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β 4x + 2) Γ· (2x β 1) 15. (2x3 β 6x2 β 7x + 6) Γ· (x β 4) 17. (4x3 β 12x2 β 5x β 1) Γ· (2x + 1) 19. (3x3 β 2x2 + x β 4) Γ· (x + 3) 21. (2x3 + 7x2 β 13x β 3) Γ· (2x β 3) 23. (4x3 β 5x2 + 13) Γ· (x + 4) 25. (x3 β 21x2 + 147x β 343) Γ· (x β 7) 27. (9x3 β x + 2) Γ· (3x β 1) 29. (x4 + x3 β 3x2 β 2x + 1) Γ· (x + 1) 31. (x4 + 2x3 β 3x2 + 2x + 6) Γ· (x + 3) 33. (x4 β 8x3 + 24x2 β 32x + 16) Γ· (x β 2) 35. (x4 β 12x3 + 54x2 β 108x + 81) Γ· (x β 3) 37. (4x4 + 2x3 β 4x2 + 2x + 2) Γ· (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 β 2, 4x3 β 3x2 β 8x + 4 41. x β 2, 4x4 β 15x2 β 4 39. x β 2, 3x4 β 6x3 β 5x + 10 42. x β 1 __, 2x4 β x3 + 2x β 1 2 40. x + 3, β4x3 + 5x2 + 8 43. x + 1 __ 3, 3x4 + x3 β 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 β 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 β6 β4 β2 2 4 6 x β6 β4 β2 2 4 6 x β6 β4 β2 2 4 6 x β6 β12 β18 β6 β12 β18 β10 β20 β30 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.4 SECTION EXERCISES 401 47. Factor is x2 + x + 1 48. Factor is x2 + 2x + 2 y 60 40 20 y 18 12 6 β6 β4 β2 2 4 6 x β6 β4 β2 2 4 6 x β20 β40 β60 β 6 β12 β18 For the following exercises, use synthetic division to find the quotient and remainder. 49. 4x3 β 33 _______ x β 2 51. 3x3 + 2x β 5 __________ x β 1 50. 2x3 + 25 _______ x + 3 53. x4 β 22 ______ x + 2 52. #β4x3 β x2 β 12 ____________ x + 4 TECHNOLOGY 54. Consider xk β 1 ______ x β 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 β k4 ______ x β 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 β 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 β i 62. x2 + 1 _____ x + i 60. x2 + 1 _____ x β i 63. x3 + 1 _____ x β 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 β 5. 66. Length is 3x β 4, area is 6x4 β 8x3 + 9x2 β 9x β 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 β 21x β 36, length is 2x + 3, width is 3x β 4. 68. Volume is 18x3 β 21x2 β 40x + 48, length is 3x β 4, width is 3x β 4. 69. Volume is 10x3 + 27x2 + 2x β 24, length is 5x β 4, 70. Volume is 10x3 + 30x2 β 8x β 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 Ο(25x3 β 65x2 β 29x β 3), radius is 5x + 1. 73. Volume is Ο(3x4 + 24x3 + 46x2 β 16x β 32), radius is x + 4. 72. Volume is Ο(4x3 + 12x2 β 15x β 50), radius is 2x + 5. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS LEARNING OBJECTIVES In this section, you will: f β’ Use synthetic diision the emainder heorem to ealuate. β’ Use the Factor heorem to find the real eros of a polynomial equation. β’ Use the ational ero heorem to list all possible rational eros. β’ Find eros of a polynomial function. β’ Use the inear Factoriation heorem to find polynomials with gien eros. 5.5 ZEROS OF POLYNOMIAL FUNCTIONS ffTh ThTh Th Evaluating a Polynomial Using the Remainder Theorem Remainder Theoremx βk kf |
k f xdx dxf xqxrx dxx βk f x=dxqx+rx fx=xβkqx+r x βkrx =k f k=kβkqk+r =Δqk+r =r f kf xx βk the Remainder Theorem f xx βkf k How Toβ¦ ff xx =kTh 1. x βk 2. Thf k Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 ZEROS OF POLYNOMIAL FUNCTIONS 403 Example 1 Using the Remainder Theorem to Evaluate a Polynomial Use the Remainder Theorem to evaluate f (x) = 6x4 β x3 β 15x2 + 2x β 7 at x = 2. Solution To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by x β 2. 2 6 β1 β15 2 β7 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 β x3 β#15x2 + 2x β 7 f (2) = 6(2)4 β (2)3 β 15(2)2 + 2(2) β 7 = 25 Try It #1 Use the Remainder Theorem to evaluate f (x) = 2x5 β 3x4 β 9x3 + 8x2 + 2 at x = β3. 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. f (x) = (x β k)q(x) + r If k is a zero, then the remainder r is f (k) = 0 and f (x) = (x β k)q(x) + 0 or f (x) = (x β k)q(x). Notice, written in this form, x β k is a factor of f (x). We can conclude if k is a zero of f (x), then x β k is a factor of f (x). Similarly, if x β k is |
a factor of f (x), then the remainder of the Division Algorithm f (x) = (x β 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 β k) is a factor of f (x). How Toβ¦ 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 β k). 2. Confirm that the remainder is 0. 3. Write the polynomial as the product of (x β 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 β 6x2 β x + 30. Find the remaining factors. Use the factors to determine the zeros of the polynomial. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Solution We can use synthetic division to show that (x + 2) is a factor of the polynomial. β2 1 β6 β1 30 β2 16 β30 1 β8 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 β 8x + 15) We can factor the quadratic factor to write the polynomial as By the Factor Theorem, the zeros of x3 β 6x2 β x + 30 are β2, 3, and 5. (x + 2)(x β 3)(x |
β 5) Try It #2 Use the Factor Theorem to find the zeros of f (x) = x3 + 4x2 β 4x β 16 given that (x β 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 β = 0 = 0 or x β 4 5 Set each factor equal to 0. 5x β 2 = 0 or 4x β 3 = 0 Multiply both sides of the equation to eliminate fractions. f (x) = (5x β 2)(4x β 3) Create the quadratic function, multiplying the factors. f (x) = 20x2 β 23x + 6 Expand the polynomial. f (x) = (5 Δ 4)x2 β 23x + (2 Δ 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 β 1 xn β 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 ZEROS OF POLYNOMIAL FUNCTIONS 405 How Toβ¦ 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 β 5x3 + x2 β 4. Solution The only possible rational zeros of f (x) are the quotients of the factors of the last term, β4, and the factors of the leading coefficient, 2. The constant term is β4; the factors of β4 are p = Β±1, Β±2, Β±4. The leading coefficient is 2; the factors of 2 are q = Β±1, Β±2. If any of the four real zeros are rational zeros, then they will be of one of the following factors of β4 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 = Β±1, Β±2, Β±4, Β± #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 β 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 Β±1 and Β± #1 _ 2. These are the _ The factors of 1 are Β±1 and the factors of 2 are Β±1 and |
Β±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 (β1) = 2(β1)3 + (β1)2 β 4(β1) + 1 = 4 3 f (1) = 2(1)3 + (1)2 β 4(11 + ( β #1 ) = 2 ( β #1 f ( β #1 __ __ __ __ ) + 1_ __ __ __ __ β Of those, β1, β#1__ 2, and #1 __ 2 are not zeros of f (x). 1 is the only rational zero of f (x). Try It #3 Use the Rational Zero Theorem to find the rational zeros of f (x) = x3 β 5x2 + 2x + 1. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 6 CHAPTER 5 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 Toβ¦ 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 β 3x β 1. p _ q is a zero of f (x), then p is a factor of β1 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 Β± #1 q are Β±1, Β± #1 _ __ __ The factors of β1 are Β±1 and the factors of 4 are Β±1, Β±2, and Β±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. Letβs begin with 1. = factor of β1__ factor of 4 1 4 4 0 β3 β1 1 4 4 0 1 4 Dividing by (x β 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 β 1)(2x + 1)2. (x β 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 = β #1 __ 2 The zeros of the function are 1 and β #1 __ with multiplicity 2. 2 Analysis Look at the graph of the function f in Figure 1. Notice, at x = β0.5, the graph bounces off the x-axis, indicating the even multiplicity (2, 4, 6β¦) for the zero β0.5. At x = 1, the graph crosses the x-axis, indicating the odd multiplicity (1, 3, 5β¦) for the zero x = 1. y Bounce 1.5 1 0.5 β2.5 β2 β1.5 0.5 1 1.5 2 2.5 x Cross β1 β0.5 β0.5 β1 β1.5 β2 β2.5 Figure 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 ZEROS OF POLYNOMIAL FUNCTIONS 407 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 β c1 and a polynomial quotient. Since x β 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 β 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 β c1)(x β c2)...(x β cn) Q & Aβ¦ Does every polynomial have at least one imaginary zero? No. Real numbers are a subset of complex numbers, but not the other way around. 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. factor of constant term ___ factor of leading coefficient factor of 3_ factor of 3 p _ q = = p _ The factors of 3 are Β±1 and Β±3. The possible values for q |
, and therefore the possible rational zeros for the function, are Β±3, Β±1, and Β± #1 __. We will use synthetic division to evaluate each possible zero until we find one that gives a remainder 3 of 0. Letβs begin with β3. β3 3 3 9 β9 0 1 3 0 β3 0 1 Dividing by (x + 3) gives a remainder of 0, so β3 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. 3x2 + 1 = 0 (x + 3)(3x2 + 1) 3 The zeros of f (x) are β3 and Β± #i β _. 3 β x2 = β #1 __ 3 x = Β± β ____ β #1 __ = Β± 3 β 3 i β _ 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Analysis Look at the graph of the function f in Figure 2. Notice that, at x = β3, the graph crosses the x-axis, indicating an odd multiplicity (1) for the zero x = β3. 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 = β3 is 1 and there are two complex solutions, which is what we found, or the multiplicity at x = β3 is three. Either way, our result is correct. y 18 12 6 Cross β6 β4 β2 2 4 6 x β6 β12 β18 Figure 2 Try It #4 Find the zeros of f (x) = 2x3 + 5x2 β 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 β c), where c is a complex number. Let f be a polynomial function with real coefficients, and suppose a + bi, b β 0, is a zero of f (x). Then, by the Factor Theorem, x β (a + bi) is a factor of f (x). For f to have real coefficients, x β (a β bi) must also be a factor of f (x). This is true because any factor other than x β (a β bi), when multiplied by x β (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 β 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 β 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 β bi, is also a zero. How Toβ¦ 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 ZEROS OF POLYNOMIAL FUNCTIONS 409 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 β |
3, 2, i, such that f (β2) = 100. Solution Because x = i is a zero, by the Complex Conjugate Theorem x = βi is also a zero. The polynomial must have factors of (x + 3), (x β 2), (x β i), and (x + i). Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Letβs begin by multiplying these factors. f (x) = a(x + 3)(x β 2)(x β i)(x + i) f (x) = a(x2 + x β 6)(x2 + 1) f (x) = a(x4 + x3 β 5x2 + x β 6) We need to find a to ensure f (β2) = 100. Substitute x = β2 and f (2) = 100 into f (x). 100 = a((β2)4 + (β2)3 β 5(β2)2 + (β2) β 6) So the polynomial function is or 100 = a(β20) β5 = a f (x) = β5(x4 + x3 β 5x2 + x β 6) f (x) = β5x4 β 5x3 + 25x2 β 5x + 30 Analysis We found that both i and βi were zeros, but only one of these zeros needed to be given. If i is a zero of a polynomial with real coefficients, then βi must also be a zero of the polynomial because βi is the complex conjugate of i. Q & Aβ¦ If 2 + 3i were given as a zero of a polynomial with real coefficients, would 2 β 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 β2i such that f (1) = 10. Using Descartesβ 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, Descartesβ 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 (βx) and the number of negative real zeros. f (βx) = (βx)4 + (βx)3 + (βx)2 + (βx) β 1 f (βx) =+x 4 β#x 3 + x 2 β#x β 1 In this case, f (βx) has 3 sign changes. This tells us that f (x) could have 3 or 1 negative real zeros. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 41 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Descartesβ Rule of Signs According to Descartesβ Rule of Signs, if we let f (x) = anxn + an β 1 xn β 1 +... + a1 x + a0 be a polynomial function with real coefficients: β’ 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. β’ The number of negative real zeros is either equal to the number of sign changes of f (βx) or is less than the number of sign changes by an even integer. Example 8 Using Descartesβ Rule of Signs Use Descartesβ Rule of Signs to determine the possible numbers of positive and negative real zeros for f (x) = βx4 β 3x3 + 6x2 β 4x β 12. Solution Begin by determining the number of sign changes. f (x) = βx 4 β 3x 3 + 6x 2 β 4x β 12 Figure 3 There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine f (βx) to determine the number of negative real roots. f (βx) = β(βx)4 β 3(βx)3 + 6(βx)2 β 4(βx) β 12 f (βx) = βx 4 + 3x 3 + 6x 2 + 4x β 12 f (βx) = βx 4 +#3 |
x 3 + 6x 2 + 4x β 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 β1 β10 β20 β30 x = β4.42 β5 β4 β3 β2 f (x) = β x4 β 3x3 + 6x2 β 4x β 12 x = β1 321 4 5 x Try It #6 Use Descartesβ Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for f (x) = 2x4 β 10x3 + 11x2 β 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. Letβs use these tools to solve the bakery problem from the beginning of the section. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.5 ZEROS OF POLYNOMIAL FUNCTIONS 411 Example 9 Solving Polynomial Equations A new bakery offers decorated sheet cakes for childrenβs 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. Letβs 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 β 1053 Subtract 1053 from both sides. Descartesβ rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are Β±3, Β±9, Β±13, Β±27, Β±39, Β±81, Β±117, Β±351, and Β±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. Letβs 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 β1048 0 β1053 21 63 21 β990 4 9 13 0 β1053 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. β’ Real Zeros, Factors, and Graphs of Polynomial Functions (http://openstaxcollege.org/l/realzeros) β’ Complex Factorization Theorem (http://openstaxcollege.org/ |
+ xβx+ xβ 42. fx= x+ x+ x+ 43. fx= xβxβx+ 44. fx= xβxβx+ x+ fx= xβxβx+ x+ 45. 46. fx= xβx+ TECHNOLOGY fi 47. fx= xβx+ 48. fx= xβxβxβ 49. fx= xβxβx+ 50. fx= x+ x+ xβx+ 51. fx= xβx+ xβx+ EXTENSIONS 52. βf= 54. 56. β βfβ= β βββfβ= β 53. 55. β f= β " βfβ= β REAL-WORLD APPLICATIONS For the following exercises, find the dimensions of the box described. 57. 59. 61. The length is twice as long as the width. The height is 2 inches greater than the width. The volume is 192 cubic inches. The length is one inch more than the width, which is one inch more than the height. The volume is 86.625 cubic inches. The length is 3 inches more than the width. The width is 2 inches more than the height. The volume is 120 cubic inches. 58. The length, width, and height are consecutive whole numbers. The volume is 120 cubic inches. 60. The length is three times the height and the height is one inch less than the width. The volume is 108 cubic inches. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 41 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS LEARNING OBJECTIVES In this section, you will: β’ Find the domains of rational functions. β’ dentify ertical asymptotes and horiontal asymptotes. β’ Graph rational functions using asymptotes intercepts and appropriate points. β’ Find xand yintercepts. β’ rite a rational function that meets specific criteria. 5.6 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 β 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 β 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 β1 β1 β2 β3 β4 β5 β5 β4 β3 β2 f (x) = 1 x 321 4 5 x β5 β4 β3 β2 Figure 1 y 5 4 3 2 1 β1 β1 β2 β3 β4 β5 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 β ββ. 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 β β. To summarize, we use arrow notation to show that x or f (x) is approaching a particular value. See Table 1. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 415 Symbol x β aβ x β a+ x β β x β ββ f (x) β β f (x) β ββ f (x) β 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 Letβs 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 β0.1 β0.01 β0.001 β0.0001 β10 β100 β1000 β10,000 We write in arrow notation Table 2 as x β 0β, f (x) β ββ 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 β 0+, f (x) β β. As x β 0 + f (x) β β y As x β ββ f (x) β 0 β5 β4 β3 β2 5 4 3 2 1 β1 β1 β2 β3 β4 β5 321 4 5 As x β β f (x) β 0 x As x β 0 β f (x) β ββ Figure 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 41 6 CHAPTER 5 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 35 β4 β3 β2 321 x = 0 β1 β1 β2 β3 β4 β5 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 β a, f (x) β β, or as x β a, f (x) β ββ. 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 β β, f (x) β 0, and as x β ββ, f (x) β 0. y β2 β3 β4 β5 As x β ββ f (x) β 0 As x β 0 + f (x) β β As x β β f (x) β 0 321 4 5 x As x β 0 β f (x) β ββ 5 4 3 2 1 β1 β1 β2 β3 β4 β5 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 β2 β3 β4 β5 y 5 4 3 2 1 321 4 5 x x = 0 β1 β1 β2 β3 β4 β5 Figure 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 417 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 β β or x β ββ, f (x) β 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 β1 β2 β4 β6 β8 β10 β12 β6 β5 β4 β3 β2 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 β 2β, f (x) β ββ, and as x β 2+, f ( |
x) β β. 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 β β, f (x) β 4 and as x β ββ, f (x) β 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 = β7 β6 β5 β4 β3 β2 β1 β1 β2 β3 y = 3 321 4 5 6 7 x Figure 7 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 41 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Notice that this function is undefined at x = β2, and the graph also is showing a vertical asymptote at x = β2. As x β β2β, f (x) β ββ, and as x β β2+, f (x) β β. 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 β Β±β, f (x) β 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 β 1 x p β 1 +... + a1 x + a0 ___ bq x q + bq β 1 x q β 1 +... + b1 x + b0, Q(x) β 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 β 0.08 > = 0.05, the concentration is greater after 12 minutes than at the beginning. 17 ___ 220 1 __ 20 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 419 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 Toβ¦ 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 β 9. Solution Begin by setting the denominator equal to zero and solving. x2 β 9 = 0 x2 = 9 x = Β±3 The denominator is equal to zero when x = Β±3. The domain of the function is all real numbers except x = Β±3. Analysis A graph of this function, as shown in Figure 8, confirms that the function is not defined when x = Β±3. y = 0 β2 β3 β4 β6 β5 y 4 3 2 1 β1 β1 β2 β3 β4 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 = β3. We will discuss these types of holes in greater detail later in this section. Try It #4 Find the domain of f (x) = 4x _____________. 5(x β 1)(x β 5) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 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 Toβ¦ 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 or "holes." Example 5 Identifying Vertical Asymptotes Find the vertical asymptotes of the graph of k(x) = 5 + 2x2 _________ 2 β x β x2. Solution First, factor the numerator and denominator. k(x) = 5 + 2x2 ________ 2 β x β x2 = 5 + 2x2 ___________ (2 + x)(1 β x) To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero: Neither x = β2 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 β x) = 0 x = β2, 1 x = β 21 3 4 5 6 x β5 β4 β3 β2 β6 y = β2 β1 β1 β2 β3 β4 β5 β6 β7 Figure 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 421 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 β 1 __________ x2 β 2x β 3 f (x) = #(x + 1)(x β 1) ____________ (x + 1)(x β 3) Notice that x + 1 is a common factor to the numerator and the denominator. The zero of this factor, x = β1, is the location of the removable discontinuity. Notice also that x β 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. [Note that removable discontinuities may not be visible when we use a graphing calculator, depending upon the window selected.] Removable discontinuity at x = β1 β3 y 6 4 2 β1 β2 β4 β6 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 β 2 _____. x2 β 4 Solution Factor the numerator and the denominator. k(x) = x β 2 ___________ (x β 2)(x + 2) Notice that there is a common factor in the numerator and the denominator, x β 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 = β2. The vertical asymptote is x = β2. See Figure 11. Download the OpenStax text for free at http |
://cnx.org/content/col11759/latest. 42 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS y x = β2 6 4 2 β8 β6 β4 β2 2 4 x β2 β4 β6 Figure 11 The graph of this function will have the vertical asymptote at x = β2, 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 β 25 ___________ x3 β 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 polynomialβs end behavior will mirror that of the leading term. Likewise, a rational functionβs end behavior will mirror that of the ratio of the function that is the ratio of the leading term. 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 β 5 4x 4 _ _ In this case, the end behavior is f (x) β 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 β8 β6 β4 β2 2 4 x β2 β4 β6 x = 1 x = β5 Figure 12 Horizontal asymptote y = 0 when f(x ) =, q(x ) β 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 β 2x + 1 __________ x β |
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) β 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 423 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 β 2x + 1 ___________ x β 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 β2 β4 β6 β4 β2 Figure 13 Slant asymptote when f( x ) = x = 1, q( x ) β 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) β 0. p(x) ____ q(x) Example: f (x) = 3x2 + 2 _________ x2 + 4x β 5 3x2 _ x2 = 3. This tells us that as the inputs grow large, this function will behave like In this case, the end behavior is f (x) β the function g(x) = 3, which is a horizontal line. As x β Β±β, f (x) β 3, resulting in a horizontal asymptote at y = 3. See Figure 14. Note that this graph crosses the horizontal asymptote. y β15 β12 β9 |
β6 β3 12 9 6 3 β3 β6 β5 Figure 14 Horizontal asymptote when f ( x ) =# # p( x ) _ q( x ), q( x ) β 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) = f (x) β 3x5 β x2 _______ x + 3 3x5 ___ x = 3x4, the end behavior of the graph would look similar to that of an even polynomial with a positive leading coefficient. x β Β±β, f (x) β β 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. β’ Degree of numerator is less than degree of denominator: horizontal asymptote at y = 0. β’ Degree of numerator is greater than degree of denominator by one: no horizontal asymptote; slant asymptote. β’ Degree of numerator is equal to degree of denominator: horizontal asymptote at ratio of leading coefficients. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Example 7 Identifying Horizontal and Slant Asymptotes For the functions listed, identify the horizontal or slant asymptote. a. g(x) = 6x3 β 10x ________ 2x3 + 5x2 b. h(x) = c. k(x) = x2 + 4x ______ x3 β 8 Solution For these solutions, we will use f (x) =, q(x) β 0. x2 β 4x + 1 _________ x |
+ 2 p(x) ____ q(x) a. g(x) = 6x3 β 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 β 4x + 1 _ x + 2 at x2 β 4x + 1_ x + 2. 2 1 β4 β2 1 β6 1 12 13 The quotient is x β 2 and the remainder is 13. There is a slant asymptote at y = x β 2. c. k(x) = : The degree of p = 2 < degree of q = 3, so there is a horizontal asymptote y = 0. x2 + 4x _ x3 β 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 β β, C(t) β 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 β 2)(x + 3) _________________ (x β 1)(x + |
2)(x β 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, β2, 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 β Β±β, f (x) β 0. This function will have a horizontal asymptote at y = 0. See Figure 15. y 6 4 2 y = 0 β6 β4 β2 2 4 6 8 x β2 β4 β6 x = β2 x = 1 x = 5 Figure 15 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 425 Try It #6 Find the vertical and horizontal asymptotes of the function: f (x) = (2x β 1)(2x + 1) _____________ (x β 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 β 2)(x + 3) __________________ (x β 1)(x + 2)(x β 5). Solution We can find the y-intercept by evaluating the function at zero f (0) = = (0 β 2)(0 + 3) _________________ (0 β 1)(0 + 2)(0 β 5) β6 ___ 10 = β #3 __ 5 The x-intercepts will occur when the function is equal to zero: = |
β0.6 0 = (x β 2)(x + 3) _________________ (x β 1)(x + 2)(x β 5) This is zero when the numerator is zero. The y-intercept is (0, β0.6), the x-intercepts are (2, 0) and (β3, 0). See Figure 16. 0 = (x β 2)(x + 3) x = 2, β3 y 6 5 4 3 2 1 (β3, 0) β4 β3 β5 β6 (2, 0) 321 4 5 6 7 8 (0, β0.61 β2 0 β1 β2 β3 β4 β5 β6 x = β2 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 6 CHAPTER 5 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 β5 β4 β3 β2 5 4 3 2 1 β1 β1 β2 β3 β4 β5 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. β5 β4 β3 β2 5 4 3 2 1 β1 β |
1 β2 β3 β4 β5 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 β 3) _____________ (x + 3)2(x β 2) y f (x) = (x + 1)2 (x β 3) (x + 3)2 (x β 23, 0) β8 β6 (β1, 0) β4 β2 2 β2 β4 β6 x = β3 x = 2 Figure 19 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 427 β’ At the x-intercept x = β1 corresponding to the (x + 1)2 factor of the numerator, the graph "bounces," consistent with the quadratic nature of the factor. β’ At the x-intercept x = 3 corresponding to the (x β 3) factor of the numerator, the graph passes through the axis as we would expect from a linear factor. β’ At the vertical asymptote x = β3 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) = β’ At the vertical asymptote x = 2, corresponding to the (x β 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 Toβ¦ 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 β 3) ____________. (x + 1)2(x β 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 β 3)__ (0 + 1)2(0 β 2) = 3 To fi nd 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 = β2 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 (β2, 0) and (3, 0). To fi nd the vertical asymptotes, we determine when the denominator is equal to zero. Th is occurs when x + 1 = 0 and when x β 2 = 0, giving us vertical asymptotes at x = β1 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS y 6 5 4 3 2 1 β6 |
β5 β4 β3 β2 β1 β1 β2 321 4 5 6 x Figure 20 The factor associated with the vertical asymptote at x = β1 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 β6 β5 β4 y 6 5 4 3 2 1 β3 β2 β1 β1 β2 β3 β4 x = β Figure 21 Try It #8 Given the function f (x) = (x + 2)2(x β 2) ______________ 2(x β 1)2 (x β 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, β¦, vm, and no xi = any vj, then the function can be written in the form: f (x) = a (x β x1) p ___ (x β v1) q 2 β¦ ( |
x β xn) p 2 β¦ (x β vm) q 1(x β x2) p 1(x β 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 RATIONAL FUNCTIONS 429 How Toβ¦ 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 βsimplestβ function with small multiplicitiesβsuch as 1 or 3βbut may be difficult for larger multiplicitiesβsuch 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. β6 β5 β4 β3 β2 y 5 4 3 2 1 β1 β1 β2 β3 β4 β5 β6 β7 321 4 5 6 x Figure 22 Solution The graph appears to have x-intercepts at x = β2 and x = 3. At both, the graph passes through the intercept, suggesting linear factors. The graph has two vertical asymptotes. The one at x = β1 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 β6 β4 β2 2 4 6 x x-intercepts β2 β4 β6 Figure 23 We can use this information to write a function of the form |
f (x) = a (x + 2)(x β 3) __ (x + 1)(x β 2)2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 43 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS To find the stretch factor, we can use another clear point on the graph, such as the y-intercept (0, β2). β2 = a β2 = a a = (0 + 2)(0 β 3)__ (0 + 1)(0 β 2)2 β6_ 4 β8 _ β6 4 _ = 3 This gives us a final function of f (x) = 4(x + 2)(x β 3) ______________ 3(x + 1)(x β 2)2. Access these online resources for additional instruction and practice with rational functions. β’ Graphing Rational Functions (http://openstaxcollege.org/l/graphrational) β’ Find the Equation of a Rational Function (http://openstaxcollege.org/l/equatrational) β’ Determining Vertical and Horizontal Asymptotes (http://openstaxcollege.org/l/asymptote) β’ Find the Intercepts, Asymptotes, and Hole of a Rational Function (http://openstaxcollege.org/l/interasymptote) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 SECTION EXERCISES 431 5.6 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 β 1 _____ x + 2 9. f (x) = x2 + 4x β 3 _________ x4 β 5x2 + 4 |
7. f (x) = x + 1 _____ x2 β 1 8. f (x) = x2 + 4 _________ x2 β 2x β 8 For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions. 10. f (x) = 4 ____ x β 1 13. f (x) = 16. f (x) = x __________ x2 + 5x β 36 x2 β 1 ___________ x3 + 9x2 + 14x 11. f (x) = 2 _____ 5x + 2 14. f (x) = 3 + x ______ x3 β 27 17. f (x) = x + 5 ______ x2 β 25 12. f (x) = x _____ x2 β 9 15. f (x) = 3x β 4 _______ x3 β 16x 18. f (x) = x β 4 _____ x β 6 19. f (x) = 4 β 2x ______ 3x β 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 β x 22. f (x) = x2 + 8x + 7 ___________ x2 + 11x + 30 23. f (x) = x2 + x + 6 ___________ x2 β 10x + 24 24. f (x) = 94 β 2x2 _______ 3x2 β 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 β 6 27. f (x) = β2x _____ x β 6 28. f (x) = x2 β 4x + 3 _________ x2 β 4x β 5 29. f (x) = 2x2 β 32 ___________ 6x2 + 13x β 5 For the following exercises, find the slant asymptote of the functions. 30. f (x) = 24x2 + 6x ________ 31. f (x) = 4x2 β 10 _______ 2x β 4 2x + 1 32. f (x) = 81x2 β 18 ________ |
3x β 2 33. f (x) = 6x3 β 5x _______ 3x2 + 4 34. f (x) = x2 + 5x + 4 _________ x β 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 43 2 CHAPTER 5 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 β 3 ______ x + 4 40. q(x) = x β 5 _____ 3x β 1 4 ______ (x β 2)2 41. s(x) = 42. r(x) = 5 ______ (x + 1)2 43. f (x) = 3x 2 β 14x β 5 __ 3x 2 + 8x β 16 44. g(x) = 2x 2 + 7x β 15 __ 3x 2 β 14 + 15 45. a(x) = x 2 + 2x β 3 _ x 2 β 1 46. b(x 47. h(x) = 2x 2 + x β 1 _ x β 4 48. k(x) = 2x 2 β 3x β 20 __ x β 5 49. w(x) = (x β 1)(x + 3)(x β 5) __ (x + 2)2(x β 4) 50. z(x) = (x + 2)2(x β 5) __ (x β 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 = β5, x-intercepts at (2, 0) and (β1, 0), y-intercept at (0, 4) 52. Vertical asymptotes at x = β |
4 and x = β1, x-intercepts at (1, 0) and (5, 0), y-intercept at (0, 7) 53. Vertical asymptotes at x = β4 and x = β5, x-intercepts at (4, 0) and (β6, 0), horizontal asymptote at y = 7 54. Vertical asymptotes at x = β3 55. Vertical asymptote at x = β1, 56. Vertical asymptote at x = 3, and x = 6, x-intercepts at (β2, 0) and (1, 0), horizontal asymptote at y = β2 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 β2 0 β1 β2 β3 β4 β5 β10 β8 β6 β4 58. 642 8 10 x β10 β8 β6 β4 y 5 4 3 2 1 β2 β1 β2 β3 β4 β5 59. 642 8 10 x β10 β8 β6 β4 y 5 4 3 2 1 β2 β1 β2 β3 β4 β5 642 8 10 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.6 SECTION EXERCISES 433 62. 642 8 10 x β10 β8 β6 β4 y 5 4 3 2 1 β2 β1 β2 β3 β4 β5 642 8 10 x 642 8 10 x 61. 642 8 10 x β10 β8 β6 β4 64. 642 8 10 x β10 β8 β6 β4 y 5 4 3 2 1 β2 β1 β2 β3 β4 β5 y 5 4 3 2 1 β2 β1 β2 β3 β4 β5 60. 63. y 5 4 3 2 1 β2 β1 β2 β3 β4 β5 y 5 4 3 2 1 β2 β1 β2 β3 β4 β5 β10 β8 β6 β4 β10 β8 β6 β4 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 β 2 66. f (x) = x _____ x β 3 67. f (x) = 2x _____ x + 4 68. f (x) = 2x ______ (x β 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 β 3 71. f (x) = 72. f (x) = 2 ___________ (x β 1)(x + 2) 73. f (x) = x + 2 ___________ (x β 1)(x β 4) 74. f (x) = (x + 3)2 ____________ (x β 1)2(x + 1) EXTENSIONS For the following exercises, identify the removable discontinuity. 75. f (x) = x2 β 4 _____ x β 2 76. f (x) = x3 + 1 _____ x + 1 78. f (x) = 2x2 + 5x β 3 __________ x + 3 79. f (x) = x3 + x2 ______ x + 1 77. f (x) = x2 + x β 6 ________ x β 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 43 4 CHAPTER 5 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 patientβs 83. The concentration C of a drug in a patientβs 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 INVERSES AND RADICAL FUNCTIONS 435 LEARNING OBJECTIVES In this section, you will: β’ β’ Find the inverse of an invertible polynomial function. Restrict the domain to ο¬nd the inverse of a polynomial function. 5.7 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 V = 1 __ Οr 2 h 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 Οr 2(2r) Οr 3 3 r =# ____ 3V ___ 2Ο β 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. Only one-to-one functions have inverses. Recall that a one-to-one function has a unique output value for each input value and passes the horizontal line test. 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 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 43 6 CHAPTER 5 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 β2 β2 β4 β6 β10 β8 β6 β4 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 = Β± β β 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 weβre looking for. y = x2 __, x > 0 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 Δ w = 36 Δ 2x = 72x = 72 β β 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 INVERSES AND RADICAL FUNCTIONS 437 Functions involving roots are often called radical functions. While it is not possible to find an inverse of most polynomial functions, some basic polynomials do have inverses. Such functions are called invertible functions, and we use the notation f β1(x). Warning: f β1(x) |
is not the same as the reciprocal of the function f (x). This use of ββ1β is reserved to denote inverse functions. To denote the reciprocal of a function f (x), we would need to write ( f (x))β1 = 1 _. f (x) An important relationship between inverse functions is that they βundoβ each other. If f β1 is the inverse of a function f, then f is the inverse of the function f β1. In other words, whatever the function f does to x, f β1 undoes itβand viceversa. More formally, we write f β1 ( f (x)) = x, for all x in the domain of f and f ( f β1 (x)) = x, for all x in the domain of f β1 Note that the inverse switches the domain and range of th original function. 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 Toβ¦ 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 β1(x). Example 1 Show that f (x) = Verifying Inverse Functions 1 _ x + 1 1 and f β1(x) = __ β 1 are inverses, for x β 0, β1. x Solution We must show that f β1( f (x)) = x and f ( f β1(x)) = x. 1 _____ ) x + 1 f β1(f (x)) = f β1 ( 1 _ 1 _____ x + 1 = (x + 1) β 1 β 1 = = x 1 __ f (f β1(x)) = f ( β 1 ) x 1 __ = 1 __ β __ x = x Therefore, f (x) = 1 _____ x + 1 1 and f β1 (x) = __ β 1 are inverses. x Try It #1 Show that f (x) = x + 5 _____ 3 and f β1(x) = 3x β 5 are invers |
es. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 43 8 CHAPTER 5 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 β 1 = 5y3 x β 1 _____ 5 = y3 f β1(x) = # β Analysis Look at the graph of f and f β1. Notice that one graph is the reflection of the other about the line y = x. This is always the case when graphing a function and its inverse function. 3 β x β 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 β1. Since (0, 1) is on the graph of f, then (1, 0) is on the graph of f β1. Similarly, since (1, 6) is on the graph of f, then (6, 1) is on the graph of f β1. See Figure 4. f (x) = 5x3 + 1 y (1, 6) y = x (6, 1) 6 4 2 (0, 1) β6 β4 β2 2 4 6 x (1, 0) β2 β4 β6 Figure 4 Try It #2 Find the inverse function of f (x) = # 3 β β 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 INVERSES AND RADICAL FUNCTIONS 439 How Toβ¦ 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 β1(x). 5. Revise the formula for f β1(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 β 4)2, x β₯ 4 b. f (x) = (x β 4)2, x β€ 4 Solution The original function f (x) = (x β 4)2 is not one-to-one, but the function is restricted to a domain of x β₯ 4 or x β€ 4 on which it is one-to-one. See Figure 5. f (x) = (x β 4)2, x β₯ 4 y 10 8 6 4 2 f (x) = (x β 4)2, x β€ 4 y 10 8 6 4 2 β10 β8 β6 β4 β2 β2 642 8 10 x β10 β8 β6 β4 β2 β2 642 8 10 x To find the inverse, start by replacing f (x) with the simple variable y. y = (x β 4)2 Interchange x and y. Figure 5 β x = (y β 4)2 Take the square root. x = y β 4 x = y Add 4 to both sides. β Β± β 4 Β± β 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 β₯ 4, so for the inverse, we should have y β₯ 4, which is what our inverse function gives. a. The domain of the original function was restricted to x β₯ 4, so the outputs of the inverse need to be the same, f (x) β₯ 4, and we must use the + case: f β1(x) = 4 + β β x b. The domain of the original function was restricted to x β€ 4, so the outputs of the inverse need to be the same, f (x) β€ 4, and we must use the β case: f β1(x) = 4 β β β 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 β1. For any coordinate pair, if (a, b) is on the graph of f, then (b, a) is on the graph of f β1. Finally, observe that the graph of f intersects the graph of f β1 on the line y = x. Points of intersection for the graphs of f and f β1 will always lie on the line y = x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS y 10 8 6 2 (0, 4) f (x) y = x f β1(x) (0, 4) 10 8 6 2 y f (x) y = x β10 β8 β6 β4 β2 β2 2 6 8 10 x β10 β8 β6 β4 β2 β2 2 6 (4, 0) (4, 0) Figure 6 f β1(x) x 8 10 Example 4 Finding the Inverse of a Quadratic Function When the Restriction Is Not Speciο¬ed Restrict the domain and then find the inverse of f (x) = (x β 2)2 β 3. Solution We can see this is a par |
abola with vertex at (2, β3) 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 β₯ 2. To fi nd 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 β 2)2 β 3 Interchange x and y. x = (y β 2)2 β 3 Add 3 to both sides. Take the square rooty β 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 β₯ 2, the outputs of the inverse should be the same, telling us to utilize the + case f β1(x) = 2 + β β 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 β₯ 2. We could just have easily opted to restrict the domain on x β€ 2, in which case f β1(x) = 2 β β 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, β3) is on the graph of f and the coordinate pair (β3, 2) is on the graph of fβ1. Observe from the graph of both functions on the same set of axes that β and domain of f = range of f β1 = [2, β) domain of f β1 = range of f = [β3, β) Finally, observe that the graph of f intersects the graph of f β1 along the line y = x. (β3, 2) β6 β8 β4 β10 y y = x f (x) f β1(x) x 8 10 642 (2, β3) 10 8 6 4 2 β2 β2 β4 β6 β8 β10 Figure 7 Download the OpenStax text for free at http://cnx. |
org/content/col11759/latest. SECTION 5.7 INVERSES AND RADICAL FUNCTIONS 441 Try It #3 Find the inverse of the function f (x) = x2 + 1, on the domain x β₯ 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 Toβ¦ 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 of the function f (x) = β β x β 4 and then find the inverse. Solution Note that the original function has range f (x) β₯ 0. Replace f (x) with y, then solve for x Replace f (x) with y. Interchange x and y. Square each side. x2 = y β 4 x2 + 4 = y f β1(x) = x2 + 4 Add 4. Rename the function fβ1(x). Recall that the domain of this function must be limited to the range of the original function. f β1(x) = x2 + 4, x β₯ 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 β1(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) = β β 2x + 3. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 2 CHAPTER 5 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 Οr 3 2 __ Find the inverse of the function V = Οr 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 Ο = 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 nor would a radius of 0. Also note the range of the function (hence, the domain of the inverse function) is V β₯ 0. Solve for r in terms of V, using the method outlined previously. Note that in real-world applications, we do not swap the variables when finding inverses. Instead, we change which variable is considered to be the independent variable. V = 2 __ Οr 3 3 r 3 = 3V ___ 2Ο ____ β 3 3V ___ 2Ο r =# Solve for r 3. Solve for r. This is the result stated in the section opener. Now evaluate this for V = 100 and Ο = 3.14. 3 r = ____ 3V ___ 2Ο _______ 3 Δ 100 ______ 2 Δ 3.14 β β 3 = Therefore, the radius is about 3.63 ft. β 3 β β 47.7707 β 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) = β ___________ (x + 2)(x β 3) ____________ # # (x β 1). Solution Because a square root is only defined when the quantity under the radical is non-negative, we need to determine β₯ 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 = β2, 1, and 3. (x + 2)(x β 3) ____________ (x β 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 (β2, 0) β3 β2 β4 β7 β6 β5 10 8 6 4 2 β1 β2 β4 β6 β8 β10 Outputs are non-negative (3, 0) 321 4 5 6 7 x Figure 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 INVERSES AND RADICAL FUNCTIONS 443 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 = β2, 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 β2 β€ x < 1 or x β₯ 3, or in interval notation, [β2, 1) βͺ [3, β). 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 in order to determine how many mL we need for a given concentration. 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 β 20 = 0.4n β Cn 100C β 20 = (0.4 β C)n Now evaluate this function at 35%, which is C = 0.35. n = 100C β 20 _________ 0.4 β C n = 100(0.35) β 20 ____________ 0.4 β 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 β 2 Access these online resources for additional instruction and practice with inverses and radical functions. β’ Graphing the Basic Square Root Function (http://openstaxcollege.org/l/graphsquareroot) β’ Find the Inverse of a Square Root Function (http://openstaxcollege.org/l/inversesquare) β’ Find the Inverse of a Rational Function (http://openstaxcollege.org/l/inverserational) β’ Find the Inverse of a Rational Function and an Inverse Function Value (http://openstaxcollege.org/l/rationalinverse) β’ Inverse Functions (http://openstaxcollege.org/l/inversefunction) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 5.7 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 β 4)2, [4, β) 6. f (x) = (x + 2)2, [β2 |
, β) 7. f (x) = (x + 1)2 β 3, [β1, β) 8. f (x) = 3x 2 + 5, (ββ, 0] 9. f (x) = 12 β x 2, [0, β) 10. f (x) = 9 β x 2, [0, β) 11. f (x) = 2x 2 + 4, [0, β) For the following exercises, find the inverse of the functions. 12. f (x) = x 3 + 5 15. f (x) = 4 β 2x 3 13. f (x) = 3x 3 + 1 14. f (x) = 4 β x 3 For the following exercises, find the inverse of the functions. 16. f (x) = β β 2x + 1 19. f (x) = β β 6x β 8 + 5 22. f (x) = 2 _____ x + 8 25. f (x) = x β 2 _____ x + 7 17. f (x) = β β 3 β 4x 20. f (x) = 9 + 2 3 β β x 23. f (x) = 3 _____ x β 4 26. f (x) = 3x + 4 ______ 5 β 4x 18. f (x) = 9 + β β 4x β 4 21. f (x) = 3 β 3 β β x 24. f (x) = x + 3 _____ x + 7 27. f (x) = 5x + 1 ______ 2 β 5x 28. f (x) = x 2 + 2x, [β1, β) 29. f (x) = x 2 + 4x + 1, [β2, β) 30. f (x) = x 2 β 6x + 3, [3, β) GRAPHICAL For the following exercises, find the inverse of the function and graph both the function and its inverse. 31. f (x) = x 2 + 2, x β₯ 0 32. f (x) = 4 β x 2, x β₯ 0 33. f (x) = (x + 3)2, x β₯ β3 34. f (x) = (x β 4)2, x β₯ 4 35. f (x) = x 3 + 3 37 |
. f (x) = x 2 + 4x, x β₯ β2 38. f (x) = x 2 β 6x + 1, x β₯ 3 36. f (x) = 1 β x 3 39. f (x) =# #2 __ x 40. f (x) = 1 __ x2, x β₯ 0 For the following exercises, use a graph to help determine the domain of the functions. 41. f (x) = β 44. f (x) = β _____________ (x + 1)(x β 1) __ x ___________ x2 β x β 20 _ x β 2 _____________ (x + 2)(x β 3) __ x β 1 42. f (x) = β 45. f (x) = β ______ 9 β x2 _ x + 4 43. f (x) = β ________ x(x + 3) _ x β 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.7 SECTION EXERCISES 445 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. 46. f (x) = x3 β x β 2, y = 1, 2, 3 47. f (x) = x3 + x β 2, y = 0, 1, 2 48. f (x) = x3 + 3x β 4, y = 0, 1, 2 49. f (x) = x3 + 8x β 4, y = β1, 0, 1 50. f (x) = x4 + 5x + 1, y = β1, 0, 1 EXTENSIONS For the following exercises, find the inverse of the functions with a, b, c positive real numbers. 51. f (x) = ax3 + b 54. f (x) = 3 β β ax + b 52. f (x) = x2 + bx 55. f (x) = ax + b ______ x + c REAL-WORLD APPLICATIONS 53. f (x) = β β ax2 + b For the following exercises, determine the function described and then use it to answer the question. 56. 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 β 4.9t 2. Express t as a function of height, h, and find the time to reach a height of 50 meters. 57. 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 β 16t 2. Express t as a function of height h, and find the time to reach a height of 400 feet. 58. The volume, V, of a sphere in terms of its radius, r, 4 __ is given by V(r) = Οr 3. Express r as a function of 3 V, and find the radius of a sphere with volume of 200 cubic feet. 59. The surface area, A, of a sphere in terms of its radius, r, is given by A(r) = 4Οr 2. Express r as a function of V, and find the radius of a sphere with a surface area of 1000 square inches. 60. 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. 61. The period T, in seconds, of a simple pendulum as a function of its length l, in feet, is given by T(l) = 2Ο β ____ l ____ 32.2. Express l as a function of T and determine the length of a pendulum with period of 2 seconds. 62. The volume of a cylinder, V, in terms of radius, r, 63. The surface area, A, of a cylinder in terms of its and height, h, is given by V = Οr 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 = 2Οr2 + 2Οrh. 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. 64. The volume of a right circular cone, V, in terms of its 1 _ radius, r, and its height, h, is given by V = Οr 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. 65. 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS LEARNING OBJECTIVES In this section, you will: β’ ole algebraic and numeric ariation problems. β’ ole application problems including direct inerse andor oint ariation. 5.8 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, Nicoleβs 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 price $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 Nicoleβs 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. We saw functions like this one when we discussed power functions. 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 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.8 MODELING USING VARIATION 447 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, or is proportional to, the nth power of x. In direct variation relationships, there is a nonzero constant ratio k = constant of variation, which help defines the relationship between the variables. y _ xn, where k is called the How Toβ¦ 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 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS Q & Aβ¦ Do the graphs of all direct variation equations look like Example 1? No. Direct variation equations are power functionsβthey 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 waterβs 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 Earthβs surface. Consider the Atlantic Ocean, which covers 22% of Earthβs surface. At a certain location, at the depth of 500 feet, the temperature may be 28Β°F. If we create Table 2, we observe that, as the depth increases, the water temperature decreases. d, depth 500 ft 1,000 ft 2,000 ft T = 14,000_ d 14,000 _ 500 = 28 14,000 _ 1,000 = 14 14,000 _ 2,000 = 7 Interpretation At a depth of 500 ft, the water temperature is 28Β° F. At a depth of 1,000 ft, the water temperature is 14Β° F. At a depth of 2,000 ft, the water temperature is 7Β° 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 Β° ( 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.8 MODELING USING VARIATION 449 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 so t =. Because time is a function of velocity, we can write t(v). 100 _ v t(v) = 100___ v = 100v β1 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. We say that the time varies inversely with velocity. How Toβ¦ 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 Δ 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 45 0 CHAPTER 5 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. Solution Begin by writing an equation to show the relationship between the variables. Substitute x = 6, y = 2, and z = 8 to find the value of the constant k. x = ky2 _ β 3 z β 6 = k22 _ β 3 8 β 6 = 4k__ 2 3 = k Now we can substitute the value of the constant into the equation for the relationship. To find x when y = 1 and |
z = 27, we will substitute values for y and z into our equation. x = 3y2 _ β 3 z β x = 3(1)2 _ β 3 27 β = 1 Try It #3 A quantity x varies directly with the square of y and inversely with z. If x = 40 when y = 4 and z = 2, find x when y = 10 and z = 25. Access these online resources for additional instruction and practice with direct and inverse variation. β’ Direct Variation (http://openstaxcollege.org/l/directvariation) β’ Inverse Variation (http://openstaxcollege.org/l/inversevariatio) β’ Direct and Inverse Variation (http://openstaxcollege.org/l/directinverse) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 5.8 SECTION EXERCISES 451 5.8 SECTION EXERCISES VERBAL 1. 2. 3. ALGEBRAIC 4. yxx=y= 6. yx 5. yxx=y= 7. yxx=y= x=y= 8. yx 9. yxx= x=y= 10. yxx=y= 12. yx x=y= y= 11. yxx = 3, y= 13. yx x=y= 14. yx 15. yx x=y= x=y= 16. yxzx= 17. yxzwx=z= z=y= w=y= 18. yx zx=z=y= 20. yxz wx=z= w=y= 19. yxz x=z=y= 21. yxzwx= z=w=y= 22. yx zw x=z=w=y= 23. yxz wtx=z=w= t=y= NUMERIC o fi 24. yxx=y= 25. yxx= yx= y=yx= 26. yxx= y=yx= 28. yx x=y=yx= 30. yxx= y=yx= 27. yx x=y=yx= 29. yxx=y= yx= 31. yxx= y=yx= 32. yx 33. yx x=y=yx |
= x=y=yx= 34. yxzx=z= 35. yxzwx=z= y=yx=z= 36. yxz x=z=y=yx= z= w=y=yx=z=w= 37. yx zx=z=y=y x=z= 38. yxzw x=z=w=y=y x=z=w= 39. yxz wx= z=w=y=yx= z=w= Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 45 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 40. yxzwtx=z= w=t=y=yx=z=w=t= TECHNOLOGY 41. yxx= 42. yxx=y= y= 43. yx 44. yxx=y= x=y= 45. yx x=y= EXTENSIONS T a 46. Ta 48. Ta 47. 49. 50. REAL-WORLD APPLICATIONS 51. 53. 55. 52. 54. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 5 REVIEW 453 CHAPTER 5 REVIEW Key Terms arrow notation a way to represent the local and end behavior of a function by using arrows to indicate that an input or output approaches a value axis of symmetry a vertical line drawn through the vertex of a parabola, that opens up or down, around which the parabola is symmetric; it is defined by x = β # b __. 2a coefficient a nonzero real number multiplied by a variable raised to an exponent constant of variation the non-zero value k that helps define the relationship between variables in direct or inverse variation continuous function a function whose graph can be drawn without lifting the pen from the paper because there are no breaks in the graph degree the highest power of the variable that occurs in a polynomial Descartesβ Rule of Signs a rule that determines the maximum possible numbers of positive and negative real zeros based on the number of sign changes of f (x) and f (βx) direct variation the relationship between two variables that are a constant multiple of each other; as one quantity increases, so does the other Division Algorithm 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) where 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). end behavior the behavior of the graph of a function as the input decreases without bound and increases without bound Factor Theorem k is a zero of polynomial function f (x) if and only if (x β k) is a factor of f (x) Fundamental Theorem of Algebra a polynomial function with degree greater than 0 has at least one complex zero general form of a quadratic function the function that describes a parabola, written in the form f (x) = ax 2 + bx + c, where a, b, and c are real numbers and a β 0. global maximum highest turning point on a graph; f (a) where f (a) β₯ f (x) for all x. global minimum lowest turning point on a graph; f (a) where f (a) β€ f (x) for all x. horizontal asymptote a horizontal line y = b where the graph approaches the line as the inputs increase or decrease without bound. Intermediate Value Theorem for two numbers a and b in the domain of f, if a < b and f (a) β f (b), then the function f takes on every value between f (a) and f (b); specifically, when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis inverse variation the relationship between two variables in which the product of the variables is a constant inversely proportional a relationship where one quantity is a constant divided by the other quantity; as one quantity increases, the other decreases invertible function any function that has an inverse function β1 imaginary number a number in the form bi where i = β β joint variation a relationship where a variable varies directly or inversely with multiple variables leading coefficient the coefficient of the leading term leading term the term containing the highest power of the variable Linear Factorization Theorem allowing for multiplicities, a polynomial function will have the same number of |
factors as its degree, and each factor will be in the form (x β c), where c is a complex number multiplicity the number of times a given factor appears in the factored form of the equation of a polynomial; if a polynomial contains a factor of the form (x β h)p, x = h is a zero of multiplicity p. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 45 4 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS polynomial function a function that consists of either zero or the sum of a finite number of non-zero terms, each of which is a product of a number, called the coefficient of the term, and a variable raised to a non-negative integer power. power function a function that can be represented in the form f (x) = kxp where k is a constant, the base is a variable, and the exponent, p, is a constant rational function a function that can be written as the ratio of two polynomials p _ Rational Zero Theorem the possible rational zeros of a polynomial function have the form q where p is a factor of the constant term and q is a factor of the leading coefficient. Remainder Theorem if a polynomial f (x) is divided by x β k, then the remainder is equal to the value f (k) removable discontinuity a single point at which a function is undefined that, if filled in, would make the function continuous; it appears as a hole on the graph of a function roots in a given function, the values of x at which y = 0, also called zeros smooth curve a graph with no sharp corners standard form of a quadratic function the function that describes a parabola, written in the form f (x) = a(x β h)2 + k, where (h, k) is the vertex. synthetic division a shortcut method that can be used to divide a polynomial by a binomial of the form x β k term of a polynomial function any aixi of a polynomial function in the form f (x) = anxn +... + a2x2 + a1x + a0 turning point the location at which the graph of a function changes direction varies directly a relationship where one quantity is a constant multiplied by the other quantity varies inversely a relationship where one quantity is a constant |
divided by the other quantity vertex the point at which a parabola changes direction, corresponding to the minimum or maximum value of the quadratic function vertex form of a quadratic function another name for the standard form of a quadratic function vertical asymptote a vertical line x = a where the graph tends toward positive or negative infinity as the inputs approach a zeros in a given function, the values of x at which y = 0, also called roots Key Equations general form of a quadratic function f (x) = ax 2 + bx + c standard form of a quadratic function f (x) = a(x β h)2 + k general form of a polynomial function f (x) = an xn +... + a2 x2 + a1x + a0 Division Algorithm f (x) = d(x)q(x) + r(x) where q(x) β 0 Rational Function Direct variation Inverse variation f (x) = P(x) ____ Q(x) =# # ap x ___ bq x pβ1 +... + a1x + a0 qβ1 +... + b1 x + b0 p + ap β 1x q + bq β 1 x, Q(x) β 0 y = kx n, k is a nonzero constant. y = k _ xn, k is a nonzero constant. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 5 REVIEW 455 Key Concepts 5.1 Quadratic Functions β’ A polynomial function of degree two is called a quadratic function. β’ The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can open either up or down. β’ The axis of symmetry is the vertical line passing through the vertex. The zeros, or x-intercepts, are the points at which the parabola crosses the x-axis. The y-intercept is the point at which the parabola crosses the y-axis. See Example 1, Example 7, and Example 8. β’ Quadratic functions are often written in general form. Standard or vertex form is useful to easily identify the vertex of a parabola. Either form can be written from a graph. See Example 2. β’ The vertex can be found from |
an equation representing a quadratic function. See Example 3. β’ The domain of a quadratic function is all real numbers. The range varies with the function. See Example 4. β’ A quadratic functionβs minimum or maximum value is given by the y-value of the vertex. β’ The minimum or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue. See Example 5 and Example 6. β’ The vertex and the intercepts can be identified and interpreted to solve real-world problems. See Example 9. 5.2 Power Functions and Polynomial Functions β’ A power function is a variable base raised to a number power. See Example 1. β’ The behavior of a graph as the input decreases beyond bound and increases beyond bound is called the end behavior. β’ The end behavior depends on whether the power is even or odd. See Example 2 and Example 3. β’ A polynomial function is the sum of terms, each of which consists of a transformed power function with positive whole number power. See Example 4. β’ The degree of a polynomial function is the highest power of the variable that occurs in a polynomial. The term containing the highest power of the variable is called the leading term. The coefficient of the leading term is called the leading coefficient. See Example 5. β’ The end behavior of a polynomial function is the same as the end behavior of the power function represented by the leading term of the function. See Example 6 and Example 7. β’ A polynomial of degree n will have at most n x-intercepts and at most n β#1 turning points. See Example 8, Example 9, Example 10, Example 11, and Example 12. 5.3 Graphs of Polynomial Functions β’ Polynomial functions of degree 2 or more are smooth, continuous functions. See Example 1. β’ To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero. See Example 2, Example 3, and Example 4. β’ Another way to find the x-intercepts of a polynomial function is to graph the function and identify the points at which the graph crosses the x-axis. See Example 5. β’ The multiplicity of a zero determines how the graph behaves at the x-intercepts. See Example 6. β’ The graph of |
a polynomial will cross the horizontal axis at a zero with odd multiplicity. β’ The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity. β’ The end behavior of a polynomial function depends on the leading term. β’ The graph of a polynomial function changes direction at its turning points. β’ A polynomial function of degree n has at most n β 1 turning points. See Example 7. β’ To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most n β 1 turning points. See Example 8 and Example 10. β’ Graphing a polynomial function helps to estimate local and global extremas. See Example 11. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 45 6 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS β’ The Intermediate Value Theorem tells us 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. See Example 9. 5.4 Dividing Polynomials β’ Polynomial long division can be used to divide a polynomial by any polynomial with equal or lower degree. See Example 1 and Example 2. β’ The Division Algorithm tells us that a polynomial dividend can be written as the product of the divisor and the quotient added to the remainder. β’ Synthetic division is a shortcut that can be used to divide a polynomial by a binomial in the form x β k. See Example 3, Example 4, and Example 5. β’ Polynomial division can be used to solve application problems, including area and volume. See Example 6. 5.5 Zeros of Polynomial Functions β’ To fi nd f (k), determine the remainder of the polynomial f (x) when it is divided by x β k. this is known as the Remainder Theorem. See Example 1. β’ According to the Factor Theorem, k is a zero of f (x) if and only if (x β k) is a factor of f (x). See Example 2. β’ According to the Rational Zero Theorem, each rational zero of a polynomial function with integer coefficients will be equal to a factor of the constant term divided by a factor of |
the leading coefficient. See Example 3 and Example 4. β’ When the leading coefficient is 1, the possible rational zeros are the factors of the constant term. β’ Synthetic division can be used to find the zeros of a polynomial function. See Example 5. β’ According to the Fundamental Theorem, every polynomial function has at least one complex zero. See Example 6. β’ Every polynomial function with degree greater than 0 has at least one complex zero. β’ Allowing for multiplicities, a polynomial function will have the same number of factors as its degree. Each factor will be in the form (x β c), where c is a complex number. See Example 7. β’ The number of positive real zeros of a polynomial function is either the number of sign changes of the function or less than the number of sign changes by an even integer. β’ The number of negative real zeros of a polynomial function is either the number of sign changes of f (βx) or less than the number of sign changes by an even integer. See Example 8. β’ Polynomial equations model many real-world scenarios. Solving the equations is easiest done by synthetic division. See Example 9. 5.6 Rational Functions 1 β’ We can use arrow notation to describe local behavior and end behavior of the toolkit functions f (x) =# #1 _ _ x and f (x) = x2. See Example 1. β’ A function that levels off at a horizontal value has a horizontal asymptote. A function can have more than one vertical asymptote. See Example 2. β’ Application problems involving rates and concentrations often involve rational functions. See Example 3. β’ The domain of a rational function includes all real numbers except those that cause the denominator to equal zero. See Example 4. β’ The vertical asymptotes of a rational function will occur where the denominator of the function is equal to zero and the numerator is not zero. See Example 5. β’ A removable discontinuity might occur in the graph of a rational function if an input causes both numerator and denominator to be zero. See Example 6. β’ A rational functionβs end behavior will mirror that of the ratio of the leading terms of the numerator and denominator functions. See Example 7, Example 8, Example 9, and Example 10. β’ Graph rational functions by finding the intercepts, behavior at the intercepts and asymptotes, and end behavior. |
See Example 11. β’ If a rational function has x-intercepts at x = x1, x2, β¦, xn, vertical asymptotes at x = v1, v2,..., vm, and no xi = any vj, then the function can be written in the form Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 5 REVIEW 457 f (x) =#a (x β x1)p ___ (x β v1)q 2...(x β xn)p n 2...(x β vm)q 1(x β x2)p 1(x β v2)q n See Example 12. 5.7 Inverses and Radical Functions β’ The inverse of a quadratic function is a square root function. β’ If f βl is the inverse of a function f, then f is the inverse of the function f βl. See Example 1. β’ While it is not possible to find an inverse of most polynomial functions, some basic polynomials are invertible. See Example 2. β’ To find the inverse of certain functions, we must restrict the function to a domain on which it will be one-to-one. See Example 3 and Example 4. β’ When finding the inverse of a radical function, we need a restriction on the domain of the answer. See Example 5 and Example 7. β’ Inverse and radical and functions can be used to solve application problems. See Example 6 and Example 8. 5.8 Modeling Using Variation β’ A relationship where one quantity is a constant multiplied by another quantity is called direct variation. See Example 1. β’ Two variables that are directly proportional to one another will have a constant ratio. β’ A relationship where one quantity is a constant divided by another quantity is called inverse variation. See Example 2. β’ Two variables that are inversely proportional to one another will have a constant multiple. See Example 3. β’ In many problems, a variable varies directly or inversely with multiple variables. We call this type of relationship joint variation. See Example 4. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 45 8 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS CHAPTER 5 REVIEW EXERCISES QUADRATIC FUNCTIONS For the following exercises, write the quadratic function |
in standard form. Then, give the vertex and axes intercepts. Finally, graph the function. 1. f (x) = x2 β 4x β 5 2. f (x) = β2x2 β 4x For the following problems, find the equation of the quadratic function using the given information. 3. The vertex is (β2, 3) and a point on the graph is (3, 6). 4. The vertex is (β3, 6.5) and a point on the graph is (2, 6). For the following exercises, complete the task. 5. A rectangular plot of land is to be enclosed by fencing. One side is along a river and so needs no fence. If the total fencing available is 600 meters, find the dimensions of the plot to have maximum area. 6. An object projected from the ground at a 45 degree angle with initial velocity of 120 feet per second has height, h, in terms of horizontal distance traveled, β32 _____ (120)2 x2 + x. Find the maximum x, given by h(x) = height the object attains. POWER FUNCTIONS AND POLYNOMIAL FUNCTIONS For the following exercises, determine if the function is a polynomial function and, if so, give the degree and leading coefficient. 7. f (x) = 4x5 β 3x3 + 2x β 1 8. f (x) = 5x + 1 β x2 9. f (x) = x2(3 β 6x + x2) For the following exercises, determine end behavior of the polynomial function. 10. f (x) = 2x4 + 3x3 β 5x2 + 7 11. f (x) = 4x3 β 6x2 + 2 12. f (x) = 2x2(1 + 3x β x2) GRAPHS OF POLYNOMIAL FUNCTIONS For the following exercises, find all zeros of the polynomial function, noting multiplicities. 13. f (x) = (x + 3)2(2x β 1)(x + 1)3 14. f (x) = x5 + 4x4 + 4x3 15. f (x) = x3 β 4x2 + x β 4 For the following exercises, based on the given graph, determine the zeros of the function and note multiplicity. 16. y 17. y |
β5 β4 β3 β2 20 16 12 8 4 β1 β4 β8 β12 β16 β20 321 4 5 x β10 β8 β6 β4 10 8 6 4 2 β2 β2 β4 β6 β8 β10 642 8 10 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 5 REVIEW 459 18. Use the Intermediate Value Theorem to show that at least one zero lies between 2 and 3 for the function f (x) = x3 β 5x + 1 DIVIDING POLYNOMIALS For the following exercises, use long division to find the quotient and remainder. 19. x3 β 2x2 + 4x + 4 ______________ x β 2 20. 3x4 β 4x2 + 4x + 8 _______________ x + 1 For the following exercises, use synthetic division to find the quotient. If the divisor is a factor, then write the factored form. 21. x3 β 2x2 + 5x β 1 x + 3 23. 2x3 + 6x2 β 11x β 12 _________________ x + 4 22. x3 + 4x + 10 __________ x β 3 24. 3x4 + 3x3 + 2x + 2 _______________ x + 1 ______________ ZEROS OF POLYNOMIAL FUNCTIONS For the following exercises, use the Rational Zero Theorem to help you solve the polynomial equation. 25. 2x3 β 3x2 β 18x β 8 = 0 26. 3x3 + 11x2 + 8x β 4 = 0 27. 2x4 β 17x3 + 46x2 β 43x + 12 = 0 28. 4x4 + 8x3 + 19x2 + 32x + 12 = 0 For the following exercises, use Descartesβ Rule of Signs to find the possible number of positive and negative solutions. 29. x3 β 3x2 β 2x + 4 = 0 30. 2x4 β x3 + 4x2 β 5x + 1 = 0 RATIONAL FUNCTIONS For the following exercises, find the intercepts and the vertical and horizontal asymptotes, and then use them to sketch a graph of the function. 31. f (x) = x + 2 _____ x β 5 33. f ( |
x) = 3x2 β 27 32. f (x) = x2 + 1 _____ x2 β 4 34. f (x) = x + 2 _____ x2 β 9 ________ x2 + x β 2 For the following exercises, find the slant asymptote. 35. f (x) = x2 β 1 _____ x + 2 36. f (x) = 2x3 β x2 + 4 __________ x2 + 1 INVERSES AND RADICAL FUNCTIONS For the following exercises, find the inverse of the function with the domain given. 37. f (x) = (x β 2)2, x β₯ 2 38. f (x) = (x + 4)2 β 3, x β₯ β4 40. f (x) = 2x3 β 3 41. f (x) = β β 4x + 5 β 3 39. f (x) = x2 + 6x β 2, x β₯ β3 42. f (x) = x β 3 _____ 2x + 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 0 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS MODELING USING VARIATION For the following exercises, find the unknown value. 43. y varies directly as the square of x. If when x = 3, y = 36, find y if x = 4. 45. y varies jointly as the cube of x and as z. If when x = 1 and z = 2, y = 6, find y if x = 2 and z = 3. 44. y varies inversely as the square root of x. If when x = 25, y = 2, find y if x = 4. 46. y varies jointly as x and the square of z and inversely as the cube of w. If when x = 3, z = 4, and w = 2, y = 48, find y if x = 4, z = 5, and w = 3. For the following exercises, solve the application problem. 47. The weight of an object above the surface of the earth varies inversely with the distance from the center of the earth.If a person weighs 150 pounds when he is on the surface of the earth (3,960 miles from center), find the weight of the person if he is 20 miles above the surface. |
48. The volume V of an ideal gas varies directly with the temperature T and inversely with the pressure P. A cylinder contains oxygen at a temperature of 310 degrees K and a pressure of 18 atmospheres in a volume of 120 liters. Find the pressure if the volume is decreased to 100 liters and the temperature is increased to 320 degrees K. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 5 PRACTICE TEST 461 CHAPTER 5 PRACTICE TEST Give the degree and leading coefficient of the following polynomial function. 1. f (x) = x3(3 β 6x2 β 2x2) Determine the end behavior of the polynomial function. 2. f (x) = 8x3 β 3x2 + 2x β 4 3. f (x) = β2x2(4 β 3x β 5x2) Write the quadratic function in standard form. Determine the vertex and axes intercepts and graph the function. 4. f (x) = x2 + 2x β 8 Given information about the graph of a quadratic function, find its equation. 5. Vertex (2, 0) and point on graph (4, 12). Solve the following application problem. 6. A rectangular field is to be enclosed by fencing. In addition to the enclosing fence, another fence is to divide the field into two parts, running parallel to two sides. If 1,200 feet of fencing is available, find the maximum area that can be enclosed. Find all zeros of the following polynomial functions, noting multiplicities. 7. f (x) = (x β 3)3(3x β 1)(x β 1)2 8. f (x) = 2x6 β 6x5 + 18x4 Based on the graph, determine the zeros of the function and multiplicities. 9. y 125 100 75 50 25 β5 β4 β3 β2 β1 β25 β50 β75 β100 β125 321 4 5 x Use long division to find the quotient. 10. 2x3 + 3x β 4 __________ x + 2 Use synthetic division to find the quotient. If the divisor is a factor, write the factored form. 11. x4 + 3x2 β 4 __________ x β 2 12. 2x3 + 5 |
x2 β 7x β 12 ________________ x + 3 Use the Rational Zero Theorem to help you find the zeros of the polynomial functions. 13. f (x) = 2x3 + 5x2 β 6x β 9 14. f (x) = 4x4 + 8x3 + 21x2 + 17x + 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 2 CHAPTER 5 POLYNOMIAL AND RATIONAL FUNCTIONS 15. f (x) = 4x4 + 16x3 + 13x2 β 15x β 18 16. f (x) = x5 + 6x4 + 13x3 + 14x2 + 12x + 8 Given the following information about a polynomial function, find the function. 17. It has a double zero at x = 3 and zeroes at x = 1 and x = β2. Itβs y-intercept is (0, 12). 18. It has a zero of multiplicity 3 at x =# #1 _ 2 and another zero at x = β3. It contains the point (1, 8). Use Descartesβ Rule of Signs to determine the possible number of positive and negative solutions. 19. 8x3 β 21x2 + 6 = 0 For the following rational functions, find the intercepts and horizontal and vertical asymptotes, and sketch a graph. 20. f (x) = x + 4 _________ x2 β 2x β 3 21. f (x) = x2 + 2x β 3 _________ x2 β 4 Find the slant asymptote of the rational function. 22. f (x) = x2 + 3x β 3 _________ x β 1 Find the inverse of the function. β x β 2 + 4 23. f (x) = β 25. f (x) = 2x + 3 ______ 3x β 1 24. f (x) = 3x3 β 4 Find the unknown value. 26. y varies inversely as the square of x and when x = 3, y = 2. Find y if x = 1. 27. y varies jointly with x and the cube root of z. If when x = 2 and z = 27, y = 12, find y if x = 5 and z = 8. Solve the following application problem |
. 28. The distance a body falls varies directly as the square of the time it falls. If an object falls 64 feet in 2 seconds, how long will it take to fall 256 feet? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Exponential and Logarithmic Functions 6 Figure 1 Electron micrograph of E. Coli bacteria (credit: βMattosaurus,β Wikimedia Commons) CHAPTER OUTLINE 6.1 Exponential Functions 6.2 Graphs of Exponential Functions 6.3 Logarithmic Functions 6.4 Graphs of Logarithmic Functions 6.5 Logarithmic Properties 6.6 Exponential and Logarithmic Equations 6.7 Exponential and Logarithmic Models 6.8 Fitting Exponential Models to Data Introduction Focus in on a square centimeter of your skin. Look closer. Closer still. If you could look closely enough, you would see hundreds of thousands of microscopic organisms. They are bacteria, and they are not only on your skin, but in your mouth, nose, and even your intestines. In fact, the bacterial cells in your body at any given moment outnumber your own cells. But that is no reason to feel bad about yourself. While some bacteria can cause illness, many are healthy and even essential to the body. Bacteria commonly reproduce through a process called binary fission, during which one bacterial cell splits into two. When conditions are right, bacteria can reproduce very quickly. Unlike humans and other complex organisms, the time required to form a new generation of bacteria is often a matter of minutes or hours, as opposed to days or years.[16] For simplicityβs sake, suppose we begin with a culture of one bacterial cell that can divide every hour. Table 1 shows the number of bacterial cells at the end of each subsequent hour. We see that the single bacterial cell leads to over one thousand bacterial cells in just ten hours! And if we were to extrapolate the table to twenty-four hours, we would have over 16 million! Hour Bacteria 0 1 1 2 2 4 3 8 5 32 4 16 Table 1 6 64 7 128 8 256 9 10 512 1024 In this chapter, we will explore exponential functions, which can be used for, among other things, modeling growth patterns such as those found in bacteria. We will also investigate logarithmic functions, which are closely related to exponential functions. Both types of functions |
have numerous real-world applications when it comes to modeling and interpreting data. 16 Todar, PhD, Kenneth. Todarβs Online Te xtbook of Bacteriology. http://te xtbookofbacteriology.net/growth_3.html. 463 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 4 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS G C F.. In this section, you will: β’ efine and eponential function and determine its domain. β’ Graph an eponential function including using transformations. β’ ntroduce. β’ rite a function that results from a gien transformation. 6.1 EXPONENTIAL FUNCTIONS India is the second most populous country in the world with a population of about 1.25 billion people in 2013. The population is growing at a rate of about 1.2% each year[17]. If this rate continues, the population of India will exceed Chinaβs population by the year 2031. When populations grow rapidly, we often say that the growth is βexponential,β meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions, which model this kind of rapid growth. Identifying Exponential Functions When exploring linear growth, we observed a constant rate of changeβa constant number by which the output increased for each unit increase in input. For example, in the equation f (x) = 3x + 4, the slope tells us the output increases by 3 each time the input increases by 1. The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people. Deο¬ning an Exponential Function A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal productsβno meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and 80% in 2050.. What exactly does it mean to grow exponentially? What does the word double have in common with percent |
increase? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media. β’ Percent change refers to a change based on a percent of the original amount. β’ Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time. β’ Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time. For us to gain a clear understanding of exponential growth, let us contrast exponential growth with linear growth. We will construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1. We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of 0, and increase each input by 1. We will add 2 to the corresponding consecutive outputs. See Table 1x) = 2x 1 2 4 8 16 32 64 Table 1 g(x) = 2x 0 2 4 6 8 10 12 17 http://www.worldometers.info/world-population/. Accessed February 24, 2014. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 465 From Table 2 we can infer that for these two functions, exponential growth dwarfs linear growth. β’ Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain. β’ Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain. Apparently, the difference between βthe same percentageβ and βthe same amountβ is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding 2 to the output whenever the input was increased by one. The general form of the exponential function is f (x) = ab x, where a is any nonzero number, b is a positive real number not equal to 1. β’ If b > 1, the function grows at a rate proportional to its size. β’ If 0 < b < 1, the function decays at a rate proportional to its size. Letβ |
s look at the function f (x) = 2x from our example. We will create a table (Table 2) to determine the corresponding outputs over an interval in the domain from β3 to 3. x f (x) = 2x β3 β2 β1 0 1 2 3 2β3 = 1 _ 8 2β2 = 1 _ 4 2β1 = 1 _ 2 Table 2 20 = 1 21 = 2 22 = 4 23 = 8 Let us examine the graph of f by plotting the ordered pairs we observe on the table in Figure 1, and then make a few observations. y f(x) = 2x (3, 8) (2, 4) (1, 2) (0, 15 β4 β3 β2 β1 β1 β2 Figure 1 Letβs define the behavior of the graph of the exponential function f (x) = 2x and highlight some its key characteristics. β’ the domain is (ββ, β), β’ the range is (0, β), β’ as x β β, f (x) β β, β’ as x β ββ, f (x) β 0, β’ f (x) is always increasing, β’ the graph of f (x) will never touch the x-axis because base two raised to any exponent never has the result of zero. β’ y = 0 is the horizontal asymptote. β’ the y-intercept is 1. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 6 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS exponential function For any real number x, an exponential function is a function with the form where β’ a is the a non-zero real number called the initial value and f (x) = ab x β’ b is any positive real number such that b β 1. β’ The domain of f is all real numbers. β’ The range of f is all positive real numbers if a > 0. β’ The range of f is all negative real numbers if a < 0. β’ The y-intercept is (0, a), and the horizontal asymptote is y = 0. Example 1 Identifying Exponential Functions Which of the following equations are not exponential functions? f (x) = 43(x β 2) g (x) = x3 j (x) = (β2)x x 1 ) h (x |
) = ( _ 3 Solution B y definition, an exponential function has a constant as a base and an independent variable as an exponent. Thus, g(x) = x3 does not represent an exponential function because the base is an independent variable. In fact, g(x) = x3 is a power function. Recall that the base b of an exponential function is always a positive constant, and b β 1. Thus, j(x) = (β2)x does not represent an exponential function because the base, β2, is less than 0. Try It #1 Which of the following equations represent exponential functions? β’ f (x) = 2x2 β 3x + 1 β’ g(x) = 0.875x β’ h(x) = 1.75x + 2 β’ j(x) = 1095.6β2x Evaluating Exponential Functions Recall that the base of an exponential function must be a positive real number other than 1. Why do we limit the base b to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive: 1 __ = β 2 1 1 ) = (β9). Then f (x) = f ( _ _ β’ Let b = β9 and x = 2 2 β9, which is not a real number. β Why do we limit the base to positive values other than 1? Because base 1 results in the constant function. Observe what happens if the base is 1: β’ Let b = 1. Then f (x) = 1x = 1 for any value of x. To evaluate an exponential function with the form f (x) = b x, we simply substitute x with the given value, and calculate the resulting power. For example: Let f (x) = 2x. What is f (3)? f (x) = 2x f (3) = 23 = 8 Substitute x = 3. Evaluate the power. To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 467 For example: Let f (x) = 30(2)x. What is f (3)? f (x) = 30(2)x f (3) = 30(2)3 = 30 |
(8) = 240 Substitute x = 3. Simplify the power first. Multiply. Note that if the order of operations were not followed, the result would be incorrect: f (3) = 30(2)3 β 603 = 216,000 Example 2 Evaluating Exponential Functions Let f (x) = 5(3)x + 1. Evaluate f (2) without using a calculator. Solution Follow the order of operations. Be sure to pay attention to the parentheses. f (x) = 5(3)x + 1 f (2) = 5(3)2 + 1 = 5(3)3 = 5(27) = 135 Substitute x = 2. Add the exponents. Simplify the power. Multiply. Try It #2 Let f (x) = 8(1.2)x β 5. Evaluate f (3) using a calculator. Round to four decimal places. Deο¬ning Exponential Growth Because the output of exponential functions increases very rapidly, the term βexponential growthβ is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth. exponential growth A function that models exponential growth grows by a rate proportional to the amount present. For any real number x and any positive real numbers a and b such that b β 1, an exponential growth function has the form where f (x) = ab x β’ a is the initial or starting value of the function. β’ b is the growth factor or growth multiplier per unit x. In more general terms, we have an exponential function, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, letβs consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function A(x) = 100 + 50x. Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function B(x) = 100(1 + 0.5)x. A few years of growth for these companies are illustrated in Table 3. Year, x 0 1 2 3 x Stores, Company A 100 + 50(0) = 100 100 + 50(1) = 150 100 |
+ 50(2) = 200 100 + 50(3) = 250 A(x) = 100 + 50x Table 3 Stores, Company B 100(1 + 0.5)0 = 100 100(1 + 0.5)1 = 150 100(1 + 0.5)2 = 225 100(1 + 0.5)3 = 337.5 B(x) = 100(1 + 0.5)x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 8 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS The graphs comparing the number of stores for each company over a five-year period are shown in Figure 2. We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth. y B(x) = 100(1 + 0.5) 500 450 400 350 300 250 200 150 100 A(x) = 100 + 50x 0 1 2 3 Years x 4 5 Figure 2 The graph shows the numbers of stores Companies A and B opened over a ο¬ve-year period. Notice that the domain for both functions is [0, β), and the range for both functions is [100, β). After year 1, Company B always has more stores than Company A. Now we will turn our attention to the function representing the number of stores for Company B, B(x) = 100(1 + 0.5)x. In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and 1 + 0.5 = 1.5 represents the growth factor. Generalizing further, we can write this function as B(x) = 100(1.5)x, where 100 is the initial value, 1.5 is called the base, and x is called the exponent. Example 3 Evaluating a Real-World Exponential Model At the beginning of this section, we learned that the population of India was about 1.25 billion in the year 2013, with an annual growth rate of about 1.2%. This situation is represented by the growth function P(t) = 1.25(1.012)t, where t is the number of years since 2013. To the nearest thousandth, what will the population of India be in 2031? Solution To estimate the population in 2031, we evaluate the models for t = 18, because 20 |
31 is 18 years after 2013. Rounding to the nearest thousandth, There will be about 1.549 billion people in India in the year 2031. P(18) = 1.25(1.012)18 β 1.549 Try It #3 The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about 0.6%. This situation is represented by the growth function P(t) = 1.39(1.006)t, where t is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 3? Finding Equations of Exponential Functions In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants a and b, and evaluate the function. How Toβ¦ Given two data points, write an exponential model. 1. If one of the data points has the form (0, a), then a is the initial value. Using a, substitute the second point into the equation f (x) = a(b)x, and solve for b. 2. If neither of the data points have the form (0, a), substitute both points into two equations with the form f (x) = a(b)x. Solve the resulting system of two equations in two unknowns to find a and b. 3. Using the a and b found in the steps above, write the exponential function in the form f (x) = a(b)x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 469 Example 4 Writing an Exponential Model When the Initial Value Is Known In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population (N) of deer over time t. Solution We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that |
by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation N(t) = 80bt to find b: N(t) = 80b t 180 = 80b 6 9 __ = b 6 4 9 __ ) b = ( 4 b β 1.1447 1 __ 6 Substitute using point (6, 180). Divide and write in lowest terms. Isolate b using properties of exponents. Round to 4 decimal places. NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section. The exponential model for the population of deer is N(t) = 80(1.1447)t. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.) We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in Figure 3 passes through the initial points given in the problem, (0, 80) and (6, 180). We can also see that the domain for the function is [0, β), and the range for the function is [80, β). N(t) 320 300 280 260 240 220 200 180 160 140 120 100 80 6, 180) (0, 80) 0 1 2 3 7 8 9 10 t 4 6 5 Years Figure 3 Graph showing the population of deer over time, N(t) = 80(1.1447)t, t years after 2006. Try It #4 A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013, the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population N of wolves over time t. Example 5 Writing an Exponential Model When the Initial Value is Not Known Find an exponential function that passes through the points (β2, 6) and (2, 1). Solution Because we donβt have the initial value, we substitute both points into an equation of the form f (x) = ab x, and then solve the system for a and b. β’ Substituting (β2, 6) gives 6 = abβ2 β’ Substituting (2, 1) gives 1 = |
ab2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 47 0 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Use the first equation to solve for a in terms of b: 6 = abβ2 6 _ bβ2 = a a = 6b 2 Substitute a in the second equation, and solve for b: Divide. Use properties of exponents to rewrite the denominator. 1 = ab 2 1 = 6b 2b 2 = 6b 4 1 1 __.6389 Substitute a. Use properties of exponents to isolate b. Round 4 decimal places. Use the value of b in the first equation to solve for the value of a: a = 6b 2 β 6(0.6389)2 β 2.4492 Thus, the equation is f (x) = 2.4492(0.6389)x. We can graph our model to check our work. Notice that the graph in Figure 4 passes through the initial points given in the problem, (β2, 6) and (2, 1). The graph is an example of an exponential decay function. f (x) 10 1 β1 β2 (β2, 6) β5 β4 β3 β2 (2, 1) 1 2 3 4 5 x Figure 4 The graph of f (x) = 2.4492(0.6389)x models exponential decay. Try It #5 Given the two points (1, 3) and (2, 4.5), find the equation of the exponential function that passes through these two points. Q & Aβ¦ Do two points always determine a unique exponential function? Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x, which in many real world cases involves time. How Toβ¦ Given the graph of an exponential function, write its equation. 1. First, identify two points on the graph. Choose the y-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error. 2. If |
one of the data points is the y-intercept (0, a), then a is the initial value. Using a, substitute the second point into the equation f (x) = a(b)x, and solve for b. 3. If neither of the data points have the form (0, a), substitute both points into two equations with the form f (x) = a(b)x. Solve the resulting system of two equations in two unknowns to find a and b. 4. Write the exponential function, f (x) = a(b)x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 471 Example 6 Writing an Exponential Function Given Its Graph Find an equation for the exponential function graphed in Figure 5. f (x) 21 18 15 12 9 6 3 β3.5 β3 β2.5 β2 β1.5 β1 β0.5 β3 0.5 1 1.5 2 2.5 3 3.5 x Figure 5 Solution We can choose the y-intercept of the graph, (0, 3), as our first point. This gives us the initial value, a = 3. Ne xt, choose a point on the curve some distance away from (0, 3) that has integer coordinates. One such point is (2, 12). Substitute the initial value 3 for a. y = ab x Write the general form of an exponential equation. y = 3b x 12 = 3b2 4 = b2 b = Β± 2 Substitute in 12 for y and 2 for x. Take the square root. Divide by 3. Because we restrict ourselves to positive values of b, we will use b = 2. Substitute a and b into the standard form to yield the equation f (x) = 3(2)x. Try It #6 Find an equation for the exponential function graphed in Figure 6. f(x) 5 4 3 2 1 (0, β2 ) β5 β4 β3 β2 β1β1 21 3 4 5 x Figure 6 How Toβ¦ Given two points on the curve of an exponential function, use a graphing calculator to find the equation. 1. Press [STAT]. 2. Clear any existing entries in columns L1 or L2. 3. In L1, enter the x-coordinates given |
. 4. In L2, enter the corresponding y-coordinates. 5. Press [STAT] again. Cursor right to CALC, scroll down to ExpReg (Exponential Regression), and press [ENTER]. 6. The screen displays the values of a and b in the exponential equation y = a β
b x Example 7 Using a Graphing Calculator to Find an Exponential Function Use a graphing calculator to find the exponential equation that includes the points (2, 24.8) and (5, 198.4). Solution Follow the guidelines above. First press [STAT], [EDIT], [1: Editβ¦], and clear the lists L1 and L2. Ne xt, in the L1 column, enter the x-coordinates, 2 and 5. Do the same in the L2 column for the y-coordinates, 24.8 and 198.4. Now press [STAT], [CALC], [0: ExpReg] and press [ENTER]. The values a = 6.2 and b = 2 will be displayed. The exponential equation is y = 6.2 β
2x. Try It #7 Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 481.07). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 47 2 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Applying the Compound-Interest Formula Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account. The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing. We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time t, principal P, APR r, and number of compounding periods in a year n: For example, observe Table 4, which shows the result of investing $1,000 at 10% |
for one year. Notice how the value of the account increases as the compounding frequency increases. r _ A(t) = P ( 1 + ) n nt Frequency Annually Semiannually Quarterly Monthly Daily Value after 1 year $1100 $1102.50 $1103.81 $1104.71 $1105.16 Table 4 the compound interest formula Compound interest can be calculated using the formula A(t nt where β’ A(t) is the account value, β’ t is measured in years, β’ P is the starting amount of the account, often called the principal, or more generally present value, β’ r is the annual percentage rate (APR) expressed as a decimal, and β’ n is the number of compounding periods in one year. Example 8 Calculating Compound Interest If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years? Solution Because we are starting with $3,000, P = 3000. Our interest rate is 3%, so r = 0.03. Because we are compounding quarterly, we are compounding 4 times per year, so n = 4. We want to know the value of the account in 10 years, so we are looking for A(10), the value when t = 10. r _ A(t) = P ( 1 + ) n nt Use the compound interest formula. A(10) = 3000 ( 1 + β $4,045.05 4 β
10 0.03 ____ ) 4 Substitute using given values. Round to two decimal places. The account will be worth about $4,045.05 in 10 years. Try It #8 An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 473 Example 9 Using the Compound Interest Formula to Solve for the Principal A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a childβs future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to $40,000 over 18 years. She believes the account will earn 6% compounded semi- |
annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now? Solution The nominal interest rate is 6%, so r = 0.06. Interest is compounded twice a year, so n = 2. We want to find the initial investment, P, needed so that the value of the account will be worth $40,000 in 18 years. Substitute the given values into the compound interest formula, and solve for P. nt r _ A(t) = P ( 1 + ) n 0.06 ____ ) 40,000 = P ( 1 + 2 2(18) Use the compound interest formula. Substitute using given values A, r, n, and t. 40,000 = P(1.03)36 40,000 ______ (1.03)36 = P Simplify. Isolate P. Lily will need to invest $13,801 to have $40,000 in 18 years. P β $13, 801 Divide and round to the nearest dollar. Try It #9 Refer to Example 9. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly? Evaluating Functions with Base e As we saw earlier, the amount earned on an account increases as the compounding frequency increases. Table 5 shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue. Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies, listed in Table 5. Frequency Annually Semiannually Quarterly Monthly Daily Hourly Once per minute Once per second 1 _ n ) A(t + 12 1 ) _ 12 ( 1 + 365 1 ) _ 365 ( 1 + 1 ) _ 8766 8766 ( 1 + 1 ) _ 525960 525960 ( 1 + 1 ) ________ 31557600 31557600 Table 5 Value $2 $2.25 $2.441406 $2.613035 $2.714567 $2.718127 $2.718279 $2.718282 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 47 4 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS These values appear to be approaching a limit as n increases without bound. In fact, as n |
gets larger and larger, the 1 expression ( 1 + _ n ) approaches a number used so frequently in mathematics that it has its own name: the letter e. This value is an irrational number, which means that its decimal expansion goes on forever without repeating. Its approximation to six decimal places is shown below. n the number e The letter e represents the irrational number ( 1 + 1 __ ) n n, as n increases without bound The letter e is used as a base for many real-world exponential models. To work with base e, we use the approximation, e β 2.718282. The constant was named by the Swiss mathematician Leonhard Euler (1707β1783) who first investigated and discovered many of its properties. Example 10 Using a Calculator to Find Powers of e Calculate e3.14. Round to five decimal places. Solution On a calculator, press the button labeled [e x]. The window shows [e^(]. Type 3.14 and then close parenthesis, [)]. Press [ENTER]. Rounding to 5 decimal places, e 3.14 β 23.10387. Caution: Many scientific calculators have an βExpβ button, which is used to enter numbers in scientific notation. It is not used to find powers of e. Try It #10 Use a calculator to find e β0.5. Round to five decimal places. Investigating Continuous Growth So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, e is used as the base for exponential functions. Exponential models that use e as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics. the continuous growth/decay formula For all real numbers t, and all positive numbers a and r, continuous growth or decay is represented by the formula where β’ a is the initial value, β’ r is the continuous growth rate per unit time, A(t) = aert β’ and t is the elapsed time. If r > 0, then the formula represents continuous growth. If r < 0, then the formula represents continuous decay. For business applications, the continuous growth formula is called the continuous compounding formula and takes the form A(t) = Pert where β’ P is the principal or the initial invested, β’ r is the growth or interest rate per unit time, β’ and t is the period or term |
of the investment. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 EXPONENTIAL FUNCTIONS 475 How Toβ¦ Given the initial value, rate of growth or decay, and time t, solve a continuous growth or decay function. 1. Use the information in the problem to determine a, the initial value of the function. 2. Use the information in the problem to determine the growth rate r. a. If the problem refers to continuous growth, then r > 0. b. If the problem refers to continuous decay, then r < 0. 3. Use the information in the problem to determine the time t. 4. Substitute the given information into the continuous growth formula and solve for A(t). Example 11 Calculating Continuous Growth A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year? Solution Since the account is growing in value, this is a continuous compounding problem with growth rate r = 0.10. The initial investment was $1,000, so P = 1000. We use the continuous compounding formula to find the value after t = 1 year: A(t) = Pert Use the continuous compounding formula. = 1000(e)0.1 Substitute known values for P, r, and t. β 1105.17 Use a calculator to approximate. The account is worth $1,105.17 after one year. Try It #11 A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years? Example 12 Calculating Continuous Decay Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days? Solution Since the substance is decaying, the rate, 17.3%, is negative. So, r = β0.173. The initial amount of radon222 was 100 mg, so a = 100. We use the continuous decay formula to find the value after t = 3 days: A(t) = aert Use the continuous growth formula. = 100eβ0.173(3) β 59.5115 Substitute known values for a, r, and t. Use a calculator to approximate. So 59.5115 mg of radon-222 will remain. |
Try It #12 Using the data in Example 12, how much radon-222 will remain after one year? Access these online resources for additional instruction and practice with exponential functions. β’ Exponential Growth Function (http://openstaxcollege.org/l/expgrowth) β’ Compound Interest (http://openstaxcollege.org/l/compoundint) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 47 6 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6.1 SECTION EXERCISES Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.1 SECTION EXERCISES 477 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 47 8 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS 479 G C F.. In this section, you will: β’ efine an eponential function and determine its domain. β’ Graph an eponential function including using transformations. β’ ntroduce. β’ rite a function that results from a gien transformation. 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS As we discussed in the previous section, exponential functions are used for many real-world applications such as finance, forensics, computer science, and most of the life sciences. Working with an equation that describes a realworld situation gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about things by seeing their pictorial representations, and that is exactly why graphing exponential equations is a powerful tool. It gives us another layer of insight for predicting future events. Graphing Exponential Functions Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form f (x) = b x whose base is greater than one. Weβll use the function f (x) = 2x. Observe how the output values in Table 1 change as the input increases by 1. x f (x) = 2x β3 1 _ 8 β |
2 1 _ 4 0 1 β1 1_ 2 Table 1 1 2 2 4 3 8 Each output value is the product of the previous output and the base, 2. We call the base 2 the constant ratio. In fact, for any exponential function with the form f (x) = ab x, b is the constant ratio of the function. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of a. Notice from the table that β’ the output values are positive for all values of x; β’ as x increases, the output values increase without bound; and β’ as x decreases, the output values grow smaller, approaching zero. Figure 1 shows the exponential growth function f (x) = 2x. f (x) 10 1, 1 2 1 4 β2, β3, 1 8 β5 β4 β3 β2 β β1 1 (3, 8) f (x) = 2x (2, 4) (1, 2) (0, 1) 21 3 4 5 x The x-axis is an asymptote. Figure 1 Notice that the graph gets close to the x-axis, but never touches it. The domain of f (x) = 2x is all real numbers, the range is (0, β), and the horizontal asymptote is y = 0. To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form f (x) = b x whose 1 ) base is between zero and one. Weβll use the function g(x) = ( _. Observe how the output values in Table 2 change as 2 the input increases by 1. x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 48 0 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS x g(x) = ( 1 __ 2 ) x β3 8 β2 4 0 1 β1 2 Table 2 1 1_ 2 2 1 _ 4 3 1 _ 8 Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or 1 _. constant ratio 2 Notice from the table that β’ the output values are positive for all values of x; β’ as x increases, the output values grow smaller, approaching zero; and β’ as x decreases |
, the output values grow without bound. x 1 ) Figure 2 shows the exponential decay function, g(x) = ( _. 2 g(x) x g(x) = 1 2 (β3, 8) (β2, 4) (β1, 2) (0, 1) 10 21, 1 42, 1 83, β5 β4 β3 β2 β1 21 3 4 5 x The x-axis is an asymptote. Figure 2 1 ) The domain of g(x) = ( _ 2 x is all real numbers, the range is (0, β), and the horizontal asymptote is y = 0. characteristics of the graph of the parent function f (x) = b x An exponential function with the form f (x) = b x, b > 0, b β 1, has these characteristics: β’ one-to-one function β’ horizontal asymptote: y = 0 β’ domain: (ββ, β) β’ range: (0, β) β’ x-intercept: none β’ y-intercept: (0, 1) β’ increasing if b > 1 β’ decreasing if b < 1 Figure 3 compares the graphs of exponential growth and decay functions. f(x) f(x) f (x) = bx b > 1 f (x) = bx 0 < b < 1 (1, b) (0, 1) x Figure 3 (0, 1) (1, b) x How Toβ¦ Given an exponential function of the form f (x) = b x, graph the function. 1. Create a table of points. 2. Plot at least 3 point from the table, including the y-intercept (0, 1). 3. Draw a smooth curve through the points. 4. State the domain, (ββ, β), the range, (0, β), and the horizontal asymptote, y = 0. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS 481 Example 1 Sketching the Graph of an Exponential Function of the Form f (x) = b x Sketch a graph of f (x) = 0.25x. State the domain, range, and asymptote. Solution Before graphing, identify the behavior and create a table of |
points for the graph. β’ Since b = 0.25 is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote y = 0. β’ Create a table of points as in Table 3. x f (x) = 0.25x β3 64 β2 16 β1 4 Table 3 0 1 1 2 3 0.25 0.0625 0.015625 β’ Plot the y-intercept, (0, 1), along with two other points. We can use (β1, 4) and (1, 0.25). Draw a smooth curve connecting the points as in Figure 4. f(x) f(x) = 0.25x (β1, 4) 6 5 4 3 2 1 β5 β4 β3 β2 β1β1 β2 β3 β4 β5 β6 (0, 1) (1, 0.25) 21 3 4 5 x The domain is (ββ, β); the range is (0, β); the horizontal asymptote is y = 0. Figure 4 Try It #1 Sketch the graph of f (x) = 4x. State the domain, range, and asymptote. Graphing Transformations of Exponential Functions Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformationsβshifts, reflections, stretches, and compressionsβto the parent function f (x) = b x without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied. Graphing a Vertical Shift The first transformation occurs when we add a constant d to the parent function f (x) = b x, giving us a vertical shift d units in the same direction as the sign. For example, if we begin by graphing a parent function, f (x) = 2x, we can then graph two vertical shifts alongside it, using d = 3: the upward shift, g(x) = 2x + 3 and the downward shift, h(x) = 2x β 3. Both vertical shifts are shown in Figure 5. y 12 10 8 6 4 2 β1β 2 β4 β6 g(x) = 2 |
x + 3 f (x) = 2 x h(x) = 2 x β 3 β6 β5 β4 β3 β2 21 3 Figure 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 48 2 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Observe the results of shifting f (x) = 2x vertically: β’ The domain, (ββ, β) remains unchanged. β’ When the function is shifted up 3 units to g(x) = 2x + 3: β¦ The y-intercept shifts up 3 units to (0, 4). β¦ The asymptote shifts up 3 units to y = 3. β¦ The range becomes (3, β). β’ When the function is shifted down 3 units to h(x) = 2x β 3: β¦ The y-intercept shifts down 3 units to (0, β2). β¦ The asymptote also shifts down 3 units to y = β3. β¦ The range becomes (β3, β). Graphing a Horizontal Shift The next transformation occurs when we add a constant c to the input of the parent function f (x) = b x, giving us a horizontal shift c units in the opposite direction of the sign. For example, if we begin by graphing the parent function f (x) = 2x, we can then graph two horizontal shifts alongside it, using c = 3: the shift left, g(x) = 2x + 3, and the shift right, h (x) = 2x β 3. Both horizontal shifts are shown in Figure 6. y 10 8 6 4 2 g(x) = 2 x + 3 f (x) = 2 x h(x 10 β10 β8 β6 β4 β2β2 42 6 8 Figure 6 Observe the results of shifting f (x) = 2x horizontally: β’ The domain, (ββ, β), remains unchanged. β’ The asymptote, y = 0, remains unchanged. β’ The y-intercept shifts such that: β¦ When the function is shifted left 3 units to g(x) = 2x + 3, the y-intercept becomes (0, 8). This is because 2x + 3 = (8)2x, so the initial value of the function is 8. 1 1 β¦ When |
the function is shifted right 3 units to h(x) = 2x β 3, the y-intercept becomes ( 0, ) 2x, ). Again, see that 2x β 3 = ( _ _ 8 8 1 _. so the initial value of the function is 8 shifts of the parent function f (x) = b x For any constants c and d, the function f (x) = b x + c + d shifts the parent function f (x) = b x β’ vertically d units, in the same direction of the sign of d. β’ horizontally c units, in the opposite direction of the sign of c. β’ The y-intercept becomes (0, bc + d). β’ The horizontal asymptote becomes y = d. β’ The range becomes (d, β). β’ The domain, (ββ, β), remains unchanged. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS 483 How Toβ¦ Given an exponential function with the form f (x) = b x + c + d, graph the translation. 1. Draw the horizontal asymptote y = d. 2. Identify the shift as (βc, d). Shift the graph of f (x) = b x left c units if c is positive, and right c units if c is negative. 3. Shift the graph of f (x) = b x up d units if d is positive, and down d units if d is negative. 4. State the domain, (ββ, β), the range, (d, β), and the horizontal asymptote y = d. Example 2 Graphing a Shift of an Exponential Function Graph f (x) = 2x + 1 β 3. State the domain, range, and asymptote. Solution We have an exponential equation of the form f (x) = b x + c + d, with b = 2, c = 1, and d = β3. Draw the horizontal asymptote y = d, so draw y = β3. Identify the shift as (βc, d), so the shift is (β1, β3). Shift the graph of f (x) = b x left 1 units and down 3 units. f (x) f (x) = 2 x + 1 β 3 |
10 8 6 4 2 (0, β1) β6 β5 β4 β3 β2 (β1, β2) β1β2 β4 β6 β8 β10 (1, 1) 3 21 4 5 6 x y = β3 The domain is (ββ, β); the range is (β3, β); the horizontal asymptote is y = β3. Figure 7 Try It #2 Graph f (x) = 2x β 1 + 3. State domain, range, and asymptote. How Toβ¦ Given an equation of the form f (x) = b x + c + d for x, use a graphing calculator to approximate the solution. 1. Press [Y=]. Enter the given exponential equation in the line headed βY1=β. 2. Enter the given value for f (x) in the line headed βY2=β. 3. Press [WINDOW]. Adjust the y-axis so that it includes the value entered for βY2=β. 4. Press [GRAPH] to observe the graph of the exponential function along with the line for the specified value of f (x). 5. To fi nd the value of x, we compute the point of intersection. Press [2ND] then [CALC]. Select βintersectβ and press [ENTER] three times. The point of intersection gives the value of x for the indicated value of the function. Example 3 Approximating the Solution of an Exponential Equation Solve 42 = 1.2(5)x + 2.8 graphically. Round to the nearest thousandth. Solution Press [Y=] and enter 1.2(5)x + 2.8 next to Y1=. Then enter 42 next to Y2=. For a window, use the values β3 to 3 for x and β5 to 55 for y. Press [GRAPH]. The graphs should intersect somewhere near x = 2. For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a different window or use a different value for Guess?) To the nearest thousandth, x β 2.166. Download the OpenStax text for free |
at http://cnx.org/content/col11759/latest. 48 4 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Try It #3 Solve 4 = 7.85(1.15)x β 2.27 graphically. Round to the nearest thousandth. Graphing a Stretch or Compression While horizontal and vertical shifts involve adding constants to the input or to the function itself, a stretch or compression occurs when we multiply the parent function f (x) = b x by a constant β£ a β£ > 0. For example, if we begin by graphing the parent function f (x) = 2x, we can then graph the stretch, using a = 3, to get g(x) = 3(2)x as shown on the 1 1 _ _, to get h(x) = left in Figure 8, and the compression, using a = (2)x as shown on the right in Figure 8. 3 3 Vertical stretch y g(x) = 3(2)x f (x) = 2x Vertical compression y f (x) = 2x h(x) = (2)x 1 3 10 8 6 4 2 β1β2 β4 (a) β5 β4 β3 β2 21 3 4 5 y = 0 x β5 β4 β3 β2 10 8 6 4 2 β1β2 β4 (b) 21 3 4 5 x y = 0 Figure 8 (a) g(x) = 3(2)x stretches the graph of f (x) = 2x vertically by a factor of 3. 1 1 __ __ (2)x compresses the graph of f (x) = 2x vertically by a factor of. (b) h(x) = 3 3 stretches and compressions of the parent function f ( x ) = b x For any factor a > 0, the function f (x) = a(b)x β’ is stretched vertically by a factor of a if β£ a β£ > 1. β’ is compressed vertically by a factor of a if β£ a β£ < 1. β’ has a y-intercept of (0, a). β’ has a horizontal asymptote at y = 0, a range of (0, β), and a domain of (ββ, β), which are unchanged from the parent function. Example 4 Graphing the Stretch of an Exponential Function |
1 ) Sketch a graph of f (x) = 4 ( _ 2 x. State the domain, range, and asymptote. Solution Before graphing, identify the behavior and key points on the graph. 1 _ β’ Since b = is between zero and one, the left tail of the graph will increase without bound as x decreases, and the 2 right tail will approach the x-axis as x increases. β’ Since a = 4, the graph of f (x) = ( 1 ) _ 2 β’ Create a table of points as shown in Table 4. x will be stretched by a factor of 4. x f (x) = 4 ( 1 __ 2 ) x β3 32 β2 16 β1 8 Table 4 0 4 1 2 2 1 3 0.5 β’ Plot the y-intercept, (0, 4), along with two other points. We can use (β1, 8) and (1, 2). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 6.2 GRAPHS OF EXPONENTIAL FUNCTIONS 485 Draw a smooth curve connecting the points, as shown in Figure 9. (β1, 8) f(x) 10 8 6 4 2 (0, 4) (1, 2) β6 β5 β4 β3 β2 β1β2 β4 321 4 5 6 Figure 9 y = 0 x f (x) = 4 1 2 x The domain is (ββ, β); the range is (0, β); the horizontal asymptote is y = 0. Try It #4 1 _ Sketch the graph of f (x) = (4)x. State the domain, range, and asymptote. 2 Graphing Reο¬ections In addition to shifting, compressing, and stretching a graph, we can also reflect it about the x-axis or the y-axis. When we multiply the parent function f (x) = b x by β1, we get a reflection about the x-axis. When we multiply the input by β1, we get a reflection about the y-axis. For example, if we begin by graphing the parent function f (x) = 2x, we can then graph the two reflections alongside it. The reflection about the x-axis, g(x) = β2x, is shown on the left side of Figure |
10, and the reflection about the y-axis h(x) = 2βx, is shown on the right side of Figure 10. Reflection about the x-axis y Reflection about the y-axis y β5 β4 β3 β2 10 8 6 4 2 β β2 1 β4 β6 β8 β10 f (x) = 2 x h(x) = 2 βx y = 0 21 3 4 5 x β5 β4 β3 β2 g(x) = β2 x f (x) = 2 x y = 0 21 3 4 5 x 10 8 6 4 2 β1β2 β4 β6 β8 β10 Figure 10 (a) g(x) = β2x reο¬ects the graph of f (x) = 2x about the x-axis. (b) g(x) = 2βx reο¬ects the graph of f (x) = 2x about the y-axis. reflections of the parent function f (x) = b x The function f (x) = βb x β’ reflects the parent function f (x) = b x about the x-axis. β’ has a y-intercept of (0, β1). β’ has a range of (ββ, 0). β’ has a horizontal asymptote at y = 0 and domain of (ββ, β), which are unchanged from the parent function. The function f (x) = bβx β’ reflects the parent function f (x) = b x about the y-axis. β’ has a y-intercept of (0, 1), a horizontal asymptote at y = 0, a range of (0, β), and a domain of (ββ, β), which are unchanged from the parent function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 48 6 CHAPTER 6 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 5 Writing and Graphing the Reο¬ection of an Exponential Function 1 ) Find and graph the equation for a function, g (x), that reflects f (x) = ( _ 4 and asymptote. 1 Solution Since we want to reflect the parent function f (x) = ( ) _ 4 g (x) = β ( 1 ) _ 4. Ne xt we create a |
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