be solved or an equation in two variables that can be graphed. 6. Solve for at least one of the unknown quantities. 7. Find all remaining unknown quantities by using the relationships given in the problem. 8. Check and interpret all quantities found in the original problem. Example 1 Number Relations The average of two real numbers is 41.125, and their product is 1683. Find the two numbers. Solution 1. Read: 2. Label: 3. Draw: 4. Translate: Two numbers are asked for. Let the numbers be a and b. A diagram is not appropriate in this problem. English Language two numbers Their average is 41.125. Mathematical Language a and b a b 2 41.125 [1] Their product is 1683. [2] 5. Consolidate: One technique to use when dealing with two unknowns is to express one in terms of the other, then substitute to obtain an equation in one variable. Solve equation [2 |
] for b. ab 1683 ab 1683 b 1683 a [2] Divide both sides by a 98 Chapter 2 Equations and Inequalities Substitute the result into equation [1] and simplify. a b 2 41.125 [1] 1 0 1 50 Figure 2.3-1 a 1683 a 2 a 1683 41.125 Substitute 1683 a for b a 82.25 Multiply both sides by 2 6. Solve: Solve the equation by using the x-Intercept Method with the graph of y1 x 1683 x 82.25, where x represents a. See Figure 2.3-1. a 1683 a 82.25 The solutions of are a 44 and a 38.25. 7. Find: Find the other number, b, from the equation ab 1683. Let a 44. Similarly, let b 44. 44b 1683 b 38.25 a 38.25 and use the same equation to find 8. Check: The average of 44 and 38.25 is 44 38.25 2 The product of 44 and 38.25 is 1683. 41.125. The two numbers are 44 and 38.25. ■ Solutions in Context When solving an application problem, it is important to interpret answers in terms of the original problem. Each solution should • make sense • satisfy the given conditions • answer the original question In particular, an equation may have several solutions, some of which may not make sense in the context of the problem. For instance, distance cannot be negative, the number of people cannot be a fraction, etc. Example 2 Dimensions of a Rectangle A rectangle is twice as wide as it is high. If it has an area of 24.5 square inches, what are its dimensions? w 2h Figure 2.3-2 h Solution 1. Read: 2. Label: 3. Draw: The dimensions of the rectangle are asked for. Let w denote the width and h denote the height. See Figure 2.3-2. 99 [3] [4] Section 2.3 Applications of Equations 4. Translate: The area of a rectangle is width English Language Width is twice height. The area is height. Mathematical Language w 2h wh 24.5 24.5 in2. 5. Consolidate: Substitute 2h for w in equation [4]. wh 24.5 h 24.5 2h 2 2h2 24.5 1 6. Solve: Solve by taking |
the square root of both sides. h2 12.25 h ± 212.25 ± 3.5 Divide by 2 Take square root of both sides Because height is never negative, only the positive root applies to this situation. Therefore, 7. Find: Find the width by using equation [3] and h 3.5 h 3.5 in.. 8. Check: The width is twice the height and the area is correct. w 2 3.5 2 1 7 in. Thus, the width is 7 inches and the height is 3.5 inches. ■ 3.5 7 1 2 24.5 in2 Example 3 Volume of a Rectangular Box A rectangular box with a square base and no top is to have a volume of 30,000 cm3. what are the dimensions of the box? If the surface area of the box is 6000 cm2, h s s Figure 2.3-3 Solution 1. Read: 2. Label: 3. Draw: 4. Translate: The quantities to be found are the length, width, and height of the box. Notice that the two sides of the base have the same length. Let s denote a side of the square base of the box. Let h denote the height of the box. See Figure 2.3-3. height The volume of a box is given by length width and the surface area is the sum of the area of the base and the area of the four sides of the box. English Language length, width, height surface area: Mathematical Language s, s, and h base surface area each side surface area total surface area Volume is Surface area is 30,000 cm3 6000 cm2 s2 sh s2 4sh s2h 30,000 s2 4sh 6,000 [5] [6] 100 Chapter 2 Equations and Inequalities 5. Consolidate: Solve equation [5] for h: h 30,000 s2 Substitute the expression for h into equation [6]. 10,000 s2 4s a 30,000 b s2 s2 120,000 s 6000 6000 6. Solve: To solve by using the Intersection Method, graph s2 120,000 s and y1 2.3-4, and find the points of intersection. Therefore, s 21.70 cm or s 64.29 cm. as shown in Figure y2 6000, 100 0 Figure 2.3-4 75 7. Find: 8. Check: h 30,000 Use to find |
h. s2 h 30,000 21.70 2 1 30,000 470.89 2 h 30,000 64.29 2 1 63.71 cm 30,000 4133.2041 2 7.26 cm Volume 63.71 21.7 2 30,000.4019 1 1 2 2 2 7.26 64.29 2 30,007.06177 2 1 1 Surface Area 2 4 21.7 6000.9180 1 2 21.7 1 21 63.71 2 2 4 64.29 6000.1857 2 1 64.29 7.26 2 2 1 1 21.7 cm 21.7 cm with a height of approxiOne base is approximately mately 63.71 cm. Another base is approximately 64.29 cm 64.29 cm with a height of approximately 7.26 cm. ■ Interest Applications Calculating interest is common in real-world applications. When an amount, P, is deposited or borrowed, P is referred to as the principal. Interest is the fee paid for the use of the money and is calculated as a percentage of the principal each year. When the duration of a loan or a bank balance is less than 1 year, simple interest is generally used. The basic rule of simple interest is I Prt P represents the principal, r represents the annual interest rate, and t represents time in years. Example 4 Stock and Savings Returns A high-risk stock pays dividends at a rate of 12% per year, and a savings account pays 6% interest per year. How much of a $9000 investment should be put in the stock and how much should be put in savings to obtain a return of 8% per year on the total investment? NOTE In Precalculus, combining basic algebraic steps to promote clarity when simplifying is encouraged as long as it does not cause confusion. Section 2.3 Applications of Equations 101 Solution Read and Label: Let s be the amount invested in stock. Then the rest of will be the amount in the sav- 9000 s, the $9000, namely ings account. Translate: Return on s dollars in stock at 12% b a Return on 9000 s dollars in savings at 6%b a 8% of $9000 > 12% of s 6% of 1 > 9000 s 2 0.08 1 > 9000 2 > > 0.12s 0.06 > 720 9000 s 0.12s 540 0.06s 720 2 1 0.12s 0.06s 720 540 0 |
.06s 180 s 180 0.06 3000 Therefore, the investment should be as follows: • $3000 in stock 9000 3000 • 2 1 $6000 in the savings account If this is done, the total return will be 12% of $3000 plus 6% of $6000, making a total return of $360 $360 $720 —which is 8% of $9000. ■ Distance Applications The basic formula for problems involving distance and a constant rate of velocity is d rt where d represents the distance traveled at rate r for time t. The units for rate should be the distance units divided by the time units, such as miles per hour. Example 5 Distance A pilot wants to make an 840-mile round trip from Cleveland to Peoria and back in 5 hours flying time. There will be a headwind of 30 mph going to Peoria, and it is estimated that there will be a 40-mph tailwind returning to Cleveland. At what constant engine speed should the plane be flown? Solution Let r be the engine speed of the plane, and note that the headwind slows the velocity by 30 and the tailwind increases the velocity by 40. 102 Chapter 2 Equations and Inequalities Distance Actual Velocity Cleveland to Peoria Peoria to Cleveland 420 420 r 30 r 40 D rP D rC Time 420 r 30 420 r 40 The time traveling to Peoria plus the time traveling back to Cleveland is the total time traveled, or 5 hours. 1 Time to Peoria 420 r 30 2 1 Time to Cleveland 420 r 40 2 5 5 Multiply both sides by the common denominator simplify. 1 r 30 r 40 2 1, and 2 420 r 40 1 r 30 r 40 2 21 r 30 r 40 r 30 5 1 2 1 r 40 420 r 30 1 2 21 420 2 84 2 r 30 2 5 2 2 2 1 2 1 420 84 r 40 r 40 r 40 1 r 40 r 30 1 r 30 1 1 r2 10r 1200 84r 3360 84r 2520 r2 158r 2040 0 0 r 170 r 12 2 1 r 170 0 or r 12 0 1 r 30 1 2 2 2 1 r 170 r 12 Obviously, the negative solution does not apply. Because both sides were multiplied by a quantity involving the variable, the positive solution, 170, should be checked in the original problem to make sure it is a solution. ■ Other Applications Example 6 Width of a Garden Walk x 40 x 24 A landscaper wants to put a cement walk of uniform width |
around a rectangular garden that measures 24 by 40 feet. She has enough cement to cover 660 square feet. How wide should the walk be in order to use all the cement? Solution Figure 2.3-5 Let x denote the width of the walk in feet and draw a picture of the situation, as shown in Figure 2.3-5. The length of the outer rectangle is the garden length plus walks on each and its width is the garden width plus walks on each end, or end, or The area of the walk is found by subtracting the area of the garden from the area of the outer rectangle. 40 2x, 24 2x. Section 2.3 Applications of Equations 103 Area of outer rectangle 1 1 2 Area of garden 2 Area of the walk 40 2x 24 2x > 2 1 > 660 960 128x 4x2 960 660 24 40 21 > 2 2 1 1 4x2 128x 660 0 x2 32x 165 0 165 2 1 2 Apply the quadratic formula x x 1 1 1 32 ± 2 2 4 32 2 1 2 1 32 ± 21684 2 2 x 4.5 or x 36.5 Only the positive solution makes sense, so the walk should be approximately 4.5 feet wide. Check the solution in the original problem. ■ Example 7 Box Construction A box with no top that has a volume of 1000 cubic inches is to be constructed from a -inch sheet of cardboard by cutting squares of equal size from each corner and folding up the flaps, as shown in Figure 2.3-6. What size square should be cut from each corner? 22 30 30 x x x 22 30 − 2x 30 − 2x x 22 − 2x Figure 2.3-6 Solution Let x represent the length of the side of the square to be cut from each corner. The dashed rectangle in Figure 2.3-6 is the bottom of the box. Its length is as shown in the figure. Similarly, the width of the box will be 30 2x, 22 2x, and its height will be x inches. Therefore, Length Width Height Volume of the box > 30 2x >>> 22 2x x 1000 1 2 Because the cardboard is 22 inches wide, x must be less than 11, and because x is a length, it must be positive. Therefore, the only meaningful solutions in this context are between 0 and 11. 2 1 104 1500 0 500 Chapter 2 Equations and Inequalities 30 2x 22 2x 1 x 2 1 |
y1 and Graph As shown in Figure 2.3-7, there are two points of intersection: one at approximately (2.23, 1000) and another at approximately (6.47, 1000), which is not identified on the graph. Because both are viable solutions, there are two boxes that meet the given conditions. y2 2 1000. Figure 2.3-7 11 Find the dimensions of the box for each possible case. Height: Length: Width: Case I 2.23 in. 30 2 22 2 1 1 25.54 in. 17.54 in. 2.23 2.23 2 2 Case II 6.47 in. 30 2 22 2 1 1 17.06 in. 9.06 in. 6.47 6.47 2 2 25.54 in 2.23 in 17.54 in Figure 2.3-8 6.47 in 9.06 in 17.06 in ■ Example 8 Mixture Problem A car radiator contains 12 quarts of fluid, 20% of which is antifreeze. Howmuch fluid should be drained and replaced with pure antifreeze so that the resulting mixture is 50% antifreeze? Solution Let x be the number of quarts of fluid to be replaced by pure antifreeze. When x quarts are drained, there are quarts of fluid left in the radiator, 20% of which is antifreeze. 12 x Amount of antifreeze in radiator after draining x quarts of fluid ± ≤ x quarts of antifreeze b a ° Amount of antifreeze in final mixture ¢ > 20% of 12 x 1 2 > x > 50% of 12 > 12 x > 0.2 2 2.4 0.2x x 6 x 0.5 1 > 12 1 2 0.8x 3.6 x 3.6 0.8 4.5 Therefore, 4.5 quarts should be drained and replaced with pure antifreeze. ■ Section 2.3 Applications of Equations 105 Exercises 2.3 In Exercises 1–4, a problem situation is given. a. Decide what is being asked for, and label the unknown quantities. b. Translate the verbal statements in the problem and the relationships between the known and unknown quantities into mathematical language, using a table like those in Examples 1–3. The table is provided in Exercises 1–2. You need not find an equation to be solved. 7. |
The diameter of a circle is 16 cm. By what amount must the radius be decreased in order to decrease the area by square centimeters? 48p 8. A corner lot has dimensions 25 by 40 yards. The city plans to take a strip of uniform width along the two sides bordering the streets in order to widen these roads. How wide should the strip be if the remainder of the lot is to have an area of 844 square yards? 1. The sum of two numbers is 15 and the difference of their squares is 5. What are the numbers? In Exercises 9–20, solve the problem. English Language the two numbers Their sum is 15. The difference of their squares is 5. Mathematical Language 2. The sum of the squares of two consecutive integers is 4513. What are the integers? English Language the two integers The integers are consecutive. The sum of their squares is 4513. Mathematical Language 3. A rectangle has a perimeter of 45 centimeters and an area of 112.5 square centimeters. What are its dimensions? 4. A triangle has an area of 96 square inches, and its height is two-thirds of its base. What are the base and height of the triangle? In Exercises 5–8, set up the problem by labeling the unknowns, translating the given information into mathematical language, and finding an equation that will produce the solution to the problem. You need not solve this equation. 5. A worker gets an 8% pay raise and now makes $1600 per month. What was the worker’s old salary? 6. A merchant has 5 pounds of mixed nuts that cost $30. He wants to add peanuts that cost $1.50 per pound and cashews that cost $4.50 per pound to obtain 50 pounds of a mixture that costs $2.90 per pound. How many pounds of peanuts are needed? 9. You have already invested $550 in a stock with an annual return of 11%. How much of an additional $1100 should be invested at 12% and how much at 6% so that the total return on the entire $1650 is 9%? 10. If you borrow $500 from a credit union at 12% annual interest and $250 from a bank at 18% annual interest, what is the effective annual interest rate (that is, what single rate of interest on $750 would result in the same total amount of interest)? 11. A radiator contains 8 quarts of fluid, 40% of which is antifreeze. |
How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 60% antifreeze? 12. A radiator contains 10 quarts of fluid, 30% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 40% antifreeze? 13. Two cars leave a gas station at the same time, one traveling north and the other south. The northbound car travels at 50 mph. After 3 hours the cars are 345 miles apart. How fast is the southbound car traveling? 14. An airplane flew with the wind for 2.5 hours and returned the same distance against the wind in 3.5 hours. If the cruising speed of the plane was a constant 360 mph in air, how fast was the wind blowing? Hint: If the wind speed is r miles per hour, then the plane travels at the wind and at mph against the wind. mph with 360 r 360 r 2 1 1 2 15. The average of two real numbers is 41.375 and their product is 1668. What are the numbers? 106 Chapter 2 Equations and Inequalities 16. A rectangle is four times as long as it is wide. If it has an area of 36 square inches, what are its dimensions? b. It is thrown downward from the top of the same building, with an initial velocity of 52 feet per second. 17. A 13-foot-long ladder leans on a wall. The bottom of the ladder is 5 feet from the wall. If the bottom is pulled out 3 feet farther from the wall, how far does the top of the ladder move down the wall? Hint: The ladder, ground, and wall form a right triangle. Draw pictures of this triangle before and after the ladder is moved. Use the Pythagorean Theorem to set up an equation. 18. A factory that makes can openers has fixed costs for building, fixtures, machinery, etc. of $26,000. The variable cost for material and labor for making one can opener is $2.75. a. What is the total cost of making 1000 can openers? 20,000? 40,000? b. What is the average cost per can opener in each case? 19. Red Riding Hood drives the 432 miles to Grandmother’s house in 1 hour less than it takes the Wolf to drive the same route. Her average speed is 6 mph faster than the Wolf |
’s average speed. How fast does each drive? 20. To get to work Sam jogs 3 kilometers to the train, then rides the remaining 5 kilometers. If the train goes 40 km per hour faster than Sam’s constant rate of jogging and the entire trip takes 30 minutes, how fast does Sam jog? In Exercises 21–24, an object is thrown upward, dropped, or thrown downward and travels in a vertical line subject only to gravity with wind resistance ignored. The height h, in feet, of the object above the ground after t seconds is given by h 16t 2 v0t h0 and v0 t 0. where is the initial height of the object at starting h0 t 0, is the initial velocity (speed) of the time object at time is taken as posiv0 t 0 tive if the object starts moving upward at time and negative if the object starts moving downward at t 0. An object that is dropped (rather than thrown downward) has initial velocity The value of 0. v0 22. You are standing on a cliff 200 feet high. How long will it take a rock to reach the ground at the bottom of the cliff in each case? a. You drop it. b. You throw it downward at an initial velocity of 40 feet per second. c. How far does the rock fall in 2 seconds if you throw it downward with an initial velocity of 40 feet per second. 23. A rocket is fired straight up from ground level with an initial velocity of 800 feet per second. a. How long does it take the rocket to rise 3200 feet? b. When will the rocket hit the ground? 24. A rocket loaded with fireworks is to be shot vertically upward from ground level with an initial velocity of 200 feet per second. When the rocket reaches a height of 400 feet on its upward trip, the fireworks will be detonated. How many seconds after lift-off will this take place? 25. The dimensions of a rectangular box are consecutive integers. If the box has volume of 13,800 cubic centimeters, what are its dimensions? 26. Find a real number that exceeds its cube by 2. 27. The lateral surface area S of the right circular cone at the left in the figure below is given by S pr2r2 h2. produce a cone of height 5 inches and lateral surface area 100 square inches? What radius should be used to h h r b 21. How long does it take an object to reach the ground in each |
case? a. It is dropped from the top of a 640-foot-high building. 28. The lateral surface area of the right square S b2b2 4h2. pyramid at the right in the figure above is given by If the pyramid has height 10 feet and lateral surface area 100 square feet, what is the length of a side b of its base? Section 2.4 Other Types of Equations 107 29. Suppose that the open-top box being made from a sheet of cardboard in Example 7 is required to have at least one of its dimensions greater than 18 inches. What size square should be cut from each corner? 30. A homemade loaf of bread turns out to be a perfect cube. Five slices of bread, each 0.6 inch thick, are cut from one end of the loaf. The remainder of the loaf now has a volume of 235 cubic inches. What were the dimensions of the original loaf? 31. A rectangular bin with an open top and volume of 38.72 cubic feet is to be built. The length of its base must be twice the width and the bin must be at least 3 feet high. Material for the base of the bin costs $12 per square foot and material for the sides costs $8 per square foot. If it costs $538.56 to build the bin, what are its dimensions? 2.4 Other Types of Equations Objectives • Solve absolute-value equations • Solve radical equations • Solve fractional equations Algebraic Definition of Absolute Value Like linear and quadratic equations, other types of equations can be solved algebraically. This section outlines procedures for solving absolute-value, radical, and fractional equations. Definition of Absolute Value The absolute value of a number c is denoted and is defined as follows. c 0 0 If c 0, then 00 If c 66 0, then c. c. c 00 c 00 00 For example, because 5 is positive, 3. 3 To determine 0 1 p 6 6 0. 3 2 0 0 Therefore, the second part of the definition applies. p 6 6 p p 6 0 0 In all cases, the absolute value of a number is nonnegative. 2 1 0 5, 5 0 p 6 0 and because, note that 3 p 3.14 is negative, and Absolute value can also be interpreted geometrically as a distance. 5 and 3 on the number line is 8 units. Observe that the distance between −10 −8 −6 −4 −2 0 |
2 4 6 8 10 8 units Figure 2.4-1 8 Notice that property of absolute value. 5 3 0 0 0 8. 0 This is an example of a key geometric 108 Chapter 2 Equations and Inequalities Absolute Value and Distance If c and d are real numbers, then c d 00 00 is the distance between c and d on the number line. For example, the number thus represents the distance between 5 and Figure 2.4-2. 5 22 can be written as 22 and A on the number line. See B 22 5 5 + √2 −10 −8 −6 −4 −2 – 2 0 2 4 6 8 10 5 Figure 2.4-2 In the special case when c 0 tance from c to 0 is 0 d 0, c 0 0 the distance formula shows that the diswhich is an alternative definition of,. c 0 0 0 Geometric Definition of Absolute Value If c is a real number, then is the distance from c to 0 on the number line. c 00 00 For example, line, as shown below. 0 0 3.5 denotes the distance from 3.5 to 0 on the number −3.5 −10 −8 −6 −4 −2 0 2 4 6 8 10 Figure 2.4-3 Properties of Absolute Value The following properties are helpful in simplifying absolute value expressions. Properties of Absolute Value Let c and d represent real numbers. 777 0 when c 0 c 1. c 00 d 00 c 00 cd 00 00 00 c d ` ` 00 00 0 and 00 c c 00 c d 00 00 00 00 00 00 2. 3. 4. 00 00, where d 0 Section 2.4 Other Types of Equations 109 Illustrations of Properties 2, 3, and 4 for absolute value are shown below. Property 2: Let 3 0 0 3 and c 0 0 3 0 0 3. CAUTION When c and d have c d opposite signs, 0 0 d c is not equal to 0 0 0 0 c 3 For example, if d 5, then and But, 5 0 0 0 0 0 0 0. Property 3: Let Property 4: Let 12 0 6 2 12 2. 0 0 0 c 3. Then 0 c 0 3 0 0 Therefore,. 3 0 0 c 6 and d 2. 12 and d 4. 0 2 0 0 0 Therefore, cd ` ` ` Therefore and The caution shows that general case is called the Triangle Inequality. d 0 0 0 0 0 0 when c |
3 and d 5. The The Triangle Inequality For any real numbers c and d, c d 00 c 00 00 00 d 00. 00 Square Root of Squares When c is a positive number, then negative. Consider the case when 2c2 c. c 3. This is not true when c is 3 2 1 2 2 29 3, which is not 3 is equal to the absolute value of c when c is any real number. 2c 2 But That is, 2 3 1 2 2 29 3 3 0. 0 Square Root of Squares For every real number c, 2c 2 c. 00 00 Solving Absolute-Value Equations Absolute-value equations can be solved by using the definitions. Some equations lend themselves to the geometric definition, while others are solved more easily using the algebraic definition. Graphing techniques can be used to check all solutions. 110 Chapter 2 Equations and Inequalities Technology Tip To compute absolute values on a calculator, use the ABS feature. The ABS feature is in the NUM submenu of the MATH menu of TI and in the NUM submenu of the OPTN menu of Casio. Example 1 Using Absolute Value and Distance Solve x 4 0 0 8 Solution using absolute value and distance. The equation x 4 0 0 8 can be interpreted as the distance from x to 4 is 8 units. See Figure 2.4-4. x −4 −2 −8 −6 0 2 4 6 8 10 12 14 16 x 8 units 8 units Figure 2.4-4 The two possible values of x that are solutions of the original equation are and 12, as shown. 4 ■ NOTE When dealing with long expressions inside absolute value bars, do the computations inside first, and then take the absolute value of the simplified expression. Extraneous Solutions As shown in Example 2 below, some solutions do not make the original equation true when checked by substitution. Such “fake” solutions are called extraneous solutions, or extraneous roots. Because extraneous solutions may occur when solving absolute-value equations, all solutions must be checked by substituting into the original equation or by graphing. Example 2 Using the Algebraic Definition of Absolute Value Solve x 4 0 0 5x 2 Solution by using the algebraic definition of absolute value. The absolute value of any quantity is either the quantity itself or the opposite of the quantity, depending on whether the quantity is positive or x 4 or negative. So, Therefore, the original equation can be rewritten as two equations |
that do not involve absolute value. is either 5x 2 4x 6 x 6 4 3 2 5x 2 0 or 5x 2 x 4 1 x 4 5x 2 2 6x 2 x 2 6 1 3 Each solution must be checked in the original equation. Section 2.4 Other Types of Equations 111 x 3 2 is a solution and checks in the original equation, as shown in Figure 2.4-5. However, do not intersect when x 1 3 x 1 3, Therefore, the only solution of 0 is an extraneous root because the graphs which can be confirmed by substitution. x 3 2 5x 2 x 4 is. 11 3 (, ) 2 2 y = 5x − 2 x −8 −4 0 4 8 −4 −8 Figure 2.4-5 Example 3 Solving an Absolute Value Equation Solve 0 x2 4x 3 2. 0 Solution Use the algebraic definition of absolute value to rewrite the original equation as two equations. x2 4x 3 0 x2 4x 3 2 x2 4x 5 0 2 0 2 x2 4x 3 1 x2 4x 3 2 x2 4x 1 0 2 The equation on the left can be solved by factoring, and the equation on the right by using the quadratic formula. x2 4x 5 0 x2 4x 1 0 x 5 1 21 x 1 2 0 x 5 or x 1 x x 4 ± 242 ± 220 2 1 2 4 ± 225 2 x 2 ± 25 y 8 4 y = x2 + 4x − 3 y = 2 x −8 −4 0 4 8 −4 −8 Figure 2.4-6 See Figure 2.4-6 to confirm that all four values are solutions. ■ Solving Radical Equations Radical equations are equations that contain expressions with a variable under a radical symbol. Although the approximate solutions of a radical equation can be found graphically, exact solutions can be found algebraically in many cases. The algebraic solution method depends on the following fact. NOTE Squaring both sides of an equation may introduce extraneous roots. If A and B are algebraic expressions and A B, for every positive integer n. An Bn then x 2 3, then is also a solution of For example, if x 2 3 x 2 2 9, 1 2 x 2 1 x 2 2 32. 2 2 9. Therefore, every solution of is a solution of However, 1 but 1 2 is not a solution of x 2 |
3. 1 112 Chapter 2 Equations and Inequalities Power Principle CAUTION Although it is always a good idea to verify solutions, solutions to radical equations must be checked in the original equation. 5 0 0 10 Figure 2.4-7 If both sides of an equation are raised to the same positive integer power, then every solution of the original equation is a solution of the new equation. However, the new equation may have solutions that are not solutions of the original one. Example 4 Solving a Radical Equation Solve 5 23x 11 x Solution Rearrange terms to get the radical expression alone on one side. 23x 11 x 5 Square both sides to remove the radical sign and solve the resulting equation. 23x 11 2 x 5 2 A B 2 1 3x 11 x2 10x 25 0 x2 13x 36 0 x 9 x 4 2 x 9 0 or If these values are solutions of the original equation, they should be x 5 x-intercepts of the graph of shows that the graph does not have an x-intercept at is an extraneous solution. But Figure 2.4-7 x 4 y 23x 11. 2 x 4. That is, 1 The graph suggests that can be confirmed by substitution. x 9 is a solution of the original equation, which ■ Sometimes the Power Principle must be applied more than once to eliminate all radicals. Example 5 Using the Power Principle Twice Solve 22x 3 2x 7 2. Solution Rearrange terms so that one side contains only a single radical term. 22x 3 2x 7 2 y 5 Square both sides and isolate the remaining radical. Section 2.4 Other Types of Equations 113 0 10 20 30 40 50 x 5 Figure 2.4-8 22 x 3 A 2 2x 7 2 2 B 2x 7 2x 3 2x 3 x 7 42x 7 4 x 14 42x 7 A A B B 2 2 22x 7 22 Square both sides again, and solve the resulting equation. 2 B 2 1 2 2 42x 7 B 2x 7 x 7 x 14 A x2 28x 196 42 A x2 28x 196 16 2 1 x2 28x 196 16x 112 x2 44x 84 0 x 2 0 x 42 2 1 x 2 0 x 2 2 or 1 x 42 0 x 42 Technology Tip Graphing calculators do not always show all solutions of a radical equation. See Example 3 in Section 2.1 for an illustration of a technological quirk. Verify by |
substitution that a solution. x 2 is an extraneous root but that x 42 is ■ Example 6 Distance Stella is standing at point A on the bank of a river that is 2.5 kilometers wide. She wants to reach point B, which is 15 kilometers downstream on the opposite bank. She plans to row downstream to point C on the opposite shore and then run to B, as shown in Figure 2.4-9. She can row downstream at a rate of 4 kilometers per hour and can run at 8 kilometers per hour. If her trip is to take 3 hours, how far from B should she land? Figure 2.4-9 Solution Refer to Figure 2.4-9. The basic formula for distance can be written in different ways. d rt or t d r 114 Chapter 2 Equations and Inequalities Let x represent the distance between C and B, t represent the time required to run from C to B, and r represent the rate Stella can run (8 kilometers per hour). Therefore, t x 8 Similarly, t d 4 denotes the time needed to run from C to B. can express the time required to row distance d. NOTE To review the Pythagorean Theorem, see the Geometry Review Appendix. 15 x Since be applied to right triangle ADC. is the distance from D to C, the Pythagorean Theorem can d2 15 x 2 1 2 2.52 or equivalently d 2 15 x 1 2 2 6.25 Therefore, the total time for the trip is given by a function of x. T x 1 2 rowing time running time d 4 x 8 x 15 4 1 2 2 6.25 2 x 8 If the trip is to take 3 hours, then x 15.25 x 8 3 Using the viewing window with function. 0 x 20 and 2 y 2, graph this x f 1 2 2 x 15 4 2 1 2 6.25 x 8 3 x 6.74 Find the zeros of f and interpret their values in the context of the problem. x 17.26. The zeros of f are These represent the distances that Stella should land from B. 6.74 represents a downstream destination from A and 17.26 represents an upstream destination from A. Therefore, Stella should land approximately 6.74 kilometers from B to make the downstream trip in 3 hours. and ■ Fractional Equations If f(x) and g(x) are algebraic expressions, the quotient is called a frac- tional expression |
with numerator f(x) and denominator g(x). As in all fractions, the denominator, g(x), cannot be zero. That is, if the frac- 0 is undefined. The following principle is used to solve fractional tion x x f 1 g 1 2 2 equations of the form 0. x x f 1 g 1 2 2 Section 2.4 Other Types of Equations 115 Solving f(x) g(x) 0 f(x) Let solutions of the equation and g (x) represent algebraic expressions. Then the are all values of x such that 0 f(x) g(x) f(x) 0 and g(x) 0. Example 7 Solving a Fractional Equation Solve 6x2 x 1 2x2 9x 5 0. Solution Find all solutions to 6x2 x 1 0. 3x 1 1 3x 1 0 x 1 3 21 6x2 x 1 0 0 2x 1 2 2x 1 0 or x 1 2 10 Discard any solution that makes 2x2 9x 5 0. 10 10 10 Figure 2.4-10a For x 1 3 : For x 1 2 : Therefore, x 1 3 is the solution of a solution. Figure 2.4-10a confirms that x 1 2 is extraneous 70 6x2 x 1 2x2 9x 5 x 1 3 5 0 0, and x 1 2 is not is a solution and that ■ Technology Tip 10.2 x 5 In Figure 2.4-10a, the vertical line shown at graph but is a result of the calculator evaluating the function just to the The calcu- left of lator erroneously connects these points with a near vertical segment. By 7.7 x 1.7 choosing a window such as vertical line will not be drawn. Using the trace feature in Figure 2.4-10b and just to the right of 2.2 y 10.2 is not part of the but not at x 5. x 5,, the near and x 5 identifies a hole at x 1 2, where the function is not defined. 7.7 1.7 2.2 Figure 2.4-10b 116 Chapter 2 Equations and Inequalities Exercises 2.4 In Exercises 1–8, rewrite each statement using the geometric definition of absolute value. Represent each on a number line, and find the value(s) of the variable. 1. The |
distance between y and 2 is 4. 2. The distance between x and 4 is 6. 3. The distance between 3w and 2 is 8. 4. The distance between 4x and 3 is 6. 5. The distance between 2x and 4 is 5. 6. The distance between 4z and 3 is 11. 7. The distance between 3x and 2 is 5. 8. The distance between 4w and 6 is 5 2. In Exercises 9–20, find all real solutions of each equation. 9 0 0 0 2x 3 6x 9 2x 3 0 x x 3 0 x2 4x 1 x2 5x 1 9. 11. 13. 15. 17. 19. 0 0 0 0 0 0 4x 1 4 3 0 0 10. 12. 14. 16. 18. 20. 0 0 0 0 0 0 3x 5 4x 5 3x 2 0 0 0 7 9 5x 4 2x 1 2x 1 0 x2 2x 9 6 0 12x2 5x 7 4 0 21. Explain why there are no real numbers that satisfy the equation 2x2 3x 12. 0 22. Describe in words the meaning of the inequality. 0 a b a b 0 0 Make sure to consider positive and negative values of a and b. 0 0 0 0 25. An instrument measures a wind speed of 20 feet per second. The true wind speed is within 5 feet per second of the measured wind speed. What are the possible values for the true wind speed? 26. For two real numbers s and t, the notation min(s, t) represents the smaller of the two numbers. When common value. It can be shown that min(s, t) can be expressed as shown. min(s, t) represents the s t, min s For each of the following, verify the equation. a. b. c. s 4 s 2 and s t 5 t 3 t 1 and 27. In statistical quality control, one needs to find the proportion of the product that is not acceptable. The upper and lower control limits (CL) are found by solving the following equation in which p is the mean percent defective, and n is the sample size for CL. CL Find CL when p 0.02 and n 200. In Exercises 28–63, find all real solutions of each equation. Find exact solutions when possible, approximate solutions otherwise. 28. 2x 7 4 29. 24x 9 5 30 |
. 23x 2 7 31. 23 5 11x 3 32. 23 6x 10 2 33. 23 x2 1 2 34. 36. 23 x 1 2 4 1 2 2x2 5x 4 2 35. 2x2 x 1 1 37. 2x 7 x 5 23. Joan weighs 120 pounds and her doctor told her 38. 2x 5 x 1 that her weight is 5 percent from her ideal weight. What are the possible values, to the nearest pound, for Joan’s ideal body weight? 24. A tightrope walker is 8 feet from one end of the rope. If he takes 2 steps and each step is 10 inches long, how far is he from the same end of the rope? Give both possible answers. 39. 23x2 7x 2 x 1 40. 24x2 10x 5 x 3 41. 23 x3 x2 4x 5 x 1 42. 23 x3 6x2 2x 3 x 1 43. 25 9 x2 x2 1 44. 24 x3 x 1 x2 1 45. 23 x5 x3 x x 2 46. 2x3 2x2 1 x3 2x 1 47. 2x2 3x 6 x4 3x2 2 48. 23 x4 x2 1 x2 x 5 49. 25x 6 3 2x 3 50. 23y 1 1 2y 4 51. 22x 5 1 2x 3 52. 2x 3 2x 5 4 53. 23x 5 22x 3 1 0 54. 220 x 29 x 3 55. 26x2 x 7 23x 2 2 56. 2x3 x2 3 2x3 x 3 1 57. 2x 2 3 58. 60. 62. x2 2x 1 x 2 0 x2 x 2 x2 5x 5 0 2x2 7x 6 3x2 5x 2 0 59. 61. 63. 2x2 3x 4 x 4 0 x2 3x 2 x2 x 6 0 3x2 4x 1 3x2 5x 2 0 In Exercises 64–67, assume that all letters represent positive numbers. Solve each equation for the required letter. 64. T 2p m g A for g 65. K 1 x2 u2 A for u 66. R 2d2 k2 for d 67. A 1 a2 b2 A for b 68. A rope is to be stretched at uniform height from a tree to |
a fence, 20 feet from the tree, and then to the side of a building, 35 ft from the tree, at a point 30 ft from the fence, as shown in the figure. a. If 63 ft of rope is to be used, how far from the building wall should the rope meet the fence? b. How far from the building wall should the rope meet the fence if as little rope as possible is to be used? Hint: What is the x value of the lowest point on the graph? Section 2.4 Other Types of Equations 117 Tree (aerial view) 30 20 Fence 69. A spotlight is to be placed on a building wall to illuminate a bench that is 32 feet from the base of the wall. The intensity of the light I at the bench is known to be I x d3, where x is the spotlight’s height above the ground and d is the distance from the bench to the spotlight. a. Express I as a function of x. It may help to draw a picture. b. How high should the spotlight be in order to provide maximum illumination at the bench? Hint: What is the x value of the highest point on the graph? Helicopter 1 mile Field Road Anne 70. Anne is standing on a straight road and wants to reach her helicopter, which is located 2 miles down the road from her and a mile off the road in a field. She can run 5 miles per hour on the road and 3 miles per hour in the field. She plans to run down the road, then cut diagonally across the field to reach the helicopter. a. Where should she leave the road in order to reach the helicopter in exactly 42 minutes, that is, 0.7 hour? b. Where should she leave the road in order to reach the helicopter as soon as possible? (see Exercise 68b) 118 Chapter 2 Equations and Inequalities 2.5 Inequalities Objectives • Use interval notation • Solve linear inequalities and compound linear inequalities • Find exact solutions of quadratic and factorable inequalities c 6 d, which is read “c is less than d,’’ means that c is to The statement the left of d on the real number line. Similarly, the statement which is read “c is greater than d,’’ means that c is to the right of d on the real number line. c 7 d, In the set of real numbers, any pair of numbers can be compared |
because the set of real numbers is ordered. That is, for any two real numbers a and b, exactly one of the following statements is true The two statements are equivalent, and both mean that c d, d c read “c is less than or equal to d,” means either c is less than d or c is equal to d. A similar statement applies to is positive. The statement c d. and The statement b 6 c 6 d, b 6 c called a compound inequality, means and simultaneously c 6 d. Interval Notation An interval of numbers is the set of all numbers lying between two fixed numbers. Such sets appear frequently enough to merit special notation. Interval Notation Let c and d be real numbers with c 66 d. [c, d] denotes the set of all real numbers x such that (c, d) denotes the set of all real numbers x such that [c, d) denotes the set of all real numbers x such that c x d. c 66 x 66 d. c x 66 d. (c, d] denotes the set of all real numbers x such that c 66 x d. All four sets above are called intervals from c to d, where c and d are the endpoints of the interval. The interval [c, d] is called the closed interval from c to d because both endpoints are included, as indicated by brackets, and (c, d) is called the open interval from c to d because neither endpoint is included, as indicated by parentheses. The last two intervals in the box above are called half-open intervals, where the bracket indicates which endpoint is included. q NOTE The symbol is read “infinity’’ but does not denote a real number. It is simply part of the notation used to denote half-lines. Section 2.5 Inequalities 119 The half-line extending to the right or left of b is also called an interval. • For the half-line to the right of b, denotes the set of all real numbers x such that denotes the set of all real numbers x such that x b. x 7 b. • For the half-line to the left of b, denotes the set of all real numbers x such that denotes the set of all real numbers x such that 4 2 x b. x 6 b. Similar notation is used for the entire number line. q, q denotes the set of all real numbers. 2 b, q b, q 2 2 |
q, b q, b 3 1 1 1 1 Basic Principles for Solving Inequalities Solving Inequalities Solutions of inequalities in one variable are all values of the variable that make the inequality true. Such solutions may be found by using algebraic, geometric, and graphical methods, each of which is discussed in this section. Whenever possible, algebra will be used to obtain exact solutions. When algebraic methods are tedious or when no algebraic method exists, approximate graphical solutions will be found. Equivalent Inequalities Like equations, two inequalities are equivalent if they have the same solutions. The basic tools for solving inequalities are as follows. Performing any of the following operations on an inequality produces an equivalent inequality. 1. Add or subtract the same quantity on both sides of the inequality. 2. Multiply or divide both sides of the inequality by the same positive quantity. 3. Multiply or divide both sides of the inequality by the same negative quantity, and reverse the direction of the inequality. Note that Principles 1 and 2 are the same as the principles used in solving linear equations, but Principle 3 states that when an inequality is multiplied or divided by a negative number, the inequality sign must be reversed. For example Multiply by 1 and reverse the inequality Solving Linear Inequalities Example 1 Solving a Compound Linear Inequality Solve 2 3x 5 6 2x 11. 120 Chapter 2 Equations and Inequalities Solution A solution of the inequality a solution of both of the following inequalities. 2 3x 5 6 2x 11 is any number that is 2 3x 5 and 3x 5 6 2x 11 Each of these inequalities can be solved by the principles listed above. Subtract 5 2 5 3x Simplify 3 3x Divide by 3 1 x Subtract 2x and Subtract 5 3x 2x 6 11 5 Simplify x 6 6 The solutions are all real numbers that satisfy both is, 1, 6 that Therefore, the solutions are the numbers in the interval as shown in Figure 2.5-1. 1 x 6 6. 1 x and x 6 6,, 3 2 −3 −2 −1 0 1 2 3 4 5 6 Figure 2.5-1 ■ Example 2 Solving a Compound Linear Inequality Solve 4 6 3 5x 6 18. Solution When a variable appears only in the middle of a compound inequality, the process can be streamlined by performing any operation on each part of the compound inequality. 4 6 35x 6 |
18 1 6 5x 6 15 1 5 7 x 7 3 Subtract 3 from each part 5 Divide each part by and reverse direction of the inequalities Intervals are usually written from the smaller to the larger, so the solution to the compound inequality is 3 6 x 6 1 5. −5 −4 −3 −2 −1 0 1 2 3 54 Figure 2.5-2 The solution of the compound inequality is the interval a 3, 1 5 b, as shown in Figure 2.5-2, where open circles indicate that the endpoints are not included in the interval. ■ Section 2.5 Inequalities 121 y f Solving Other Inequalities Although the basic principles play a role in the solution of nonlinear inequalities, geometrically the key to solving such inequalities is the following fact. x x −2 3 g Figure 2.5-3a f − g y −2 3 Figure 2.5-3b The solutions of an inequality of the form intervals on the x-axis where the graph of f is below the graph of g. consist of f(x) 66 g(x) f(x) 77 g(x) The solutions of where the graph of f is above the graph of g. consist of intervals on the x-axis In Figure 2.5-3a, the blue portion of the graph of f is below the graph of correspond to the intervals on the x-axis g. The solutions of f denoted in red. In Figure 2.5-3a, and when x 7 3. 2 6 x 6 1 when The graph of f(x) g(x) 77 0 Although the above procedure can always be used, solutions of an 6 g x are often easier to find by using an inequality expressed as 1 x f equivalent inequality in the form y f(x) g(x) 2 1 lies above the x-axis when 6 0 and below the x-axis when y f Figure 2.5-3b shows the graph of the difference of the two functions shown in Figure 2.5-3a. This graph is below the x-axis in the same intervals where the graph of f is below the graph of g. Therefore, 6 0. the solution of is the same as the solution of (x) g(x) 66 0 can be rewritten in the equivalent Any inequality of the form 2 1 1 g x x by subtracting from both sides of the inequalform 2 2 1 6 g x |
x x ity. The procedure for solving 2 2 and find the intervals on the x-axis where the graph is below the x-axis. A similar procedure applies when the inequality sign is reversed, except that the solution is determined by x-intervals where the graph is above the x-axis. is to graph Example 3 Solving an Inequality Solve x4 10x3 21x2 7 40x 80. y Solution Rewrite the inequality as x4 10x3 21x2 x4 10x3 21x2 40x 80 1 1 2 40x 80 7 0. 2 1 f x is shown in Figure The graph of 2 1 2.5-4. The graph shows that and the other near 2. The portion of the graph above the x-axis is shown in red. has two zeros, one between and x f 1 2 2 x 2 4 6 100 80 60 40 20 0 −2 −40 −60 −80 −100 −6 −4 Figure 2.5-4 122 Chapter 2 Equations and Inequalities Graphing Exploration Use the graphical root finder of a calculator to find approximate values of the x-intercepts. The Exploration shows that the graph of f is above the x-axis approxiso the approximate mately when x 6 1.53 solutions of the original inequality are all numbers x such that or x 6 1.53 x 7 1.89, x 7 1.89. and when ■ Quadratic and Factorable Inequalities The preceding example shows that solving an inequality depends only on knowing the zeros of a function and the places where its graph is above or below the x-axis. In the case of quadratic inequalities or completely factored expressions, exact solutions can by found algebraically. Example 4 Solving a Quadratic Inequality Solve 2x2 3x 4 0. Solution y 8 4 −8 −4 0 4 8 −4 −8 Figure 2.5-5 2x2 3x 4 0 The solutions of f by using the quadratic formula. 2x2 3x 4 x 2 1 are the numbers x where the graph of lies on or below the x-axis. The zeros of f can be found x 3 ± 232 4 2 2 1 2 4 2 1 21 2 3 ± 241 4 x As shown in Figure 2.5-5, the graph lies below the x-axis between the two zeros. Therefore, the solutions of the original inequality are |
all numbers x such that 3 241 4 2.35 x 0.85 x 3 241 4 Exact solution Approximate solution ■ Example 5 Solving an Inequality Solve 21 1 0. Solution 1 f x 2 2 5, x 5 x are easily read from the facThe zeros of 1 2, and 8. Therefore, you need only determine where tored form to be the graph of f is on or below the x-axis. A partial graph of f that clearly shows all three x-intercepts is shown in Figure 2.5-6a. A complete graph, which does not clearly show the x-intercept at 2, is shown in Figure x 8 21 2 1 2 6 Section 2.5 Inequalities 123 2.5-6b. Using both graphs, or using the trace feature on either one, conx 5 and x 8. firms that the graph is on or below the x-axis between Therefore, the solutions of the inequality are all numbers x such that 5 x 8. 5 200,000 10 10 10 10 5 Figure 2.5-6a 1,000,000 Figure 2.5-6b ■ The procedures used in the previous examples may be summarized as follows. Solving Inequalities 1. Write the inequality in one of the following forms. f(x) 77 0 f(x) 0 f(x) 66 0 f(x) 0 2. Determine the zeros of f, exactly if possible, approximately otherwise. 3. Determine the interval, or intervals, on the x-axis where the graph of f is above (or below) the x-axis. Applications Example 6 Solving a Cost Inequality A computer store has determined the cost C of ordering and storing x laser printers. C 2x 300,000 x If the delivery truck can bring at most 450 printers per order, how many printers should be ordered at a time to keep the cost below $1600? Solution To find the values of x that make C less than 1600, solve the inequality 6 1600 or equivalently, 2x 300,000 2x 300,000 1600 6 0. x x 124 500 Chapter 2 Equations and Inequalities In this context, the only solutions that make sense are those between 0 and 450. Therefore, choose a viewing window, such as the one shown in Figure 2.5-7, and graph 0 450 2x 300,000 x f x 1 2 1600. 500 Figure 2.5-7 |
The graph in Figure 2.5-7 shows that the zero of f is and the graph of C is negative, i.e., below the x-axis, for values greater than 300. Therefore, to keep costs under $1600, between 300 and 450 printers should be ordered per delivery. x 300, ■ Exercises 2.5 In Exercises 1–4, express the given statement in symbols. 1. x is nonnegative. 2. t is positive. 3. c is at most 3. 4. z is at least 17. In Exercises 5–10, represent the given interval on a number line. 29. 30. 31. 32 2x 7 3x 3x 2 2 x 1 1 2 2 3x x 5 3 2x 2x 1 3 2 5. (0, 8] 8. 1 1, 1 2 6. 9. 1 0, q 2 q, 0 1 7. 10. 2, 1 2, 7 4 2 3 3 4 In Exercises 11–16, use interval notation to denote the set of all real numbers x that satisfy the given inequality. 11. 5 x 8 12. 2 x 7 13. 3 6 x 6 14 14. 7 6 x 6 135 15. x 8 16. x 12 In Exercises 17–36, solve the inequality and express your answer in interval notation. 17. 2x 4 7 18. 3x 5 7 6 19. 3 5x 6 13 20. 2 3x 6 11 21. 6x 3 x 5 22. 5x 3 2x 7 23. 5 7x 6 2x 4 24. 5 3x 7 7x 3 25. 2 6 3x 4 6 8 26. 1 6 5x 6 6 9 27. 0 6 5 2x 11 28. 4 7 3x 6 0 33. 2x 3 5x 6 6 3x 7 34. 4x 2 6 x 8 6 9x 1 35. 3 x 6 2x 1 3x 4 36. 2x 5 4 3x 6 1 4x In Exercises 37–40, the constants a, b, c, and d are positive. Solve each inequality for x. 37. ax b 6 c 38. d cx 7 a 39. 0 6 x c 6 a 40. d 6 x c 6 d In Exercises 41 – 70, solve the inequality. Find exact solutions when possible, and approximate them otherwise. 41. x2 4x 3 0 |
42. x2 7x 10 0 43. x2 9x 15 0 44. x2 8x 20 0 45. 8 x x2 0 46. 4 3x x2 0 47. x3 x 0 48. x3 2x2 x 7 0 49. x3 2x2 3x 6 0 50. x4 14x3 48x2 0 51. x4 5x2 4 6 0 52. x4 10x2 9 0 53. x3 2x2 5x 7 2x 1 54. x4 6x3 2x2 6 5x 2 55. 2x4 3x3 6 2x2 4x 2 56. x5 5x4 7 4x3 3x2 2 57. 59. 61. 63. 3x 1 2x 4 7 0 x2 x 2 x2 2x 65. 2 x 3 1 x 1 58. 60. 62. 64. 66. 2x 1 5x 3 0 2x2 x 1 x2 4x 4 0 x 5 2x 3 2 2x 67. 68. 69. x3 3x2 5x 29 x2 7 7 3 x4 3x3 2x2 2 x 2 7 15 2x2 6x 8 2x2 5x 3 6 1 Be alert for hidden behavior. 70. 1 x2 In Exercises 71–73, read the solution of the inequality from the given graph. 71. 3 2x 6 0.8x 7 y y = 3 − 2x (−1.43, 5.86) y = 0.8x + 7 8 6 4 2 x 2 4 −4 −2 0 −2 Section 2.5 Inequalities 125 72. 8 7 5x 7 − 5x| (0.4, 3) y = 3 (2.4, 3) x −8 −4 0 4 8 −4 −8 73. x2 3x 1 4 y = x2 + 3x + 1 y 10 (−3.79, 4) 8 6 4 2 (0.79, 4) y = 4 x −4 −2 0 −2 2 4 74. The graphs of the revenue and cost functions for a manufacturing firm are shown in the figure. a. What is the break-even point? b. Which region represents profit? 60,000 40,000 20,000 Cost Revenue 1000 2000 3000 4000 75. One freezer costs $623.95 and uses 90 kilowatt |
hours (kwh) of electricity each month. A second freezer costs $500 and uses 100 kwh of electricity each month. The expected life of each freezer is 12 years. What is the minimum electric rate in cents per kwh for which the 12-year total cost (purchase price freezer? electricity costs) will be less for the first 126 Chapter 2 Equations and Inequalities 76. A business executive leases a car for $300 per month. She decides to lease another brand for $250 per month, but has to pay a penalty of $1000 for breaking the first lease. How long must she keep the second car in order to come out ahead? 77. One salesperson is paid a salary of $1000 per month plus a commission of 2% of her total sales. A second salesperson receives no salary, but is paid a commission of 10% of her total sales. What dollar amount of sales must the second salesperson have in order to earn more per month than the first? 78. A developer subdivided 60 acres of a 100-acre tract, leaving 20% of the 60 acres as a park. Zoning laws require that at least 25% of the total tract be set aside for parks. For financial reasons the developer wants to have no more than 30% of the tract as parks. How many one-quarter-acre lots can the developer sell in the remaining 40 acres and still meet the requirements for the whole tract? 79. If $5000 is invested at 8%, how much more should be invested at 10% in order to guarantee a total annual interest income between $800 and $940? 80. How many gallons of a 12% salt solution should be added to 10 gallons of an 18% salt solution in order to produce a solution whose salt content is between 14% and 16%? 81. Find all pairs of numbers that satisfy these two conditions: Their sum is 20 and the sum of their squares is less than 362. 82. The length of a rectangle is 6 inches longer than its width. What are the possible widths if the area of the rectangle is at least 667 square inches? 83. It costs a craftsman $5 in materials to make a medallion. He has found that if he sells the 50 x dollars each, where x is medallions for the number of medallions produced each week, then he can sell all that he makes. His fixed costs are $350 per week. If he wants to sell all he makes and show a profit each |
week, what are the possible numbers of medallions he should make? 84. A retailer sells file cabinets for 80 x dollars each, where x is the number of cabinets she receives from the supplier each week. She pays $10 for each file cabinet and has fixed costs of $600 per week. How many file cabinets should she order from the supplier each week in order to guarantee that she makes a profit? In Exercises 85 – 88, you will need the following formula for the height h of an object above the ground at time t seconds, where denotes initial velocity and h0 v0 h 16t 2 v0t h0 denotes initial height. 85. A toy rocket is fired straight up from ground level with an initial velocity of 80 feet per second. During what time interval will it be at least 64 feet above the ground? 86. A projectile is fired straight up from ground level with an initial velocity of 72 feet per second. During what time interval is it at least 37 feet above the ground? 87. A ball is dropped from the roof of a 120-foot-high building. During what time period will it be strictly between 56 feet and 39 feet above the ground? 88. A ball is thrown straight up from a 40-foot-high tower with an initial velocity of 56 feet per second. a. During what time interval is the ball at least 8 feet above the ground? b. During what time interval is the ball between 53 feet and 80 feet above the ground? Section 2.5A Excursion: Absolute-Value Inequalities 127 2.5.A Excursion: Absolute-Value Inequalities Objectives • Solve absolute-value inequalities by the Intersection Method • Solve absolute-value inequalities by the xIntercept Method y 50 40 30 20 f Polynomial and rational inequalities involving absolute value can be solved by either of two graphing methods. Intersection Method • Graph the expressions on each side of the inequality. • Determine the intervals on the x-axis where the graph of the expression on one side of the inequality is above or below the graph of the expression on the other side of the inequality. x-Intercept Method • Rewrite the inequality in an equivalent form with 0 on one side of the inequality. • Graph the function given by the nonzero side of the inequality. • Determine the x-values where the graph is above or below the x-axis. Example 1 Solving an Absolute-Value Inequality Using the Intersection Method g |
Solve 0 x4 2x2 x 2 6 11x. 0 Solution x −4 −2 0 2 4 −20 Figure 2.5.A-1 x4 2x2 x 2 The solutions of the x-intervals for which the graph of the graph of 11x. g x 0 0 6 11x f x 1 2 1 2 0 can be found be determining is below x4 2x2 x 2 0 x 0.17 A graphical intersection finder shows that the points of intersection occur and the graph of f is below the graph of g when between them, as shown in Figure 2.5.A-1. Therefore, approximate solutions of the original inequality are all x such that x 1.92, and 0.17 6 x 6 1.92. ■ Example 2 Solving an Absolute-Value Inequality Using the x-Intercept Method Solve x 4 x 2 ` ` 7 3. Solution Rewrite x 4 x 2 ` ` 7 3 as x 4 x 2 ` ` 3 7 0, graph, and find the intervals on the x-axis where the graph is above the x-axis. 128 Chapter 2 Equations and Inequalities y 8 4 −8 −4 0 4 8 x −4 −8 Figure 2.5.A-2 The graph of f is above the x-axis between the two zeros, which can be found algebraically or graphically to be function is not defined at denominator. Therefore, the solutions of x 2 x 1 2 and x 5. Notice that the because a fraction cannot have a zero x 4 x 2 ` ` 3 7 0 and (2, 5). The solution can also be written out. 6 x 6 2 or 2 6 x 6 5 ■ are the x-intervals 1 2 a, 2 b 1 2 Algebraic Methods Most linear and quadratic inequalities that contain absolute values can be solved exactly by using algebra. In fact, this is often the easiest way to solve such inequalities. The key to the algebraic method is to interpret the absolute value of a number as distance on the number line. For example, the inequality 5 states that r 0 0 the distance from r to 0 is less than or equal to 5 units. A glance at the number line in Figure 2.5.A-3a shows that these are the numbers r such that 5 r 5. −8 −6 −4 −2 0 2 4 6 8 5 units 5 units Figure 2.5.A- |
3a Similarly, the inequality 5 states that r 0 0 the distance from r to 0 is greater than or equal to 5 units. These values are the numbers r such that Figure 2.5.A-3b. r 5 or r 5, as shown in −8 −6 −4 −2 0 2 4 6 8 5 units 5 units Figure 2.5.A-3b Similar conclusions hold in the general cases, with 5 replaced by any number k. Section 2.5A Excursion: Absolute-Value Inequalities 129 Absolute-Value Inequalities Let k be a positive real number and r any real number. k k r r 00 00 00 00 is equivalent to is equivalent to k r k. r k or r k. Example 3 Solving an Absolute-Value Inequality Solve 0 3x 7 0 11. Solution Apply the first fact in the box above, with 11 place of k, and conclude that 3x 7 3x 7 is equivalent to in place of r and 11 in 0 0 11 3x 7 11 4 3x 18 4 3 6 x Add 7 Divide by 3 Therefore, the solution to 4 3 that is, all x such that 3x is all numbers in the interval 11 x 6. ■ Example 4 Solving an Absolute-Value Inequality Solve 0 5x 2 0 7 3. Solution Apply the second fact in the box with k, and in place of 7. 5x 2 in place of r, 3 in place of 5x 2 6 3 or 5x Therefore, the solutions of the original inequality are the numbers in either of the intervals q, 1 1 or 2 1 5 a, q b, that is, x 6 1 or x 7 1 5. ■ Example 5 Solving an Absolute-Value Inequality Solve 0 x2 x 4 2. 0 Solution Rewrite the absolute-value inequality as two quadratic inequalities using the algebraic definition. 130 Chapter 2 Equations and Inequalities The inequality 2 x2 x 4 0 x2 x 4 2 x2 x 2 0 0 is equivalent to two inequalities. or x2 x 4 2 x2 x 6 0 The solutions are all numbers that are solutions of either one of the two inequalities shown above. The solutions are the intervals on the x-axis that are determined by the following x2 x 2 x2 x 6 is on or below the x-axis is on or above the x-axis y 8 4 f(x) = x2 − x − 2 |
y 8 4 g(x) = x2 − x − 6 −8 −4 0 4 8 −8 −4 0 4 8 x x −4 −8 −4 −8 Figure 2.5.A-4a Figure 2.5.A-4b As shown in Figure 2.5.A-4a, the graph of the x-axis when f x 1 2 x2 x 2 is on or below 1 x 2. As shown in Figure 2.5.A-4b, the graph of above the x-axis when g x 2 1 x2 x 6 is on or x 2 or x 3. Therefore, the solutions of the original inequality are all numbers x such that x 2 or 1 x 2 or x 3, as shown in Figure 2.5.A-4c. −2 −1 2 3 Solutions of x2 − x − 2 ≤ 0 Solutions of x2 − x − 6 ≥ 0 Solutions of either one Figure 2.5.A-4c ■ Section 2.5A Excursion: Absolute-Value Inequalities 131 Example 6 Interpreting an Absolute-Value Inequality Let a and d represent real numbers with positive. d 0 0 6 d geometrically. x a a. Interpret b. Draw the interval represented. c. Write the equivalent simplified extended inequality. d. Interpret the last inequality. Solution a. Geometrically, x a 0 0 6 d means that the distance from x to a is less than d. b. c. a − δ a a + δ δ units δ units Figure 2.5.A- Add a to each term d. The solutions of the inequality are all numbers strictly between a d and a d. Exercises 2.5.A In Exercises 1–32, solve the inequality. Find exact solutions when possible, and approximate values otherwise. 1. 3. 5. 7. 9. 11. 13. 3x 2 3 2x 2x 3 5x 12 5 2x 2x. 4. 6. 8. 5x 1 4 5x 3x 1 2 3x 10. 12. 14. ` ` ` 5 6 3x 6 7 6 ` x 1 3x 15. 17. 19. 21. 23. 25. 27. 28. 29. 31. ■ 3x 1 1 2x ` ` 2 x2 4 3 0 1 x2 1 ` ` 2 x2 x 4 0 |
x2 3x 4 2 6 6 0 7 1 4x x3 0 1 4x 2 3x ` ` 6 1 x2 2 x2 2 6 1 7 4 0 0 x2 x 1 1 0 3x2 8x 2 x5 x3 1 0 6 2 0 6 2 16. 18. 20. 22. 24. 26. 0 0 0 0 x4 x3 x2 x 1 7 4 x3 6x2 4x 30. 2x2 2x 12 x3 x2 x 2 ` 7 2 32. 0 ` 0 ` x2 9 x2 4 0 x2 x 2 x2 x 2 ` 6 2 7 3 132 Chapter 2 Equations and Inequalities 33. Critical Thinking Let E be a fixed real number. Show that every solution of solution of 5x 4 11 0 2 0 1 x 3 0 6 E. 6 E 5 0 is also a with a 6 b. Show that the solutions of 34. Critical Thinking Let a and b be fixed real numbers are all x such that a 6 x 6 b Important Concepts Section 2.1 Complete graphs.......................... 82 The intersection method.................... 82 Zeros, x-intercepts, and solutions............. 83 The x-intercept method..................... 84 Technological quirks....................... 84 Section 2.2 Section 2.3 Section 2.4 Section 2.5 Definition of a quadratic equation............ 88 Techniques used to solve quadratic equations... 88 Solving a quadratic equation by factoring...... 89 The Zero Product Property.................. 89 Solutions of........................ 89 Completing the square..................... 91 Solving ax2 bx c 0 by completing x2 k |
the square............................. 92 The quadratic formula..................... 92 The discriminant.......................... 93 Polynomial equations in quadratic form....... 94 Applied problems guideline................. 97 Solutions in context....................... 98 Interest applications...................... 100 Distance applications..................... 101 Other applications........................ 102 Algebraic definition of absolute value........ 107 Geometric definition of absolute value........ 108 Properties of absolute value................ 108 Square root of squares..................... 109 Solving absolute-value equations............ 109 Extraneous solutions...................... 110 Solving radical equations.................. 111 Power Principle.......................... 112 Fractional equations...................... 114 Interval notation......................... 118 Equivalent inequalities.................... 119 Solving linear inequalities.................. 119 Solving |
other inequalities.................. 121 Quadratic and factorable inequalities......... 122 133 134 Chapter Review Section 2.5.A The intersection method................... 127 The x-intercept method.................... 127 Algebraic methods....................... 128 Important Facts and Formulas To solve an equation of the form Method, use two steps. f x 1 2 g x 1 2 with the Intersection 1. Graph 2. Find the x-coordinate of each point of intersection and y2 y1 x x 1 1 2 2 g f When f is a function and a is a real number, the following are equivalent statements: • a is a zero of the function 2 • a is an x-intercept of the graph of f • a is a solution, or root, of the equation x 1 y f 0 f x 1 2 To solve an equation by the x-Intercept Method, use three steps. 0 f x 1. Rewrite the equation in the form 2. Graph f 3. Find the x-intercepts of the graph. The x-intercepts of the graph 0. of f are the solutions of 1 The only number whose square root is zero is zero itself. x f 1 2 2 A fraction is zero only when its numerator is zero and its denominator is nonzero. Quadratic Formula If a 0, then the solutions of ax2 bx c 0 are x b ± 2b2 4ac 2 a a 0, then the number of real solutions of If 1, or 2, depending on whether the discriminant, tive, zero, or positive, respectively. ax2 bx c 0 b2 4ac is 0,, is nega- Absolute Value if x 0 x x if x 6 0 x x 0 0 0 0 represents the distance between c and d on the number line. c d 0 represents the distance between c and 0 on the number line. c 0 0 0 Review Questions In Questions 1–8, solve the equation graphically. You need only find solutions in the given interval. Chapter Review 135 0, q 2 q, 0 2 0, q |
2 q, 1 2 10, q 2 0, q 2 0, q 2 5 Section 2.1 1. x3 2x2 11x 6; 2. x3 2x2 11x 6; 3. x4 x3 10x2 8x 16; 4. 2x4 x3 2x2 6x 2 0; 5. 6. x3 2x2 3x 4 x2 2x 15 0; 3x4 x3 6x2 2x x5 x3 2 0; 7. 2x3 2x2 3x 5 0; 8. 21 2x 3x2 4x3 x4 0; Section 2.2 9. Solve for x: 3x2 2x 5 0 10. Solve for y: 3y2 2y 5 11. Solve for z: 5z2 6z 7 12. Solve for x: 325x2 17x 127 0 13. Solve for x: x4 11x2 18 0 14. Solve for x: x6 4x3 4 0 15. Find the number of real solutions of the equation 20x2 12 31x. 16. For what value of k does the equation kt2 5t 2 0 have exactly one real solution for t? Section 2.3 17. A jeweler wants to make a 1-ounce ring composed of gold and silver, using $200 worth of metal. If gold costs $600 per ounce and silver $50 per ounce, how much of each metal should she use? 18. A calculator is on sale for 15% less than the list price. The sale price, plus a 5% shipping charge, totals $210. What is the list price? 19. Karen can do a job in 5 hours and Claire can do the same job in 4 hours. How long will it take them to do the job together? 20. A car leaves the city traveling at 54 mph. A half hour later, a second car leaves from the same place and travels at 63 mph along the same road. How long will it take for the second car to catch up to the first? 21. A 12-foot rectangular board is cut into two pieces so that one piece is four times as long as the other. How long is the bigger piece? 22. George owns 200 shares of stock, 40% of which are in the computer industry. How many more shares must he buy in order to have 50% of his total shares in computers |
? 136 Chapter Review 23. A square region is changed into a rectangular one by making it 2 feet longer and twice as wide. If the area of the rectangular region is three times larger than the area of the original square region, what was the length of a side of the square before it was changed? 24. The radius of a circle is 10 inches. By how many inches should the radius be increased so that the area increases by 5p square inches? 25. If c x 2 1 is the cost of producing x units, then c x 1 x 2 is the average cost per unit. The cost of manufacturing x caseloads of ballpoint pens is given by 600x2 600x x2 1 c x 2 1 c x where order to have an average cost of $25? 2 1 is in dollars. How many caseloads should be manufactured in 26. An open-top box with a rectangular base is to be constructed. The box is to be at least 2 inches wide, twice as long as it is wide, and must have a volume of 150 cubic inches. What should be the dimensions of the box if the surface area is 90 square inches? Section 2.4 27. Simplify: b2 2b 1 0 0 In Exercises 28–40, find all real exact solutions. 28. 30. 32 2x 1 3 x 4 0 34. 26x2 7x 5 0 36. x2 x 2 x 2 0 38. 23 1 t2 2 0 0 29. 31. 33. 35. x 3 5 2 0 3x 1 4 0 2x2 x 2 0 x2 6x 8 x 1 0 37. 2x 1 2 x 39. 2x 1 2x 1 1 40. 23 x4 2x3 6x 7 x 3 Section 2.5 41. Express in interval notation: 8 a. The set of all real numbers that are strictly greater than b. The set of all real numbers that are less than or equal to 5. 42. Express in interval notation: a. The set of all real numbers that are strictly between b. The set of all real numbers that are greater than or equal to 5, but and 9; 6 strictly less than 14. 43. Solve for x: 44. Solve for x: 3 x 4 5 x. 1 2 4 6 2x 5 6 9. 45. On which intervals is 2x 1 3x 1 6 1? Chapter Review Chapter Review 137 137 46. On |
which intervals is 2 x 1 6 x? 47. Solve for x: 48. Solve for x: x 1 2 x2 1 1 2 1 x2 x 7 12. x 0. 2 49. If a. c. x 3 2x 3 x 3 2x 3 3 2x x 3 7 1, then which of these statements is true? 6 1, 7 1 6 1 2x 3 x 3 2x 3 6 x 3 b. d. e. None of these 50. If a. c. e. 0 6 r s t then which of these statements is false? b. 51. Solve and express your answer in interval notation: 2x 3 5x 9 6 3x 4. In Questions 52–61, solve the inequality. 52. x2 x 20 7 0 53. x 2 x 4 3 54. 55. 57 x2 x 9 x 3 6 1 x2 x 5 x2 2 7 2 Section 2.5.A 59. y 2 3 ` 3x 2 61. 0 5 2 ` 0 56. 58. x2 x 6 x 3 7 1 x4 3x2 2x 3 x2 4 6 1 60. 1 1 x2 ` ` 12 Maximum Area ctions There are two related branches of calculus: differential calculus and integral calculus. Differential calculus is a method of calculating the changes in one variable produced by changes in a related variable. It is often used to find maximum or minimum values of a function. Integral calculus is used to calculate quantities like distance, area, and volume. This Can Do Calculus finds the maximum area of the triangle formed by folding a piece of paper using different methods. The Maximum Area of a Triangle Problem -inch piece of paper is folded over to the oppoOne corner of an site side, as shown in Figure 2.C-1. A triangle is formed, and its area 8.5 11 formula is A 1 2 (base)(height). The following Example will find the length of the base that will produce the maximum area of the triangle using numerical, graphical, and algebraic methods. Example 1 Numerical Method 8.5 11 One corner of an -inch piece of paper is folded over to the opposite side, as shown in Figure 2.C-1. The area of the darkly shaded triangle at the lower left is the focus of this problem. a. Determine the shortest and the longest base that will produce a trian- gle by folding the paper. |
b. Measure the height when x has the lengths given in the chart, and cal- culate the area in each case. c. Create a scatter plot of the data. d. Estimate the length of the base that produces the maximum area, and state the approximate maximum area. Solution a. The base must be greater than 0 and less than 8.5 inches, and nt in the chart indicates that no triangle can be formed with a base length of 9 inches. b. The values shown in the chart may vary from your data. Base – 1 Height 4.25 Area 2.125 – 2 4.1 4.1 – 3 3.6 5.4 – 4 3.3 6.6 – 5 2.75 6.875 – 6 2 6 – 7 1.5 – 8 0.5 5.25 2 – 9 nt - x Figure 2.C-1 138 7 0 0 7 0 0 c. The graph of the data is shown in Figure 2.C-2. d. A maximum area of 6.875 in2 appears to occur when the base length Figure 2.C-2 8.5 y y x Figure 2.C-3 is 5 in. ■ 9 In Example 1, all calculations were accomplished with measurements. where y is the height of the trianNotice that the hypotenuse is gle. (Why?) 8.5 y, Because each triangle formed was a right triangle, the Pythagorean Theorem can be used to find an expression that gives the height as a function of the length of the base. That function can then be used to write a function that gives area in terms of the length of the base. Example 2 Algebraic Method Find a function of the base to represent the area of the triangle described in Example 1, graph the function along with the scatter plot of the data found in Example 1. Find the length of the base that produces maximum area. What is the maximum area? Solution The Pythagorean Theorem yields the following equation. 8. 72.25 17y y 2 y 2 72.25 17y x y 72.25 x 2 17 Height as a function of the base Hence, the area is represented by A 1 2 1 x 2 a 2 72.25 x 17 b. Using the max- imum finder on a calculator indicates that the maximum area of occurs at approximately x 4.9 in. 6.95 in2 ■ 9 Figure 2.C-4 To get exact values |
of x and the area, differential calculus is needed. However, graphing technology can provide very good approximations. Exercises In each problem, find the maximum by using a numerical method like the one shown in Example 1, and then by using an analytical and graphical method like the one shown in Example 2. Answer all questions given in the two examples. 1. Ten yards of wire is to be used to create a rectangle. What is the maximum possible area of the rectangle? 2. A rectangle is bounded by the x- and y-axes and the semicircle dimensions of the rectangle with maximum area? What are the y 236 x 2. 3. A rectangle is bounded by the x- and y-axes and the line y 1 2. What are the dimensions of 4 x 2 the rectangle with maximum area? 139 C H A P T E R 3 Functions and Graphs This is rocket science! If a rocket is fired straight up from the ground, its height is a function of time. This function can be adapted to give the height of any object that is falling or thrown along a vertical path. The shape of the graph of the function, a parabola, appears in applications involving motion, revenue, communications, and many other topics. See Exercise 50 of Section 3.3. 140 Chapter Outline 3.1 Functions 3.2 Graphs of Functions 3.3 Quadratic Functions 3.4 Graphs and Transformations 3.4.A Excursion: Symmetry 3.5 Operations on Functions 3.5.A Excursion: Iterations and Dynamical Systems 3.6 Inverse Functions 3.7 Rates of Change Chapter Review can do calculus Instantaneous Rates of Change Interdependence of Sections 3.1 > 3.2 > > > 3.3 3.4 3.5 > > 3.6 3.7 T he concept of a function and function notation are central to mod- ern mathematics and its applications. In this chapter you will review functions, operations on functions, and how to use function notation. Then you will develop skill in constructing and interpreting graphs of functions. To understand the origin of the concept of a function it may help to consider some “real-life” situations in which one numerical quantity depends on, corresponds to, or determines another. Example 1 Determining Inputs and Outputs of Functions Describe the set of inputs, the set of outputs, and the rule for the following functions: a. The amount of |
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