1. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. 4.1Nonlinear Functions and Their graphs 4.2Polynomial Functions and Models 4.3Real Zeros of Polynomial Functions 4.4The Fundamental Theorem of Algebra 4.5Rational Functions and Models 4.6Polynomial and Rational Inequalities 4.7 Power Functions and Radical Equations Nonlinear Functions and Equations 4

3. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Nonlinear Functions and Their Graphs ♦ Learn terminology about polynomial functions ♦ Identify intervals where a function is increasing or decreasing ♦ Find extrema of a function ♦ Identify symmetry in a graph of a function ♦ Determine if a function is odd, even, or neither 4.1

4. Slide 4- 4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Polynomial Functions Polynomial functions are frequently used to approximate data. A polynomial function of degree 2 or higher is a nonlinear function.

5. Slide 4- 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Increasing or Decreasing Functions The concept of increasing and decreasing relate to whether the graph of a function rises or falls. Moving from left to right along a graph of an increasing function would be uphill. Moving from left to right along a graph of a decreasing function would be downhill. We speak of a function f increasing or decreasing over an interval of its domain.

6. Slide 4- 6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Increasing or Decreasing Functions continued

7. Slide 4- 7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use the graph of shown below and interval notation to identify where f is increasing or decreasing. Solution Moving from left to right on the graph of f, the y-values decreases until x = 0, increases until x = 2, and decreases thereafter. Thus, in interval notation f is decreasing on

8. Slide 4- 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Extrema of Nonlinear Functions Graphs of polynomial functions often have “hills” or “valleys”. The “highest hill” on the graph is located at (–2, 12.7). This is the absolute maximum of g. There is a smaller peak located at the point (3, 2.25). This is called the local maximum.

9. Slide 4- 9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Extrema of Nonlinear Functions continued Maximum and minimum values that are either absolute or local are called extrema. A function may have several local extrema, but at most one absolute maximum and one absolute minimum. It is possible for a function to assume an absolute extremum at two values of x. The absolute maximum is 11. It is a local maximum as well, because near x = –2 and x = 2 it is the largest y-value.

10. Slide 4- 10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Absolute and Local Extrema

11. Slide 4- 11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example The monthly average ocean temperature in degrees Fahrenheit at Bermuda can be modeled by where x = 1 corresponds to January and x = 12 to December. The domain of f is D = {x|1 }. (Source: J. Williams, The Weather Almanac 1995.) a)Graph f in [1, 12, 1] by [50, 90, 10]. b)Estimate the absolute extrema. Interpret the results. Solution

12. Slide 4- 12 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b)Many graphing calculators have the capability to find maximum and minimum y-values. An absolute minimum of about 61.5 corresponds to the point (2.01, 61.5). This means the monthly average ocean temperature is coldest during February, when it reaches An absolute maximum of about 82 corresponds to the point (7.61, 82.0), meaning that the warmest average ocean temperature occurs in August when it reaches a maximum of

13. Slide 4- 13 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Symmetry If a graph was folded along the y-axis, and the right and left sides would match, the graph would be symmetric with respect to the y-axis. A function whose graph satisfies this characteristic is called an even function.

14. Slide 4- 14 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Symmetry continued Another type of of symmetry occurs in respect to the origin. If the graph could rotate, the original graph would reappear after half a turn. This represents an odd function.

15. Slide 4- 15 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Identify whether the function is even or odd. Solution Since f is a polynomial containing only odd powers of x, it is an odd function. This also can be shown symbolically as follows.

16. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Polynomial Functions and Models ♦ Understand the graphs of polynomial functions. ♦ Evaluate and graph piecewise-defined functions 4.2

17. Slide 4- 17 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphs of Polynomial Functions A polynomial function f of degree n can be expressed as f(x) = a n x n + … + a 2 x 2 + a 1 x + a 0, where each coefficient a k is a real number, a n  0, and n is a nonnegative integer. A turning point occurs whenever the graph of a polynomial function changes from increasing to decreasing or from decreasing to increasing. Turning points are associated with “hills” or “valleys” on a graph.

18. Slide 4- 18 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Constant Polynomial Function Has no x-intercepts or turning points

19. Slide 4- 19 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Linear Polynomial Function Degree 1 and one x-intercept and no turning points.

20. Slide 4- 20 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Quadratic Polynomial Functions Degree 2, parabola that opens up or down. Can have zero, one or two x-intercepts. Has exactly one turning point, which is also the vertex.

21. Slide 4- 21 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Cubic Polynomial Functions Degree 3, can have zero or two turning points.

22. Slide 4- 22 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Quartic Polynomial Functions Degree 4, can have up to four x-intercepts and three turning points.

23. Slide 4- 23 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Quintic Polynomial Functions Degree 5, may have up to five x-intercepts and four turning points.

24. Slide 4- 24 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Degree, x-intercepts, and turning points The graph of a polynomial function of degree n  1 has at most n x-intercepts and at most n  1 turning points.

25. Slide 4- 25 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use the graph of the polynomial function shown. a) How many turning points and x-intercepts are there? b) Is the leading coefficient a positive or negative? Is the degree odd or even? c) Determine the minimum degree of f. Solution a) There are three turning points corresponding to the one “hill” and two “valleys”. There appear to be 4 x-intercepts.

26. Slide 4- 26 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b) Is the leading coefficient a positive or negative? Is the degree odd or even? The left side and the right side rise. Therefore, a > 0 and the polynomial function has even degree. c) Determine the minimum degree of f. The graph has three turning points. A polynomial of degree n can have at most n  1 turning points. Therefore, f must be at least degree 4.

27. Slide 4- 27 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Graph f(x) = 2x 3  5x 2  5x + 7, and then complete the following. a) Identify the x-intercepts. b) Approximate the coordinates of any turning points to the nearest hundredth. c) Use the turning points to approximate any local extrema. Solution a) The graph appears to intersect the x-axis at the points (  1.3, 0), (0.89, 0) and (2.9, 0)

28. Slide 4- 28 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b) There are two turning points. From the graphs their coordinates are approximately (  0.40, 8.1) and (2.07,  7.04) c) There is a local maximum of about 8.07 and a local minimum of about  7.04.

29. Slide 4- 29 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Let f(x) = 3x 4 + 5x 3  2x 2. a) Give the degree and leading coefficient. b) State the end behavior of the graph of f. Solution a) The term with the highest degree is 3x 4 so the degree is 4 and the leading coefficient is 3. b) The degree is even and the leading coefficient is positive. Therefore the graph of f rises to the left and right. More formally,

30. Slide 4- 30 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Piecewise-Defined Polynomial Functions Example Evaluate f(x) at  6, 0, and 4. Solution To evaluate f(  6) we use the formula  5x because  6 is <  5. f(  6) =  5(  6) = 30 Similarly, f(0) = x 3 + 1 = (0) 3 + 1 = 1 f(4) = 3  x 2 = 3  (4) 2 =  13

31. Slide 4- 31 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Complete the following. a) Sketch the graph of f. b) Determine if f is continuous on its domain. c) Evaluate f(1). Solution a) Graph as shown to the right.

32. Slide 4- 32 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b) The domain is not continuous since there is a break in the graph. c) f(1) = 4  x 2 = 4 – (1) 2 = 3

33. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Real Zeros of Polynomial Functions ♦ Divide Polynomials ♦ Understand the division algorithm, remainder theorem, and factor theorem ♦ Factor higher degree polynomials ♦ Analyze polynomials with multiple zeros ♦ Find rational zeros ♦ Solve polynomial equations 4.3

34. Slide 4- 34 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Divide x 3 + 2x 2  5x  6 by x  3. Check the result. Solution The quotient is x 2 + 5x + 10 with a remainder of 24.

35. Slide 4- 35 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Check

36. Slide 4- 36 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Synthetic Division A short cut called synthetic division can be used to divide x – k into a polynomial. Steps 1. Write k to the left and the coefficients of f(x) to the right in the top row. If any power does not appear in f(x), include a 0 for that term. 2. Copy the leading coefficient of f(x) into the third row and multiply it by k. Write the result below the next coefficient of f(x) in the second row. Add the numbers in the second column and place the result in the third row. Repeat the process. 3. The last number in the third row is the remainder. If the remainder is 0, then the binomial x – k is a factor of f(x). The other numbers in the third row are the coefficients of the quotient in descending powers.

37. Slide 4- 37 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use synthetic division to divide 2x 3 + 7x 2 – 5 by x + 3. Solution Let k = –3 and perform the following. The remainder is 4 and the quotient is 2x 2 + x – 3. The result can be expressed as –3270–5 –6–39 21 4

38. Slide 4- 38 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

39. Slide 4- 39 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use the graph of f(x) = x 3  x 2 – 9x + 9 and the factor theorem to list the factors of f(x). Solution The graph shows that the zeros or x-intercepts of f are  3, 1and 3. Since f(  3) = 0, the factor theorem states that(x + 3) is a factor, and f(1) = 0 implies that (x  1) is a factor and f(3) = 0 implies (x  3) is a factor. Thus the factors are (x + 3)(x  1), and (x  3).

40. Slide 4- 40 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

41. Slide 4- 41 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write the complete factorization for the polynomial 6x 3 + 19x 2 + 2x – 3 with given zeros –3, –1/2 and 1/3. Solution Leading coefficient is 6 Zeros are –3, –1/2 and 1/3 The complete factorization:

42. Slide 4- 42 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example The polynomial f(x) = 2x 3  3x 2  17x + 30 has a zero of 2. Express f(x) in complete factored form. Solution If 2 is a zero, by the factor theorem x  2 is a factor. Use synthetic division. 22– 3–1730 42–30 21–150

43. Slide 4- 43 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rational Zeros

44. Slide 4- 44 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find all rational zeros of f(x) = 6x 4 + 7x 3  12x 2  3x + 2. Write in complete factored form. Solution Any rational zero must occur in the list

45. Slide 4- 45 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Evaluate f(x) at each value in the list. From the table there are four rational zeros of 1,  2,  1/2, and 1/3. The complete factored form is: xf(x)f(x)xf(x)f(x)xf(x)f(x) 10½  5/4 1/61.20 11 88 ½½ 0  1/6 2.14 21001/302/3  2.07 22 0  1/3 1.48  2/3  2.22

46. Slide 4- 46 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find all real solutions to the equation 4x 4 – 17x 2 – 50 = 0. Solution The expression can be factored similar to a quadratic equation. The only solutions are since the equation x 2 = –2 has no real solutions.

47. Slide 4- 47 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the equation x 3 – 2.1x 2 – 7.1x + 0.9 = 0 graphically. Round any solutions to the nearest hundredth. Solution Since there are three x-intercepts the equation has three real solutions. x .012,  1.89, and 3.87

48. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Fundamental Theorem of Algebra ♦ Perform arithmetic operations on complex numbers ♦ Solve quadratic equations having complex solutions ♦ Apply the fundamental theorem of algebra ♦ Factor polynomials having complex zeros ♦ Solve polynomial equations having complex solutions 4.4

49. Slide 4- 49 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Complex Numbers A complex number can be written in standard form as a + bi, where a and b are real numbers. The real part is a and the imaginary part is b.

50. Slide 4- 50 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Write each expression in standard form. Support your results using a calculator. a) (  4 + 2i) + (6  3i)b) (  9i)  (4  7i) c) (  2 + 5i) 2 d) Solution a) (  4 + 2i) + (6  3i) =  4 + 6 + 2i  3i = 2  i b) (  9i)  (4  7i) =  4  9i + 7i =  4  2i

51. Slide 4- 51 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued c) (  2 + 5i) 2 = (  2 + 5i)(  2 + 5i) = 4 – 10i – 10i + 25i 2 = 4  20i + 25(  1) =  21  20i d)

52. Slide 4- 52 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Quadratic Equations with Complex Solutions We can use the quadratic formula to solve quadratic equations if the discriminant is negative. There are no real solutions, and the graph does not intersect the x-axis. The solutions can be expressed as imaginary numbers.

53. Slide 4- 53 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve the quadratic equation 4x 2 – 12x = –11. Solution Rewrite the equation: 4x 2 – 12x + 11 = 0 a = 4, b = –12, c = 11

54. Slide 4- 54 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Fundamental Theorem of Algebra The polynomial f(x) of degree n  1 has at least one complex zero. Number of Zeros Theorem A polynomial of degree n has at most n distinct zeros.

55. Slide 4- 55 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Represent a polynomial of degree 4 with leading coefficient 3 and zeros of  2, 4, i and  i in complete factored form and expanded form. Solution Let a n = 3, c 1 =  2, c 2 = 4, c 3 = i, and c 4 =  i. f(x) = 3(x + 2)(x  4)(x  i)(x + i) Expanded: 3(x + 2)(x  4)(x  i)(x + i) = 3(x + 2)(x  4)(x 2 + 1) = 3(x + 2)(x 3  4x 2 + x  4) = 3(x 4  2x 3  7x 2  2x  8) = 3x 4  6x 3 – 21x 2 – 6x – 24

56. Slide 4- 56 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Conjugate Zeros Theorem If a polynomial f(x) has only real coefficients and if a + bi is a zero of f(x), then the conjugate a  bi is also a zero of f(x).

57. Slide 4- 57 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Find the zeros of f(x) = x 4 + 5x 2 + 4 given one zero is  i. Solution By the conjugate zeros theorem it follows that i must be a zero of f(x). (x + i) and (x  i) are factors (x + i)(x  i) = x 2 + 1, using long division we can find another quadratic factor of f(x).

58. Slide 4- 58 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Long division The solution is x 4 + 5x 2 + 4 = (x 2 + 4)(x 2 + 1)

59. Slide 4- 59 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve x 3 = 2x 2  5x + 10. Solution Rewrite the equation: f(x) = 0, where f(x) = x 3  2x 2 + 5x  10 We can use factoring by grouping or graphing to find one real zero. The graph shows a zero at 2. So, x  2 is a factor. Use synthetic division.

60. Slide 4- 60 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Synthetic division x 3  2x 2 + 5x  10 = (x  2)(x 2 + 5) The solutions are 2 and 21–25–10 2010 1050

61. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rational Functions and Models ♦ Identify a rational function and state its domain ♦ Find and interpret vertical asymptotes ♦ Find and interpret horizontal asymptotes ♦ Solve rational equations ♦ Solve applications involving rational equations ♦ Solve applications involving variation 4.5

62. Slide 4- 62 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

63. Slide 4- 63 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

64. Slide 4- 64 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use the graph of to sketch the graph of Include all asymptotes in your graph. Write g(x) in terms of f(x). Solution g(x) is a translation of f(x) left one unit and down 2 units. The vertical asymptote is x =  1 The horizontal asymptote is y =  2 g(x) = f(x + 1)  2

65. Slide 4- 65 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

66. Slide 4- 66 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example For each rational function, determine any horizontal or vertical asymptotes. a)b)c) Solution a) Horizontal Asymptote: Degree of numerator equals the degree of the denominator. y = a/b is asymptote, so y = 2/4 = 1/2 Vertical Asymptote: 4x  8 = 0, x = 2

67. Slide 4- 67 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b) Horizontal Asymptote: Degree: numerator < denominator y = 0 Vertical Asymptote: x 2  9 = 0 x =  3

68. Slide 4- 68 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued c) Horizontal Asymptote: Degree: numerator > denominator no horizontal asymptotes Vertical Asymptote: no vertical asymptotes The graph is the line y = x + 2 with the point (2, 4) missing.

69. Slide 4- 69 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slant or Oblique Asymptote A third type of asymptote is neither horizontal or vertical. Occurs when the numerator of a rational function has a degree one more than the degree of the denominator.

70. Slide 4- 70 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Let a) Use a calculator to graph f. b) Identify any asymptotes. c) Sketch a graph of f that includes the asymptotes. Solution a)

71. Slide 4- 71 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b) Asymptotes: The function is undefined when x  2 = 0 or when x = 2. Vertical asymptote at x = 2 Oblique asymptote at y = x + 2 c)

72. Slide 4- 72 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve Solution SymbolicGraphicalNumerical

73. Slide 4- 73 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve Solution Multiply by the LCD to clear the fractions. When 1 is substituted for x, two expressions in the given equation are undefined. There are no solutions.

74. Slide 4- 74 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

75. Slide 4- 75 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example At a distance of 3 meters, a 100-watt bulb produces an intensity of 0.88 watt per square meter. a) Find the constant of proportionality k. b) Determine the intensity at a distance of 2.5 meters. Solution a) Substitute d = 3 and I = 0.88 into the equation and solve for k.

76. Slide 4- 76 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b) Let and d = 2.5. The intensity at 2.5 meters is 1.27 watts per square meter.

77. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Polynomial and Rational Inequalities ♦ Solve polynomial inequalities ♦ Solve rational inequalities 4.6

78. Slide 4- 78 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Introduction An inequality says that one expression is greater than, greater than or equal to, less than, or less than or equal to, another expression. Solving Inequalities Boundary numbers (x-values) are found where the inequality holds. A graph or a table of test values can be used to determine the intervals where the inequality holds.

79. Slide 4- 79 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Polynomial Inequalities

80. Slide 4- 80 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve symbolically and graphically. Solution Symbolically Step 1: Write the inequality as Step 2: Replace the inequality symbol with an equal sign and solve. The boundary numbers are –5, –2, and 0.

81. Slide 4- 81 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Step 3: The boundary numbers separate the number line into four disjoint intervals: 5-7-5-3135-8-404-62-2-8

82. Slide 4- 82 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Step 4: Complete a table of test values. The solution set is IntervalTest Value xx 3 + 7x 2 + 10xPositive/Negative –6–24Negative –48Positive –1–4Negative 118Positive

83. Slide 4- 83 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Graphically

84. Slide 4- 84 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rational Inequalities Inequalities involving rational expressions are called rational inequalities.

85. Slide 4- 85 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Health Care: There is a formula for determining the amount of a child's particular medication dosage based on an adults dosage. Solve the inequality to determine the age of the child to receive the dose of 8 mg when an adult receives 24 mg. (Source: Olsen, Medical Dosage Calculations, 6 th ed.)

86. Slide 4- 86 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Solution Solving this graphically, we find this dose is acceptable when the child is at least 6 years of age.

87. Slide 4- 87 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solve Solution Step 1: Rewrite the inequality in the form

88. Slide 4- 88 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Step 2: Find the zeros of the numerator and the denominator. Step 3: The boundary numbers are – 4 and 1, which separate the number line into three disjoint intervals: NumeratorDenominator 1 – x = 0 x = 1 x + 4 = 0 x = –4

89. Slide 4- 89 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Step 4: Use a table to solve the inequality. The interval notation is (–4, 1]. IntervalTest Value x(1–x)/(x + 4)Positive/Negative –5–6Negative –23/2Positive 2–1/6Negative

90. Slide 4- 90 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rational Inequality Caution: When solving a rational inequality, it is essential not to multiply or divide each side of the inequality by the LCD if the LCD contains a variable. This techniques often leads to an incorrect solution set.

91. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Power Functions and Radical Equations ♦ Learn properties of rational exponents ♦ Learn radical notation ♦ Understand properties and graphs of power functions ♦ Use power functions to model data ♦ Solve equations involving rational exponents ♦ Solve equations involving radical expressions 4.7

92. Slide 4- 92 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Rational Exponents and Radical Notation Expressions with rational exponents can be simplified with the following properties.

93. Slide 4- 93 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Simplify each expression by hand. a)8 2/3 b) (–32) –4/5 Solutions

94. Slide 4- 94 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Use positive rational exponents to write each expression. a) b) Solutions a) b)

95. Slide 4- 95 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Power Functions and Models Power functions typically have rational exponents. A special type of power function is a root function. Examples of power functions include: f 1 (x) = x 2, f 2 (x) = x 3/4, f 3 (x) = x 0.4, and f 4 (x) =

96. Slide 4- 96 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Often, the domain of a power function f is restricted to nonnegative numbers. Suppose the rational number p/q is written in lowest terms. The the domain of f(x) = x p/q is all real numbers whenever q is odd and all nonnegative numbers whenever q is even. The following graphs show 3 common power functions.

97. Slide 4- 97 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Modeling Wing Size of a Bird: Heavier birds have larger wings with more surface areas than do lighter birds. For some species the relationship can be modeled by S(w) = 0.2w 2/3, where w is the weight of the bird in kilograms and S is surface area of the wings in square meters. (Source: C. Pennycuick, Newton Rules Biology.) a)Approximate S(0.75) and interpret the result. b)What weight corresponds to a surface area of 0.45 square meter?

98. Slide 4- 98 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example continued Solution a)S(0.75) = 0.2(0.75) 2/3 The wings of a bird that weighs about 0.75 kilogram have the surface area of about 0.165 square meter. b)To answer this, we must solve the equation 0.2w 2/3 = 0.45.

99. Slide 4- 99 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Since w must be positive, the wings of a 3.4 kilogram bird must have a surface area of about 0.45 square meter.

100. Slide 4- 100 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations Involving Rational Exponents Example Solve 4x 3/2 – 6 = 6. Approximate the answer to the nearest hundredth, and give graphical support. Solutions Symbolic SolutionGraphical Solution 4x 3/2 – 6 = 6 4x 3/2 = 12 (x 3/2 ) 2 = 3 2 x 3 = 9 x = 9 1/3 x = 2.08

101. Slide 4- 101 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations Having Negative Exponents Example Solve Solution Since

102. Slide 4- 102 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations Involving Radicals When solving equations that contain square roots, it is common to square each side of an equation. Example Solve Solution

103. Slide 4- 103 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued Check Substituting these values in the original equation shows that the value of 1 is an extraneous solution because it does not satisfy the given equation. Therefore, the only solution is 6.

104. Slide 4- 104 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Some equations may contain a cube root. Solve Solution Both solutions check, so the solution set is