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a0. A ▲ similar proof shows that q is a factor of an. We see from the Rational Zeros Theorem that if the leading coefﬁcient is 1 or 1, then the rational zeros must be factors of the constant term. E X AM P L E 1 \| Using the Rational Zeros Theorem Find the rational zeros of P x x3 3x 2. 1 2 ▼ SO LUTI O N Since the leading coefﬁcient is 1, any rational zero must be a divisor of the constant term 2. So the possible rational zeros are 1 and 2. We test each of these possibilities The rational zeros of P are 1 and 2. ✎ Practice what you’ve learned: Do Exercise 15. 1 2 2 2 1 ▲ SE CTI ON 4.4 \| Real Zeros of Polynomials 323 The following box explains how we use the Rational Zeros Theorem with synthetic di- vision to factor a polynomial. FINDING THE RATIONAL ZEROS OF A POLYNOMIAL 1. List Possible Zeros. List all possible rational zeros, using the Rational Zeros Theorem. 2. Divide. Use synthetic division to evaluate the polynomial at each of the candidates for the rational zeros that you found in Step 1. When the remainder is 0, note the quotient you have obtained. 3. Repeat. Repeat Steps 1 and 2 for the quotient. Stop when you reach a quo- tient that is quadratic or factors easily, and use the quadratic formula or factor to ﬁnd the remaining zeros. E X AM P L E 2 \| Finding Rational Zeros Factor the polynomial P x 2x 3 x 2 13x 6, and ﬁnd all its zeros. 1 2 ▼ SO LUTI O N By the Rational Zeros Theorem the rational zeros of P are of the form possible rational zero of P factor of constant term factor of leading coefﬁcient The constant term is 6 and the leading coefﬁcient is 2, so possible rational zero of P factor of 6 factor of 2 The factors of 6 are 1, 2, 3, 6 and the factors of 2 are 1, 2. Thus, the possible rational zeros of P are Simplifying the fractions and eliminating duplicates, we get the following list of possible rational zeros: 1, 2
, 3, 6, 1 2, 3 2 Evariste Galois (1811–1832) is one of the very few mathematicians to have an entire theory named in his honor. Not yet 21 when he died, he completely settled the central problem in the theory of equations by describing a criterion that reveals whether a polynomial equation can be solved by algebraic operations. Galois was one of the greatest mathematicians in the world at that time, although no one knew it but him. He repeatedly sent his work to the eminent mathemati cians Cauchy and Poisson, who either lost his letters or did not understand his ideas. Galois wrote in a terse style and included few details, which probably played a role in his failure to pass the entrance exams at the Ecole Polytechnique in Paris. A political radical, Galois spent several months in prison for his revolutionary activities. His brief life came to a tragic end when he was killed in a duel over a love affair. The night before his duel, fearing that he would die, Galois wrote down the essence of his ideas and entrusted them to his friend Auguste Chevalier. He concluded by writing “there will, I hope, be people who will ﬁnd it to their advantage to decipher all this mess.” The mathematician Camille Jordan did just that, 14 years later. 324 CHAPTER 4 \| Polynomial and Rational Functions To check which of these possible zeros actually are zeros, we need to evaluate P at each of these numbers. An efﬁcient way to do this is to use synthetic division. 1 Test whether 1 is a zero 2 11 13 16 3 10 2 3 10 4 2 Test whether 2 is a zero 2 2 11 13 6 4 10 6 2 5 3 0 Remainder is not 0, so 1 is not a zero. Remainder is 0, so 2 is a zero. From the last synthetic division we see that 2 is a zero of P and that P factors as P x 1 2 2x 3 x 2 13x 6 2x 2 5x 3 2 x 3 2x From synthetic division Factor 2x2 + 5x – 3 2 From the factored form we see that the zeros of P are 2, ✎ Practice what you’ve learned: Do Exercise 27. 1 2, and –3. ▲ E X AM P L E 3 \| Using the Rational Zeros Theorem and the Quadratic Formula
x 4 5x 3 5x 2 23x 10. 1 2 x P Let (a) Find the zeros of P. (b) Sketch the graph of P. ▼ SO LUTI O N (a) The leading coefﬁcient of P is 1, so all the rational zeros are integers: They are divi- sors of the constant term 10. Thus, the possible candidates are 1, 2, 5, 10 Using synthetic division (see the margin), we ﬁnd that 1 and 2 are not zeros but that 5 is a zero and that P factors as x 4 5x 3 5x 2 23x 10 x 5 x 3 5x 2 1 2 1 2 We now try to factor the quotient x 3 5x 2. Its possible zeros are the divisors of 2, namely, 1, 2 Since we already know that 1 and 2 are not zeros of the original polynomial P, we don’t need to try them again. Checking the remaining candidates 1 and 2, we see that 2 is a zero (see the margin), and P factors as x 4 5x 3 5x 2 23x 10 x 3 5x 2 x 2 2 x 2 2x Now we use the quadratic formula to obtain the two remaining zeros of P 12 The zeros of P are 5, 2, 1 12, and 1 12. 1 2 5 1 5 5 23 10 1 4 9 14 14 1 4 9 24 1 5 5 23 2 6 22 1 1 3 11 10 2 12 1 5 5 5 0 5 2 10 23 0 25 10 50 (b) Now that we know the zeros of P, we can use the methods of Section 4.2 to sketch SE CTI ON 4.4 \| Real Zeros of Polynomials 325 _3 6 _50 FIGURE 1 P x 1 2 x 4 5x 3 5x 2 23x 10 Polynomial x 2 4x 1 2x 3 x 6 x 4 3x 2 x 4 Variations in sign 0 1 2 the graph. If we want to use a graphing calculator instead, knowing the zeros allows us to choose an appropriate viewing rectangle—one that is wide enough to contain all the x-intercepts of P. Numerical approximations to the zeros of P are So in this case we choose the rectangle shown in Figure 1. 3 5, 2, 2.4, and 0.4
50, 50 3, 6 by 4 3 4 and draw the graph ✎ Practice what you’ve learned: Do Exercises 45 and 55. ▲ ■ Descartes’ Rule of Signs and Upper and Lower Bounds for Roots In some cases, the following rule—discovered by the French philosopher and mathematician René Descartes around 1637 (see page 245)—is helpful in eliminating candidates from lengthy lists of possible rational roots. To describe this rule, we need the concept of variation in sign. If is a polynomial with real coefﬁcients, written with descending powers of x (and omitting powers with coefﬁcient 0), then a variation in sign occurs whenever adjacent coefﬁcients have opposite signs. For example, 5x 7 3x 5 x 4 2x has three variations in sign. 2 DESCARTES’ RULE OF SIGNS Let P be a polynomial with real coefﬁcients. 1. The number of positive real zeros of P tions in sign in 1 2. The number of negative real zeros of P ations in sign in or is less than that by an even whole number. either is equal to the number of varia- or is less than that by an even whole number. either is equal to the number of vari- E X AM P L E 4 \| Using Descartes’ Rule Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros of the polynomial 3x 6 4x 5 3x 3 x 3 P x 1 2 ▼ SO LUTI O N The polynomial has one variation in sign, so it has one positive zero. Now 3x 6 4x 5 3x has three variations in sign. Thus, P So making a total of either two or four real zeros. ✎ Practice what you’ve learned: Do Exercise 65. x 1 2 has either three or one negative zero(s), ▲ We say that a is a lower bound and b is an upper bound for the zeros of a polynomial if every real zero c of the polynomial satisﬁes a c b. The next theorem helps us to ﬁnd such bounds for the zeros of a polynomial. 326 CHAPTER 4 \| Polynomial and Rational Functions THE UPPER AND LOWER BOUNDS THEOREM
Let P be a polynomial with real coefﬁcients. 1. If we divide 2 1 x P by x b (with b 0) using synthetic division and if the row that contains the quotient and remainder has no negative entry, then b is an upper bound for the real zeros of P. x by x a (with a 0) using synthetic division and if the row P 2. If we divide that contains the quotient and remainder has entries that are alternately nonpositive and nonnegative, then a is a lower bound for the real zeros of P. 1 2 A proof of this theorem is suggested in Exercise 95. The phrase “alternately nonpositive and nonnegative” simply means that the signs of the numbers alternate, with 0 considered to be positive or negative as required. E X AM P L E 5 \| Upper and Lower Bounds for Zeros of a Polynomial Show that all the real zeros of the polynomial and 2. P x 1 2 x 4 3x 2 2x 5 lie between 3 ▼ SO LUTI O N We divide by x 2 and x 3 using synthetic division All entries positive 3 9 18 1 3 6 16 48 43 Entries alternate in sign. By the Upper and Lower Bounds Theorem, 3 is a lower bound and 2 is an upper bound for the zeros. Since neither 3 nor 2 is a zero (the remainders are not 0 in the division table), all the real zeros lie between these numbers. ✎ Practice what you’ve learned: Do Exercise 69. ▲ E X AM P L E 6 \| Factoring a Fifth-Degree Polynomial Factor completely the polynomial P x 2x 5 5x 4 8x 3 14x 2 6x 9 2, 1, ▼ SO LUTI O N The possible rational zeros of P are check the positive candidates ﬁrst, beginning with the smallest. 1 2 1 3 2, 3, 9 2, and 9. We 1 2 2 5 8 14 5 2 1 3 2 6 5 33 2 6 33 4 9 4 9 9 8 63 8 1 2 5 8 14 6 9 2 7 1 15 9 2 7 1 15 9 0 P(1) = 0 is not a 1 2 zero So 1 is a zero, and. We continue by factoring the quotient. We still have the same list of possible zeros except that has been eliminated. 2x
4 7x 3 x 2 15x 15 9 8 7 9 8 7 16 2 9 2 SE CTI ON 4.4 \| Real Zeros of Polynomials 327 3 2 2 7 1 15 9 9 15 21 3 1 is not a zero. 2 10 14 6 0 3 2B 0 P A, all entries nonnegative 3 2 is both a zero and an upper bound for the zeros of, so we do not need We see that to check any further for positive zeros, because all the remaining candidates are greater 3 than. 2x 3 2 1 2 1 2x 3 10x 2 14x 6 2 From synthetic division x 3 5x 2 7x 3 2 2 1 Factor 2 from last factor, multiply into second factor By Descartes’ Rule of Signs, x3 5x 2 7x 3 has no positive zero, so its only possible rational zeros are 1 and 3(–1) = 0 2 Therefore 2x 3 x 1 2 1 2 1 2x 4x 3 2 x 3 2 From synthetic division Factor quadratic, 1, and 3. The graph of the polynomial is shown 3 2 This means that the zeros of P are 1, in Figure 2. ✎ Practice what you’ve learned: Do Exercise 77. ▲ 40 9 _20 _4 FIGURE 2 P x 1 2 2x 5 5x 4 8x 3 2x 3 1 x 1 2 1 2 1 14x 2 6x 9 x 3 2 x 1 2 1 2 ■ Using Algebra and Graphing Devices to Solve Polynomial Equations In Section 2.3 we used graphing devices to solve equations graphically. We can now use the algebraic techniques that we’ve learned to select an appropriate viewing rectangle when solving a polynomial equation graphically. E X AM P L E 7 \| Solving a Fourth-Degree Equation Graphically Find all real solutions of the following equation, correct to the nearest tenth. 3x 4 4x 3 7x 2 2x 3 0 ▼ SO LUTI O N To solve the equation graphically, we graph 3x 4 4x 3 7x 2 2x 3 P x 1 2 We use the Upper and Lower Bounds Theorem to see where the solutions can be found. First we use the Upper and Lower Bounds Theorem to ﬁnd two numbers between which all the solutions must lie. This allows us to choose a viewing rectangle that
is certain to contain all the x-intercepts of P. We use synthetic division and proceed by trial and error. 328 CHAPTER 4 \| Polynomial and Rational Functions To ﬁnd an upper bound, we try the whole numbers, 1, 2, 3,..., as potential candidates. We see that 2 is an upper bound for the solutions. 3 7 2 2 4 6 10 3 20 13 26 24 3 48 45 All positive 20 Now we look for a lower bound, trying the numbers 1, 2, and 3 as potential can- didates. We see that 3 is a lower bound for the solutions. _3 2 _20 FIGURE 3 y 3x 4 4x 3 7x 2 2x 3 Volume of a cylinder: V pr 2h Volume of a sphere: V 4 3 pr 3 150 0 50 FIGURE 5 y 4 3 px 3 4px 2 3 and y 100 3 3 4 7 2 15 24 8 26 3 78 75 Entries alternate in sign. 9 3 5 4 Thus, all the solutions lie between 3 and 2. So the viewing rectangle by 20, 20 contains all the x-intercepts of P. The graph in Figure 3 has two x-intercepts, one 3 between 3 and 2 and the other between 1 and 2. Zooming in, we ﬁnd that the solutions of the equation, to the nearest tenth, are 2.3 and 1.3. ✎ Practice what you’ve learned: Do Exercise 91. 3, 2 ▲ 3 4 E X AM P L E 8 \| Determining the Size of a Fuel Tank A fuel tank consists of a cylindrical center section that is 4 ft long and two hemispherical end sections, as shown in Figure 4. If the tank has a volume of 100 ft3, what is the radius r shown in the ﬁgure, correct to the nearest hundredth of a foot? 4 ft r FIGURE 4 ▼ SO LUTI O N Using the volume formula listed on the inside back cover of this book, we see that the volume of the cylindrical section of the tank is p # r 2 # 4 The two hemispherical parts together form a complete sphere whose volume is 3 4 3 pr Because the total volume of the tank is 100 ft3, we get the following equation: 3 pr 3 4pr 2 100 4 2 3 px 3 4px A negative solution for r would be meaningless
in this physical situation, and by substitution we can verify that r 3 leads to a tank that is over 226 ft3 in volume, much larger than the required 100 ft3. Thus, we know the correct radius lies somewhere between 0 to graph the function and 3 ft, so we use a viewing rectangle of by y 4, as shown in Figure 5. Since we want the value of this function to be 100, we also graph the horizontal line y 100 in the same viewing rectangle. The correct radius will be the x-coordinate of the point of intersection of the curve and the line. Using the cursor and zooming in, we see that at the point of intersection x 2.15, correct to two decimal places. Thus, the tank has a radius of about 2.15 ft. ✎ Practice what you’ve learned: Do Exercise 97. 50, 150 0, 3 ▲ 4 3 3 4 SE CTI ON 4.4 \| Real Zeros of Polynomials 329 Note that we also could have solved the equation in Example 8 by ﬁrst writing it as and then ﬁnding the x-intercept of the function 3 px 3 4px 2 100. 4 3 pr 3 4pr 2 100 0 y 4 4. ▼ CONCE PTS 1. If the polynomial function x P anxn an1xn1 p a1x a0 has integer coefﬁcients, then the only numbers that could possibly be rational zeros of P are all of the 1 2 form where p is a factor of, p q and q is a factor of. The possible rational zeros of 6x 3 5x 2 19x 10 are P x 1 2. 2. Using Descartes’ Rule of Signs, we can tell that the polynomial x 5 3x 4 2x 3 x 2 8x 8 has, P x 1 2, or positive real zeros and negative real zeros. 3. True or false? If c is a real zero of the polynomial P, then all the other zeros of P are zeros of. True or false? If a is an upper bound for the real zeros of the polynomial P, then a is necessarily a lower bound for the real zeros of P. ▼ SKI LLS 5–10 ■ List all possible rational zeros given by the Rational Zeros Theorem (but don’t check to see which actually are
zeros). 5. 6. P 1 Q x 2 x 7. 9. 10. x 3 4x 2 3 x 4 3x 3 6x 8 2x 5 3x 3 4x 2 8 6x 4 x 2 2x 12 4x 4 2x 2 7 12x 5 6x 3 2x 8 11–14 ■ A polynomial function P and its graph are given. (a) List all possible rational zeros of P given by the Rational Zeros Theorem. (b) From the graph, determine which of the possible rational zeros actually turn out to be zeros. 3 x 2 5x 1 5x 11 12. P x 1 2 3x 3 4x 2 x 2 y 1 0 1 x 13. P x 1 2 2x 4 9x 3 9x 2 x 3 y 1 0 1 x 14. P x 1 2 4x 4 x 3 4x 1 y 1 0 1 x 15–44 ■ Find all rational zeros of the polynomial, and write the polynomial in factored form. ✎ 15. 16. 17. 18. 19. 20. 21. 22. P P P P P P P P 23. P x 3 3x 2 4 x 3 7x 2 14x 8 x 3 3x 2 x 3 4x 2 3x 18 x 3 6x 2 12x 8 x 3 x 2 8x 12 x 3 4x 2 x 6 x 3 4x 2 7x 10 x 3 3x ✎ ✎ ✎ 330 CHAPTER 4 \| Polynomial and Rational Functions 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44 x3 4x2 11x 30 x 4 5x 2 4 x 4 2x 3 3x 2 8x 4 x 4 6x 3 7x 2 6x 8 x 4 x 3 23x 2 3x 90 4x 4 25x 2 36 2x4 x3 19x2 9x 9 3x 4 10x 3 9x 2 40x 12 2x 3 7x 2 4x 4 4x 3 4x 2 x 1 2x 3 3x 2 2x 3 4x 3 7x 3 8x 3 10x 2 x 3 4x 3 8x 2 11x 15 6x 3 11x 2 3x 2 2x 4 7x 3 3x
2 8x 4 6x 4 7x 3 12x 2 3x 2 x 5 3x 4 9x 3 31x 2 36 x 5 4x 4 3x 3 22x 2 4x 24 3x 5 14x 4 14x 3 36x 2 43x 10 2x 6 3x 5 13x 4 29x 3 27x 2 32x 12 45–54 ■ Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example 3(a). 45. 46. 47. 48. 49. 50. 51. 52. 53. 54 4x 2 3x 2 x 3 5x 2 2x 12 x 4 6x 3 4x 2 15x 4 x 4 2x 3 2x 2 3x 2 x 4 7x 3 14x 2 3x 9 x 5 4x 4 x 3 10x 2 2x 4 4x 3 6x 2 1 3x 3 5x 2 8x 2 2x 4 15x 3 17x 2 3x 1 4x 5 18x 4 6x 3 91x 2 60x 9 55–62 ■ A polynomial P is given. (a) Find all the real zeros of P. (b) Sketch the graph of P. 55. 56. 57. 58. 59. 60. 61. P P P P P P P 62 3x 2 4x 12 x 3 2x 2 5x 6 2x 3 7x 2 4x 4 3x 3 17x 2 21x 9 x 4 5x 3 6x 2 4x 8 x 4 10x 2 8x 8 x 5 x 4 5x 3 x 2 8x 4 x 5 x 4 6x 3 14x 2 11x 3 63–68 ■ Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros the polynomial can have. Then determine the possible total number of real zeros. 63. 64. 65. 66. 67. 68 2x 3 x 2 4x 7 2x 6 5x 4 x 3 5x 1 x 4 x 3 x 2 x 12 x 5 4x 3 x 2 6x 69. 69–72 ■ Show that the given values for a and b are lower and upper bounds for the real zeros of the polynomial. 2x 3 5x 2 x 2; a 3, b 1 x 4 2x 3 9x 2 2x 8; a 3, b
5 8x 3 10x 2 39x 9; a 3, b 2 3x 4 17x 3 24x 2 9x 1; a 0, b 6 72. 71. 70 ✎ ✎ 73–76 ■ Find integers that are upper and lower bounds for the real zeros of the polynomial. 73. 74. 75. 76 3x 2 4 2x 3 3x 2 8x 12 x 4 2x 3 x 2 9x 2 x 5 x 4 1 77–82 ■ Find all rational zeros of the polynomial, and then ﬁnd the irrational zeros, if any. Whenever appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes’ Rule of Signs, the quadratic formula, or other factoring techniques. ✎ 77. 78. 79. 80. 81. 82 2x 4 3x 3 4x 2 3x 2 2x 4 15x 3 31x 2 20x 4 4x 4 21x 2 5 6x 4 7x 3 8x 2 5x x 5 7x 4 9x 3 23x 2 50x 24 8x 5 14x 4 22x 3 57x 2 35x 6 83–86 ■ Show that the polynomial does not have any rational zeros. 83. 84. 85. 86 2x 4 x 3 x 2 3x 3 x 2 6x 12 x 50 5x 25 x 2 1 87–90 ■ The real solutions of the given equation are rational. List all possible rational roots using the Rational Zeros Theorem, and then graph the polynomial in the given viewing rectangle to determine which values are actually solutions. (All solutions can be seen in the given viewing rectangle.) 87. x 3 3x 2 4x 12 0; 88. x 4 5x 2 4 0; 89. 2x 4 5x 3 14x 2 5x 12 0; 3, 3 90. 3x 3 8x 2 5x 2 0; 4 2, 5 by 4 10, 10 by 3 30, 30 3 4, 4 3 40, 40 15, 15 4, 4 3 by by ✎ 91–94 ■ Use a graphing device to ﬁnd all real solutions of the equation, correct to two decimal places. 91. x 4 x 4 0 92. 2x 3 8x 2 9x 9 0 93. 4.00x 4 4.00x 3 10.96x
2 5.88x 9.09 0 94. x 5 2.00x 4 0.96x 3 5.00x 2 10.00x 4.80 0 95. Let P x be a polynomial with real coefﬁcients and let b 0. 1 2 Use the Division Algorithm to write # Suppose that r 0 and that all the coefﬁcients in nonnegative. Let z b. 0 (a) Show that. (b) Prove the ﬁrst part of the Upper and Lower Bounds Q P x z 2 1 2 1 2 are Theorem. (c) Use the ﬁrst part of the Upper and Lower Bounds Theorem satisﬁes x to prove the second part. [Hint: Show that if the second part of the theorem, then ﬁrst part.] x P 2 1 P 2 satisﬁes the 1 96. Show that the equation x 5 x 4 x 3 5x 2 12x 6 0 has exactly one rational root, and then prove that it must have either two or four irrational roots. ✎ ▼ APPLICATIONS 97. Volume of a Silo A grain silo consists of a cylindrical main section and a hemispherical roof. If the total volume of the silo (including the part inside the roof section) is 15,000 ft3 and the cylindrical part is 30 ft tall, what is the radius of the silo, correct to the nearest tenth of a foot? SE CTI ON 4.4 \| Real Zeros of Polynomials 331 99. Depth of Snowfall Snow began falling at noon on Sunday. The amount of snow on the ground at a certain location at time t was given by the function h t 1 2 11.60t 12.41t 2 6.20t 1.58t 4 0.20t 3 5 0.01t 6 1 2 t h where t is measured in days from the start of the snowfall is the depth of snow in inches. Draw a graph of and this function, and use your graph to answer the following questions. (a) What happened shortly after noon on Tuesday? (b) Was there ever more than 5 in. of snow on the ground? If so, on what day(s)? (c) On what day and at what time (to the nearest hour) did the snow disappear completely? 100. Volume of a
Box An open box with a volume of 1500 cm3 is to be constructed by taking a piece of cardboard 20 cm by 40 cm, cutting squares of side length x cm from each corner, and folding up the sides. Show that this can be done in two different ways, and ﬁnd the exact dimensions of the box in each case. 40 cm x 20 cm x 101. Volume of a Rocket A rocket consists of a right circular cylinder of height 20 m surmounted by a cone whose height and diameter are equal and whose radius is the same as that of the cylindrical section. What should this radius be (correct to two decimal places) if the total volume is to be 500p/3 m3? 30 ft 20 m 98. Dimensions of a Lot A rectangular parcel of land has an area of 5000 ft2. A diagonal between opposite corners is measured to be 10 ft longer than one side of the parcel. What are the dimensions of the land, correct to the nearest foot? x x+10 102. Volume of a Box A rectangular box with a volume of ft3 has a square base as shown below. The diagonal of 2 12 the box (between a pair of opposite corners) is 1 ft longer than each side of the base. (a) If the base has sides of length x feet, show that x 6 2x 5 x 4 8 0 332 CHAPTER 4 \| Polynomial and Rational Functions (b) Show that two different boxes satisfy the given conditions. Find the dimensions in each case, correct to the nearest hundredth of a foot. 105. The Depressed Cubic The most general cubic (third- degree) equation with rational coefﬁcients can be written as x 3 ax 2 bx c 0 (a) Show that if we replace x by X a /3 and simplify, we end up with an equation that doesn’t have an X 2 term, that is, an equation of the form X 3 pX q 0 This is called a depressed cubic, because we have “depressed” the quadratic term. (b) Use the procedure described in part (a) to depress the equation x 3 6x 2 9x 4 0. 106. The Cubic Formula The quadratic formula can be used to solve any quadratic (or second-degree) equation. You might have wondered whether similar formulas exist for cubic (third-degree), quartic (fourth-degree), and higher-degree equations.
For the depressed cubic x 3 px q 0, Cardano (page 340) found the following formula for one solution: x q 2 3 C q2 4 B p3 27 q 2 3 C q2 4 B p3 27 A formula for quartic equations was discovered by the Italian mathematician Ferrari in 1540. In 1824 the Norwegian mathematician Niels Henrik Abel proved that it is impossible to write a quintic formula, that is, a formula for ﬁfth-degree equations. Finally, Galois (page 323) gave a criterion for determining which equations can be solved by a formula involving radicals. Use the cubic formula to ﬁnd a solution for the following equations. Then solve the equations using the methods you learned in this section. Which method is easier? (a) x 3 3x 2 0 (b) x 3 27x 54 0 (c) x 3 3x 4 0 x x 103. Girth of a Box A box with a square base has length plus girth of 108 in. (Girth is the distance “around” the box.) What is the length of the box if its volume is 2200 in3? b l b ▼ DISCOVE RY • DISCUSSION • WRITI NG 104. How Many Real Zeros Can a Polynomial Have? Give examples of polynomials that have the following properties, or explain why it is impossible to ﬁnd such a polynomial. (a) A polynomial of degree 3 that has no real zeros (b) A polynomial of degree 4 that has no real zeros (c) A polynomial of degree 3 that has three real zeros, only one of which is rational (d) A polynomial of degree 4 that has four real zeros, none of which is rational What must be true about the degree of a polynomial with integer coefﬁcients if it has no real zeros? DISCOVERY PR OJECT ZEROING IN ON A ZERO We have seen how to ﬁnd the zeros of a polynomial algebraically and graphically. Let’s work through a numerical method for ﬁnding the zeros. With this method we can ﬁnd the value of any real zero to as many decimal places as we wish. a The Intermediate Value Theorem states: If P is a polynomial and if b 2
are of opposite sign, then P has a zero between a and b. (See page 305.) The Intermediate Value Theorem is an example of an existence theorem—it tells us that a zero exists but doesn’t tell us exactly where it is. Nevertheless, we can use the theorem to zero in on the zero. and P P P 0 For example, consider the polynomial 0. By the Intermediate Value Theorem P must have a zero between 2 and and 3. To “trap” the zero in a smaller interval, we evaluate P at successive tenths between 2 and 3 until we ﬁnd where P changes sign, as in Table 1. From the table we see that the zero we are looking for lies between 2.2 and 2.3, as shown in Figure 1. x 3 8x 30. Notice that TABLE 1 x 2.1 2.2 2.3 y 1 0 _1 P x 2 1 3.94 1.75 0.57 }change of sign y=P(x) 2.2 2.3 x TABLE 2 x 2.26 2.27 2.28 y 0.1 0 _0.1 P x 2 1 0.38 0.14 0.09 }change of sign y=P(x) 2.275 2.27 2.28 x FIGURE 1 FIGURE 2 We can repeat this process by evaluating P at successive 100ths between 2.2 and 2.3, as in Table 2. By repeating this process over and over again, we can get a numerical value for the zero as accurately as we want. From Table 2 we see that the zero is between 2.27 and 2.28. To see whether it is closer to 2.27 or 2.28, we check the value of P halfway between these two numbers:. Since this value is negative, 2.275 the zero we are looking for lies between 2.275 and 2.28, as illustrated in Figure 2. Correct to the nearest 100th, the zero is 2.28. 0.03 P 2 1 1. (a) Show that P x x 2 2 has a zero between 1 and 2. 1 2 (b) Find the zero of P to the nearest tenth. (c) Find the zero of P to the nearest 100th. (d) Explain why the zero you found is an approximation to process several times to obtain pare your results to 12 2. Find a polynomial that has 13
5 zero in on to four decimal places. obtained by a calculator. 13 5 12 correct to three decimal places. Com- 12. Repeat the as a zero. Use the process described here to (CONTINUES) 333 ZEROING IN ON A ZERO (CONTINUED) 3. Show that the polynomial has a zero between the given integers, and then zero in on that zero, correct to two decimals. (a) (b) (c) (d 2x 4 4x 2 1 2x 4 4x 2 1 ; between 1 and 2 ; between 2 and 3 ; between 1 and 2 ; between 1 and. Find the indicated irrational zero, correct to two decimals. x 2 x 2 (a) The positive zero of P 1 (b) The negative zero of P 1 x 4 2x 3 x 2 1 x 4 2x 3 x 2 1 5. In a passageway between two buildings, two ladders are propped up from the base of each building to the wall of the other so that they cross, as shown in the ﬁgure. h a c b x k If the ladders have lengths a 3 m and b 2 m and the crossing point is at height c 1 m, then it can be shown that the distance x between the buildings is a solution of the equation x 8 22x 6 163x 4 454x 2 385 0 (a) This equation has two positive solutions, which lie between 1 and 2. Use the technique of “zeroing in” to ﬁnd both of these correct to the nearest tenth. (b) Draw two scale diagrams, like the ﬁgure, one for each of the two values of x that you found in part (a). Measure the height of the crossing point on each. Which value of x seems to be the correct one? (c) Here is how to get the above equation. First, use similar triangles to show that 1 c 1 h 1 k Then use the Pythagorean Theorem to rewrite this as 1 c 1 2a2 x 2 1 2b2 x 2 Substitute a 3, b 2, and c 1, then simplify to obtain the desired equation. [Note that you must square twice in this process to eliminate both square roots. This is why you obtain an extraneous solution in part (a). (See the Warning on page 105.)] 334 SE CTI ON 4. 5 \| Complex Z
eros and the Fundamental Theorem of Algebra 335 Complex Zeros and the Fundamental Theorem of Algebra 4.5 LEARNING OBJECTIVES After completing this section, you will be able to: ■ State the Fundamental Theorem of Algebra ■ Factor a polynomial completely (into linear factors) over the complex numbers ■ Determine the multiplicity of a zero of a polynomial ■ Use the Conjugate Roots Theorem to ﬁnd polynomials with speciﬁed zeros ■ Factor a polynomial completely (into linear and quadratic factors) over the real numbers We have already seen that an nth-degree polynomial can have at most n real zeros. In the complex number system an nth-degree polynomial has exactly n zeros and so can be factored into exactly n linear factors. This fact is a consequence of the Fundamental Theorem of Algebra, which was proved by the German mathematician C. F. Gauss in 1799 (see page 338). ■ The Fundamental Theorem of Algebra and Complete Factorization The following theorem is the basis for much of our work in factoring polynomials and solving polynomial equations. FUNDAMENTAL THEOREM OF ALGEBRA Every polynomial an xn an1xn1... a1x a0 P x 1 2 n 1, an 0 1 2 with complex coefﬁcients has at least one complex zero. Because any real number is also a complex number, the theorem applies to polynomi- als with real coefﬁcients as well. The Fundamental Theorem of Algebra and the Factor Theorem together show that a polynomial can be factored completely into linear factors, as we now prove. COMPLETE FACTORIZATION THEOREM is a polynomial of degree n 1, then there exist complex numbers x P If a, c1, c2,..., cn (with a 0) such that 2 1 P x 1 2 a x c12 1 x c22 1 p x cn 2 1 ▼ P RO O F By the Fundamental Theorem of Algebra, P has at least one zero. Let’s call x P it c1. By the Factor Theorem (see page 319), 2 1 x c12 can be factored as # Q11 P x x 2 2 1 1 336 CHAPTER 4
\| Polynomial and Rational Functions where 2 gives us the factorization Q11 x is of degree n 1. Applying the Fundamental Theorem to the quotient Q11 x 2 # x x P # Q21 x c22 2 is of degree n 2 and c2 is a zero of Q11 where x Qn1 steps, we get a ﬁnal quotient 2 means that P has been factored as a x c12 Q21. Continuing this process for n of degree 0, a nonzero constant that we will call a. This x c22 To actually ﬁnd the complex zeros of an nth-degree polynomial, we usually ﬁrst factor as much as possible, then use the quadratic formula on parts that we can’t factor further. x c12 1 x cn2 ▲ 2 1 1 1 E X AM P L E 1 \| Factoring a Polynomial Completely 1 2 x P x3 3x2 x 3. Let (a) Find all the zeros of P. (b) Find the complete factorization of P. ▼ SO LUTI O N (a) We ﬁrst factor P as follows. P x 1 2 x3 3x2 x 3 x2 1 2 x2 Given 2 Group terms Factor x 3 We ﬁnd the zeros of P by setting each factor equal to 0 x2 1 2 This factor is 0 when x 3. This factor is 0 when x i or i. Setting x 3 0, we see that x 3 is a zero. Setting x 2 1 0, we get x 2 1, so x i. So the zeros of P are 3, i, and i. (b) Since the zeros are 3, i, and i, by the Complete Factorization Theorem P factors as x P x 1 x i ✎ Practice what you’ve learned: Do Exercise 5 AM P L E 2 \| Factoring a Polynomial Completely 1 2 x P x 3 2x 4 Let. (a) Find all the zeros of P. (b) Find the complete factorization of P ▼ SO LUTI O N (a) The possible rational zeros are the factors of 4, which are 1, 2, 4. Using synthetic division (see the margin), we ﬁnd that 2 is a zero, and the polynomial factors as 2x
2 x 2 This factor is 0 when x 2. Use the quadratic formula to ﬁnd when this factor is 0. SE CTI ON 4. 5 \| Complex Zeros and the Fundamental Theorem of Algebra 337 To ﬁnd the zeros, we set each factor equal to 0. Of course, x 2 0 means that x 2. We use the quadratic formula to ﬁnd when the other factor is 0. x x2 2x 2 0 2 14 8 2 2 2i 2 x 1 i x Set factor equal to 0 Quadratic Formula Take square root Simplify So the zeros of P are 2, 1 i, and 1 i. (b) Since the zeros are 2, 1 i, and 1 i, by the Complete Factorization Theorem P factors as ✎ Practice what you’ve learned: Do Exercise 19 ▲ ■ Zeros and Their Multiplicities In the Complete Factorization Theorem the numbers c1, c2,..., cn are the zeros of P. These zeros need not all be different. If the factor x c appears k times in the complete factor, then we say that c is a zero of multiplicity k (see page 309). For example, ization of 2 the polynomial P x 1 has the following zeros multiplicity 3 1 1, 2 1 2 multiplicity 2, 3 1 2 multiplicity 5 2 The polynomial P has the same number of zeros as its degree: It has degree 10 and has 10 zeros, provided we count multiplicities. This is true for all polynomials, as we prove in the following theorem. ZEROS THEOREM Every polynomial of degree n 1 has exactly n zeros, provided that a zero of multiplicity k is counted k times. ▼ P RO O F Let P be a polynomial of degree n. By the Complete Factorization Theorem x cn2 Now suppose that c is a zero of P other than c1, c2,..., cn. Then x c12 1 x c2 c22 Thus, by the Zero-Product Property one of the factors c ci must be 0, so c ci for some ▲ i. It follows that P has exactly the n zeros c1, c2,..., cn. c c12 1 c cn2 AM P L E
3 \| Factoring a Polynomial with Complex Zeros Find the complete factorization and all ﬁve zeros of the polynomial 3x 5 24x 3 48x P x 1 2 338 CHAPTER 4 \| Polynomial and Rational Functions ▼ SO LUTI O N Since 3x is a common factor, we have P x 1 2 3x 3x 1 1 x4 8x2 16 x2 4 2 2 2 This factor is 0 when x 0. This factor is 0 when x 2i or x 2i. To factor x 2 4, note that 2i and 2i are zeros of this polynomial. Thus, x 2i x 2i, so x2 4 1 2 1 2 P x 1 2 3x 3x x 2i 3 1 x 2i x 2i 2 1 x 2i is a zero of multiplicity 1. 2i is a zero of multiplicity 2. 2i is a zero of multiplicity 2. The zeros of P are 0, 2i, and 2i. Since the factors x 2i and x 2i each occur twice in the complete factorization of P, the zeros 2i and 2i are of multiplicity 2 (or double zeros). Thus, we have found all ﬁve zeros. ✎ Practice what you’ve learned: Do Exercise 29. ▲ The following table gives further examples of polynomials with their complete factor- izations and zeros. Degree Polynomial Zero(s) Number of zeros x2 10x 25 x 5 x 5 2 1 2 x3 x4 18x2 81 x 3i x 3i 2 2 1 x5 2x4 x3 x3 x 1 2 1 2 5 multiplicity 2 2 1 0, i, i 2 2 3i, multiplicity 2 2 1 3i multiplicity 2 2 1 0 1, multiplicity 3 2 1 multiplicity 2 2 1 1 2 3 4 5 Carl Friedrich Gauss (1777–1855) is considered the greatest mathematician of modern times. His contemporaries called him the “Prince of Mathematics.” He was born into a poor family; his father made a living as a mason. As a very small child, Gauss found a calculation error in his father’s accounts, the ﬁrst of many incidents that gave evidence of his mathematical precocity. (See also page 613.) At 19 Ga
uss I S B R O C © demonstrated that the regular 17-sided polygon can be constructed with straight-edge and compass alone. This was remarkable because, since the time of Euclid, it had been thought that the only regular polygons constructible in this way were the triangle and pentagon. Because of this discovery Gauss decided to pursue a career in mathematics instead of languages, his other passion. In his doctoral dissertation, written at the age of 22, Gauss proved the Fundamental Theorem of Algebra: A polynomial of degree n with complex coefﬁcients has n roots. His other accomplishments range over every branch of mathematics, as well as physics and astronomy. SE CTI ON 4. 5 \| Complex Zeros and the Fundamental Theorem of Algebra 339 E X AM P L E 4 \| Finding Polynomials with Speciﬁed Zeros (a) Find a polynomial (b) Find a polynomial multiplicity 3. P 1 Q x 2 x 1 2 of degree 4, with zeros i, i, 2, and 2, and with P of degree 4, with zeros 2 and 0, where 2 is a zero of 3 2 1 25. ▼ SO LUTI O N (a) The required polynomial has the form x2 1 x i 1 x2 4 1 2 1 x4 3x2 4 a x 2 x 2 1 2 1 22 22 1 2 1 P We know that 3 1 2 a 1 2 34 3 # 32 4 1 P x 2 2x4 3 1 2 50a 25 2x2 2 Difference of squares Multiply, so a 1 2. Thus, (b) We require 3x 2 x3 6x2 12x 8 2 x4 6x3 12x2 8x x 1 1 Special Product Formula 4 (Section P.5) 2 Since we are given no information about Q other than its zeros and their multiplicity, we can choose any number for a. If we use a 1, we get x4 6x3 12x2 8x Q x 1 2 ✎ Practice what you’ve learned: Do Exercise 35. ▲ E X AM P L E 5 \| Finding All the Zeros of a Polynomial Find all four zeros of P x 3x4 2x3 x2 12x 4. 1 2 ▼ SO LUTI O N Using the Rational Zeros Theorem from Section 4
.4, we obtain the 4 following list of possible rational zeros: 1, 2, 4,. Checking these 3 are zeros, and we get the following facusing synthetic division, we ﬁnd that 2 and torization. 2 3 1 3 1 3,, 40 3x4 2x3 x2 12x 4 P x 1 2 _2 4 x 2 1 2 1 3x3 4x2 7x 2 2 3x2 3x 3B1 x 1 1 2 A x2 x 2 2 3B1 Factor x 2 Factor x 1 3 Factor 3 2 _20 The zeros of the quadratic factor are FIGURE 1 P x 1 2 3x 4 2x 3 x 2 12x 4 Figure 1 shows the graph of the polynomial P in Example 5. The x-intercepts correspond to the real zeros of P. The imaginary zeros cannot be determined from the graph. x 1 11 8 2 1 2 i 17 2 Quadratic formula so the zeros of P are x 1 2 2, 1 3, 1 2 i 17 2, and 1 2 i 17 2 ✎ Practice what you’ve learned: Do Exercise 45. ▲ 340 CHAPTER 4 \| Polynomial and Rational Functions ■ Complex Zeros Come in Conjugate Pairs As you might have noticed from the examples so far, the complex zeros of polynomials with real coefﬁcients come in pairs. Whenever a bi is a zero, its complex conjugate a bi is also a zero. CONJUGATE ZEROS THEOREM If the polynomial P has real coefﬁcients and if the complex number z is a zero of P, then its complex conjugate is also a zero of P. z ▼ P RO O F Let an xn an1xn1... a1x a0 P x 1 2 P. We where each coefﬁcient is real. Suppose that use the facts that the complex conjugate of a sum of two complex numbers is the sum of the conjugates and that the conjugate of a product is the product of the conjugates.. We must prove that an11 an1 2 an zn an1 zn1... a1 z a0 n1... a1z a0 z 2 an zn an1 zn1... a1z
a0 Because the coefﬁcients are real anzn an1zn1... a1z a0 P 0 0 z 2 This shows that 1 z is also a zero of P(x),which proves the theorem. ▲ E X AM P L E 6 \| A Polynomial with a Speciﬁed Complex Zero P Find a polynomial ▼ SO LUTI O N Since 3 i is a zero, then so is 3 i by the Conjugate Zeros Theorem. This means that of degree 3 that has integer coefﬁcients and zeros and 3 i. must have the following form Gerolamo Cardano (1501–1576) is certainly one of the most colorful ﬁgures in the history of mathematics. He was the best-known physician in Europe in his day, yet throughout his life he was plagued by numerous maladies, including ruptures, hemorrhoids, and an irrational fear of encountering rabid dogs. He was a doting father, but his beloved sons broke his heart—his favorite was eventually beheaded for murdering his own wife. Cardano was also a compulsive gambler; indeed, this vice might have driven him to write the Book on Games of Chance, the ﬁrst study of probability from a mathematical point of view. In Cardano’s major mathematical work, the Ars Magna, he detailed the solution of the general third- and fourth-degree poly- nomial equations. At the time of its publication, mathematicians were uncomfortable even with negative numbers, but Cardano’s formulas paved the way for the acceptance not just of negative numbers, but also of imaginary numbers, because they occurred naturally in solving polynomial equations. For example, for the cubic equation x 3 15x 4 0 one of his formulas gives the solution x 23 2 1121 23 2 1121 (See page 332, Exercise 106). This value for x actually turns out to be the integer 4, yet to ﬁnd it, Cardano had to use the imaginary number 1121 11i. SE CTI ON 4. 5 \| Complex Zeros and the Fundamental Theorem of Algebra 341 2B 3 2B i2 x 3 x 1 2 x2 6x 10 x 1 2B1 2 2 x2 13x 5 x3 13 2B 3 1 4 A A A B Regroup Difference of Squares Formula Expand Expand To make all coef�
��cients integers, we set a 2 and get 2x3 13x2 26x 10 P x 1 2 Any other polynomial that satisﬁes the given requirements must be an integer multiple of this one. ✎ Practice what you’ve learned: Do Exercise 39. ▲ ■ Linear and Quadratic Factors We have seen that a polynomial factors completely into linear factors if we use complex numbers. If we don’t use complex numbers, then a polynomial with real coefﬁcients can always be factored into linear and quadratic factors. We use this property in Section 6.4 when we study partial fractions. A quadratic polynomial with no real zeros is called irreducible over the real numbers. Such a polynomial cannot be factored without using complex numbers. LINEAR AND QUADRATIC FACTORS THEOREM Every polynomial with real coefﬁcients can be factored into a product of linear and irreducible quadratic factors with real coefﬁcients. ▼ P RO O F We ﬁrst observe that if c a bi is a complex number, then bi 2 4 bi 2 4 3 a bi bi 2ax bi a2 b2 1 2 The last expression is a quadratic with real coefﬁcients. Now, if P is a polynomial with real coefﬁcients, then by the Complete Factorization Theorem P x 1 2 a x c12 1 x c2 2 1 p x cn2 1 Since the complex roots occur in conjugate pairs, we can multiply the factors corresponding to each such pair to get a quadratic factor with real coefﬁcients. This results in P being ▲ factored into linear and irreducible quadratic factors. E X AM P L E 7 \| Factoring a Polynomial into Linear and Quadratic Factors 1 2 x P x4 2x2 8. Let (a) Factor P into linear and irreducible quadratic factors with real coefﬁcients. (b) Factor P completely into linear factors with complex coefﬁcients. 342 CHAPTER 4 \| Polynomial and Rational Functions ▼ SO LUTI O N (a) 1 2 x P x 4 2x 2 8 x 2 4 1 The factor x 2
4 is irreducible, since it has no real zeros. x 2 4 x 12 x 2 2 2 1 x 12 2 1 2 1 2 1 2 (b) To get the complete factorization, we factor the remaining quadratic factor. x P x 2 4 x 2i 1 ✎ Practice what you’ve learned: Do Exercise 65. x 12 1 x 12 x 12 x 12 2i 2 ▲ 4. ▼ CONCE PTS 1. The polynomial P 3 x x 5 1 2. It has zeros 5, 3, and 1 x 3 3 2 1 x 2 has degree 2 2 1. The zero 5 has mul- tiplicity, and the zero 3 has multiplicity. 2. (a) If a is a zero of the polynomial P, then must be a factor of P(x). (b) If a is a zero of multiplicity m of the polynomial P, then must be a factor of P(x) when we factor P completely. 3. A polynomial of degree has exactly zero of multiplicity m is counted m times. n 1 zeros if a 4. If the polynomial function P has real coefﬁcients and if a bi ✎ is a zero of P, then is also a zero of P. ✎ 2 1 x P 5. ▼ SKI LLS 5–16 ■ A polynomial P is given. (a) Find all zeros of P, real and complex. (b) Factor P completely. x 4 4x 2 x3 2x2 2x x4 2x2 1 x 4 16 x 3 8 x 6 1 x 5 9x 6x 2 9 x 3 8 x 6 7x 3 8 11. 12. 13. 14. 10. 16. 15. 8. 9. 7. 6 17–34 ■ Factor the polynomial completely, and ﬁnd all its zeros. State the multiplicity of each zero ✎ 17. 19. 21. 23. 25. P x 2 25 x2 2x 2 x 3 4x x 4 1 16x 4 81 18. 20. 22. 24. 26. P 4x 2 9 x 2 8x 17 x 3 x 2 x x 4 625 x 3 64 ✎ 27. 29. 31. 33 9x 9 x4 2x2 1 x 4 3x 2 4 x 5 6x 3 9
x 28. 30. 32. 34. x 6 729 x 4 10x 2 25 x 5 7x 3 x 6 16x 3 64 35–44 ■ Find a polynomial with integer coefﬁcients that satisﬁes the given conditions. 35. P has degree 2 and zeros 1 i and 1 i. ✎ 36. P has degree 2 and zeros and 37. Q has degree 3 and zeros 3, 2i, and 2i. 1 i12 1 i12. 38. Q has degree 3 and zeros 0 and i. 39. P has degree 3 and zeros 2 and i. 40. Q has degree 3 and zeros 3 and 1 i. 41. R has degree 4 and zeros 1 2i and 1, with 1 a zero of multi- plicity 2. 42. S has degree 4 and zeros 2i and 3i. 43. T has degree 4, zeros i and 1 i, and constant term 12. 44. U has degree 5, zeros, 1, and i, and leading coefﬁcient 4; 1 2 the zero 1 has multiplicity 2 47. 48. 49. 45. 46. 45–62 ■ Find all zeros of the polynomial. x3 2x2 4x 8 x 3 7x 2 17x 15 x3 2x2 2x 1 x 3 7x 2 18x 18 x 3 3x 2 3x 2 x 3 x 6 2x3 7x2 12x 9 2x3 8x2 9x 9 x 4 x 3 7x 2 9x 18 x4 2x3 2x2 2x 3 54. P 50. 52. 51. 53 55. 56. 57. 58. 59. 60. 61. 62Hint: Factor by grouping.] x5 x4 7x3 7x2 12x 12 x 5 x 3 8x 2 8 x 4 6x 3 13x 2 24x 36 x4 x2 2x 2 4x 4 4x 3 5x 2 4x 1 4x4 2x3 2x2 3x 1 x5 3x4 12x3 28x2 27x 9 x5 2x4 2x3 4x2 x 2 63–68 ■ A polynomial P is given. (a) Factor P into linear and irreducible quadratic factors with real coefﬁcients. (
b) Factor P completely into linear factors with complex coefﬁcients. ✎ 63. 64. 65. 67 5x 2 4x 20 x3 2x 4 x 4 8x 2 9 x 6 64 66. 68 8x 2 16 x 5 16x 69. By the Zeros Theorem, every nth-degree polynomial equation has exactly n solutions (including possibly some that are repeated). Some of these may be real, and some may be imaginary. Use a graphing device to determine how many real and imaginary solutions each equation has. (a) x 4 2x 3 11x 2 12x 0 (b) x 4 2x 3 11x 2 12x 5 0 (c) x 4 2x 3 11x 2 12x 40 0 70–72 ■ So far, we have worked only with polynomials that have real coefﬁcients. These exercises involve polynomials with real and imaginary coefﬁcients. SE CT IO N 4.6 \| Rational Functions 343 70. Find all solutions of the equation. (a) 2x 4i 1 (c) x 2 2ix 1 0 (b) x 2 ix 0 (d) ix 2 2x i 0 71. (a) Show that 2i and 1 i are both solutions of the equation x 2 x but that their complex conjugates 2i and 1 i are not. (b) Explain why the result of part (a) does not violate the Con- 2 2i 1 i 0 1 1 2 2 jugate Zeros Theorem. 72. (a) Find the polynomial with real coefﬁcients of the smallest possible degree for which i and 1 i are zeros and in which the coefﬁcient of the highest power is 1. (b) Find the polynomial with complex coefﬁcients of the smallest possible degree for which i and 1 i are zeros and in which the coefﬁcient of the highest power is 1. ▼ DISCOVE RY • DISCUSSION • WRITI NG 73. Polynomials of Odd Degree The Conjugate Zeros Theorem says that the complex zeros of a polynomial with real coefﬁcients occur in complex conjugate pairs. Explain how this fact proves that a polynomial with real coefﬁcients and odd degree has at least
one real zero. 74. Roots of Unity There are two square roots of 1, namely, 1 and 1. These are the solutions of x 2 1. The fourth roots of 1 are the solutions of the equation x 4 1 or x 4 1 0. How many fourth roots of 1 are there? Find them. The cube roots of 1 are the solutions of the equation x 3 1 or x 3 1 0. How many cube roots of 1 are there? Find them. How would you ﬁnd the sixth roots of 1? How many are there? Make a conjecture about the number of nth roots of 1. 4.6 Rational Functions LEARNING OBJECTIVES After completing this section, you will be able to: ■ Find the vertical asymptotes of a rational function ■ Find the horizontal asymptote of a rational function ■ Use asymptotes to graph a rational function ■ Find the slant asymptote of a rational function A rational function is a function of the form P 1 Q 1 where P and Q are polynomials. We assume that P have no factor in common. Even though rational functions are constructed from polynomials, their graphs look quite different than the graphs of polynomial functions. 2 2 x and 344 CHAPTER 4 \| Polynomial and Rational Functions Domains of rational expressions are discussed in Section P.7. ■ Rational Functions and Asymptotes The domain of a rational function consists of all real numbers x except those for which the denominator is zero. When graphing a rational function, we must pay special attention to the behavior of the graph near those x-values. We begin by graphing a very simple rational function. E X AM P L E 1 \| A Simple Rational Function 1 x Sketch a graph of the rational function x f. 2 1 ▼ SO LUTI O N The function f is not deﬁned for x 0. The following tables show that when x is close to zero, the value of is large, and the closer x gets to zero, the larger f gets For positive real numbers, 1 BIG NUMBER 1 small number small number BIG NUMBER x 0.1 0.01 0.00001 ff x 1 2 10 100 100,000 x 0.1 0.01 0.00001 ff x 1 2 10 100 100,000 Approaching 0 Approaching Approaching 0 Approaching We describe this behavior in words and in symbols as follows. The ﬁr
st table shows that as y f x approaches 0 from the left, the values of decrease without bound. In symbols, x 1 2 q as x 0 f x 1 2 “y approaches negative inﬁnity as x approaches 0 from the left” The second table shows that as x approaches 0 from the right, the values of without bound. In symbols, f x 1 2 increase q as x 0 f x 1 2 “y approaches inﬁnity as x approaches 0 from the right” The next two tables show how f x 1 2 changes as x 0 0 becomes large. x 10 100 100,000 2 ff x 1 0.1 0.01 0.00001 x 10 100 100,000 ff x 1 2 0.1 0.01 0.00001 Approaching Approaching 0 Approaching Approaching 0 These tables show that as 0 zero. We describe this situation in symbols by writing becomes large, the value of x 0 f x 1 2 gets closer and closer to f x 1 2 0 as x q and f 0 as x q x 1 2 SE CTION 4.6 \| Rational Functions 345 Using the information in these tables and plotting a few additional points, we obtain the graph shown in Figure 1(x) → ` as x → 0+ f(x) → 0 as x → ` 2 x f(x) → 0 as x → _` f(x) → _` as x → 0_ FIGURE 1 1 f x x 1 2 ✎ Practice what you’ve learned: Do Exercise 7. ▲ In Example 1 we used the following arrow notation. Symbol Meaning x a x a x q x q x approaches a from the left x approaches a from the right x goes to negative inﬁnity; that is, x decreases without bound x goes to inﬁnity; that is, x increases without bound The line x 0 is called a vertical asymptote of the graph in Figure 1, and the line y 0 is a horizontal asymptote. Informally speaking, an asymptote of a function is a line to which the graph of the function gets closer and closer as one travels along that line. DEFINITION OF VERTICAL AND HORIZONTAL ASYMPTOTES 1. The line x a is a vertical asymptote of the function y f or left. if y approaches q as x approaches a from the right →
` as x → a+ y → ` as x → a− y → −` as x → a+ y → −` as x → a− 2. The line y b is a horizontal asymptote of the function y f if y approaches b as x approaches q as x → ` y → b as x → −` A rational function has vertical asymptotes where the function is undeﬁned, that is, where the denominator is zero. 346 CHAPTER 4 \| Polynomial and Rational Functions ■ Transformations of A rational function of the form y 1 x ax b cx d r x 1 2 shown in can be graphed by shifting, stretching, and/or reﬂecting the graph of Figure 1, using the transformations studied in Section 3.5. (Such functions are called linear fractional transformations.) 2 1 f x 1 x E X AM P L E 2 \| Using Transformations to Graph Rational Functions y 1 0 Vertical asymptote x = 3 r(x)= 2 x-3 3 x Horizontal asymptote y = 0 FIGURE 2 3 x 23x 5 3x 6 1 Sketch a graph of each rational function. (a) r x 1 2 2 x 3 (b) s x 1 2 3x 5 x 2 ▼ SO LUTI O N (a) Let f x 1 x 1 2. Then we can express r in terms of f as follows Factor 2 Since f(x) = 1 x 22 From this form we see that the graph of r is obtained from the graph of f by shifting 3 units to the right and stretching vertically by a factor of 2. Thus, r has vertical asymptote x 3 and horizontal asymptote y 0. The graph of r is shown in Figure 2. 1 1 (b) Using long division (see the margin), we get s s in terms of f as follows: 3 1 x 2. Thus, we can express Rearrange terms f x 2 1 2 3 Since f(x) = 1 x From this form we see that the graph of s is obtained from the graph of f by shifting 2 units to the left, reﬂecting in the x-axis, and shifting upward 3 units. Thus, s has vertical asymptote x 2 and horizontal asymptote y 3. The graph of s is shown in Figure 3. Vertical asymptote x = −2 y Horizontal as
ymptote y = 3 3 s(x)= 3x+5 x+2 _2 0 x FIGURE 3 ✎ Practice what you’ve learned: Do Exercises 35 and 37. ▲ SE CTION 4.6 \| Rational Functions 347 ■ Asymptotes of Rational Functions The methods of Example 2 work only for simple rational functions. To graph more complicated ones, we need to take a closer look at the behavior of a rational function near its vertical and horizontal asymptotes. E X AM P L E 3 \| Asymptotes of a Rational Function Graph the rational function r 2x 2 4x 5 x 2 2x 1. x 1 2 ▼ SO LUTI O N Vertical asymptote: We ﬁrst factor the denominator 2x2 4x 5 x 1 The line x 1 is a vertical asymptote because the denominator of r is zero when x 1. To see what the graph of r looks like near the vertical asymptote, we make tables of values for x-values to the left and to the right of 1. From the tables shown below we see that x r 2 1 2 1 2 y q as x 1 and y q as x 1 x 1 x 0 0.5 0.9 0.99 y 5 14 302 30,002 x 1 x 2 1.5 1.1 1.01 y 5 14 302 30,002 Approaching 1– Approaching Approaching 1+ Approaching Thus, near the vertical asymptote x 1, the graph of r has the shape shown in Figure 4. y → ` as + x → 1 Horizontal asymptote: The horizontal asymptote is the value y approaches as x q. To help us ﬁnd this value, we divide both numerator and denominator by x 2, the highest power of x that appears in the expression: y 2x2 4x 5 x2 2x 1 # 1 x2 1 x2 2 4 x 1 2 x 5 x2 1 x2 2 The fractional expressions x page 19). So as x q, we have 5 x2 4 x,,, and 1 x2 all approach 0 as x q (see Exercise 67, These terms approach 0 These terms approach 0. Thus, the horizontal asymptote is the line y 2. y → ` as − x → 1 y 5 1 −1 0 1 2 x FIGURE 4 348
CHAPTER 4 \| Polynomial and Rational Functions y 5 1 y → 2 as x → −` y → 2 as x → ` −1 0 1 2 x FIGURE 5 2x 2 4x 5 x 2 2x 1 r x 1 2 Since the graph must approach the horizontal asymptote, we can complete it as in Figure 5. ✎ Practice what you’ve learned: Do Exercise 45. ▲ From Example 3 we see that the horizontal asymptote is determined by the leading coefﬁcients of the numerator and denominator, since after dividing through by x 2 (the highest power of x) all other terms approach zero. In general, if r and the degrees of P and Q are the same (both n, say), then dividing both numerator and denominator by x n shows that the horizontal asymptote is P / leading coefﬁcient of P leading coefﬁcient of Q The following box summarizes the procedure for ﬁnding asymptotes. FINDING ASYMPTOTES OF RATIONAL FUNCTIONS Let r be the rational function r x 1 2 an xn an1xn1... a1x a0 bm xm bm1xm1... b1x b0 1. The vertical asymptotes of r are the lines x a, where a is a zero of the denominator. 2. (a) If n m, then r has horizontal asymptote y 0. an bm (b) If n m, then r has horizontal asymptote y (c) If n m, then r has no horizontal asymptote.. E X AM P L E 4 \| Asymptotes of a Rational Function Find the vertical and horizontal asymptotes of r 3x2 2x 1 2x2 3x 2. x 1 2 SE CTION 4.6 \| Rational Functions 349 ▼ SO LUTI O N Vertical asymptotes: We ﬁrst factor 3x2 2x 1 x 2 2x 1 r x 1 2 1 2 1 2 10 This factor is O when x =.1 2 This factor is O when x 2. The vertical asymptotes are the lines x 1 2 and x 2. _6 3 Horizontal asymptote: The degrees of the numerator and denominator are the same,
and _10 leading coefﬁcient of numerator leading coefﬁcient of denominator 3 2 FIGURE 6 r x 3x 2 2x 1 2x 2 3x 2 2 1 Graph is drawn using dot mode to avoid extraneous lines. A fraction is 0 if and only if its numerator is 0. Thus, the horizontal asymptote is the line y 3 2. To conﬁrm our results, we graph r using a graphing calculator (see Figure 6). ✎ Practice what you’ve learned: Do Exercises 23 and 25. ▲ ■ Graphing Rational Functions We have seen that asymptotes are important when graphing rational functions. In general, we use the following guidelines to graph rational functions. SKETCHING GRAPHS OF RATIONAL FUNCTIONS 1. Factor. Factor the numerator and denominator. 2. Intercepts. Find the x-intercepts by determining the zeros of the numerator and the y-intercept from the value of the function at x 0. 3. Vertical Asymptotes. Find the vertical asymptotes by determining the zeros of the denominator, and then see whether y q or y q on each side of each vertical asymptote by using test values. 4. Horizontal Asymptote. Find the horizontal asymptote (if any), using the procedure described in the box on page 348. 5. Sketch the Graph. Graph the information provided by the ﬁrst four steps. Then plot as many additional points as needed to ﬁll in the rest of the graph of the function. E X AM P L E 5 \| Graphing a Rational Function Graph the rational function r 2x2 7x 4 x2 x 2. x 1 2 ▼ SO LUTI O N We factor the numerator and denominator, ﬁnd the intercepts and asymptotes, and sketch the graph. Factor: y 1 2x 350 CHAPTER 4 \| Polynomial and Rational Functions When choosing test values, we must make sure that there is no x-intercept between the test point and the vertical asymptote. x-Intercepts: The x-intercepts are the zeros of the numerator, y-Intercept: To ﬁnd the y-intercept, we substitute x 0 into the original form of the function. and
x 4 The y-intercept is 2 Vertical asymptotes: The vertical asymptotes occur where the denominator is 0, that is, where the function is undeﬁned. From the factored form we see that the vertical asymptotes are the lines x 1 and x 2. Behavior near vertical asymptotes: We need to know whether y q or y q on each side of each vertical asymptote. To determine the sign of y for x-values near the vertical asymptotes, we use test values. For instance, as x 1, we use a test value close to check whether y is positive or negative to the left to and to the left of of x 1. x 0.9, say 1 2 1 2 y 1 0.9 1 0.9 2 1 1 11 2 2 11 2 11 0.9 0.9 2 4 2 2 whose sign is 1 1 2 2 2 2 1 2 1 negative 1 2 So y q as x 1. On the other hand, as x 1, we use a test value close to and to the right of 1 x 1.1, say So y q as x 1. The other entries in the following table are calculated similarly. 2 2 whose sign is 1 1 2 2 2 1 2 1 positive 1 2 1 2 11 1.1 1 1.1 2 1 1 2 11 2 11, to get 4 2 2 1.1 1.1 2 y 1 2 2 As x 2 2 1 1 the sign of y 1 2x is so y 2 1 Horizontal asymptote: The degrees of the numerator and denominator are the same, and leading coefﬁcient of numerator leading coefﬁcient of denominator 2 1 2 Thus, the horizontal asymptote is the line y 2. Graph: We use the information we have found, together with some additional values, to sketch the graph in Figure 7. x 6 3 1 1.5 2 3 y 0.93 1.75 4.50 6.29 4.50 3.50 FIGURE 7 2x2 7x 4 x2 ✎ Practice what you’ve learned: Do Exercise 53. ▲ SE CTION 4.6 \| Rational Functions 351 E X AM P L E 6 \| Graphing a Rational Function Graph the rational function r x 1 2 5x 21 x 2 10x 25. ▼ SO LUTI O N
Factor: 2 y 5x 21 x 5 2 1 21 5 x-Intercept:, from 5x 21 0 y-Intercept: 21 25, because r 0 1 2 5 # 0 21 02 10 # 0 25 21 25 Vertical asymptote: x 5, from the zeros of the denominator Behavior near vertical asymptote: the sign of As x y 5x 21 x 5 2 1 2 is 1 so Horizontal asymptote: of the denominator y 0, because the degree of the numerator is less than the degree MATH IN THE MODERN WORLD Unbreakable Codes If you read spy novels, you know about secret codes and how the hero “breaks” the code. Today secret codes have a much more common use. Most of the information that is stored on computers is coded to prevent unauthorized use. For example, your banking records, medical records, and school records are coded. Many cellular and cordless phones code the signal carrying your voice so that no one can listen in. Fortunately, because of recent advances in mathematics, today’s codes are “unbreakable.” Modern codes are based on a simple principle: Factoring is much harder than multiplying. For example, try multiplying 78 and 93; now try factoring 9991. It takes a long time to factor 9991 because it is a product of two primes 97 103, so to factor it, we have to ﬁnd one of these primes. Now imagine trying to factor a number N that is the product of two primes p and q, each about 200 digits long. Even the fastest computers would take many millions of years to factor such a number! But the same computer would take less than a second to multiply two such numbers. This fact was used by Ron Rivest, Adi Shamir, and Leonard Adleman in the 1970s to devise the RSA code. Their code uses an extremely large number to encode a message but requires us to know its factors to decode it. As you can see, such a code is practically unbreakable. The RSA code is an example of a “public key encryption” code. In such codes, anyone can code a message using a publicly known procedure based on N, but to decode the message, they must know p and q, the factors of N. When the RSA code was developed, it was thought that a carefully selected 80-digit number would provide an unbreakable code. But interestingly
, recent advances in the study of factoring have made much larger numbers necessary. 352 CHAP TER 4 \| Polynomial and Rational Functions Graph: We use the information we have found, together with some additional values, to sketch the graph in Figure 8. x 15 10 3 1 3 5 10 y 0.5 1.2 1.5 1.0 0.6 0.5 0.3 FIGURE 8 r x 1 2 5x 21 x2 10x 25 y 1 0 5 x ✎ Practice what you’ve learned: Do Exercise 55. ▲ From the graph in Figure 8 we see that, contrary to the common misconception, a graph may cross a horizontal asymptote. The graph in Figure 8 crosses the x-axis (the horizontal asymptote) from below, reaches a maximum value near x 3, and then approaches the x-axis from above as x q. E X AM P L E 7 \| Graphing a Rational Function Graph the rational function r x2 3x 4 2x2 4x. x 1 2 ▼ SO LUTI O N Factor: y 1 x 4 x 1 2x 2 1 x 2 2 2 x-Intercepts: 1 and 4, from x 1 0 and x 4 0 1 y-Intercept: None, because r is undeﬁned 0 1 2 Vertical asymptotes: x 0 and x 2, from the zeros of the denominator Behavior near vertical asymptotes: As x 2 2 0 0 the sign of y 1 x 4 x 1 2x 2 1 x 2 1 2 so y 2 is Horizontal asymptote: 2 denominator are the same and, because the degree of the numerator and the degree of the leading coefﬁcient of numerator leading coefﬁcient of denominator 1 2 SE CTION 4.6 \| Rational Functions 353 Graph: We use the information we have found, together with some additional values, to sketch the graph in Figure 9. y 2 3 x 3 2.5 0.5 1 3 5 y 2.33 3.90 1.50 1.00 0.13 0.09 FIGURE 9 x2 3x 4 2x2 4x r x 1 2 ✎ Practice what you’ve learned: Do Exercise 57. x ▲ ■ Slant Asymptotes and End Behavior r If is a rational function in which the degree of the numerator is
one more than the degree of the denominator, we can use the Division Algorithm to express the function in the form P / ax where the degree of R is less than the degree of Q and a 0. This means that as x q, approaches the graph of the line y ax b. In this situation we say that y ax b is a slant asymptote, or an oblique asymptote., so for large values of the graph of y r 0 / AM P L E 8 \| A Rational Function with a Slant Asymptote Graph the rational function r x 2 4x 5 x 3. x 1 2 ▼ SO LUTI O N Factor-Intercepts: 1 and 5, from x 1 0 and x 5 0 y-Intercepts: 5 3, because r 0 1 2 02 4 # 0 5 0 3 5 3 Horizontal asymptote: None, because the degree of the numerator is greater than the degree of the denominator Vertical asymptote: x 3, from the zero of the denominator Behavior near vertical asymptote: y q as x 3 and y q as x 3 Slant asymptote: Since the degree of the numerator is one more than the degree of the denominator, the function has a slant asymptote. Dividing (see the margin), we obtain r x 1 2 x 1 8 x 3 Thus, y x 1 is the slant asymptote. x 1 x 3x2 4x 5 x2 3x x 5 x 3 8 354 CHAPTER 4 \| Polynomial and Rational Functions Graph: We use the information we have found, together with some additional values, to sketch the graph in Figure 10. x 2 1 2 4 6 y 1.4 4 9 5 2.33 y 5 Slant asymptote 2 y=x-1 r(x)= ≈-4x-5 x-3 FIGURE 10 ✎ Practice what you’ve learned: Do Exercise 65. x ▲ So far, we have considered only horizontal and slant asymptotes as end behaviors for rational functions. In the next example we graph a function whose end behavior is like that of a parabola. E X AM P L E 9 \| End Behavior of a Rational Function Graph the rational function x3 2x2 3 x 2 r x 1 2 and describe its end behavior. ▼
SO LUTI O N Factor: y 1 x 1 x 2 3x 3 2 1 x 2 2 x-Intercepts: 1, from x 1 0 (The other factor in the numerator has no real zeros.) 03 2 # 02 3 0 2 y-Intercepts:, because 3 2 3 2 0 r 1 2 Vertical asymptote: x 2, from the zero of the denominator Behavior near vertical asymptote: y q as x 2 and y q as x 2 Horizontal asymptote: None, because the degree of the numerator is greater than the degree of the denominator x2 x 2x3 2x2 0x 3 x3 2x2 3 End behavior: Dividing (see the margin), we get / x 2 is large. That is, 2 This shows that the end behavior of r is like that of the parabola y x 2 because 0 is small when of r will be close to the graph of y x 2 for large x 2 2 as x q. This means that the graph x 0 In Figure 11(a) we graph r in a small viewing rectangle; we can see the intercepts, Graph: the vertical asymptotes, and the local minimum. In Figure 11(b) we graph r in a larger viewing rectangle; here the graph looks almost like the graph of a parabola. In Figure 11(c) and y x 2; these graphs are very close to each other except near x we graph both 1 the vertical asymptote. y r 3/. 1 2 2 1 0 SE CTION 4.6 \| Rational Functions 355 20 200 _4 4 _30 30 _8 _20 (a) _200 (b) FIGURE 11 x 3 2x 2 3 x 2 r x 1 2 ✎ Practice what you’ve learned: Do Exercise 73. 20 _5 (c) y=≈ 8 ▲ ■ Applications Rational functions occur frequently in scientiﬁc applications of algebra. In the next example we analyze the graph of a function from the theory of electricity. E X AM P L E 10 \| Electrical Resistance When two resistors with resistances R1 and R2 are connected in parallel, their combined resistance R is given by the formula R R1R2 R2 R1 8 ohms x FIGURE 12 FIGURE 13 8x 8 x R x 1 2 Suppose that a ﬁxed 8-
ohm resistor is connected in parallel with a variable resistor, as shown in Figure 12. If the resistance of the variable resistor is denoted by x, then the combined resistance R is a function of x. Graph R, and give a physical interpretation of the graph. ▼ SO LUTI O N Substituting R1 x into the formula gives the function 8 and R2 R x 1 2 8x 8 x Since resistance cannot be negative, this function has physical meaning only when x 0.. The The function is graphed in Figure 13(a) using the viewing rectangle function has no vertical asymptote when x is restricted to positive values. The combined resistance R increases as the variable resistance x increases. If we widen the viewing rect, we obtain the graph in Figure 13(b). For large x the combined angle to 4 resistance R levels off, getting closer and closer to the horizontal asymptote R 8. No matter how large the variable resistance x, the combined resistance is never greater than 8 ohms. 0,100 3 0, 10 0, 20 0, 10 by by 4 4 3 3 3 4 10 0 10 20 0 100 (a) (b) ✎ Practice what you’ve learned: Do Exercise 83. ▲ 356 CHAPTER 4 \| Polynomial and Rational Functions 4. ▼ CONCE PTS 1. If the rational function x 2, x 2, then as y r 1 x 2 y either y r has the vertical asymptote or y 17.. 2. If the rational function y 2, then y x 2 x q. 1 as has the horizontal asymptote 3–6 ■ The following questions are about the rational function. The function r has x-intercepts 4. The function r has y-intercept 2 2. and. 19. 5. The function r has vertical asymptotes x = and x =. 6. The function r has horizontal asymptote y =. 17–20 ■ From the graph, determine the x- and y-intercepts and the vertical and horizontal asymptotes. 18. 20. 4 x y 4 0 y 2 −3 10 3 x y 2 0 1 x −4 y 2 0 −6 4 x ▼ SKI LLS 7–10 ■ A rational function is given. (a) Complete each table for the function. (b) Describe the behavior of the function near its vertical asymptote, based on
Tables 1 and 2. (c) Determine the horizontal asymptote, based on Tables 3 and 4. TABLE 1 TABLE.5 1.9 1.99 1.999 TABLE 3 x 10 50 100 1000.5 2.1 2.01 2.001 TABLE 4 x 10 50 100 1000 ✎ 7. r 9 3x 10 x 2 2 1 2 8. r 10. r x x 1 1 2 2 4x 1 x 2 3x 2 1 x 2 2 1 2 11–16 ■ Find the x- and y-intercepts of the rational function. 11. r x 1 2 13. t x 1 2 15 12. s 14. r 16 3x x 5 2 x 2 3x 4 x 3 8 x 2 4 21–32 ■ Find all horizontal and vertical asymptotes (if any). 21. r x 1 2 5 x 2 ✎ 23. r ✎ 25. s 27. s 29 6x x 2 2 6x 2 1 2 1 2x 2 x 1 5x 1 3x 1 x 1 x 2 2 1 6x 3 2 2x 3 5x 2 6x 2 1 1 ✎ 31 22. r 24. r 26. s 2 28. s 30. r 32. r 2x 3 x 2 1 2x 4 x2 x 1 8x 2 1 2 1 4x 2 2x 6 2x 1 3x 1 x 3 x 4 2 1 5x 3 x 3 2x 2 5x 2 2 1 1 x 3 3x 33–40 ■ Use transformations of the graph of tional function, as in Example 2. y 1 x to graph the ra- 33. r ✎ 35. s x x 1 1 2 2 ✎ 37. t x 1 2 39 2x 3 x 2 x 2 x 3 34. r 36. s x x 1 1 2 2 38. t x 1 2 40 3x 3 x 2 2x 9 x 4 41–64 ■ Find the intercepts and asymptotes, and then sketch a graph of the rational function. Use a graphing device to conﬁrm your answer. 41. r x 1 2 4x 4 x 2 42. r x 1 2 2x 6 6x 3 43. s ✎ 45. r 47. s 49 51. t x 1 2 ✎ 53. r ✎ 55. r ✎ 57. r 59. r 61. r 63 3x x 7
18 x 3 1 2 2 4x 5x 6 2 3x 6 2 2 1 2 1 1 1 x2 2x 8 x 2 x 1 x 3 x 1 x2 2x 1 x2 2x 1 2x2 10x 12 x2 x 6 2 x 2 x 6 x 2 3x 3x2 6 x2 2x 3 x2 2x 1 x3 3x2 44. s 46. r 48. s 50 52. t x 1 2 54. r 56. r 58. r 60. r 62 2x 2x 2x 4 x2 x 2 x 1 x 2 x 2 4x 2x 4x2 x2 2x 3 2x2 2x 4 x2 x x 2 3x x 2 x 6 5x 2 5 x 2 4x 4 64 3x 2 65–72 ■ Find the slant asymptote, the vertical asymptotes, and sketch a graph of the function. ✎ 65. r 67. r 69. r 71. r x 2 x 2 x2 2x 8 x x 2 5x 66. r 68. r 70. r 72 x2 2x x 1 3x x2 2x 2 x3 4 2x2 x 1 2x3 2x x2 1 73–76 ■ Graph the rational function f, and determine all vertical asymptotes from your graph. Then graph f and g in a sufﬁciently large viewing rectangle to show that they have the same end behavior. ✎ 73. f 74. f 75. f 76 2x2 6x 6 x 3 x3 6x2 5 x2 2x x3 2x2 16 x 2 x4 2x3 2x x 1 2 1 2, g 2x x 1 2, g x2 x 1 2, g x 1 2 1 x2 77–82 ■ Graph the rational function, and ﬁnd all vertical asymptotes, x- and y-intercepts, and local extrema, correct to the nearest decimal. Then use long division to ﬁnd a polynomial that has the same end behavior as the rational function, and graph both functions in a sufﬁciently large viewing rectangle to verify that the end behaviors of the polynomial and the rational function are the same. SE CT IO N 4.6 \| Rational Functions 357 y 2x2 5x 2x 3 y x4
3x3 x2 3x 3 x2 3x y x5 x3 1 x4 3x3 6 x 3 81. r x 1 2 y x4 80. x2 2 4 x2 x4 x2 1 82. r x 1 2 77. 78. 79. ▼ APPLICATIONS 83. Population Growth Suppose that the rabbit population on ✎ Mr. Jenkins’ farm follows the formula 3000t t 1 p t 1 2 where t 0 is the time (in months) since the beginning of the year. (a) Draw a graph of the rabbit population. (b) What eventually happens to the rabbit population? 84. Drug Concentration After a certain drug is injected into a patient, the concentration c of the drug in the bloodstream is monitored. At time t 0 (in minutes since the injection), the concentration (in mg/L) is given by 30t t 2 2 c t 2 1 (a) Draw a graph of the drug concentration. (b) What eventually happens to the concentration of drug in the bloodstream? 85. Drug Concentration A drug is administered to a patient, and the concentration of the drug in the bloodstream is monitored. At time t 0 (in hours since giving the drug), the concentration (in mg/L) is given by Graph the function c with a graphing device. (a) What is the highest concentration of drug that is reached in the patient’s bloodstream? (b) What happens to the drug concentration after a long period of time? (c) How long does it take for the concentration to drop below 0.3 mg/L? 86. Flight of a Rocket Suppose a rocket is ﬁred upward from the surface of the earth with an initial velocity √ (measured in m/s). Then the maximum height h (in meters) reached by the rocket is given by the function, 5t t 2 1 358 CHAPTER 4 \| Polynomial and Rational Functions R√ 2 2gR √ h √ 1 2 2 moves close to the lens? (c) What happens to the focusing distance y as the object where R 6.4 106 m is the radius of the earth and g 9.8 m/s2 is the acceleration due to gravity. Use a graphing device to draw a graph of the function h. (Note that h and √ must both be positive, so the viewing rectangle need not contain negative values.) What does the vertical asymptote represent physically
? x F y ▼ DISCOVE RY • DISCUSSION • WRITI NG 89. Constructing a Rational Function from Its Asymptotes Give an example of a rational function that has vertical asymptote x 3. Now give an example of one that has vertical asymptote x 3 and horizontal asymptote y 2. Now give an example of a rational function with vertical asymptotes x 1 and x 1, horizontal asymptote y 0, and x-intercept 4. 90. A Rational Function with No Asymptote Explain how you 87. The Doppler Effect As a train moves toward an observer can tell (without graphing it) that the function (see the ﬁgure), the pitch of its whistle sounds higher to the observer than it would if the train were at rest, because the crests of the sound waves are compressed closer together. This phenomenon is called the Doppler effect. The observed pitch P is a function of the speed √ of the train and is given by s0 √ b 1 2 √ P s0 P0 a where P0 is the actual pitch of the whistle at the source and 332 m/s is the speed of sound in air. Suppose that a s0 train has a whistle pitched at P0 tion asymptote of this function be interpreted physically? 440 Hz. Graph the func- using a graphing device. How can the vertical y P √ 2 1 88. Focusing Distance For a camera with a lens of ﬁxed focal length F to focus on an object located a distance x from the lens, the ﬁlm must be placed a distance y behind the lens, where F, x, and y are related by 1 x 1 y 1 F (See the ﬁgure.) Suppose the camera has a 55-mm lens (F 55). (a) Express y as a function of x and graph the function. (b) What happens to the focusing distance y as the object moves far away from the lens? r x 1 2 x 6 10 x 4 8x 2 15 has no x-intercept and no horizontal, vertical, or slant asymptote. What is its end behavior?’ 91. Graphs with Holes In this chapter we adopted the convention that in rational functions, the numerator and denominator don’t share a common factor. In this exercise we
consider the graph of a rational function that does not satisfy this rule. (a) Show that the graph of 3x 2 3x 6 x 2 r x 1 2 is the line y 3x 3 with the point [Hint: Factor. What is the domain of r?] 2, 9 1 2 removed. (b) Graph the rational functions: x2 x 20 x 5 2x2 x2 2x x u 1 2 92. Transformations of y 1/x 2 In Example 2 we saw that some simple rational functions can be graphed by shifting, stretching, or reﬂecting the graph of y 1/x. In this exercise we consider rational functions that can be graphed by transforming the graph of y 1/x 2, shown on the following page. (a) Graph the function by transforming the graph of y 1/x 2. (b) Use long division and factoring to show that the function 2x2 4x 5 x2 2x 1 x s 2 1 can be written as Then graph s by transforming the graph of y 1/x 2. (c) One of the following functions can be graphed by trans- 2 forming the graph of y 1/x 2; the other cannot. Use transformations to graph the one that can be, and explain why this method doesn’t work for the other one. CHAPTER 4 \| REVIEW ▼ P R O P E RTI LAS Quadratic Functions (pp. 292–295) A quadratic function is a function of the form ax2 bx c f x 1 2 It can be expressed in the standard form 1 by completing the square The graph of a quadratic function in standard form is a parabola with vertex a 0, then the quadratic function f has the minimum value k at h, k. 1 2 If x h. a 0, If x h. then the quadratic function f has the maximum value k at Polynomial Functions (p. 300) A polynomial function of degree n is a function P of the form an x n an1x n1 p a1x a0 P x 1 2 The numbers ai are the coefﬁcients of the polynomial; an is the leading coefﬁcient, and a0 is the constant coefﬁcient (or constant term). The graph of a polynomial function is a smooth, continuous curve. Real Z
eros of Polynomials (p. 304) 0. A zero of a polynomial P is a number c for which The following are equivalent ways of describing real zeros of polynomials: P c 1 2 1. c is a real zero of P. 2. 3. x c x c is a solution of the equation P 0. x 1 2 is a factor of P x. 2 1 4. c is an x-intercept of the graph of P. Multiplicity of a Zero (p. 309) A zero c of a polynomial P has multiplicity m if m is the highest power for which is a factor of CHAPTER 4 \| Review 359 2 3x2 x2 4x 4 p x 1 2 q 12x 3x2 x2 4x 4 x 1 2 y 1 0 y= 1 ≈ 1 x Local Maxima and Minima (p. 310) A polynomial function P of degree n has extrema (i.e., local maxima and minima). n 1 or fewer local Division of Polynomials (p. 315) D If P and D are any polynomials (with then we can divide P by D using either long division or (if D is linear) synthetic division. The result of the division can be expressed in one of the following equivalent forms In this division, P is the dividend, D is the divisor, Q is the quotient, and R is the remainder. When the division is continued to its completion, the degree of R will be less than the degree of D (or R 0 x. 1 2 2 Remainder Theorem (p. 318) When P(x) is divided by the linear divisor mainder is the constant P(c). So one way to evaluate a polynomial function P at c is to use synthetic division to divide P(x) by and observe the value of the remainder. x c, x c the re- D x 2 1 Rational Zeros of Polynomials (pp. 322–323) If the polynomial P given by an x n an1x n1 p a1x a0 P x 1 2 has integer coefﬁcients, then all the rational zeros of P have the form x ; p q where p is a divisor of the constant term a0 and q is a divisor of the leading coefﬁcient an. So to �
��nd all the rational zeros of a polynomial, we list all the possible rational zeros given by this principle and then check to see which actually are zeros by using synthetic division. 360 CHAPTER 4 \| Polynomial and Rational Functions Descartes’ Rule of Signs (p. 325) Let P be a polynomial with real coefﬁcients. Then: The number of positive real zeros of P either is the number of changes of sign in the coefﬁcients of an even number. or is less than that by P x 1 2 Linear and Quadratic Factors Theorem (p. 341) Every polynomial with real coefﬁcients can be factored into linear and irreducible quadratic factors with real coefﬁcients. Rational Functions (p. 344) A rational function r is a quotient of polynomial functions: The number of negative real zeros of P either is the number of changes of sign in the coefﬁcients of by an even number. or is less than that Upper and Lower Bounds Theorem (p. 326) Suppose we divide the polynomial P by the linear expression and arrive at the result # Q and the coefﬁcients of Q, followed by r, are all nonnega If tive, then c is an upper bound for the zeros of P. c 0 x c c 0 and the coefﬁcients of Q, followed by r (including zero If coefﬁcients), are alternately nonnegative and nonpositive, then c is a lower bound for the zeros of P. The Fundamental Theorem of Algebra, Complete Factorization, and the Zeros Theorem (pp. 335–337) Every polynomial P of degree n with complex coefﬁcients has exactly n complex zeros, provided that each zero of multiplicity m is counted m times. P factors into n linear factors as follows: x c12 1 x c22 where a is the leading coefﬁcient of P and zeros of P. x cn2 c1,..., 1 c1, a P x 1 2 1 # # # Conjugate Zeros Theorem (p. 340) If the polynomial P has real coefﬁcients and if P, then its complex conjugate a bi is also a
zero of P. a bi cn are the is a zero of We generally assume that the polynomials P and Q have no factors in common. Asymptotes (pp. 344–345) The line is a vertical asymptote of the function y f x a if x 1 2 or y q as x a or is a horizontal asymptote of the function y b as x q or x q x a y q y b if The line y f x 1 2 Asymptotes of Rational Functions (pp. 345–348) Let be a rational function. 2 2 The vertical asymptotes of r are the lines zero of Q. x a where a is a If the degree of P is less than the degree of Q, then the horizontal asymptote of r is the line. y 0 If the degrees of P and Q are the same, then the horizontal asymptote of r is the line, where y b b leading coefficient of P leading coefficient of Q If the degree of P is greater than the degree of Q, then r has no horizontal asymptote. ▼ CO N C E P T S U M MARY Section 4.1 ■ 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 Section 4.2 ■ 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 Section 4.3 ■ 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 Review Exercises 1(a)–4(a) 1(b)–4(b) 5–6 7–8 Review Exercises 9–14 15–20, 39–46 15–20, 39–46 15–16, 39–46 17–20 Review Exercises 29–30 23–28 31–32, 35–36 33–34 47–50 CH
APTER 4 \| Review 361 Review Exercises 37(a)–38(a) 37(b)–38(b) 51–60 61–64 Review Exercises 51–60 39–46, 51–60 48–49 65–66 Review Exercises 67–76 67–76 67–76 75–76 Section 4.4 ■ 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 real zeros of a polynomial ■ Use algebra and graphing devices to solve polynomial equations Section 4.5 ■ State the Fundamental Theorem of Algebra ■ Factor a polynomial completely (into linear factors) over the complex numbers ■ Determine the multiplicity of a zero of a polynomial ■ Use the Conjugate Roots Theorem to ﬁnd polynomials with speciﬁed zeros ■ Factor a polynomial completely (into linear and quadratic factors) over the real numbers Section 4.6 ■ Find the vertical asymptotes of a rational function ■ Find the horizontal asymptote of a rational function ■ Use asymptotes to graph a rational function ■ Find the slant asymptote of a rational function ▼ E X E RC I S E S 1–4 ■ A quadratic function is given. (a) Express the function in standard form. (b) Graph the function. 1. 3 4x 1 1 8x x 2 2. 4. f g x x 1 1 2 2 2x 2 12x 12 6x 3x 2 5–6 ■ Find the maximum or minimum value of the quadratic function. 5. f x 1 2 2x 2 4x 5 6. A stone is thrown upward from the top of a building. Its height (in feet) above the ground after t seconds is given by the funct tion 2 the stone reach?. What maximum height does 16t 2 48t 32 h 1 8. The proﬁt P (in dollars) generated by selling x units of a cer- tain commodity is given by the function 1500 12x 0.004x 2 P x 1 2 What is the maximum proﬁt, and how
many units must be sold to generate it? 9–14 ■ Graph the polynomial by transforming an appropriate graph of the form y x n. Show clearly all x- and y-intercepts. 9. 11. 13 64 2 x 1 1 32 2 x 1 4 32 5 2 1 10. 12. 14 2x3 16 81 96 2 15–16 ■ A polynomial function P is given. (a) Determine the multiplicity of each zero of P. (b) Sketch a graph of P. 15 16 17–20 ■ Use a graphing device to graph the polynomial. Find the x- and y-intercepts and the coordinates of all local extrema, correct to the nearest decimal. Describe the end behavior of the polynomial. 17. P x 1 2 x 3 4x 1 18. P x 1 2 2x3 6x2 2 19. 20. P P x x 1 1 2 2 3x 4 4x 3 10x 1 x 5 x 4 7x 3 x 2 6x 3 21. The strength S of a wooden beam of width x and depth y is given by the formula S 13.8xy 2. A beam is to be cut from a log of diameter 10 in., as shown in the ﬁgure. (a) Express the strength S of this beam as a function of x only. (b) What is the domain of the function S? (c) Draw a graph of S. (d) What width will make the beam the strongest? 22. A small shelter for delicate plants is to be constructed of thin plastic material. It will have square ends and a rectangular top and back, with an open bottom and front, as shown in the ﬁgure. The total area of the four plastic sides is to be 1200 in2. (a) Express the volume V of the shelter as a function of the depth x. (b) Draw a graph of V. (c) What dimensions will maximize the volume of the shelter? y x x 362 CHAPTER 4 \| Polynomial and Rational Functions 23–30 ■ Find the quotient and remainder. 50. Prove that the equation 3x 4 5x 2 2 0 has no real root. 23. 25. 27. 29. x 2 3x 5 x 2 x 3 x 2 11x 2 x 4 x 4 8x 2 2x 7 x 5 2x 3 x 2 8x 15
x 2 2x 1 24. 26. 28. 30. x 2 x 12 x 3 x 3 2x 2 10 x 3 2x 4 3x 3 12 x 4 x 4 2x 2 7x x 2 x 3 31–32 ■ Find the indicated value of the polynomial using the Remainder Theorem. 31. 32. P 1 Q x 2 x 2x3 9x2 7x 13 ; ﬁnd x 4 4x 3 7x 2 10x 15 P 5 1 2 ; ﬁnd 2 1 33. Show that 1 2 is a zero of the polynomial Q 3 1 2 2x4 x3 5x2 10x 4 P x 1 2 34. Use the Factor Theorem to show that x 4 is a factor of the polynomial P x 1 2 x5 4x4 7x3 23x2 23x 12 35. What is the remainder when the polynomial x500 6x201 x2 2x 4 P x 1 2 is divided by x 1? 36. What is the remainder when x101 x4 2 is divided by x 1? 37–38 ■ A polynomial P is given. (a) List all possible rational zeros (without testing to see whether they actually are zeros). (b) Determine the possible number of positive and negative real zeros using Descartes’ Rule of Signs. 37. 38. P P x x 1 1 2 2 x5 6x3 x2 2x 18 6x 4 3x 3 x 2 3x 4 39–46 ■ A polynomial P is given. (a) Find all real zeros of P, and state their multiplicities. (b) Sketch the graph of P. 39. 41. 43. 44. 45. 46 3x 2 4x x 4 5x 2 4 2 1 x x P 42. 40. x 3 16x x4 x3 2x2 P x4 2x3 7x2 8x 12 x4 2x3 2x2 8x 8 2x4 x3 2x2 3x 2 9x 5 21x 4 10x 3 6x 2 3x 1 1 2 47. Find a polynomial of degree 3 with constant coefﬁcient 12 and zeros 1 2, 2, and 3 53. 52. 51. 51–60 ■ Find all rational, irrational, and complex zeros (and state their multiplic
ities). Use Descartes’ Rule of Signs, the Upper and Lower Bounds Theorem, the Quadratic Formula, or other factoring techniques to help you whenever possible. x 3 3x 2 13x 15 2x3 5x2 6x 9 x 4 6x 3 17x 2 28x 20 x 4 7x 3 9x 2 17x 20 x5 3x4 x3 11x2 12x 4 x 4 81 x 6 64 18x 3 3x 2 4x 1 6x 4 18x 3 6x 2 30x 36 x 4 15x 2 54 57. 58. 56. 55. 59. 60. 54 61–64 ■ Use a graphing device to ﬁnd all real solutions of the equation. 61. 2x 2 5x 3 62. x 3 x 2 14x 24 0 63. x 4 3x 3 3x 2 9x 2 0 64. x 5 x 3 65–66 ■ A polynomial function P is given. Find all the real zeros of P, and factor P completely into linear and irreducible quadratic factors with real coefﬁcients. 65. P x 1 2 x 3 2x 4 66. P x 1 2 x4 3x2 4 67–72 ■ Graph the rational function. Show clearly all x- and y-intercepts and asymptotes. 3x 12 x 1 68. 67. x x r r 2 2 1 1 69. r 71 2x 8 x 2 9 2x 2 1 70. r 72 2x 2 6x 7 x 4 x 3 27 x 4 73–76 ■ Use a graphing device to analyze the graph of the rational function. Find all x- and y-intercepts and all vertical, horizontal, and slant asymptotes. If the function has no horizontal or slant asymptote, ﬁnd a polynomial that has the same end behavior as the rational function. x 3 2x 6 x 3 8 2x 7 x2 9 2x3 x2 x 1 x 2 x 2 76. 74. 73. 75 48. Find a polynomial of degree 4 that has integer coefﬁcients and 77. Find the coordinates of all points of intersection of the zeros 3i and 4, with 4 a double zero. graphs of 49. Does there exist a polynomial of degree 4 with integer coefﬁ
cients that has zeros i, 2i, 3i, and 4i? If so, ﬁnd it. If not, explain why. y x4 x2 24x and y 6x3 20 ■ CHAPTER 4 \| TEST 1. Express the quadratic function f x 1 2 x 2 x 6 in standard form, and sketch its graph. 2. Find the maximum or minimum value of the quadratic function g 2x 2 6x 3. x 1 2 3. A cannonball ﬁred out to sea from a shore battery follows a parabolic trajectory given by the graph of the equation 10x 0.01x 2 h x 1 2 x where h 2 1 distance of x feet. is the height of the cannonball above the water when it has traveled a horizontal (a) What is the maximum height that the cannonball reaches? (b) How far does the cannonball travel horizontally before splashing into the water? h(x) x 3 27 4. Graph the polynomial 5. (a) Use synthetic division to ﬁnd the quotient and remainder when x 4 4x 2 2x 5 is, showing clearly all x- and y-intercepts. x 2 P x 2 1 2 1 divided by x 2. (b) Use long division to ﬁnd the quotient and remainder when 2x 5 4x 4 x 3 x 2 7 is divided by 2x 2 1. 6. Let x P 2x3 5x2 4x 3. (a) List all possible rational zeros of P. 1 2 (b) Find the complete factorization of P. (c) Find the zeros of P. (d) Sketch the graph of P. 7. Find all real and complex zeros of P 2 x 1 x x 3 x 2 4x 6. x4 2x3 5x2 8x 4. 8. Find the complete factorization of 9. Find a fourth-degree polynomial with integer coefﬁcients that has zeros 3i and 1, with 1 P 1 2 a zero of multiplicity 2. 10. Let P x 2x4 7x3 x2 18x 3. 1 2 (a) Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros P can have. (b) Show that 4 is an upper bound and 1 is a lower bound for the real zeros
of P. (c) Draw a graph of P, and use it to estimate the real zeros of P, correct to two decimal places. (d) Find the coordinates of all local extrema of P, correct to two decimals. 11. Consider the following rational functions: 1 r x 2x 1 x3 27 x2 4 x2 x 6 x3 9x x 2 x2 25 (a) Which of these rational functions has a horizontal asymptote? x2 b) Which of these functions has a slant asymptote? (c) Which of these functions has no vertical asymptote? (d) Graph y u may have. x 1 2, showing clearly any asymptotes and x- and y-intercepts the function (e) Use long division to ﬁnd a polynomial P that has the same end behavior as t. Graph both P and t on the same screen to verify that they have the same end behavior. 363 FITTING POLYNOMIAL CURVES TO DATA We have learned how to ﬁt a line to data (see Focus on Modeling, page 192). The line models the increasing or decreasing trend in the data. If the data exhibit more variability, such as an increase followed by a decrease, then to model the data, we need to use a curve rather than a line. Figure 1 shows a scatter plot with three possible models that appear to ﬁt the data. Which model ﬁts the data best? y y y x x x Linear model Quadratic model Cubic model FIGURE 1 Polynomial Functions as Models Polynomial functions are ideal for modeling data where the scatter plot has peaks or valleys (that is, local maxima or minima). For example, if the data have a single peak as in Figure 2(a), then it may be appropriate to use a quadratic polynomial to model the data. The more peaks or valleys the data exhibit, the higher the degree of the polynomial needed to model the data (see Figure 2). y y y x x x (a) (b) (c) FIGURE 2 Graphing calculators are programmed to ﬁnd the polynomial of best ﬁt of a speciﬁed degree. As is the case for lines (see page 193), a polynomial of a given degree ﬁts the data
best if the sum of the squares of the distances between the graph of the polynomial and the data points is minimized. E X AM P L E 1 \| Rainfall and Crop Yield Rain is essential for crops to grow, but too much rain can diminish crop yields. The data give rainfall and cotton yield per acre for several seasons in a certain county. (a) Make a scatter plot of the data. What degree polynomial seems appropriate for mod- eling the data? (b) Use a graphing calculator to ﬁnd the polynomial of best ﬁt. Graph the polynomial on the scatter plot. 364 c) Use the model that you found to estimate the yield if there are 25 in. of rainfall. Fitting Polynomial Curves to Data 365 Season Rainfall (in.) Yield (kg/acre 10 23.3 20.1 18.1 12.5 30.9 33.6 35.8 15.5 27.6 34.5 5311 4382 3950 3137 5113 4814 3540 3850 5071 3881 ▼ SO LUTI O N (a) The scatter plot is shown in Figure 3. The data appear to have a peak, so it is appro- priate to model the data by a quadratic polynomial (degree 2). 6000 10 1500 40 FIGURE 3 Scatter plot of yield vs. rainfall data (b) Using a graphing calculator, we ﬁnd that the quadratic polynomial of best ﬁt is y 12.6x2 651.5x 3283.2 The calculator output and the scatter plot, together with the graph of the quadratic model, are shown in Figure 4. 6000 10 1500 40 (b) (a) FIGURE 4 (c) Using the model with x 25, we get y 12.6 25 2 651.5 25 3283.2 5129.3 1 We estimate the yield to be about 5130 kg per acre. 2 2 1 ▲ 366 Focus on Modeling Cod Redfish Hake Otoliths for several ﬁsh species. E X AM P L E 2 \| Length-at-Age Data for Fish Otoliths (“earstones”) are tiny structures that are found in the heads of ﬁsh. Microscopic growth rings on the otoliths, not unlike growth rings on a tree,
record the age of a ﬁsh. The table gives the lengths of rock bass caught at different ages, as determined by the otoliths. Scientists have proposed a cubic polynomial to model this data. (a) Use a graphing calculator to ﬁnd the cubic polynomial of best ﬁt for the data. (b) Make a scatter plot of the data and graph the polynomial from part (a). (c) A ﬁsherman catches a rock bass 20 in. long. Use the model to estimate its age. Age (yr) Length (in.) Age (yr) Length (in.).8 8.8 8.0 7.9 11.9 14.4 14.1 15.8 15.6 17.8 9 9 10 10 11 12 12 13 14 14 18.2 17.1 18.8 19.5 18.9 21.7 21.9 23.8 26.9 25.1 ▼ SO LUTI O N (a) Using a graphing calculator (see Figure 5(a)), we ﬁnd the cubic polynomial of best ﬁt: y 0.0155x3 0.372x2 3.95x 1.21 (b) The scatter plot of the data and the cubic polynomial are graphed in Figure 5(b). 30 0 0 15 (b) (a) FIGURE 5 (c) Moving the cursor along the graph of the polynomial, we ﬁnd that y 20 when x 10.8. Thus, the ﬁsh is about 11 years old. ▲ Problems 1. Tire Inﬂation and Treadwear Car tires need to be inﬂated properly. Overinﬂation or underinﬂation can cause premature treadwear. The data and scatter plot on the next page show tire life for different inﬂation values for a certain type of tire. (a) Find the quadratic polynomial that best ﬁts the data. (b) Draw a graph of the polynomial from part (a) together with a scatter plot of the data. (c) Use your result from part (b) to estimate the pressure that gives the longest tire life. Fitting Polynomial Curves to Data 367 Pressure (lb/in2) Tire life (mi) 26 28 31 35 38 42
45 50,000 66,000 78,000 81,000 74,000 70,000 59,000 y (mi) 80,000 70,000 60,000 50,000 0 0 25 30 35 40 45 50 x (lb/in2) 2. Too Many Corn Plants per Acre? The more corn a farmer plants per acre, the greater is the yield the farmer can expect, but only up to a point. Too many plants per acre can cause overcrowding and decrease yields. The data give crop yields per acre for various densities of corn plantings, as found by researchers at a university test farm. (a) Find the quadratic polynomial that best ﬁts the data. (b) Draw a graph of the polynomial from part (a) together with a scatter plot of the data. (c) Use your result from part (b) to estimate the yield for 37,000 plants per acre. Density (plants/acre) Crop yield (bushels/acre) 15,000 20,000 25,000 30,000 35,000 40,000 45,000 50,000 43 98 118 140 142 122 93 67 3. How Fast Can You List Your Favorite Things? If you are asked to make a list of objects in a certain category, how fast you can list them follows a predictable pattern. For example, if you try to name as many vegetables as you can, you’ll probably think of several right away— for example, carrots, peas, beans, corn, and so on. Then after a pause you might think of ones you eat less frequently—perhaps zucchini, eggplant, and asparagus. Finally, a few more exotic vegetables might come to mind—artichokes, jicama, bok choy, and the like. A psychologist performs this experiment on a number of subjects. The table below gives the average number of vegetables that the subjects named by a given number of seconds. (a) Find the cubic polynomial that best ﬁts the data. (b) Draw a graph of the polynomial from part (a) together with a scatter plot of the data. (c) Use your result from part (b) to estimate the number of vegetables that subjects would be able to name in 40 seconds. (d) According to the model, how long (to the nearest 0.1 s) would it take a person to name ﬁ
ve vegetables? Seconds Number of Vegetables 1 2 5 10 15 20 25 30 2 6 10 12 14 15 18 21 368 Focus on Modeling 4. Clothing Sales Are Seasonal Clothing sales tend to vary by season, with more clothes sold in spring and fall. The table gives sales ﬁgures for each month at a certain clothing store. (a) Find the quartic (fourth-degree) polynomial that best ﬁts the data. (b) Draw a graph of the polynomial from part (a) together with a scatter plot of the data. (c) Do you think that a quartic polynomial is a good model for these data? Explain. Month Sales ($) January February March April May June July August September October November December 8,000 18,000 22,000 31,000 29,000 21,000 22,000 26,000 38,000 40,000 27,000 15,000 5. Height of a Baseball A baseball is thrown upward, and its height measured at 0.5-second intervals using a strobe light. The resulting data are given in the table. (a) Draw a scatter plot of the data. What degree polynomial is appropriate for modeling the data? (b) Find a polynomial model that best ﬁts the data, and graph it on the scatter plot. (c) Find the times when the ball is 20 ft above the ground. (d) What is the maximum height attained by the ball? Time (s) Height (ft) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 4.2 26.1 40.1 46.0 43.9 33.7 15.8 6. Torricelli’s Law Water in a tank will ﬂow out of a small hole in the bottom faster when the t tank is nearly full than when it is nearly empty. According to Torricelli’s Law, the height of water remaining at time t is a quadratic function of t. h 1 2 A certain tank is ﬁlled with water and allowed to drain. The height of the water is measured at different times as shown in the table. (a) Find the quadratic polynomial that best ﬁts the data. (b) Draw a graph of the polynomial from part (a) together with a scatter plot of the data. (c) Use
your graph from part (b) to estimate how long it takes for the tank to drain completely. Time (min) Height (ft) 0 4 8 12 16 5.0 3.1 1.9 0.8 0.2 CHAPTER 5 Exponential and Logarithmic Functions 5.1 Exponential Functions 5.2 Logarithmic Functions 5.3 Laws of Logarithms 5.4 Exponential and Logarithmic Equations 5.5 Modeling with Exponential and Logarithmic Functions © Population explosion? Cities and towns all over the world have experienced huge increases in population over the past century. The above photos of Hollywood in the 1920s and in 2000 tell the population story quite dramatically. But to ﬁnd out where population is really headed, we need a mathematical model. Population grows in much the same way as money grows in a bank account: the more money in the account, the more interest is paid. In the same way, the more people there are in the world, the more babies are born. This type of growth is modeled by exponential functions. According to some exponential models, world population will double in successive forty-year periods, resulting in a population explosion. However, different exponential models, which take into account the limited resources available for growth, predict that world population will eventually stabilize at a level that our planet can support (see Exercise 65, Section 5.1, and Focus on Modeling: Fitting Exponential and Power Curves to Data, page 431). 369369 369 370 CHAPTER 5 \| Exponential and Logarithmic Functions 5.1 Exponential Functions LEARNING OBJECTIVES After completing this section, you will be able to: ■ Evaluate exponential functions ■ Graph exponential functions ■ Evaluate and graph the natural exponential function ■ Find compound interest ■ Find continuously compounded interest In this chapter we study a new class of functions called exponential functions. For example, 2x f x 1 2 is an exponential function (with base 2). Notice how quickly the values of this function increase: f 3 1 10 f 1 23 8 210 1024 2 2 f 30 230 1,073,741,824 1 Compare this with the function. The point is that when the variable is in the exponent, even a small change in the variable can cause a dramatic change in the value of the function. 302 900, where x2 2 x 30 g g 2 1 2 1 In spite of this incomprehensibly huge growth, exponential functions are
appropriate for modeling population growth for all living things, from bacteria to elephants. To understand how a population grows, consider the case of a single bacterium, which divides every hour. After one hour we would have 2 bacteria, after two hours 22 or 4 bacteria, after three hours 23 or 8 bacteria, and so on. After x hours we would have 2x bacteria. This leads us to model the bacteria population by the function 2x The principle governing population growth is the following: The larger the population, the greater the number of offspring. This same principle is present in many other real-life situations. For example, the larger your bank account, the more interest you get. So we also use exponential functions to ﬁnd compound interest. To study exponential functions, we must ﬁrst deﬁne what we mean by the exponential expression a x when x is any real number. ■ Exponential Functions In Section P.5 we deﬁned a x for a 0 and x a rational number, but we have not yet deﬁned or 2p? To deﬁne a x when x is irrational, we apirrational powers. So what is meant by proximate x by rational numbers. 513 The Laws of Exponents are listed on page 21. SE CTI ON 5.1 \| Exponential Functions 371 For example, since 13 1.73205... is an irrational number, we successively approximate a13 by the following rational powers: a1.7, a1.73, a1.732, a1.7320, a1.73205,... a13 Intuitively, we can see that these rational powers of a are getting closer and closer to. It can be shown by using advanced mathematics that there is exactly one number that these powers approach. We deﬁne to be this number. a13 For example, using a calculator, we ﬁnd 513 51.732 16.2411... The more decimal places of of 513. It can be proved that the Laws of Exponents are still true when the exponents are real 13 we use in our calculation, the better our approximation numbers. EXPONENTIAL FUNCTIONS The exponential function with base a is deﬁned for all real numbers x by where a 0 and a 1. ax f x 1 2 We assume that a 1 because the function x 1 Here are some examples
of exponential functions: f 2 1x 1 is just a constant function. f x 1 2 2x g x 1 2 3x h 10 x x 1 2 Base 2 Base 3 Base 10 E X AM P L E 1 \| Evaluating Exponential Functions 3x Let (a) f 1 f x 2 2, and evaluate the following: p (b) (c ▼ SO LUTI O N We use a calculator to obtain the values of f. 2 1 1 (d) f 12 Calculator keystrokes 2^3 ENTER Output 9 (^3 _ ( ) )32 ENTER 0.4807498 32 9 32/3 0.4807 3p 31.544 (a) (b) (c 12 P^3 ENTER A f 312 4.7288 (d) ✎ Practice what you’ve learned: Do Exercise 5. 21^3 ENTER B 31.5442807 4.7288043 ▲ ■ Graphs of Exponential Functions We ﬁrst graph exponential functions by plotting points. We will see that the graphs of such functions have an easily recognizable shape. 372 CHAPTER 5 \| Exponential and Logarithmic Functions E X AM P L E 2 \| Graphing Exponential Functions by Plotting Points Draw the graph of each function. (a) f x 1 2 3x (b ▼ SO LUTI O N We calculate values of graphs in Figure 1. f x 1 2 and g x 1 2 and plot points to sketch the ff x 1 2 gg x 1 2 1 27 1 9 1 3 1 3 9 27 x 1 3 B A 27 9 3 1 1 3 1 9 1 27 y=! @˛1 3 y 1 0 y=3˛ 1 x FIGURE 1 Notice that g x 1 2 x 1 3 b 1 3x a 3x f x 1 2 Reﬂecting graphs is explained in Section 3.5. so we could have obtained the graph of g from the graph of f by reﬂecting in the y-axis. ✎ Practice what you’ve learned: Do Exercise 17. ▲ 1 2 f x in- 2x To see just how quickly creases, let’s perform the following thought experiment. Suppose we start with a piece of paper that is a thousandth of an inch thick, and we fold it in half 50 times. Each time we fold the paper, the thickness of the paper
stack doubles, so the thickness of the resulting stack would be 250/1000 inches. How thick do you think that is? It works out to be more than 17 million miles! FIGURE 2 A family of exponential functions See Section 4.6, page 345, where the “arrow notation” used here is explained. f Figure 2 shows the graphs of the family of exponential functions 0, 1 x for various 2 because a0 1 for values of the base a. All of these graphs pass through the point a 0. You can see from Figure 2 that there are two kinds of exponential functions: If 0 a 1, the exponential function decreases rapidly. If a 1, the function increases rapidly (see the margin note). 1 2 1 ax y=! @˛1 3 y=! @˛1 2 y=! @˛1 10 y=! @˛1 5 y=10 ˛ y=5˛ y 2 0 y=3˛ y=2˛ 1 x The x-axis is a horizontal asymptote for the exponential function f. This is because when a 1, we have a x 0 as x q, and when 0 a 1, we have a x 0 as x 2 1 ax SE CTI ON 5.1 \| Exponential Functions 373 x q (see Figure 2). Also, a x 0 for all and range x ax. These observations are summarized in the following box. f, so the function 0, q x 1 2 has domain 1 2 GRAPHS OF EXPONENTIAL FUNCTIONS The exponential function f x ax a 0, a 1 2 1 0, q has domain tote of f. The graph of f has one of the following shapes. 2. The line y 0 (the x-axis) is a horizontal asymp- and range 1 1 2 y 0 (0, 1) x y 0 (0, 1) x Ï=a˛ for a>1 Ï=a˛ for 0<a<1 E X AM P L E 3 \| Identifying Graphs of Exponential Functions Find the exponential function (a) f x 1 2 a x whose graph is given. (b) y 5 0 _1 (2, 25) 1 2 x _3 y 1 0 1!3, @ 8 3 x f ▼ SO LUTI O N (a) Since 2 (b) Since. So 3 ✎ Practice what you’
ve learned: Do Exercise 21., we see that the base is a 5. So f a2 25 a3 1 8, we see that the base is 5x. x 1. 2B A ▲ In the next example we see how to graph certain functions, not by plotting points, but by taking the basic graphs of the exponential functions in Figure 2 and applying the shifting and reﬂecting transformations of Section 3.5 © The Gateway Arch in St. Louis, Missouri, is shaped in the form of the graph of a combination of exponential functions (not a parabola, as it might ﬁrst appear). Speciﬁcally, it is a catenary, which is the graph of an equation of the form ebx ebx y a 1 2 (see Exercise 53). This shape was chosen because it is optimal for distributing the internal structural forces of the arch. Chains and cables suspended between two points (for example, the stretches of cable between pairs of telephone poles) hang in the shape of a catenary. 374 CHAPTER 5 \| Exponential and Logarithmic Functions E X AM P L E 4 \| Transformations of Exponential Functions 2x x 2 to sketch the graph of each function. Use the graph of (a) x f 1 1 2x 2x 2x1 (b) (c Shifting and reﬂecting of graphs is explained in Section 3.5. ▼ SO LUTI O N (a) To obtain the graph of it upward 1 unit. Notice from Figure 3(a) that the line y 1 is now a horizontal asymptote. 1 2 1 2x, we start with the graph of f x 2x and shift g x 1 2 the graph of (b) Again we start with the graph of x h 2 (c) This time we start with the graph of k x 2 shown in Figure 3(b). 2x f x 2 shown in Figure 3(c). get the graph of 2x1 2x x f 1 1 1 2x 1 2, but here we reﬂect in the x-axis to get and shift it to the right by 1 unit to y y=1+2˛ Horizontal asymptote 2 y=2˛ 0 1 x y 1 0 _1 y=2˛ x 1 y=_2˛ y 1 0 y=2˛ y=2˛–
¡ 1 x FIGURE 3 (a) (b) (c) ✎ Practice what you’ve learned: Do Exercises 27, 29, and 33. ▲ E X AM P L E 5 \| Comparing Exponential and Power Functions Compare the rates of growth of the exponential function g function 1 rectangles. (a) and the power by drawing the graphs of both functions in the following viewing x 2 0, 3 0, 8 by x 2 2 1 f x 2x 3 4 (b) (c) 3 3 0, 6 4 0, 20 4 3 4 0, 25 by 3 by 4 0, 1000 3 4 ▼ SO LUTI O N (a) Figure 4(a) shows that the graph of g 1 at x 2. 2x than, the graph of f x 1 2 takes that of g x x 2 when x 4. 1 2 much larger than g x 1 2 x 2. (b) The larger viewing rectangle in Figure 4(b) shows that the graph of (c) Figure 4(c) gives a more global view and shows that when x is large, x 2 x 2 catches up with, and becomes higher 2x over- f x 1 2 2x is f x 1 2 SE CTI ON 5.1 \| Exponential Functions 375 8 0 FIGURE 4 Ï=2x ˝=≈ 3 25 0 ˝=≈ Ï=2x 1000 Ï=2x ˝=≈ 6 0 (a) (b) (c) ✎ Practice what you’ve learned: Do Exercise 47. 20 ▲ ■ The Natural Exponential Function Any positive number can be used as the base for an exponential function, but some bases are used more frequently than others. We will see in the remaining sections of this chapter that the bases 2 and 10 are convenient for certain applications, but the most important base is the number denoted by the letter e. The number e is deﬁned as the value that approaches as n becomes large. (In calculus this idea is made more precise through the concept of a limit. See Exercise 51.) The table in the margin shows the values of the expression for increasingly large values of n. It appears that, correct to ﬁve decimal places, e 2.71828; in fact, the approximate value to 20 decimal places is 1 1/n 1 2 1 2 n 1 1/n n
e 2.71828182845904523536 It can be shown that e is an irrational number, so we cannot write its exact value in decimal form. Why use such a strange base for an exponential function? It might seem at ﬁrst that a base such as 10 is easier to work with. We will see, however, that in certain applications the number e is the best possible base. In this section we study how e occurs in the description of compound interest. THE NATURAL EXPONENTIAL FUNCTION The natural exponential function is the exponential function with base e. It is often referred to as the exponential function 10 100 1000 10,000 100,000 1,000,000 n 1 1 n b a 2.00000 2.48832 2.59374 2.70481 2.71692 2.71815 2.71827 2.71828 The notation e was chosen by Leonhard Euler (see page 100), probably because it is the ﬁrst letter of the word exponential. y y=3˛ y=2˛ y=e ˛ 1 0 1 of x Since 2 e 3, the graph of the natural exponential function lies between the graphs y 2x, as shown in Figure 5. Scientiﬁc calculators have a special key for the function. We use this key in y 3x e x and x f 1 2 FIGURE 5 Graph of the natural exponential function the next example. E X AM P L E 6 \| Evaluating the Exponential Function Evaluate each expression correct to ﬁve decimal places. (a) 2e0.53 e4.8 (b) (c) e3 key on a calculator to evaluate the exponential eX ▼ SO LUTI O N We use the function. (a) e3 20.08554 ✎ Practice what you’ve learned: Do Exercise 9. (b) 2e0.53 1.17721 (c) e4.8 121.51042 ▲ 376 CHAPTER 5 \| Exponential and Logarithmic Functions E X AM P L E 7 \| Transformations of the Exponential Function y 1 0 y=e–˛ FIGURE 6 y=e ˛ 1 x Sketch the graph of each function. (a) ex (b) x x g f 3e0.5x 1 2 ▼ SO LUTI O N (a) We
start with the graph of y e x and reﬂect in the y-axis to obtain the graph of 2 1 y ex as in Figure 6. (b) We calculate several values, plot the resulting points, then connect the points with a smooth curve. The graph is shown in Figure 7. x 3 2 1 0 1 2 3 3e0.5x ff x 1 2 0.67 1.10 1.82 3.00 4.95 8.15 13.45 y 12 9 6 3 0 y=3e0.5x 3 x _3 FIGURE 7 ✎ Practice what you’ve learned: Do Exercise 15. ▲ E X AM P L E 8 \| An Exponential Model for the Spread of a Virus An infectious disease begins to spread in a small city of population 10,000. After t days, the number of people who have succumbed to the virus is modeled by the function √ t 1 2 10,000 5 1245e0.97t (a) How many infected people are there initially (at time t 0)? (b) Find the number of infected people after one day, two days, and ﬁve days. (c) Graph the function √, and describe its behavior. ▼ SO LUTI O N (a) Since 0 √ 1 2 initially have the disease. (b) Using a calculator, we evaluate 10,000/ 5 1245e0 1 10,000/1250 8, we conclude that 8 people, and √ 5 1 2 and then round off to obtain the 3000 following values. 12 0 FIGURE 8 √ t 1 2 10,000 5 1245e0.97t Days Infected people 1 2 5 21 54 678 (c) From the graph in Figure 8 we see that the number of infected people ﬁrst rises slowly, then rises quickly between day 3 and day 8, and then levels off when about 2000 people are infected. ✎ Practice what you’ve learned: Do Exercise 63. ▲ The graph in Figure 8 is called a logistic curve or a logistic growth model. Curves like it occur frequently in the study of population growth. (See Exercises 63–66.) SE CTI ON 5.1 \| Exponential Functions 377 ■ Compound Interest Exponential functions occur in calculating compound interest. If an amount of money P, called the principal, is invested at an interest rate i per
time period, then after one time period the interest is Pi, and the amount A of money is A P Pi P 1 i 1 2 1 i If the interest is reinvested, then the new principal is 1 i other time period is A P. In general, after k periods the amount is the amount is A P 1 1 i P, and the amount after an1 2. Similarly, after a third time period Notice that this is an exponential function with base 1 i. A P 1 i 1 k 2 If the annual interest rate is r and if interest is compounded n times per year, then in each time period the interest rate is i r/n, and there are nt time periods in t years. This leads to the following formula for the amount after t years. COMPOUND INTEREST Compound interest is calculated by the formula nt r is often referred to as the nominal annual interest rate. where A 1 amount after t years t 2 P principal r interest rate per year n number of times interest is compounded per year t number of years E X AM P L E 9 \| Calculating Compound Interest A sum of $1000 is invested at an interest rate of 12% per year. Find the amounts in the account after 3 years if interest is compounded annually, semiannually, quarterly, monthly, and daily. ▼ SO LUTI O N We use the compound interest formula with P $1000, r 0.12, and t 3. Compounding n Amount after 3 years Annual Semiannual Quarterly 1 2 4 1000 1000 1000 Monthly 12 1000 Daily 365 1000 1 0.12 1 b a 1 0.12 2 b a 1 0.12 4 b a 1 0.12 12 b a 1 0.12 a 365 12 3 1 2 365 3 1 2 $1404.93 $1418.52 $1425.76 $1430.77 $1433.24 ✎ Practice what you’ve learned: Do Exercise 69. ▲ 378 CHAPTER 5 \| Exponential and Logarithmic Functions We see from Example 9 that the interest paid increases as the number of compounding periods n increases. Let’s see what happens as n increases indeﬁnitely. If we let m n/r, then /r a nt rt rt m t d d 2 1 Recall that as m becomes large, the quantity approaches the number e. Thus, 1 the amount approaches A Pe rt. This expression gives the
amount when the interest is compounded at “every instant.” 2 1 1/m m CONTINUOUSLY COMPOUNDED INTEREST Continuously compounded interest is calculated by the formula Pe rt A t 1 2 where A 1 amount after t years t 2 P principal r interest rate per year t number of years E X AM P L E 10 \| Calculating Continuously Compounded Interest Find the amount after 3 years if $1000 is invested at an interest rate of 12% per year, compounded continuously. ▼ SO LUTI O N We use the formula for continuously compounded interest with P $1000, r 0.12, and t 3 to get A 3 1000e1 0.12 3 1000e0.36 $1433.33 2 1 2 Compare this amount with the amounts in Example 9. ✎ Practice what you’ve learned: Do Exercise 71. ▲ 5. ▼ CONCE PTS x 1. The function f 2 1 2 5x ; f 1 2 f 2 1 2, and f 6 1 2 is an exponential function with base, f 0 1 2., 2. Match the exponential function with its graph. (a) f x 1 2 2x (b) f x 1 2 2x (c) f x 1 2 2x I y 2 0 1 x II y 2 0 1 x III y 2 0 3. The function f x 1 2 ex is called the exponential function. The number e is approximately equal to. 4. In the formula A Pert t 1 2 for continuously compound interest, the letters P, r, and t stand for,, and t 2 1 $100 is invested at an interest rate of 6% compounded continu-, respectively, and stands for. So if A ously, then the amount after 2 years is. 1 x ▼ SKI LLS 5–10 ■ Use a calculator to evaluate the function at the indicated values. Round your answers to three decimals. ✎ 5. 6. 7. 8. f f g g ✎ 9. h 10.5 4x; f 1 3x1; f, f 1 2 1.5 1 x1; g A 2 3B 3 4B A ex; h 2x; g 1 1.3 1 0.7 2, g 1 2 0.23 12, f 2, g A 1 3B, f p, f, f 1 2 13 1 15 2, f 2, g e 2
1 2p A, g 2 1 17/2 1, g 2 1/p A, g 2 1 2 5 4B 1 2B 2 3B A 3, h 1 e2x 22 SE CTI ON 5.1 \| Exponential Functions 379 25–26 ■ Match the exponential function with one of the graphs labeled I or II. 25. x 1 2 f I 5x1 y x 1 2 26. f II 1 0 1 x 5x 1 y 1 0 1 x 27–40 ■ Graph the function, not by plotting points, but by starting from the graphs in Figures 2 and 5. State the domain, range, and asymptote. 11–16 ■ Sketch the graph of the function by making a table of values. Use a calculator if necessary. 11. 13. f f ✎ 15 2x x 1 3B A 3ex 12. g 14. h 16 8x x 1.1 2 1 2e0.5x 17–20 ■ Graph both functions on one set of axes 27. 31. 29. 3x 2x 3 4 1 2B A 10x3 ex x 35. 2 37. y ex 1 x2 33. x f f 1 1 2 39 30. 28. 10x 2x3 6 3x x A x 34. 2 36. y 1 e x 32. 1 5B h x f 1 1 2 ✎ 17. 18. 19. 20 2x and g 1 3x and g 1 4x and g x 1 x and g 2 3B A 2x x 1 3B A 2 7x x 4 3B A 2 2 x 2 x 1 21–24 ■ Find the exponential function given. f x 1 2 ax whose graph is ✎ 21. 22. y 1 0 y 1 0 (2, 9) 3 x y 1 0 1!_1, @ 5 _3 24. (_3, 8) 1!2, @ 16 3 x _3 y 1 0 _3 23. _3 1 f x ex 38. 2 40. y e x3 4 2x 3 and g x 1 2 2x. 2 1 41. (a) Sketch the graphs of f x (b) How are the graphs related? 1 2 42. (a) Sketch the graphs of f x 9x/2 and g x 3x. 2 (b) Use the Laws of Exponents to explain the relationship 1 1 2 between these graphs. 43. Compare the
functions x3 both of them for x 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 15, and 20. Then draw the graphs of f and g on the same set of axes. by evaluating 3x and g x x f 2 2 1 1 44. If f x 1 2 10x, show that 10x 10h 1 h a. b 45. The hyperbolic cosine function is deﬁned by cosh x 1 2 ex ex 2 (a) Sketch the graphs of the functions y 1 on the same axes, and use graphical addition (see Section 3.6) to sketch the graph of y cosh y 1 and 2 ex x. (b) Use the deﬁnition to show that cosh(x) cosh(x). 1 2 2 ex 3 x 46. The hyperbolic sine function is deﬁned by sinh x 1 2 ex ex 2 (a) Sketch the graph of this function using graphical addition as in Exercise 45. (b) Use the deﬁnition to show that sinh(x) sinh(x) ✎ 47. (a) Compare the rates of growth of the functions f 2x x 1 2 x5 x g and the following viewing rectangles. 2 1 by drawing the graphs of both functions in (i) (ii) (iii) 0, 5 4 0, 25 0, 50 by 3 by by 0, 20 4 0, 107 0, 108 b) Find the solutions of the equation 2x x5, correct to one 3 x decimal place. 380 CHAPTER 5 \| Exponential and Logarithmic Functions 48. (a) Compare the rates of growth of the functions f x x4 x g and the following viewing rectangles: by drawing the graphs of both functions in 3x 1 2 (ii) 0, 10 3 4 by 3 0, 5000 4 2 1 4, 4 3 0, 20 3 4 (i) (iii) by 4 by 0, 20 3 4 0, 105 4 3 (b) Find the solutions of the equation 3 x x 4, correct to two decimal places. 49–50 ■ Draw graphs of the given family of functions for c 0.25, 0.5, 1, 2, 4. How are the graphs related? 49. 50. f f x x 1 1 2 2 c2x 2cx 51. Illust
rate the deﬁnition of the number e by graphing the curve and the line y e on the same screen, using. 4 1 1/x y the viewing rectangle 0, 40 3 0, 4 by 1 2 4 3 x 52. Investigate the behavior of the function f x x 1 1 x b 1 a as x q by graphing f and the line y 1/e on the same screen, using the viewing rectangle 0, 20 0, 1 by 2 3 53. (a) Draw the graphs of the family of functions ex/a ex/a 2 for a 0.5, 1, 1.5, and 2. (b) How does a larger value of a affect the graph? 54–55 ■ 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. 60. Radioactive Decay Doctors use radioactive iodine as a tracer in diagnosing certain thyroid gland disorders. This type of iodine decays in such a way that the mass remaining after t days is given by the function m t 6e0.087t 2 t m 1 is measured in grams. where (a) Find the mass at time t 0. (b) How much of the mass remains after 20 days? 2 1 61. Sky Diving A sky diver jumps from a reasonable height above the ground. The air resistance she experiences is proportional to her velocity, and the constant of proportionality is 0.2. It can be shown that the downward velocity of the sky diver at time t is given by √ t 1 2 80 1 1 e0.2t 2 2 1 t √ is measured in feet where t is measured in seconds and per second (ft/s). (a) Find the initial velocity of the sky diver. (b) Find the velocity after 5 s and after 10 s. (c) Draw a graph of the velocity function t √ (d) The maximum velocity of a falling object with wind resistance is called its terminal velocity. From the graph in part (c) ﬁnd the terminal velocity of this sky diver.. 2 1 54. 55 ex e3x 1 2 √(t)=80(1-e_º.™t) 56–57 ■ Find, correct to two decimal places, (a) the intervals on which the function is increasing or decreasing and (b) the range of the function. y 10xx2 56
. 57. y xex ▼ APPLICATIONS 58. Medical Drugs When a certain medical drug is administered to a patient, the number of milligrams remaining in the patient’s bloodstream after t hours is modeled by 50e0.2t D t 1 2 How many milligrams of the drug remain in the patient’s bloodstream after 3 hours? 59. Radioactive Decay A radioactive substance decays in such a way that the amount of mass remaining after t days is given by the function m t 13e0.015t 2 is measured in kilograms. 1 t m where (a) Find the mass at time t 0. (b) How much of the mass remains after 45 days? 1 2 62. Mixtures and Concentrations A 50-gallon barrel is ﬁlled completely with pure water. Salt water with a concentration of 0.3 lb/gal is then pumped into the barrel, and the resulting mixture overﬂows at the same rate. The amount of salt in the barrel at time t is given by Q t 1 2 15 1 1 e0.04t 2 is measured in pounds. Q where t is measured in minutes and (a) How much salt is in the barrel after 5 min? (b) How much salt is in the barrel after 10 min? (c) Draw a graph of the function (d) Use the graph in part (c) to determine the value that the Q t t. 1 2 2 1 amount of salt in the barrel approaches as t becomes large. Is this what you would expect? Q(t)=15(1-e_º.º¢ t) ✎ 63. Logistic Growth Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions the population follows a logistic growth model: P t 1 2 d 1 kect where c, d, and k are positive constants. For a certain ﬁsh population in a small pond d 1200, k 11, c 0.2, and t is measured in years. The ﬁsh were introduced into the pond at time t 0. (a) How many ﬁsh were originally put in the pond? (b) Find the population after 10, 20, and 30 years. (c) Evaluate for large values of t. What value does the population approach as t q? Does the graph shown conﬁrm your calculations? P t 1 2 P 1200 1000 800 600
400 200 0 10 20 30 40 t 64. Bird Population The population of a certain species of bird is limited by the type of habitat required for nesting. The population behaves according to the logistic growth model n t 1 2 5600 0.5 27.5e0.044t where t is measured in years. (a) Find the initial bird population. (b) Draw a graph of the function n (c) What size does the population approach as time t. 2 1 goes on? 65. World Population The relative growth rate of world popula- tion has been decreasing steadily in recent years. On the basis of this, some population models predict that world population will eventually stabilize at a level that the planet can support. One such logistic model is ✎ P t 1 2 73.2 6.1 5.9e0.02t where t 0 is the year 2000 and population is measured in billions. (a) What world population does this model predict for the year 2200? For 2300? (b) Sketch a graph of the function P for the years 2000 to 2500. (c) According to this model, what size does the world popula- tion seem to approach as time goes on? 66. Tree Diameter For a certain type of tree the diameter D (in feet) depends on the tree’s age t (in years) according to the logistic growth model ✎ D t 1 2 5.4 1 2.9e0.01t SE CTI ON 5.1 \| Exponential Functions 381 Find the diameter of a 20-year-old tree. D 5 4 3 2 1 0 100 300 500 700 t 67–68 ■ Compound Interest An investment of $5000 is deposited into an account in which interest is compounded monthly. Complete the table by ﬁlling in the amounts to which the investment grows at the indicated times or interest rates. 67. r 4% Time (years) Amount 1 2 3 4 5 6 68. t 5 years Rate per year Amount 1% 2% 3% 4% 5% 6% 69. Compound Interest If $10,000 is invested at an interest rate of 10% per year, compounded semiannually, ﬁnd the value of the investment after the given number of years. (a) 5 years (b) 10 years (c) 15 years 70. Compound Interest If $4000 is borrowed at a rate of 16% interest per year, compounded quarterly,
ﬁnd the amount due at the end of the given number of years. (a) 4 years (b) 6 years (c) 8 years 71. Compound Interest If $3000 is invested at an interest rate of 9% per year, ﬁnd the amount of the investment at the end of 5 years for the following compounding methods. (a) Annual (d) Weekly (g) Continuously (b) Semiannual (e) Daily (c) Monthly (f) Hourly 382 CHAPTER 5 \| Exponential and Logarithmic Functions (c) Use the graph of A will amount to $25,000. to determine when this investment t 1 2 ▼ DISCOVE RY • DISCUSSION • WRITI NG 77. Growth of an Exponential Function Suppose you are of- fered a job that lasts one month, and you are to be very well paid. Which of the following methods of payment is more proﬁtable for you? (a) One million dollars at the end of the month (b) Two cents on the ﬁrst day of the month, 4 cents on the second day, 8 cents on the third day, and, in general, 2 n cents on the nth day 78. The Height of the Graph of an Exponential Function Your mathematics instructor asks you to sketch a graph of the exponential function f x 2x 2 for x between 0 and 40, using a scale of 10 units to one inch. What are the dimensions of the sheet of paper you will need to sketch this graph? 1 72. Compound Interest If $4000 is invested in an account for which interest is compounded quarterly, ﬁnd the amount of the investment at the end of 5 years for the following interest rates. (a) 6% (c) 7% 1 (b) 6 % 2 (d) 8% 73. Compound Interest Which of the given interest rates and compounding periods would provide the best investment? (i) 8 % per year, compounded semiannually (ii) 8 % per year, compounded quarterly (iii) 8% per year, compounded continuously 1 2 1 4 74. Compound Interest Which of the given interest rates and compounding periods would provide the better investment? (i) 9 % per year, compounded semiannually (ii) 9% per year, compounded continuously 1 4 75. Present Value The present value of a sum of money is the amount that
must be invested now, at a given rate of interest, to produce the desired sum at a later date. (a) Find the present value of $10,000 if interest is paid at a rate of 9% per year, compounded semiannually, for 3 years. (b) Find the present value of $100,000 if interest is paid at a rate of 8% per year, compounded monthly, for 5 years. 76. Investment A sum of $5000 is invested at an interest rate of 9% per year, compounded semiannually. t A of the investment after t years. (a) Find the value 2 1.A (b) Draw a graph of t 2 1 DISCOVERY PR OJECT EXPONENTIAL EXPLOSION To help us grasp just how explosive exponential growth is, let’s try a thought experiment. Suppose you put a penny in your piggy bank today, two pennies tomorrow, four pennies the next day, and so on, doubling the number of pennies you add to the bank each day (see the table). How many pennies will you put in your piggy bank on day 30? The answer is 230 pennies. That’s simple, but can you guess how many dollars that is? 230 pennies is more than 10 million dollars! Day Pennies 16 o 2n o $10,000,000 in pennies! As you can see, the exponential function f x 2x 2 grows extremely fast. This is the principle behind atomic explosions. An atom splits releasing two neutrons, which cause two atoms to split, each releasing two neutrons, causing four atoms to split, and so on. At the nth stage 2n atoms split—an exponential explosion! 1 Populations also grow exponentially. Let’s see what this means for a type of bacteria that splits every minute. Suppose that at 12:00 noon a single bacterium colonizes a discarded food can. The bacterium and his descendants are all happy, but they fear the time when the can is completely full of bacteria—doomsday. 1. How many bacteria are in the can at 12:05? At 12:10? 2. The can is completely full of bacteria at 1:00 P.M. At what time was the can only half full of bacteria? 3. When the can is exactly half full, the president of the bacteria colony reassures his constituents that doomsday is far away—after all, there is as
much room left in the can as has been used in the entire previous history of the colony. Is the president correct? How much time is left before doomsday? 4. When the can is one-quarter full, how much time remains till doomsday? 5. A wise bacterium decides to start a new colony in another can and slow down splitting time to 2 minutes. How much time does this new colony have? 383 384 CHAPTER 5 \| Exponential and Logarithmic Functions 5.2 Logarithmic Functions LEARNING OBJECTIVES After completing this section, you will be able to: ■ Evaluate logarithmic functions ■ Change between logarithmic and exponential forms of an expression ■ Use basic properties of logarithms ■ Graph logarithmic functions ■ Use common and natural logarithms In this section we study the inverses of exponential functions. ■ Logarithmic Functions, with a 0 and a 1, is a one-to-one function by the ax Every exponential function Horizontal Line Test (see Figure 1 for the case a 1) and therefore has an inverse function. The inverse function f 1 is called the logarithmic function with base a and is denoted by loga. Recall from Section 3.7 that f 1 is deﬁned by x f 2 1 y 0 f(x)=a ˛, a>1 x FIGURE 1 f x 1 2 ax is one-to-one This leads to the following deﬁnition of the logarithmic function DEFINITION OF THE LOGARITHMIC FUNCTION Let a be a positive number with a 1. The logarithmic function with base a, denoted by log a, is deﬁned by We read loga x y as “log base a of x is y.” loga x y 3 ay x So loga x is the exponent to which the base a must be raised to give x. By tradition the name of the logarithmic function is loga, not just a single letter. Also, we usually omit the parentheses in the function notation and write x loga1 2 loga x When we use the deﬁnition of logarithms to switch back and forth between the logarithmic form logax y and the exponential form a y x, it is helpful to notice that, in both forms, the base is the same: Log
arithmic form Exponential form Exponent Exponent loga x y a y x Base Base E X AM P L E 1 \| Logarithmic and Exponential Forms The logarithmic and exponential forms are equivalent equations: If one is true, then so is the other. So we can switch from one form to the other as in the following illustrations. SE CTI ON 5.2 \| Logarithmic Functions 385 Logarithmic form Exponential form log10 100,000 5 log28 3 log2!1 @ 3 8 log5 s r 105 100,000 23 8 23 1 8 5r s ✎ Practice what you’ve learned: Do Exercise 5. ▲ It is important to understand that loga x is an exponent. For example, the numbers in the right column of the table in the margin are the logarithms (base 10) of the numbers in the left column. This is the case for all bases, as the following example illustrates. x 104 103 102 10 1 101 102 103 104 log10 AM P L E 2 \| Evaluating Logarithms because 103 1000 (a) log101000 3 because 25 32 (b) log2 32 5 because 101 0.1 (c) log10 0.1 1 because 161/2 4 log16 4 1 (d) ✎ Practice what you’ve learned: Do Exercises 7 and 9. 2 Inverse Function Property: f 1 x x f 1 f 1 22 x x f 1 1 1 22 When we apply the Inverse Function Property described on page 267 to 1 f x 1 2 loga x, we get loga1 x x ax aloga x x x 0 2 ▲ ax and f x 1 2 We list these and other properties of logarithms discussed in this section. PROPERTIES OF LOGARITHMS Property 1. loga 1 0 2. loga a 1 3. loga a x x aloga x x 4. Reason We must raise a to the power 0 to get 1. We must raise a to the power 1 to get a. We must raise a to the power x to get ax. loga x is the power to which a must be raised to get x. E X AM P L E 3 \| Applying Properties of Logarithms y 1 y=a˛, a>1 y=log a x 1 x y=x FIGURE 2 Graph of the logarithmic
loga x function f x 1 2 We illustrate the properties of logarithms when the base is 5. log5 5 1 5log5 12 12 log5 1 0 log5 58 8 Property 3 ✎ Practice what you’ve learned: Do Exercises 19 and 25. Property 1 Property 2 Property 4 ▲ ■ Graphs of Logarithmic Functions Recall that if a one-to-one function f has domain A and range B, then its inverse function f 1 has domain B and range A. Since the exponential function with a 1 has f loga x, we conclude that its inverse function, domain and range, has do1 0, q main f The graph of in the line y x. Figure 2 shows the case a 1. The fact that y a x (for a 1) is a very rapidly 0, q 2. loga x f is obtained by reﬂecting the graph of and range 1 ax x x 2 1 1 f ax 386 CHAPTER 5 \| Exponential and Logarithmic Functions Arrow notation is explained on page 345. increasing function for x 0 implies that y loga x is a very slowly increasing function for x 1 (see Exercise 88). Since loga 1 0, the x-intercept of the function y loga x is 1. The y-axis is a vertical asymptote of y loga x because loga x q as x 0. E X AM P L E 4 \| Graphing a Logarithmic Function by Plotting Points Sketch the graph of f x log2 x. 1 2 ▼ SO LUTI O N To make a table of values, we choose the x-values to be powers of 2 so that we can easily ﬁnd their logarithms. We plot these points and connect them with a smooth curve as in Figure 3. x 23 22 2 1 21 22 23 24 log2 _1 _2 _3 _4 f(x)=log¤ x 1 2 4 6 8 x FIGURE 3 ✎ Practice what you’ve learned: Do Exercise 41. ▲ Figure 4 shows the graphs of the family of logarithmic functions with bases 2, 3, 5, and y 10x 10. These graphs are drawn by reﬂecting the graphs of (see Figure 2 in Section 5.1) in the line y x. We can also plot points as an aid to sketching these graphs, as
illustrated in Example 4. y 2x, y 3x, y 5x, and y 1 0 1 y=log¤ x y=log‹ x y=logﬁ x y=log⁄‚ x x FIGURE 4 A family of logarithmic functions In the next two examples we graph logarithmic functions by starting with the basic graphs in Figure 4 and using the transformations of Section 3.5. SE CTI ON 5.2 \| Logarithmic Functions 387 E X AM P L E 5 \| Reﬂecting Graphs of Logarithmic Functions 1 1 2 2 x x (b) g h Sketch the graph of each function. (a) log2 x log21 x 2 ▼ SO LUTI O N (a) We start with the graph of log2 x (b) We start with the graph of log21 x 2 in Figure 5(a). in Figure 5(b). log2 x log2 x and reﬂect in the x-axis to get the graph of and reﬂect in the y-axis to get the graph of y 1 0 y f(x)=log¤ x 1 f(x)=log¤ x 1 x 0 _1 1 x g(x)=_log¤ x h(x)=log¤(_x) FIGURE 5 (b) ✎ Practice what you’ve learned: Do Exercise 55. (a) ▲ E X AM P L E 6 \| Shifting Graphs of Logarithmic Functions Find the domain of each function, and sketch the graph. (a) x g h 1 2 x 2 log5 x log101 x 3 (b) 1 2 ▼ SO LUTI O N (a) The graph of g is obtained from the graph of 2 upward 2 units (see Figure 6 on the next page). The domain of f is log5 x f x 1 2 (Figure 4) by shifting 0, q. 1 2 MATHEMATICS IN THE MODERN WORLD Law Enforcement © Bettmann /CORBIS Mathematics aids law enforcement in numerous and surprising ways, from the reconstruction of bullet trajectories to determining the time of death to calculating the probability that a DNA sample is from a particular person. One interesting use is in the search for missing persons. If a person has been missing for several years, that person might look quite different from his or her most recent available
photograph. This is particularly true if the missing person is a child. Have you ever wondered what you will look like 5, 10, or 15 years from now? © Hulton-Deutsch Collection /CORBIS 2 5 1 3 in a child in an adult. By collecting data and analyzing the Researchers have found that different parts of the body grow at different rates. For example, you have no doubt noticed that a baby’s head is much larger relative to its body than an adult’s. As another example, the ratio of arm length to height is but about graphs, researchers are able to determine the functions that model growth. As in all growth phenomena, exponential and logarithmic functions play a crucial role. For instance, the formula that relates arm length l to height h is l aekh where a and k are constants. By studying various physical characteristics of a person, mathematical biologists model each characteristic by a function that describes how it changes over time. Models of facial characteristics can be programmed into a computer to give a picture of how a person’s appearance changes over time. These pictures aid law enforcement agencies in locating missing persons. 388 CHAPTER 5 \| Exponential and Logarithmic Functions John Napier (1550–1617) was a Scottish landowner for whom mathematics was a hobby. We know him today because of his key invention: logarithms, which he published in 1614 under the title A Description of the Marvelous Rule of Logarithms. In Napier’s time, logarithms were used exclusively for simplifying complicated calculations. For example, to multiply two large numbers, we would write them as powers of 10. The exponents are simply the logarithms of the numbers. For instance, 4532 57783 103.65629 104.76180 108.41809 261,872,564 The idea is that multiplying powers of 10 is easy (we simply add their exponents). Napier produced extensive tables giving the logarithms (or exponents) of numbers. Since the advent of calculators and computers, logarithms are no longer used for this purpose. The logarithmic functions, however, have found many applications, some of which are described in this chapter. Napier wrote on many topics. One of his most colorful works is a book entitled A Plaine Discovery of the Whole Revelation of Saint John, in which he predicted that the world would end in the year 1700
. y 3 2 1 0 g(x)=2+logﬁ x f(x)=logﬁ x 1 x FIGURE 6 (b) The graph of h is obtained from the graph of f x log10 x (Figure 4) by shifting to the right 3 units (see Figure 7 below). The line x 3 is a vertical asymptote. Since log10 x is deﬁned only when x 0, the domain of is x 3 0 log101 2 1 3 Asymptote x = 3 f(x)=log⁄‚ x h(x)=log⁄‚(x-3) 1 4 x FIGURE 7 ✎ Practice what you’ve learned: Do Exercises 53 and 57. ▲ ■ Common Logarithms We now study logarithms with base 10. COMMON LOGARITHM The logarithm with base 10 is called the common logarithm and is denoted by omitting the base: log x log10 x From the deﬁnition of logarithms we can easily ﬁnd that log 10 1 and log 100 2 But how do we ﬁnd log 50? We need to ﬁnd the exponent y such that 10 y 50. Clearly, 1 is too small and 2 is too large. So 1 log 50 2 To get a better approximation, we can experiment to ﬁnd a power of 10 closer to 50. Fortunately, scientiﬁc calculators are equipped with a key that directly gives values of common logarithms. LOG SE CTI ON 5.2 \| Logarithmic Functions 389 E X AM P L E 7 \| Evaluating Common Logarithms log x f Use a calculator to ﬁnd appropriate values of graph. x 1 2 and use the values to sketch the ▼ SO LUTI O N We make a table of values, using a calculator to evaluate the function at those values of x that are not powers of 10. We plot those points and connect them by a smooth curve as in Figure 8. x 0.01 0.1 0.5 1 4 5 10 log x 2 1 0.301 0 0.602 0.699 1 y 2 1 0 _1 f(x)=log x 2 4 6 8 10 12 x FIGURE 8 ✎ Practice what you’ve learned: Do Exercise 43. ▲ Scientists model
human response to stimuli (such as sound, light, or pressure) using logarithmic functions. For example, the intensity of a sound must be increased manyfold before we “feel” that the loudness has simply doubled. The psychologist Gustav Fechner formulated the law as S k log I I0 b a Human response to sound and light intensity is logarithmic. where S is the subjective intensity of the stimulus, I is the physical intensity of the stimulus, I0 stands for the threshold physical intensity, and k is a constant that is different for each sensory stimulus. We study the decibel scale in more detail in Section 5.5. E X AM P L E 8 \| Common Logarithms and Sound The perception of the loudness B (in decibels, dB) of a sound with physical intensity I (in W/m2) is given by B 10 log I I0 b a where I0 is the physical intensity of a barely audible sound. Find the decibel level (loudness) of a sound whose physical intensity I is 100 times that of I0. ▼ SO LUTI O N We ﬁnd the decibel level B by using the fact that I 100I0. B 10 log I I0 b a 10 log 100I0 a I0 b 10 log 100 10 # 2 20 Deﬁnition of B I = 100I0 Cancel I0 Deﬁnition of log The loudness of the sound is 20 dB. ✎ Practice what you’ve learned: Do Exercise 83. ▲ 390 CHAPTER 5 \| Exponential and Logarithmic Functions ■ Natural Logarithms Of all possible bases a for logarithms, it turns out that the most convenient choice for the purposes of calculus is the number e, which we deﬁned in Section 5.1. NATURAL LOGARITHM The notation ln is an abbreviation for the Latin name logarithmus naturalis. The logarithm with base e is called the natural logarithm and is denoted by ln: ln x loge x The natural logarithmic function y ln x is the inverse function of the exponential function y e x. Both functions are graphed in Figure 9. By the deﬁnition of inverse functions we have ln x y 3 ey x y 1 y=e˛ y=ln x
1 x FIGURE 9 Graph of the natural logarithmic function y=x If we substitute a e and write “ln” for “loge” in the properties of logarithms men- tioned earlier, we obtain the following properties of natural logarithms. PROPERTIES OF NATURAL LOGARITHMS Property 1. ln 1 0 2. ln e 1 3. ln e x x 4. eln x x Reason We must raise e to the power 0 to get 1. We must raise e to the power 1 to get e. We must raise e to the power x to get e x. ln x is the power to which e must be raised to get x. Calculators are equipped with an LN key that directly gives the values of natural logarithms. E X AM P L E 9 \| Evaluating the Natural Logarithm Function (a) ln e8 8 Deﬁnition of natural logarithm 1 e2 b (b) ln ln e2 2 a (c) ln 5 1.609 key on calculator ✎ Practice what you’ve learned: Do Exercise 39. Use LN Deﬁnition of natural logarithm ▲ SE CTI ON 5.2 \| Logarithmic Functions 391 E X AM P L E 10 \| Finding the Domain of a Logarithmic Function Find the domain of the function ▼ SO LUTI O N As with any logarithmic function, ln x is deﬁned when x 0. Thus, the domain of f is. 1 2 1 2 f x ln 4 x2 x 5 0 4 x2 0 6 x 5 0 x2 4 x 5 6 2 x 2 x 6 ✎ Practice what you’ve learned: Do Exercise 63, 2 2 ▲ 3 _3 3 FIGURE 10 y x ln 4 x2 1 _3 2 E X AM P L E 11 \| Drawing the Graph of a Logarithmic Function y x ln 4 x 2 Draw the graph of the function cal maximum and minimum values. 1 2, and use it to ﬁnd the asymptotes and lo- 3 ▼ SO LUTI O N As in Example 10 the domain of this function is the interval 3, 3 we choose the viewing rectangle 4 from it we see that the lines x
2 and x 2 are vertical asymptotes., so. The graph is shown in Figure 10, and 3, 3 4 3 by 2, 2 The function has a local maximum point to the right of x 1 and a local minimum point to the left of x 1. By zooming in and tracing along the graph with the cursor, we ﬁnd that the local maximum value is approximately 1.13 and this occurs when x 1.15. Similarly (or by noticing that the function is odd), we ﬁnd that the local minimum value is about 1.13, and it occurs when x 1.15. ✎ Practice what you’ve learned: Do Exercise 69. ▲ 2 1 5. ▼ CONCE PTS 1. log x is the exponent to which the base 10 must be raised to get. So we can complete the following table for log x. x 103 102 101 100 101 102 103 101/2 ▼ SKI LLS 5–6 ■ Complete the table by ﬁnding the appropriate logarithmic or exponential form of the equation, as in Example 1. 5.✎ Logarithmic form Exponential form log x 2. The function f base x log9 x f 9 2 1. So 2 1 81 2 1 9 2 f f 1 3. (a) 53 125, so log, 1 (b) log5 25 2, so is the logarithm function with, f 1 2 1, and f 3 1 2,. log8 8 1 log8 64 2 log8 1 8B A 1 82/3 4 83 512 82 1 64 4. Match the logarithmic function with its graph. (a) f x 1 2 log2 x (b) f x 1 2 log2 1 x 2 I y 1 II y 1 (c) f x 1 2 III log2x 392 CHAPTER 5 \| Exponential and Logarithmic Functions 6. Logarithmic form Exponential form log 4 2 1 2 log4 log4 2 1 2 1 16B 1 2B A A 43 64 43/2 8 45/2 1 32 ✎ ✎ 7–12 ■ Express the equation in exponential form. (b) log5 1 0 7. (a) log5 25 2 8. (a) log10 0.1 1 (b) log8 512 3 log8 2 1 3 1 9. (a) (b) log2A 8
B 10. (a) log3 81 4 log8 4 2 11. (a) ln 5 x x 1 (b) (b) ln y 5 x 1 12. (a) 2 (b) ln ln 3 3 1 4 2 1 2 13–18 ■ Express the equation in logarithmic form. 13. (a) 5 3 125 14. (a) 10 3 1000 (b) 104 0.0001 (b) 811/2 9 23 1 (b) 8 (b) 73 343 (b) e 3 y (b) e0.5x t 81 1 15. (a) 8 16. (a) 43/2 0.125 17. (a) e x 2 18. (a) e x1 0.5 19–28 ■ Evaluate the expression. 37–40 ■ Use a calculator to evaluate the expression, correct to four decimal places. 37. (a) log 2 38. (a) log 50 ✎ 39. (a) ln 5 40. (a) ln 27 (b) log 35.2 (b) log 12 (b) ln 25.3 (b) ln 7.39 (c) (c) log log A 2 3B 3 12 1 2 1 13 2 (c) ln 1 (c) ln 54.6 41–44 ■ Sketch the graph of the function by plotting points. ✎ ✎ 41. 43. f f x x 1 1 2 2 log3 x 2 log x 42. 44. g g x x 1 1 2 2 log4 x 1 log x 45–48 ■ Find the function of the form y loga x whose graph is given. 45. y 1 46. y (5, 1) 0 1 5 x 1 0 _1 1 1!, _1@ 2 47. y 1 0 48. y 1 0!3, @1 2 1 3 x x (9, 2) 1 3 6 9 x ✎ 19. (a) log3 3 20. (a) log5 54 21. (a) log6 36 23. (a) 22. (a) log2 32 1 27 B 24. (a) log5 125 log3A ✎ 25. (a) 2log2 37 26. (a) eln p (b) log3 1 (b) log4 64 (b) log9 81 (b) log
8 817 (b) log10 110 (b) log 49 7 3log3 8 (b) (b) 10 log 5 (c) log3 32 (c) log3 9 (c) log7 710 (c) log6 1 (c) log5 0.2 log9 13 eln15 (c) (c) (c) 10 log 87 2 1 ln (c) 1 2B (b) 1/e 28. (a) (c) log4 8 27. (a) log8 0.25 log4 12 (b) ln e4 log4A 29–36 ■ Use the deﬁnition of the logarithmic function to ﬁnd x. 29. (a) log2 x 5 30. (a) log5 x 4 31. (a) log3 243 x 32. (a) log4 2 x 33. (a) log10 x 2 34. (a) logx 1000 3 35. (a) logx 16 4 logx 6 1 36. (a) (b) log2 16 x (b) log10 0.1 x (b) log3 x 3 (b) log4 x 2 (b) log5 x 2 (b) logx 25 2 logx 8 3 (b) (b) logx 3 1 2 2 3 49–50 ■ Match the logarithmic function with one of the graphs labeled I or II. 2 ln x 50. 49 II y ln 1 x 2 2 x=2 (3, 0) y 2 0 (1, 2) 1 x 0 1 3 x 51. Draw the graph of y 4x, then use it to draw the graph of y log4 x. 52. Draw the graph of y 3x, then use it to draw the graph of y log3 x. 53–62 ■ Graph the function, not by plotting points, but by starting from the graphs in Figures 4 and 9. State the domain, range, and asymptote. x 4 log21 log51 x log10 x ln x 2 56. g 54. 55. 53 ✎ ✎ ✎ ✎ 57. y 2 log3 x 59. y 1 log10 x 61. y ln x 0 0 SE CTI ON 5.2 \| Logarithmic Functions 393 2 2 2 where I0 is the intensity of the incident light and I
is the intensity of light that emerges. Find the concentration of the substance if the intensity I is 70% of I0. 58. 60. 62. x 1 x y log31 y 1 ln y ln x 1 0 0 63. 63–68 ■ Find the domain of the function. x 3 x 2 1 64. 66. 65 log101 log31 2 2 x ln x ln 2 1x 2 log51 1 1 2 67. 68. h h x x 1 1 2 2 10 x 2 I0 I 2 8 2x log51 ln 1 2 2 x x 2 2 1 69–74 ■ Draw the graph of the function in a suitable viewing rectangle, and use it to ﬁnd the domain, the asymptotes, and the local maximum and minimum values. ✎ 69. 71. y log101 y x ln x y ln x x 1 x 2 2 70. x 2 x y ln y x 1 ln x 2 2 72. 73. 2 1 y x log101 75. Compare the rates of growth of the functions f 74. x 10 2 ln x and x 1 2 1x 1 2 x g the viewing rectangle by drawing their graphs on a common screen using 1, 6 3 1, 30 3 by. 4 4 76. (a) By drawing the graphs of the functions 1 ln and g 1 x x f x 1x 1 1 2 2 in a suitable viewing rectangle, show that even when a logarithmic function starts out higher than a root function, it is ultimately overtaken by the root function. 1 2 (b) Find, correct to two decimal places, the solutions of the 1 x 1x 1 ln equation. 1 2 77–78 ■ A family of functions is given. (a) Draw graphs of the family for c 1, 2, 3, and 4. (b) How are the graphs in part (a) related? 77. f x 1 2 log cx 1 2 78. f x 1 2 c log x 79–80 ■ A function tion f. (b) Find the inverse function of f. x f 2 1 is given. (a) Find the domain of the func- ✎ 83. Carbon Dating The age of an ancient artifact can be determined by the amount of radioactive carbon-14 remaining in it. If D0 is the original amount of carbon-14 and D is the amount remaining, then the artifact’s age A
(in years) is given by A 8267 ln D D0 b a Find the age of an object if the amount D of carbon-14 that remains in the object is 73% of the original amount D0. 84. Bacteria Colony A certain strain of bacteria divides every three hours. If a colony is started with 50 bacteria, then the time t (in hours) required for the colony to grow to N bacteria is given by t 3 log N/50 1 log 2 2 Find the time required for the colony to grow to a million bacteria. 85. Investment The time required to double the amount of an investment at an interest rate r compounded continuously is given by t ln 2 r 22 Find the time required to double an investment at 6%, 7%, and 8%. 79. f x 1 2 log21 log10 x 2 80. f x 1 2 81. (a) Find the inverse of the function f x 1 2 ln ln 1 1 2x. 1 2x ln x (b) What is the domain of the inverse function? ▼ APPLICATIONS 82. Absorption of Light A spectrophotometer measures the concentration of a sample dissolved in water by shining a light through it and recording the amount of light that emerges. In other words, if we know the amount of light that is absorbed, we can calculate the concentration of the sample. For a certain substance the concentration (in moles/liter) is found by using the formula C 2500 ln I I0 b a 86. Charging a Battery The rate at which a battery charges is slower the closer the battery is to its maximum charge C0. The time (in hours) required to charge a fully discharged battery to a charge C is given by t k ln 1 C C0 b a where k is a positive constant that depends on the battery. For a certain battery, k 0.25. If this battery is fully discharged, how long will it take to charge to 90% of its maximum charge C0? 87. Difﬁculty of a Task The difﬁculty in “acquiring a target” (such as using your mouse to click on an icon on your computer screen) depends on the distance to the target and the 394 CHAPTER 5 \| Exponential and Logarithmic Functions size of the target. According to Fitts’s Law, the index of difﬁculty (ID) is
given by ID log 2A/W 1 log 2 2 where W is the width of the target and A is the distance to the center of the target. Compare the difﬁculty of clicking on an icon that is 5 mm wide to clicking on one that is 10 mm wide. In each case, assume that the mouse is 100 mm from the icon. ▼ DISCOVE RY • DISCUSSION • WRITI NG 88. The Height of the Graph of a Logarithmic Function Suppose that the graph of y 2x is drawn on a coordinate plane where the unit of measurement is an inch. (a) Show that at a distance 2 ft to the right of the origin the height of the graph is about 265 mi. (b) If the graph of y log2 x is drawn on the same set of axes, how far to the right of the origin do we have to go before the height of the curve reaches 2 ft? 89. The Googolplex A googol is 10100, and a googolplex is 10googol. Find log log 1 1 googol 22 and log log 1 1 log 1 googolplex 222 90. Comparing Logarithms Which is larger, log4 17 or log5 24? Explain your reasoning. 91. The Number of Digits in an Integer Compare log 1000 to the number of digits in 1000. Do the same for 10,000. How many digits does any number between 1000 and 10,000 have? Between what two values must the common logarithm of such a number lie? Use your observations to explain why the number of digits in any positive integer x is “log x‘ 1. (The symbol “n‘ is the greatest integer function deﬁned in Section 3.2.) How many digits does the number 2100 have? 5.3 Laws of Logarithms LEARNING OBJECTIVES After completing this section, you will be able to: ■ Use the Laws of Logarithms to evaluate logarithmic expressions ■ Use the Laws of Logarithms to expand logarithmic expressions ■ Use the Laws of Logarithms to combine logarithmic expressions ■ Use the Change of Base Formula In this section we study properties of logarithms. These properties give logarithmic functions a wide range of applications, as we will see in Section 5.5. ■ Laws of Logarith
ms Since logarithms are exponents, the Laws of Exponents give rise to the Laws of Logarithms. 7LAWS OF LOGARITHMS Let a be a positive number, with a 1. Let A, B, and C be any real numbers with A 0 and B 0. Law Description 1. 2. 3. loga1 loga a AB 2 A B b loga A loga B loga A loga B AC loga1 2 C loga A The logarithm of a product of numbers is the sum of the logarithms of the numbers. The logarithm of a quotient of numbers is the difference of the logarithms of the numbers. The logarithm of a power of a number is the exponent times the logarithm of the number. SE CTIO N 5.3 \| Laws of Logarithms 395 ▼ P RO O F We make use of the property logaax x from Section 5.2. Law 1 Let loga A u and loga B √. When written in exponential form, these equations become Thus loga1 Law 2 Using Law 1, we have au A and a√ B loga1 loga1 2 2 u √ loga A loga B au√ aua√ AB 2 loga A loga c a A B b B d loga a A B b loga B so loga a A B b loga A loga B Law 3 Let loga A u. Then au A, so loga1 loga1 AC au 2 2 C loga1 auC 2 uC C loga A ▲ E X AM P L E 1 \| Using the Laws of Logarithms to Evaluate Expressions Evaluate each expression. (a) log4 2 log4 32 (b) log2 80 log2 5 1 (c) 3 log 8 ▼ SO LUTI O N (a) log4 2 log4 32 log41 2 # 32 2 log4 64 3 (b) log2 80 log2 5 log2A 80 5 B log2 16 4 (c) 1 3 log 8 log 81/3 log A 0.301 1 2B Law 1 Because 64 = 43 Law 2 Because 16 = 24 Law 3 Property of negative exponents Calculator ✎ Practice what you’ve learned: Do Exercises 7, 9, and 11. ▲
■ Expanding and Combining Logarithmic Expressions The Laws of Logarithms allow us to write the logarithm of a product or a quotient as the sum or difference of logarithms. This process, called expanding a logarithmic expression, is illustrated in the next example. E X AM P L E 2 \| Expanding Logarithmic Expressions Use the Laws of Logarithms to expand each expression. (a) log21 6x 2 (b) x3y6 log51 2 ▼ SO LUTI O N (a) log21 6x 2 log2 6 log2 x (c) ln ab 13 c b a Law 1 396 CHAPTER 5 \| Exponential and Logarithmic Functions (b) x3y6 log51 2 log5 x3 log5 y6 3 log5 x 6 log5 y (c) ln ab 13 c b a ln ab 1 2 ln 13 c Law 1 Law 3 Law 2 Law 1 ln a ln b ln c1/3 ln a ln b 1 3 ln c ✎ Practice what you’ve learned: Do Exercises 19, 21, and 33. Law 3 ▲ The Laws of Logarithms also allow us to reverse the process of expanding that was done in Example 2. That is, we can write sums and differences of logarithms as a single logarithm. This process, called combining logarithmic expressions, is illustrated in the next example. E X AM P L E 3 \| Combining Logarithmic Expressions x 1 into a single logarithm. 3 log x 1 Combine 2 log 1 2 ▼ SO LUTI O N 3 log x 1 2 log x 1 1 2 log x3 log log x 1 x3 1 1 x 1 1/2 2 1 1/2 2 2 Law 3 Law 1 ✎ Practice what you’ve learned: Do Exercise 47. E X AM P L E 4 \| Combining Logarithmic Expressions Combine 3 ln s 1 2 ln t 4 ln ▼ SO LUTI O N 2 1 t 1 2 into a single logarithm. 3 ln s 1 2 ln t 4 ln 2 1 t 1 2 ln s3 ln t1/2 ln ln ln s3t1/ ln a t 1 2
s31t 2 1 1 4 b 2 ✎ Practice what you’ve learned: Do Exercise 49. ▲ ▲ 4 2 Law 3 Law 1 Law 2 WARNING Although the Laws of Logarithms tell us how to compute the logarithm of a product or a quotient, there is no corresponding rule for the logarithm of a sum or a difference. For instance, In fact, we know that the right side is equal to quotients or powers of logarithms. For instance, loga1 x y 2 loga x loga y loga1 xy 2. Also, don’t improperly simplify log 6 log 2 log 6 2 b a and log2 x 1 2 3 3 log2 x Logarithmic functions are used to model a variety of situations involving human behavior. One such behavior is how quickly we forget things we have learned. For example, if you learn algebra at a certain performance level (say, 90% on a test) and then don’t use algebra for a while, how much will you retain after a week, a month, or a year? Hermann Ebbinghaus (1850–1909) studied this phenomenon and formulated the law described in the next example. E X AM P L E 5 \| The Law of Forgetting SE CTIO N 5.3 \| Laws of Logarithms 397 Ebbinghaus’ Law of Forgetting states that if a task is learned at a performance level P0, then after a time interval t the performance level P satisﬁes t 1 c log log P log P0 1 2 where c is a constant that depends on the type of task and t is measured in months. (a) Solve for P. (b) If your score on a history test is 90, what score would you expect to get on a similar test after two months? After a year? (Assume that c 0.2.) Forgetting what we’ve learned depends logarithmically on how long ago we learned it. ▼ SO LUTI O N (a) We ﬁrst combine the right-hand side. log P log P0 log P log P0 c log log 1 t 1 t 1 Given equation 2 c 2 Law 3 Law 2 1 P0 t 1 c 2 log P log 1 P0 t 1 P 1 c 2 Because log is one-to-one (b) Here P
0 90, c 0.2, and t is measured in months. In two months: t 2 and P In one year: t 12 and P 90 2 1 0.2 2 72 90 12 1 0.2 2 54 1 1 Your expected scores after two months and one year are 72 and 54, respectively. ✎ Practice what you’ve learned: Do Exercise 69. ▲ ■ Change of Base For some purposes, we ﬁnd it useful to change from logarithms in one base to logarithms in another base. Suppose we are given loga x and want to ﬁnd logb x. Let y logb x We write this in exponential form and take the logarithm, with base a, of each side. by x Exponential form by loga x loga1 2 y loga b loga x Take loga of each side Law 3 y loga x loga b Divide by loga b We may write the Change of Base Formula as logb x 1 loga b b a loga x So logb x is just a constant multiple of loga x; the constant is 1 loga b. This proves the following formula. CHANGE OF BASE FORMULA logb x loga x loga b 398 CHAPTER 5 \| Exponential and Logarithmic Functions In particular, if we put x a, then loga a 1, and this formula becomes logb a 1 loga b We can now evaluate a logarithm to any base by using the Change of Base Formula to express the logarithm in terms of common logarithms or natural logarithms and then using a calculator. E X AM P L E 6 \| Evaluating Logarithms with the Change of Base Formula Use the Change of Base Formula and common or natural logarithms to evaluate each logarithm, correct to ﬁve decimal places. We get the same answer whether we use log10 or ln: log8 5 ln 5 ln 8 0.77398 (a) log8 5 (b) log9 20 ▼ SO LUTI O N (a) We use the Change of Base Formula with b 8 and a 10: log8 5 log10 5 log10 8 0.77398 (b) We use the Change of Base Formula with b 9 and a e: log9 20 ln 20 ln 9 1.36342 ✎ Practice
what you’ve learned: Do Exercises 55 and 57. ▲ E X AM P L E 7 \| Using the Change of Base Formula to Graph a Logarithmic Function log6 x Use a graphing calculator to graph ▼ SO LUTI O N Calculators don’t have a key for log6, so we use the Change of Base Formula to write x f. 1 2 36 f x 1 2 log6 x ln x ln 6 FIGURE 1 f log6 x ln x ln 6 x 1 2 Since calculators do have an graph it. The graph is shown in Figure 1. ✎ Practice what you’ve learned: Do Exercise 63. LN key, we can enter this new form of the function and 2 0 _1 5. ▼ CONCE PTS 1. The logarithm of a product of two numbers is the same as the log51 # 25 125 2 of the logarithms of these numbers. So. 2. The logarithm of a quotient of two numbers is the same as the log51 25 125 2 of the logarithms of these numbers. So. 3. The logarithm of a number raised to a power is the same as the power log51 2510 2 # the logarithm of the number. So ▲. 4. (a) We can expand log to get a (b) We can combine 2 log x log y log z to get x 2 y z b. 5. Most calculators can ﬁnd logarithms with base and base the. To ﬁnd logarithms with different bases, we use Formula. To ﬁnd log7 12, we write log log log7 12 6. True or false? We get the same answer if we do the calculation. in Exercise 5 using ln in place of log. ✎ ✎ ✎ ▼ SKI LLS 7–18 ■ Evaluate the expression. log3 127 7. 8. log2 160 log2 5 9. log 4 log 25 10. log 1 11000 11. log4 192 log4 3 12. log12 9 log12 16 13. log2 6 log2 15 log2 20 14. log3 100 log3 18 log3 50 15. log4 16100 17. log 1 log 1010,000 2 16. log2 833 ln ee200 18. ln
1 2 19–44 ■ Use the Laws of Logarithms to expand the expression. ✎ 19. log21 2x 2 ✎ 22 1 x 21. x 1 log21 23. log 610 log21 log31 log5 23 x2 1 2 x 1y AB2 25. 29. 27. 2 31. ln 1ab ✎ 33. log x3y4 z6 b a 35. log2 a 37. ln a x B 2 x2 1 x 1 2x2 1 b y z b 20. 22. 24. 26. 28. 30. 32. 2 5y log31 x 2 log5 ln 1z log6 14 17 xy log21 10 2 x2 yz3 b loga a ln 23 3r2s a2 1c b x 1 x 1 34. log a b4 36. log5 B 3x2 x 1 38. ln 1 2 x 13 1 x b 39. log 24 x2 y2 40. log a x2 4 x3 7 41. 43. log B x3 ln a x2 1 2 1 1 1x 1 3x 4 b 2 2 42. log 3x2y1z 44. log 10x x2 1 a x 1 2 1 x4 2 b 2 45–54 ■ Use the Laws of Logarithms to combine the expression. 45. log3 5 5 log3 2 log 12 1 2 log 7 log 2 46. 47. log2 A log2 B 2 log2 C x 1 48. ✎ ✎ 49. x2 1 log51 4 log x 1 a b log51 2 x2 1 3 log 1 ln a b ln 50. 51. ln 5 2 ln x 3 ln 2 1 1 2 2 x2 5 1 2 2 2 log 1 2 ln c x 1 2 SE CTIO N 5.3 \| Laws of Logarithms 399 52. 2 log5 x 2 log5 y 3 log5 z 3 1 2 3 2 log x 4 log x 2 1 1 3 log 1 x 2 x 6 53. 2 54. loga b c loga d r loga s 55–62 ■ Use the Change of Base Formula and a calculator to evaluate the logarithm, correct to six decimal places. Use either natural or common logarithms. 2 1 4 2 ✎ ✎ 55. log2 5 57. log3 16 59
. log7 2.61 61. log4 125 56. log5 2 58. log6 92 60. log6 532 62. log12 2.5 ✎ 63. Use the Change of Base Formula to show that log3 x ln x ln 3 Then use this fact to draw the graph of the function f log3 x x. 1 2 64. Draw graphs of the family of functions y loga x for a 2, e, 5, and 10 on the same screen, using the viewing rectangle 0, 5 by. How are these graphs related? 3 4 3, 3 3 4 65. Use the Change of Base Formula to show that log e 1 ln 10 66. Simplify: log2 5 1 ln 67. Show that log5 7 2 1 x 2x2 1 2 1 ln 1 2 x 2x2 1. 2 ▼ APPLICATIONS 68. Forgetting Use Ebbinghaus’ Law of Forgetting (Example 5) to estimate a student’s score on a biology test two years after he got a score of 80 on a test covering the same material. Assume that c 0.3 and t is measured in months. observed that most of the wealth of a country is owned by a few members of the population. Pareto’s Principle is log P log c k log W where W is the wealth level (how much money a person has) and P is the number of people in the population having that much money. (a) Solve the equation for P. (b) Assume that k 2.1, c 8000, and W is measured in millions of dollars. Use part (a) to ﬁnd the number of people who have $2 million or more. How many people have $10 million or more? 70. Biodiversity Some biologists model the number of species S in a ﬁxed area A (such as an island) by the species-area relationship log S log c k log A where c and k are positive constants that depend on the type of species and habitat. (a) Solve the equation for S. 10 ✎ 69. Wealth Distribution Vilfredo Pareto (1848–1923) 400 CHAPTER 5 \| Exponential and Logarithmic Functions (b) Use part (a) to show that if k 3, then doubling the area increases the number of species eightfold. 71. Magnitude of Stars The magnitude
M of a star is a measure of how bright a star appears to the human eye. It is deﬁned by M 2.5 log B B0 b a where B is the actual brightness of the star and B0 is a constant. (a) Expand the right-hand side of the equation. (b) Use part (a) to show that the brighter a star, the less its magnitude. (c) Betelgeuse is about 100 times brighter than Albiero. Use part (a) to show that Betelgeuse is 5 magnitudes less bright than Albiero. ▼ DISCOVE RY • DISCUSSION • WRITI NG 72. True or False? Discuss each equation and determine whether it is true for all possible values of the variables. (Ignore values of the variables for which any term is undeﬁned.) (a) log x y b a log x log y (b) log21 x y 2 log2 x log2 y (c) (d) (e) (f) (g) (h) log5 a 2 log5 b a log5 a b2 b log 2z z log 2 log Q 2 2 1 log P log Q log a log b log P 1 log a log b x x log2 7 log2 7 1 2 loga aa a (i) log x y 2 1 log x log y (j) ln 1 A b a ln A 73. Find the Error What is wrong with the following argument? log 0.1 2 log 0.1 log 0.1 2 1 log 0.01 log 0.1 log 0.01 0.1 0.01 2 2 1 f f x x 2 4f. Show that 74. Shifting, Shrinking, and Stretching Graphs of Functions 2x Let 2 shows that shrinking the graph of f horizontally has the same effect as stretching it vertically. Then use the identities e2x ex e2e x and 2 h x izontal shift is the same as a vertical stretch and for a horizontal shrinking is the same as a vertical shift. ln 2 ln x to show that for, and explain how this a hor ln x 2x ln.4 Exponential and Logarithmic Equations LEARNING OBJECTIVES After completing this section, you will be able to: ■ Solve exponential equations ■ Solve logarithmic equations ■ Solve
problems involving compound interest ■ Calculate annual percentage yield In this section we solve equations that involve exponential or logarithmic functions. The techniques that we develop here will be used in the next section for solving applied problems. ■ Exponential Equations An exponential equation is one in which the variable occurs in the exponent. For example, 2x 7 SE CTI ON 5. 4 \| Exponential and Logarithmic Equations 401 The variable x presents a difﬁculty because it is in the exponent. To deal with this difﬁculty, we take the logarithm of each side and then use the Laws of Logarithms to “bring down x” from the exponent. 2x 7 ln 2x ln 7 x ln 2 ln 7 Given equation Take ln of each side Law 3 (bring down exponent) x ln 7 ln 2 2.807 Solve for x Calculator Recall that Law 3 of the Laws of Logarithms says that loga AC C loga A. The method that we used to solve 2x 7 is typical of how we solve exponential equa- tions in general. GUIDELINES FOR SOLVING EXPONENTIAL EQUATIONS 1. Isolate the exponential expression on one side of the equation. 2. Take the logarithm of each side, then use the Laws of Logarithms to “bring down the exponent.” 3. Solve for the variable. E X AM P L E 1 \| Solving an Exponential Equation Find the solution of the equation 3x2 7 ▼ SO LUTI O N We take the common logarithm of each side and use Law 3., correct to six decimal places. We could have used natural logarithms instead of common logarithms. In fact, using the same steps, we get x ln 7 ln 3 2 0.228756 3x2 7 3x2 log 7 log 3 log 7 2 log x 2 1 2 1 x 2 x log 7 log 3 log 7 log 3 Given equation Take log of each side Law 3 (bring down exponent) Divide by log 3 2 Subtract 2 Calculator ✎ Practice what you’ve learned: Do Exercise 7. 0.228756 ▲ Check Your Answer Substituting x 0.228756 into the original equation and using a calculator, we get 0.228756 31 2 7
2 ✔ E X AM P L E 2 \| Solving an Exponential Equation Solve the equation 8e2x 20. ▼ SO LUTI O N We ﬁrst divide by 8 to isolate the exponential term on one side of the equation. 402 CHAPTER 5 \| Exponential and Logarithmic Functions 8e2x 20 e2x 20 8 ln e2x ln 2.5 2x ln 2.5 x ln 2.5 2 Given equation Divide by 8 Take ln of each side Property of ln Divide by 2 Calculator ✎ Practice what you’ve learned: Do Exercise 9. 0.458 ▲ Check Your Answer Substituting x 0.458 into the original equation and using a calculator, we get 8e2 1 0.458 2 20 ✔ E X AM P L E 3 \| Solving an Exponential Equation Algebraically and Graphically Solve the equation e32x 4 algebraically and graphically. ▼ SO LUTI O N 1: Algebraic Since the base of the exponential term is e, we use natural logarithms to solve this equation. e32x 4 e32x ln 4 2 1 3 2x ln 4 ln 2x 3 ln 4 x 1 21 3 ln 4 2 0.807 Given equation Take ln of each side Property of ln Subtract 3 Multiply by 1 2 You should check that this answer satisﬁes the original equation. ▼ SO LUTI O N 2: Graphical We graph the equations y e32x and y 4 in the same viewing rectangle as in Figure 1. The solutions occur where the graphs intersect. Zooming in on the point of intersection of the two graphs, we see that x 0.81. ✎ Practice what you’ve learned: Do Exercise 11. ▲ 5 0 y=4 y=e3_2x 2 FIGURE 1 2 Radiocarbon dating is a method archeologists use to determine the age of ancient objects. The carbon dioxide in the atmosphere always contains a ﬁxed fraction of radioactive carbon, carbon-14 14C, with a half-life of about 5730 years. Plants absorb carbon 1 dioxide from the atmosphere, which then makes its way to animals through the food chain. Thus, all living creatures contain the same ﬁxed proportions of 14C to nonradioactive 12C
as the atmosphere. After an organism dies, it stops assimilating 14C, and the amount of 14C in it begins to decay exponentially. We can then determine the time elapsed since the death of the organism by measuring the amount of 14C left in it. For example, if a don- key bone contains 73% as much 14C as a living donkey and it died t years ago, then by the formula for radioactive decay (Section 5.5), 0.73 1.00 1 2 e 1 t ln 2 2 /5730 We solve this exponential equation to ﬁnd t is about 2600 years old. 2600, so the bone SE CTI ON 5. 4 \| Exponential and Logarithmic Equations 403 E X AM P L E 4 \| An Exponential Equation of Quadratic Type Solve the equation e2x ex 6 0. ▼ SO LUTI O N To isolate the exponential term, we factor. If we let „ ex, we get the quadratic equation „2 „ 6 0 which factors as „ 3 1 „ 2 0 2 2 1 e2x ex 6 0 ex 2 ex 6 0 2 1 ex 3 ex 2 0 1 2 1 ex 3 0 or ex 2 0 2 Given equation Law of Exponents Factor (a quadratic in ex) Zero-Product Property Check Your Answers x 0 : 3 0 e0 02e0 0 2 1 x 3: 3 3 1 e3 2e3 3 1 2 2 9e3 9e3 0 ex 3 ex 2 The equation e x 3 leads to x ln 3. But the equation e x 2 has no solution because e x 0 for all x. Thus, x ln 3 1.0986 is the only solution. You should check that this answer satisﬁes the original equation. ✎ Practice what you’ve learned: Do Exercise 29. ▲ E X AM P L E 5 \| Solving an Exponential Equation 3xex x2ex 0. Solve the equation ▼ SO LUTI O N First we factor the left side of the equation. 3xex x2ex 0 Given equation ✔ ✔ x 1 3 x ex 0 2 3 x x 1 0 2 Factor out common factors Divide by ex (because ex ≠ 0) x 0 or 3 x 0 Zero-Product Property Thus, the solutions are x 0 and x 3. ✎ Practice what you’ve learned
: Do Exercise 33. ▲ ■ Logarithmic Equations A logarithmic equation is one in which a logarithm of the variable occurs. For example, To solve for x, we write the equation in exponential form. log21 x 2 2 5 x 2 25 Exponential form x 32 2 30 Solve for x Another way of looking at the ﬁrst step is to raise the base, 2, to each side of the equation. 2log21 x2 2 25 x 2 25 Raise 2 to each side Property of logarithms x 32 2 30 Solve for x The method used to solve this simple problem is typical. We summarize the steps as follows. 404 CHAPTER 5 \| Exponential and Logarithmic Functions GUIDELINES FOR SOLVING LOGARITHMIC EQUATIONS 1. Isolate the logarithmic term on one side of the equation; you might ﬁrst need to combine the logarithmic terms. 2. Write the equation in exponential form (or raise the base to each side of the equation). 3. Solve for the variable. E X AM P L E 6 \| Solving Logarithmic Equations Solve each equation for x. (a) ln x 8 (b) ▼ SO LUTI O N (a) log21 25 x 3 2 ln x 8 x e8 Given equation Exponential form Therefore, x e8 2981. We can also solve this problem another way: ln x 8 eln x e8 x e8 Given equation Raise e to each side Property of ln (b) The ﬁrst step is to rewrite the equation in exponential form. log21 25 x 3 2 25 x 23 Given equation Exponential form (or raise 2 to each side) Check Your Answer If x 17, we get 25 x 8 x 25 8 17 log21 25 17 2 log2 8 3 ✔ ✎ Practice what you’ve learned: Do Exercises 37 and 41. ▲ E X AM P L E 7 \| Solving a Logarithmic Equation Solve the equation 4 3 log 16. 2x 1 2 ▼ SO LUTI O N We ﬁrst isolate the logarithmic term. This allows us to write the equation in exponential form. Check Your Answer If x 5000, we get 4 3 log 2 5000 1 2 4 3 log 10,000 4 3 4

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