Copyright 2006 Pearson Education Inc Publishing as Pearson
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
4 4. 1 4. 2 4. 3 4. 4 4. 5 4. 6 4. 7 Nonlinear Functions and Equations Nonlinear Functions and Their graphs Polynomial Functions and Models Real Zeros of Polynomial Functions The Fundamental Theorem of Algebra Rational Functions and Models Polynomial and Rational Inequalities Power Functions and Radical Equations Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
4. 1 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 ♦ 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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5
Increasing or Decreasing Functions continued Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 6
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 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 7
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 8
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9
Absolute and Local Extrema Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 10
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. ) Graph f in [1, 12, 1] by [50, 90, 10]. b) Estimate the absolute extrema. Interpret the results. Solution a) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 11
Solution continued Many graphing calculators have the capability to find maximum and minimum y-values. b) • • 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 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 12
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 13
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 14
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 15
4. 2 Polynomial Functions and Models ♦ Understand the graphs of polynomial functions. ♦ Evaluate and graph piecewise-defined functions 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) = anxn + … + a 2 x 2 + a 1 x + a 0, where each coefficient ak is a real number, an 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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 17
Constant Polynomial Function Has no x-intercepts or turning points Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 18
Linear Polynomial Function Degree 1 and one x-intercept and no turning points. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 19
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 20
Cubic Polynomial Functions Degree 3, can have zero or two turning points. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 21
Quartic Polynomial Functions Degree 4, can have up to four x-intercepts and three turning points. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 22
Quintic Polynomial Functions Degree 5, may have up to five x-intercepts and four turning points. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 23
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 24
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 25
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 26
Example Graph f(x) = 2 x 3 5 x 2 5 x + 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) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 27
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 28
Example Let f(x) = 3 x 4 + 5 x 3 2 x 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 3 x 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, Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 29
Piecewise-Defined Polynomial Functions Example Evaluate f(x) at 6, 0, and 4. Solution To evaluate f( 6) we use the formula 5 x 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 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 30
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 31
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 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 32
4. 3 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 ♦ Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example Divide x 3 + 2 x 2 5 x 6 by x 3. Check the result. Solution The quotient is x 2 + 5 x + 10 with a remainder of 24. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 34
Solution continued Check Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 35
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 36
Example Use synthetic division to divide 2 x 3 + 7 x 2 – 5 by x + 3. Solution Let k = – 3 and perform the following. – 3 2 7 0 – 5 – 6 – 3 9 2 1 – 3 4 The remainder is 4 and the quotient is 2 x 2 + x – 3. The result can be expressed as Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 37
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Example Use the graph of f(x) = x 3 x 2 – 9 x + 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, 1 and 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). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 39
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Example Write the complete factorization for the polynomial 6 x 3 + 19 x 2 + 2 x – 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: Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 41
Example The polynomial f(x) = 2 x 3 3 x 2 17 x + 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. 2 2 2 – 3 – 17 30 4 2 – 30 1 – 15 0 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 42
Rational Zeros Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 43
Example Find all rational zeros of f(x) = 6 x 4 + 7 x 3 12 x 2 3 x + 2. Write in complete factored form. Solution Any rational zero must occur in the list Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 44
Solution continued Evaluate f(x) at each value in the list. x 1 1 2 f(x) 0 8 100 x ½ ½ 1/3 f(x) 5/4 0 0 x 1/6 2/3 f(x) 1. 20 2. 14 2. 07 2 0 1/3 1. 48 2/3 2. 22 From the table there are four rational zeros of 1, 2, 1/2, and 1/3. The complete factored form is: Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 45
Example Find all real solutions to the equation 4 x 4 – 17 x 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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 46
Example Solve the equation x 3 – 2. 1 x 2 – 7. 1 x + 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 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 47
4. 4 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 ♦ 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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 49
Example Write each expression in standard form. Support your results using a calculator. a) ( 4 + 2 i) + (6 3 i) b) ( 9 i) (4 7 i) c) ( 2 + 5 i)2 d) Solution a) ( 4 + 2 i) + (6 3 i) = 4 + 6 + 2 i 3 i = 2 i b) ( 9 i) (4 7 i) = 4 9 i + 7 i = 4 2 i Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 50
Solution continued c) ( 2 + 5 i)2 = ( 2 + 5 i) = 4 – 10 i + 25 i 2 = 4 20 i + 25( 1) = 21 20 i d) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 51
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 52
Example Solve the quadratic equation 4 x 2 – 12 x = – 11. Solution Rewrite the equation: 4 x 2 – 12 x + 11 = 0 a = 4, b = – 12, c = 11 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 53
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 54
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 an = 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 4 x 2 + x 4) = 3(x 4 2 x 3 7 x 2 2 x 8) = 3 x 4 6 x 3 – 21 x 2 – 6 x – 24 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 55
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). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 56
Example Find the zeros of f(x) = x 4 + 5 x 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). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 57
Solution continued Long division The solution is x 4 + 5 x 2 + 4 = (x 2 + 4)(x 2 + 1) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 58
Example Solve x 3 = 2 x 2 5 x + 10. Solution Rewrite the equation: f(x) = 0, where f(x) = x 3 2 x 2 + 5 x 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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 59
Solution continued Synthetic division 2 1 – 2 2 1 0 5 – 10 0 10 5 0 x 3 2 x 2 + 5 x 10 = (x 2)(x 2 + 5) The solutions are 2 and Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 60
4. 5 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 variation ♦ Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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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 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 64
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 65
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: 4 x 8 = 0, x = 2 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 66
Solution continued b) Horizontal Asymptote: Degree: numerator < denominator y=0 Vertical Asymptote: x 2 9 = 0 x = 3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 67
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 68
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 69
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) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 70
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) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 71
Example Solve Solution Symbolic Graphical Numerical Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 72
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 73
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 74
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 75
Solution continued b) Let and d = 2. 5. The intensity at 2. 5 meters is 1. 27 watts per square meter. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 76
4. 6 Polynomial and Rational Inequalities Solve polynomial inequalities ♦ Solve rational inequalities ♦ 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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 78
Polynomial Inequalities Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 79
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 80
Solution continued Step 3: The boundary numbers separate the number line into four disjoint intervals: -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 81
Solution continued Step 4: Complete a table of test values. Interval Test Value x x 3 + 7 x 2 + 10 x Positive/Negative – 6 – 24 Negative – 4 8 Positive – 1 – 4 Negative 1 18 Positive The solution set is Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 82
Solution continued Graphically Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 83
Rational Inequalities • Inequalities involving rational expressions are called rational inequalities. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 84
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 ed. ) th Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 85
Example continued Solution Solving this graphically, we find this dose is acceptable when the child is at least 6 years of age. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 86
Example Solve Solution Step 1: Rewrite the inequality in the form Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 87
Solution continued Step 2: Find the zeros of the numerator and the denominator. Numerator 1–x=0 x=1 Denominator x+4=0 x = – 4 Step 3: The boundary numbers are – 4 and 1, which separate the number line into three disjoint intervals: Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 88
Solution continued Step 4: Use a table to solve the inequality. Interval Test Value x (1–x)/(x + 4) Positive/Negative – 5 – 6 Negative – 2 3/2 Positive 2 – 1/6 Negative The interval notation is (– 4, 1]. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 89
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 90
4. 7 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 ♦ 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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 92
Example Simplify each expression by hand. a) 82/3 b) (– 32)– 4/5 Solutions • • Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 93
Example Use positive rational exponents to write each expression. a) b) Solutions a) b) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 94
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) = Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 95
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) = xp/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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 96
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. 2 w 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) b) Approximate S(0. 75) and interpret the result. What weight corresponds to a surface area of 0. 45 square meter? Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 97
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. 2 w 2/3 = 0. 45. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 98
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 99
Equations Involving Rational Exponents Example Solve 4 x 3/2 – 6 = 6. Approximate the answer to the nearest hundredth, and give graphical support. Solutions Symbolic Solution Graphical Solution 4 x 3/2 – 6 = 6 4 x 3/2 = 12 (x 3/2)2 = 32 x 3 = 9 x = 91/3 x = 2. 08 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 100
Equations Having Negative Exponents Example Solve Solution Since Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 101
Equations Involving Radicals When solving equations that contain square roots, it is common to square each side of an equation. Example Solve Solution Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 102
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. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 103
Example Some equations may contain a cube root. Solve Solution Both solutions check, so the solution set is Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 104
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