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20 b a At what speed can the greatest number of cars travel the highway safely? 54. Volume of Water Between 0°C and 30°C, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is given by the formula V 999.87 0.06426T 0.0085043T 2 0.0000679T 3 Find the temperature at which the volume of 1 kg of water is a minimum. 55. Coughing When a foreign object that is lodged in the trachea (windpipe) forces a person to cough, the diaphragm thrusts upward, causing an increase in pressure in the lungs. At the same time, the trachea contracts, causing the expelled air to move faster and increasing the pressure on the foreign object. According to a mathematical model of coughing, the velocity √ of the airstream through an average-sized person’s trachea is related to the radius r of the trachea (in centimeters) by the function √ r 1 2 3. Determine the value of r for which √ is a maximum. ▼ DISCOVE RY • DISCUSSION • WRITI NG 56. Functions That Are Always Increasing or Decreasing Sketch rough graphs of functions that are deﬁned for all real numbers and that exhibit the indicated behavior (or explain why the behavior is impossible). (a) f is always increasing, and x f 2 1 (b) f is always decreasing, and x f 2 1 (c) f is always increasing, and x f 2 1 (d) f is always decreasing, and x f 2 1 for all x for all x for all x for all x 0 0 0 0 57. Maxima and Minima In Example 7 we saw a real-world situation in which the maximum value of a function is important. Name several other everyday situations in which a maximum or minimum value is important. 58. Minimizing a Distance When we seek a minimum or maximum value of a function, it is sometimes easier to work with a simpler function instead. 1f (a) Suppose 0 for all x. Explain, where why the local minima and maxima of f and g occur at the same values of xb) Let be the distance between the point (3, 0) and the on the graph of the parabola y x 2. Express g x g 2 1 x, x 2 point 1 as a function of x. 2
(c) Find the minimum value of the function g that you found in part (b). Use the principle described in part (a) to simplify your work. 236 CHAPTER 3 \| Functions 3.4 Average Rate of Change of a Function LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find the average rate of change of a function ■ Interpret average rate of change in real-world situations ■ Recognize that a function with constant average rate of change is linear Functions are often used to model changing quantities. In this section we learn how to ﬁnd the rate at which the values of a function change as the input variable changes. ■ Average Rate of Change We are all familiar with the concept of speed: If you drive a distance of 120 miles in 2 hours, then your average speed, or rate of travel, is 60 mi/h. 120 mi 2 h Now suppose you take a car trip and record the distance that you travel every few min- utes. The distance s you have traveled is a function of the time t: total distance traveled at time t s t 1 2 We graph the function s as shown in Figure 1. The graph shows that you have traveled a total of 50 miles after 1 hour, 75 miles after 2 hours, 140 miles after 3 hours, and so on. To ﬁnd your average speed between any two points on the trip, we divide the distance traveled by the time elapsed. Let’s calculate your average speed between 1:00 P.M. and 4:00 P.M. The time elapsed is 4 1 3 hours. To ﬁnd the distance you traveled, we subtract the distance at 1:00 P.M. from the distance at 4:00 P.M., that is, 200 50 150 mi. Thus, your average speed is 150 mi average speed distance traveled time elapsed 150 mi 3 h 50 mi/h The average speed that we have just calculated can be expressed by using function notation: s (mi) 200 100 3 h 0 1 2 3 4 t (h) average speed 200 50 3 50 mi/h FIGURE 1 Average speed Note that the average speed is different over different time intervals. For example, between 2:00 P.M. and 3:00 P.M. we ﬁnd that average speed 140 75 1 65 mi/h Finding average rates of change is important in many contexts. For instance, we might be interested in knowing how quickly
the air temperature is dropping as a storm approaches or how fast revenues are increasing from the sale of a new product. So we need to know how to determine the average rate of change of the functions that model these quantities. In fact, the concept of average rate of change can be deﬁned for any function. SE CTI O N 3. 4 \| Average Rate of Change of a Function 237 AVERAGE RATE OF CHANGE The average rate of change of the function y f x between x a and x b is average rate of change 1 change in y change in The average rate of change is the slope of the secant line between x a and x b on the graph of f, that is, the line that passes through and b, f a, f a b. 1 1 22 1 1 22 y f(b) f(a) average rate of change= f(b)-f(a) b-a y=Ï f(b)-f(a) b-a 0 a b x E X AM P L E 1 \| Calculating the Average Rate of Change 2, x 3 x f For the function 2 of change between the following points: (a) x 1 and x 3 1 2 1 (b) x 4 and x 7 whose graph is shown in Figure 2, ﬁnd the average rate ▼ SO LUTI O N (a) Average rate of change Deﬁnition Use f(x) (x 3)2 y 16 9 1 0 1 3 4 7 x FIGURE b) Average rate of change Deﬁnition 16 1 3 5 Use f(x) (x 3)2 ✎ Practice what you’ve learned: Do Exercise 11. ▲ E X AM P L E 2 \| Average Speed of a Falling Object If an object is dropped from a high cliff or a tall building, then the distance it has fallen after t seconds is given by the function Find its average speed (average rate of change) over the following intervals: (a) Between 1 s and 5 s (b) Between t a and t a h 16t2. d t 1 2 238 CHAPTER 3 \| Functions d(t) = 16t2 ▼ SO LUTI O N (a) Average rate of change 16 2 16 2 5 1 400 16 4 96 ft/s Deﬁnition 2 1 1 2 Use d(t) 16t2 Function: In t seconds
the stone falls 16t2 ft. (b) Average rate of change Deﬁnition 1 16 2 a h 2 16 a 2 a h 1 2 1 a2 2ah h2 a2 h 16 16 1 1 16h 16 1 2ah h2 h 2a h h 2a h 1 2 2 2 a 1 2 Use d(t) 16t2 2 Expand and factor 16 2 Simplify numerator Factor h Simplify ✎ Practice what you’ve learned: Do Exercise 15. ▲ The average rate of change calculated in Example 2(b) is known as a difference quotient. In calculus we use difference quotients to calculate instantaneous rates of change. An example of an instantaneous rate of change is the speed shown on the speedometer of your car. This changes from one instant to the next as your car’s speed changes. The graphs in Figure 3 show that if a function is increasing on an interval, then the average rate of change between any two points is positive, whereas if a function is decreasing on an interval, then the average rate of change between any two points is negative. y y=Ï y y=Ï Slope>0 Slope< increasing Average rate of change positive ƒ decreasing Average rate of change negative FIGURE 3 °F 70 60 50 40 30 0 Time Temperature (°F) 8:00 A.M. 9:00 A.M. 10:00 A.M. 11:00 A.M. 12:00 NOON 1:00 P.M. 2:00 P.M. 3:00 P.M. 4:00 P.M. 5:00 P.M. 6:00 P.M. 7:00 P.M. 38 40 44 50 56 62 66 67 64 58 55 51 SE CTI O N 3. 4 \| Average Rate of Change of a Function 239 E X AM P L E 3 \| Average Rate of Temperature Change The table gives the outdoor temperatures observed by a science student on a spring day. Draw a graph of the data, and ﬁnd the average rate of change of temperature between the following times: (a) 8:00 A.M. and 9:00 A.M. (b) 1:00 P.M. and 3:00 P.M. (c) 4:00 P.M. and 7:00 P.M. ▼ SO LUTI O N A graph of the
temperature data is shown in Figure 4. Let t represent time, measured in hours since midnight (so, for example, 2:00 P.M. corresponds to t 14). Deﬁne the function F by (a) Average rate of change 2 t temperature at time t F 1 temperature at 9 A.M. temperature at 8 A.M. 9 8 40 38 The average rate of change was 2°F per hour. (b) Average rate of change 8 10 12 14 16 18 h temperature at 3 P.M. temperature at 1 P.M. 15 13 67 62 2 F 15 1 15 13 2.5 13 F 2 2 1 FIGURE 4 The average rate of change was 2.5°F per hour. (c) Average rate of change temperature at 7 P.M. temperature at 4 P.M. 19 16 51 64 3 F 19 1 19 16 4.3 16 F 1 2 2 The average rate of change was about 4.3°F per hour during this time interval. The negative sign indicates that the temperature was dropping. ✎ Practice what you’ve learned: Do Exercise 25. ▲ x mx b ■ Linear Functions Have Constant Rate of Change f For a linear function the average rate of change between any two points is the same constant m. This agrees with what we learned in Section 2.4: that the slope of a is the average rate of change of y with respect to x. On the other hand, if line a function f has constant average rate of change, then it must be a linear function. You are asked to prove this fact in Exercise 34. In the next example we ﬁnd the average rate of change for a particular linear function. y mx b 2 1 E X AM P L E 4 \| Linear Functions Have Constant Rate of Change. Find the average rate of change of f between the following points. 2 1 f x 3x 5 Let (a) x 0 and x 1 (b) x 3 and x 7 (c) x a and x a h What conclusion can you draw from your answers? 240 CHAPTER 3 \| Functions ▼ SO LUTI O N (a) Average rate of change 16 3a 5 4 Average rate of change (b) (c) Average rate of change 3a 3h 5 3a 5 h 3h h 3 It appears that the average rate of change is always 3 for this function. In fact, part (c) proves that the rate
of change between any two arbitrary points x a and x a h is 3. ✎ Practice what you’ve learned: Do Exercise 21. ▲ 3. ▼ CONCE PTS 1. If you travel 100 miles in two hours, then your average speed for the trip is average speed 2. The average rate of change of a function f between x a and x b is average rate of change 3. The average rate of change of the function f x 1 and x 5 is x 2 between x 1 2 average rate of change 4. (a) The average rate of change of a function f between x a and a, f x b a is the slope of the b b, f. 2 2 (b) The average rate of change of the linear function and 2 2 1 1 1 1 line between 3x 5 between any two points is. f x 1 2 ▼ SKI LLS 5–8 ■ The graph of a function is given. Determine the average rate of change of the function between the indicated points on the graph. 5. 6. 7. y 4 2 _1 0 5 x 9–20 ■ A function is given. Determine the average rate of change of the function between the given values of the variable. ✎ ✎ 9. 10. 11. 12. 13. 14. 15. 16 3x 2; x 2, x 3 5 1 2 x; x 1, x 5 t 2 2t; t 1, t 4 1 3z2; z 2, z 0 x3 4x2; x 0, x 10 x x4; x 1, x 3 3x2; x 2, x 2 h 4 x2; x 1, x 1 h 1 x ; x 1, x a 2 x 1 ; x 0, x h 2 t ; t a, t a h 1t; t a, t a h 17. g 18. g 19. f 20 21–22 ■ A linear function is given. (a) Find the average rate of. (b) Show change of the function between that the average rate of change is the same as the slope of the line. x a h x a and SE CTI O N 3. 4 \| Average Rate of Change of a Function 241 (a) What was the average rate of change of P between x 20 and x 40? (b) Interpret the value of the average rate of change that you found in part (a). P (thousands) 50
40 30 20 10 0 10 20 30 40 50 x (years) ✎ 25. Population Growth and Decline The table gives the population in a small coastal community for the period 1997–2006. Figures shown are for January 1 in each year. (a) What was the average rate of change of population be- tween 1998 and 2001? (b) What was the average rate of change of population be- tween 2002 and 2004? (c) For what period of time was the population increasing? (d) For what period of time was the population decreasing? Year 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 Population 624 856 1,336 1,578 1,591 1,483 994 826 801 745 ✎ 21. f x 1 2 1 2 x 3 22. g x 1 2 4x 2 ▼ APPLICATIONS 23. Changing Water Levels The graph shows the depth of water W in a reservoir over a one-year period as a function of the number of days x since the beginning of the year. What was the average rate of change of W between x 100 and x 200? W (ft) 100 75 50 25 0 100 200 300 x (days) 24. Population Growth and Decline The graph shows the population P in a small industrial city from 1950 to 2000. The variable x represents the number of years since 1950. 26. Running Speed A man is running around a circular track that is 200 m in circumference. An observer uses a stopwatch to record the runner’s time at the end of each lap, obtaining the data in the following table. (a) What was the man’s average speed (rate) between 68 s and 152 s? (b) What was the man’s average speed between 263 s and 412 s? (c) Calculate the man’s speed for each lap. Is he slowing down, speeding up, or neither? Time (s) Distance (m) 32 68 108 152 203 263 335 412 200 400 600 800 1000 1200 1400 1600 242 CHAPTER 3 \| Functions 27. CD Player Sales The table shows the number of CD players sold in a small electronics store in the years 1993–2003. (a) What was the average rate of change of sales between 1993 and 2003? (b) What was the average rate of change of sales between 1993 and 1994? 30. Farms in the United States The graph gives the number of farms in the United States from 1850 to 2000. (
a) Estimate the average rate of change in the number of farms between (i) 1860 and 1890 and (ii) 1950 and 1970. (b) In which decade did the number of farms experience the (c) What was the average rate of change of sales between greatest average rate of decline? 1994 and 1996? (d) Between which two successive years did CD player sales increase most quickly? Decrease most quickly? Year 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 CD players sold 512 520 413 410 468 510 590 607 732 612 584 28. Book Collection Between 1980 and 2000, a rare book col- lector purchased books for his collection at the rate of 40 books per year. Use this information to complete the following table. (Note that not every year is given in the table.) Year 1980 1981 1982 1985 1990 1992 1995 1997 1998 1999 2000 Number of books 420 460 1220 29. Cooling Soup When a bowl of hot soup is left in a room, the soup eventually cools down to room temperature. The temperature T of the soup is a function of time t. The table below gives the temperature (in °F) of a bowl of soup t minutes after it was set on the table. Find the average rate of change of the temperature of the soup over the ﬁrst 20 minutes and over the next 20 minutes. Which interval had the higher average rate of change? y 7000 6000 5000 4000 3000 2000 1860 1880 1900 1920 1940 1960 1980 2000 x ▼ DISCOVE RY • DISCUSSION • WRITI NG 31. 100-Meter Race A 100-m race ends in a three-way tie for ﬁrst place. The graph shows distance as a function of time for each of the three winners. (a) Find the average speed for each winner. (b) Describe the differences between the ways in which the three runners ran the race. d (m) 100 50 0 A C B 5 10 t (s) t (min) T (°F) t (min) T (°F) 0 5 10 15 20 25 30 200 172 150 133 119 108 100 35 40 50 60 90 120 150 94 89 81 77 72 70 70 32. Changing Rates of Change: Concavity The two tables and graphs on the next page give the distances traveled by a racing car during two different 10-s portions of a race. In each case, calculate the average speed at which the car is
traveling between the observed data points. Is the speed increasing or decreasing? In other words, is the car accelerating or decelerating on each of these intervals? How does the shape of the graph tell you whether the car is accelerating or decelerating? (The ﬁrst graph is said to be concave up, and the second graph is said to be concave down.) (a) Time (s) Distance (ft) 0 2 4 6 8 10 0 34 70 196 490 964 (b) Time (s) Distance (ft) 30 32 34 36 38 40 5208 5734 6022 6204 6352 6448 (ft) d 800 600 400 200 0 42 6 8 10 (s) t d (ft) 6400 6000 5600 5200 0 30 t (s) 40 SE CTI ON 3. 5 \| Transformations of Functions 243 33. Linear Functions Have Constant Rate of Change If mx b x f change of f between any two real numbers is a linear function, then the average rate of and 1 2 x2 is average rate of change f 1 x2 2 x2 x1 f 1 x1 x12 Calculate this average rate of change to show that it is the same as the slope m. 34. Functions with Constant Rate of Change Are Linear If the function f has the same average rate of change c between any two points, then for the points a and x we have Rearrange this expression to show that f x cx f a ca 2 and conclude that f is a linear function. 1 1 2 1 2 3.5 Transformations of Functions LEARNING OBJECTIVES After completing this section, you will be able to: ■ Shift graphs vertically ■ Shift graphs horizontally ■ Stretch or shrink graphs vertically ■ Stretch or shrink graphs horizontally ■ Determine whether a function is odd or even In this section we study how certain transformations of a function affect its graph. This will give us a better understanding of how to graph functions. The transformations that we study are shifting, reﬂecting, and stretching. ■ Vertical Shifting Adding a constant to a function shifts its graph vertically: upward if the constant is positive and downward if it is negative. In general, suppose we know the graph of graphs of y f. How do we obtain from it the x 1 2 Recall that the graph of the function f is the same as the graph of the equation and y f 1 x c c 0 2 The y-coordinate of each point on the graph of
y-coordinate of the corresponding point on the graph of y f x of 1 obtain the graph of x is c units above the 2 1 y f. So we obtain the graph 1 x upward c units. Similarly, we 2 y f by shifting the graph of simply by shifting the graph of downward c units 244 CHAPTER 3 \| Functions VERTICAL SHIFTS OF GRAPHS Suppose c 0. To graph To graph, shift the graph of, shift the graph of upward c units. downward c units. y 0 y=f(x)+c c y=f(x) x y 0 y=f(x) c y=f(x)-c x E X AM P L E 1 \| Vertical Shifts of Graphs x2 x to sketch the graph of each function. Use the graph of (a) f 1 x2 3 g x 2 1 2 ▼ SO LUTI O N sketched again in Figure 1. (a) Observe that (b) h x x2 2 1 x 2 f 1 2 x2 The function was graphed in Example 1(a), Section 3.2. It is g x 1 2 x2 3 f 3 x 1 2 So the y-coordinate of each point on the graph of g is 3 units above the corresponding point on the graph of f. This means that to graph g we shift the graph of f upward 3 units, as in Figure 1. (b) Similarly, to graph h, we shift the graph of f downward 2 units, as shown. y 2 0 g(x) = x 2 + 3 f(x) = x2 h(x) = x2 – 2 2 x FIGURE 1 ✎ Practice what you’ve learned: Do Exercises 11 and 13. ▲ ■ Horizontal Shifting Suppose that we know the graph of y f x c y f x 1 and y f 2 x c. How do we use it to obtain the graphs of The value of 2 to the left of x, it follows that the graph of f 1 x c 2 1 at x is the same as the value of 1 x f 2 1 x c at x c. Since x c is c units is just the graph of SE CTI ON 3. 5 \| Transformations of Functions 245 shifted to the right c units. Similar reasoning shows that the graph of graph of is the shifted to the left c units. The following box summarizes these facts HORIZONTAL SHIF
TS OF GRAPHS Suppose c 0. To graph To graph, shift the graph of, shift the graph of y=f(x-c) to the right c units. to the left c units=f(x+c) y=Ï c x c 0 y=Ï x y 0 E X AM P L E 2 \| Horizontal Shifts of Graphs x2 Use the graph of (a to sketch the graph of each function. x 2 (b ▼ SO LUTI O N (a) To graph g, we shift the graph of f to the left 4 units. (b) To graph h, we shift the graph of f to the right 2 units. The graphs of g and h are sketched in Figure 2. 1 2 g(x) = (x + 4)2 f (x) = x 2 ™ FIGURE 2 _ 4 y 1 0 h(x) = (x – 2)2 1 x ✎ Practice what you’ve learned: Do Exercises 15 and 17. ▲ René Descartes (1596–1650) was born in the town of La Haye in southern France. From an early age Descartes liked mathematics because of “the certainty of its results and the clarity of its reasoning.” He believed that to arrive at truth, one must begin by doubting everything, including one’s own existence; this led him to formulate perhaps the best-known sentence in all of philosophy: “I think, therefore I am.” In his book Discourse on Method he described what is now called the Cartesian plane. This idea of combining algebra and geometry enabled mathematicians for the ﬁrst time to “see” the equations they were studying. The philosopher John Stuart Mill called this invention “the greatest single step ever made in the progress of the exact sciences.” Descartes liked to get up late and spend the morning in bed thinking and writing. He invented the coordinate plane while lying in bed watching a ﬂy crawl on the ceiling, reasoning that he could describe the exact location of the ﬂy by knowing its distance from two perpendicular walls. In 1649 Descartes became the tutor of Queen Christina of Sweden. She liked her lessons at 5 o’clock in the morning, when, she said, her mind was sharpest. However, the change from his usual habits and the ice-
cold library where they studied proved too much for Descartes. In February 1650, after just two months of this, he caught pneumonia and died. 246 CHAPTER 3 \| Functions MATHEMATICS IN THE MODERN WORLD Computers For centuries machines have been designed to perform speciﬁc tasks. For example, a washing machine washes clothes, a weaving machine weaves cloth, an adding machine adds numbers, and so on. The computer has changed all that. The computer is a machine that does nothing—until it is given instructions on what to do. So your computer can play games, draw pictures, or calculate p to a million decimal places; it all depends on what program (or instructions) you give the computer. The computer can do all this because it is able to accept instructions and logically change those instructions based on incoming data. This versatility makes computers useful in nearly every aspect of human endeavor. The idea of a computer was de- scribed theoretically in the 1940s by the mathematician Allan Turing (see page 160) in what he called a universal machine. In 1945 the mathematician John Von Neumann, extending Turing’s ideas, built one of the ﬁrst electronic computers. Mathematicians continue to develop new theoretical bases for the design of computers. The heart of the computer is the “chip,” which is capable of processing logical instructions. To get an idea of the chip’s complexity, consider that the Pentium chip has over 3.5 million logic circuits! E X AM P L E 3 \| Combining Horizontal and Vertical Shifts 1x 3 4. Sketch the graph of x f 2 1 ▼ SO LUTI O N We start with the graph of it to the right 3 units to obtain the graph of upward 4 units to obtain the graph of x f y 1x y 1x 3 1x 3 4 1 2 (Example 1(c), Section 3.2) and shift. Then we shift the resulting graph shown in Figure 3. y 4 0 f (x) = x – 3 + 4 (3, 4 FIGURE 3 ✎ Practice what you’ve learned: Do Exercise 27. ▲ 1 x and y f y f ■ Reﬂecting Graphs Suppose we know the graph of. How do we use it to obtain the graphs of y f is simply the negative of the y-coordinate of the corresponding point on the graph of y f in the x-axis.
On at x, so the other hand, the value of 1 in the y-axis. The followthe desired graph here is the reﬂection of the graph of ing box summarizes these observations.. So the desired graph is the reﬂection of the graph of x x 2 1 y f at x is the same as the value of? The y-coordinate of each point on the graph of REFLECTING GRAPHS To graph To graph y f y f x 1 x 1, reﬂect the graph of, reﬂect the graph of in the x-axis. in the y-axis. y 0 y=Ï x y=_Ï y=f(_x) y 0 y=Ï x E X AM P L E 4 \| Reﬂecting Graphs Sketch the graph of each function. (a) f x x2 (b) g x 1x 1 2 ▼ SO LUTI O N (a) We start with the graph of y x 2. The graph of 1 2 reﬂected in the x-axis (see Figure 4). x 2 is the graph of =x™ 2 x f(x)=_x™ SE CTI ON 3. 5 \| Transformations of Functions 247 (b) We start with the graph of y 1x (Example 1(c) in Section 3.2). The graph of x 1x g the domain of the function is the graph of g 2 1 y 1x x 1x is 1 2 x 5 0 x 0. 6 reﬂected in the y-axis (see Figure 5). Note that g(x)=œ∑_x y 1 0 y=œ∑x 1 x FIGURE 5 FIGURE 4 ✎ Practice what you’ve learned: Do Exercises 19 and 21. ▲ ■ Vertical Stretching and Shrinking Suppose we know the graph of y cf? The y-coordinate of x 1 y-coordinate of x of vertically stretching or shrinking the graph by a factor of c.. How do we use it to obtain the graph of at x is the same as the corresponding x multiplied by c. Multiplying the y-coordinates by c has the effect y f x 1 y cf y f 2 1 2 2 1 2 VERTICAL STRETCHING AND SHRINKING OF GRAPHS To graph : y
cf x 2 1 If c 1, stretch the graph of If 0 c 1, shrink the graph of y f x 2 1 y f y y=c Ï vertically by a factor of c. vertically by a factor of c. x 1 2 y y=Ï 0 y=Ï x 0 x y=c Ï c >1 0 <c <1 E X AM P L E 5 \| Vertical Stretching and Shrinking of Graphs Use the graph of (a) 3x2 g x f x 1 2 x2 (b) h x 1 3 x2 to sketch the graph of each function. 1 2 ▼ SO LUTI O N (a) The graph of g is obtained by multiplying the y-coordinate of each point on the 1 2 graph of f by 3. That is, to obtain the graph of g, we stretch the graph of f vertically by a factor of 3. The result is the narrower parabola in Figure 6. (b) The graph of h is obtained by multiplying the y-coordinate of each point on the graph of f by. That is, to obtain the graph of h, we shrink the graph of f vertically by a factor of. The result is the wider parabola in Figure 6. 1 3 1 3 y 4 f (x) = x 2 g(x) = 3x 2 h(x) = x 2 1 3 0 1 x ✎ Practice what you’ve learned: Do Exercises 23 and 25. ▲ FIGURE 6 We illustrate the effect of combining shifts, reﬂections, and stretching in the following example. 248 CHAPTER 3 \| Functions E X AM P L E 6 \| Combining Shifting, Stretching, and Reﬂecting Sketch the graph of the function x2 ▼ SO LUTI O N the graph of the graph of 1 2 f x y Starting with the graph of x 3, we ﬁrst shift to the right 3 units to get 2. Then we reﬂect in the x-axis and stretch by a factor of 2 to get 2 x 3 2. Finally, we shift upward 1 unit to get the graph of 1 shown in Figure 7 = x2 y = (x – 3)2 (3, 1) 1 x f (x) = 1 – 2(x – 3)2 FIGURE 7 ✎ Practice what you’ve learned
: Do Exercise 29. y = –2(x – 3)2 ▲ x 2 1 y f, then how is the graph of ■ Horizontal Stretching and Shrinking Now we consider horizontal shrinking and stretching of graphs. If we know the graph of y f cx 2 at cx. Thus, the x-coordinates in the graph at x is the same as the y-coordinate of multiplied by c. of Looking at this the other way around, we see that the x-coordinates in the graph of y f multiplied by 1/c. In other words,, we must shrink (or stretch) the to change the graph of graph horizontally by a factor of 1/c, as summarized in the following box. related to it? The y-coordinate of 2 x 1 are the x-coordinates in the graph of to the graph of correspond to the x-coordinates in the graph of y f 1 y f cx cx cx cx HORIZONTAL SHRINKING AND STRETCHING OF GRAPHS To graph y f : cx 1 y f If c 1, shrink the graph of If 0 c 1, stretch the graph of 2 x 2 1 y f horizontally by a factor of 1/c. horizontally by a factor of 1/c. x 1 2 y y=f(cx) y y=f(cx) 0 x y=Ï 0 x y=Ï c >1 0 <c <1 E X AM P L E 7 \| Horizontal Stretching and Shrinking of Graphs y f x 1 2 is shown in Figure 8 on the next page. Sketch the graph of each The graph of function. y f (a) 2x 1 2 (b) y f 1 2x Sonya Kovalevsky (1850–1891) is considered the most important woman mathematician of the 19th century. She was born in Moscow to an aristocratic family. While a child, she was exposed to the principles of calculus in a very unusual fashion: Her bedroom was temporarily wallpapered with the pages of a calculus book. She later wrote that she “spent many hours in front of that wall, trying to understand it.” Since Russian law forbade women from studying in universities, she entered a marriage of convenience, which allowed her to travel to Germany and obtain a doctorate in mathematics from the University of Göttingen. She eventually was awarded a full professorship
at the University of Stockholm, where she taught for eight years before dying in an inﬂuenza epidemic at the age of 41. Her research was instrumental in helping to put the ideas and applications of functions and calculus on a sound and logical foundation. She received many accolades and prizes for her research work. SE CTI ON 3. 5 \| Transformations of Functions 249 y 1 0 1 x FIGURE 8 y f x 1 2 ▼ SO LUTI O N Using the principles described in the preceding box, we obtain the graphs shown in Figures 9 and 10. y 1 0 1 2 1 x _1 y 1 0 y f 2x FIGURE 9 FIGURE 10 ✎ Practice what you’ve learned: Do Exercise 53. 2 1 y f 1 2x B A 1 2 x ▲ ■ Even and Odd Functions f If a function f satisﬁes even function. For instance, the function f x x 1 2 2 1 for every number x in its domain, then f is called an x2 is even because 2x2 x2 f x 1 2 The graph of an even function is symmetric with respect to the y-axis (see Figure 11). This means that if we have plotted the graph of f for x 0, then we can obtain the entire graph simply by reﬂecting this portion in the y-axis. for every number x in its domain, then f is called an odd If f satisﬁes x 1 function. For example, the function f x f 2 2 1 x3 is odd because 3x3 x3 f x 1 2 The graph of an odd function is symmetric about the origin (see Figure 12). If we have plotted the graph of f for x 0, then we can obtain the entire graph by rotating this portion through 180° about the origin. (This is equivalent to reﬂecting ﬁrst in the x-axis and then in the y-axis.) y Ï=x™ y Ï=x£ _x 0 x x _x 0 x x FIGURE 11 function. f x 1 2 x2 is an even FIGURE 12 function. f x 1 2 x3 is an odd 250 CHAPTER 3 \| Functions EVEN AND ODD FUNCTIONS Let f be a function. x f is even if 1 x f is odd if for all x in the domain of f. for all x in the domain of f. 2
f(_x) Ï _x 0 x x _x f(_x) y 0 Ï x x The graph of an even function is symmetric with respect to the y-axis. The graph of an odd function is symmetric with respect to the origin. E X AM P L E 8 \| Even and Odd Functions Determine whether the functions are even, odd, or neither even nor odd. (a) x f x5 x 1 x4 2x x2 1 2 x g h (b) x (c) 2 1 2 1 ▼ SO LUTI O N x (a) f 1 (b) (c) g x 1 1 2 So g is even. 2 h 1 2 x h Since 1 nor odd x5 x f x 1 x5 x 1 2 2 Therefore, f is an odd function. 1 x 1 2 4 1 x4 and 2 2x x2 h x h 1 2 x, we conclude that h is neither even 2 1 ✎ Practice what you’ve learned: Do Exercises 65, 67, and 69. ▲ The graphs of the functions in Example 8 are shown in Figure 13. The graph of f is symmetric about the origin, and the graph of g is symmetric about the y-axis. The graph of h is not symmetric either about the y-axis or the origin. 2.5 Ï=x∞+x 2.5 2.5 h(x)=2x-x™ _1.75 1.75 _2 2 _1 3 FIGURE 13 _2.5 (a) _2.5 (b) ˝=1-x¢ _2.5 (c) 3. ▼ CONCE PTS 1. Fill in the blank with the appropriate direction (left, right, up, or down). (a) The graph of y f x 1 2 (b) The graph of y f 1 by shifting y f 1 by shifting y f x 1 2 x 3 2 is obtained from the graph of x 3 3 units. is obtained from the graph of 2 3 units. 2. Fill in the blank with the appropriate direction (left, right, up, or down). (a) The graph of y f x 1 2 (b) The graph of y f 1 by shifting y f 1 by shifting y f x 1 2 x 3 2 is obtained from the graph of x 3 3 units. is obtained from the graph of 2
3 units. 3. Fill in the blank with the appropriate axis (x-axis or y-axis). x is obtained from the graph of (a) The graph of y f 1 2 (b) The graph of y f x by reﬂecting in the. is obtained from the graph of 1 by reﬂecting in the. Match the graph with the functionb) (da) (c II IV III ▼ SKI LLS 5–8 ■ Explain how the graph of g is obtained from the graph of f. x 2 x2 2 x 4 x3 4 5. (a) (b) 6. (a) (b x2, g x2, g x3, g x 1 x3. (a) f (b SE CTI ON 3. 5 \| Transformations of Functions 251 f f x x 1x, g 1x, g 8. (a) (b) 2 1 2 1 9. Use the graph of 1 1 x f x x 2 2 x2 1x 1 1x 1 in Example 3 to graph the 2 1 x2 1 x 1 1 x2 following. (a) g (b) g (c) g (d) g 1 1 1 1 10. Use the graph of following. (a) g (b) g (c) g (d) g x x x x 1 2 1x 2 1x 1 1x 2 2 1x 1 1 1 1 1 2 2 2 2 1x in Example 4 to graph the 11–34 ■ Sketch the graph of the function, not by plotting points, but by starting with the graph of a standard function and applying transformations. ✎ ✎ ✎ ✎ ✎ ✎ ✎ ✎ ✎ ✎ 11. 13. 15. 17. 19. 21. 23. 25. 27. 29. 31. 33 x2 1 1x 1 x 5 2 1x 4 x3 1 2 2 1 y 24 x y 1 4 x2 1x 4 3 y 1 2 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34 x2 23 x x y 52x y 1 2 0 0 y 1x 4 3 y 2 1x 35–44 ■ A function f is given, and the indicated transformations are applied to its graph (in the given order). Write the equation for the ﬁnal transformed graph. ; shift upward
3 units ; shift downward 1 unit ; shift 2 units to the left ; shift 1 unit to the right x2 x3 1x 23 x x 0 0 x f 2 2 units 1 f f x x 1 1 2 2 24 x x2 x2 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. ; shift 3 units to the right and shift upward 1 unit ; shift 4 units to the left and shift downward x 0 0 ; reﬂect in the y-axis and shift upward 1 unit ; shift 2 units to the left and reﬂect in the x-axis x f 2 units, and shift 3 units to the right 1 2 ; stretch vertically by a factor of 2, shift downward x x f 1 unit, and shift upward 3 units 2 1 0 0 ; shrink vertically by a factor of, shift to the left 1 2 252 CHAPTER 3 \| Functions 45–50 ■ The graphs of f and g are given. Find a formula for the function g. f(x) = x2 45. 46. 47. 48. 49(x) = x3 1 x g f(x)=\| x \| 1 x g f(x)=\| x \| 1 x y 1 0 f(x)= x 1 x g 50. y f(x)= x2 2 0 2 x g 51–52 ■ The graph of its graph. y f x 1 2 is given. Match each equation with 51. (a) (c) 52. (a) (c) y f 1 y 2f x 4 2 x 6 1 (b) (d) y f x 1 y f 3 2x 2 2 1 2 ➁ _6 _3 _3 ➃ (b) (d) 3 2 y 1 3f _6 _3 ➂ _3 ➀ Ï ➁ 3 6 x ✎ 53. The graph of f is given. Sketch the graphs of the following functions. y f (a) 1 y 2f (c) y f (eb) (d) (f 2f 54. The graph of g is given. Sketch the graphs of the following x y g 1 y g x 1 y 2g 2 x 1 2 2 2 functions. y g (a) 1 y g (c) 1 y g (eb) (d) (f) y 2 0 2 x 64. If f 55. The graph of g is given
. Use it to graph each of the following functions. y g (a) 2x 1 2 (b 56. The graph of h is given. Use it to graph each of the following SE CTI O N 3. 5 \| Transformations of Functions 253 6, 6 by 4 62. Viewing rectangle y 1 1x (a) 3 (c) y 1 2 1x 3 4, 4 3 (b) 4 y (d) y 1 1x 3 1 2 1x 3 3 63. If f x 22x x2, graph the following functions in the 1 2 5, 5 viewing rectangle 3 to the graph in part (a)? (a) x y f (b) 1 2 22x x2 x 1 2 5, 5 viewing rectangle 3 to the graph in part (a)? y f (a) x 2 1 y f (c) 1 y f (e) x 1 2x 2 A B by 4, 4 3 4. How is each graph related 4 y f 2x 1 2 (c) y f A 1 2x B, graph the following functions in the by 4, 4 3 4. How is each graph related 4 (b) (d) y f y f x 2 2x 1 1 2 65–72 ■ Determine whether the function f is even, odd, or neither. If f is even or odd, use symmetry to sketch its graph. ✎ ✎ ✎ 65. 67. 69. f f f 71 x4 x2 x x3 x 1 13 x 66. 68. 70. f f f 72 x3 x4 4x2 3x3 2x2 1 x 1 x 73–74 ■ The graph of a function deﬁned for x 0 is given. Complete the graph for x 0 to make (a) an even function and (b) an odd function. functions. y h (a) 3x 1 2 _3 (b) y h 1 3x A B 73 74 is 75–78 ■ These exercises show how the graph of obtained from the graph of x. y f x2 4 2 1 and 75. The graphs of are shown. Explain how the graph of g is obtained from the graph of f x2 4 y 8 4 0 _4 _2 y 8 4 2 x _2 0 _4 2 x x 4 6 2 Ï=≈-4 ˝=\| ≈-4 \| 57–58 ■ Use the graph
of 2 to graph the indicated function. x f 1 “ x‘ described on page 218 57. y “ 2x‘ 58. y “ 1 4x‘ 59–62 ■ Graph the functions on the same screen using the given viewing rectangle. How is each graph related to the graph in part (a)? y 14 x 5 y 4 214 x 5 59. Viewing rectangle y 14 x y 214 x 5 (a) (c) 3 8, 8 by 4 60. Viewing rectangle y x 0 0 y 3 (a) (c) x 0 0 61. Viewing rectangle y x6 y 1 3 x6 (a) (c) 3 8, 8 3 4 by 4, 6 4 by 4 2, 8 3 (b) (d) 6, 6 3 (b) (d) 4b) (d) y 1 31 is shown. Use this graph to 80. Changing Temperature Scales The temperature on a cer- 254 CHAPTER 3 \| Functions 76. The graph of f x sketch the graph of 1 2 x4 4x2 g x 1 2 0 x4 4x2. 0 y 4 2 _3 _1 1 3 x _4 77–78 ■ Sketch the graph of each function. 77. (a) 78. (a) f f x x 1 1 2 2 4x x2 x3 (b) (b 4x x2 0 x3 0 ▼ APPLICATIONS 79. Sales Growth The annual sales of a certain company can be, where t represents modeled by the function t years since 1990 and (a) What shifting and shrinking operations must be performed t f 1 is measured in millions of dollars. 4 0.01t2 f 2 on the function to obtain the function y f 2 1 y t2? t 1 2 (b) Suppose you want t to represent years since 2000 instead of 1990. What transformation would you have to apply to the function function t to accomplish this? Write the new 1 that results from this transformation. y f t y g 2 1 2 tain afternoon is modeled by the function t C 1 2 t2 2 where t represents hours after 12 noon measured in °C. (a) What shifting and shrinking operations must be performed 0 t 6 and C is 2 2 1 1 on the function? (b) Suppose you want to measure the temperature in °F in- to obtain the function 2 1 t y C y t2 t y C stead. What transformation
would you have to apply to the function 2 1 the relationship between Celsius and Fahrenheit degrees is 5C 32 given by that results from this transformation. to accomplish this? (Use the fact that.) Write the new function y F F 9 t 2 1 ▼ DISCOVE RY • DISCUSSION • WRITI NG 81. Sums of Even and Odd Functions f g functions, is sum necessarily odd? What can you say about the sum if one is odd and one is even? In each case, prove your answer. necessarily even? If both are odd, is their If f and g are both even 82. Products of Even and Odd Functions Answer the same questions as in Exercise 81, except this time consider the product of f and g instead of the sum. 83. Even and Odd Power Functions What must be true about the integer n if the function f x x n 2 is an even function? If it is an odd function? Why do you think the names “even” and “odd” were chosen for these function properties? 1 3.6 Combining Functions LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find sums, differences, products, and quotients of functions ■ Add functions graphically ■ Find the composition of two functions ■ Express a given function as a composite function In this section we study different ways to combine functions to make new functions. The sum of f and g is deﬁned by The name of the new function is “f g.” So this sign stands for the operation of addition of functions. The sign on the right side, however, stands for addition of the numbers x f 1 and g x. 1 2 2 ■ Sums, Differences, Products, and Quotients f g f/g Two functions f and g can be combined to form new functions, in a manner similar to the way we add, subtract, multiply, and divide real numbers. For example, we deﬁne the function f g, f g, and fg by f g x f x g x 2 is called the sum of the functions f and g; its value at x is g and. Of course, the sum on the right-hand side makes sense only if both are deﬁned, that is, if x belongs to the domain of f and also to the domain of g. So if The new function SE CTI ON 3.6 \| Combining Functions 255 f
g the domain of f is A and the domain of g is B, then the domain of. Similarly, we can deﬁne the difference of these domains, that is, uct fg, and the quotient f/g of the functions f and g. Their domains are case of the quotient we must remember not to divide by 0. A B is the intersection f g, the prodA B, but in the ALGEBRA OF FUNCTIONS Let f and g be functions with domains A and B. Then the functions fg, and f g, f g, 1 f/ are deﬁned as follows fg Domain A B Domain A B Domain A B Domain AM P L E 1 \| Combinations of Functions and Their Domains and g x 1x. Let f x 1 2 1 x 2 (a) Find the functions (b) Find f g 4, 1 2 1 2 1 ▼ SO LUTI O N (a) The domain of f is 2 1 f g, f g of the domains of f and g is f g, fg 4, 2 1 2 1 fg, and f/g, and 4 2 2 1 1 and their domains. f/, and the domain of g is 6 x 5 0 x 0 6. The intersection x 5 0 x 0 and x 2 0, 2 3 2 2, q 1 2 6 To divide fractions, invert the denominator and multiply: 1/ 1 x 2 1x 2 2 1/ x 2 1 1x/ 1x 1x 2 Thus, we have fg 1x x 2 1 x 2 2 f/g 1x 1x Domain 1x Domain Domain Domain and x 2 x 0 and x 2 6 6 x 0 and x 2 x 0 and x 2 6 6 Note that in the domain of we exclude 0 because g 0 0. 1 2 (b) Each of these values exist because x 4 is in the domain of each function fg 14 5 2 14 14 ✎ Practice what you’ve learned: Do Exercise 5. 2 14 a 1 ▲ 256 CHAPTER 3 \| Functions The graph of the function can be obtained from the graphs of f and g by graphical addition. This means that we add corresponding y-coordinates, as illustrated in the next example. f g E X AM P L E 2 \| Using Graphical Addition y y=˝ The graphs of f and g are shown in Figure 1. Use
graphical addition to graph the function f g. ▼ SO LUTI O N We obtain the graph of to of PR to obtain the point S on the graph of x 2 as shown in Figure 2. This is implemented by copying the line segment PQ on top. f by “graphically adding” the value of =Ï x y FIGURE 1 FIGURE 2 Graphical addition y=( f+g)(x) y=˝ f (x) g(x) y=Ï f (x) x S R Q P ✎ Practice what you’ve learned: Do Exercise 15. ▲ 1 2 ■ Composition of Functions Now let’s consider a very important way of combining two functions to get a new function. Suppose 1x and x f x g 1 f 2 g x2 1 f x. We may deﬁne a function h as x2 1 2x2 1 h x 1 1 2 1 22 1 The function h is made up of the functions f and g in an interesting way: Given a number x, we ﬁrst apply to it the function g, then apply f to the result. In this case, f is the rule “take the square root,” g is the rule “square, then add 1,” and h is the rule “square, then add 1, then take the square root.” In other words, we get the rule h by applying the rule g and then the rule f. Figure 3 shows a machine diagram for h. 2 x input g x2+1 f ≈+1 œ∑∑∑∑∑ output FIGURE 3 The h machine is composed of the g machine (ﬁrst) and then the f machine. In general, given any two functions f and g, we start with a number x in the domain of is in the domain of f, we can then calculate g and ﬁnd its image the value of that is obtained by substituting g into f. It is called the composition (or composite) of f and g and is denoted by f g g. The result is a new function (“f composed with g”).. If this number f 22 22 SE CTI ON 3.6 \| Combining Functions 257 COMPOSITION OF FUNCTIONS Given two functions f and g, the composite function sition of f and g) is de
ﬁned by f g (also called the compo 22 f g The domain of of f. In other words, can picture f g 1 is the set of all x in the domain of g such that g g is deﬁned whenever both and f g g x x f is in the domain are deﬁned. We x 1 x 1 2 22 1 using an arrow diagram (Figure 4). 2 1 2 1 2 f$g g f x g(x) fÓ˝Ô FIGURE 4 Arrow diagram for f g E X AM P L E 3 \| Finding the Composition of Functions Let f x x2 and g x 2 1 2 (a) Find the functions f g (b) Find and and 1 2 1 2 1 2 1 2 and their domains. In Example 3, f is the rule “square” and g is the rule “subtract 3.” The function f g ﬁrst subtracts 3 and then squares; the function g f ﬁrst squares and then subtracts 3. ▼ SO LUTI O N (a) We have f g 1 x 2 2 1 and 22 22 1 g g 1 f 1 x2 2 1 x2 3 Deﬁnition of f g Deﬁnition of g Deﬁnition of f Deﬁnition of g f Deﬁnition of f Deﬁnition of g The domains of both f g and g f are. (b) We have 22 22 f g 2 2 49 1 1 22 4 49 3 46 2 ✎ Practice what you’ve learned: Do Exercises 21 and 35. ▲ You can see from Example 3 that, in general, f g g f. Remember that the notation f g means that the function g is applied ﬁrst and then f is applied second. 258 CHAPTER 3 \| Functions The graphs of f and g of Example 4, as well as f g, g f, f f, and g g, are shown below. These graphs indicate that the operation of composition can produce functions that are quite different from the original functions. g f f$g g$f f$f g$g E X AM P L E 4 \| Finding the Composition of Functions f If (a) 1x x 2 1 f g g and (b) 1 x 2 g
f 12 x (c) f f (d) g g, ﬁnd the following functions and their domains. ▼ SO LUTI O N f g (a 22 1 12 x 1 2 312 x 14 2 x is f g x 5 0 Deﬁnition of f g Deﬁnition of g Deﬁnition of f 2 x 0 x 0 6 Deﬁnition of g f 5 Deﬁnition of f x 2 1 6 q, 2. 4 The domain of (b 22 1 g 1x 32 1x 1 2 to be deﬁned, we must have 1x 2, that is, is the closed interval For have main of f f 1x 2 1x 0 g f f f x f 22 1 1x 2 1 x 1 2 1 Deﬁnition of g x 0, or 0, 4 x 4. 4 Deﬁnition of f f 3 (c) 2 1 31x 14 x Deﬁnition of f Deﬁnition of f. For 32 1x. Thus, we have to be deﬁned, we must 0 x 4, so the do- The domain of f f is 0, q 3. 2 (d 22 1 12 x 1 1 2 32 12 x Deﬁnition of g g Deﬁnition of g Deﬁnition of g This expression is deﬁned when both x 2 equality means x 2 or and, and the second is equivalent to 2 x 2 g g ✎ Practice what you’ve learned: Do Exercise 41., so the domain of. Thus, 2 x 0 2 12 x 0 12 x 2, or 2, 2 is. 4 3. The ﬁrst in2 x 4, ▲ It is possible to take the composition of three or more functions. For instance, the com- posite function f g h is found by ﬁrst applying h, then g, and then f as follows 222 1 E X AM P L E 5 \| A Composition of Three Functions x 1 f g h x10, and Find x/, g if ▼ SO LUTI 222 1 1 1 x 3 g 22 1 x 3 10 2 x 3 11 2 10 1 1 x 3 2 10 1 2 Deﬁnition of f g h De
ﬁnition of h Deﬁnition of g Deﬁnition of f ✎ Practice what you’ve learned: Do Exercise 45. ▲ SE CTI ON 3.6 \| Combining Functions 259 So far, we have used composition to build complicated functions from simpler ones. But in calculus it is useful to be able to “decompose” a complicated function into simpler ones, as shown in the following example. E X AM P L E 6 \| Recognizing a Composition of Functions F f g., ﬁnd functions f and g such that 14 x 9 Given F x 1 2 ▼ SO LUTI O N Since the formula for F says to ﬁrst add 9 and then take the fourth root, we let g x 1 2 x 9 and f 14 x x 1 2 Then 22 1 x 9 f 2 1 14 x 9 Deﬁnition of f g Deﬁnition of g Deﬁnition of f ✎ Practice what you’ve learned: Do Exercise 49. ▲ F x 1 2 E X AM P L E 7 \| An Application of Composition of Functions A ship is traveling at 20 mi/h parallel to a straight shoreline. The ship is 5 mi from shore. It passes a lighthouse at noon. time=noon 5 mi s d (a) Express the distance s between the lighthouse and the ship as a function of d, the distance the ship has traveled since noon; that is, ﬁnd f so that s f d. 2 1 (b) Express d as a function of t, the time elapsed since noon; that is, ﬁnd g so that d g (c) Find t. 2 1 f g. What does this function represent? ▼ SO LUTI O N We ﬁrst draw a diagram as in Figure 5. (a) We can relate the distances s and d by the Pythagorean Theorem. Thus, s can be ex- t time= pressed as a function of d by s f d 1 2 225 d 2 FIGURE 5 (b) Since the ship is traveling at 20 mi/h, the distance d it has traveled is a function of t distance rate time as follows: (c) We have d g t 1 2 20t 20t 22 f 2 1 225 Deﬁnition of f g Deﬁn
ition of g Deﬁnition of f 20t 1 2 2 The function of time. f g gives the distance of the ship from the lighthouse as a function ✎ Practice what you’ve learned: Do Exercise 63. ▲ 260 CHAPTER 3 \| Functions 3. ▼ CONCE PTS 1. From the graphs of f and g in the ﬁgure, we ﬁnd f g 2 2 2 1 fg 16. By deﬁnition, 12, then. So if g. 5 and 3. If the rule of the function f is “add one” and the rule of the function g is “multiply by 2,” then the rule of f g is “,” and the rule of g f is “.” 4. We can express the functions in Exercise 3 algebraically as, and f/g and their domains. ▼ SKI LLS 5–10 ■ Find ✎ 5. 6. 7. 8. f f f f 9. f 10, x2 2x f g fg, x2 g x 1, g 24 x2, 2 x 1 g 2 x 1 2 3x2 1 11 x 2x2 4 29 x2 on a common screen 17. 17–20 ■ Draw the graphs of f, g, and to illustrate graphical addition. 11 x x2 x2 x g 2 1 1x 1 18. 19 11 x3 1 2 1 2 20. f x 1 2 14 1 x, g x 1 2 1 x2 9 B 21–26 ■ Use expression. f x 1 2 3x 5 and g x 1 2 2 x2 to evaluate the ✎ 21. (a) f 0 g 1 1 22 (b) g 0 f 1 1 22. (a) 23. (a) 24. (a) 25. (a) 26. (a) 2 1 1 22 b) (b) (b) (b) (b 22 22 27–32 ■ Use the given graphs of f and g to evaluate the expression. y g 2 f 0 2 x 11–14 ■ Find the domain of the function. 11. f x 1 2 1x 11 x 12. g x 1 2 1x 1 1 x x h 13. 1x 3 x 1 15–16 ■ Use graphical addition to sketch the graph of x 3 14. 1/. ✎ 15. y g
f 0 x 27. f 2 g 1 1 22 22 28. 29. 30. 31. 32. 1 2 1 2 33–44 ■ Find the functions domains. f g, g f, f f, and g g and their 33. 34. f f x x 1 1 2 2 2x 3, g 6x 5, g x 1 x 1 4x 1 x 2 2 2 x x2 g, 1 x3 2, 2 x 1 13 x g x 1 2 SE CTI ON 3.6 \| Combining Functions 261 59. Use the fact that revenue price per item number of items sold, 1 x x2 1x 13 x, 2 x, 2x 4 g x 1 2, g 1x 3 2x 2x x2 4x g g x 1 1 2 x 14 x 2 x x 2 ✎ 35. 36. f f 37. f 38. 39. 40. f f f ✎ 41. f 42. f 43. f 44 ✎ 45. f 46. f 47. f 48 45–48 ■ Find f g h. x 1, 1x, x4 1 1 1x, x x3, 2 h x 1 2 x2, 1x 13 x 49–54 ■ Express the function in the form f g. ✎ 49. 50. F F x x 1 1 2 2 51. G x 1 2 52. G 53. H 54 1x 1 x2 x2 4 1 x 3 1 x3 0 0 31 1x 55–58 ■ Express the function in the form 55. F x 1 2 1 x2 1 56. 57. F 1 G x 2 x 1 2 33 1x 1 4 13 x 9 1 2 x to express 2 product of two functions of x. R 1, the revenue from an order of x stickers, as a 60. Use the fact that proﬁt revenue cost x to express 1 ence of two functions of x. P 2, the proﬁt on an order of x stickers, as a differ- 61. Area of a Ripple A stone is dropped in a lake, creating a cir- cular ripple that travels outward at a speed of 60 cm/s. (a) Find a function g that models the radius as a function of time. (b) Find a function f that models the area of the circle as a function of the radius. (c) Find f g. What does this function represent? 62
. Inﬂating a Balloon A spherical balloon is being inﬂated. The radius of the balloon is increasing at the rate of 1 cm/s. (a) Find a function f that models the radius as a function of time. (b) Find a function g that models the volume as a function of the radius. g f (c) Find. What does this function represent? f g h. ✎ 63. Area of a Balloon A spherical weather balloon is being inﬂated. The radius of the balloon is increasing at the rate of 2 cm/s. Express the surface area of the balloon as a function of time t (in seconds). 58. G x 1 2 2 3 1x 1 2 2 r ▼ APPLICATIONS 59–60 ■ Revenue, Cost, and Proﬁt A print shop makes bumper stickers for election campaigns. If x stickers are ordered (where x 10,000 and the total cost of producing the order is dollars. ), then the price per sticker is 0.095x 0.0000005x2 0.15 0.000002x dollars, 262 CHAPTER 3 \| Functions 64. Multiple Discounts You have a $50 coupon from the manufacturer good for the purchase of a cell phone. The store where you are purchasing your cell phone is offering a 20% discount on all cell phones. Let x represent the regular price of the cell phone. (a) Suppose only the 20% discount applies. Find a function f that models the purchase price of the cell phone as a function of the regular price x. (b) Suppose only the $50 coupon applies. Find a function g that models the purchase price of the cell phone as a function of the sticker price x. (c) If you can use the coupon and the discount, then the purx chase price is either, depending on the 2 order in which they are applied to the price. Find both f g x 1 price?. Which composition gives the lower f g g f g f and or x x 2 1 1 2 1 2 ▼DISCOVE RY • DISCUSSION • WRITI NG 67. Compound Interest A savings account earns 5% interest compounded annually. If you invest x dollars in such an account, A then the amount initial investment plus 5%; that is, Find of the investment after one year is the x 0.05x 1.05x. What do these compositions represent? Find a formula for what you get when
you compose n copies of A. 65. Multiple Discounts An appliance dealer advertises a 68. Composing Linear Functions The graphs of the functions 10% discount on all his washing machines. In addition, the manufacturer offers a $100 rebate on the purchase of a washing machine. Let x represent the sticker price of the washing machine. (a) Suppose only the 10% discount applies. Find a function f that models the purchase price of the washer as a function of the sticker price x. (b) Suppose only the $100 rebate applies. Find a function g that models the purchase price of the washer as a function of the sticker price x. (c) Find f g and Which is the better deal? g f. What do these functions represent? 66. Airplane Trajectory An airplane is ﬂying at a speed of 350 mi/h at an altitude of one mile. The plane passes directly above a radar station at time t 0. (a) Express the distance s (in miles) between the plane and the radar station as a function of the horizontal distance d (in miles) that the plane has ﬂown. (b) Express d as a function of the time t (in hours) that the plane has ﬂown. (c) Use composition to express s as a function of t. d s 1 mi m1x b1 m2x b2 f g x x 1 1 2 2 are lines with slopes m1 and m2, respectively. Is the graph of f g a line? If so, what is its slope? 69. Solving an Equation for an Unknown Function Suppose that 2x 1 4x2 4x Find a function f such that. (Think about what operations you would have to perform on the formula for g to end up with the formula for h.) Now suppose that 3x 5 3x2 3x 2 f h x x 1 1 2 2 Use the same sort of reasoning to ﬁnd a function g such that f g h. 70. Compositions of Odd and Even Functions Suppose that h f g If g is an even function, is h necessarily even? If g is odd, is h odd? What if g is odd and f is odd? What if g is odd and f is even? DISCOVERY PR OJECT ITERATION AND CHAOS The iterates of a function f at a point x 0 are f write x02 1, f f x022 1
1, f f f x022 2 1 1 1, and so on. We x1 x2 x3 o f f f o 1 1 1 x02 f 1 f 1 o x022 f x0222 1 The ﬁrst iterate The second iterate The third iterate x2 1 2 x, then the iterates of f at 2 are f For example, if, and so on. (Check this.) Iterates can be described graphically as in Figure 1. Start with x0 on the x-axis, move vertically to the graph of f, then horizontally to the line, then vertically to the graph of f, and so on. The x-coordinates of the points on the graph of f are the iterates of f at x0. y x x1 4, x2 16, x3 256 y=x y=Ï f(x›) y f(x¤) f(x‹) f(x⁄) f(x‚) 10 11 12 xn 0.1 0.234 0.46603 0.64700 0.59382 0.62712 0.60799 0.61968 0.61276 0.61694 0.61444 0.61595 0.61505 x‚ x⁄ x¤ x› x‹ x FIGURE 1 Iterates are important in studying the logistic function f x 1 2 kx 1 1 x 2 10% 1 which models the population of a species with limited potential for growth (such as rabbits on an island or ﬁsh in a pond). In this model, the maximum population that the environment can support is 1 (that is, 100%). If we start with a fraction of that population,, then the iterates of f at 0.1 give the population after each time interval say, 0.1 2 (days, months, or years, depending on the species). The constant k depends on the rate of growth of the species being modeled; it is called the growth constant. For example, for k 2.6 and x0 0.1 the iterates shown in the table to the left give the population of the species for the ﬁrst 12 time intervals. The population seems to be stabilizing around 0.615 (that is, 61.5% of maximum). (CONTINUES) 263 263 ITERATION AND CHAOS (CONTINUED) In the
three graphs in Figure 2, we plot the iterates of f at 0.1 for different values of the growth constant k. For k 2.6 the population appears to stabilize at a value 0.615 of maximum, for k 3.1 the population appears to oscillate between two values, and for k 3.8 no obvious pattern emerges. This latter situation is described mathematically by the word chaos. 1 1 21 0 21 0 21 k=2.6 k=3.1 k=3.8 1 0 FIGURE 2 1. Use the graphical procedure illustrated in Figure 1 to ﬁnd the ﬁrst ﬁve iterates f x of 2x 1 x 2 2. Find the iterates of f 1 2 1 3. Find the iterates of f at x 2 x 2 1 1 x 0.1. x2 at 1 x f x 1 2 x 1. at x 2. 4. Find the ﬁrst six iterates of iterate of f at 2? 1/ 1 1 x 2 at x 2. What is the 1000th 5. Find the ﬁrst 10 iterates of the logistic function at x 0.1 for the given value of k. Does the population appear to stabilize, oscillate, or is it chaotic? (a) k 2.1 (b) k 3.2 (c) k 3.9 6. It’s easy to ﬁnd iterates using a graphing calculator. The following steps show on a TI-83 calcu- kx how to ﬁnd the iterates of lator. (The procedure can be adapted for any graphing calculator.) at 0.1 for Y1 K * X * 3 S K 0.1 S X Y1 S X 0.27 0.5913 0.72499293 0.59813454435 Enter f as Y1 on the graph list Store 3 in the variable K Store 0.1 in the variable X Evaluate f at X and store result back in X Press and obtain ﬁrst iterate ENTER Keep pressing ENTER to re-execute the command and obtain successive iterates x kx You can also use the program in the margin to graph the iterates and study them visually. Use a graphing calculator to experiment with how the value of k affects the iterates at 0.1. Find several different values of k that make the iterates staf of
bilize at one value, oscillate between two values, and exhibit chaos. (Use values of k between 1 and 4.) Can you ﬁnd a value of k that makes the iterates oscillate between four values? 1 x 1 1 2 2 The following TI-83 program draws the ﬁrst graph in Figure 2. The other graphs are obtained by choosing the appropriate value for K in the program. PROGRAM:ITERATE :ClrDraw :2.6 S K :0.1 S X :For(N, 1, 20) :KX(1-X) S Z :Pt-On(N, Z, 2) :Z S X :End 264 SE CTI O N 3. 7 \| One-to-One Functions and Their Inverses 265 3.7 One-to-One Functions and Their Inverses LEARNING OBJECTIVES After completing this section, you will be able to: ■ Determine whether a function is one-to-one ■ Find the inverse function of a one-to-one function ■ Draw the graph of an inverse function The inverse of a function is a rule that acts on the output of the function and produces the corresponding input. So the inverse “undoes” or reverses what the function has done. Not all functions have inverses; those that do are called one-to-one. ■ One-to-One Functions Let’s compare the functions f and g whose arrow diagrams are shown in Figure 1. Note that f never takes on the same value twice (any two numbers in A have different images), whereas g does take on the same value twice (both 2 and 3 have the same image, 4). In x2. Functions that have this latter x12 symbols, property are called one-to-one. whenever x1 g f x22 but 10 7 4 2 A 4 3 2 1 B 10 4 2 f g f is one-to-one g is not one-to-one FIGURE 1 DEFINITION OF A ONE-TO-ONE FUNCTION A function with domain A is called a one-to-one function if no two elements of A have the same image, that is, x12 whenever x1 x2 f x22 f 1 1 An equivalent way of writing the condition for a one-to-one function is this: y y=Ï f(x⁄) f(x¤)
f If f x12 x22 If a horizontal line intersects the graph of f at more than one point, then we see from Figure 2 that there are numbers x1. This means that f is not oneto-one. Therefore, we have the following geometric method for determining whether a function is one-to-one. x2 such that f, then x1 f x12 x22. 1 1 1 1 x2 0 x⁄ x¤ x HORIZONTAL LINE TEST FIGURE 2 This function is not x2 2 one-to-one because f x12 f 1 1 A function is one-to-one if and only if no horizontal line intersects its graph more than once.. 266 CHAPTER 3 \| Functions E X AM P L E 1 \| Deciding Whether a Function Is One-to-One y 1 0 1 x f Is the function 1 ▼ SO LUTI O N 1 cube). Therefore, ▼ SO LUTI O N 2 x3 f x x x3 one-to-one? 2 If x1 x x3 f 1 is one-to-one. x2, then x3 1 x3 2 (two different numbers cannot have the same 2 From Figure 3 we see that no horizontal line intersects the graph of more than once. Therefore, by the Horizontal Line Test, f is one-to-one. 2 1 ✎ Practice what you’ve learned: Do Exercise 13. ▲ FIGURE 3 f x 1 2 x3 is one-to-one. Notice that the function f of Example 1 is increasing and is also one-to-one. In fact, it can be proved that every increasing function and every decreasing function is oneto-one. E X AM P L E 2 \| Deciding Whether a Function Is One-to-One y 1 0 1 x FIGURE 4 one-to-one. f x 1 2 x2 is not y 1 0 1 x g Is the function 1 ▼ SO LUTI O N 1 x x2 one-to-one? 2 This function is not one-to-one because, for instance, g 1 1 2 1 and g 1 1 2 1 so 1 and 1 have the same image. From Figure 4 we see that there are horizontal lines that intersect the ▼ SO LUTI O N 2 graph of g more than once. Therefore, by the Horizontal Line Test, g is not one-to-
one. ✎ Practice what you’ve learned: Do Exercise 15. ▲ Although the function g in Example 2 is not one-to-one, it is possible to restrict its do- main so that the resulting function is one-to-one. In fact, if we deﬁne then h is one-to-one, as you can see from Figure 5 and the Horizontal Line Test. x2 x 0 h x 1 2 E X AM P L E 3 \| Showing That a Function Is One-to-One 3x 4 Show that the function is one-to-one. x f 1 2 ▼ SO LUTI O N Suppose there are numbers x1 and x2 such that f x12 1 f x22 1. Then 3x1 4 3x2 3x2 x2 3x1 x1 4 Suppose f(x1) = f(x2) Subtract 4 Divide by 3 FIGURE 5 one-to-one. f x 1 2 x2 1 x 0 is 2 Therefore, f is one-to-one. ✎ Practice what you’ve learned: Do Exercise 11. ▲ ■ The Inverse of a Function One-to-one functions are important because they are precisely the functions that possess inverse functions according to the following deﬁnition. SE CTI O N 3. 7 \| One-to-One Functions and Their Inverses 267 DEFINITION OF THE INVERSE OF A FUNCTION Let f be a one-to-one function with domain A and range B. Then its inverse function f1 has domain B and range A and is deﬁned by for any y in B. f1 This deﬁnition says that if f takes x to y, then f1 takes y back to x. (If f were not oneto-one, then f1 would not be deﬁned uniquely.) The arrow diagram in Figure 6 indicates that f1 reverses the effect of f. From the deﬁnition we have domain of f1 range of f range of f1 domain of f E X AM P L E 4 \| Finding f 1 for Speciﬁc Values A x f f _1 B y=Ï FIGURE 6 Don’t mistake the 1 in f1 for an exponent. f 1 does not mean 1 x f 2 is written as 1 The reciprocal 1
f. x 1/, and f1 10 If f1, f1, and, ﬁnd 5, f 1 2 ▼ SO LUTI O N 7 10 2 From the deﬁnition of f1 we have 1 because f 3 because f 8 because f f1 f1 7 1 10 1 Figure 7 shows how f1 reverses the effect of f in this case. f1 10 A 1 3 8 B 5 7 _10 A 1 3 8 B 5 7 _10 f f _1 FIGURE 7 ✎ Practice what you’ve learned: Do Exercise 21. ▲ By deﬁnition the inverse function f1 undoes what f does: If we start with x, apply f, and then apply f1, we arrive back at x, where we started. Similarly, f undoes what f1 does. In general, any function that reverses the effect of f in this way must be the inverse of f. These observations are expressed precisely as follows. INVERSE FUNCTION PROPERTY Let f be a one-to-one function with domain A and range B. The inverse function f1 satisﬁes the following cancellation properties 22 22 x for every x in A x for every x in B Conversely, any function f1 satisfying these equations is the inverse of f. 268 CHAPTER 3 \| Functions These properties indicate that f is the inverse function of f1, so we say that f and f1 are inverses of each other. E X AM P L E 5 \| Verifying That Two Functions Are Inverses x1/3 are inverses of each other. Show that x3 and g x x f 1 2 1 2 ▼ SO LUTI O N Note that the domain and range of both f and g is. We have g f x 1 g 1 x f 22 g f x 3 2 1 x 1/3 1 x 3 1/3 x 3 x 2 x 1/3 1 1 1 22 2 So by the Property of Inverse Functions, f and g are inverses of each other. These equations simply say that the cube function and the cube root function, when composed, cancel each other. ✎ Practice what you’ve learned: Do Exercise 27. ▲ 2 1 Now let’s examine how we compute inverse functions. We ﬁrst observe from the deﬁni- tion of f1 that
y f x 3 f1 y x 1 1 and if we are able to solve this equation for x in terms of y, then we must 2 y. If we then interchange x and y, we have, which is the desired 1 y f1 f1 So if have equation. HOW TO FIND THE INVERSE OF A ONE-TO-ONE FUNCTION y f x. 1. Write 2. Solve this equation for x in terms of y (if possible). 3. Interchange x and y. The resulting equation is 1 2 y f1 x. 2 1 Note that Steps 2 and 3 can be reversed. In other words, we can interchange x and y ﬁrst and then solve for y in terms of x. E X AM P L E 6 \| Finding the Inverse of a Function Find the inverse of the function f x 3x 2. ▼ SO LUTI O N First we write 2 1 y f 2. x 1 y 3x 2 Then we solve this equation for x. 3x y 2 y 2 3 x Add 2 Divide by 3 Finally, we interchange x and y. ✔ Therefore, the inverse function is y x 2 3 x 2 3 2. f1 x 1 ✎ Practice what you’ve learned: Do Exercise 35. ▲ In Example 6 note how f1 reverses the effect of f. The function f is the rule “multiply by 3, then subtract 2,” whereas f1 is the rule “add 2, then divide by 3.” Check Your Answer We use the Inverse Function Property. f 1 3x 2 f 1 x f 1 1 22 2 2 1 3x 22 3x SE CTI O N 3. 7 \| One-to-One Functions and Their Inverses 269 E X AM P L E 7 \| Finding the Inverse of a Function In Example 7 note how f1 reverses the effect of f. The function f is the rule “Take the ﬁfth power, subtract 3, then divide by 2,” whereas f1 is the rule “Multiply by 2, add 3, then take the ﬁfth root.” Check Your Answer We use the Inverse Function Property. f 1 x f 1 1 22 f 1 f 1 x 1 22 1/5 3 d b b 1/5 2 1//5 x 2 2x 3 11 2x 3 3 1 2 2
2x 3 3 2 1 f 2 1/5 x 5 2x 2 x ✔ Find the inverse of the function f x 1 2 y ▼ SO LUTI O N We ﬁrst write y x5 3 2 2y x5 3 x5 2y 3 2y 3 x 1/5. x5 3 2 x5 3 1 2 /2 and solve for x. Equation deﬁning function Multiply by 2 Add 3 (and switch sides) 1 2 Then we interchange x and y to get f1 ✎ Practice what you’ve learned: Do Exercise 47. 2x 3 y 1/5 x. 2 1 2 1 1 2 Take ﬁfth root of each side 2x 3 1/5. Therefore, the inverse function is ▲ ■ Graphing the Inverse of a Function The principle of interchanging x and y to ﬁnd the inverse function also gives us a method for obtaining the graph of f1 from the graph of f. If a f, then. Thus, the is on the graph of f1. But a, b point by reﬂecting in the line y x (see Figure 8). we get the point Therefore, as Figure 9 illustrates, the following is true. a 2 is on the graph of f if and only if the point 1 from the point b b, a f1 b, a a The graph of f1 is obtained by reﬂecting the graph of f in the line y x. y y=x (b, a) (a, b) x y=x y f _¡ x f FIGURE 8 FIGURE 9 E X AM P L E 8 \| Graphing the Inverse of a Function y=x y=f –¡(x) y 2 2 y=Ï=œ∑x-2 FIGURE 10 1x 2 2. x (a) Sketch the graph of f 1 (b) Use the graph of f to sketch the graph of f1. (c) Find an equation for f1. ▼ SO LUTI O N (a) Using the transformations from Section 3.5, we sketch the graph of plotting the graph of the function it to the right 2 units. y 1x by (Example 1(c) in Section 3.2) and moving y 1x 2 (b) The graph of f1 is obtained from the
graph of f in part (a) by reﬂecting it in the line y x, as shown in Figure 10. x (c) Solve y 1x 2 for x, noting that y 0. 1x 2 y x 2 y 2 Square each side x y 2 2 y 0 Add 2 270 CHAPTER 3 \| Functions In Example 8 note how f1 reverses the effect of f. The function f is the rule “Subtract 2, then take the square root,” whereas f1 is the rule “Square, then add 2.” Interchange x and y Thus, This expression shows that the graph of f1 is the right half of the parabola y x2 2, and from the graph shown in Figure 10, this seems reasonable. ✎ Practice what you’ve learned: Do Exercise 57. ▲ 1 2 3. ▼ CONCE PTS 1. A function f is one-to-one if different inputs produce outputs. You can tell from the graph that a function is one-to-one by using the Test. 2. (a) For a function to have an inverse, it must be. So which one of the following functions has an inverse? f x 1 2 x2 g x3 x 1 2 9. y 0 10. x y 0 x (b) What is the inverse of the function that you chose in 11–20 ■ Determine whether the function is one-to-one. part (a)? 3. A function f has the following verbal description: “Multiply by 3, add 5, and then take the third power of the result.” (a) Write a verbal description for f1. (b) Find algebraic formulas that express f and f1 in terms of the input x. 4. True or false? (a) If f has an inverse, then (b) If f has an inverse, then is the same as x. x 22 ✎ ✎ ✎ 11. 13. 15. 17. 18. f g h f f 19 2x 4 1x x2 2x x4 5 x4 5, 0 x 2 12. 14. 16 3x 2 x 0 0 x3 8 1 x2 20 21–22 ■ Assume that f is a one-to-one function. ▼ SKI LLS 5–10 ■ The graph of a function f is given. Determine whether f is one
-to-one. 5. 7. y 0 y 0 6. 8. x x y 0 y 0 x x ✎ 21. (a) If (b) If 22. (a) If (b) If 23. If f 24. If 18. 2., ﬁnd 1 f 2 1 f1 f 5 1 f1 2 1 7 3 2 18 2 2 4 1, ﬁnd, ﬁnd f1 7 1, ﬁnd f f1 1 f 2 1 f1 2 5 2x x2 4x 2, ﬁnd 3 2 1 with x 2, ﬁnd. g1 5. 2 1 25–34 ■ Use the Inverse Function Property to show that f and g are inverses of each other. 25. f 26. f ✎ 27. f 28. f 29. f 30; g 3x 2x 5 4x 3 x 4 ; g 1 x x5 15 x SE CTI ON 3. 7 \| One-to-One Functions and Their Inverses 271 31. f g 32. f 33. f 34, x 0 ; 1x 4, x 4 69. f x 1 2 4 x2 x3 1/3 2 1 x 24 x2, 0 x 2 ; 24 x2, 0 x 2 1 70 35–54 ■ Find the inverse function of f. ✎ 35. 37. f f 39. f 41. f 43. f 45. ✎ 47. 49. 51. 52. 53 2x 1 4x 7 x 2 1 x 2 1 3x 5 2x 22 5x 4 x2, x 0 4 13 x 1 21 x 36. 38. f f 40. f 42. f 44. f 46. 48. 50 5x 1 x2, x 0 x 2 x 2 5 4x3 x2 x, x 1 2 22x 1 2 x3 1 5 2 29 x2, 0 x 3 x4, x 0 54. f x 1 2 1 x3 55–58 ■ A function f is given. (a) Sketch the graph of f. (b) Use the graph of f to sketch the graph of f1. (c) Find f1. 55. 57 3x 6 2x 1 56. 58. f f x x 1 1 2 2 16 x2, x 0 x3 1 59–64 ■ Draw the graph of f and use it
to determine whether the function is one-to-one. 59. f 61. f 63 x3 x x 12 x 6 x 0 0 60. f 62. f 64 x3 x 2x3 4x 1 x # x 0 0 65–68 ■ A one-to-one function is given. (a) Find the inverse of the function. (b) Graph both the function and its inverse on the same screen to verify that the graphs are reﬂections of each other in the line y x. 65. 67 2x 3 66. 68 2x x2 1, x 0 69–72 ■ The given function is not one-to-one. Restrict its domain so that the resulting function is one-to-one. Find the inverse of the function with the restricted domain. (There is more than one correct answer.) 71 72 0_1 x y 1 0 1 x 73–74 ■ Use the graph of f to sketch the graph of f1. 74. y 73 ▼ APPLICATIONS 75. Fee for Service For his services, a private investigator requires a $500 retention fee plus $80 per hour. Let x represent the number of hours the investigator spends working on a case. (a) Find a function f that models the investigator’s fee as a function of x. (b) Find f 1. What does f 1 represent? (c) Find f 1 1220 1. What does your answer represent? 2 76. Toricelli’s Law A tank holds 100 gallons of water, which drains from a leak at the bottom, causing the tank to empty in 40 minutes. Toricelli’s Law gives the volume of water remaining in the tank after t minutes as V t 100 a 2 1 t 2 1 40 b (a) Find V 1. What does V 1 represent? (b) Find V 1 15 1. What does your answer represent? 2 77. Blood Flow As blood moves through a vein or artery, its velocity √ is greatest along the central axis and decreases as the distance r from the central axis increases (see the ﬁgure 272 CHAPTER 3 \| Functions below). For an artery with radius 0.5 cm, √ is given as a function of r by f function represent? 1 x 2 7 2x. Find f1. What does the function f1 √ 18,500 r 2 (a) Find √1. What does √1 represent? (
b) Find √1 30 0.25 r2 1 1 2. What does your answer represent? 2 1 r ▼DISCOVE RY • DISCUSSION • WRITI NG 84. Determining When a Linear Function Has an Inverse For mx b the linear function to be one-to-one, what must be true about its slope? If it is one-to-one, ﬁnd its inverse. Is the inverse linear? If so, what is its slope? x f 2 1 78. Demand Function The amount of a commodity that is sold is called the demand for the commodity. The demand D for a certain commodity is a function of the price given by 85. Finding an Inverse “in Your Head” In the margin notes in this section we pointed out that the inverse of a function can be found by simply reversing the operations that make up the function. For instance, in Example 6 we saw that the inverse of p 3p 150 D 1 (a) Find D1. What does D1 represent? (b) Find D1 1 30 2 2. What does your answer represent? 79. Temperature Scales The relationship between the Fahren- heit (F) and Celsius (C) scales is given by C 9 F (a) Find F 1. What does F 1 represent? (b) Find F 1 1 5C 32 86 2 2 1. What does your answer represent? 80. Exchange Rates The relative value of currencies ﬂuctuates every day. When this problem was written, one Canadian dollar was worth 1.0573 U.S. dollar. (a) Find a function f that gives the U.S. dollar value of x f 1 2 x Canadian dollars. (b) Find f1. What does f1 represent? (c) How much Canadian money would $12,250 in U.S. cur- rency be worth? 81. Income Tax In a certain country, the tax on incomes less than or equal to €20,000 is 10%. For incomes that are more than €20,000, the tax is €2000 plus 20% of the amount over €20,000. (a) Find a function f that gives the income tax on an income x. Express f as a piecewise deﬁned function. (b) Find f1. What does f1 represent? (c) How much income would require paying a tax of €10,
000? 82. Multiple Discounts A car dealership advertises a 15% discount on all its new cars. In addition, the manufacturer offers a $1000 rebate on the purchase of a new car. Let x represent the sticker price of the car. (a) Suppose only the 15% discount applies. Find a function f that models the purchase price of the car as a function of the sticker price x. (b) Suppose only the $1000 rebate applies. Find a function g that models the purchase price of the car as a function of the sticker price x. (c) Find a formula for H f g. (d) Find H1. What does H1 represent? (e) Find H1 13,000 1. What does your answer represent? 2 83. Pizza Cost Marcello’s Pizza charges a base price of $7 for a large pizza plus $2 for each topping. Thus, if you order a large pizza with x toppings, the price of your pizza is given by the 3x 2 is f1 because the “reverse” of “Multiply by 3 and subtract 2” is “Add 2 and divide by 3.” Use the same procedure to ﬁnd the inverse of the following functions. 2x 1 5 2x3 2 (a) (cb) (d 2x 5 1 3 2 Now consider another function: x3 2x 6 f x 1 2 Is it possible to use the same sort of simple reversal of operations to ﬁnd the inverse of this function? If so, do it. If not, explain what is different about this function that makes this task difﬁcult. 86. The Identity Function The function I x x is called the 1 2 identity function. Show that for any function f we have f I f, I f f that the identity function I behaves for functions and composition just the way the number 1 behaves for real numbers and multiplication.) f f1 f1 f I. (This means, and 87. Solving an Equation for an Unknown Function In Exercise 69 of Section 3.6 you were asked to solve equations in which the unknowns were functions. Now that we know about inverses and the identity function (see Exercise 86), we can use algebra to solve such equations. For instance, to f g h for the unknown function f, we perform the solve following steps: f g h f g g1 h g1 f I h
g1 f h g1 Problem: Solve for f Compose with on the right f h g1 for the indicated unknown function.. Use this technique to solve the So the solution is f g h equation (a) Solve for f, where g x 1 4x2 4x 7 x 1 3x2 3x 2 f 2x 1 and 3x 5 and 2 2 (b) Solve for g, where h x 1 2 h x 1 2 CHAPTER 3 \| REVIEW ▼ P R O P E RTI LAS Function Notation (p. 205) If a function is given by the formula pendent variable and denotes the input; y is the dependent variable and denotes the output; the domain is the set of all possible inputs x; the range is the set of all possible outputs y., then x is the inde- x 2 1 y f The Graph of a Function (p. 215) The graph of a function f is the graph of the equation deﬁnes f. y f that x 1 2 The Vertical Line Test (p. 219) A curve in the coordinate plane is the graph of a function if and only if no vertical line intersects the graph more than once. Increasing and Decreasing Functions (p. 229) CHAPTER 3 \| Review 273 Vertical and Horizontal Stretching and Shrinking of Graphs (pp. 247, 248) y cf, stretch the graph of, then to graph c 1 y f x If vertically by a factor of c. 1 2 x 1 2 If y f 0 c 1 x, then to graph y cf 1 vertically by a factor of c. 2 1 c 1 cx, then to graph If horizontally by a factor of 1/c. y f 1, shrink the graph of x 2, shrink the graph of y f 2 x 1 2 If y f 0 c 1 x 1 2, then to graph y f cx, stretch the graph of horizontally by a factor of 1/c. 1 2 Even and Odd Functions (p. 250) A function f is A function f is increasing on an interval if x1 in the interval. x2 f x12 1 f x22 1 whenever even if f A function f is decreasing on an interval if x1 in the interval. x2 f x12 1 f x2 2 1 whenever odd if f 2 for every x in the domain of f. 1 2 1 Local Maximum and Minimum Values (p. 230) Composition of
Functions (p. 257) The function value f x f if 1 local maximum at x a. a f 2 2 1 f a is a local maximum value of the function for all x near a. In this case we also say that f has a 1 2 f The function value f f if 1 local minimum at x b is a local minimum value of the function for all x near b. In this case we also say that f has a Average Rate of Change (p. 237) The average rate of change of the function f between x a and x b is the slope of the secant line between b, f and a, f 22 b a : 1 1 1 1 22 average rate of change Vertical and Horizontal Shifts of Graphs (p. 244) Let c be a positive constant. To graph c units. To graph c units. To graph c units. To graph c units, shift the graph of y f c, shift the graph of, shift the graph of y f, shift the graph of y f upward by downward by to the right by to the left by Reﬂecting Graphs (p. 246) To graph y f To graph y f 1 x 1 x, reﬂect the graph of, reﬂect the graph of in the x-axis. in the y-axis. Given two functions f and g, the composition of f and g is the function deﬁned by 22 is the set of all x for which both g and x 1 2 The domain of f f g are deﬁned. g x 1 22 1 One-to-One Functions (p. 265) A function f is one-to-one if different elements of the domain of f. x1 2 f 1 f x22 1 whenever x1 and x2 are To see from its graph whether a function is one-to-one, use the Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more than once. Inverse of a Function (p. 267) Let f be a one-to-one function with domain A and range B. The inverse of f is the function f1 deﬁned by The inverse function f1 y x 3 f x y 1 f1 2 has domain B and range A. 2 1 The functions f and properties: f1 satisfy the following cancellation f 1 f 1 f 1 f 1 x for every
x in A x for every x in B x x 1 1 22 22 274 CHAPTER 3 \| Functions ▼ CO N C E P T S U M MARY Section 3.1 ■ Recognize functions in the real world ■ Work with function notation ■ Find domains of functions ■ Represent functions verbally, algebraically, graphically, and numerically Section 3.2 ■ Graph a function by plotting points ■ Graph a function with a graphing calculator ■ Graph piecewise deﬁned functions ■ Use the Vertical Line Test ■ Determine whether an equation deﬁnes a function Section 3.3 ■ Find function values from a graph ■ Find the domain and range of a function from a graph ■ Find where a function is increasing or decreasing from a graph ■ Find local maxima and minima of functions from a graph Section 3.4 ■ Find the average rate of change of a function ■ Interpret average rate of change in real-world situations ■ Recognize that a function with constant average rate of change is linear Section 3.5 ■ Shift graphs vertically ■ Shift graphs horizontally ■ Stretch or shrink graphs vertically ■ Stretch or shrink graphs horizontally ■ Determine whether a function is odd or even Section 3.6 ■ Find sums, difference, products, and quotients of functions ■ Add functions graphically ■ Find the composition of two functions ■ Express a given function as a composite function Section 3.7 ■ Determine whether a function is one-to-one ■ Find the inverse function of a one-to-one function ■ Draw the graph of an inverse function ▼ E X E RC I S E S 1–2 ■ A verbal description of a function f is given. Find a formula that expresses f in function notation. 1. “Square, then subtract 5.” 2. “Divide by 2, then add 9.” 3–4 ■ A formula for a function f is given. Give a verbal description of the function. 3. f 4 10 1 2 16x 10 Review Exercises 7–8 9–10 13–22 1–6, 23–40 Review Exercises 23–40 45–50 37–40 11 41–44 Review Exercises 12(a) 12(b), 12(c) 2(d), 53–54 12(e), 67–72 Review Exercises 55–58 59–60 61–62 Review Exercises 63–64 63–64 63–64 63–64 65–66 Review Exercises 75
73–74 75–78 79–80 Review Exercises 11, 12(f), 81–86 87–90 91–92 5–6 ■ Complete the table of values for the given function. 5. g x 1 2 x2 4x 6. h x 1 2 3x2 2x 5 gg(x) x 1 0 1 2 3 h(x) x 2 1 0 1 2 C 7. A publisher estimates that the cost C(x) of printing a run of x copies of a certain mathematics textbook is given by the function (a) Find C(1000) and C(10,000). (b) What do your answers in part (a) represent? (c) Find C(0). What does this number represent? 5000 30x 0.001x2. x 2 1 8. Reynalda works as a salesperson in the electronics division of a department store. She earns a base weekly salary plus a commission based on the retail price of the goods she has sold. If she sells x dollars worth of goods in a week, her earnings for that week are given by the function (a) Find E(2000) and E(15,000). (b) What do your answers in part (a) represent? (c) Find E(0). What does this number represent? (d) From the formula for E, determine what percentage 400 0.03x E x. 1 2 Reynalda earns on the goods that she sells. If, ﬁnd x x2 4x 6 2x 2f, and 2 2 2 1 4 13x 6, ﬁnd x2 x 2, and 1 2 1 10. If f of the functions are one-to-one? (a) y (b) 11. Which of the following ﬁgures are graphs of functions? Which 0 x y y 0 x (d) x 0 x (c) y 0 CHAPTER 3 \| Review 275 13–14 ■ Find the domain and range of the function. 13. f x 1 2 1x 3 14. F t 1 2 t2 2t 5 15–22 ■ Find the domain of the function. 15. f 17. f 19. f 21 7x 15 1x 4 1 x 1 x 1 1 x 2 14 x 2x2 1 16. f 18. f x x 1 1 2 2 20. g x 1 2 22. f x 1 2 2x 1 2
x 1 2 3x 1x 1 2x2 5x 3 2x2 5x 3 23 2x 1 23 2x 2 23–40 ■ Sketch the graph of the function. 26. 28. 30. 32. 34 2t t 3 8x 2x2 0 x 0 1x 3 x3 3x2 36 23. 24. 25. 27. 29. 31. 33 35 2x 1 x 5 3 1 1 1 2 t x2 6x 6 1 1x 1 2 x3 13 x 1 x2 37. f 38. f 39 40 2x 2x 1 x 6 x2 e x x 2 1 if x 0 if x 0 if x 0 if x 0 if x 2 if x 2 if x 0 if 0 x 2 if x 2 12. The graph of a function f is given. 2 1 f f 2 and 2 (a) Find. 2 (b) Find the domain of f. (c) Find the range of f. (d) On what intervals is f increasing? On what intervals is f 1 41–44 ■ Determine whether the equation deﬁnes y as a function of x. c 41. 43. x y2 14 x3 y3 27 42. 44. 3x 2y 8 2x y4 16 45. Determine which viewing rectangle produces the most appropriate graph of the function decreasing? (e) What are the local maximum values of f? (f) Is f one-to-one? ii(i) y 2 0 f 2 x f x 6x3 15x2 4x 1. 2 1 2, 2 12, 12 4 by 3 by 2, 2 4 3 4, 4 3 (iii) (ii) (iv) 3 8, 8 by 3 4 100, 100 8, 8 4 100, 100 by 4 4 4 46. Determine which viewing rectangle produces the most 2100 x3. x f 3 4 3 3 4 by appropriate graph of the function ii(i) i(ii) (iii) (iv) 4, 4 3 4 10, 10 by 4 3 10, 40 by 3 4 100, 100 by 3 4, 4 3 10, 10 4 3 10, 10 3 4 100, 100 3 4 4 1 2 47–50 ■ Draw the graph of the function in an appropriate viewing rectangle. 47. f 48. f x x 1 1 2 2 x2 25x 173 1.1x3 9.6x2 1.
4x 3.2 276 CHAPTER 3 \| Functions 49. f 50. f x x 1 1 2 2 x 2x2 16 51. Find, approximately, the domain of the function 2 52. Find, approximately, the range of the function 1 f x 2x3 4x 1. x4 x3 x2 3x 6. f x 1 2 53–54 ■ Draw a graph of the function f, and determine the intervals on which f is increasing and on which f is decreasing. 53. f x 1 2 x3 4x2 54. f x 1 2 0 x4 16 0 55–58 ■ Find the average rate of change of the function between the given points. 55. f 56. f 57. f 58 x2 3x; x 0, x 2 1 x 2 ; x 4, x 8 ; x 3; x a, x a h 2 t P 3000 200t 0.1t2 59. The population of a planned seaside community in Florida is, where t given by the function represents the number of years since the community was incorporated in 1985. (a) Find (b) Find the average rate of change of P between t 10 and 2 t 20. What does this number represent?. What do these values represent? and 10 20 P P 2 1 2 1 1 60. Ella is saving for her retirement by making regular deposits into a 401(k) plan. As her salary rises, she ﬁnds that she can deposit increasing amounts each year. Between 1995 and 2008, the annual amount (in dollars) that she deposited was given 3500 15t2 by the function the year of the deposit measured from the start of the plan (so 1995 corresponds to t 0 and 1996 corresponds to t 1, and so on). (a) Find D (b) Assuming that her deposits continue to be modeled by the. What do these values represent?, where t represents and 15 function D, in what year will she deposit $17,000? (c) Find the average rate of change of D between t 0 and t 15. What does this number represent? 61–62 ■ A function f is given. (a) Find the average rate of change of f between x 0 and x 2, and the average rate of change of f between x 15 and x 50. (b) Were the two average rates of change that you found in part (a) the same? Explain why or why not. 61. f x
1 2 1 2x 6 62. f x 1 2 8 3x 63. Suppose the graph of f is given. Describe how the graphs of the following functions can be obtained from the graph of f. (a) (c) (e) (g) y f (b) 1 y f (d) 1 y f (f) (h) y f1 y f x 1 y 1 2f 64. The graph of f is given. Draw the graphs of the following functions. y f (a) 1 y 3 f (c) y f 1 (eb) (d) (f 65. Determine whether f is even, odd, or neither. (a) (c) f f x x 1 1 2 2 2x5 3x2 2 1 x2 1 x2 (b) (d) f f x x 1 1 2 2 x3 x7 1 x 2 66. Determine whether the function in the ﬁgure is even, odd, or neither. (a) (c) y 0 y 0 (b) (d) x x 67. Find the minimum value of the function 68. Find the maximum value of the function y 0 y x 0 x 2x2 4x 5. 1 x x2. g 1 f x 2 x 1 2 69. A stone is thrown upward from the top of a building. Its height (in feet) above the ground after t seconds is given by t h 2 reach? 16t2 48t 32. What maximum height does it 1 70. The proﬁt P (in dollars) generated by selling x units of a cer- tain commodity is given by 1500 12x 0.0004x2 P x 1 2 What is the maximum proﬁt, and how many units must be sold to generate it? 71–72 ■ Find the local maximum and minimum values of the function and the values of x at which they occur. State each answer correct to two decimal places. 71. f x 1 2 3.3 1.6x 2.5x3 72. f x 1 2 x2/3 6 x 1/3 2 1 f x 2 1 functions. (a) f g (d) f/g 73–74 ■ Two functions, f and g, are given. Draw graphs of f, g, and f g on the same graphing calculator screen to illustrate the concept of graphical addition. x2 x 2, g
1 x2 1, g 3 x2 73. 74 75. If 2 x2 3x 2 1 and g x 1 2 4 3x, ﬁnd the following (b) f g f g (e) g f (c) fg (f) 1x 1 (c) (f), ﬁnd the following, g f, f f, and g g and their 1 x2 76. If f (a) (d) 1 x 2 f g f f 2 and (b) (e 77–78 ■ Find the functions domains. 77. f 78. f x x 1 1 2 2 3x 1, g 1x, g x 1 2 x 2x x2 2 1 2 x 4 79. Find h x f g h 1 1x, where. 1 2 f x 1 2 11 x, g x 1 2 1 x2, and 80. If T x 1 2 1 31 2x f g h T., ﬁnd functions f, g, and h such that CHAPTER 3 \| Review 277 81–86 ■ Determine whether the function is one-to-one. 81. 82. f g x 1 x 1 83. h x 1 2 2 2 3 x3 2 2x x2 1 x4 84. 85. 86 1x 3 3.3 1.6x 2.5x3 3.3 1.6x 2.5x3 87–90 ■ Find the inverse of the function. 87. f x 1 2 3x 2 88. f x 1 89 91. (a) Sketch the graph of the function 1 90. f x 2x 1 3 1 15 b) Use part (a) to sketch the graph of f1. (c) Find an equation for f1. 92. (a) Show that the function f x 1 14 x is one-to-one. (b) Sketch the graph of f. (c) Use part (b) to sketch the graph of f1. (d) Find an equation for f1. 1 2 ■ CHAPTER 3 \| TEST 1. Which of the following are graphs of functions? If the graph is that of a function, is it one-to-one? (a) (c) y 0 y (b) (d. Let f x 2 1 (a) Evaluate 1x 1 x 3, f f 1 2 (b) Find the domain of f. 5,
and f 1 2 3. A function f has the following verbal description: “Subtract 2, then cube the result.” (a) Find a formula that expresses f algebraically. (b) Make a table of values of f, for the inputs 1, 0, 1, 2, 3, and 4. (c) Sketch a graph of f, using the table of values from part (b) to help you. (d) How do we know that f has an inverse? Give a verbal description for f 1 (e) Find a formula that expresses algebraically. f 1. 4. A school fund-raising group sells chocolate bars to help ﬁnance a swimming pool for their physical education program. The group ﬁnds that when they set their price at x dollars per bar (where x R ), their total sales revenue (in dollars) is given by the function 500x2 3000x 0 x 5. 1 2 (a) Evaluate R(2) and R(4). What do these values represent? (b) Use a graphing calculator to draw a graph of R. What does the graph tell us about what happens to revenue as the price increases from 0 to 5 dollars? (c) What is the maximum revenue, and at what price is it achieved? 5. Determine the average rate of change for the function f t2 2t between t 2 and t 5. t 1 2 6. (a) Sketch the graph of the function f (b) Use part (a) to graph the function 1 g x 2 x x3.. (a) How is the graph of obtained from the graph of f? (b) How is the graph of obtained from the graph of f if x 1 if x 1 8. Let f x 1 2 1 x 2x 1 (a) Evaluate f and f 2 1 2 (b) Sketch the graph of f. b x2 1 and g x x f 9. If 1 2 2 1 f g (a) 278 f 2 (c) g 22 1 (e, ﬁnd the following. (b) g f (d) g 2 f 1 1 22 CHAPTER 3 \| Test 279 13 x, ﬁnd the inverse function f1. 10. (a) If f x 1 2 (b) Sketch the graphs of f and f1 on the same coordinate axes. 11. The graph of a function f is given. (a
) Find the domain and range of f. (b) Sketch the graph of (c) Find the average rate of change of f between x 2 and x 6. f1. y 1 0 1 x 12. Let f x 1 2 3x4 14x2 5x 3. (a) Draw the graph of f in an appropriate viewing rectangle. (b) Is f one-to-one? (c) Find the local maximum and minimum values of f and the values of x at which they occur. State each answer correct to two decimal places. (d) Use the graph to determine the range of f. (e) Find the intervals on which f is increasing and on which f is decreasing MODELING WITH FUNCTIONS Many of the processes that are studied in the physical and social sciences involve understanding how one quantity varies with respect to another. Finding a function that describes the dependence of one quantity on another is called modeling. For example, a biologist observes that the number of bacteria in a certain culture increases with time. He tries to model this phenomenon by ﬁnding the precise function (or rule) that relates the bacteria population to the elapsed time. In this Focus we will learn how to ﬁnd models that can be constructed using geometric or algebraic properties of the object under study. Once the model is found, we use it to analyze and predict properties of the object or process being studied. Modeling with Functions We begin with a simple real-life situation that illustrates the modeling process. E X AM P L E 1 \| Modeling the Volume of a Box A breakfast cereal company manufactures boxes to package their product. For aesthetic reasons, the box must have the following proportions: Its width is 3 times its depth, and its height is 5 times its depth. (a) Find a function that models the volume of the box in terms of its depth. (b) Find the volume of the box if the depth is 1.5 in. (c) For what depth is the volume 90 in3? (d) For what depth is the volume greater than 60 in3? Thinking About the Problem Let’s experiment with the problem. If the depth is 1 in., then the width is 3 in. and the height is 5 in. So in this case, the volume is V 1 3 5 15 in3. The table gives other values. Notice that all the boxes have the same shape, and the greater the depth the greater the volume. Depth Volume 1 2 3
4 1 3 5 15 2 6 10 120 3 9 15 405 4 12 20 960 3x 5x x ▼ SO LUTI O N (a) To ﬁnd the function that models the volume of the box, we use the following steps. 280 Modeling with Functions 281 ■ EXPRESS THE MODEL IN WORDS We know that the volume of a rectangular box is volume depth width height ■ CHOOSE THE VARIABLE There are three varying quantities: width, depth, and height. Because the function we want depends on the depth, we let Then we express the other dimensions of the box in terms of x. x depth of the box In Words In Algebra Depth Width Height x 3x 5x ■ SET UP THE MODEL The model is the function V that gives the volume of the box in terms of the depth x. width height volume V V x x 1 1 2 2 depth x # 3x # 5x 15x 3 The volume of the box is modeled by the function V graphed in Figure 1. 15x 3. The function V is x 1 2 ■ USE THE MODEL We use the model to answer the questions in parts (b), (c), and (d). (b) If the depth is 1.5 in., the volume is (c) We need to solve the equation V 1.5 1 90 15 2 or 1.5 V x 1 2 3 50.625 in3. 15x 2 1 3 90 3 6 x x 23 6 1.82 in. The volume is 90 in3 when the depth is about 1.82 in. (We can also solve this equation graphically, as shown in Figure 2.) (d) We need to solve the inequality 60 or V x 15x 2 1 3 60 3 4 x x 23 4 1.59 The volume will be greater than 60 in3 if the depth is greater than 1.59 in. (We can also ▲ solve this inequality graphically, as shown in Figure 3.) The steps in Example 1 are typical of how we model with functions. They are summa- rized in the following box. 400 0 FIGURE 1 400 y=15x£ y=90 0 15x £=90 FIGURE 2 400 0 y=15x£ y=60 15x £≥60 FIGURE 3 3 3 3 282 Focus on Modeling GUIDELINES FOR MODELING WITH FUNCTIONS 1. Express the Model in Words. Ident
ify the quantity you want to model, and express it, in words, as a function of the other quantities in the problem. 2. Choose the Variable. Identify all the variables that are used to express the function in Step 1. Assign a symbol, such as x, to one variable, and express the other variables in terms of this symbol. 3. Set up the Model. Express the function in the language of algebra by writing it as a function of the single variable chosen in Step 2. 4. Use the Model. Use the function to answer the questions posed in the problem. (To ﬁnd a maximum or a minimum, use the methods described in Section 3.3.) E X AM P L E 2 \| Fencing a Garden A gardener has 140 feet of fencing to fence in a rectangular vegetable garden. (a) Find a function that models the area of the garden she can fence. (b) For what range of widths is the area greater than 825 ft2? (c) Can she fence a garden with area 1250 ft2? (d) Find the dimensions of the largest area she can fence. Thinking About the Problem If the gardener fences a plot with width 10 ft, then the length must be 60 ft, because 10 10 60 60 140. So the area is A width length 10 # 60 600 ft2 The table shows various choices for fencing the garden. We see that as the width increases, the fenced area increases, then decreases. Width Length Area 10 20 30 40 50 60 60 50 40 30 20 10 600 1000 1200 1200 1000 600 width length ▼ SO LUTI O N (a) The model that we want is a function that gives the area she can fence. ■ EXPRESS THE MODEL IN WORDS We know that the area of a rectangular garden is area width length ■ CHOOSE THE VARIABLE There are two varying quantities: width and length. Because the function we want depends on only one variable, we let x width of the garden l l x FIGURE 4 x Maximum values of functions are discussed on page 230. Modeling with Functions 283 Then we must express the length in terms of x. The perimeter is ﬁxed at 140 ft, so the length is determined once we choose the width. If we let the length be l, as in Figure 4, then 2x 2l 140, so l 70 x. We summarize these facts. In Words Width Length In Algebra x 70 x ■
SET UP THE MODEL The model is the function A that gives the area of the garden for any width x. area width length 70 x x 1 70x The area that she can fence is modeled by the function A 70x x 2. x 1 2 ■ USE THE MODEL We use the model to answer the questions in parts (b)–(d). (b) We need to solve the inequality 825 y 70x x2 and y 825 in the same viewing rectangle (see Figure 5). We see that 15 x 55.. To solve graphically, we graph A x 1 2 (c) From Figure 6 we see that the graph of so an area of 1250 ft2 is never attained. A x 1 2 always lies below the line y 1250, (d) We need to ﬁnd where the maximum value of the function The function is graphed in Figure 7. Using the tor, we ﬁnd that the function achieves its maximum value at. So the maximum area that she can fence is that when the garden’s width is 35 ft and its length is ft. The maximum area then is 35 35 1225 ft2 70 35 35 TRACE. x A 70x x2 occurs. feature on a graphing calcula- 2 1 x 35 1500 1500 y=825 y=70x-≈ y=70x-≈ _5 _100 FIGURE 5 75 _5 _100 FIGURE 6 y=1250 75 1500 (35, 1225) y=70x-x™ _5 _100 FIGURE 7 75 284 Focus on Modeling Pythagoras (circa 580–500 B.C.) founded a school in Croton in southern Italy, devoted to the study of arithmetic, geometry, music, and astronomy. The Pythagoreans, as they were called, were a secret society with peculiar rules and initiation rites. They wrote nothing down and were not to reveal to anyone what they had learned from the Master. Although women were barred by law from attending public meetings, Pythagoras allowed women in his school, and his most famous student was Theano (whom he later married). According to Aristotle, the Pythagoreans were convinced that “the principles of mathematics are the principles of all things.” Their motto was “Everything is Number,” by which they meant whole numbers. The outstanding contribution of Pythagoras is the theorem that bears his name: In a right triangle the
area of the square on the hypotenuse is equal to the sum of the areas of the square on the other two sides. a c b c™=a™+b™ The converse of Pythagoras’s Theorem is also true; that is, a triangle whose sides a, b, and c satisfy a2 b2 c2 is a right triangle. E X AM P L E 3 \| Minimizing the Metal in a Can A manufacturer makes a metal can that holds 1 L (liter) of oil. What radius minimizes the amount of metal in the can? Thinking About the Problem To use the least amount of metal, we must minimize the surface area of the can, that is, the area of the top, bottom, and the sides. The area of the top and bottom is 2pr2 and the area of the sides is 2prh (see Figure 8), so the surface area of the can is S 2pr 2 2prh The radius and height of the can must be chosen so that the volume is exactly 1 L, or 1000 cm3. If we want a small radius, say r 3, then the height must be just tall enough to make the total volume 1000 cm3. In other words, we must have 2h 1000 h 1000 9p Volume of the can is pr 2h 35.4 cm Solve for h p 3 1 2 Now that we know the radius and height, we can ﬁnd the surface area of the can: 2 2p surface area 2p 723.8 cm3 35.4 3 3 1 2 1 2 1 2 If we want a different radius, we can ﬁnd the corresponding height and surface area in a similar fashion. 2πr r r r h h FIGURE 8 ▼ SO LUTI O N The model that we want is a function that gives the surface area of the can. ■ EXPRESS THE MODEL IN WORDS We know that for a cylindrical can surface area area of top and bottom area of sides ■ CHOOSE THE VARIABLE There are two varying quantities: radius and height. Because the function we want depends on the radius, we let r radius of can Next, we must express the height in terms of the radius r. Because the volume of a cylindrical can is V pr 2h and the volume must be 1000 cm3, we have pr 2h 1000 h 1000 pr 2 Volume of can is 1000 cm3 Solve for
h 1000 0 FIGURE 9 S 2pr 2 2000 r We can now express the areas of the top, bottom, and sides in terms of r only. Modeling with Functions 285 In Words Radius of can Height of can Area of top and bottom Area of sides (2prh) In Algebra r 1000 2 pr 2pr 2 2pr 1000 pr 2 b a ■ SET UP THE MODEL The model is the function S that gives the surface area of the can as a function of the radius r. surface area area of top and bottom area of sides S S r r 1 1 2 2 2pr 2pr 2 2pr a 2 2000 r 1000 pr 2 b 15 ■ USE THE MODEL We use the model to ﬁnd the minimum surface area of the can. We graph S in Figure 9 and zoom in on the minimum point to ﬁnd that the minimum value of S is about 554 cm2 and ▲ occurs when the radius is about 5.4 cm. Problems 1–18 ■ In these problems you are asked to ﬁnd a function that models a real-life situation. Use the principles of modeling described in this Focus to help you. 1. Area A rectangular building lot is three times as long as it is wide. Find a function that models its area A in terms of its width „. 2. Area A poster is 10 inches longer than it is wide. Find a function that models its area A in terms of its width „. 3. Volume A rectangular box has a square base. Its height is half the width of the base. Find a function that models its volume V in terms of its width „. 4. Volume The height of a cylinder is four times its radius. Find a function that models the volume V of the cylinder in terms of its radius r. 5. Area A rectangle has a perimeter of 20 ft. Find a function that models its area A in terms of the length x of one of its sides. 6. Perimeter A rectangle has an area of 16 m2. Find a function that models its perimeter P in terms of the length x of one of its sides. 7. Area Find a function that models the area A of an equilateral triangle in terms of the length x of one of its sides. 8. Area Find a function that models the surface area S of a cube in terms of its volume V. 9. Radius Find a function that models the radius r of a circle in terms of its area A.
10. Area Find a function that models the area A of a circle in terms of its circumference C. 11. Area A rectangular box with a volume of 60 ft3 has a square base. Find a function that mod- els its surface area S in terms of the length x of one side of its base. 286 Focus on Modeling 12. Length A woman 5 ft tall is standing near a street lamp that is 12 ft tall, as shown in the ﬁgure. Find a function that models the length L of her shadow in terms of her distance d from the base of the lamp. 12 ft 5 ft L d 13. Distance Two ships leave port at the same time. One sails south at 15 mi/h, and the other sails east at 20 mi/h. Find a function that models the distance D between the ships in terms of the time t (in hours) elapsed since their departure. D 14. Product The sum of two positive numbers is 60. Find a function that models their product P in terms of x, one of the numbers. 15. Area An isosceles triangle has a perimeter of 8 cm. Find a function that models its area A in terms of the length of its base b. 16. Perimeter A right triangle has one leg twice as long as the other. Find a function that models its perimeter P in terms of the length x of the shorter leg. 17. Area A rectangle is inscribed in a semicircle of radius 10, as shown in the ﬁgure. Find a function that models the area A of the rectangle in terms of its height h. h A h 10 18. Height The volume of a cone is 100 in3. Find a function that models the height h of the cone in terms of its radius r. 19–32 ■ In these problems you are asked to ﬁnd a function that models a real-life situation, and then use the model to answer questions about the situation. Use the guidelines on page 282 to help you. 19. Maximizing a Product Consider the following problem: Find two numbers whose sum is 19 and whose product is as large as possible. (a) Experiment with the problem by making a table like the one following, showing the product of different pairs of numbers that add up to 19. On the basis of the evidence in your table, estimate the answer to the problem. Modeling with Functions 287 First number Second number Product 1 2 3 o 18 17 16 o 18 34 48 o (
b) Find a function that models the product in terms of one of the two numbers. (c) Use your model to solve the problem, and compare with your answer to part (a). 20. Minimizing a Sum Find two positive numbers whose sum is 100 and the sum of whose squares is a minimum. 21. Fencing a Field Consider the following problem: A farmer has 2400 ft of fencing and wants to fence off a rectangular ﬁeld that borders a straight river. He does not need a fence along the river (see the ﬁgure). What are the dimensions of the ﬁeld of largest area that he can fence? (a) Experiment with the problem by drawing several diagrams illustrating the situation. Calculate the area of each conﬁguration, and use your results to estimate the dimensions of the largest possible ﬁeld. (b) Find a function that models the area of the ﬁeld in terms of one of its sides. (c) Use your model to solve the problem, and compare with your answer to part (a). x A x 22. Dividing a Pen A rancher with 750 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle (see the ﬁgure). (a) Find a function that models the total area of the four pens. (b) Find the largest possible total area of the four pens. 23. Fencing a Garden Plot A property owner wants to fence a garden plot adjacent to a road, as shown in the ﬁgure. The fencing next to the road must be sturdier and costs $5 per foot, but the other fencing costs just $3 per foot. The garden is to have an area of 1200 ft2. (a) Find a function that models the cost of fencing the garden. (b) Find the garden dimensions that minimize the cost of fencing. (c) If the owner has at most $600 to spend on fencing, ﬁnd the range of lengths he can fence along the road. x 288 Focus on Modeling 24. Maximizing Area A wire 10 cm long is cut into two pieces, one of length x and the other of length 10 x, as shown in the ﬁgure. Each piece is bent into the shape of a square. (a) Find a function that models the total area enclosed by the two
squares. (b) Find the value of x that minimizes the total area of the two squares. 10 cm x 10-x 25. Light from a Window A Norman window has the shape of a rectangle surmounted by a semicircle, as shown in the ﬁgure to the left. A Norman window with perimeter 30 ft is to be constructed. (a) Find a function that models the area of the window. (b) Find the dimensions of the window that admits the greatest amount of light. x 26. Volume of a Box A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 12 in. by 20 in. by cutting out equal squares of side x at each corner and then folding up the sides (see the ﬁgure). (a) Find a function that models the volume of the box. (b) Find the values of x for which the volume is greater than 200 in3. (c) Find the largest volume that such a box can have. 12 in. x x x x 20 in. x x x x x 27. Area of a Box An open box with a square base is to have a volume of 12 ft3. (a) Find a function that models the surface area of the box. (b) Find the box dimensions that minimize the amount of material used. 28. Inscribed Rectangle Find the dimensions that give the largest area for the rectangle shown in the ﬁgure. Its base is on the x-axis and its other two vertices are above the x-axis, lying on the parabola y 8 x2. y 0 y=8-≈ (x, y) x 29. Minimizing Costs A rancher wants to build a rectangular pen with an area of 100 m2. (a) Find a function that models the length of fencing required. (b) Find the pen dimensions that require the minimum amount of fencing. Modeling with Functions 289 30. Minimizing Time A man stands at a point A on the bank of a straight river, 2 mi wide. To reach point B, 7 mi downstream on the opposite bank, he ﬁrst rows his boat to point P on the opposite bank and then walks the remaining distance x to B, as shown in the ﬁgure. He can row at a speed of 2 mi/h and walk at a speed of 5 mi/h. (a) Find a
function that models the time needed for the trip. (b) Where should he land so that he reaches B as soon as possible? 7 mi x P B A 31. Bird Flight A bird is released from point A on an island, 5 mi from the nearest point B on a straight shoreline. The bird ﬂies to a point C on the shoreline and then ﬂies along the shoreline to its nesting area D (see the ﬁgure). Suppose the bird requires 10 kcal/mi of energy to ﬂy over land and 14 kcal/mi to ﬂy over water. (a) Use the fact that energy used energy per mile miles flown to show that the total energy used by the bird is modeled by the function E x 1 2 142x2 25 10 12 x 1 2 (b) If the bird instinctively chooses a path that minimizes its energy expenditure, to what point does it ﬂy? Island A 5 mi B C D x 12 mi Nesting area 5 x 5 x 12 12 32. Area of a Kite A kite frame is to be made from six pieces of wood. The four pieces that form its border have been cut to the lengths indicated in the ﬁgure. Let x be as shown in the ﬁgure. (a) Show that the area of the kite is given by the function (b) How long should each of the two crosspieces be to maximize the area of the kite? A x 1 2 x A 225 x 2 2144 x 2 B This page intentionally left blank 4.1 Quadratic Functions and Models 4.2 Polynomial Functions and Their Graphs 4.3 Dividing Polynomials 4.4 Real Zeros of Polynomials 4.5 Complex Zeros and the Fundamental Theorem of Algebra 4.6 Rational Functions © CHAPTER 4 Polynomial and Rational Functions Beautiful curves? An artist can draw a beautiful curve with a graceful sweep of pencil on paper. But to construct such a curve in steel and concrete, on the grand scale of the famous Disney Center, an architect needs a more precise description of the curve. In this chapter we will see that graphs of polynomial functions have gentle curves that are more varied the higher the degree of the polynomial. Such graphs can be pieced together to precisely reproduce any artist’s curve to any scale! We will also see how polynomial graphs, with their
many peaks and valleys, are used to model real-world quantities that vary in more intricate ways than we’ve studied so far—from the seasonal demand for different types of clothing to the uneven growth of living things over their lifetime (see Focus on Modeling: Fitting Polynomial Curves to Data, pages 364–368). 291291 291 292 CHAPTER 4 \| Polynomial and Rational Functions 4.1 Quadratic Functions and Models LEARNING OBJECTIVES After completing this section, you will be able to: ■ Express a quadratic function in standard form ■ Graph a quadratic function using its standard form ■ Find maximum and minimum values of quadratic functions ■ Model with quadratic functions A polynomial function is a function that is deﬁned by a polynomial expression. So a polynomial function of degree n is a function of the form anx n an1x n1 p a1x a 0 P x 1 2 Polynomial expressions are deﬁned in Section P.5. We have already studied polynomial functions of degree 0 and 1. These are functions of the form respectively, whose graphs are lines. In this section we study polynomial functions of degree 2. These are called quadratic functions. a1x a0, a 0 and P P x x 1 2 1 2 QUADRATIC FUNCTIONS A quadratic function is a polynomial function of degree 2. So a quadratic function is a function of the form f x 1 2 ax 2 bx c a 0 1 2 We see in this section how quadratic functions model many real-world phenomena. We begin by analyzing the graphs of quadratic functions. ■ Graphing Quadratic Functions Using the Standard Form in the quadratic function If we take b c 0 a 1 and we get the quadratic function ax 2 bx whose graph is the parabola graphed in Example 1 of Section 3.2. In fact, the graph of any quadratic function is a parabola; it can be obtained from the graph of by the transformations given in Section 3.5. x 2 x f 1 2 STANDARD FORM OF A QUADRATIC FUNCTION A quadratic function f ax x 1 2 2 bx can be expressed in the standard form by completing the square. The graph of f is a parabola with vertex parabola opens
upward if a 0 or downward if a 0. h, k 1 2 ; the y k Vertex (h, k) 0 h Ï=a(x-h)™+k, a>0 x y k 0 Vertex (h, k) h x Ï=a(x-h)™+k, a<0 SE CTI O N 4. 1 \| Quadratic Functions and Models 293 E X AM P L E 1 \| Standard Form of a Quadratic Function f x 2x Let (a) Express f in standard form. 2 12x 23. 2 1 (b) Sketch the graph of f. Completing the square is discussed in Section 1.3. ▼ SO LUTI O N (a) Since the coefﬁcient of x 2 is not 1, we must factor this coefﬁcient from the terms involving x before we complete the square Vertex is 3, 5 1 2 1 2 2x2 12x 23 2 23 x2 6x x2 6x The standard form is 2 23 2 # 9 2 x 3 2 5. Factor 2 from the x-terms Complete the square: Add 9 inside # parentheses, subtract 2 9 outside Factor and simplify Galileo Galilei (1564–1642) was born in Pisa, Italy. He studied medicine, but later abandoned this in favor of science and mathematics. At the age of 25 he demonstrated that light objects fall at the same rate as heavier ones, by dropping cannonballs of various sizes from the Leaning Tower of Pisa. This contradicted the then-accepted view of Aristotle that heavier objects fall more quickly. He also showed that the distance an object falls is proportional to the square of the time it has been falling, and from this was able to prove that the path of a projectile is a parabola. Galileo constructed the ﬁrst telescope, and using it, discovered the moons of Jupiter. His advocacy of the Copernican view that the earth revolves around the sun (rather than being stationary) led to his being called before the Inquisition. By then an old man, he was forced to recant his views, but he is said to have muttered under his breath “the earth nevertheless does move.” Galileo revolutionized science by expressing scientiﬁc principles in the language of mathematics. He said, “The great book of nature is written in mathematical symbols.” 2 1 2 (
b) The standard form tells us that we get the graph of f by taking the parabola y x 2, shifting it to the right 3 units, stretching it by a factor of 2, and moving it upward and the parabola opens upward. We 5 units. The vertex of the parabola is at sketch the graph in Figure 1 after noting that the y-intercept is f 0 23 3 25 23 15 5 Ï=2(x-3)™+5 Vertex (3, 5) FIGURE 1 x ✎ Practice what you’ve learned: Do Exercise 13. 3 0 ▲ ■ Maximum and Minimum Values of Quadratic Functions If a quadratic function has vertex, then the function has a minimum value at the vertex if its graph opens upward and a maximum value at the vertex if its graph opens downward. For example, the function graphed in Figure 1 has minimum value 5 when x 3, since the vertex is the lowest point on the graph. 3, 5 h, k 1 2 1 2 MAXIMUM OR MINIMUM VALUE OF A QUADRATIC FUNCTION Let f be a quadratic function with standard form mum or minimum value of f occurs at x h. If a 0, then the minimum value of f is f 1 If a 0, then the maximum value of f is f 2 h h k. k. The maxi- y k 0 Minimum h x y k 0 Maximum h x Ï=a(x-h)™+k, a>0 Ï=a(x-h)™+k, a<0 294 CHAPTER 4 \| Polynomial and Rational Functions Ï=5(x-3)™+4 (3, 4) 3 Minimum value 4 x y 49 4 0 E X AM P L E 2 \| Minimum Value of a Quadratic Function f Consider the quadratic function (a) Express f in standard form. (b) Sketch the graph of f. (c) Find the minimum value of f. 5x2 30x 49. x 2 1 ▼ SO LUTI O N (a) To express this quadratic function in standard form, we complete the square. f x 1 2 5x2 30x 49 5 49 x2 6x x2 6x 49 5 # 9 Factor 5 from the x-terms Complete the square: Add 9 inside parentheses, subtract 5 9 outside Factor and simplify (b)
The graph is a parabola that has its vertex at Figure 2. 3, 4 1 2 and opens upward, as sketched in FIGURE 2 f 3 1 2 4. (c) Since the coefﬁcient of x 2 is positive, f has a minimum value. The minimum value is y 1 0 _1 1!, @ 2 9 4 Maximum value 9 4 1 2 x FIGURE 3 Graph of x2 x 2 f x 1 2 ✎ Practice what you’ve learned: Do Exercise 25. ▲ E X AM P L E 3 \| Maximum Value of a Quadratic Function f Consider the quadratic function (a) Express f in standard form. (b) Sketch the graph of f. (c) Find the maximum value of f. x2 x 2. x 2 1 ▼ SO LUTI O N (a) To express this quadratic function in standard form, we complete the square. y x2 x 2 1 A A x2 x 2 2 x2 x 1 4B 2 9 4 x 1 2B 2 1 1 4 2 1 Factor 1 from the x-terms 1 Complete the square: Add 4 1 parentheses, subtract 1 2 inside 1 4 outside Factor and simplify (b) From the standard form we see that the graph is a parabola that opens downward 9. As an aid to sketching the graph, we ﬁnd the intercepts. The 4B 2. To ﬁnd the x-intercepts, we set and factor the 0 x f and has vertex y-intercept is f 2 resulting equation. 1 2, A 0 1 1 2 x2 x 2 0 x2 x 2 0 0 x 1 x 2 Set y = 0 Multiply by –1 Factor 1 2 1 2 Thus, the x-intercepts are x 2 and x 1. The graph of f is sketched in Figure 3. 9 (c) Since the coefﬁcient of x 2 is negative, f has a maximum value, which is 4 ✎ Practice what you’ve learned: Do Exercise 27. 1 2B f A. ▲ Expressing a quadratic function in standard form helps us to sketch its graph as well as ﬁnd its maximum or minimum value. If we are interested only in ﬁnding the maximum or SE CTI O N 4. 1 \| Quadratic Functions and Models 295 minimum value, then
a formula is available for doing so. This formula is obtained by completing the square for the general quadratic function as follows: ax 2 bx b2 4a2 b c a b2 4a2 b a a a x b 2a b 2 c b2 4a Factor Factor a from the x-terms Complete the square: Add inside parentheses, b2 4a2 subtract a b2 4a2 b a outside 4a This equation is in standard form with 1 imum or minimum value occurs at x h, we have the following result. and 2a 1 2 k c b2/ h b/. Since the max- 2 MAXIMUM OR MINIMUM VALUE OF A QUADRATIC FUNCTION The maximum or minimum value of a quadratic function occurs at f x 1 2 ax2 bx c x b 2a If a 0, then the minimum value is f If a 0, then the maximum value is f a b 2a b. a b 2a b. E X AM P L E 4 \| Finding Maximum and Minimum Values of Quadratic Functions Find the maximum or minimum value of each quadratic function. (b) (a) 2x2 4x 5 x2 4x g x x f 1 2 ▼ SO LUTI O N (a) This is a quadratic function with a 1 and b 4. Thus, the maximum or minimum 1 2 value occurs at x 4 2 # 1 Since a 0, the function has the minimum value 2 2 4 b 2a 2 2 f 2 4 2 (b) This is a quadratic function with a 2 and b 4. Thus, the maximum or mini- 1 1 2 1 2 mum value occurs at 4 2 2 Since a 0, the function has the maximum value x b 2a 2 # 1 1 2 ✎ Practice what you’ve learned: Do Exercises 33 and 35 _5 _6 The minimum value occurs at x = _2. 1 _2 2 4 _6 The maximum value occurs at x = 1. 296 CHAPTER 4 \| Polynomial and Rational Functions ■ Modeling with Quadratic Functions We study some examples of real-world phenomena that are modeled by quadratic functions. These examples and the Application exercises for this section show some of the variety of situations that are naturally modeled by quadratic functions. E X AM P L E 5 \| Maximum Gas Mileage for a Car Most cars get their
best gas mileage when traveling at a relatively modest speed. The gas mileage M for a certain new car is modeled by the function M s 1 2 1 28 s2 3s 31, 15 s 70 where s is the speed in mi/h and M is measured in mi/gal. What is the car’s best gas mileage, and at what speed is it attained? ▼ SO LUTI O N its maximum value occurs when The function M is a quadratic function with a 1 28 and b 3. Thus, s b 2a 2 3 42 The maximum is 1 is 32 mi/gal, when it is traveling at 42 mi/h. ✎ Practice what you’ve learned: Do Exercise 67. A 31 32 3 1 28 1 42 1 28B 42 42 M 2 2 2 2 1. So the car’s best gas mileage ▲ 40 15 0 70 The maximum gas mileage occurs at 42 mi/h. E X AM P L E 6 \| Maximizing Revenue from Ticket Sales A hockey team plays in an arena that has a seating capacity of 15,000 spectators. With the ticket price set at $14, average attendance at recent games has been 9500. A market survey indicates that for each dollar the ticket price is lowered, the average attendance increases by 1000. (a) Find a function that models the revenue in terms of ticket price. (b) Find the price that maximizes revenue from ticket sales. (c) What ticket price is so high that no one attends and so no revenue is generated? ▼ SO LUTI O N (a) The model that we want is a function that gives the revenue for any ticket price. Express the model in words revenue ticket price attendance There are two varying quantities: ticket price and attendance. Since the function we want depends on price, we let Choose the variable x ticket price Next, we express attendance in terms of x. In Words In Algebra Ticket price Amount ticket price is lowered Increase in attendance Attendance x 14 x 14 x 1000 2 1 9500 1000 1 14 x 2 SE CTI O N 4. 1 \| Quadratic Functions and Models 297 The model that we want is the function R that gives the revenue for a given ticket price x. Set up the model revenue ticket price attendance Use the model Use the model 14 9500 1000 x 23,500 1000x x 23,500x 1000x a 1000 1 2 2 (b) Since R is a quadratic function with and b
23,500, the maximum occurs at x b 2a 23,500 1000 1 2 2 11.75 So a ticket price of $11.75 gives the maximum revenue. (c) We want to ﬁnd the ticket price for which R 2 0 2 0 0 23,500x 1000x 23.5x x 23.5 x x 1 x 0 or x 23.5 2 0 : x 1 2 Set R(x) = 0 Divide by 1000 Factor Solve for x 25 So according to this model, a ticket price of $23.50 is just too high; at that price, no one attends to watch this team play. (Of course, revenue is also zero if the ticket price is zero.) 150,000 0 0 Maximum attendance occurs when ticket price is $11.75. ✎ Practice what you’ve learned. Do Exercises 75 and 77. ▲ 4. ▼ CONCE PTS 1. To put the quadratic function f form, we complete the 2. The quadratic function is in standard f is a parabola with vertex the graph of opens f is the f value of. the graph of opens f is the f value of.,. 1 2. In this case. In this case is a parabola that opens, with its vertex at, 1, and 2 is the (minimum/maximum) (b) If form. (a) The graph of a 0, k h 2 a 0, k 1 (c) If h f f 1 3. The graph of 2 3 f f value of. 1 2 4. The graph of f 1 opens x 2 x 3 1 2, with its vertex at 2 2 5 is a parabola that, 1, and 2 ax 2 bx c in standard x 1 2. f 3 2 1 f of. is the (minimum/maximum) value ▼ SKI LLS 5–8 ■ The graph of a quadratic function coordinates of the vertex. (b) Find the maximum or minimum value of f. (c) Find the domain and range of f. is given. (a) Find the f 5. f x 1 2 y x2 6x 5 6. f x 1 2 1 2 x2 2x 298 CHAPTER 4 \| Polynomial and Rational Functions 7. f x 1 2 2x2 4x 1 8. f x 1 2 3x2 6x 1 y 1 0 1 x y 1 0
correct to two decimal places. (b) Find the exact maximum or minimum value of f, and compare it with your answer to part (a). 49. 50. f f x x 1 1 2 2 x2 1.79x 3.21 1 x 12x2 1 x 51–54 ■ Find all local maximum and minimum values of the function whose graph is shown. 51. 52. 9–22 ■ A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its x- and y-intercept(s). (c) Sketch its graph. ✎ 9. 11. 13. 15. 17. 19. 21 x2 6x 2x2 6x x2 4x 3 x2 6x 4 2x2 4x 3 2x2 20x 57 4x2 16x 3 10. 12. 14. 16. 18. 20. 22 x2 8x x2 10x x2 2x 2 x2 4x 4 3x2 6x 2 2x2 x 6 6x2 12x 5 ✎ ✎ ✎ ✎ 2 1 f x 23. 23–32 ■ A quadratic function is given. (a) Express the quadratic function in standard form. (b) Sketch its graph. (c) Find its maximum or minimum value. x2 2x 1 3x2 6x 1 x2 3x 3 3x2 12x 13 1 x x2 x2 8x 8 5x 2 30x 4 1 6x x2 2x2 8x 11 3 4x 4x2 25. 29. 30. 27. 26. 32. 31. 28. 24 33–42 ■ Find the maximum or minimum value of the function. 33. 35. 37. f f f 39. h 41. f 1 1 1 1 1 x2 x 1 100 49t 7t2 s2 1.2s 16 x 2 t 2 s 2 1 2 x2 2x 6 3 x 1 2 x2 x x 2 2 34. 36. f f 38. g 40. f 42 3x x2 10t2 40t 113 100x2 1500x x2 3 x 4 2x 7 2x 7 1 2 43. Find a function whose graph is a parabola with vertex and that passes through the point 4, 16. 2 1 44. Find a function whose graph is a
parabola with vertex and that passes through the point 1, 8 1. 2 45–48 ■ Find the domain and range of the function. 1, 2 2 3, 4 2 1 1 45. 47. f f x x 1 1 2 2 x2 4x 3 2x2 6x 7 46. 48. f f x x 1 1 2 2 x2 2x 3 3x2 6x 4 49–50 ■ A quadratic function is given. (a) Use a graphing device to ﬁnd the maximum or minimum value of the quadratic function f, y 1 0 1 x 53. y 54 55–62 ■ Find the local maximum and minimum values of the function and the value of x at which each occurs. State each answer correct to two decimal places. 55. f x 1 2 57. 59. g 1 U x 2 x 1 2 61. V x 1 2 x3 x x4 2x3 11x2 x16 x 1 x2 x3 56. f x 1 2 58. 60 x2 x3 x5 8x3 20x x2x x2 62. V x 1 2 1 x2 x 1 ▼ APPLICATIONS 63. Height of a Ball If a ball is thrown directly upward with a velocity of 40 ft/s, its height (in feet) after t seconds is given by y 40t 16t 2. What is the maximum height attained by the ball? 64. Path of a Ball A ball is thrown across a playing ﬁeld from a height of 5 ft above the ground at an angle of 45º to the horizontal at a speed of 20 ft/s. It can be deduced from physical principles that the path of the ball is modeled by the function y 2 x 5 2 x 32 20 2 where x is the distance in feet that the ball has traveled horizontally. 1 (a) Find the maximum height attained by the ball. (b) Find the horizontal distance the ball has traveled when it hits the ground. 5 ft x 65. Revenue A manufacturer ﬁnds that the revenue generated by x 80x 0.4x2 selling x units of a certain commodity is given by the function R is measured in, where the revenue dollars. What is the maximum revenue, and how many units should be manufactured to obtain this maximum? R x 1 2 1 2 66. Sales A soft-drink vendor at a popular beach analyzes his sales records and ﬁ
nds that if he sells x cans of soda pop in one day, his proﬁt (in dollars) is given by 0.001x2 3x 1800 P x 1 2 What is his maximum proﬁt per day, and how many cans must he sell for maximum proﬁt? ✎ 67. Advertising The effectiveness of a television commercial depends on how many times a viewer watches it. After some experiments an advertising agency found that if the effectiveness E is measured on a scale of 0 to 10, then E n 1 2 2 3 n 1 90 n2 where n is the number of times a viewer watches a given commercial. For a commercial to have maximum effectiveness, how many times should a viewer watch it? 68. Pharmaceuticals When a certain drug is taken orally, the concentration of the drug in the patient’s bloodstream after t minutes is given by t 2 0 t 240 and the concentration is measured in mg/L. When is the maximum serum concentration reached, and what is that maximum concentration? 0.06t 0.0002t 2, where C 1 69. Agriculture The number of apples produced by each tree in an apple orchard depends on how densely the trees are planted. If n trees are planted on an acre of land, then each tree produces 900 9n apples. So the number of apples produced per acre is A n 1 2 n 1 900 9n 2 How many trees should be planted per acre to obtain the maximum yield of apples? ✎ SE CTI O N 4. 1 \| Quadratic Functions and Models 299 70. Agriculture At a certain vineyard it is found that each grape vine produces about 10 pounds of grapes in a season when about 700 vines are planted per acre. For each additional vine that is planted, the production of each vine decreases by about 1 percent. So the number of pounds of grapes produced per acre is modeled by A n 1 2 1 700 n 2 1 10 0.01n 2 where n is the number of additional vines planted. Find the number of vines that should be planted to maximize grape production. 71–74 ■ Use the formulas of this section to give an alternative solution to the indicated problem in Focus on Modeling: Modeling with Functions on pages 287–288. 71. Problem 21 73. Problem 25 72. Problem 22 74. Problem 24 ✎ 75. Fencing a Horse Corral Carol has 1200 ft of fencing to fence in a rectangular horse corral. (a) Find a function that
models the area of the corral in terms of the width x of the corral. (b) Find the dimensions of the rectangle that maximize the area of the corral. x 1200 – 2x 76. Making a Rain Gutter A rain gutter is formed by bending up the sides of a 30-inch-wide rectangular metal sheet as shown in the ﬁgure. (a) Find a function that models the cross-sectional area of the gutter in terms of x. (b) Find the value of x that maximizes the cross-sectional area of the gutter. (c) What is the maximum cross-sectional area for the gutter? x 30 in. 77. Stadium Revenue A baseball team plays in a stadium that holds 55,000 spectators. With the ticket price at $10, the average attendance at recent games has been 27,000. A market survey indicates that for every dollar the ticket price is lowered, attendance increases by 3000. (a) Find a function that models the revenue in terms of ticket price. (b) Find the price that maximizes revenue from ticket sales. (c) What ticket price is so high that no revenue is generated? 300 CHAPTER 4 \| Polynomial and Rational Functions 78. Maximizing Proﬁt A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The materials for each feeder cost $6, and the society sells an average of 20 per week at a price of $10 each. The society has been considering raising the price, so it conducts a survey and ﬁnds that for every dollar increase, it loses 2 sales per week. (a) Find a function that models weekly proﬁt in terms of price per feeder. (b) What price should the society charge for each feeder to maximize proﬁts? What is the maximum weekly proﬁt? ▼ DISCOVE RY • DISCUSSION • WRITI NG 79. Vertex and x-Intercepts We know that the graph of the 2 1 1 f x 2 1 x n is a parabola. x m quadratic function Sketch a rough graph of what such a parabola would look like. What are the x-intercepts of the graph of Can you tell from your graph the x-coordinate of the vertex in terms of m and n? (Use the symmetry of the parab
ola.) Conﬁrm your answer by expanding and using the formulas of this section. f? 2 80. Maximum of a Fourth-Degree Polynomial Find the maxi- mum value of the function 3 4x 2 x 4 f x 1 2 [Hint: Let t x 2.] 4.2 Polynomial Functions and Their Graphs LEARNING OBJECTIVES After completing this section, you will be able to: ■ Graph basic polynomial functions ■ Use end behavior of a polynomial function to help sketch its graph ■ Use the zeros of a polynomial function to sketch its graph ■ Use the multiplicity of a zero to help sketch the graph of a polynomial function ■ Find local maxima and minima of polynomial functions In this section we study polynomial functions of any degree. But before we work with polynomial functions, we must agree on some terminology. POLYNOMIAL FUNCTIONS A polynomial function of degree n is a function of the form an x n an1x n1... a1x a0 P x 1 2 an where n is a nonnegative integer and 0. The numbers a0, a1, a2, p, an are called the coefﬁcients of the polynomial. The number a0 is the constant coefﬁcient or constant term. The number an, the coefﬁcient of the highest power, is the leading coefﬁcient, and the term a nx n is the leading term. We often refer to polynomial functions simply as polynomials. The following polynomial has degree 5, leading coefﬁcient 3, and constant term 6. Leading coefﬁcient 3 Degree 5 Constant term 6 Leading term 3x 5 3x 5 6x 4 2x 3 x 2 7x 6 Coefﬁcients 3, 6, 2, 1, 7, and 6 SE CTI ON 4. 2 \| Polynomial Functions and Their Graphs 301 Here are some more examples of polynomials. 3 4x 7 x2 x 2x3 6x2 10 Degree 0 Degree 1 Degree 2 Degree 3 2 2 1 x x Q and x 3 6x 5 are monomials. If a polynomial consists of just a single term, then it is called a monomial. For example, P 1 ■ Graphing
Basic Polynomial Functions The graphs of polynomials of degree 0 or 1 are lines (Section 2.4), and the graphs of polynomials of degree 2 are parabolas (Section 4.1). The greater the degree of the polynomial, the more complicated its graph can be. However, the graph of a polynomial function is always a smooth curve; that is, it has no breaks or corners (see Figure 1). The proof of this fact requires calculus. y y y smooth and continuous y smooth and continuous cusp corner break hole x x x x FIGURE 1 Not the graph of a polynomial function Not the graph of a polynomial function Graph of a polynomial function Graph of a polynomial function The simplest polynomial functions are the monomials P, whose graphs are shown in Figure 2. As the ﬁgure suggests, the graph of has the same general shape as y x 2 when n is even and the same general shape as y x 3 when n is odd. However, as the degree n becomes larger, the graphs become ﬂatter around the origin and steeper elsewherea) y=x (b) y=≈ (c) y=x£ (d) y=x¢ (e) y=x∞ FIGURE 2 Graphs of monomials E X AM P L E 1 \| Transformations of Monomials Sketch the graphs of the following functions. x Q (a) (b) P x x 3 2x 5 4 (c ▼ SO LUTI O N We use the graphs in Figure 2 and transform them using the techniques of Section 3.5. 302 CHAPTER 4 \| Polynomial and Rational Functions (a) The graph of 1 2 x x 3 P shown in Figure 3(a) below. x 2 Q shown in Figure 3(b). x 1 1 2 (b) The graph of is the reﬂection of the graph of y x 3 in the x-axis, as 4 is the graph of y x 4 shifted to the right 2 units, as 2 (c) We begin with the graph of y x 5. The graph of y 2x 5 is obtained by stretching the graph vertically and reﬂecting it in the x-axis (see the dashed blue graph in Figure 3(c)). Finally, the graph of 4 units (see the red graph in Figure 3(c)). is obtained
by shifting upward 2x 5 4 R x 2 1 P(x)=_x£ y 1 0 Q(x)=(x-2)¢ y 16 8 y 8 4 R(x)=_2x∞+4 x1 _2 0 2 4 x _1 0 1 x (a) (b) (c) FIGURE 3 ✎ Practice what you’ve learned: Do Exercise 5. ▲ ■ End Behavior and the Leading Term The end behavior of a polynomial is a description of what happens as x becomes large in the positive or negative direction. To describe end behavior, we use the following notation: xq means “x becomes large in the positive direction” xq means “x becomes large in the negative direction” MATHEMATICS IN THE MODERN WORLD Splines A spline is a long strip of wood that is curved while held ﬁxed at certain points. In the old days shipbuilders used splines to create the curved shape of a boat’s hull. Splines are also used to make the curves of a piano, a violin, or the spout of a teapot. Mathematicians discovered that the shapes of splines can be obtained by piecing together parts of polynomials. For example, the graph of a cubic polynomial can be made to ﬁt speciﬁed points by adjusting the coefﬁcients of the polynomial (see Example 10, page 311). Curves obtained in this way are called cubic splines. In modern computer design programs, such as Adobe Illustrator or Microsoft Paint, a curve can be drawn by ﬁxing two points, then using the mouse to drag one or more anchor points. Moving the anchor points amounts to adjusting the coefﬁcients of a cubic polynomial. SE CTI ON 4. 2 \| Polynomial Functions and Their Graphs 303 For example, the monomial y x 2 in Figure 2(b) has the following end behavior: y q as x q and y q as x q The monomial y x 3 in Figure 2(c) has the following end behavior: y q as x q and y q as x q For any polynomial, the end behavior is determined by the term that contains the highest power of x, because when x is large, the other terms are relatively insigniﬁcant in size. The following
box shows the four possible types of end behavior, based on the highest power and the sign of its coefﬁcient. END BEHAVIOR OF POLYNOMIALS The end behavior of the polynomial sign of the leading coefﬁcient an, as indicated in the following graphs. anx n an1x n1... a1x a0 P x 1 2 is determined by the degree n and the P has odd degree y → ` as x → ` y → ` as x → _` y P has even degree y → ` as x → ` y → ` as x → _` as x → _` y → _` as x → ` y → _` as x → _` y → _` as x → ` Leading coefﬁcient positive Leading coefﬁcient negative Leading coefﬁcient positive Leading coefﬁcient negative E X AM P L E 2 \| End Behavior of a Polynomial Determine the end behavior of the polynomial 2x 4 5x 3 4x 7 P x 1 2 ▼ SO LUTI O N The polynomial P has degree 4 and leading coefﬁcient 2. Thus, P has even degree and negative leading coefﬁcient, so it has the following end behavior: y q as x q and y q as x q The graph in Figure 4 illustrates the end behavior of P. _3 y → _` as x → _` 30 _50 5 y → _` as x → ` ✎ Practice what you’ve learned: Do Exercise 11. ▲ FIGURE 4 P x 1 2 2x 4 5x 3 4x 7 304 CHAPTER 4 \| Polynomial and Rational Functions E X AM P L E 3 \| End Behavior of a Polynomial _1 Q P 1 _1 FIGURE 5 P Q x x 3x5 5x3 2x 3x5 1 1 2 2 x 15 30 50 P x 1 2 2,261,280 72,765,060 936,875,100 P (a) Determine the end behavior of the polynomial 3x 5 (b) Conﬁrm that P and its leading term Q x x 3x 5 5x 3 2x. 2 1 have the same end behavior by graphing them together. 1 2 ▼ SO LUTI O N (a) Since P has odd
degree and positive leading coefﬁcient, it has the following end behavior: y q as x q and y q as x q (b) Figure 5 shows the graphs of P and Q in progressively larger viewing rectangles. The larger the viewing rectangle, the more the graphs look alike. This conﬁrms that they have the same end behavior. 2 Q P 50 Q P 10,000 PQ 1 _2 2 _3 3 _10 _2 _50 _10,000 ✎ Practice what you’ve learned: Do Exercise 41. 10 ▲ Q x 1 2 2,278,125 72,900,000 937,500,000 To see algebraically why P and Q in Example 3 have the same end behavior, factor P as follows and compare with Q. 3x 5 P x 1 2 1 5 3x 2 a 2 3x 4 b 3x 5 Q x 1 2 When x is large, the terms 5/3x 2 and 2/3x 4 are close to 0 (see Exercise 67 on page 19). So for large x, we have P x 1 2 3x 5 1 1 0 0 2 3x 5 Q x 1 2 So when x is large, P and Q have approximately the same values. We can also see this numerically by making a table like the one in the margin. By the same reasoning we can show that the end behavior of any polynomial is deter- mined by its leading term. ■ Using Zeros to Graph Polynomials P If P is a polynomial function, then c is called a zero of P if. In other words, the 0 x P zeros of P are the solutions of the polynomial equation, then the graph of P has an x-intercept at x c, so the x-intercepts of the graph are the zeros of the function. 0. Note that if REAL ZEROS OF POLYNOMIALS If P is a polynomial and c is a real number, then the following are equivalent: 1. c is a zero of P. 2. x c is a solution of the equation 3. x c is a factor of 4. c is an x-intercept of the graph of P SE CTI ON 4. 2 \| Polynomial Functions and Their Graphs 305 To ﬁnd the zeros of a polynomial P, we factor and then use the Zero-Product Property x 2 x
6, we factor P to get (see page 87). For example, to ﬁnd the zeros of x 2 1 From this factored form we easily see that. 2 is a zero of P. 2. x 2 is a solution of the equation x 2 x 6 0. 3. x 2 is a factor of x 2 x 6. 4. 2 is an x-intercept of the graph of P. The same facts are true for the other zero, 3. The following theorem has many important consequences. (See, for instance, the Dis- covery Project on page 333.) Here we use it to help us graph polynomial functions. INTERMEDIATE VALUE THEOREM FOR POLYNOMIALS If P is a polynomial function and ists at least one value c between a and b for which and P P b a 1 2 1 2 have opposite signs, then there ex- 0. P c 1 2 We will not prove this theorem, but Figure 6 shows why it is intuitively plausible. One important consequence of this theorem is that between any two successive zeros the values of a polynomial are either all positive or all negative. That is, between two successive zeros the graph of a polynomial lies entirely above or entirely below the x-axis. To see why, suppose c1 and c2 are successive zeros of P. If P has both positive and negative values between c1 and c2, then by the Intermediate Value Theorem P must have another zero between c1 and c2. But that’s not possible because c1 and c2 are successive zeros. This observation allows us to use the following guidelines to graph polynomial functions. GUIDELINES FOR GRAPHING POLYNOMIAL FUNCTIONS 1. Zeros. Factor the polynomial to ﬁnd all its real zeros; these are the x-intercepts of the graph. 2. Test Points. Make a table of values for the polynomial. Include test points to determine whether the graph of the polynomial lies above or below the x-axis on the intervals determined by the zeros. Include the y-intercept in the table. 3. End Behavior. Determine the end behavior of the polynomial. 4. Graph. Plot the intercepts and other points you found in the table. Sketch a smooth curve that passes through these points and exhibits the required end behavior. E X
AM P L E 4 \| Using Zeros to Graph a Polynomial Function x 1 Sketch the graph of the polynomial function. 2 ▼ SO LUTI O N The zeros are x 2, 1, and 3. These determine the intervals 2, 1 3, q 1 the following sign diagram (see Section 1.6)., 2. Using test points in these intervals, we get the information in q, 2 x 3 x 2, and 1=P(x) c b x y P(b) a 0 P(a) FIGURE 6 306 CHAPTER 4 \| Polynomial and Rational Functions Sign of Graph of P Test point x = –3 P (–3) < 0 Test point x = –1 P (–1) > 0 Test point x = 2 P(2) < 0 Test point x = 4 P(3) > 0 _2 1 3 - below x-axis + above x-axis - + below x-axis above x-axis Plotting a few additional points and connecting them with a smooth curve helps us to complete the graph in Figure 7. Test point Test point Test point Test point x P x 1 2 3 24 18 4 y Test point P(4) > 0 Test point P(–1) > 0 5 0 1 x Test point P(–3) < 0 Test point P(2) < 0 x 2 ✎ Practice what you’ve learned: Do Exercise 17. FIGURE AM P L E 5 \| Finding Zeros and Graphing a Polynomial Function 2 1 x P x 3 2x 2 3x. Let (a) Find the zeros of P. (b) Sketch the graph of P. MATHEMATICS IN THE MODERN WORLD Automotive Design. Computer-aided design (CAD) has completely changed the way in which car companies design and manufacture cars. Before the 1980s automotive engineers would build a full-scale “nuts and bolts” model of a proposed new car; this was really the only way to tell whether the design was feasible. Today automotive engineers build a mathematical model, one that exists only in the memory of a computer. The model incorporates all the main design features of the car. Certain polynomial curves, called splines, are used in shaping the body of the car. The resulting “mathematical car” can be tested for structural stability, handling, aerodynamics, suspension response, and more. All this
testing is done before a prototype is built. As you can imagine, CAD saves car manufacturers millions of dollars each year. More importantly, CAD gives automotive engineers far more ﬂexibility in design; desired changes can be created and tested within seconds. With the help of computer graphics, designers can see how good the “mathematical car” looks before they build the real one. Moreover, the mathematical car can be viewed from any perspective; it can be moved, rotated, or seen from the inside. These manipulations of the car on the computer monitor translate mathematically into solving large systems of linear equations. SE CTI ON 4. 2 \| Polynomial Functions and Their Graphs 307 ▼ SO LUTI O N (a) To ﬁnd the zeros, we factor completely. x 3 2x 2 3x x x 2 2x 3 2 x 1 x 3 x Thus, the zeros are x 0, x 3, and x 1 Factor x Factor quadratic (b) The x-intercepts are x 0, x 3, and x 1. The y-intercept is P 0. We 0 1 2 make a table of values of to the right and left of) successive zeros. P x 1 2, making sure that we choose test points between (and Since P is of odd degree and its leading coefﬁcient is positive, it has the following end behavior: y q as x q and y q as x q We plot the points in the table and connect them by a smooth curve to complete the graph, as shown in Figure 8. x P x 2 1 2 10 20 Test point Test point Test point Test point y 5 0 1 x FIGURE 8 ✎ Practice what you’ve learned: Do Exercise 27. P x 1 2 x 3 2x 2 3x ▲ E X AM P L E 6 \| Finding Zeros and Graphing a Polynomial Function Let (a) Find the zeros of P. (b) Sketch the graph of P. 2x 4 x 3 3x 2. P x 2 1 ▼ SO LUTI O N (a) To ﬁnd the zeros, we factor completely. P x 1 2 2x 4 x 3 3x 2 x 2 2x 2 x 3 2x 3 x 2 1 Thus, the zeros are x 0, (b) The x-intercepts are x 0, P make a
table of values of to the right and left of) successive zeros. 1 2 1 x 3 2 x x Factor x 2 Factor quadratic 2 x 1 2 1, and x 1., and x 1. The y-intercept is 3 2 2, making sure that we choose test points between (and 0. We P 0 1 2 Since P is of even degree and its leading coefﬁcient is negative, it has the follow- ing end behavior: y q as x q and y q as x q 308 CHAPTER 4 \| Polynomial and Rational Functions We plot the points from the table and connect the points by a smooth curve to complete the graph in Figure 9. A table of values is most easily calculated by using a programmable calculator or a graphing calculator. x 2 1.5 1 0.5 0 0.5 1 1.5 2 P x 1 12 0 2 0.75 0 0.5 0 6.75 y 0 2 1 x _12 x P 1 ✎ Practice what you’ve learned: Do Exercise 31. FIGURE 9 2 2x 4 x 3 3x 2 ▲ E X AM P L E 7 \| Finding Zeros and Graphing a Polynomial Function Let (a) Find the zeros of P. x3 2x2 4x 8. (b) Sketch the graph of P. P x 2 1 2 1 x P ▼ SO LUTI O N (a) To ﬁnd the zeros, we factor completely. x 3 2x 2 4x 8 x 2 1 Thus, the zeros are x 2 and x 2 Group and factor Factor x 2 Difference of squares 2 Simplify (b) The x-intercepts are x 2 and x 2. The y-intercept is P 8. The table gives 0 1 2 additional values of P x. 1 2 Since P is of odd degree and its leading coefﬁcient is positive, it has the following end behavior: y q as x q and y q as x q We connect the points by a smooth curve to complete the graph in Figure 10. x P x 1 2 3 25 ✎ Practice what you’ve learned: Do Exercise 33. FIGURE 10 P 1 x 3 2x 2 4x 8 ▲ SE CTI ON 4. 2 \| Polynomial Functions and Their Graphs 309 ■ Shape of the Graph Near a Zero Although x 2 is a zero of
the polynomial in Example 7, the graph does not cross the x-axis at the x-intercept 2. This is because the factor corresponding to that zero is raised to an even power, so it doesn’t change sign as we test points on either side of 2. In the same way the graph does not cross the x-axis at x 0 in Example 6. x 2 In general, if c is a zero of P and the corresponding factor x c occurs exactly m times in the factorization of P then we say that c is a zero of multiplicity m. By considering test points on either side of the x-intercept c, we conclude that the graph crosses the x-axis at c if the multiplicity m is odd and does not cross the x-axis if m is even. Moreover, it can be shown by using calculus that near x c the graph has the same general shape as A x c m. 1 2 2 1 2 SHAPE OF THE GRAPH NEAR A ZERO OF MULTIPLICITY m If c is a zero of P of multiplicity m, then the shape of the graph of P near c is as follows. Multiplicity of c Shape of the graph of P near the x-intercept c m odd, m 1 m even, m 1 y y y y c c OR OR x x c x c x E X AM P L E 8 \| Graphing a Polynomial Function Using Its Zeros x 2 Graph the polynomial The zeros of P are 1, 0, and 2 with multiplicities 2, 4, and 3, ▼ SO LUTI O N respectively. 0 is a zero of multiplicity 4. 2 is a zero of multiplicity 3. –1 is a zero of multiplicity 2 The zero 2 has odd multiplicity, so the graph crosses the x-axis at the x-intercept 2. But the zeros 0 and 1 have even multiplicity, so the graph does not cross the x-axis at the x-intercepts 0 and 1. Since P is a polynomial of degree 9 and has positive leading coefﬁcient, it has the following end behavior: y q as x q and y q as x q With this information and a table of values, we sketch the graph in Figure 11. 310 CHAPTER 4 \| Polynomial and Rational Functions y 5 0 Even multiplicities x P x 1 2 1.3
9.2 1 0.5 3.9 0 0 1 2 2.3 0 4 0 8.2 x 4 ✎ Practice what you’ve learned: Do Exercise 25. FIGURE 11 P x 2 1 1 x Odd multiplicity ▲ ■ Local Maxima and Minima of Polynomials Recall from Section 3.3 that if the point f within some viewing rectangle, then the lowest point on the graph of f within a viewing rectangle, then f value (see Figure 12). We say that such a point graph and that imum points on the graph of a function is called its local extrema. is the highest point on the graph of f is 1 is a local minimum is a local maximum point on the is a local minimum point. The set of all local maximum and min- a, f 1 is a local maximum value of f, and if b, f a, f b, f 22 22 22 22 Óa, f(a)Ô Local maximum point y=Ï Ób, f(b)Ô Local minimum point a b x FIGURE 12 For a polynomial function the number of local extrema must be less than the degree, as the following principle indicates. (A proof of this principle requires calculus.) LOCAL EXTREMA OF POLYNOMIALS x P anx n an1x n1... a1x a0 If the graph of P has at most n 1 local extrema. 2 1 is a polynomial of degree n, then x 5 A polynomial of degree n may in fact have less than n 1 local extrema. For example, x P (graphed in Figure 2) has no local extrema, even though it is of degree 5. The preceding principle tells us only that a polynomial of degree n can have no more than n 1 local extrema. 2 1 SE CTI ON 4. 2 \| Polynomial Functions and Their Graphs 311 E X AM P L E 9 \| The Number of Local Extrema Determine how many local extrema each polynomial has. (a) 2 2 x x (c) (b) x 4 x 3 16x 2 4x 48 x 5 3x 4 5x 3 15x 2 4x 15 7x 4 3x 2 10x P11 P21 P31 2 ▼ SO LUTI O N (a) P1 has two local minimum
points and one local maximum point, for a total of three The graphs are shown in Figure 13. x local extrema. (b) P2 has two local minimum points and two local maximum points, for a total of four local extrema. (c) P3 has just one local extremum, a local minimum. 100 100 100 _5 5 _5 5 _5 _100 (a) _100 (b) _100 (c) FIGURE 13 P⁄(x)=x¢+x£-16≈-4x+48 P¤(x)=x∞+3x¢-5x£-15≈+4x-15 P‹(x)=7x¢+3≈-10x ✎ Practice what you’ve learned: Do Exercises 61 and 63. 5 ▲ With a graphing calculator we can quickly draw the graphs of many functions at once, on the same viewing screen. This allows us to see how changing a value in the deﬁnition of the functions affects the shape of its graph. In the next example we apply this principle to a family of third-degree polynomials. E X AM P L E 10 \| A Family of Polynomials P Sketch the family of polynomials ing the value of c affect the graph? x 3 cx 2 x 2 1 for c 0, 1, 2, and 3. How does chang- c=0 c=1 c=2 10 c=3 ▼ SO LUTI O N The polynomials _2 4 _10 FIGURE 14 A family of polynomials P x 3 cx 2 x 1 2 x 3 x 3 2x 2 x x P01 P21 2 2 x 3 x 2 x 3 3x 2 x x P11 P31 2 2 are graphed in Figure 14. We see that increasing the value of c causes the graph to develop an increasingly deep “valley” to the right of the y-axis, creating a local maximum at the origin and a local minimum at a point in Quadrant IV. This local minimum moves lower and farther to the right as c increases. To see why this happens, factor. 2 The polynomial P has zeros at 0 and c, and the larger c gets, the farther to the right the minimum between 0 and c will be ✎ Practice what you’ve learned: Do Exercise 71.
▲ 312 CHAPTER 4 \| Polynomial and Rational Functions 4. ▼ CONCE PTS 1. Only one of the following graphs could be the graph of a polynomial function. Which one? Why are the others not graphs of polynomials? I y II y x III y x I 8. (a) (cb) (d 64 5 16 2 9–14 ■ Match the polynomial function with one of the graphs I–VI. Give reasons for your choice 10. 9. Q P x x ✎ 11. R 13 12. 14. II x 5 5x 3 4x x 4 2x 2x 4 x 3 2x 2 y 1 0 1 x x III y IV y 2. Every polynomial has one of the following behaviors: and and y q as y q as y q y q x q x q as as x q x q y q x q as y q as y q as y q x q x q as x q and and (i) (ii) (iii) (iv) For each polynomial, choose the appropriate description of its end behavior from the list above. y x 3 8x 2 2x 15 y 2x4 12x 100: : end behavior end behavior (b) (a).. 3. If c is a zero of the polynomial P, which of the following state- c P ments must be true? 0. (a) (c) x c is a factor of (d) c is the y-intercept of the graph of P. (b. 4. Which of the following statements couldn’t possibly be true about the polynomial function P? (a) P has degree 3, two local maxima, and two local minima. (b) P has degree 3 and no local maxima or minima. (c) P has degree 4, one local maximum, and no local minima. ▼ SKI LLS 5–8 ■ Sketch the graph of each function by transforming the graph of an appropriate function of the form y x n from Figure 2. Indicate all x- and y-intercepts on each graph. ✎ 5. (a) (c) 6. (a) (c) 7. (a) (c 2x 2 2 x 4 16 16 Q S (b) (d) (b) (db) Q (d 27 VI y 1 0 1 x 15–
26 ■ Sketch the graph of the polynomial function. Make sure your graph shows all intercepts and exhibits the proper end behavior 3x 2 2 1 2 2 15. P 16. P ✎ 17. P 18. P 19. P 20. P 21. P 23. P ✎ 25 2x 12 22. P 24. P 26 27–40 ■ Factor the polynomial and use the factored form to ﬁnd the zeros. Then sketch the graph. x 3 2x 2 8x 2x 3 x 2 x x 5 9x 3 x 3 3x 2 4x 12 ✎ 27. P ✎ ✎ 29. P 31. P 33. P 35. P 36. P 37. P 38. P 39. P 40 6x x 3 x 2 12x x 4 3x 3 2x 2 28. P 30. P 32. P x 3 x 2 x 1 34 2x 3 x 2 18x 9 2x 4 3x 3 16x 24 1 8 1 2 2 x 4 2x 3 8x 16 x 4 2x 3 8x 16 x 4 3x 2 4 x 6 2x 3 1 41–46 ■ Determine the end behavior of P. Compare the graphs of P and Q on large and small viewing rectangles, as in Example 3(b). ✎ ✎ ✎ 41. P 42. P 43. P 44. P 45. P 46 3x 3 x 2 5x 1 12x; Q 3x 7x 2 5x 5; Q x x 4 x 5 2x 2 x; Q x x 11 9x 9; Q 1 2x 2 x 12 11 x 12 47–50 ■ The graph of a polynomial function is given. From the graph, ﬁnd (a) the x- and y-intercepts, and (b) the coordinates of all local extrema. 47. P x 1 2 x 2 4x 48 49 50 SE CTI ON 4. 2 \| Polynomial Functions and Their Graphs 313 51–58 ■ Graph the polynomial in the given viewing rectangle. Find the coordinates of all local extrema. State each answer correct to two decimal places. 51. y x 2 8x, 52. y x 3 3x 2, 4, 12 3 2, 5 3 4 by by 50, 30 3 4 10, 10 4 3 5, 5 4 30, 30 53. y x
3 12x 9, 4 54. y 2x 3 3x 2 12x 32, 3 4 by 60, 30 3 4 by 3 5, 5 3 30, 30 3 5, 5 by by 4 3 3, 3 3 3, 3 4 by 3 5, 10 4 by 3 5, 10 4 4 3 4 4 100, 100 4 55. y x 4 4x 3, 5, 5 3 4 56. y x 4 18x 2 32, 57. y 3x 5 5x 3 3, 58. y x 5 5x 2 6, 3 59–68 ■ Graph the polynomial and determine how many local maxima and minima it has. 59. y 2x 2 3x 5 60. y x 3 12x 61. y x 3 x 2 x 62. y 6x 3 3x 1 63. y x 4 5x 2 4 64. y 1.2x 5 3.75x 4 7x 3 15x 2 18x 65. y x 2 5 32 1 2 67. y x 8 3x 4 x 66. 68 17x 2 7 y 1 69–74 ■ Graph the family of polynomials in the same viewing rectangle, using the given values of c. Explain how changing the value of c affects the graph. 69. P 70. P ✎ 71. P 72. P 73. P 74 cx 3; c 1, 2, 5, 1 2 x c 4; c 1, 0, 1, 2 1 2 x 4 c; c 1, 0, 1, 2 x 3 cx; c 2, 0, 2, 4 x 4 cx; c 0, 1, 8, 27 x c; c 1, 3, 5, 7 75. (a) On the same coordinate axes, sketch graphs (as accurately as possible) of the functions y x3 2x2 x 2 and y x2 5x 2 (b) On the basis of your sketch in part (a), at how many points do the two graphs appear to intersect? (c) Find the coordinates of all intersection points. 76. Portions of the graphs of y x 2, y x 3, y x 4, y x 5, and y x 6 are plotted in the ﬁgures. Determine which function belongs to each graph 314 CHAPTER 4 \| Polynomial and Rational Functions 77. Recall that a function f is odd if f x 1 2 f x 1 2 or even if x
f 1 1 2 x for all real x. f 2 (a) Show that a polynomial P 1 ers of x is an odd function. P that contains only odd pow- x 2 that contains only even pow- (b) Show that a polynomial x 1 ers of x is an even function. (c) Show that if a polynomial P contains both odd and even powers of x, then it is neither an odd nor an even function. x 2 1 2 (d) Express the function x 5 6x 3 x 2 2x 5 P x 1 2 as the sum of an odd function and an even function. 78. (a) Graph the function x 4 2 ﬁnd all local extrema, correct to the nearest tenth and 2 (b) Graph the function and use your answers to part (a) to ﬁnd all local extrema, correct to the nearest tenth. 79. (a) Graph the function x 4 P 1 determine how many local extrema it has and 2 2 1 (b) If a b c, explain why the function x c x b x a P x 2 must have two local extrema. 2 1 1 1 2 1 2 80. (a) How many x-intercepts and how many local extrema does 1 2 x P have? the polynomial x 3 4x (b) How many x-intercepts and how many local extrema does x 3 4x (c) If a 0, how many x-intercepts and how many local exand the polynomial x 3 ax trema does each of the polynomials Q x 3 ax have? Explain your answer. have ▼ APPLICATIONS 81. Market Research A market analyst working for a smallappliance manufacturer ﬁnds that if the ﬁrm produces and sells x blenders annually, the total proﬁt (in dollars) is 8x 0.3x 2 0.0013x 3 372 P x 1 2 Graph the function P in an appropriate viewing rectangle and use the graph to answer the following questions. (a) When just a few blenders are manufactured, the ﬁrm loses 263.3, money (proﬁt is negative). (For example, so the ﬁrm loses $263.30 if it produces and sells only 10 blenders.) How many blenders must the �
��rm produce to break even? 10 P 1 2 (b) Does proﬁt increase indeﬁnitely as more blenders are pro- duced and sold? If not, what is the largest possible proﬁt the ﬁrm could have? 82. Population Change The rabbit population on a small island is observed to be given by the function 120t 0.4t 4 1000 P t 1 2 where t is the time (in months) since observations of the island began. (a) When is the maximum population attained, and what is that maximum population? (b) When does the rabbit population disappear from the island? P 0 t 83. Volume of a Box An open box is to be constructed from a piece of cardboard 20 cm by 40 cm by cutting squares of side length x from each corner and folding up the sides, as shown in the ﬁgure. (a) Express the volume V of the box as a function of x. (b) What is the domain of V? (Use the fact that length and vol- ume must be positive.) (c) Draw a graph of the function V, and use it to estimate the maximum volume for such a box. 40 cm x 20 cm x 84. Volume of a Box A cardboard box has a square base, with each edge of the base having length x inches, as shown in the ﬁgure. The total length of all 12 edges of the box is 144 in. (a) Show that the volume of the box is given by the function V x 1 2 2x 2 18 x 1. 2 (b) What is the domain of V? (Use the fact that length and volume must be positive.) (c) Draw a graph of the function V and use it to estimate the maximum volume for such a box. x x ▼ DISCOVE RY • DISCUSSION • WRITI NG 85. Graphs of Large Powers Graph the functions y x 2, y x 3, y x 4, and y x 5, for 1 x 1, on the same coordinate axes. What do you think the graph of y x100 would look like on this same interval? What about y x101? Make a table of values to conﬁrm your answers. 86. Maximum Number of Local Extrema What is the smallest possible degree that the polynomial whose graph is shown can have? Explain. y 0 x SE
CTI O N 4.3 \| Dividing Polynomials 315 87. Possible Number of Local Extrema Is it possible for a third-degree polynomial to have exactly one local extremum? Can a fourth-degree polynomial have exactly two local extrema? How many local extrema can polynomials of third, fourth, ﬁfth, and sixth degree have? (Think about the end behavior of such polynomials.) Now give an example of a polynomial that has six local extrema. 88. Impossible Situation? Is it possible for a polynomial to have two local maxima and no local minimum? Explain. 4.3 Dividing Polynomials LEARNING OBJECTIVES After completing this section, you will be able to: ■ Use long division to divide polynomials ■ Use synthetic division to divide polynomials ■ Use the Remainder Theorem to ﬁnd values of a polynomial ■ Use the Factor Theorem to factor a polynomial ■ Find a polynomial with speciﬁed zeros So far in this chapter we have been studying polynomial functions graphically. In this section we begin to study polynomials algebraically. Most of our work will be concerned with factoring polynomials, and to factor, we need to know how to divide polynomials. ■ Long Division of Polynomials Dividing polynomials is much like the familiar process of dividing numbers. When we divide 38 by 7, the quotient is 5 and the remainder is 3. We write Dividend 38 7 5 3 7 Quotient Divisor Remainder To divide polynomials, we use long division, as follows. DIVISION ALGORITHM To write the division algorithm another way, divide through by and If 2 1 Q polynomials x D degree of are polynomials, with 2 x R x and 1 1, such that, where is either 0 or of degree less than the, then there exist unique Remainder Dividend Divisor Quotient The polynomials x Q tively, 2 2 is the quotient, and 1 1 and D x P x are called the dividend and divisor, respecR is the remainder. x 1 2 1 2 316 CHAPTER 4 \| Polynomial and Rational Functions E X AM P L E 1 \| Long
Division of Polynomials Divide 6x 2 26x 12 by x 4. ▼ SO LUTI O N arranging them as follows: The dividend is 6x 2 26x 12 and the divisor is x 4. We begin by x 46x 2 26x 12 Next we divide the leading term in the dividend by the leading term in the divisor to get the ﬁrst term of the quotient: 6x 2/x 6x. Then we multiply the divisor by 6x and subtract the result from the dividend. 6x x 46x 2 26x 12 6x 2 24x 2x 12 Divide leading terms: 6x 2 x 6x Multiply: x 4 Subtract and “bring down” 12 6x 2 24x 6x 2 1 We repeat the process using the last line 2x 12 as the dividend. 6x 2 2oo x 46x 2 26x 12 6x 2 24x 2x 12 2x 8 4 Divide leading terms: 2x x 2 Multiply: Subtract 2 1 x 4 2 2x 8 The division process ends when the last line is of lesser degree than the divisor. The last line then contains the remainder, and the top line contains the quotient. The result of the division can be interpreted in either of two ways. Dividend 6x 2 26x 12 x 4 Divisor Quotient 6x 2 4 x 4 Remainder or 6x 2 26x 12 x 4 1 2 1 6x 2 2 4 Remainder Dividend Divisor Quotient ✎ Practice what you’ve learned: Do Exercise 3. ▲ E X AM P L E 2 \| Long Division of Polynomials Let x R 1 2 x P 1 such that 8x 4 6x 2 3x and 1 R x 2x 2 x 2. Find polynomials Q and ▼ SO LUTI O N We use long division after ﬁrst inserting the term 0x 3 into the dividend to ensure that the columns line up correctly. 4x 2 2x 2x 2 x 28x 4 0x 3 6x 2 3x 1 8x 4 4x 3 8x 2 4x 3 2x 2 3x 4x 3 2x 2 4x Multiply divisor by 4x 2 Subtract Multiply divisor by 2x 7x 1 Subtract SE CTI O
N 4.3 \| Dividing Polynomials 317 The process is complete at this point because 7x 1 is of lesser degree than the divisor 2x 2 x 2. From the above long division we see that and R 4x 2 2x 7x 1, so Q x x 1 2 1 2 8x 4 6x 2 3x 1 2x 2 x 2 4x 2 2x 1 2 7x 1 2 2 1 1 ✎ Practice what you’ve learned: Do Exercise 19. ▲ ■ Synthetic Division Synthetic division is a quick method of dividing polynomials; it can be used when the divisor is of the form x c. In synthetic division we write only the essential parts of the long division. Compare the following long and synthetic divisions, in which we divide 2x 3 7x 2 5 by x 3. (We’ll explain how to perform the synthetic division in Example 3.) Long Division 2x 2 x 3 x 32x 3 7x 2 0x 5 Quotient 2x 3 6x 2 x 2 0x x 2 3x 3x 5 3x 9 4 Remainder Synthetic Division 144424443 Quotient Remainder Note that in synthetic division we abbreviate 2x 3 7x 2 5 by writing only the coefﬁcients: 2 7 0 5, and instead of x 3, we simply write 3. (Writing 3 instead of 3 allows us to add instead of subtract, but this changes the sign of all the numbers that appear in the gold boxes.) The next example shows how synthetic division is performed. E X AM P L E 3 \| Synthetic Division Use synthetic division to divide 2x 3 7x 2 5 by x 3. ▼ SO LUTI O N We begin by writing the appropriate coefﬁcients to represent the divisor and the dividend. Divisor x – 3 2 3 7 0 5 Dividend 2x 3 – 7x 2 + 0x + 5 We bring down the 2, multiply 3 2 6, and write the result in the middle row. Then we add. 3 2 -7 0 5 6 2 -1 Multiply: 3 · 2 = 6 Add: –7 + 6 = –1 We repeat this process of multiplying and then adding until the table is complete. 5 3 2 2 −7 6 −1 0 −3 −3 Multiply: 3(–1) = –3 Add:
0 + (–3) = –3 318 CHAPTER 4 \| Polynomial and Rational Functions 3 2 2 −7 6 −1 0 −3 −3 5 −9 −4 Quotient 2x2 – x – 3 Remainder –4 Multiply: 3(–3) = –9 Add: 5 + (–9) = – 4 From the last line of the synthetic division we see that the quotient is 2x 2 x 3 and the remainder is 4. Thus 2x 3 7x 2 5 ✎ Practice what you’ve learned: Do Exercise 31. x 3 2 1 2x 2 x 3 1 4 2 ▲ ■ The Remainder and Factor Theorems The next theorem shows how synthetic division can be used to evaluate polynomials easily. REMAINDER THEOREM If the polynomial P x 1 2 is divided by x c, then the remainder is the value P c. 2 1 If the divisor in the Division Algorithm is of the form x c for some real ▼ P RO O F number c, then the remainder must be a constant (since the degree of the remainder is less than the degree of the divisor). If we call this constant r, then 1 Replacing x by c in this equation, we get P is the remainder r, that is, ▲ E X AM P L E 4 \| Using the Remainder Theorem to Find the Value of a Polynomial 2 1 x P 3x 5 5x 4 4x 3 7x 3. Let (a) Find the quotient and remainder when P (b) Use the Remainder Theorem to ﬁnd is divided by x 2. x P 1 2 2 2. 1 ▼ SO LUTI O N (a) Since x 2 x following form. 2 1 2, the synthetic division for this problem takes the, Remainder is 5, so P(–2) = 5. (b) By the Remainder Theorem, The quotient is 3x 4 x 3 2x 2 4x 1, and the remainder is 5. 2 1 x 2. From part (a) the remainder is 5, so ✎ Practice what you’ve learned: Do Exercise 39. is the remainder when is divided by. ▲ SE CTI O N 4.3 \| Dividing Polynomials 319 The next theorem says that zeros of polynomials correspond to factors
; we used this fact in Section 4.2 to graph polynomials. FACTOR THEOREM c is a zero of P if and only if x c is a factor of P x. 2 1 ▼ P RO O F If P x 1 2 factors as, then #, then by the Remainder Theorem # Conversely, if P c 1 2 1 so x c is a factor of 1x 3 0x 2 7x 6 x 3 x 2 x 2 7x x 2 x 6x 6 6x 6 0 E X AM P L E 5 \| Factoring a Polynomial Using the Factor Theorem Let P x x 3 7x 6. Show that P 1 1 2 ▼ SO LUTI O N Substituting, we see that Theorem this means that x 1 is a factor of P in the margin), we see that 2 1 0 1 2 P, and use this fact to factor 1 13 7 # 1 6 0. By the Factor 1 2. Using synthetic or long division (shown completely 7x See margin Factor quadratic x2 + x – 6 2 ✎ Practice what you’ve learned: Do Exercises 53 and 57. ▲ E X AM P L E 6 \| Finding a Polynomial with Speciﬁed Zeros Find a polynomial of degree 4 that has zeros 3, 0, 1, and 5. y 10 _3 0 1 5 x ▼ SO LUTI O N By the Factor Theorem factors of the desired polynomial, so let x4 3x3 13x2 15x, x 1, and x 5 must all be 1 x P Since lem must be a constant multiple of change the degree. 2 1 2 is of degree 4, it is a solution of the problem. Any other solution of the prob, since only multiplication by a constant does not P x FIGURE 1 x 1 x 3 x) P 1 1 zeros 3, 0, 1, and 5. x 1 2 ✎ Practice what you’ve learned: Do Exercise 59. ▲ x 5 has 2 21 The polynomial P of Example 6 is graphed in Figure 1. Note that the zeros of P corre- spond to the x-intercepts of the graph. 320 CHAPTER 4 \| Polynomial and Rational Functions 4. ▼ CONCE PTS 1. If we divide the polynomial P by the factor the equation and
we obtain x c then we say that is the divisor, 1 Q 2 x 1 is the 2 1 2 2 1, and R x is the 1 2. (a) If we divide the polynomial x c we obtain a remainder of 0, then we know that c is a 2 by the factor P x 1 2 2 1. and of P. (b) If we divide the polynomial x c and we obtain a remainder of k, then we know that by the factor P x 2 1 P c 1 2. ▼ SKI LLS 3–8 ■ Two polynomials P and D are given. Use either synthetic or x D long division to divide x P, and express P in the form P R by. 3x 2 5x 4, 2 1 x 3 4x 2 6x 1, 2 2x 3 3x 2 2x, 2x 3 2 2x 1 x D 1 4x 3 7x 9, 1 x 4 x 3 4x 2, D x 2 D 2 2x 5 4x 4 4x 3 x 3 1 x 2 1 3. P 4. P 5. P 6. P 7. P 8–14 ■ Two polynomials P and D are given. Use either synthetic or long division to divide, and express the quotient by P in the form D / 4x 8, x 3 6x 2x 1 4x 2 3x 7, 2 1 6x 3 x 2 12x 5, 2x 4 x 3 9x 2, 2 x 5 x 4 2x 3 x 1, 1 D x 9. P 10. P 11. P 12. P 13. P 14 3x 15–24 ■ Find the quotient and remainder using long division. 15. 17. ✎ 19. 21. x 2 6x 8 x 4 4x 3 2x 2 2x 3 2x 1 x 3 6x 3 x 2 2x 2 6x 3 2x 2 22x 2x 2 5 16. 18. 20. 22. x 3 x 2 2x 6 x 2 x 3 3x 2 4x 3 3x 6 3x 4 5x 3 20x 5 x 2 x 3 9x 2 x 5 3x 2 7x ✎ 23 24. 2x 5 7x 4 13 4x 2 6x 8 25–38 ■ Find the quotient and remainder using synthetic division. 25. 27. 29. ✎ 31. 33.
35. 36. 37. x 2 5x 4 x 3 3x 2 5x x 6 x 3 2x 2 2x 1 x 2 x 3 8x 2 x 3 x 5 3x 3 6 x 1 26. 28. 30. 32. 34. x 2 5x 4 x 1 4x 2 3 x 5 3x 3 12x 2 9x 9x 2 27x 27 x 3 2x 3 3x 2 2x 1 x 1 2 6x 4 10x 3 5x 2 x 1 x 2 3 x 3 27 x 3 38. x 4 16 x 2 39–51 ■ Use synthetic division and the Remainder Theorem to evaluate 4x 2 12x 5, 2x 2 9x 1, x 3 3x 2 7x 2x 2 7 c 2, 2x 3 21x 2 9x 200, 5x 4 30x 3 40x 2 36x 14, c 2 6x 5 10x 3 x 1, c 3 x 7 3x 2 1 2x 6 7x 5 40x 4 7x 2 10x 112, 3x 3 4x 2 2x 2x 2 3x 8, c 0.1 c 11 c 2 3, c 7 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50 51. P x 1 52. Let 6x 7 40x 6 16x 5 200x 4 60x 3 69x 2 13x 139 Calculate tuting x 7 into the polynomial and evaluating directly. by (a) using synthetic division and (b) substi- P 7 2 1 53–56 ■ Use the Factor Theorem to show that x c is a factor of P x 2 for the given value(s) of c. x 3 3x 2 3x 1, x P 1 53. c 1 1 2 54. 55. 56 2x 2 3x 10 c 2, c 1 2x 3 7x 2 6x 5, 2 x 4 3x 3 16x 2 27x 63, c 3, 3 57–58 ■ Show that the given value(s) of c are zeros of. ﬁnd all other zeros of x P, and 11x 15 c 3, 3x 4 x 3 21x 2 11x 6, ✎ 57. 58, 2 59–62 ■ Find a polynomial of the speciﬁed degree that has the given zeros. ✎
59. Degree 3; 60. Degree 4; 61. Degree 4; zeros 1, 1, 3 zeros 2, 0, 2, 4 zeros 1, 1, 3, 5 zeros 2, 1, 0, 1, 2 62. Degree 5; 63. Find a polynomial of degree 3 that has zeros 1, 2, and 3 and in which the coefﬁcient of x 2 is 3. 64. Find a polynomial of degree 4 that has integer coefﬁcients and zeros 1, 1, 2, and. 1 2 SE CTI ON 4.4 \| Real Zeros of Polynomials 321 67. Degree 4 68. Degree ▼ DISCOVE RY • DISCUSSION • WRITI NG 69. Impossible Division? Suppose you were asked to solve the following two problems on a test: A. Find the remainder when 6x1000 17x 562 12x 26 is divided by x 1. Is x 1 a factor of x 567 3x 400 x 9 2? B. Obviously, it’s impossible to solve these problems by dividing, because the polynomials are of such large degree. Use one or more of the theorems in this section to solve these problems without actually dividing. 65–68 ■ Find the polynomial of the speciﬁed degree whose graph is shown. 70. Nested Form of a Polynomial Expand Q to prove that the polynomials P and Q are the same. 65. Degree 3 66. Degree 3x 4 5x 3 x 2 3x 5 3x and 11 1 2 2 in your head, using the 2 2 Try to evaluate 2 1 forms given. Which is easier? Now write the polynomial R like the polynomial Q. Use the nested form to ﬁnd your head. x 5 2x 4 3x 3 2x 2 3x 4 in “nested” form, R in 3 x 1 2 2 1 2 2 Do you see how calculating with the nested form follows the same arithmetic steps as calculating the value of a polynomial using synthetic division? 4.4 Real Zeros of Polynomials LEARNING OBJECTIVES After completing this section, you will be able to: ■ Use the Rational Zeros Theorem to ﬁnd the rational zeros of a polynomial ■ Use Descartes’ Rule of Signs to determine
the number of positive and negative zeros of a polynomial ■ Use the Upper and Lower Bounds Theorem to ﬁnd upper and lower bounds for the zeros of a polynomial ■ Use algebra and graphing devices to solve polynomial equations The Factor Theorem tells us that ﬁnding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we study some algebraic methods that help us to ﬁnd the real zeros of a polynomial and thereby factor the polynomial. We begin with the rational zeros of a polynomial. 322 CHAPTER 4 \| Polynomial and Rational Functions ■ Rational Zeros of Polynomials To help us understand the next theorem, let’s consider the polynomial x3 x2 14x 24 2 1 2 Factored form Expanded form From the factored form we see that the zeros of P are 2, 3, and 4. When the polynomial is expanded, the constant 24 is obtained by multiplying. This means that the zeros of the polynomial are all factors of the constant term. The following generalizes this observation. 4 2 3 1 2 2 1 RATIONAL ZEROS THEOREM an x If the polynomial coefﬁcients, then every rational zero of P is of the form n an1x n1... a1x a0 P x 2 1 has integer p q where and p is a factor of the constant coefﬁcient a0 q is a factor of the leading coefﬁcient an. ▼ P RO O F If p/q is a rational zero, in lowest terms, of the polynomial P, then we have an a n1 n p q b an1 a... a1 a p q b an pn an1 pn1q... a1pqn1 a0qn 0 an pn1 an1 pn2q... a1qn1 p q b a0 0 p 1 a0qn 2 Multiply by qn Subtract a0qn and factor LHS Now p is a factor of the left side, so it must be a factor of the right side as well. Since p/q is in lowest terms, p and q have no factor in common, and so p must be a factor of

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