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Identifying Polynomial Functions Which of the following are polynomial functions? f (x) = 2x 3 ⋅ 3x + 4 g(x) = −x(x2 − 4) h(x) = 5 √ — x + 2 Solution The first two functions are examples of polynomial functions because they can be written in the form f (x) = an xn +... + a2 x2 + a1 x + a0, where the powers are non-negative integers and the coefficients are real numbers. SECTION 3.3 power Functions and polynomial Functions 229 • f (x) can be written as f (x) = 6x4 + 4. • g(x) can be written as g(x) = −x3 + 4x. • h(x) cannot be written in this form and is therefore not a polynomial function. Identifying the Degree and leading Coefficient of a Polynomial Function Because of the form of a polynomial function, we can see an infinite variety in the number of terms and the power of the variable. Although the order of the terms in the polynomial function is not important for performing operations, we typically arrange the terms in descending order of power, or in general form. The degree of the polynomial is the highest power of the variable that occurs in the polynomial; it is the power of the first variable if the function is in general form. The leading term is the term containing the highest power of the variable, or the term with the highest degree. The leading coefficient is the coefficient of the leading term. terminology of polynomial functions We often rearrange polynomials so that the powers are descending. Leading coefficient Degree f (x) = an x n + … + a2 x 2 + a1x + a0 Leading term When a polynomial is written in this way, we say that it is in general form. How To… Given a polynomial function, identify the degree and leading coefficient. 1. Find the highest power of x to determine the degree function. 2. Identify the term containing the highest power of x to find the leading term. 3. Identify the coefficient of the leading term. Example 5 Identifying the Degree and Leading Coefficient of a Polynomial Function Identify the degree, leading term, and leading coefficient of the following polynomial functions. f (x) = 3 + 2x 2 −
4x 3 g(t) = 5t 5 − 2t 3 + 7t h(p) = 6p − p 3 − 2 Solution For the function f (x), the highest power of x is 3, so the degree is 3. The leading term is the term containing that degree, −4x3. The leading coefficient is the coefficient of that term, −4. For the function g(t), the highest power of t is 5, so the degree is 5. The leading term is the term containing that degree, 5t5. The leading coefficient is the coefficient of that term, 5. For the function h(p), the highest power of p is 3, so the degree is 3. The leading term is the term containing that degree, −p3; the leading coefficient is the coefficient of that term, −1. Try It #3 Identify the degree, leading term, and leading coefficient of the polynomial f (x) = 4x 2 − x 6 + 2x − 6. 230 CHAPTER 3 polynomial and rational Functions Identifying End Behavior of Polynomial Functions Knowing the degree of a polynomial function is useful in helping us predict its end behavior. To determine its end behavior, look at the leading term of the polynomial function. Because the power of the leading term is the highest, that term will grow significantly faster than the other terms as x gets very large or very small, so its behavior will dominate the graph. For any polynomial, the end behavior of the polynomial will match the end behavior of the term of highest degree. See Table 3. Polynomial Function Leading Term Graph of Polynomial Function f (x) = 5x4 + 2x3 − x − 4 5x4 –5 –4 –3 –2 f (x) = −2x6 − x5 + 3x4 + x3 −2x6 –5 –4 –3 –2 f (x) = 3x5 − 4x4 + 2x2 + 1 3x5 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 y 21 3 4 5 x 21 3 4 5 x 21 3
4 5 x f (x) = −6x3 + 7x2 + 3x + 1 −6x3 x Table 3 SECTION 3.3 power Functions and polynomial Functions 231 Example 6 Identifying End Behavior and Degree of a Polynomial Function Describe the end behavior and determine a possible degree of the polynomial function in Figure 7. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 21 3 4 5 6 x Figure 7 Solution As the input values x get very large, the output values f (x) increase without bound. As the input values x get very small, the output values f (x) decrease without bound. We can describe the end behavior symbolically by writing as x → −∞, as x → ∞, f (x) → −∞ f (x) → ∞ In words, we could say that as x values approach infinity, the function values approach infinity, and as x values approach negative infinity, the function values approach negative infinity. We can tell this graph has the shape of an odd degree power function that has not been reflected, so the degree of the polynomial creating this graph must be odd and the leading coefficient must be positive. Try It #4 Describe the end behavior, and determine a possible degree of the polynomial function in Figure 8. y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 21 3 4 5 6 x Figure 8 Example 7 Identifying End Behavior and Degree of a Polynomial Function Given the function f (x) = −3x2(x − 1)(x + 4), express the function as a polynomial in general form, and determine the leading term, degree, and end behavior of the function. Solution Obtain the general form by expanding the given expression for f (x). f (x) = −3x2(x − 1)(x + 4) = −3x2(x2 + 3x − 4) = −3x4 − 9x3 + 12x2 The general form is f (x) = −3x4 − 9x3 + 12x2. The leading term is −3x4; therefore, the degree of the polynomial is 4. The degree is even (4) and the leading coefficient is negative (−
3), so the end behavior is as x → −∞, f (x) → −∞ as x → ∞, f (x) → −∞ 232 CHAPTER 3 polynomial and rational Functions Try It #5 Given the function f (x) = 0.2(x − 2)(x + 1)(x − 5), express the function as a polynomial in general form and determine the leading term, degree, and end behavior of the function. Identifying Local Behavior of Polynomial Functions In addition to the end behavior of polynomial functions, we are also interested in what happens in the “middle” of the function. In particular, we are interested in locations where graph behavior changes. A turning point is a point at which the function values change from increasing to decreasing or decreasing to increasing. We are also interested in the intercepts. As with all functions, the y-intercept is the point at which the graph intersects the vertical axis. The point corresponds to the coordinate pair in which the input value is zero. Because a polynomial is a function, only one output value corresponds to each input value so there can be only one y-intercept (0, a0). The x-intercepts occur at the input values that correspond to an output value of zero. It is possible to have more than one x-intercept. See Figure 9. y Turning point x-intercepts x Turning point ←y-intercept Figure 9 intercepts and turning points of polynomial functions A turning point of a graph is a point at which the graph changes direction from increasing to decreasing or decreasing to increasing. The y-intercept is the point at which the function has an input value of zero. The x-intercepts are the points at which the output value is zero. How To… Given a polynomial function, determine the intercepts. 1. Determine the y-intercept by setting x = 0 and finding the corresponding output value. 2. Determine the x-intercepts by solving for the input values that yield an output value of zero. Example 8 Determining the Intercepts of a Polynomial Function Given the polynomial function f (x) = (x − 2)(x + 1)(x − 4), written in factored form for your convenience, determine the y- and x-intercepts. Solution The y-intercept occurs when the input is zero so
substitute 0 for x. The y-intercept is (0, 8). f (0) = (0 − 2)(0 + 1)(0 − 4) = (−2)(1)(− 4) = 8 SECTION 3.3 power Functions and polynomial Functions 233 The x-intercepts occur when the output is zero. 0 = (x − 2)(x + 1)(x − 4) The x-intercepts are (2, 0), (–1, 0), and (4, 0). x = 2 or x − 2 = 0 or x + 1 = 0 x = −1 or or x − 4 = 0 x = 4 We can see these intercepts on the graph of the function shown in Figure 10. y-intercept (0, 85 –4 –3 –2 21 3 –1 –1 –2 –3 –4 Figure 10 x 4 5 x-intercepts (−1, 0), (2, 0), and (4, 0) Example 9 Determining the Intercepts of a Polynomial Function with Factoring Given the polynomial function f (x) = x4 − 4x2 − 45, determine the y- and x-intercepts. Solution The y-intercept occurs when the input is zero. The y-intercept is (0, −45). f (0) = (0)4 − 4(0)2 − 45 = −45 The x-intercepts occur when the output is zero. To determine when the output is zero, we will need to factor the polynomial. f (x) = x4 − 4x2 − 45 = (x2 − 9)(x2 + 5) = (x − 3)(x + 3)(x2 + 5) 0 = (x − 3)(x + 3)(x2 + 5) x − 3 = 0 or x + 3 = 0 or x2 + 5 = 0 x = 3 or x = −3 or (no real solution) The x-intercepts are (3, 0) and (−3, 0). We can see these intercepts on the graph of the function shown in Figure 11. We can see that the function is even because f (x) = f (−x). y 120 100 80 60 40 20 –2 –1 –20 –40 –60 –80 –100 –120 –5 –4 –3 x-intercepts (−3, 0)
and (3, 0) x 21 3 4 5 y-intercept (0, −45) Try It #6 Given the polynomial function f (x) = 2x3 − 6x2 − 20x, determine the y- and x-intercepts. Figure 11 234 CHAPTER 3 polynomial and rational Functions Comparing Smooth and Continuous Graphs The degree of a polynomial function helps us to determine the number of x-intercepts and the number of turning points. A polynomial function of nth degree is the product of n factors, so it will have at most n roots or zeros, or x-intercepts. The graph of the polynomial function of degree n must have at most n − 1 turning points. This means the graph has at most one fewer turning point than the degree of the polynomial or one fewer than the number of factors. A continuous function has no breaks in its graph: the graph can be drawn without lifting the pen from the paper. A smooth curve is a graph that has no sharp corners. The turning points of a smooth graph must always occur at rounded curves. The graphs of polynomial functions are both continuous and smooth. intercepts and turning points of polynomials A polynomial of degree n will have, at most, n x-intercepts and n − 1 turning points. Example 10 Determining the Number of Intercepts and Turning Points of a Polynomial Without graphing the function, determine the local behavior of the function by finding the maximum number of x-intercepts and turning points for f (x) = −3x10 + 4x7 − x4 + 2x3. Solution The polynomial has a degree of 10, so there are at most 10 x-intercepts and at most 10 − 1 = 9 turning points. Try It #7 Without graphing the function, determine the maximum number of x-intercepts and turning points for f (x) = 108 − 13x9 − 8x4 + 14x12 + 2x3 Example 11 Drawing Conclusions about a Polynomial Function from the Graph What can we conclude about the polynomial represented by the graph shown in Figure 12 based on its intercepts and turning points? y 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 12 Solution The end behavior of the graph
tells us this is the graph of an even-degree polynomial. See Figure 13. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 x-intercepts 21 3 4 5 x Turning points Figure 13 The graph has 2 x-intercepts, suggesting a degree of 2 or greater, and 3 turning points, suggesting a degree of 4 or greater. Based on this, it would be reasonable to conclude that the degree is even and at least 4. SECTION 3.3 power Functions and polynomial Functions 235 Try It #8 What can we conclude about the polynomial represented by the graph shown in Figure 14 based on its intercepts and turning points? y –5 –4 –3 –2 10 8 6 4 2 –1 –2 –4 –6 –8 –10 21 3 4 5 x Figure 14 Example 12 Drawing Conclusions about a Polynomial Function from the Factors Given the function f (x) = −4x(x + 3)(x − 4), determine the local behavior. Solution The y-intercept is found by evaluating f (0). f (0) = −4(0)(0 + 3)(0 − 4) = 0 The y-intercept is (0, 0). The x-intercepts are found by determining the zeros of the function. 0 = −4x(x + 3)(x − 4) x = 0 or x + 3 = 0 or x − 4 = 0 x = 0 or x = −3 or x = 4 The x-intercepts are (0, 0), (−3, 0), and (4, 0). The degree is 3 so the graph has at most 2 turning points. Try It #9 Given the function f (x) = 0.2(x − 2)(x + 1)(x − 5), determine the local behavior. Access these online resources for additional instruction and practice with power and polynomial functions. • Find Key Information about a Given Polynomial Function (http://openstaxcollege.org/l/keyinfopoly) • end Behavior of a Polynomial Function (http://openstaxcollege.org/l/endbehavior) • Turning Points and x-Intercepts of Polynomial Functions (http://openstaxcollege.org/l/turningpoints) • least Possible Degree of a Polynomial Function (http://openstax
college.org/l/leastposdegree) 236 CHAPTER 3 polynomial and rational Functions 3.3 SeCTIOn exeRCISeS VeRBAl 1. Explain the difference between the coefficient 2. If a polynomial function is in factored form, what of a power function and its degree. 3. In general, explain the end behavior of a power function with odd degree if the leading coefficient is positive. 5. What can we conclude if, in general, the graph of a polynomial function exhibits the following end behavior? As x → −∞, f (x) → −∞ and as x → ∞, f (x) → −∞. AlGeBRAIC would be a good first step in order to determine the degree of the function? 4. What is the relationship between the degree of a polynomial function and the maximum number of turning points in its graph? For the following exercises, identify the function as a power function, a polynomial function, or neither. 6. f (x) = x5 9. f (x) = x2 _____ x2 − 1 7. f (x) = (x2)3 8. f (x) = x − x4 10. f (x) = 2x(x + 2)(x − 1)2 11. f (x) = 3x + 1 For the following exercises, find the degree and leading coefficient for the given polynomial. 12. −3x 15. x(4 − x2)(2x + 1) 13. 7 − 2x2 16. x 2 (2x − 3)2 14. −2x2 − 3x5 + x − 6 For the following exercises, determine the end behavior of the functions. 17. f (x) = x4 20. f (x) = −x9 18. f (x) = x3 19. f (x) = −x4 21. f (x) = −2x4 − 3x2 + x − 1 22. f (x) = 3x2 + x − 2 23. f (x) = x2(2x3 − x + 1) 24. f (x) = (2 − x)7 For the following exercises, find the intercepts of the functions. 25. f (t) = 2(t − 1)(t + 2)(t − 3)
26. g(n) = −2(3n − 1)(2n + 1) 27. f (x) = x4 − 16 28. f (x) = x3 + 27 29. f (x) = x(x2 − 2x − 8) 30. f (x) = (x + 3)(4x2 − 1) GRAPHICAl For the following exercises, determine the least possible degree of the polynomial function shown. 31. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 32. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 33. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x SECTION 3.3 section exercises 237 34. 37. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 –5 –4 –3 –2 35. 21 3 4 5 x –5 –4 –3 –2 38. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 36. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x 21 3 4 5 x For the following exercises, determine whether the graph of the function provided is a graph of a polynomial function. If so, determine the number of turning points and the least possible degree for the function. 42. y x 39. 43. y y y y 40. 44. x x 41. 45. x x y y x x 238 CHAPTER 3 polynomial and rational Functions nUMeRIC For the following exercises, make a table to confirm the end behavior of the function. 46. f (x) = −x3 49. f (x) = (x − 1)(x − 2)(3 − x) 47. f (x) =
x4 − 5x2 50. f (x) = x5 __ 10 − x4 48. f (x) = x2(1 − x)2 TeCHnOlOGY For the following exercises, graph the polynomial functions using a calculator. Based on the graph, determine the intercepts and the end behavior. 51. f (x) = x3(x − 2) 52. f (x) = x(x − 3)(x + 3) 53. f (x) = x(14 − 2x)(10 − 2x) 54. f (x) = x(14 − 2x)(10 − 2x)2 55. f (x) = x3 − 16x 56. f (x) = x3 − 27 57. f (x) = x4 − 81 60. f (x) = x3 − 0.01x exTenSIOnS 58. f (x) = −x3 + x2 + 2x 59. f (x) = x3 − 2x2 − 15x For the following exercises, use the information about the graph of a polynomial function to determine the function. Assume the leading coefficient is 1 or −1. There may be more than one correct answer. 61. The y-intercept is (0, −4). The x-intercepts are (−2, 0), (2, 0). Degree is 2. End behavior: as x → −∞, f (x) → ∞, as x → ∞, f (x) → ∞. 62. The y-intercept is (0, 9). The x-intercepts are (−3, 0), (3, 0). Degree is 2. End behavior: as x → −∞, f (x) → −∞, as x → ∞, f (x) → −∞. 63. The y-intercept is (0, 0). The x-intercepts are (0, 0), (2, 0). Degree is 3. End behavior: as x → −∞, f (x) → −∞, as x → ∞, f (x) → ∞. 64. The y-intercept is (0, 1). The x-intercept is (1, 0). Degree is 3. End behavior: as x → −∞, f (x)
→ ∞, as x → ∞, f (x) → −∞. 65. The y-intercept is (0, 1). There is no x-intercept. Degree is 4. End behavior: as x → −∞, f (x) → ∞, as x → ∞, f (x) → ∞. ReAl-WORlD APPlICATIOnS For the following exercises, use the written statements to construct a polynomial function that represents the required information. 66. An oil slick is expanding as a circle. The radius of the circle is increasing at the rate of 20 meters per day. Express the area of the circle as a function of d, the number of days elapsed. 67. A cube has an edge of 3 feet. The edge is increasing at the rate of 2 feet per minute. Express the volume of the cube as a function of m, the number of minutes elapsed. 68. A rectangle has a length of 10 inches and a width of 6 inches. If the length is increased by x inches and the width increased by twice that amount, express the area of the rectangle as a function of x. 69. An open box is to be constructed by cutting out square corners of x-inch sides from a piece of cardboard 8 inches by 8 inches and then folding up the sides. Express the volume of the box as a function of x. 70. A rectangle is twice as long as it is wide. Squares of side 2 feet are cut out from each corner. Then the sides are folded up to make an open box. Express the volume of the box as a function of the width (x). SECTION 3.4 graphs oF polynomial Functions 239 leARnInG OBjeCTIVeS In this section, you will: • • • • • • • Recognize characteristics of graphs of polynomial functions. Use factoring to find zeros of polynomial functions. Identify zeros and their multiplicities. Determine end behavior. Understand the relationship between degree and turning points. Graph polynomial functions. Use the Intermediate Value Theorem. 3.4 GRAPHS OF POlYnOMIAl FUnCTIOnS The revenue in millions of dollars for a fictional cable company from 2006 through 2013 is shown in Table 1. Year 2006 2007 2008 2009 2010 2011 2012 2013 Revenues 52.4 52.8 51.2 49.5 48.6
48.6 48.7 47.1 The revenue can be modeled by the polynomial function Table 1 R(t) = −0.037t4 + 1.414t3 − 19.777t2 + 118.696t − 205.332 where R represents the revenue in millions of dollars and t represents the year, with t = 6 corresponding to 2006. Over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing? These questions, along with many others, can be answered by examining the graph of the polynomial function. We have already explored the local behavior of quadratics, a special case of polynomials. In this section we will explore the local behavior of polynomials in general. Recognizing Characteristics of Graphs of Polynomial Functions Polynomial functions of degree 2 or more have graphs that do not have sharp corners; recall that these types of graphs are called smooth curves. Polynomial functions also display graphs that have no breaks. Curves with no breaks are called continuous. Figure 1 shows a graph that represents a polynomial function and a graph that represents a function that is not a polynomial. y y f x Figure 1 x f 240 CHAPTER 3 polynomial and rational Functions Example 1 Recognizing Polynomial Functions Which of the graphs in Figure 2 represents a polynomial function? y y f h y x x Figure 2 y g k x x Solution The graphs of f and h are graphs of polynomial functions. They are smooth and continuous. The graphs of g and k are graphs of functions that are not polynomials. The graph of function g has a sharp corner. The graph of function k is not continuous. Q & A… Do all polynomial functions have as their domain all real numbers? Yes. Any real number is a valid input for a polynomial function. Using Factoring to Find Zeros of Polynomial Functions Recall that if f is a polynomial function, the values of x for which f (x) = 0 are called zeros of f. If the equation of the polynomial function can be factored, we can set each factor equal to zero and solve for the zeros. We can use this method to find x-intercepts because at the x-intercepts we find the input values when the output value is zero. For general polynomials, this can be a challenging prospect
. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. Consequently, we will limit ourselves to three cases in this section: 1. The polynomial can be factored using known methods: greatest common factor and trinomial factoring. 2. The polynomial is given in factored form. 3. Technology is used to determine the intercepts. How To… Given a polynomial function f, find the x-intercepts by factoring. 1. Set f (x) = 0. 2. If the polynomial function is not given in factored form: a. Factor out any common monomial factors. b. Factor any factorable binomials or trinomials. 3. Set each factor equal to zero and solve to find the x-intercepts. SECTION 3.4 graphs oF polynomial Functions 241 Example 2 Finding the x-Intercepts of a Polynomial Function by Factoring Find the x-intercepts of f (x) = x 6 − 3x 4 + 2x2. Solution We can attempt to factor this polynomial to find solutions for f (x) = 0. x 6 − 3x 4 + 2x 2 = 0 Factor out the greatest common factor. x 2(x 4 − 3x2 + 2) = 0 Factor the trinomial. x 2(x 2 − 1)(x 2 − 2) = 0 Set each factor equal to zero. (x 2 − 1) = 0 (x 2 − 2) = 0 x 2 = 0 or x 2 = 1 or This gives us five x-intercepts: (0, 0), (1, 0), (−1, 0),  √ even function. — 2, 0 , and  − √ — 2, 0 . See Figure 3. We can see that this is an (–, 0) √2 –2 f (x) = x6 – 3x4 + 2x2 y 2 1 (, 0)√2 x 2 (–1, 0) –1 (0, 0) (1, 0) –2 Figure 3 Example 3 Finding the x-Intercepts of a Polynomial Function by Factoring
Find the x-intercepts of f (x) = x 3 − 5x 2 − x + 5. Solution Find solutions for f (x) = 0 by factoring. x 3 − 5x 2 − x + 5 = 0 Factor by grouping. x 2(x − 5) − (x − 5) = 0 Factor out the common factor. (x 2 − 1)(x − 5) = 0 Factor the difference of squares. (x + 1)(x − 1)(x − 5) = 0 Set each factor equal to zero. x + 1 = 0 or x − 1 = 0 or x − 5 = 0 x = −1 x = 1 x = 5 There are three x-intercepts: (−1, 0), (1, 0), and (5, 0). See Figure 4. y 18 12 6 f (x) = x3 − 5x2 − x + 5 –6 –4 –2 2 4 6 x –6 –12 –18 Figure 4 242 CHAPTER 3 polynomial and rational Functions Example 4 Finding the y- and x-Intercepts of a Polynomial in Factored Form Find the y- and x-intercepts of g(x) = (x − 2)2(2x + 3). Solution The y-intercept can be found by evaluating g(0). g(0) = (0 − 2)2(2(0) + 3) = 12 So the y-intercept is (0, 12). The x-intercepts can be found by solving g(x) = 0. (x − 2)2(2x + 3) = 0 (x − 2)2 = 0 (2x + 3) = 0 x − 2 = 0 or x = 2 x = − 3 __ 2 So the x-intercepts are (2, 0) and  − 3, 0 . __ 2 Analysis We can always check that our answers are reasonable by using a graphing calculator to graph the polynomial as shown in Figure 5. y 15 (0, 12) 12 9 6 3 (–1.5, 0) g(x) = (x − 2)2(2x + 3) (2, 0) –4 –3 –2 –1 1 2 3 4 x –3 Figure 5 Example 5 Finding the x-Intercepts of a Polynomial Function Using a Graph Find the
x-intercepts of h(x) = x3 + 4x2 + x − 6. Solution This polynomial is not in factored form, has no common factors, and does not appear to be factorable using techniques previously discussed. Fortunately, we can use technology to find the intercepts. Keep in mind that some values make graphing difficult by hand. In these cases, we can take advantage of graphing utilities. Looking at the graph of this function, as shown in Figure 6, it apears that there are x-intercepts at x = −3, −2, and 1. h(x) = x3 + 4x2 + x − 6 y 2 –4 –2 2 4 x –2 –4 –6 –8 Figure 6 SECTION 3.4 graphs oF polynomial Functions 243 We can check whether these are correct by substituting these values for x and verifying that Since h(x) = x 3 + 4x 2 + x − 6, we have: h(−3) = h(−2) = h(1) = 0. h(−3) = (−3)3 + 4(−3)2 + (−3) − 6 = −27 + 36 − 3 − 6 = 0 h(−2) = (−2)3 + 4(−2)2 + (−2) − 6 = −8 + 16 − 2 − 6 = 0 h(1) = (1)3 + 4(1)2 + (1 Each x-intercept corresponds to a zero of the polynomial function and each zero yields a factor, so we can now write the polynomial in factored form. h(x) = x3 + 4x2 + x − 6 = (x + 3)(x + 2)(x − 1) Try It #1 Find the y- and x-intercepts of the function f (x) = x 4 − 19x 2 + 30x. Identifying Zeros and Their Multiplicities Graphs behave differently at various x-intercepts. Sometimes, the graph will cross over the horizontal axis at an intercept. Other times, the graph will touch the horizontal axis and bounce off. Suppose, for example, we graph the function f (x) = (x + 3)(x − 2)2(x + 1)3. Notice in Figure 7 that the behavior of the function at each of the x-
intercepts is different. f (x) = (x + 3)(x − 2)2(x + 1)3 2 4 x y 40 30 20 10 –10 –20 –30 –40 –4 –2 Figure 7 Identifying the behavior of the graph at an x-intercept by examining the multiplicity of the zero. The x-intercept x = −3 is the solution of equation (x + 3) = 0. The graph passes directly through the x-intercept at x = −3. The factor is linear (has a degree of 1), so the behavior near the intercept is like that of a line—it passes directly through the intercept. We call this a single zero because the zero corresponds to a single factor of the function. The x-intercept x = 2 is the repeated solution of equation (x − 2)2 = 0. The graph touches the axis at the intercept and changes direction. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept. (x − 2)2 = (x − 2)(x − 2) The factor is repeated, that is, the factor (x − 2) appears twice. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. The zero associated with this factor, x = 2, has multiplicity 2 because the factor (x − 2) occurs twice. 244 CHAPTER 3 polynomial and rational Functions The x-intercept x = −1 is the repeated solution of factor (x + 1)3 = 0. The graph passes through the axis at the intercept, but flattens out a bit first. This factor is cubic (degree 3), so the behavior near the intercept is like that of a cubic— with the same S-shape near the intercept as the toolkit function f (x) = x 3. We call this a triple zero, or a zero with multiplicity 3. For zeros with even multiplicities, the graphs touch or are tangent to the x-axis. For zeros with odd multiplicities, the graphs cross or intersect the x-axis. See Figure 8 for examples of graphs of polynomial functions with multiplicity 1, 2, and 3. y y y p = Single zero Zero with multiplicity 2 Zero with multiplicity 3 Figure 8 For higher even
powers, such as 4, 6, and 8, the graph will still touch and bounce off of the horizontal axis but, for each increasing even power, the graph will appear flatter as it approaches and leaves the x-axis. For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the x-axis. graphical behavior of polynomials at x-intercepts If a polynomial contains a factor of the form (x − h)p, the behavior near the x-intercept h is determined by the power p. We say that x = h is a zero of multiplicity p. The graph of a polynomial function will touch the x-axis at zeros with even multiplicities. The graph will cross the x-axis at zeros with odd multiplicities. The sum of the multiplicities is the degree of the polynomial function. How To… Given a graph of a polynomial function of degree n, identify the zeros and their multiplicities. 1. If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero. 2. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. 3. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity. 4. The sum of the multiplicities is n. Example 6 Identifying Zeros and Their Multiplicities Use the graph of the function of degree 6 in Figure 9 to identify the zeros of the function and their possible multiplicities. SECTION 3.4 graphs oF polynomial Functions 245 y 240 160 80 –6 –4 –2 2 4 6 x –80 –160 –240 Figure 9 Solution The polynomial function is of degree n. The sum of the multiplicities must be n. Starting from the left, the first zero occurs at x = −3. The graph touches the x-axis, so the multiplicity of the zero must be even. The zero of −3 has multiplicity 2. The next zero occurs at x = −1. The graph looks almost linear at this point. This is a single zero of multiplicity 1. The last zero occurs at x = 4. The graph crosses the x-axis, so the multipl
icity of the zero must be odd. We know that the multiplicity is likely 3 and that the sum of the multiplicities is likely 6. Try It #2 Use the graph of the function of degree 5 in Figure 10 to identify the zeros of the function and their multiplicities. y 60 40 20 –6 –4 –2 2 4 6 x –20 –40 –60 Figure 10 Determining end Behavior As we have already learned, the behavior of a graph of a polynomial function of the form f (x) = an x n +... + a2 x 2 + a1 x + a0 will either ultimately rise or fall as x increases without bound and will either rise or fall as x decreases without bound. This is because for very large inputs, say 100 or 1,000, the leading term dominates the size of the output. The same is true for very small inputs, say −100 or −1,000. Recall that we call this behavior the end behavior of a function. As we pointed out when discussing quadratic equations, when the leading term of a polynomial function, anxn, is an even power function, as x increases or decreases without bound, f (x) increases without bound. When the leading term is an odd power function, as x decreases without bound, f (x) also decreases without bound; as x increases without bound, f (x) also increases without bound. If the leading term is negative, it will change the direction of the end behavior. Figure 11 summarizes all four cases. 246 CHAPTER 3 polynomial and rational Functions Even Degree Odd Degree Positive Leading Coefficient, an > 0 y Positive Leading Coefficient, an > 0 y x x End Behavior: x → ∞, f (x) → ∞ x → −∞, f (x) → ∞ End Behavior: x → ∞, f (x) → ∞ x → −∞, f (x) → ∞ Negative Leading Coefficient, an < 0 y Negative Leading Coefficient, an < 0 y x x End Behavior: x → ∞, f (x) → −∞ x → −∞, f (x) → −∞ End Behavior: x → ∞, f (x) → −∞ x → −∞, f (x) → ∞ Figure 11 Understanding the Relationship Between Degree and Turning Points In addition to the end behavior, recall that we can analyze a
polynomial function’s local behavior. It may have a turning point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). Look at the graph of the polynomial function f (x) = x4 − x3 − 4x2 + 4x in Figure 12. The graph has three turning points. y Increasing Decreasing Increasing Decreasing x Turning points Figure 12 This function f is a 4th degree polynomial function and has 3 turning points. The maximum number of turning points of a polynomial function is always one less than the degree of the function. SECTION 3.4 graphs oF polynomial Functions 247 interpreting turning points A turning point is a point of the graph where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising). A polynomial of degree n will have at most n − 1 turning points. Example 7 Finding the Maximum Number of Turning Points Using the Degree of a Polynomial Function Find the maximum number of turning points of each polynomial function. b. f (x) = −(x − 1)2(1 + 2x2) a. f (x) = −x 3 + 4x 5 − 3x 2 + 1 Solution a. f (x) = −x3 + 4x 5 − 3x 2 + 1 First, rewrite the polynomial function in descending order: f (x) = 4x 5 − x 3 − 3x 2 + 1 Identify the degree of the polynomial function. This polynomial function is of degree 5. The maximum number of turning points is 5 − 1 = 4. b. f (x) = −(x − 1)2(1 + 2x2) First, identify the leading term of the polynomial function if the function were expanded. f (x) = −(x − 1)2(1 + 2x2) an = −(x 2)(2x 2) − 2x 4 Then, identify the degree of the polynomial function. This polynomial function is of degree 4. The maximum number of turning points is 4 − 1 = 3. Graphing Polynomial Functions We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Let us put this all together and look at the steps required to graph polynomial functions
. How To… Given a polynomial function, sketch the graph. 1. Find the intercepts. 2. Check for symmetry. If the function is an even function, its graph is symmetrical about the y-axis, that is, f (−x) = f (x). If a function is an odd function, its graph is symmetrical about the origin, that is, f (−x) = −f (x). 3. Use the multiplicities of the zeros to determine the behavior of the polynomial at the x-intercepts. 4. Determine the end behavior by examining the leading term. 5. Use the end behavior and the behavior at the intercepts to sketch a graph. 6. Ensure that the number of turning points does not exceed one less than the degree of the polynomial. 7. Optionally, use technology to check the graph. Example 8 Sketching the Graph of a Polynomial Function Sketch a graph of f (x) = −2(x + 3)2(x − 5). Solution This graph has two x-intercepts. At x = −3, the factor is squared, indicating a multiplicity of 2. The graph will bounce at this x-intercept. At x = 5, the function has a multiplicity of one, indicating the graph will cross through the axis at this intercept. The y-intercept is found by evaluating f (0). The y-intercept is (0, 90). f (0) = −2(0 + 3)2(0 − 5) = −2 ⋅ 9 ⋅ (−5) = 90 248 CHAPTER 3 polynomial and rational Functions Additionally, we can see the leading term, if this polynomial were multiplied out, would be −2x3, so the end behavior is that of a vertically reflected cubic, with the outputs decreasing as the inputs approach infinity, and the outputs increasing as the inputs approach negative infinity. See Figure 13. y Figure 13 x To sketch this, we consider that: • As x → −∞ the function f (x) → ∞, so we know the graph starts in the second quadrant and is decreasing toward the x-axis. • Since f (−x) = −2(−x + 3)2 (−x − 5) is not equal to f (x), the graph does not display symmetry. • At (−3, 0), the graph bounces off of the x
-axis, so the function must start increasing. At (0, 90), the graph crosses the y-axis at the y-intercept. See Figure 14. y (0, 90) (−3, 0) x Figure 14 Somewhere after this point, the graph must turn back down or start decreasing toward the horizontal axis because the graph passes through the next intercept at (5, 0). See Figure 15. y (0, 90) (−3, 0) (5, 0) x Figure 15 As x → ∞ the function f (x) → −∞, so we know the graph continues to decrease, and we can stop drawing the graph in the fourth quadrant. SECTION 3.4 graphs oF polynomial Functions 249 Using technology, we can create the graph for the polynomial function, shown in Figure 16, and verify that the resulting graph looks like our sketch in Figure 15. f (x) = −2(x + 3)2(x − 5) y 180 120 60 –6 –4 –2 2 4 6 x –60 –120 –180 Figure 16 The complete graph of the polynomial function f (x ) = −2(x + 3)2(x − 5) Try It #3 1 __ x(x − 1)4(x + 3)3. Sketch a graph of f (x) = 4 Using the Intermediate Value Theorem In some situations, we may know two points on a graph but not the zeros. If those two points are on opposite sides of the x-axis, we can confirm that there is a zero between them. Consider a polynomial function f whose graph is smooth and continuous. The Intermediate Value Theorem states that for two numbers a and b in the domain of f, if a < b and f (a) ≠ f (b), then the function f takes on every value between f (a) and f (b). We can apply this theorem to a special case that is useful in graphing polynomial functions. If a point on the graph of a continuous function f at x = a lies above the x-axis and another point at x = b lies below the x-axis, there must exist a third point between x = a and x = b where the graph crosses the x-axis. Call this point (c, f (c)). This means that we are assured there is a solution c where f (c) = 0. In other words
, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis. Figure 17 shows that there is a zero between a and b. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 f (b) is positive f (c) = 0 x 5 21 3 4 f(a) is negative Figure 17 Using the Intermediate Value Theorem to show there exists a zero Intermediate Value Theorem Let f be a polynomial function. The Intermediate Value Theorem states that if f (a) and f (b) have opposite signs, then there exists at least one value c between a and b for which f (c) = 0. 250 CHAPTER 3 polynomial and rational Functions Example 9 Using the Intermediate Value Theorem Show that the function f (x) = x3 − 5x2 + 3x + 6 has at least two real zeros between x = 1 and x = 4. Solution As a start, evaluate f (x) at the integer values x = 1, 2, 3, and 4. See Table 2. x f (x) 1 5 2 0 3 −3 4 2 Table 2 We see that one zero occurs at x = 2. Also, since f (3) is negative and f (4) is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4. We have shown that there are at least two real zeros between x = 1 and x = 4. Analysis We can also see on the graph of the function in Figure 18 that there are two real zeros between x = 1 and x = 4. –5 –4 –3 –2 y 10 8 6 4 2 –1 –2 –4 –6 –8 –10 f (1) = 5 is positive f (4) = 2 is positive 21 3 4 5 f (3) = −3 is negative Figure 18 Try It #4 Show that the function f (x) = 7x5 − 9x4 − x2 has at least one real zero between x = 1 and x = 2. Writing Formulas for Polynomial Functions Now that we know how to find zeros of polynomial functions, we can use them to write formulas based on graphs. Because a polynomial function written in factored form will have an x-intercept where
each factor is equal to zero, we can form a function that will pass through a set of x-intercepts by introducing a corresponding set of factors. factored form of polynomials If a polynomial of lowest degree p has horizontal intercepts at x = x1, x2, …, xn, then the polynomial can be written (x − x2)p2 … (x − xn)pn where the powers pi on each factor can be determined in the factored form: f (x) = a(x − x1)p1 by the behavior of the graph at the corresponding intercept, and the stretch factor a can be determined given a value of the function other than the x-intercept. How To… Given a graph of a polynomial function, write a formula for the function. 1. Identify the x-intercepts of the graph to find the factors of the polynomial. 2. Examine the behavior of the graph at the x-intercepts to determine the multiplicity of each factor. 3. Find the polynomial of least degree containing all the factors found in the previous step. 4. Use any other point on the graph (the y-intercept may be easiest) to determine the stretch factor. SECTION 3.4 graphs oF polynomial Functions 251 Example 10 Writing a Formula for a Polynomial Function from the Graph Write a formula for the polynomial function shown in Figure 19. y 6 4 2 –6 –4 –2 2 4 6 x –2 –4 –6 Figure 19 Solution This graph has three x-intercepts: x = −3, 2, and 5. The y-intercept is located at (0, −2). At x = −3 and x = 5, the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. At x = 2, the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). Together, this gives us f (x) = a(x + 3)(x − 2)2(x − 5) To determine the stretch factor, we utilize another point on the graph. We will use the y-intercept (0, −2), to solve for a. f (0) = a(0 + 3)(0 − 2)2(0 − 5) −2 = a(
0 + 3)(0 − 2)2(0 − 5) −2 = −60a The graphed polynomial appears to represent the function f (x) = (x + 3)(x − 2)2 (x − 5). a = 1 __ 30 1 __ 30 Try It #5 Given the graph shown in Figure 20, write a formula for the function shown. y 12 8 4 –6 –4 –2 2 4 6 x –4 –8 –12 Figure 20 Using Local and Global Extrema With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph. Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum. These are also referred to as the absolute maximum and absolute minimum values of the function. 252 CHAPTER 3 polynomial and rational Functions local and global extrema A local maximum or local minimum at x = a (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around x = a. If a function has a local maximum at a, then f (a) ≥ f (x) for all x in an open interval around x = a. If a function has a local minimum at a, then f (a) ≤ f (x) for all x in an open interval around x = a. A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at a, then f (a) ≥ f (x) for all x. If a function has a global minimum at a, then f (a) ≤ f (x) for all x. We can see the difference between local and global extrema in Figure 21. –6 –5 –4 –3 –2 y Global maximum Local maximum 21 3 4 5 6 x Local minimum 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6 Figure 21 Q & A… Do all polynomial functions have
a global minimum or maximum? No. Only polynomial functions of even degree have a global minimum or maximum. For example, f (x) = x has neither a global maximum nor a global minimum. Example 11 Using Local Extrema to Solve Applications An open-top box is to be constructed by cutting out squares from each corner of a 14 cm by 20 cm sheet of plastic then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box. Solution We will start this problem by drawing a picture like that in Figure 22, labeling the width of the cut-out squares with a variable, w. w w w w Figure 22 w w w w Notice that after a square is cut out from each end, it leaves a (14 − 2w) cm by (20 − 2w) cm rectangle for the base of the box, and the box will be w cm tall. This gives the volume V(w) = (20 − 2w)(14 − 2w)w = 280w − 68w2 + 4w3 SECTION 3.4 graphs oF polynomial Functions 253 Notice, since the factors are w, 20 − 2w and 14 − 2w, the three zeros are 10, 7, and 0, respectively. Because a height of 0 cm is not reasonable, we consider the only the zeros 10 and 7. The shortest side is 14 and we are cutting off two squares, so values w may take on are greater than zero or less than 7. This means we will restrict the domain of this function to 0 < w < 7. Using technology to sketch the graph of V(w) on this reasonable domain, we get a graph like that in Figure 23. We can use this graph to estimate the maximum value for the volume, restricted to values for w that are reasonable for this problem—values from 0 to 7. V(w) V(w) = 280w − 68w2 + 4w3 400 300 200 100 –2 2 –100 –200 4 68 1 0 12 w Figure 23 From this graph, we turn our focus to only the portion on the reasonable domain, [0, 7]. We can estimate the maximum value to be around 340 cubic cm, which occurs when the squares are about 2.75 cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure
24. V(w) 340 339 338 337 336 335 334 333 332 331 330 2.4 2.6 2.8 3 Figure 24 w From this zoomed-in view, we can refine our estimate for the maximum volume to about 339 cubic cm, when the squares measure approximately 2.7 cm on each side. Try It #6 Use technology to find the maximum and minimum values on the interval [−1, 4] of the function f (x) = −0.2(x − 2)3(x + 1)2(x − 4). Access the following online resource for additional instruction and practice with graphing polynomial functions. • Intermediate Value Theorem (http://openstaxcollege.org/l/ivt) 254 CHAPTER 3 polynomial and rational Functions 3.4 SeCTIOn exeRCISeS VeRBAl 1. What is the difference between an x-intercept and a zero of a polynomial function f? 2. If a polynomial function of degree n has n distinct zeros, what do you know about the graph of the function? 3. Explain how the Intermediate Value Theorem can 4. Explain how the factored form of the polynomial assist us in finding a zero of a function. helps us in graphing it. 5. If the graph of a polynomial just touches the x-axis and then changes direction, what can we conclude about the factored form of the polynomial? AlGeBRAIC For the following exercises, find the x- or t-intercepts of the polynomial functions. 6. C(t) = 2(t − 4)(t + 1)(t − 6) 7. C(t) = 3(t + 2)(t − 3)(t + 5) 8. C(t) = 4t(t − 2)2(t + 1) 9. C(t) = 2t(t − 3)(t + 1)2 10. C(t) = 2t4 − 8t3 + 6t2 11. C(t) = 4t4 + 12t3 − 40t2 12. f (x) = x4 − x2 13. f (x) = x3 + x2 − 20x 14. f (x) = x3 + 6x2 − 7x 15. f (x) = x3 + x2 − 4x
− 4 16. f (x) = x3 + 2x2 − 9x − 18 17. f (x) = 2x3 − x2 − 8x + 4 18. f (x) = x6 − 7x3 − 8 19. f (x) = 2x4 + 6x2 − 8 20. f (x) = x3 − 3x2 − x + 3 21. f (x) = x6 − 2x4 − 3x2 22. f (x) = x6 − 3x4 − 4x2 23. f (x) = x5 − 5x3 + 4x For the following exercises, use the Intermediate Value Theorem to confirm that the given polynomial has at least one zero within the given interval. 24. f (x) = x3 − 9x, between x = −4 and x = −2. 25. f (x) = x3 − 9x, between x = 2 and x = 4. 26. f (x) = x5 − 2x, between x = 1 and x = 2. 27. f (x) = −x4 + 4, between x = 1 and x = 3. 28. f (x) = −2x3 − x, between x = −1 and x = 1. 29. f (x) = x3 − 100x + 2, between x = 0.01 and x = 0.1 For the following exercises, find the zeros and give the multiplicity of each. 30. f (x) = (x + 2)3(x − 3)2 32. f (x) = x3 (x − 1)3(x + 2) 31. f (x) = x2(2x + 3)5(x − 4)2 33. f (x) = x2(x2 + 4x + 4) 34. f (x) = (2x + 1)3(9x2 − 6x + 1) 35. f (x) = (3x + 2)5(x2 − 10x + 25) 36. f (x) = x(4x2 − 12x + 9)(x2 + 8x + 16) 37. f (x) = x6 − x5 − 2x4 38. f (x) = 3x4 + 6x3 + 3x2 40.
f (x) = 2x4(x3 − 4x2 + 4x) 39. f (x) = 4x5 − 12x4 + 9x3 41. f (x) = 4x4(9x4 − 12x3 + 4x2) GRAPHICAl For the following exercises, graph the polynomial functions. Note x- and y-intercepts, multiplicity, and end behavior. 42. f (x) = (x + 3)2(x − 2) 43. g(x) = (x + 4)(x − 1)2 44. h(x) = (x − 1)3(x + 3)2 45. k(x) = (x − 3)3(x − 2)2 46. m(x) = −2x(x − 1)(x + 3) 47. n(x) = −3x(x + 2)(x − 4) SECTION 3.4 section exercises 255 For the following exercises, use the graphs to write the formula for a polynomial function of least degree. 48. f(x) 49. f(x) 50. f(x) 5 4 3 2 1 –1–1 –2 –3 –4 –5 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 51. f(x) 52. f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, use the graph to identify zeros and multiplicity. 54. 21 3 4 5 x –5 –4 –3 –2 55. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x 53. 56. y 5
4 3 2 1 –1–1 –2 –3 –4 –5 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 –5 –4 –3 –2 For the following exercises, use the given information about the polynomial graph to write the equation. 57. Degree 3. Zeros at x = −2, x = 1, and x = 3. 58. Degree 3. Zeros at x = −5, x = −2, and x = 1. y-intercept at (0, −4). y-intercept at (0, 6) 59. Degree 5. Roots of multiplicity 2 at x = 3 and x = 1, and a root of multiplicity 1 at x = −3. y-intercept at (0, 9) 60. Degree 4. Root of multiplicity 2 at x = 4, and roots of multiplicity 1 at x = 1 and x = −2. y-intercept at (0, −3). 256 CHAPTER 3 polynomial and rational Functions 61. Degree 5. Double zero at x = 1, and triple zero at 62. Degree 3. Zeros at x = 4, x = 3, and x = 2. x = 3. Passes through the point (2, 15). y-intercept at (0, −24). 63. Degree 3. Zeros at x = −3, x = −2 and x = 1. y-intercept at (0, 12). 1 _ 65. Degree 4. Roots of multiplicity 2 at x = and roots 2 of multiplicity 1 at x = 6 and x = −2. y-intercept at (0,18). 64. Degree 5. Roots of multiplicity 2 at x = −3 and x = 2 and a root of multiplicity 1 at x = −2. y-intercept at (0, 4). 66. Double zero at x = −3 and triple zero at x = 0. Passes through the point (1, 32). TeCHnOlOGY For the following exercises, use a calculator to approximate local minima and maxima or the global minimum and maximum. 67. f (x) = x3 − x − 1 68. f (x) = 2x3 − 3x − 1 69. f (x) = x4 + x 70. f (x) = −
x4 + 3x − 2 71. f (x) = x4 − x3 + 1 exTenSIOnS For the following exercises, use the graphs to write a polynomial function of least degree. (0, 50,000,000) 100 200 300 400 500 600 700 x 6·107 5·107 4·107 3·107 2·107 1·107 –100 0 –1·107 –2·107 –3·107 –4·107 –5·107 –6·107 –7·107 72. f (x) 24 16 73. f (x), 0 2 3 (0, 8) 4 3, 0 –6 –4 –2 2 4 6 x –300 –200 –8 –16 –24 74. (–300, 0) f(x) 2·105 1·105 (100, 0) x 100 200 (0, –90,000) –400 –300 –200 –100 –1·105 –2·105 –3·105 –4·105 ReAl-WORlD APPlICATIOnS For the following exercises, write the polynomial function that models the given situation. 75. A rectangle has a length of 10 units and a width of 8 units. Squares of x by x units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a polynomial function in terms of x. 77. A square has sides of 12 units. Squares x + 1 by x + 1 units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume of the box as a function in terms of x. 79. A right circular cone has a radius of 3x + 6 and a height 3 units less. Express the volume of the cone as a polynomial function. The volume of a cone is 1 __ V = πr2h for radius r and height h. 3 76. Consider the same rectangle of the preceding problem. Squares of 2x by 2x units are cut out of each corner. Express the volume of the box as a polynomial in terms of x. 78. A cylinder has a radius of x + 2 units and a height of 3 units greater. Express the volume of the cylinder as a polynomial function. SECTION 3.5 dividing polynomials 257 leARnInG OBje
CTIVeS In this section, you will: • Use long division to divide polynomials. • Use synthetic division to divide polynomials. 3.5 DIVIDInG POlYnOMIAlS Figure 1 lincoln Memorial, Washington, D.C. (credit: Ron Cogswell, Flickr) The exterior of the Lincoln Memorial in Washington, D.C., is a large rectangular solid with length 61.5 meters ( m ), width 40 m, and height 30 m.[15] We can easily find the volume using elementary geometry. V = l ⋅ w ⋅ h = 61.5 ⋅ 40 ⋅ 30 = 73,800 So the volume is 73,800 cubic meters (m3). Suppose we knew the volume, length, and width. We could divide to find the height. h = V ____ l ⋅ w = 73,800 _______ 61.5 ⋅ 40 = 30 As we can confirm from the dimensions above, the height is 30 m. We can use similar methods to find any of the missing dimensions. We can also use the same method if any or all of the measurements contain variable expressions. For example, suppose the volume of a rectangular solid is given by the polynomial 3x4 − 3x3 − 33x2 + 54x. The length of the solid is given by 3x; the width is given by x − 2. To find the height of the solid, we can use polynomial division, which is the focus of this section. Using long Division to Divide Polynomials We are familiar with the long division algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division. Long Division 5 × 3 = 15 and 17 − 15 = 2 Step 1: Step 2: Bring down the 8 9 × 3 = 27 and 28 − 27 = 1 Step 3: 1 __ Answer: 59 R 1 or 59 3 59 3)178 −15 28 −27 1 15. National Park Service. “Lincoln Memorial Building Statistics.” http://www.nps.gov/linc/historyculture/lincoln-memorial-building-statistics.htm. Accessed 4/3/2014/ 258 CHAPTER 3 po
lynomial and rational Functions Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic. dividend = (divisor ⋅ quotient) + remainder 178 = (3 ⋅ 59) + 1 = 177 + 1 = 178 We call this the Division Algorithm and will discuss it more formally after looking at an example. Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide 2x3 − 3x2 + 4x + 5 by x + 2 using the long division algorithm, it would look like this: Set up the division problem. 2x3 divided by x is 2x2. Multiply x + 2 by 2x2. Subtract. Bring down the next term. −7x2 divided by x is −7x. Multiply x + 2 by −7x. Subtract. Bring down the next term. x + 2)2x3 − 3x2 + 4x + 5 2x2 x + 2)2x3 − 3x2 + 4x + 5 2x2 x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x 2x2 − 7x x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x −(−7x2 + 14x) 18x + 5 2x2 − 7x + 18 x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x −(−7x2 + 14x) 18x + 5 −18x + 36 −31 18x divided by x is 18. Multiply x + 2 by 18. Subtract. We have found or 2x3 − 3x2 + 4x + 5 __ x + 2 = 2x2 − 7x +
18 − 31 _ x + 2 2x3 − 3x2 + 4x + 5 __ x + 2 = (x + 2)(2x2 − 7x + 18) − 31 We can identify the dividend, the divisor, the quotient, and the remainder. 2x3 – 3x2 + 4x + 5 = (x + 2) (2x2 – 7x + 18) + (–31) Dividend Divisor Quotient Remainder Writing the result in this manner illustrates the Division Algorithm. SECTION 3.5 dividing polynomials 259 the Division Algorithm The Division Algorithm states that, given a polynomial dividend f (x) and a non-zero polynomial divisor d(x) where the degree of d(x) is less than or equal to the degree of f (x), there exist unique polynomials q(x) and r(x) such that f (x) = d(x)q(x) + r(x) q(x) is the quotient and r(x) is the remainder. The remainder is either equal to zero or has degree strictly less than d(x). If r(x) = 0, then d(x) divides evenly into f (x). This means that, in this case, both d(x) and q(x) are factors of f (x). How To… Given a polynomial and a binomial, use long division to divide the polynomial by the binomial. 1. Set up the division problem. 2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor. 3. Multiply the answer by the divisor and write it below the like terms of the dividend. 4. Subtract the bottom binomial from the top binomial. 5. Bring down the next term of the dividend. 6. Repeat steps 2–5 until reaching the last term of the dividend. 7. If the remainder is non-zero, express as a fraction using the divisor as the denominator. Example 1 Using Long Division to Divide a Second-Degree Polynomial Divide 5x2 + 3x − 2 by x + 1. Solution x + 1)5x2 + 3x − 2 5x Set up division problem. x + 1)5x2 + 3x
− 2 5x2 divided by x is 5x. 5x x + 1)5x2 + 3x − 2 −(5x2 + 5x) −2x − 2 5x − 2 x + 1)5x2 + 3x − 2 −(5x2 + 5x) −2x − 2 −(−2x − 2) 0 Multiply x + 1 by 5x. Subtract. Bring down the next term. −2x divided by x is −2. Multiply x + 1 by −2. Subtract. The quotient is 5x − 2. The remainder is 0. We write the result as or 5x2 + 3x − 2 __________ x + 1 = 5x − 2 5x2 + 3x − 2 = (x + 1)(5x − 2) Analysis This division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend. 260 CHAPTER 3 polynomial and rational Functions Example 2 Using Long Division to Divide a Third-Degree Polynomial Divide 6x3 + 11x2 − 31x + 15 by 3x − 2. Solution 2x2 + 5x − 7 3x − 2)6x3 + 11x2 − 31x + 1 −(6x3 − 4x2) 15x2 − 31x −(15x2 − 10x) −21x + 15 −(−21x + 14) 1 6x3 divided by 3x is 2x2. Multiply 3x − 2 by 2x2. Subtract. Bring down the next term. 15x2 divided by 3x is 5x. Multiply 3x − 2 by 5x. Subtract. Bring down the next term. −21x divided by 3x is −7. Multiply 3x − 2 by −7. Subtract. The remainder is 1. There is a remainder of 1. We can express the result as: 6x3 + 11x2 − 31x + 15 __________________ 3x − 2 = 2x2 + 5x − 7 + 1 _____ 3x − 2 Analysis We can check our work by using the Division Algorithm to rewrite the solution. Then multiply. (3x − 2)(2x2 + 5x − 7)
+ 1 = 6x3 + 11x2 − 31x + 15 Notice, as we write our result, • the dividend is 6x3 + 11x2 − 31x + 15 • the divisor is 3x − 2 • the quotient is 2x2 + 5x − 7 • the remainder is 1 Try It #1 Divide 16x3 − 12x2 + 20x − 3 by 4x + 5. Using Synthetic Division to Divide Polynomials As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1. To illustrate the process, recall the example at the beginning of the section. Divide 2x3 − 3x2 + 4x + 5 by x + 2 using the long division algorithm. The final form of the process looked like this: 2x2 + x + 18 x + 2)2x3 − 3x2 + 4x + 5 −(2x3 + 4x2) −7x2 + 4x −(−7x2 − 14x) 18x + 5 −(18x + 36) −31 There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem. SECTION 3.5 dividing polynomials 261 2)2 −3 4 5 −2 −4 −7 14 18 −36 −31 Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the “divisor” to −2, multiply and add. The process starts by bringing down the leading coefficient. −2 2 2 −3 −4 −7 4 5 14 −36 18 −31 We then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is 2x2 −
7x + 18 and the remainder is −31. The process will be made more clear in Example 3. synthetic division Synthetic division is a shortcut that can be used when the divisor is a binomial in the form x − k. In synthetic division, only the coefficients are used in the division process. How To… Given two polynomials, use synthetic division to divide. 1. Write k for the divisor. 2. Write the coefficients of the dividend. 3. Bring the lead coefficient down. 4. Multiply the lead coefficient by k. Write the product in the next column. 5. Add the terms of the second column. 6. Multiply the result by k. Write the product in the next column. 7. Repeat steps 5 and 6 for the remaining columns. 8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so on. Example 3 Using Synthetic Division to Divide a Second-Degree Polynomial Use synthetic division to divide 5x2 − 3x − 36 by x − 3. Solution Begin by setting up the synthetic division. Write k and the coefficients. 3 5 −3 −36 Bring down the lead coefficient. Multiply the lead coefficient by k. 3 5 −3 −36 15 5 Continue by adding the numbers in the second column. Multiply the resulting number by k. Write the result in the next column. Then add the numbers in the third column. The result is 5x + 12. The remainder is 0. So x − 3 is a factor of the original polynomial. 3 5 −3 −36 15 −36 12 0 5 262 CHAPTER 3 polynomial and rational Functions Analysis remainder. Just as with long division, we can check our work by multiplying the quotient by the divisor and adding the (x − 3)(5x + 12) + 0 = 5x2 − 3x − 36 Example 4 Using Synthetic Division to Divide a Third-Degree Polynomial Use synthetic division to divide 4x 3 + 10x 2 − 6x − 20 by x + 2. Solution The binomial divisor is x + 2 so k = −2. Add each column, multiply the result by −2, and repeat until the last column is reached. −2 4 4 −6 −
20 10 −8 −4 20 2 −10 0 The result is 4x 2 + 2x − 10. The remainder is 0. Thus, x + 2 is a factor of 4x3 + 10x2 − 6x − 20. Analysis The graph of the polynomial function f (x) = 4x3 + 10x2 − 6x − 20 in Figure 2 shows a zero at x = k = −2. This confirms that x + 2 is a factor of 4x 3 + 10x2 − 6x − 20. −2 −1.8 –5 –4 –3 –2 y 14 12 10 8 6 4 2 –1 –2 –4 –6 –8 –10 –12 –14 –16 –18 –20 –22 321 4 5 x Figure 2 Example 5 Using Synthetic Division to Divide a Fourth-Degree Polynomial Use synthetic division to divide −9x4 + 10x3 + 7x2 − 6 by x − 1. Solution Notice there is no x-term. We will use a zero as the coefficient for that term. 1 −9 −9 10 −9 1 7 1 8 0 −6 8 8 2 8 The result is −9x3 + x2 + 8x + 8 + 2 _. x − 1 Try It #2 Use synthetic division to divide 3x4 + 18x3 − 3x + 40 by x + 7. Using Polynomial Division to Solve Application Problems Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. We looked at an application at the beginning of this section. Now we will solve that problem in the following example. SECTION 3.5 dividing polynomials 263 Example 6 Using Polynomial Division in an Application Problem The volume of a rectangular solid is given by the polynomial 3x4 − 3x3 − 33x2 + 54x. The length of the solid is given by 3x and the width is given by x − 2. Find the height of the solid. Solution There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in Figure 3. Height Width x − 2 Length 3x Figure 3 We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid. V = l ⋅ w ⋅ h 3
x4 − 3x3 − 33x2 + 54x = 3x ⋅ (x − 2) ⋅ h To solve for h, first divide both sides by 3x. 3x ⋅ (x − 2) ⋅ h ____________ 3x = 3x4 − 3x3 − 33x2 + 54x ___________________ 3x Now solve for h using synthetic division. (x − 2)h = x3 − x2 − 11x + 18 h = x3 − x2 − 11x + 18 ________________ x − 2 2 1 −1 −11 2 1 18 2 −18 −9 0 1 The quotient is x2 + x − 9 and the remainder is 0. The height of the solid is x2 + x − 9. Try It #3 The area of a rectangle is given by 3x3 + 14x2 − 23x + 6. The width of the rectangle is given by x + 6. Find an expression for the length of the rectangle. Access these online resources for additional instruction and practice with polynomial division. • Dividing a Trinomial by a Binomial Using long Division (http://openstaxcollege.org/l/dividetribild) • Dividing a Polynomial by a Binomial Using long Division (http://openstaxcollege.org/l/dividepolybild) • ex 2: Dividing a Polynomial by a Binomial Using Synthetic Division (http://openstaxcollege.org/l/dividepolybisd2) • ex 4: Dividing a Polynomial by a Binomial Using Synthetic Division (http://openstaxcollege.org/l/dividepolybisd4) 264 CHAPTER 3 polynomial and rational Functions 3.5 SeCTIOn exeRCISeS VeRBAl 1. If division of a polynomial by a binomial results in a remainder of zero, what can be conclude? 2. If a polynomial of degree n is divided by a binomial of degree 1, what is the degree of the quotient? AlGeBRAIC For the following exercises, use long division to divide. Specify the quotient and the remainder. 3. (x2 + 5x − 1) ÷ (x − 1) 6. (4x2 − 10x + 6) ÷ (4x
+ 2) 9. (2x2 − 3x + 2) ÷ (x + 2) 12. (x3 − 3x2 + 5x − 6) ÷ (x − 2) 4. (2x2 − 9x − 5) ÷ (x − 5) 7. (6x2 − 25x − 25) ÷ (6x + 5) 10. (x3 − 126) ÷ (x − 5) 13. (2x3 + 3x2 − 4x + 15) ÷ (x + 3) 5. (3x2 + 23x + 14) ÷ (x + 7) 8. (−x2 − 1) ÷ (x + 1) 11. (3x2 − 5x + 4) ÷ (3x + 1) For the following exercises, use synthetic division to find the quotient. 14. (3x3 − 2x2 + x − 4) ÷ (x + 3) 16. (6x3 − 10x2 − 7x − 15) ÷ (x + 1) 18. (9x3 − 9x2 + 18x + 5) ÷ (3x − 1) 20. (−6x3 + x2 − 4) ÷ (2x − 3) 22. (3x3 − 5x2 + 2x + 3) ÷ (x + 2) 24. (x3 − 3x + 2) ÷ (x + 2) 26. (x3 − 15x2 + 75x − 125) ÷ (x − 5) 28. (6x3 − x2 + 5x + 2) ÷ (3x + 1) 30. (x4 − 3x2 + 1) ÷ (x − 1) 32. (x4 − 10x3 + 37x2 − 60x + 36) ÷ (x − 2) 34. (x4 + 5x3 − 3x2 − 13x + 10) ÷ (x + 5) 36. (4x4 − 2x3 − 4x + 2) ÷ (2x − 1) 15. (2x3 − 6x2 − 7x + 6) ÷ (x − 4) 17. (4x3 − 12x2 − 5x − 1) ÷ (2x + 1) 19. (3
x3 − 2x2 + x − 4) ÷ (x + 3) 21. (2x3 + 7x2 − 13x − 3) ÷ (2x − 3) 23. (4x3 − 5x2 + 13) ÷ (x + 4) 25. (x3 − 21x2 + 147x − 343) ÷ (x − 7) 27. (9x3 − x + 2) ÷ (3x − 1) 29. (x4 + x3 − 3x2 − 2x + 1) ÷ (x + 1) 31. (x4 + 2x3 − 3x2 + 2x + 6) ÷ (x + 3) 33. (x4 − 8x3 + 24x2 − 32x + 16) ÷ (x − 2) 35. (x4 − 12x3 + 54x2 − 108x + 81) ÷ (x − 3) 37. (4x4 + 2x3 − 4x2 + 2x + 2) ÷ (2x + 1) For the following exercises, use synthetic division to determine whether the first expression is a factor of the second. If it is, indicate the factorization. 38. x − 2, 4x3 − 3x2 − 8x + 4 41. x − 2, 4x4 − 15x2 − 4 39. x − 2, 3x4 − 6x3 − 5x + 10 42. x − 1 __, 2x4 − x3 + 2x − 1 2 40. x + 3, −4x3 + 5x2 + 8 43. x + 1 __ 3, 3x4 + x3 − 3x + 1 GRAPHICAl For the following exercises, use the graph of the third-degree polynomial and one factor to write the factored form of the polynomial suggested by the graph. The leading coefficient is one. 44. Factor is x2 − x + 3 46. Factor is x2 + 2x + 5 45. Factor is x2 + 2x + 4 y 18 12 6 y 18 12 6 y 30 20 10 –6 –4 –2 2 4 6 x –6 –4 –2 2 4 6 x –6 –4 –2 2 4 6 x –6 –12 –18 –6 –12 –18 –10 –20 –30 SECTION 3.5 section
exercises 265 47. Factor is x2 + x + 1 48. Factor is x2 + 2x + 2 y 60 40 20 y 18 12 6 –6 –4 –2 2 4 6 x –6 –4 –2 2 4 6 x –20 –40 –60 – 6 –12 –18 For the following exercises, use synthetic division to find the quotient and remainder. 49. 4x3 − 33 _______ x − 2 51. 3x3 + 2x − 5 __________ x − 1 50. 2x3 + 25 _______ x + 3 53. x4 − 22 ______ x + 2 ____________ 52. −4x3 − x2 − 12 x + 4 TeCHnOlOGY 54. Consider xk − 1 ______ x − 1 with k = 1, 2, 3. What do you For the following exercises, use a calculator with CAS to answer the questions. xk + 1 ______ x + 1 the result to be if k = 7? xk _____ x + 1 the result to be if k = 4? expect the result to be if k = 4? expect the result to be if k = 4? for k = 1, 2, 3. What do you x4 − k4 ______ x − k 55. Consider 56. Consider 57. Consider for k = 1, 3, 5. What do you expect with k = 1, 2, 3. What do you expect 58. Consider xk _____ x − 1 the result to be if k = 4? with k = 1, 2, 3. What do you expect exTenSIOnS For the following exercises, use synthetic division to determine the quotient involving a complex number. 59. x + 1 _____ x − i 62. x2 + 1 _____ x + i 60. x2 + 1 _____ x − i 63. x3 + 1 _____ x − i 61. x + 1 _____ x + i ReAl-WORlD APPlICATIOnS For the following exercises, use the given length and area of a rectangle to express the width algebraically. 64. Length is x + 5, area is 2x2 + 9x − 5. 66. Length is 3x − 4, area is 6x4 − 8x3 + 9x2 − 9x − 4 65. Length is 2x + 5, area is 4x3 + 10x2 + 6x + 15 For
the following exercises, use the given volume of a box and its length and width to express the height of the box algebraically. 67. Volume is 12x3 + 20x2 − 21x − 36, length is 2x + 3, width is 3x − 4. 68. Volume is 18x3 − 21x2 − 40x + 48, length is 3x − 4, width is 3x − 4. 69. Volume is 10x3 + 27x2 + 2x − 24, length is 5x − 4, 70. Volume is 10x3 + 30x2 − 8x − 24, length is 2, width is 2x + 3. width is x + 3. For the following exercises, use the given volume and radius of a cylinder to express the height of the cylinder algebraically. 71. Volume is π(25x3 − 65x2 − 29x − 3), radius is 5x + 1. 73. Volume is π(3x4 + 24x3 + 46x2 − 16x − 32), radius is x + 4. 72. Volume is π(4x3 + 12x2 − 15x − 50), radius is 2x + 5. 266 CHAPTER 3 polynomial and rational Functions leARnInG OBjeCTIVeS In this section, you will: • Evaluate a polynomial using the Remainder Theorem. • Use the Factor Theorem to solve a polynomial equation. • Use the Rational Zero Theorem to find rational zeros. • Find zeros of a polynomial function. • Use the Linear Factorization Theorem to find polynomials with given zeros. • Use Decartes' Rule of Signs. • Solve real-world applications of polynomial equations. 3.6 ZeROS OF POlYnOMIAl FUnCTIOnS A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be? This problem can be solved by writing a cubic function and solving a cubic equation for the volume of the cake. In this section, we
will discuss a variety of tools for writing polynomial functions and solving polynomial equations. evaluating a Polynomial Using the Remainder Theorem In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem. If the polynomial is divided by x − k, the remainder may be found quickly by evaluating the polynomial function at k, that is, f (k) Let’s walk through the proof of the theorem. Recall that the Division Algorithm states that, given a polynomial dividend f (x) and a non-zero polynomial divisor d(x) where the degree of d(x) is less than or equal to the degree of f (x), there exist unique polynomials q(x) and r(x) such that If the divisor, d(x), is x − k, this takes the form f (x) = d(x)q(x) + r(x) f (x) = (x − k)q(x) + r Since the divisor x − k is linear, the remainder will be a constant, r. And, if we evaluate this for x = k, we have f (k) = (k − k)q(k) + r = 0 ⋅ q(k) + r = r In other words, f (k) is the remainder obtained by dividing f (x) by x − k. the Remainder Theorem If a polynomial f (x) is divided by x − k, then the remainder is the value f (k). How To… Given a polynomial function f, evaluate f (x) at x = k using the Remainder Theorem. 1. Use synthetic division to divide the polynomial by x − k. 2. The remainder is the value f (k). SECTION 3.6 Zeros oF polynomial Functions 267 Example 1 Using the Remainder Theorem to Evaluate a Polynomial Use the Remainder Theorem to evaluate f (x) = 6x4 − x3 − 15x2 + 2x − 7 at x = 2. Solution To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by x − 2. 2 6 −1 −15 2 −7 12 11 22 7 14 16
32 25 6 The remainder is 25. Therefore, f (2) = 25. Analysis We can check our answer by evaluating f (2). f (x) = 6x4 − x3 − 15x2 + 2x − 7 f (2) = 6(2)4 − (2)3 − 15(2)2 + 2(2) − 7 = 25 Try It #1 Use the Remainder Theorem to evaluate f (x) = 2x5 − 3x4 − 9x3 + 8x2 + 2 at x = −3. Using the Factor Theorem to Solve a Polynomial equation The Factor Theorem is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors. Recall that the Division Algorithm tells us f (x) = (x − k)q(x) + r. If k is a zero, then the remainder r is f (k) = 0 and f (x) = (x − k)q(x) + 0 or f (x) = (x − k)q(x). Notice, written in this form, x − k is a factor of f (x). We can conclude if k is a zero of f (x), then x − k is a factor of f (x). Similarly, if x − k is a factor of f (x), then the remainder of the Division Algorithm f (x) = (x − k)q(x) + r is 0. This tells us that k is a zero. This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree n in the complex number system will have n zeros. We can use the Factor Theorem to completely factor a polynomial into the product of n factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial. the Factor Theorem According to the Factor Theorem, k is a zero of f (x) if and only if (x − k) is a factor of f (x). How To… Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial. 1. Use synthetic division to divide the polynomial by (x − k). 2. Confirm that the remainder is 0. 3. Write the polynomial as
the product of (x − k) and the quadratic quotient. 4. If possible, factor the quadratic. 5. Write the polynomial as the product of factors. Example 2 Using the Factor Theorem to Solve a Polynomial Equation Show that (x + 2) is a factor of x3 − 6x2 − x + 30. Find the remaining factors. Use the factors to determine the zeros of the polynomial. 268 CHAPTER 3 polynomial and rational Functions Solution We can use synthetic division to show that (x + 2) is a factor of the polynomial. −2 1 −6 −1 30 −2 16 −30 1 −8 15 0 The remainder is zero, so (x + 2) is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient: (x + 2)(x2 − 8x + 15) We can factor the quadratic factor to write the polynomial as By the Factor Theorem, the zeros of x3 − 6x2 − x + 30 are −2, 3, and 5. (x + 2)(x − 3)(x − 5) Try It #2 Use the Factor Theorem to find the zeros of f (x) = x3 + 4x2 − 4x − 16 given that (x − 2) is a factor of the polynomial. Using the Rational Zero Theorem to Find Rational Zeros Another use for the Remainder Theorem is to test whether a rational number is a zero for a given polynomial. But first we need a pool of rational numbers to test. The Rational Zero Theorem helps us to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial. 3 2 __ __ Consider a quadratic function with two zeros, x = and x =. By the Factor Theorem, these zeros have factors 4 5 associated with them. Let us set each factor equal to 0, and then construct the original quadratic function absent its stretching factor. 3 2 __ __ x − = 0 = 0 or x − 4 5 5x − 2 = 0 or 4x − 3 = 0 Set each factor equal to 0. Multiply both sides of the equation to eliminate fractions. f (x) =
(5x − 2)(4x − 3) Create the quadratic function, multiplying the factors. f (x) = 20x2 − 23x + 6 Expand the polynomial. f (x) = (5 ⋅ 4)x2 − 23x + (2 ⋅ 3) Notice that two of the factors of the constant term, 6, are the two numerators from the original rational roots: 2 and 3. Similarly, two of the factors from the leading coefficient, 20, are the two denominators from the original rational roots: 5 and 4. We can infer that the numerators of the rational roots will always be factors of the constant term and the denominators will be factors of the leading coefficient. This is the essence of the Rational Zero Theorem; it is a means to give us a pool of possible rational zeros. the Rational Zero Theorem The Rational Zero Theorem states that, if the polynomial f (x) = anxn + an − 1 xn − 1 +... + a1 x + a0 has integer p _ q where p is a factor of the constant term a0 and q is a coefficients, then every rational zero of f (x) has the form factor of the leading coefficient an. When the leading coefficient is 1, the possible rational zeros are the factors of the constant term. SECTION 3.6 Zeros oF polynomial Functions 269 How To… Given a polynomial function f (x), use the Rational Zero Theorem to find rational zeros. 1. Determine all factors of the constant term and all factors of the leading coefficient. p _ 2. Determine all possible values of q, where p is a factor of the constant term and q is a factor of the leading coefficient. Be sure to include both positive and negative candidates. p q . 3. Determine which possible zeros are actual zeros by evaluating each case of f  _ Example 3 Listing All Possible Rational Zeros List all possible rational zeros of f (x) = 2x4 − 5x3 + x2 − 4. Solution The only possible rational zeros of f (x) are the quotients of the factors of the last term, −4, and the factors of the leading coefficient, 2. The constant term is −4; the factors of −4 are p = ±1, ±2, ±4. The leading coefficient is 2; the factors
of 2 are q = ±1, ±2. If any of the four real zeros are rational zeros, then they will be of one of the following factors of −4 divided by one of the factors of 2. p 1 1 __ __ __ __ __ __ __ __ = 2, which have already been listed. So we can shorten our list. = 1 and Note that 2 2 p _ q = Factors of the last __ Factors of the first = ±1, ±2, ±4, ± 1 __ 2 Example 4 Using the Rational Zero Theorem to Find Rational Zeros Use the Rational Zero Theorem to find the rational zeros of f (x) = 2x3 + x2 − 4x + 1. p _ Solution The Rational Zero Theorem tells us that if q is a zero of f (x), then p is a factor of 1 and q is a factor of 2. p _ q = factor of constant term ___ factor of leading coefficient = factor of 1 _ factor of 2 p q are ±1 and ± 1 _ 2. These are the _ The factors of 1 are ±1 and the factors of 2 are ±1 and ±2. The possible values for possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for x in f (x). f (−1) = 2(−1)3 + (−1)2 − 4(−1) + 1 = 4 3 f (1) = 2(1)3 + (1)2 − 4(1 __ __ __ __  + __ __ __ __ − Of those, −1, − 1 __ 2 1 __ are not zeros of f (x). 1 is the only rational zero of f (x)., and 2 Try It #3 Use the Rational Zero Theorem to find the rational zeros of f (x) = x3 − 5x2 + 2x + 1. 270 CHAPTER 3 polynomial and rational Functions Finding the Zeros of Polynomial Functions The Rational Zero Theorem helps us to narrow down the list of possible rational zeros for a polynomial function. Once we have done this, we can use synthetic division repeatedly to determine all of the zeros of a polynomial function. How To… Given a polynomial function f, use synthetic division to find its zeros. 1. Use the Rational Zero Theorem to list all possible rational zeros of
the function. 2. Use synthetic division to evaluate a given possible zero by synthetically dividing the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is not zero, discard the candidate. 3. Repeat step two using the quotient found with synthetic division. If possible, continue until the quotient is a quadratic. 4. Find the zeros of the quadratic function. Two possible methods for solving quadratics are factoring and using the quadratic formula. Example 5 Finding the Zeros of a Polynomial Function with Repeated Real Zeros Find the zeros of f (x) = 4x3 − 3x − 1. p _ q is a zero of f (x), then p is a factor of −1 and q is a factor of 4. Solution The Rational Zero Theorem tells us that if p _ q = factor of constant term ___ factor of leading coefficient p, and ± 1 q are ±1, ± 1 _ __ __ The factors of −1 are ±1 and the factors of 4 are ±1, ±2, and ±4. The possible values for. These 4 2 are the possible rational zeros for the function. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder of 0. Let’s begin with 1. = factor of −1 __ factor of 4 1 4 4 0 −3 −1 1 4 4 0 1 4 Dividing by (x − 1) gives a remainder of 0, so 1 is a zero of the function. The polynomial can be written as The quadratic is a perfect square. f (x) can be written as (x − 1)(2x + 1)2. (x − 1)(4x2 + 4x + 1). We already know that 1 is a zero. The other zero will have a multiplicity of 2 because the factor is squared. To find the other zero, we can set the factor equal to 0. 2x + 1 = 0 x = − 1 __ 2 The zeros of the function are 1 and − 1 __ with multiplicity 2. 2 Analysis Look at the graph of the function f in Figure 1. Notice, at x = −0.5, the graph bounces off the x-axis, indicating the even multiplicity (2, 4, 6…) for the zero −0.5. At x = 1, the graph crosses the
x-axis, indicating the odd multiplicity (1, 3, 5…) for the zero x = 1. y Bounce 1.5 1 0.5 –2.5 –2 –1.5 0.5 1 1.5 2 2.5 x Cross –1 –0.5 –0.5 –1 –1.5 –2 –2.5 Figure 1 SECTION 3.6 Zeros oF polynomial Functions 271 Using the Fundamental Theorem of Algebra Now that we can find rational zeros for a polynomial function, we will look at a theorem that discusses the number of complex zeros of a polynomial function. The Fundamental Theorem of Algebra tells us that every polynomial function has at least one complex zero. This theorem forms the foundation for solving polynomial equations. Suppose f is a polynomial function of degree four, and f (x) = 0. The Fundamental Theorem of Algebra states that there is at least one complex solution, call it c1. By the Factor Theorem, we can write f (x) as a product of x − c1 and a polynomial quotient. Since x − c1 is linear, the polynomial quotient will be of degree three. Now we apply the Fundamental Theorem of Algebra to the third-degree polynomial quotient. It will have at least one complex zero, call it c2. So we can write the polynomial quotient as a product of x − c2 and a new polynomial quotient of degree two. Continue to apply the Fundamental Theorem of Algebra until all of the zeros are found. There will be four of them and each one will yield a factor of f (x). The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra states that, if f (x) is a polynomial of degree n > 0, then f (x) has at least one complex zero. We can use this theorem to argue that, if f (x) is a polynomial of degree n > 0, and a is a non-zero real number, then f (x) has exactly n linear factors where c1, c2,..., cn are complex numbers. Therefore, f (x) has n roots if we allow for multiplicities. f (x) = a(x − c1)(x − c2)...(x − cn) Q & A
… Does every polynomial have at least one imaginary zero? No. A complex number is not necessarily imaginary. Real numbers are also complex numbers. Example 6 Finding the Zeros of a Polynomial Function with Complex Zeros Find the zeros of f (x) = 3x3 + 9x2 + x + 3. p _ Solution The Rational Zero Theorem tells us that if q is a zero of f (x), then p is a factor of 3 and q is a factor of 3. p _ q = factor of constant term ___ factor of leading coefficient p _ The factors of 3 are ±1 and ±3. The possible values for q, and therefore the possible rational zeros for the function, are ±3, ±1, and ± 1 __. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder 3 of 0. Let’s begin with −3. = factor of 3 _ factor of 3 −3 3 3 9 −9 0 1 3 0 −3 0 1 Dividing by (x + 3) gives a remainder of 0, so −3 is a zero of the function. The polynomial can be written as We can then set the quadratic equal to 0 and solve to find the other zeros of the function. (x + 3)(3x2 + 1) 3x2 + 1 = 0 3 The zeros of f (x) are −3 and ± i √ _. 3 — x2 = − 1 __ 3 x = ± √ ____ − 1 __ = ± 3 — 3 i √ _ 3 272 CHAPTER 3 polynomial and rational Functions Analysis Look at the graph of the function f in Figure 2. Notice that, at x = −3, the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero x = −3. Also note the presence of the two turning points. This means that, since there is a 3rd degree polynomial, we are looking at the maximum number of turning points. So, the end behavior of increasing without bound to the right and decreasing without bound to the left will continue. Thus, all the x-intercepts for the function are shown. So either the multiplicity of x = −3 is 1 and there are two complex solutions, which is what we found, or the multiplicity at x = −3 is three. Either way, our result is correct. y 18
12 6 Cross –6 –4 –2 2 4 6 x –6 –12 –18 Figure 2 Try It #4 Find the zeros of f (x) = 2x3 + 5x2 − 11x + 4. Using the linear Factorization Theorem to Find Polynomials with Given Zeros A vital implication of the Fundamental Theorem of Algebra, as we stated above, is that a polynomial function of degree n will have n zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into n factors. The Linear Factorization Theorem tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form (x − c), where c is a complex number. Let f be a polynomial function with real coefficients, and suppose a + bi, b ≠ 0, is a zero of f (x). Then, by the Factor Theorem, x − (a + bi) is a factor of f (x). For f to have real coefficients, x − (a − bi) must also be a factor of f (x). This is true because any factor other than x − (a − bi), when multiplied by x − (a + bi), will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function f with real coefficients has a complex zero a + bi, then the complex conjugate a − bi must also be a zero of f (x). This is called the Complex Conjugate Theorem. complex conjugate theorem According to the Linear Factorization Theorem, a polynomial function will have the same number of factors as its degree, and each factor will be in the form (x − c), where c is a complex number. If the polynomial function f has real coefficients and a complex zero in the form a + bi, then the complex conjugate of the zero, a − bi, is also a zero. How To… Given the zeros of a polynomial function f and a point (c, f (c)) on the graph of f, use the Linear Factorization Theorem to find the polynomial function. 1. Use the zeros to construct the linear factors of the polynomial. 2. Multiply
the linear factors to expand the polynomial. 3. Substitute (c, f (c)) into the function to determine the leading coefficient. 4. Simplify. SECTION 3.6 Zeros oF polynomial Functions 273 Example 7 Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros Find a fourth degree polynomial with real coefficients that has zeros of −3, 2, i, such that f (−2) = 100. Solution Because x = i is a zero, by the Complex Conjugate Theorem x = −i is also a zero. The polynomial must have factors of (x + 3), (x − 2), (x − i), and (x + i). Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let’s begin by multiplying these factors. f (x) = a(x + 3)(x − 2)(x − i)(x + i) f (x) = a(x2 + x − 6)(x2 + 1) f (x) = a(x4 + x3 − 5x2 + x − 6) We need to find a to ensure f (−2) = 100. Substitute x = −2 and f (2) = 100 into f (x). 100 = a((−2)4 + (−2)3 − 5(−2)2 + (−2) − 6) So the polynomial function is or 100 = a(−20) −5 = a f (x) = −5(x4 + x3 − 5x2 + x − 6) f (x) = −5x4 − 5x3 + 25x2 − 5x + 30 Analysis We found that both i and −i were zeros, but only one of these zeros needed to be given. If i is a zero of a polynomial with real coefficients, then −i must also be a zero of the polynomial because −i is the complex conjugate of i. Q & A… If 2 + 3i were given as a zero of a polynomial with real coefficients, would 2 − 3i also need to be a zero? Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.
Try It #5 Find a third degree polynomial with real coefficients that has zeros of 5 and −2i such that f (1) = 10. Using Descartes’ Rule of Signs There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order, Descartes’ Rule of Signs tells us of a relationship between the number of sign changes in f (x) and the number of positive real zeros. For example, the polynomial function below has one sign change. f (x This tells us that the function must have 1 positive real zero. There is a similar relationship between the number of sign changes in f (−x) and the number of negative real zeros. f (−x) = (−x)4 + (−x)3 + (−x)2 + (−x) − 1 f (−x) =+ In this case, f (−x) has 3 sign changes. This tells us that f (x) could have 3 or 1 negative real zeros. 274 CHAPTER 3 polynomial and rational Functions Descartes’ Rule of Signs According to Descartes’ Rule of Signs, if we let f (x) = anxn + an − 1 xn − 1 +... + a1 x + a0 be a polynomial function with real coefficients: • The number of positive real zeros is either equal to the number of sign changes of f (x) or is less than the number of sign changes by an even integer. • The number of negative real zeros is either equal to the number of sign changes of f (−x) or is less than the number of sign changes by an even integer. Example 8 Using Descartes’ Rule of Signs Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for f (x) = −x4 − 3x3 + 6x2 − 4x − 12. Solution Begin by determining the number of sign changes. f (x) = −x 4 − 3x 3 + 6x 2 − 4x − 12 Figure 3 There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine f (−x) to determine the number of negative real roots. f (−x) = −(−x)4 − 3(−x)3 + 6(−x)
2 − 4(−x) − 12 f (−x) = −x 4 + 3x 3 + 6x 2 + 4x − 12 f (−x) = −x 4 + 3x 3 + 6x 2 + 4x − 12 Figure 4 Again, there are two sign changes, so there are either 2 or 0 negative real roots. There are four possibilities, as we can see in Table 1. Positive Real Zeros 2 2 0 0 Negative Real Zeros 2 0 2 0 Table 1 Complex Zeros 0 2 2 4 Total Zeros 4 4 4 4 Analysis We can confirm the numbers of positive and negative real roots by examining a graph of the function. See Figure 5. We can see from the graph that the function has 0 positive real roots and 2 negative real roots. y 60 50 40 30 20 10 0 –1 –10 –20 –30 x = −4.42 –5 –4 –3 –2 f (x) = − x4 − 3x3 + 6x2 − 4x − 12 x = −1 321 4 5 x Try It #6 Use Descartes’ Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for f (x) = 2x4 − 10x3 + 11x2 − 15x + 12. Use a graph to verify the numbers of positive and negative real zeros for the function. Figure 5 Solving Real-World Applications We have now introduced a variety of tools for solving polynomial equations. Let’s use these tools to solve the bakery problem from the beginning of the section. SECTION 3.6 Zeros oF polynomial Functions 275 Example 9 Solving Polynomial Equations A new bakery offers decorated sheet cakes for children’s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be? Solution Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by V = lwh. We were given that the length must be four inches longer than the width, so we can express the length of the cake as l = w + 4. We were given that the height of the cake is one-third of
the width, so we can express the height of the cake as h = 1 __ w. Let’s write the volume of the cake in terms of width of the cake. 3 1 __ w  V = (w + 4)(w)  3 1 4 __ __ V = w 3 + w 2 3 3 Substitute the given volume into this equation. 1 4 __ __ 351 = w 3 + w 2 3 3 1053 = w 3 + 4w 2 Substitute 351 for V. Multiply both sides by 3. 0 = w 3 + 4w 2 − 1053 Subtract 1053 from both sides. Descartes’ rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are ±1, ±3, ±9, ±13, ±27, ±39, ±81, ±117, ±351, and ±1053. We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let’s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check x = 1. Since 1 is not a solution, we will check x = 3. 1 3 Since 3 is not a solution either, we will test x = 91053 5 5 5 −1048 0 −1053 21 63 21 −990 4 9 13 0 −1053 1053 117 0 117 Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan. The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches = 13 and h = 1 __ __ (9) = 3 w = 3 3 Try It #7 A shipping container in the shape of a rectangular solid must have a volume of 84 cubic meters. The client tells the manufacturer that, because of the contents, the length of the container must be one meter longer than the width, and the height must be one meter greater than twice the width. What should the dimensions of the container be? Access these online resources for additional instruction and practice with zeros of polynomial functions. • Real Zeros, Factors, and Graphs of Polynomial Functions (http://openstaxcollege.org/l/realzeros
) • Complex Factorization Theorem (http://openstaxcollege.org/l/factortheorem) • Find the Zeros of a Polynomial Function (http://openstaxcollege.org/l/findthezeros) • Find the Zeros of a Polynomial Function 2 (http://openstaxcollege.org/l/findthezeros2) • Find the Zeros of a Polynomial Function 3 (http://openstaxcollege.org/l/findthezeros3) 276 CHAPTER 3 polynomial and rational Functions 3.6 SeCTIOn exeRCISeS VeRBAl 1. Describe a use for the Remainder Theorem. 3. What is the difference between rational and real zeros? 5. If synthetic division reveals a zero, why should we try that value again as a possible solution? AlGeBRAIC 2. Explain why the Rational Zero Theorem does not guarantee finding zeros of a polynomial function. 4. If Descartes’ Rule of Signs reveals a no change of signs or one sign of changes, what specific conclusion can be drawn? For the following exercises, use the Remainder Theorem to find the remainder. 6. (x4 − 9x2 + 14) ÷ (x − 2) 8. (x4 + 5x3 − 4x − 17) ÷ (x + 1) 7. (3x3 − 2x2 + x − 4) ÷ (x + 3) 9. (−3x2 + 6x + 24) ÷ (x − 4) 10. (5x5 − 4x4 + 3x3 − 2x2 + x − 1) ÷ (x + 6) 11. (x4 − 1) ÷ (x − 4) 12. (3x3 + 4x2 − 8x + 2) ÷ (x − 3) 13. (4x3 + 5x2 − 2x + 7) ÷ (x + 2) For the following exercises, use the Factor Theorem to find all real zeros for the given polynomial function and one factor. 14. f (x) = 2x3 − 9x2 + 13x − 6; x − 1 15. f (x) = 2x3 + x2 − 5x + 2; x + 2 16. f (x) = 3
x3 + x2 − 20x + 12; x + 3 17. f (x) = 2x3 + 3x2 + x + 6; x + 2 18. f (x) = −5x3 + 16x2 − 9; x − 3 20. 4x3 − 7x + 3; x − 1 19. x3 + 3x2 + 4x + 12; x + 3 21. 2x3 + 5x2 − 12x − 30, 2x + 5 For the following exercises, use the Rational Zero Theorem to find all real zeros. 22. x3 − 3x2 − 10x + 24 = 0 23. 2x3 + 7x2 − 10x − 24 = 0 24. x3 + 2x2 − 9x − 18 = 0 25. x3 + 5x2 − 16x − 80 = 0 26. x3 − 3x2 − 25x + 75 = 0 27. 2x3 − 3x2 − 32x − 15 = 0 28. 2x3 + x2 − 7x − 6 = 0 29. 2x3 − 3x2 − x + 1 = 0 30. 3x3 − x2 − 11x − 6 = 0 31. 2x3 − 5x2 + 9x − 9 = 0 32. 2x3 − 3x2 + 4x + 3 = 0 33. x4 − 2x3 − 7x2 + 8x + 12 = 0 34. x4 + 2x3 − 9x2 − 2x + 8 = 0 35. 4x4 + 4x3 − 25x2 − x + 6 = 0 36. 2x4 − 3x3 − 15x2 + 32x − 12 = 0 37. x4 + 2x3 − 4x2 − 10x − 5 = 0 38. 4x3 − 3x + 1 = 0 39. 8x4 + 26x3 + 39x2 + 26x + 6 For the following exercises, find all complex solutions (real and non-real). 40. x3 + x2 + x + 1 = 0 41. x3 − 8x2 + 25x − 26 = 0 42. x3 + 13x2 + 57x + 85 = 0 43. 3x3 − 4x2 + 11x + 10 = 0 44. x4 + 2x3 + 22x2 + 50x −
75 = 0 45. 2x3 − 3x2 + 32x + 17 = 0 GRAPHICAl For the following exercises, use Descartes’ Rule to determine the possible number of positive and negative solutions. Then graph to confirm which of those possibilities is the actual combination. 46. f (x) = x3 − 1 47. f (x) = x4 − x2 − 1 48. f (x) = x3 − 2x2 − 5x + 6 49. f (x) = x3 − 2x2 + x − 1 50. f (x) = x4 + 2x3 − 12x2 + 14x − 5 51. f (x) = 2x3 + 37x2 + 200x + 300 SECTION 3.6 section exercises 277 52. f (x) = x3 − 2x2 − 16x + 32 53. f (x) = 2x4 − 5x3 − 5x2 + 5x + 3 54. f (x) = 2x4 − 5x3 − 14x2 + 20x + 8 55. f (x) = 10x4 − 21x2 + 11 nUMeRIC For the following exercises, list all possible rational zeros for the functions. 56. f (x) = x4 + 3x3 − 4x + 4 57. f (x) = 2x3 + 3x2 − 8x + 5 58. f (x) = 3x3 + 5x2 − 5x + 4 59. f (x) = 6x4 − 10x2 + 13x + 1 60. f (x) = 4x5 − 10x4 + 8x3 + x2 − 8 TeCHnOlOGY For the following exercises, use your calculator to graph the polynomial function. Based on the graph, find the rational zeros. All real solutions are rational. 61. f (x) = 6x3 − 7x2 + 1 62. f (x) = 4x3 − 4x2 − 13x − 5 63. f (x) = 8x3 − 6x2 − 23x + 6 64. f (x) = 12x4 + 55x3 + 12x2 − 117x + 54 65. f (x) = 16x4 − 24x3 + x2 − 15x + 25 exTenSIOnS For the following exercises
, construct a polynomial function of least degree possible using the given information. 66. Real roots: −1, 1, 3 and (2, f (2)) = (2, 4) 67. Real roots: −1, 1 (with multiplicity 2 and 1) and 1 __ 68. Real roots: −2, (with multiplicity 2) and 2 (−3, f (−3)) = (−3, 5) 70. Real roots: −4, −1, 1, 4 and (−2, f (−2)) = (−2, 10) (2, f (2)) = (2, 4) 69. Real roots: − 1 1 __ __ and (−2, f (−2)) = (−2, 6), 0, 2 2 ReAl-WORlD APPlICATIOnS For the following exercises, find the dimensions of the box described. 71. The length is twice as long as the width. The height is 2 inches greater than the width. The volume is 192 cubic inches. 72. The length, width, and height are consecutive whole numbers. The volume is 120 cubic inches. 73. The length is one inch more than the width, which is 74. The length is three times the height and the height is one inch more than the height. The volume is 86.625 cubic inches. one inch less than the width. The volume is 108 cubic inches. 75. The length is 3 inches more than the width. The width is 2 inches more than the height. The volume is 120 cubic inches. For the following exercises, find the dimensions of the right circular cylinder described. 76. The radius is 3 inches more than the height. The 77. The height is one less than one half the radius. volume is 16π cubic meters. The volume is 72π cubic meters. 78. The radius and height differ by one meter. The 79. The radius and height differ by two meters. radius is larger and the volume is 48π cubic meters. 1 __ 80. The radius is meter greater than the height. The 3 98 ___ π cubic meters. 9π volume is The height is greater and the volume is 28.125π cubic meters. 278 CHAPTER 3 polynomial and rational Functions leARnInG OBjeCTIVeS In this section, you will: • Use arrow notation. • Solve applied problems involving rational functions. • Find the domains of rational functions. • • Identify vertical asym
ptotes. Identify horizontal asymptotes. • Graph rational functions. 3.7 RATIOnAl FUnCTIOnS Suppose we know that the cost of making a product is dependent on the number of items, x, produced. This is given by the equation C(x) = 15,000x − 0.1x2 + 1000. If we want to know the average cost for producing x items, we would divide the cost function by the number of items, x. The average cost function, which yields the average cost per item for x items produced, is f (x) = 15,000x − 0.1x2 + 1000 __________________ x Many other application problems require finding an average value in a similar way, giving us variables in the denominator. Written without a variable in the denominator, this function will contain a negative integer power. In the last few sections, we have worked with polynomial functions, which are functions with non-negative integers for exponents. In this section, we explore rational functions, which have variables in the denominator. Using Arrow notation We have seen the graphs of the basic reciprocal function and the squared reciprocal function from our study of toolkit functions. Examine these graphs, as shown in Figure 1, and notice some of their features. Graphs of Toolkit Functions y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 f (x) = 1 x 321 4 5 x –5 –4 –3 –2 Figure 1 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 f (x) = 1 x2 321 4 5 x 1 _ Several things are apparent if we examine the graph of f (x) = x. 1. On the left branch of the graph, the curve approaches the x-axis (y = 0) as x → −∞. 2. As the graph approaches x = 0 from the left, the curve drops, but as we approach zero from the right, the curve rises. 3. Finally, on the right branch of the graph, the curves approaches the x-axis (y = 0) as x → ∞. To summarize, we use arrow notation to show that x or f (x) is approaching a particular value. See Table 1. SECTION 3.7 rational Functions 279 Symbol x → a− x → a+ x → ∞ x → −�
� f (x) → ∞ f (x) → −∞ f (x) → a Meaning x approaches a from the left (x < a but close to a) x approaches a from the right (x > a but close to a) x approaches infinity (x increases without bound) x approaches negative infinity (x decreases without bound) The output approaches infinity (the output increases without bound) The output approaches negative infinity (the output decreases without bound) The output approaches a Table 1 Arrow Notation 1 __ Local Behavior of f (x ) = x Let’s begin by looking at the reciprocal function, f (x) = 1 _ x. We cannot divide by zero, which means the function is undefined at x = 0; so zero is not in the domain. As the input values approach zero from the left side (becoming very small, negative values), the function values decrease without bound (in other words, they approach negative infinity). We can see this behavior in Table 2. x f (x) = 1 __ x −0.1 −0.01 −0.001 −0.0001 −10 −100 −1000 −10,000 We write in arrow notation Table 2 as x → 0−, f (x) → −∞ As the input values approach zero from the right side (becoming very small, positive values), the function values increase without bound (approaching infinity). We can see this behavior in Table 3. x f (x) = 1 __ x 0.1 10 0.01 100 Table 3 0.001 0.0001 1000 10,000 We write in arrow notation See Figure 2. As x → 0+, f (x) → ∞. As x → 0+ f (x) → ∞ y As x → −∞ f (x) → 0 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 321 4 5 x As x → ∞ f (x) → 0 As x → 0– f (x) → −∞ Figure 2 280 CHAPTER 3 polynomial and rational Functions This behavior creates a vertical asymptote, which is a vertical line that the graph approaches but never crosses. In this case, the graph is approaching the vertical line x = 0 as the input becomes close to zero. See Figure 3. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –
4 –3 –2 321 4 5 x x = 0 Figure 3 vertical asymptote A vertical asymptote of a graph is a vertical line x = a where the graph tends toward positive or negative infinity as the inputs approach a. We write As x → a, f (x) → ∞, or as x → a, f (x) → −∞. 1 _ x End Behavior of f (x ) = As the values of x approach infinity, the function values approach 0. As the values of x approach negative infinity, the function values approach 0. See Figure 4. Symbolically, using arrow notation As x → ∞, f (x) → 0, and as x → −∞, f (x) → 0. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –3 –2 –4 –5 As x → −∞ f (x) → 0 As x → 0+ f (x) → ∞ As x → ∞ f (x) → 0 x 321 4 5 As x → 0– f (x) → −∞ Figure 4 Based on this overall behavior and the graph, we can see that the function approaches 0 but never actually reaches 0; it seems to level off as the inputs become large. This behavior creates a horizontal asymptote, a horizontal line that the graph approaches as the input increases or decreases without bound. In this case, the graph is approaching the horizontal line y = 0. See Figure 5. y = 0 –2 –3 –4 –5 y 5 4 3 2 1 321 4 5 x x = 0 –1 –1 –2 –3 –4 –5 Figure 5 SECTION 3.7 rational Functions 281 horizontal asymptote A horizontal asymptote of a graph is a horizontal line y = b where the graph approaches the line as the inputs increase or decrease without bound. We write As x → ∞ or x → −∞, f (x) → b. Example 1 Using Arrow Notation Use arrow notation to describe the end behavior and local behavior of the function graphed in Figure 6. y 12 10 8 6 4 2 –1 –2 –4 –6 –8 –10 –12 –6 –5 –4 –3 –2 321 4 5 6 x Figure 6 Solution Notice that the graph is showing a vertical asymptote at x = 2, which tells us that the function is undefined at x
= 2. As x → 2−, f (x) → −∞, and as x → 2+, f (x) → ∞. And as the inputs decrease without bound, the graph appears to be leveling off at output values of 4, indicating a horizontal asymptote at y = 4. As the inputs increase without bound, the graph levels off at 4. As x → ∞, f (x) → 4 and as x → −∞, f (x) → 4. Try It #1 Use arrow notation to describe the end behavior and local behavior for the reciprocal squared function. Example 2 Using Transformations to Graph a Rational Function Sketch a graph of the reciprocal function shifted two units to the left and up three units. Identify the horizontal and vertical asymptotes of the graph, if any. Solution Shifting the graph left 2 and up 3 would result in the function 1 _____ x + 2 or equivalently, by giving the terms a common denominator, f (x) = + 3 The graph of the shifted function is displayed in Figure 7. f (x) = 3x + 7 ______ x + 2 x = −7 –6 –5 –4 –3 –2 –1 –1 –2 –3 y = 3 321 4 5 6 7 x Figure 7 282 CHAPTER 3 polynomial and rational Functions Notice that this function is undefined at x = −2, and the graph also is showing a vertical asymptote at x = −2. As x → −2−, f (x) → −∞, and as x → −2+, f (x) → ∞. As the inputs increase and decrease without bound, the graph appears to be leveling off at output values of 3, indicating a horizontal asymptote at y = 3. As x → ±∞, f (x) → 3. Analysis Notice that horizontal and vertical asymptotes are shifted left 2 and up 3 along with the function. Try It #2 Sketch the graph, and find the horizontal and vertical asymptotes of the reciprocal squared function that has been shifted right 3 units and down 4 units. Solving Applied Problems Involving Rational Functions In Example 2, we shifted a toolkit function in a way that resulted in the function f (x) = 3x + 7 ______ x + 2. This is an example of a rational function. A rational function is a function that can be written as the quotient
of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions. rational function A rational function is a function that can be written as the quotient of two polynomial functions P(x) and Q(x). f (x) = P(x) ____ Q(x) = ap x p + ap − 1 x p − 1 +... + a1 x + a0 ___ bq x q + bq − 1 x q − 1 +... + b1 x + b0, Q(x) ≠ 0 Example 3 Solving an Applied Problem Involving a Rational Function A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning? Solution Let t be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each: water: W(t) = 100 + 10t in gallons sugar: S(t) = 5 + 1t in pounds The concentration, C, will be the ratio of pounds of sugar to gallons of water The concentration after 12 minutes is given by evaluating C(t) at t = 12. C(t) = 5 + t ________ 100 + 10t This means the concentration is 17 pounds of sugar to 220 gallons of water. C(12) = 5 + 12 __________ 100 + 10(12) 17 ___ 220 = At the beginning, the concentration is C(0) = 5 + 0 _________ 100 + 10(0) 1 __ 20 = Since ≈ 0.08 > = 0.05, the concentration is greater after 12 minutes than at the beginning. 17 ___ 220 1 __ 20 SECTION 3.7 rational Functions 283 Analysis To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator: 1 __ 10 = 0.1 Notice
the horizontal asymptote is y = 0.1. This means the concentration, C, the ratio of pounds of sugar to gallons of water, will approach 0.1 in the long term. Try It #3 There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m. Finding the Domains of Rational Functions A vertical asymptote represents a value at which a rational function is undefined, so that value is not in the domain of the function. A reciprocal function cannot have values in its domain that cause the denominator to equal zero. In general, to find the domain of a rational function, we need to determine which inputs would cause division by zero. domain of a rational function The domain of a rational function includes all real numbers except those that cause the denominator to equal zero. How To… Given a rational function, find the domain. 1. Set the denominator equal to zero. 2. Solve to find the x-values that cause the denominator to equal zero. 3. The domain is all real numbers except those found in Step 2. Example 4 Finding the Domain of a Rational Function Find the domain of f (x) = x + 3 _ x2 − 9. Solution Begin by setting the denominator equal to zero and solving. x2 − 9 = 0 x2 = 9 x = ±3 The denominator is equal to zero when x = ±3. The domain of the function is all real numbers except x = ±3. Analysis A graph of this function, as shown in Figure 8, confirms that the function is not defined when x = ±3. y = 0 –2 –3 –4 –6 –5 y 4 3 2 1 –1 –1 –2 –3 –4 321 4 5 6 x x = 3 Figure 8 There is a vertical asymptote at x = 3 and a hole in the graph at x = −3. We will discuss these types of holes in greater detail later in this section. 284 CHAPTER 3 polynomial and rational Functions Try It #4 Find the domain of f (x) = 4x _____________. 5(x − 1)(x − 5) Identifying Vertical Asymptotes of Rational Functions By looking at the graph of a rational function, we can investigate
its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and calculate their location. Vertical Asymptotes The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Vertical asymptotes occur at the zeros of such factors. How To… Given a rational function, identify any vertical asymptotes of its graph. 1. Factor the numerator and denominator. 2. Note any restrictions in the domain of the function. 3. Reduce the expression by canceling common factors in the numerator and the denominator. 4. Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur. 5. Note any restrictions in the domain where asymptotes do not occur. These are removable discontinuities. Example 5 Identifying Vertical Asymptotes Find the vertical asymptotes of the graph of k(x) = Solution First, factor the numerator and denominator. 5 + 2x2 _________ 2 − x − x2. k(x) = 5 + 2x2 ________ 2 − x − x2 5 + 2x2 ___________ (2 + x)(1 − x) = To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero: Neither x = −2 nor x = 1 are zeros of the numerator, so the two values indicate two vertical asymptotes. The graph in Figure 9 confirms the location of the two vertical asymptotes. (2 + x)(1 − x) = 0 x = −2, 1 x = − 21 3 4 5 6 x –6 –5 –4 –3 –2 y = −2 –1 –1 –2 –3 –4 –5 –6 –7 Figure 9 SECTION 3.7 rational Functions 285 Removable Discontinuities Occasionally, a graph will contain a hole: a single point where the graph is not defined, indicated by an open circle. We call such a hole a removable discontinuity. For example, the function f (x) = may be re-written by factoring the numerator and the denominator. x2
− 1 __________ x2 − 2x − 3 f (x) = (x + 1)(x − 1) ____________ (x + 1)(x − 3) Notice that x + 1 is a common factor to the numerator and the denominator. The zero of this factor, x = −1, is the location of the removable discontinuity. Notice also that x − 3 is not a factor in both the numerator and denominator. The zero of this factor, x = 3, is the vertical asymptote. See Figure 10. Removable discontinuity at x = −1 –3 y 6 4 2 –1 –2 –4 –6 1 3 5 7 9 x Vertical asymptote at x = 3 Figure 10 removable discontinuities of rational functions A removable discontinuity occurs in the graph of a rational function at x = a if a is a zero for a factor in the denominator that is common with a factor in the numerator. We factor the numerator and denominator and check for common factors. If we find any, we set the common factor equal to 0 and solve. This is the location of the removable discontinuity. This is true if the multiplicity of this factor is greater than or equal to that in the denominator. If the multiplicity of this factor is greater in the denominator, then there is still an asymptote at that value. Example 6 Identifying Vertical Asymptotes and Removable Discontinuities for a Graph Find the vertical asymptotes and removable discontinuities of the graph of k(x) = x − 2 _____. x2 − 4 Solution Factor the numerator and the denominator. k(x) = x − 2 ___________ (x − 2)(x + 2) Notice that there is a common factor in the numerator and the denominator, x − 2. The zero for this factor is x = 2. This is the location of the removable discontinuity. Notice that there is a factor in the denominator that is not in the numerator, x + 2. The zero for this factor is x = −2. The vertical asymptote is x = −2. See Figure 11. 286 CHAPTER 3 polynomial and rational Functions y x = −2 6 4 2 –8 –6 –4 –2 2 4 x –2 –4 –6 Figure 11 The graph of this function will have the vertical asympt
ote at x = −2, but at x = 2 the graph will have a hole. Try It #5 Find the vertical asymptotes and removable discontinuities of the graph of f (x) = x2 − 25 ___________ x3 − 6x2 + 5x. Identifying Horizontal Asymptotes of Rational Functions While vertical asymptotes describe the behavior of a graph as the output gets very large or very small, horizontal asymptotes help describe the behavior of a graph as the input gets very large or very small. Recall that a polynomial’s end behavior will mirror that of the leading term. Likewise, a rational function’s end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions. There are three distinct outcomes when checking for horizontal asymptotes: Case 1: If the degree of the denominator > degree of the numerator, there is a horizontal asymptote at y = 0. Example: f (x) = 4x + 2 _________ x2 + 4x − 5 4x 4 _ _ In this case, the end behavior is f (x) ≈ x2 = x. This tells us that, as the inputs increase or decrease without bound, this 4 _ function will behave similarly to the function g(x) = x, and the outputs will approach zero, resulting in a horizontal asymptote at y = 0. See Figure 12. Note that this graph crosses the horizontal asymptote. y 6 4 2 y = 0 –8 –6 –4 –2 2 4 x –2 –4 –6 x = 1 x = −5 Figure 12 Horizontal Asymptote y = 0 when f(x ) =, q(x ) ≠ 0 where degree of p < degree of q. p(x ) ____ q(x ) Case 2: If the degree of the denominator < degree of the numerator by one, we get a slant asymptote. Example: f (x) = 3x2 − 2x + 1 __________ x − 1 3x2 _ x = 3x. This tells us that as the inputs increase or decrease without bound, this In this case, the end behavior is f (x) ≈ function will behave similarly to the function g(x) = 3x. As the inputs grow large, the outputs will grow and not
level off, so this graph has no horizontal asymptote. However, the graph of g(x) = 3x looks like a diagonal line, and since f will behave similarly to g, it will approach a line close to y = 3x. This line is a slant asymptote. SECTION 3.7 rational Functions 287 To find the equation of the slant asymptote, divide slant asymptote is the graph of the line g(x) = 3x + 1. See Figure 13. 3x2 − 2x + 1 ___________ x − 1. The quotient is 3x + 1, and the remainder is 2. The y = 3x + 1 2 4 6 8 x y 14 12 10 8 6 4 2 –2 –4 –6 –4 –2 Figure 13 Slant Asymptote when f( x ) = x = 1, q( x ) ≠ 0 where degree of p > degree of q by 1. p( x ) ____ q( x ) an _ Case 3: If the degree of the denominator = degree of the numerator, there is a horizontal asymptote at y =, where an and bn bn are the leading coefficients of p(x) and q(x) for f (x) =, q(x) ≠ 0. p(x) ____ q(x) Example: f (x) = 3x2 + 2 _________ x2 + 4x − 5 3x2 _ In this case, the end behavior is f (x) ≈ x2 = 3. This tells us that as the inputs grow large, this function will behave like the function g(x) = 3, which is a horizontal line. As x → ±∞, f (x) → 3, resulting in a horizontal asymptote at y = 3. See Figure 14. Note that this graph crosses the horizontal asymptote. y 12 9 6 3 –3 –6 –9 –15 –12 –9 –6 –5 Figure 14 Horizontal Asymptote when f ( x ) = p( x ) _ q( x ), q( x ) ≠ 0 where degree of p = degree of q. Notice that, while the graph of a rational function will never cross a vertical asymptote, the graph may or may not cross a horizontal or slant asymptote. Also
, although the graph of a rational function may have many vertical asymptotes, the graph will have at most one horizontal (or slant) asymptote. It should be noted that, if the degree of the numerator is larger than the degree of the denominator by more than one, the end behavior of the graph will mimic the behavior of the reduced end behavior fraction. For instance, if we had the function with end behavior f (x) = 3x5 − x2 _______ x + 3 f (x) ≈ = 3x4, 3x5 ___ x the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient. x → ±∞, f (x) → ∞ horizontal asymptotes of rational functions The horizontal asymptote of a rational function can be determined by looking at the degrees of the numerator and denominator. • Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0. • Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote. • Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients. 288 CHAPTER 3 polynomial and rational Functions Example 7 Identifying Horizontal and Slant Asymptotes For the functions below, identify the horizontal or slant asymptote. a. g(x) = 6x3 − 10x ________ 2x3 + 5x2 b. h(x) = x2 − 4x + 1 _________ x + 2 c. k(x) = x2 + 4x ______ x3 − 8 Solution For these solutions, we will use f (x) =, q(x) ≠ 0. p(x) ____ q(x) a. g(x) = 6x3 − 10x _ 2x3 + 5x2 : The degree of p = degree of q = 3, so we can find the hori zontal asymptote by taking the ratio of the leading terms. There is a horizontal asymptote at y = 6 _ 2 or y = 3. : The degree of p = 2 and degree of q = 1. Since p > q by 1, there is a slant asymptote found b. h(x) =
x2 − 4x + 1 _ x + 2 at x2 − 4x + 1 _ x + 2. 2 1 −4 −2 1 −6 1 12 13 The quotient is x − 2 and the remainder is 13. There is a slant asymptote at y = x − 2. c. k(x) = : The degree of p = 2 < degree of q = 3, so there is a horizontal asymptote y = 0. x2 + 4x _ x3 − 8 Example 8 Identifying Horizontal Asymptotes In the sugar concentration problem earlier, we created the equation C(t) = 5 + t ________ 100 + 10t. Find the horizontal asymptote and interpret it in context of the problem. Solution Both the numerator and denominator are linear (degree 1). Because the degrees are equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the numerator, the leading term is t, with coefficient 1. In the denominator, the leading term is 10t, with coefficient 10. The horizontal asymptote will be at the ratio of these values: This function will have a horizontal asymptote at y = 1 __. 10 t → ∞, C(t) → 1 __ 10 This tells us that as the values of t increase, the values of C will approach. In context, this means that, as more time goes by, the concentration of sugar in the tank will approach one-tenth of a pound of sugar per gallon of water or pounds per gallon. 1 __ 10 1 __ 10 Example 9 Identifying Horizontal and Vertical Asymptotes Find the horizontal and vertical asymptotes of the function f (x) = (x − 2)(x + 3) _________________ (x − 1)(x + 2)(x − 5) Solution First, note that this function has no common factors, so there are no potential removable discontinuities. The function will have vertical asymptotes when the denominator is zero, causing the function to be undefined. The denominator will be zero at x = 1, −2, and 5, indicating vertical asymptotes at these values. The numerator has degree 2, while the denominator has degree 3. Since the degree of the denominator is greater than the degree of the numerator, the denominator will grow faster than the numerator, causing the outputs to
tend towards zero as the inputs get large, and so as x → ±∞, f (x) → 0. This function will have a horizontal asymptote at y = 0. See Figure 15. y 6 4 2 y = 0 –6 –4 –2 2 4 6 8 x –2 –4 –6 x = −2 x = 1 x = 5 Figure 15 SECTION 3.7 rational Functions 289 Try It #6 Find the vertical and horizontal asymptotes of the function: f (x) = (2x − 1)(2x + 1) _____________ (x − 2)(x + 3) intercepts of rational functions A rational function will have a y-intercept when the input is zero, if the function is defined at zero. A rational function will not have a y-intercept if the function is not defined at zero. Likewise, a rational function will have x-intercepts at the inputs that cause the output to be zero. Since a fraction is only equal to zero when the numerator is zero, x-intercepts can only occur when the numerator of the rational function is equal to zero. Example 10 Finding the Intercepts of a Rational Function Find the intercepts of f (x) = (x − 2)(x + 3) __________________ (x − 1)(x + 2)(x − 5). Solution We can find the y-intercept by evaluating the function at zero f (0) = = (0 − 2)(0 + 3) _________________ (0 − 1)(0 + 2)(0 − 5) −6 ___ 10 = − 3 __ 5 The x-intercepts will occur when the function is equal to zero: = −0.6 0 = (x − 2)(x + 3) _________________ (x − 1)(x + 2)(x − 5) This is zero when the numerator is zero. The y-intercept is (0, −0.6), the x-intercepts are (2, 0) and (−3, 0). See Figure 16. 0 = (x − 2)(x + 3) x = 2, −3 y 6 5 4 3 2 1 (−3, 0) –4 –3 –5 –6 (2, 0) y = 0 x 321 4 5 6 7 8 (0, –0.6) x = 1 x = 5 –1 –2 0 –1 –2 –
3 –4 –5 –6 x = −2 Figure 16 Try It #7 Given the reciprocal squared function that is shifted right 3 units and down 4 units, write this as a rational function. Then, find the x- and y-intercepts and the horizontal and vertical asymptotes. 290 CHAPTER 3 polynomial and rational Functions Graphing Rational Functions In Example 9, we see that the numerator of a rational function reveals the x-intercepts of the graph, whereas the denominator reveals the vertical asymptotes of the graph. As with polynomials, factors of the numerator may have integer powers greater than one. Fortunately, the effect on the shape of the graph at those intercepts is the same as we saw with polynomials. The vertical asymptotes associated with the factors of the denominator will mirror one of the two toolkit reciprocal functions. When the degree of the factor in the denominator is odd, the distinguishing characteristic is that on one side of the vertical asymptote the graph heads towards positive infinity, and on the other side the graph heads towards negative infinity. See Figure 17. y –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 1 y = x 321 4 5 x x = 0 Figure 17 When the degree of the factor in the denominator is even, the distinguishing characteristic is that the graph either heads toward positive infinity on both sides of the vertical asymptote or heads toward negative infinity on both sides. See Figure 18. –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y = 1 x2 321 4 5 x x = 0 Figure 18 For example, the graph of f (x) = is shown in Figure 19. (x + 1)2(x − 3) _____________ (x + 3)2(x − 2) f (x) = (x + 1)2 (x − 3) (x + 3)2 (x − 23, 0) –8 –6 (−1, 0) –4 –2 2 –2 –4 –6 x = −3 x = 2 Figure 19 SECTION 3.7 rational Functions 291 • At the x-intercept x = −1 corresponding to the (x + 1)2 factor of the numerator, the graph bounces, consistent with the quadratic nature of the
factor. • At the x-intercept x = 3 corresponding to the (x − 3) factor of the numerator, the graph passes through the axis as we would expect from a linear factor. • At the vertical asymptote x = −3 corresponding to the (x + 3)2 factor of the denominator, the graph heads towards positive infinity on both sides of the asymptote, consistent with the behavior of the function f (x) = • At the vertical asymptote x = 2, corresponding to the (x − 2) factor of the denominator, the graph heads towards positive infinity on the left side of the asymptote and towards negative infinity on the right side, consistent with the 1 __ behavior of the function f (x) =. x 1 _ x 2. How To… Given a rational function, sketch a graph. 1. Evaluate the function at 0 to find the y-intercept. 2. Factor the numerator and denominator. 3. For factors in the numerator not common to the denominator, determine where each factor of the numerator is zero to find the x-intercepts. 4. Find the multiplicities of the x-intercepts to determine the behavior of the graph at those points. 5. For factors in the denominator, note the multiplicities of the zeros to determine the local behavior. For those factors not common to the numerator, find the vertical asymptotes by setting those factors equal to zero and then solve. 6. For factors in the denominator common to factors in the numerator, find the removable discontinuities by setting those factors equal to 0 and then solve. 7. Compare the degrees of the numerator and the denominator to determine the horizontal or slant asymptotes. 8. Sketch the graph. Example 11 Graphing a Rational Function Sketch a graph of f (x) = (x + 2)(x − 3) ____________. (x + 1)2(x − 2) Solution We can start by noting that the function is already factored, saving us a step. Next, we will find the intercepts. Evaluating the function at zero gives the y-intercept: f (0) = (0 + 2)(0 − 3) __ (0 + 1)2(0 − 2) = 3 To find the x-intercepts, we determine when the numerator of the function is zero
. Setting each factor equal to zero, we find x-intercepts at x = −2 and x = 3. At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept. We have a y-intercept at (0, 3) and x-intercepts at (−2, 0) and (3, 0). To find the vertical asymptotes, we determine when the denominator is equal to zero. This occurs when x + 1 = 0 and when x − 2 = 0, giving us vertical asymptotes at x = −1 and x = 2. There are no common factors in the numerator and denominator. This means there are no removable discontinuities. Finally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at y = 0. To sketch the graph, we might start by plotting the three intercepts. Since the graph has no x-intercepts between the vertical asymptotes, and the y-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in Figure 20. 292 CHAPTER 3 polynomial and rational Functions y 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1 –1 –2 321 4 5 6 x Figure 20 The factor associated with the vertical asymptote at x = −1 was squared, so we know the behavior will be the same on both sides of the asymptote. The graph heads toward positive infinity as the inputs approach the asymptote on the right, so the graph will head toward positive infinity on the left as well. For the vertical asymptote at x = 2, the factor was not squared, so the graph will have opposite behavior on either side of the asymptote. See Figure 21. After passing through the x-intercepts, the graph will then level off toward an output of zero, as indicated by the horizontal asymptote. y = 0 –6 –5 –4 y 6 5 4 3 2 1 –3 –2 –1 –1 –2 –3 –4 x = – Figure 21 Try It #8 Given the function f (x) = (x + 2)2(x − 2) ______________ 2(x − 1)2 (x − 3
), use the characteristics of polynomials and rational functions to describe its behavior and sketch the function. Writing Rational Functions Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. A rational function written in factored form will have an x-intercept where each factor of the numerator is equal to zero. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x-intercepts by introducing a corresponding set of factors. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors. writing rational functions from intercepts and asymptotes If a rational function has x-intercepts at x = x1, x2,..., xn, vertical asymptotes at x = v1, v2, …, vm, and no xi = any vj, then the function can be written in the form: f (x) = a (x − x1) p ___ (x − v1) q 2 … (x − xn) p 2 … (x − vm) q 1(x − x2) p 1(x − v2) q n n where the powers pi or qi on each factor can be determined by the behavior of the graph at the corresponding intercept or asymptote, and the stretch factor a can be determined given a value of the function other than the x-intercept or by the horizontal asymptote if it is nonzero. SECTION 3.7 rational Functions 293 How To… Given a graph of a rational function, write the function. 1. Determine the factors of the numerator. Examine the behavior of the graph at the x-intercepts to determine the zeroes and their multiplicities. (This is easy to do when finding the “simplest” function with small multiplicities—such as 1 or 3—but may be difficult for larger multiplicities—such as 5 or 7, for example.) 2. Determine the factors of the denominator. Examine the behavior on both sides of each vertical asymptote to determine the factors and their
powers. 3. Use any clear point on the graph to find the stretch factor. Example 12 Writing a Rational Function from Intercepts and Asymptotes Write an equation for the rational function shown in Figure 22. –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –7 321 4 5 6 x Figure 22 Solution The graph appears to have x-intercepts at x = −2 and x = 3. At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at x = −1 seems to exhibit the basic behavior 1 __ similar to, with the graph heading toward positive infinity on one side and heading toward negative infinity on the x 1 _ other. The asymptote at x = 2 is exhibiting a behavior similar to x 2, with the graph heading toward negative infinity on both sides of the asymptote. See Figure 23. y 6 4 2 Vertical asymptotes –6 –4 –2 2 4 6 x x-intercepts –2 –4 –6 Figure 23 We can use this information to write a function of the form f (x) = a (x + 2)(x − 3) __ (x + 1)(x − 2)2 294 CHAPTER 3 polynomial and rational Functions To find the stretch factor, we can use another clear point on the graph, such as the y-intercept (0, −2). −2 = a −2 = a (0 + 2)(0 − 3) __ (0 + 1)(0 − 2)2 −6 _ 4 −8 _ −6 4 _ = 3 a = This gives us a final function of f (x) = 4(x + 2)(x − 3) ______________ 3(x + 1)(x − 2)2. Access these online resources for additional instruction and practice with rational functions. • Graphing Rational Functions (http://openstaxcollege.org/l/graphrational) • Find the equation of a Rational Function (http://openstaxcollege.org/l/equatrational) • Determining Vertical and Horizontal Asymptotes (http://openstaxcollege.org/l/asymptote) • Find the Intercepts, Asymptotes, and Hole of a Rational Function (http://openstaxcollege.org/l/
interasymptote) SECTION 3.7 section exercises 295 3.7 SeCTIOn exeRCISeS VeRBAl 1. What is the fundamental difference in the algebraic representation of a polynomial function and a rational function? 3. If the graph of a rational function has a removable discontinuity, what must be true of the functional rule? 5. Can a graph of a rational function have no x-intercepts? If so, how? 2. What is the fundamental difference in the graphs of polynomial functions and rational functions? 4. Can a graph of a rational function have no vertical asymptote? If so, how? AlGeBRAIC For the following exercises, find the domain of the rational functions. 6. f (x) = x − 1 _____ x + 2 9. f (x) = x2 + 4x − 3 _________ x4 − 5x2 + 4 7. f (x) = x + 1 _____ x2 − 1 8. f (x) = x2 + 4 _________ x2 − 2x − 8 For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions. 10. f (x) = 4 ____ x − 1 13. f (x) = 16. f (x) = x __________ x2 + 5x − 36 x2 − 1 ___________ x3 + 9x2 + 14x 11. f (x) = 2 _____ 5x + 2 14. f (x) = 3 + x ______ x3 − 27 17. f (x) = x + 5 ______ x2 − 25 12. f (x) = x _____ x2 − 9 15. f (x) = 3x − 4 _______ x3 − 16x 18. f (x) = x − 4 _____ x − 6 19. f (x) = 4 − 2x ______ 3x − 1 For the following exercises, find the x- and y-intercepts for the functions. 20. f (x) = x + 5 _____ x2 + 4 21. f (x) = x _____ x2 − x 22. f (x) = x2 + 8x + 7 ___________ x2 + 11x + 30 23. f (x) = x2 + x + 6 ___________ x2
− 10x + 24 24. f (x) = 94 − 2x2 _______ 3x2 − 12 For the following exercises, describe the local and end behavior of the functions. 25. f (x) = x _____ 2x + 1 26. f (x) = 2x _____ x − 6 27. f (x) = −2x _____ x − 6 28. f (x) = x2 − 4x + 3 _________ x2 − 4x − 5 29. f (x) = 2x2 − 32 ___________ 6x2 + 13x − 5 For the following exercises, find the slant asymptote of the functions. 30. f (x) = 24x2 + 6x ________ 31. f (x) = 4x2 − 10 _______ 2x − 4 2x + 1 32. f (x) = 81x2 − 18 ________ 3x − 2 33. f (x) = 6x3 − 5x _______ 3x2 + 4 34. f (x) = x2 + 5x + 4 _________ x − 1 296 CHAPTER 3 polynomial and rational Functions GRAPHICAl For the following exercises, use the given transformation to graph the function. Note the vertical and horizontal asymptotes. 35. The reciprocal function shifted up two units. 36. The reciprocal function shifted down one unit and left three units. 37. The reciprocal squared function shifted to the right 38. The reciprocal squared function shifted down 2 units 2 units. and right 1 unit. For the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph. 39. p(x) = 2x − 3 ______ x + 4 40. q(x) = x − 5 _____ 3x − 1 4 ______ (x − 2)2 41. s(x) = 42. r(x) = 5 ______ (x + 1)2 43. f (x) = 3x 2 − 14x − 5 __ 3x 2 + 8x − 16 44. g(x) = 2x 2 + 7x − 15 __ 3x 2 − 14 + 15 45. a(x) = x 2 + 2x − 3 _ x 2 − 1 46. b(x 47. h(
x) = 2x 2 + x − 1 _ x − 4 48. k(x) = 2x 2 − 3x − 20 __ x − 5 49. w(x) = (x − 1)(x + 3)(x − 5) __ (x + 2)2(x − 4) 50. z(x) = (x + 2)2(x − 5) __ (x − 3)(x + 1)(x + 4) For the following exercises, write an equation for a rational function with the given characteristics. 51. Vertical asymptotes at x = 5 and x = −5, x-intercepts at (2, 0) and (−1, 0), y-intercept at (0, 4) 52. Vertical asymptotes at x = −4 and x = −1, x-intercepts at (1, 0) and (5, 0), y-intercept at (0, 7) 53. Vertical asymptotes at x = −4 and x = −5, x-intercepts at (4, 0) and (−6, 0), horizontal asymptote at y = 7 54. Vertical asymptotes at x = −3 55. Vertical asymptote at x = −1, 56. Vertical asymptote at x = 3, and x = 6, x-intercepts at (−2, 0) and (1, 0), horizontal asymptote at y = −2 double zero at x = 2, y-intercept at (0, 2) double zero at x = 1, y-intercept at (0, 4) For the following exercises, use the graphs to write an equation for the function. 57. y 5 4 3 2 1 –2 0 –1 –2 –3 –4 –5 –10 –8 –6 –4 58. 642 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 59. 642 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 642 8 10 x SECTION 3.7 section exercises 297 62. 642 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 642 8 10 x
642 8 10 x 61. 642 8 10 x –10 –8 –6 –4 64. 642 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 60. 63. y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 y 5 4 3 2 1 –2 –1 –2 –3 –4 –5 –10 –8 –6 –4 –10 –8 –6 –4 nUMeRIC For the following exercises, make tables to show the behavior of the function near the vertical asymptote and reflecting the horizontal asymptote. 65. f (x) = 1 _____ x − 2 66. f (x) = x _____ x − 3 67. f (x) = 2x _____ x + 4 68. f (x) = 2x ______ (x − 3)2 69. f (x) = x2 _________ x2 + 2x + 1 TeCHnOlOGY For the following exercises, use a calculator to graph f (x). Use the graph to solve f (x) > 0. 70. f (x) = 2 _____ x + 1 4 _____ 2x − 3 71. f (x) = 72. f (x) = 2 ___________ (x − 1)(x + 2) 73. f (x) = x + 2 ___________ (x − 1)(x − 4) 74. f (x) = (x + 3)2 ____________ (x − 1)2(x + 1) exTenSIOnS For the following exercises, identify the removable discontinuity. 75. f (x) = x2 − 4 _____ x − 2 76. f (x) = x3 + 1 _____ x + 1 78. f (x) = 2x2 + 5x − 3 __________ x + 3 79. f (x) = x3 + x2 ______ x + 1 77. f (x) = x2 + x − 6 ________ x − 2 298 CHAPTER 3 polynomial and rational Functions ReAl-WORlD APPlICATIOnS For the following exercises, express a rational function that describes the situation. 80. A large mixing tank currently
contains 200 gallons of water, into which 10 pounds of sugar have been mixed. A tap will open, pouring 10 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 3 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after t minutes. 81. A large mixing tank currently contains 300 gallons of water, into which 8 pounds of sugar have been mixed. A tap will open, pouring 20 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 2 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after t minutes. For the following exercises, use the given rational function to answer the question. 82. The concentration C of a drug in a patient’s 83. The concentration C of a drug in a patient’s bloodstream t hours after injection in given by C(t) = 2t _ 3 + t2. What happens to the concentration of the drug as t increases? bloodstream t hours after injection is given by C(t) =. Use a calculator to approximate the 100t _ 2t2 + 75 time when the concentration is highest. For the following exercises, construct a rational function that will help solve the problem. Then, use a calculator to answer the question. 84. An open box with a square base is to have a volume of 108 cubic inches. Find the dimensions of the box that will have minimum surface area. Let x = length of the side of the base. 85. A rectangular box with a square base is to have a volume of 20 cubic feet. The material for the base costs 30 cents/square foot. The material for the sides costs 10 cents/square foot. The material for the top costs 20 cents/square foot. Determine the dimensions that will yield minimum cost. Let x = length of the side of the base. 86. A right circular cylinder has volume of 100 cubic inches. Find the radius and height that will yield minimum surface area. Let x = radius. 87. A right circular cylinder with no top has a volume of 50 cubic meters. Find the radius that will yield minimum surface area. Let x = radius. 88. A right circular cylinder is to have a volume of 40 cubic inches. It costs 4 cents/square inch to construct the top and bottom and 1 cent/square inch to construct the rest of the cylinder. Find the radius to yield minimum cost
. Let x = radius. SECTION 3.8 inverses and radical Functions 299 leARnInG OBjeCTIVeS In this section, you will: • • Find the inverse of a polynomial function. Restrict the domain to find the inverse of a polynomial function. 3.8 InVeRSeS AnD RADICAl FUnCTIOnS A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume is found using a formula from elementary geometry. Figure 1 πr 2 h V = 1 __ 3 = 1 __ 3 = 2 __ 3 We have written the volume V in terms of the radius r. However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula πr 2(2r) πr 3 3 r = ____ 3V ___ 2π √ This function is the inverse of the formula for V in terms of r. In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process. Finding the Inverse of a Polynomial Function Two functions f and g are inverse functions if for every coordinate pair in f, (a, b), there exists a corresponding coordinate pair in the inverse function, g, (b, a). In other words, the coordinate pairs of the inverse functions have the input and output interchanged. For a function to have an inverse function the function to create a new function that is one-to-one and would have an inverse function. For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in Figure 2. We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water. 12 in 18 in 3 ft Figure 2 30 0 CHAPTER 3 polynomial and rational Functions Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with x measured horizontally and y measured vertically, with the origin at the vertex of the parabola. See Figure 3. y 18 16 14 12 10 8 6 4 2 –
2 –2 –4 –6 –10 –8 –6 –4 642 8 10 x Figure 3 From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form y(x) = ax2. Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor a. Our parabolic cross section has the equation 18 = a62 a = 18 __ 36 = 1 __ 2 y(x) = 1 __ x2 2 We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth y the width will be given by 2x, so we need to solve the equation above for x and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative. To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive x values. On this domain, we can find an inverse by solving for the input variable: 1 __ y = x2 2 2y = x2 x = ± √ — 2y This is not a function as written. We are limiting ourselves to positive x values, so we eliminate the negative solution, giving us the inverse function we’re looking for. y =, x > 0 x2 __ 2 Because x is the distance from the center of the parabola to either side, the entire width of the water at the top will be 2x. The trough is 3 feet (36 inches) long, so the surface area will then be: Area = l ⋅ w = 36 ⋅ 2x = 72x = 72 √ — 2y This example illustrates two important points: 1. When finding the inverse of a quadratic, we have to limit ourselves to a domain on which the function is one-to-one. 2. The inverse of a quadratic function is a square root function. Both are toolkit functions and different types of power functions. SECTION 3.8 inverses and radical Functions 301 Functions involving roots are often called radical functions. While it is not possible to find an inverse of most polyn
omial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation f −1(x). Warning: f −1(x) is not the same as the reciprocal of the function f (x). This use of “−1” is reserved to denote inverse functions. To denote the reciprocal of a function f (x), we would need to write ( f (x))−1 = 1 _. f (x) An important relationship between inverse functions is that they “undo” each other. If f −1 is the inverse of a function f, then f is the inverse of the function f −1. In other words, whatever the function f does to x, f −1 undoes it—and viceversa. More formally, we write and f −1 ( f (x)) = x, for all x in the domain of f f ( f −1 (x)) = x, for all x in the domain of f −1 verifying two functions are inverses of one another Two functions, f and g, are inverses of one another if for all x in the domain of f and g. g( f (x)) = f ( g(x)) = x How To… Given a polynomial function, find the inverse of the function by restricting the domain in such a way that the new function is one-to-one. 1. Replace f (x) with y. 2. Interchange x and y. 3. Solve for y, and rename the function f −1(x). Example 1 Show that f (x) = Verifying Inverse Functions 1 _ x + 1 1 and f −1(x) = __ − 1 are inverses, for x ≠ 0, −1. x Solution We must show that f −1( f (x)) = x and f ( f −1(x)) = x. 1 _____  x + 1 f −1(f (x)) = f −1  1 _ 1 _____ x + 1 = (x + 1) − 1 − 1 = = x 1 __ f (f −1(x)) = f  − 1  x 1 __ = 1 __ − __ x = x Therefore, f (x) = 1 _____ x + 1 1 and f −1 (x) = __ − 1 are
inverses. x Try It #1 Show that f (x) = x + 5 _____ 3 and f −1(x) = 3x − 5 are inverses. 30 2 CHAPTER 3 polynomial and rational Functions Example 2 Finding the Inverse of a Cubic Function Find the inverse of the function f (x) = 5x3 + 1. Solution This is a transformation of the basic cubic toolkit function, and based on our knowledge of that function, we know it is one-to-one. Solving for the inverse by solving for x. y = 5x3 + 1 x = 5y3 + 1 x − 1 = 5y3 x − 1 _____ 5 = y3 f −1(x) = √ Analysis Look at the graph of f and f −1. Notice that the two graphs are symmetrical about the line y = x. This is always the case when graphing a function and its inverse function. 3 — x − 1 _____ 5 Also, since the method involved interchanging x and y, notice corresponding points. If (a, b) is on the graph of f, then (b, a) is on the graph of f −1. Since (0, 1) is on the graph of f, then (1, 0) is on the graph of f −1. Similarly, since (1, 6) is on the graph of f, then (6, 1) is on the graph of f −1. See Figure 4. y f (x) = 5x3 + 1 (1, 6) y = x (6, 1) 6 4 2 (0, 1) –6 –4 –2 2 4 6 x (1, 0) –2 –4 –6 Figure 4 Try It #2 Find the inverse function of f (x) = 3 √ — x + 4 Restricting the Domain to Find the Inverse of a Polynomial Function So far, we have been able to find the inverse functions of cubic functions without having to restrict their domains. However, as we know, not all cubic polynomials are one-to-one. Some functions that are not one-to-one may have their domain restricted so that they are one-to-one, but only over that domain. The function over the restricted domain would then have an inverse function. Since quadratic functions are not one-to-one, we must restrict their domain in
order to find their inverses. restricting the domain If a function is not one-to-one, it cannot have an inverse. If we restrict the domain of the function so that it becomes one-to-one, thus creating a new function, this new function will have an inverse. SECTION 3.8 inverses and radical Functions 303 How To… Given a polynomial function, restrict the domain of a function that is not one-to-one and then find the inverse. 1. Restrict the domain by determining a domain on which the original function is one-to-one. 2. Replace f (x) with y. 3. Interchange x and y. 4. Solve for y, and rename the function or pair of functions f −1(x). 5. Revise the formula for f −1(x) by ensuring that the outputs of the inverse function correspond to the restricted domain of the original function. Example 3 Restricting the Domain to Find the Inverse of a Polynomial Function Find the inverse function of f : a. f (x) = (x − 4)2, x ≥ 4 b. f (x) = (x − 4)2, x ≤ 4 Solution The original function f (x) = (x − 4)2 is not one-to-one, but the function is restricted to a domain of x ≥ 4 or x ≤ 4 on which it is one-to-one. See Figure 5. y 10 8 6 4 2 f (x) = (x − 4)2, x ≥ 4 y 10 8 6 4 2 f (x) = (x − 4)2, x ≤ 4 –10 –8 –6 –4 –2 –2 642 8 10 –10 –8 –6 –4 –2 –2 642 8 10 x x Figure 5 To find the inverse, start by replacing f (x) with the simple variable y. y = (x − 4)2 Interchange x and y. — x = (y − 4)2 Take the square root. x = y − 4 x = y Add 4 to both sides. — ± √ 4 ± √ This is not a function as written. We need to examine the restrictions on the domain of the original function to determine the inverse. Since we reversed the roles of x and y for the original f (x), we looked at the domain: the values x could assume. When we reversed the roles of x
and y, this gave us the values y could assume. For this function, x ≥ 4, so for the inverse, we should have y ≥ 4, which is what our inverse function gives. a. The domain of the original function was restricted to x ≥ 4, so the outputs of the inverse need to be the same, f (x) ≥ 4, and we must use the + case: f −1(x) = 4 + √ — x b. The domain of the original function was restricted to x ≤ 4, so the outputs of the inverse need to be the same, f (x) ≤ 4, and we must use the − case: f −1(x) = 4 − √ — x Analysis On the graphs in Figure 6, we see the original function graphed on the same set of axes as its inverse function. Notice that together the graphs show symmetry about the line y = x. The coordinate pair (4, 0) is on the graph of f and the coordinate pair (0, 4) is on the graph of f −1. For any coordinate pair, if (a, b) is on the graph of f, then (b, a) is on the graph of f −1. Finally, observe that the graph of f intersects the graph of f −1 on the line y = x. Points of intersection for the graphs of f and f −1 will always lie on the line y = x. 30 4 CHAPTER 3 polynomial and rational Functions y 10 8 6 2 (0, 4) f (x) y = x f –1(x) (0, 4) 10 8 6 2 y f (x) y = x –10 –8 –6 –4 –2 –2 2 6 8 10 (4, 0) x Figure 6 –10 –8 –6 –4 –2 –2 2 6 (4, 0) f –1(x) x 8 10 Example 4 Finding the Inverse of a Quadratic Function When the Restriction Is Not Specified Restrict the domain and then find the inverse of f (x) = (x − 2)2 − 3. Solution We can see this is a parabola with vertex at (2, −3) that opens upward. Because the graph will be decreasing on one side of the vertex and increasing on the other side, we can restrict this function to a domain on which it will be one-to-one by limiting the domain to
x ≥ 2. To find the inverse, we will use the vertex form of the quadratic. We start by replacing f (x) with a simple variable, y, then solve for x. y = (x − 2)2 − 3 Interchange x and y. x = (y − 2)2 − 3 Add 3 to both sides. Take the square rooty − 2)1(x Add 2 to both sides. Rename the function. Now we need to determine which case to use. Because we restricted our original function to a domain of x ≥ 2, the outputs of the inverse should be the same, telling us to utilize the + case f −1(x) = 2 + √ — x + 3 If the quadratic had not been given in vertex form, rewriting it into vertex form would be the first step. This way we may easily observe the coordinates of the vertex to help us restrict the domain. Analysis Notice that we arbitrarily decided to restrict the domain on x ≥ 2. We could just have easily opted to restrict the domain on x ≤ 2, in which case f −1(x) = 2 − √ x + 3. Observe the original function graphed on the same set of axes as its inverse function in Figure 7. Notice that both graphs show symmetry about the line y = x. The coordinate pair (2, −3) is on the graph of f and the coordinate pair (−3, 2) is on the graph of f −1. Observe from the graph of both functions on the same set of axes that — and domain of f = range of f −1 = [2, ∞) domain of f −1 = range of f = [−3, ∞) Finally, observe that the graph of f intersects the graph of f −1 along the line y = x. (−3, 2) –10 –8 –6 –4 y y = x f (x) f –1(x) x 8 10 642 (2, −3) 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 7 SECTION 3.8 inverses and radical Functions 305 Try It #3 Find the inverse of the function f (x) = x2 + 1, on the domain x ≥ 0. Solving Applications of Radical Functions Notice that the functions from previous examples were all polynomials, and their inverses were radical functions. If we want to find the
inverse of a radical function, we will need to restrict the domain of the answer because the range of the original function is limited. How To… Given a radical function, find the inverse. 1. Determine the range of the original function. 2. Replace f (x) with y, then solve for x. 3. If necessary, restrict the domain of the inverse function to the range of the original function. Example 5 Finding the Inverse of a Radical Function Restrict the domain and then find the inverse of the function f (x) = √ — x − 4. Solution Note that the original function has range f (x) ≥ 0. Replace f (x) with y, then solve for x Replace f (x) with y. Interchange x and y. Square each side. x2 = y − 4 x2 + 4 = y f −1(x) = x2 + 4 Add 4. Rename the function f−1(x). Recall that the domain of this function must be limited to the range of the original function. f −1(x) = x2 + 4, x ≥ 0 Analysis Notice in Figure 8 that the inverse is a reflection of the original function over the line y = x. Because the original function has only positive outputs, the inverse function has only positive inputs. 12 10 8 6 4 2 y f –1(x) y = x f (x) x 2 4 6 8 10 12 Figure 8 Try It #4 Restrict the domain and then find the inverse of the function f (x) = √ — 2x + 3. 30 6 CHAPTER 3 polynomial and rational Functions Solving Applications of Radical Functions Radical functions are common in physical models, as we saw in the section opener. We now have enough tools to be able to solve the problem posed at the start of the section. Example 6 Solving an Application with a Cubic Function A mound of gravel is in the shape of a cone with the height equal to twice the radius. The volume of the cone in terms of the radius is given by V = 2 __ 3 πr 3 2 __ Find the inverse of the function V = πr 3 that determines the volume V of a cone and is a function of the radius r. 3 Then use the inverse function to calculate the radius of such a mound of gravel measuring 100 cubic feet. Use π = 3.14. Solution Start with the given function for V. Notice that the meaningful domain for the function is
r ≥ 0 since negative radii would not make sense in this context. Also note the range of the function (hence, the domain of the inverse function) is V ≥ 0. Solve for r in terms of V, using the method outlined previously. πr 3 V = 2 __ 3 r 3 = 3V ___ 2π ____ √ 3V ___ 2π r = 3 Solve for r 3. Solve for r. This is the result stated in the section opener. Now evaluate this for V = 100 and π = 3.14. r = 3 ____ 3V ___ 2π _______ 3 ⋅ 100 ______ 2 ⋅ 3.14 √ √ 3 = Therefore, the radius is about 3.63 ft. ≈ 3 √ — 47.7707 ≈ 3.63 Determining the Domain of a Radical Function Composed with Other Functions When radical functions are composed with other functions, determining domain can become more complicated. Example 7 Finding the Domain of a Radical Function Composed with a Rational Function Find the domain of the function f (x) = √ ___________ (x + 2)(x − 3) ____________ (x − 1). Solution Because a square root is only defined when the quantity under the radical is non-negative, we need to determine ≥ 0. The output of a rational function can change signs (change from positive to negative or vice where versa) at x-intercepts and at vertical asymptotes. For this equation, the graph could change signs at x = −2, 1, and 3. (x + 2)(x − 3) ____________ (x − 1) To determine the intervals on which the rational expression is positive, we could test some values in the expression or sketch a graph. While both approaches work equally well, for this example we will use a graph as shown in Figure 9. y x = 1 Outputs are non-negative (−2, 0) –3 –2 –4 –5 –7 –6 10 8 6 4 2 –1 –2 –4 –6 –8 –10 Outputs are non-negative (3, 0) 321 4 5 6 7 x Figure 9 SECTION 3.8 inverses and radical Functions 307 This function has two x-intercepts, both of which exhibit linear behavior near the x-intercepts. There is one vertical asymptote, corresponding to a linear factor
; this behavior is similar to the basic reciprocal toolkit function, and there is no horizontal asymptote because the degree of the numerator is larger than the degree of the denominator. There is a y-intercept at (0, √ — 6 ). From the y-intercept and x-intercept at x = −2, we can sketch the left side of the graph. From the behavior at the asymptote, we can sketch the right side of the graph. From the graph, we can now tell on which intervals the outputs will be non-negative, so that we can be sure that the original function f (x) will be defined. f (x) has domain −2 ≤ x < 1 or x ≥ 3, or in interval notation, [−2, 1) ∪ [3, ∞). Finding Inverses of Rational Functions As with finding inverses of quadratic functions, it is sometimes desirable to find the inverse of a rational function, particularly of rational functions that are the ratio of linear functions, such as in concentration applications. Example 8 Finding the Inverse of a Rational Function The function C = represents the concentration C of an acid solution after n mL of 40% solution has been 20 + 0.4n ________ 100 + n added to 100 mL of a 20% solution. First, find the inverse of the function; that is, find an expression for n in terms of C. Then use your result to determine how much of the 40% solution should be added so that the final mixture is a 35% solution. Solution We first want the inverse of the function. We will solve for n in terms of C. C = 20 + 0.4n ________ 100 + n C(100 + n) = 20 + 0.4n 100C + Cn = 20 + 0.4n 100C − 20 = 0.4n − Cn 100C − 20 = (0.4 − C)n Now evaluate this function for C = 0.35 (35%). n = 100C − 20 _________ 0.4 − C n = 100(0.35) − 20 ____________ 0.4 − 0.35 = 15 ___ 0.05 = 300 We can conclude that 300 mL of the 40% solution should be added. Try It #5 Find the inverse of the function f (x) = x + 3 _____. x − 2 Access these online resources for additional instruction and
practice with inverses and radical functions. • Graphing the Basic Square Root Function (http://openstaxcollege.org/l/graphsquareroot) • Find the Inverse of a Square Root Function (http://openstaxcollege.org/l/inversesquare) • Find the Inverse of a Rational Function (http://openstaxcollege.org/l/inverserational) • Find the Inverse of a Rational Function and an Inverse Function Value (http://openstaxcollege.org/l/rationalinverse) • Inverse Functions (http://openstaxcollege.org/l/inversefunction) 30 8 CHAPTER 3 polynomial and rational Functions 3.8 SeCTIOn exeRCISeS VeRBAl 1. Explain why we cannot find inverse functions for all 2. Why must we restrict the domain of a quadratic polynomial functions. function when finding its inverse? 3. When finding the inverse of a radical function, what 4. The inverse of a quadratic function will always take restriction will we need to make? what form? AlGeBRAIC For the following exercises, find the inverse of the function on the given domain. 5. f (x) = (x − 4)2, [4, ∞) 6. f (x) = (x + 2)2, [−2, ∞) 7. f (x) = (x + 1)2 − 3, [−1, ∞) 8. f (x. f (x) = 3x 2 + 5, (−∞, 0] 10. f (x) = 12 − x 2, [0, ∞) 11. f (x) = 9 − x 2, [0, ∞) 12. f (x) = 2x 2 + 4, [0, ∞) For the following exercises, find the inverse of the functions. 13. f (x) = x 3 + 5 16. f (x) = 4 − 2x 3 14. f (x) = 3x 3 + 1 15. f (x) = 4 − x 3 For the following exercises, find the inverse of the functions. 17. f (x) = √ — 2x + 1 18. f (x) = √ — 3 − 4x 19. f (x) = 9 + √ — 4x
− 4 20. f (x) = √ — 6x − 8 + 5 21. f (x) = 9 + 2 3 √ — x 23. f (x) = 2 _____ x + 8 26. f (x) = x − 2 _____ x + 7 24. f (x) = 3 _____ x − 4 27. f (x) = 3x + 4 ______ 5 − 4x 22. f (x) = 3 − 3 √ — x 25. f (x) = x + 3 _____ x + 7 28. f (x) = 5x + 1 ______ 2 − 5x 29. f (x) = x 2 + 2x, [−1, ∞) 30. f (x) = x 2 + 4x + 1, [−2, ∞) 31. f (x) = x 2 − 6x + 3, [3, ∞) GRAPHICAl For the following exercises, find the inverse of the function and graph both the function and its inverse. 32. f (x) = x 2 + 2, x ≥ 0 33. f (x) = 4 − x 2, x ≥ 0 34. f (x) = (x + 3)2, x ≥ −3 35. f (x) = (x − 4)2, x ≥ 4 36. f (x) = x 3 + 3 38. f (x) = x 2 + 4x, x ≥ −2 39. f (x) = x 2 − 6x + 1, x ≥ 3 37. f (x) = 1 − x 3 40. f (x) = 2 __ x 41. f (x) = 1 __ x2, x ≥ 0 For the following exercises, use a graph to help determine the domain of the functions. 42. f (x) = √ 45. f (x) = √ _____________ (x + 1)(x − 1) __ x ___________ x2 − x − 20 _ x − 2 _____________ (x + 2)(x − 3) __ x − 1 43. f (x) = √ 46. f (x) = √ ______ 9 − x2 _ x + 4 44. f (x) = √ ________ x(x + 3) _ x − 4 SECTION 3.8 section exercises 309 TeCHnOlOGY For the following
exercises, use a calculator to graph the function. Then, using the graph, give three points on the graph of the inverse with y-coordinates given. 47. f (x) = x3 − x − 2, y = 1, 2, 3 48. f (x) = x3 + x − 2, y = 0, 1, 2 49. f (x) = x3 + 3x − 4, y = 0, 1, 2 50. f (x) = x3 + 8x − 4, y = −1, 0, 1 51. f (x) = x4 + 5x + 1, y = −1, 0, 1 exTenSIOnS For the following exercises, find the inverse of the functions with a, b, c positive real numbers. 52. f (x) = ax3 + b 53. f (x) = x2 + bx 54. f (x) = √ — ax2 + b 55. f (x) = 3 √ — ax + b 56. f (x) = ax + b ______ x + c ReAl-WORlD APPlICATIOnS For the following exercises, determine the function described and then use it to answer the question. 57. An object dropped from a height of 200 meters has a height, h(t), in meters after t seconds have lapsed, such that h(t) = 200 − 4.9t 2. Express t as a function of height, h, and find the time to reach a height of 50 meters. 58. An object dropped from a height of 600 feet has a height, h(t), in feet after t seconds have elapsed, such that h(t) = 600 − 16t 2. Express t as a function of height h, and find the time to reach a height of 400 feet. 59. The volume, V, of a sphere in terms of its radius, r, 4 __ is given by V(r) = πr 3. Express r as a function of 3 V, and find the radius of a sphere with volume of 200 cubic feet. 60. The surface area, A, of a sphere in terms of its radius, r, is given by A(r) = 4πr 2. Express r as a function of V, and find the radius of a sphere with a surface area of 1000 square inches. 61. A container holds 100 ml of a
solution that is 25 ml acid. If n ml of a solution that is 60% acid is added, the function C(n) = gives the 25 + 0.6n ________ 100 + n concentration, C, as a function of the number of ml added, n. Express n as a function of C and determine the number of mL that need to be added to have a solution that is 50% acid. 62. The period T, in seconds, of a simple pendulum as a function of its length l, in feet, is given by T(l) = 2π √ ____ l ____ 32.2. Express l as a function of T and determine the length of a pendulum with period of 2 seconds. 63. The volume of a cylinder, V, in terms of radius, r, 64. The surface area, A, of a cylinder in terms of its and height, h, is given by V = πr 2h. If a cylinder has a height of 6 meters, express the radius as a function of V and find the radius of a cylinder with volume of 300 cubic meters. radius, r, and height, h, is given by A = 2πr2 + 2πrh. If the height of the cylinder is 4 feet, express the radius as a function of V and find the radius if the surface area is 200 square feet. 65. The volume of a right circular cone, V, in terms of its 1 _ radius, r, and its height, h, is given by V = πr 2h. 3 Express r in terms of h if the height of the cone is 12 feet and find the radius of a cone with volume of 50 cubic inches. 66. Consider a cone with height of 30 feet. Express the radius, r, in terms of the volume, V, and find the radius of a cone with volume of 1000 cubic feet. 31 0 CHAPTER 3 polynomial and rational Functions leARnInG OBjeCTIVeS In this section, you will: • • • Solve direct variation problems. Solve inverse variation problems. Solve problems involving joint variation. 3.9 MODelInG USInG VARIATIOn A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,
600, she will earn $736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate. Solving Direct Variation Problems In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula e = 0.16s tells us her earnings, e, come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See Table 1. s, sales prices $4,600 $9,200 $18,400 e = 0.16s e = 0.16(4,600) = 736 Interpretation A sale of a $4,600 vehicle results in $736 earnings. e = 0.16(9,200) = 1,472 A sale of a $9,200 vehicle results in $1472 earnings. e = 0.16(18,400) = 2,944 A sale of a $18,400 vehicle results in $2944 earnings. Table 1 Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation. Each variable in this type of relationship varies directly with the other. Figure 1 represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula y = kxn is used for direct variation. The value k is a nonzero constant greater than zero and is called the constant of variation. In this case, k = 0.16 and n = 1. 5,000 4,000 3,000 2,000 1,000 $, 18,400, 2,944) (9,200, 1,472) (4,600, 736) 6,000 12,000 18,000 24,000 30,000 s, Sales Prices in Dollars Figure 1 SECTION 3.9 modeling using variation 311 direct variation If x and y are related by an equation
of the form y = kx n then we say that the relationship is direct variation and y varies directly with the nth power of x. In direct y _ variation relationships, there is a nonzero constant ratio k = xn, where k is called the constant of variation, which help defines the relationship between the variables. How To… Given a description of a direct variation problem, solve for an unknown. 1. Identify the input, x, and the output, y. 2. Determine the constant of variation. You may need to divide y by the specified power of x to determine the constant of variation. 3. Use the constant of variation to write an equation for the relationship. 4. Substitute known values into the equation to find the unknown. Example 1 Solving a Direct Variation Problem The quantity y varies directly with the cube of x. If y = 25 when x = 2, find y when x is 6. Solution The general formula for direct variation with a cube is y = kx 3. The constant can be found by dividing y by the cube of x. k = y _ x3 25 __ 23 25 __ 8 = = Now use the constant to write an equation that represents this relationship. Substitute x = 6 and solve for y. y = 25 __ x3 8 (6)3 y = 25 __ 8 = 675 Analysis The graph of this equation is a simple cubic, as shown in Figure 2. y 800 600 400 200 (6, 675) (2, 25) 0 2 x 8 10 4 6 Figure 2 31 2 CHAPTER 3 polynomial and rational Functions Q & A… Do the graphs of all direct variation equations look like Example 1? No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through (0,0). Try It #1 The quantity y varies directly with the square of x. If y = 24 when x = 3, find y when x is 4. Solving Inverse Variation Problems 14,000 ______ d Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, the formula T = gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be
28°F. If we create Table 2, we observe that, as the depth increases, the water temperature decreases. d, depth 500 ft 350 ft 250 ft T = 14,000 _ d 14,000 _ 500 = 28 14,000 _ 350 = 40 14,000 _ 250 = 56 Interpretation At a depth of 500 ft, the water temperature is 28° F. At a depth of 350 ft, the water temperature is 40° F. At a depth of 250 ft, the water temperature is 56° F. Table 2 We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations. For our example, Figure 3 depicts the inverse variation. We say the water temperature varies inversely with the depth k _ of the water because, as the depth increases, the temperature decreases. The formula y = x for inverse variation in this case uses k = 14,000 ° ( 42 35 28 21 14 7 0 (500, 28) (1000, 14) (2000, 7) 1,000 2,000 3,000 4,000 inverse variation If x and y are related by an equation of the form Depth, d (ft) Figure 3 y = k __ xn where k is a nonzero constant, then we say that y varies inversely with the nth power of x. In inversely proportional relationships, or inverse variations, there is a constant multiple k = xny. SECTION 3.9 modeling using variation 313 Example 2 Writing a Formula for an Inversely Proportional Relationship A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives. Solution Recall that multiplying speed by time gives distance. If we let t represent the drive time in hours, and v represent the velocity (speed or rate) at which the tourist drives, then vt = distance. Because the distance is fixed at 100 miles, vt = 100. Solving this relationship for the time gives us our function. t(v) = 100 ___ v = 100v −1 We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction. How To… Given a description of an indirect variation problem, solve for an unknown. 1. Identify the input
, x, and the output, y. 2. Determine the constant of variation. You may need to multiply y by the specified power of x to determine the constant of variation. 3. Use the constant of variation to write an equation for the relationship. 4. Substitute known values into the equation to find the unknown. Example 3 Solving an Inverse Variation Problem A quantity y varies inversely with the cube of x. If y = 25 when x = 2, find y when x is 6. Solution The general formula for inverse variation with a cube is y = y by the cube of x. k = x3 y k __ x3. The constant can be found by multiplying Now we use the constant to write an equation that represents this relationship. = 23 ⋅ 25 = 200 Substitute x = 6 and solve for y. y = k __ x3, k = 200 200 ___ x3 200 ___ 63 25 __ 27 y = y = = Analysis The graph of this equation is a rational function, as shown in Figure 4. y 30 25 20 15 10 5 0 (2, 25) 6, 25 27 x 2 4 6 8 10 Figure 4 Try It #2 A quantity y varies inversely with the square of x. If y = 8 when x = 3, find y when x is 4. 31 4 CHAPTER 3 polynomial and rational Functions Solving Problems Involving joint Variation Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c, cost, varies jointly with the number of students, n, and the distance, d. joint variation Joint variation occurs when a variable varies directly or inversely with multiple variables. For instance, if x varies directly with both y and z, we have x = kyz. If x varies directly with y and inversely with z, we have x =. Notice that we only use one constant in a joint variation equation. ky __ z Example 4 Solving Problems Involving Joint Variation A quantity x varies directly with the square of y and inversely with the cube root of z. If x = 6 when y = 2 and z = 8, find x when y = 1 and z = 27

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