Polynomial and Rational Functions 79. P() 9 9 8. P() 6 6 8 7 8 8. The solutions to the equation are all the cube roots of. (A) How many cube roots of are there? (B) is obviously a cube root of ; find all others. 8. The solutions to the equation 8 are all the cube roots of 8. (A) How many cube roots of 8 are there? (B) is obviously a cube root of 8; find all others. 8. If P is a polynomial function with real coefficients of degree n, with n odd, then what is the maimum number of times the graph of y P() can cross the ais? What is the minimum number of times? 8. Answer the questions in Problem 8 for n even. 8. Given P() i with i a zero, show that i is not a zero of P(). Does this contradict Theorem? Eplain. 86. If P() and Q() are two polynomials of degree n, and if P() Q() for more than n values of, then how are P() and Q() related? APPLICATIONS Find all rational solutions eactly, and find irrational solutions to two decimal places. 87. Storage. A rectangular storage unit has dimensions by by feet. If each dimension is increased by the same amount, how much should this amount be to create a new storage unit with volume ten times the old? 88. Construction. A rectangular bo has dimensions by by feet. If each dimension is increased by the same amount, how much should this amount be to create a new bo with volume si times the old? 89. Packaging. An open bo is to be made from a rectangular piece of cardboard that measures 8 by inches, by cutting out squares of the same size from each corner and bending up the sides (see the figure). If the volume of the bo is to be cubic inches, how large a square should be cut from each corner? [Hint: Determine the domain of from physical considerations before starting.] 9. Fabrication. An open metal chemical tank is to be made from a rectangular piece of stainless steel that measures by 8 feet, by cutting out squares of the same size from each corner and bending up the sides (see the figure). If the volume of the tank is to be 8 cubic feet, how large a square should be cut from each corner? SECTION - Approimating Real Zeros of Polynomials Locating Real Zeros The Bisection Method Approimating Real Zeros Using a Graphing Utility Application The strategy for finding zeros discussed in the preceding section is designed to find as many eact real and imaginary zeros as possible. But there are zeros that cannot be found by using the strategy. For eample, the polynomial P() must have at least one real zero (Theorem in Section -). Since the only possible rational zeros are and neither of these turns out to be a zero, P() must have at least one irrational zero. We cannot find the eact value of this zero, but it can be approimated using various well-known methods.
- Approimating Real Zeros of Polynomials In this section we will develop two important tools for locating real zeros, the location theorem and the upper and lower bound theorem. Net we will discuss how the location theorem forms the basis for the method of bisection, a popular method that is used by most graphing utilities to approimate real zeros. Finally, we will see how the upper and lower bound theorem can aid in approimating real zeros with a graphing utility. We will restrict our attention to the real zeros of polynomials with real coefficients. Locating Real Zeros Let us return to the polynomial function P() P() As we found above, P() has no rational zeros and at least one irrational zero. The graph of P() is shown in Figure. Note that P() and P(). Since the graph of a polynomial function is continuous, the graph of P() must cross the ais at least once between and. This observation is the basis for Theorem and leads to an effective method for locating zeros. FIGURE P(). Theorem Location Theorem If f is continuous on an interval I, a and b are two numbers in I, and f(a) and f(b) are of opposite sign, then there is at least one intercept between a and b. We will find Theorem very useful when we are searching for real zeros, hence the name location theorem. It is important to remember that at least in Theorem means one or more. Notice in Figure (a) that f(), f(), and f has one zero between and. In Figure (b), f() and f(), but this time there are three zeros between and. FIGURE The location theorem. f() f() f() (a) f () (b) f() (c) f() The converse to the location theorem (Theorem ) is false; that is, if c is a zero of f, then f may or may not change sign at c. Compare Figure (a) and (c). Both functions have a zero at, but the first changes sign at and the second does not.
Polynomial and Rational Functions EXAMPLE Locating Real Zeros Let P() 6 9. Use a synthetic division table to locate the zeros of P() between successive integers. Solution We construct a synthetic division table and look for sign changes. 6 9 6 9 addbddcaddbddc Sign change Sign change addbddc Sign change According to Theorem, P() must have a real zero in each of the intervals (, ), (, ), and (, ). Since P() is a cubic polynomial, we have located all its zeros. Matched Problem Let P() 8. Use a synthetic division table to locate the zeros of P() between successive integers. In the solution to Eample, we located three zeros in a relatively few number of steps and could stop searching because we knew that a cubic polynomial could not have more than three zeros. But what if we had not found three zeros? Some cubic polynomials have only one real zero. How can we tell when we have searched far enough? The net theorem tells us how to find upper and lower bounds for the real zeros of a polynomial. Any number that is greater than or equal to the largest zero of a polynomial is called an upper bound of the zeros of the polynomial. Similarly, any number that is less than or equal to the smallest zero of the polynomial is called a lower bound of the zeros of the polynomial. Theorem, based on the synthetic division process, enables us to determine upper and lower bounds of the real zeros of a polynomial with real coefficients. Theorem Upper and Lower Bounds of Real Zeros Given an nth-degree polynomial P() with real coefficients, n, a n, and P() divided by r using synthetic division:. Upper Bound. If r and all numbers in the quotient row of the synthetic division, including the remainder, are nonnegative, then r is an upper bound of the real zeros of P().. Lower Bound. If r and all numbers in the quotient row of the synthetic division, including the remainder, alternate in sign, then r is a lower bound of the real zeros of P().
addbddc addbddc - Approimating Real Zeros of Polynomials [Note: In the lower-bound test, if appears in one or more places in the quotient row, including the remainder, the sign in front of it can be considered either positive or negative, but not both. For eample, the numbers,, can be considered to alternate in sign, while,, cannot.] We sketch a proof of part of Theorem. The proof of part is similar, only a little more difficult. Proof If all the numbers in the quotient row of the synthetic division are nonnegative after dividing P() by r, then P() ( r)q() R where the coefficients of Q() are nonnegative and R is nonnegative. If r, then r and Q() ; hence, P() ( r)q() R Thus, P() cannot be for any greater than r, and r is an upper bound for the real zeros of P(). EXAMPLE Bounding Real Zeros Let P() 9. Find the smallest positive integer and the largest negative integer that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). Also note the location of any zeros discovered in the process of building the synthetic division table. Solution An easy way to locate the upper and lower bounds is to test r,,,...until the quotient row turns nonnegative; then test r,,,...until the quotient row alternates in sign. It is also useful to include r in the table to detect any sign changes between r and r. 9 9 9 6 7 9 UB 7 6 7 8 7 This quotient row is nonnegative; hence, is an upper bound (UB). 78 6 LB 6 7 6 8 6 This quotient row alternates in sign; hence, is a lower bound (LB).
Polynomial and Rational Functions Because of Theorem, we conclude that all real zeros of P() 9 must lie between and. We also note that there must be at least one zero in (, ) and at least one in (, ). Matched Problem Let P() 7. Find the smallest positive integer and the largest negative integer that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). Also note the location of any zeros discovered in the process of building the synthetic division table. The Bisection Method Now that we know how to locate real zeros of a polynomial, we turn to the problem of actually approimating a real zero. Eplore-Discuss provides an introduction to the repeated systematic application of the location theorem (Theorem ) called the bisection method. This is the method for approimating real zeros that is programmed into many graphing utilities. EXPLORE-DISCUSS Let P(). Since P() and P(), the location theorem implies that P() must have at least one zero in (, ). (A) Is P(.) positive or negative? Is there a zero in (,.) or in (., )? (B) Let m be the midpoint of the interval from part A that contains a zero. Is P(m) positive or negative? What does this tell you about the location of the zero? (C) Eplain how this process could be used repeatedly to approimate a zero to any desired accuracy. The bisection method used to approimate real zeros is straightforward: Let P() be a polynomial with real coefficients. If P() has opposite signs at the endpoints of the interval (a, b), then a real zero r lies in this interval. We bisect this interval [find the midpoint m (a b)/], check the sign of P(m), and choose the interval (a, m) or (m, b) on which P() has opposite signs at the endpoints. We repeat this bisecting process (producing a set of nested intervals, each half the size of the preceding one and each containing the real zero r) until we get the desired decimal accuracy for the zero approimation. At any point in the process if P(m), we stop, since m is a real zero. An eample will help clarify the process. EXAMPLE Approimating Real Zeros by Bisection For the polynomial P() 9 in Eample, we found that all the real zeros lie between and and that each of the intervals (, ) and (, ) contained at least one zero. Use bisection to approimate a real zero on the interval (, ) to one decimal place accuracy.
- Approimating Real Zeros of Polynomials Solution We start the process with a synthetic division table: 9 7 9 P() 8 P() TABLE Bisection Approimation Sign change interval Midpoint Sign of P (a, b) m P(a) P(m) P(b) (, ). (., ).7 (.,.7).6 (.,.6).6 (.6,.6) We stop here Since the sign of P() changes at the endpoints of the interval (.6,.6), we conclude that a real zero lies on this interval and is given by r.6 to one decimal place accuracy (each endpoint rounds to.6). Figure illustrates the nested intervals produced by the bisection method in Table. Match each step in Table with an interval in Figure. Note how each interval that contains a zero gets smaller and smaller and is contained in the preceding interval that contained the zero. FIGURE Nested intervals produced by the bisection method in Table..6.6 ( (( ) ) )..7 If we had wanted two decimal place accuracy, we would use four decimal places for the values of and continue the process in Table until the endpoints of a sign change interval rounded to the same two decimal place number. Matched Problem Use the bisection method to approimate to one decimal place accuracy a zero on the interval (, ) for the polynomial in Eample. Approimating Real Zeros Using a Graphing Utility The bisection method is easy to understand but tedious to carry out, especially if the approimation must be accurate to more than two decimal places. Fortunately, this is the type of repetitive calculation that a graphing utility can be programmed to carry out. In fact, we have been using a graphing utility for some time now to find the zeros of a function (see Section in Graphs and Functions ). Now we will see how the upper and lower bound theorem can be used in conjunction with the zero approimation routine on a graphing utility to approimate all the real zeros of a polynomial.
6 Polynomial and Rational Functions EXAMPLE Approimating Real Zeros Using a Graphing Utility Given the polynomial P() : (A) Form a synthetic division table to find upper and lower bounds for any real zeros, and locate real zeros between successive integers. (B) Graph P() in a graphing utility, and approimate any real zeros to four decimal places using a built-in zero approimation routine. Solution (A) Form a synthetic division table: UB afdbfdc Real zero LB From the table we see that all real zeros of P() are between and, and a real zero lies on the interval (, ). (B) Enter P() in a graphing utility, and set the window dimensions with the synthetic division table in part A as a guide. Figure (a) shows the graph of P(), and Figure (b) shows the zero approimation using a built-in routine. FIGURE (a) (b) It is clear from the graph and the upper and lower bounds of the zeros found in part A that P() has only one real zero, which is, to four decimal places,.79. Matched Problem Given the polynomial P() : (A) Form a synthetic division table to find upper and lower bounds for any real zeros, and locate real zeros between successive integers. (B) Graph P() in a graphing utility and approimate any real zeros to four decimal places using a built-in zero approimation routine. Earlier in this section and in Section. we saw that a calculator is a useful tool for constructing a synthetic division table. A graphing utility that can store and eecute programs is even more useful. Table shows a simple program for a graphing calculator that will perform synthetic division. The table also shows the output generated when we use this program to construct the synthetic division table in Eample.
- Approimating Real Zeros of Polynomials 7 TABLE Program SYNDIV* TI-8/TI-8 TI-8/TI-86 Output Lbl A Prompt R I dim(l) N {} L N dim(l) L() L() Lbl B L(I-)*R+L(I) L(I) +I I If I N Goto B Pause L Goto A Synthetic Division on a Graphing Utility Lbl A Prompt R I diml L N {} L N diml L L() L() Lbl B L(I-)*R+L(I) L(I) +I I If I N Goto B Pause L Goto A *Available for download at www.mhhe.com/barnett. EXPLORE-DISCUSS If you have a TI-8, TI-8, TI-8, or TI-86 graphing calculator, enter the appropriate version of SYNDIV in your calculator eactly as shown in Table. To use the program, store the coefficients of the polynomial in L (see the first line of output in Table ) and eecute the program. Press ENTER to continue after each line is displayed. Press QUIT at the R? prompt to terminate the program. If you have some other graphing utility that can store and eecute programs, consult your manual and modify the statements in SYNDIV so that the program works on your graphing utility. EXAMPLE Approimating Real Zeros with a Graphing Utility Let P() 7 7: (A) Find the smallest positive integer multiple of and the largest negative integer multiple of that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). (B) Use a graphing utility to approimate the real zeros of P() to two decimal places. Solution (A) We construct a synthetic division table to search for bounds for the zeros of P(). The size of the coefficients in P() indicates that we can speed up this search by choosing larger increments between test values.
8 Polynomial and Rational Functions 7 7 7 7 78 UB 7 7, LB 67 7,7 Thus, all real zeros of P() 7 7 must lie between and. (B) Graphing P() for (Fig. ) shows that P() has three zeros. The approimate values of these zeros (details omitted) are.8,.8, and.. FIGURE P() 7 7. Matched Problem Let P() 7 7. (A) Find the smallest positive integer multiple of and the largest negative integer multiple of that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). (B) Use a graphing utility to approimate the real zeros of P() to two decimal places. Remark: One of the most frequently asked questions concerning graphing utilities is how to determine the correct viewing window. The upper and lower bound theorem provides an answer to this question for polynomial functions. As Eample illustrates, the upper and lower bound theorem and the zero approimation routine on a graphing utility are two important mathematical tools that work very well together. Application EXAMPLE 6 Construction An oil tank is in the shape of a right circular cylinder with a hemisphere at each end (see Fig. 6). The cylinder is inches long, and the volume of the tank is, cubic inches (approimately cubic feet). Let denote the common radius of the hemispheres and the cylinder.
- Approimating Real Zeros of Polynomials 9 (A) Find a polynomial equation that must satisfy. (B) Approimate to one decimal place. FIGURE 6 inches Solution (A) If is the common radius of the hemispheres and the cylinder in inches, then Volume of Volume of two Volume of tank hemispheres cylinder, Multiply by /., 6 6, Thus, must be a positive zero of P() 6, (B) Since the coefficients of P() are large, we use larger increments in the synthetic division table: 6,,, UB,9 6, Graphing y P() for (Fig. 7), we see that. inches (to one decimal place). [If you do not have a graphing utility, construct a table like Table to approimate the zero of P().] FIGURE 7 P() 6,. 7, 7, Matched Problem 6 Repeat Eample 6 if the volume of the tank is, cubic inches.
Polynomial and Rational Functions Answers to Matched Problems. Intervals containing zeros: (, ), (, ), (, 6). Lower bound: ; Upper bound: 6 Intervals containing zeros: (, ), (, )... (A) Lower bound: ; Upper bound: Intervals containing zeros: (, ) (B) Real zero:.887. (A) Lower bound: ; Upper bound: (B) Real zeros:.,.6,. 6. (A) P() 6, (B).7 in. EXERCISE - A In Problems, use the table of values for the polynomial function P to discuss the possible locations of the intercepts of the graph of y P()..... P() P() P() P() 7 9 8 6 7 In Problems 8, use a synthetic division table and Theorem to locate each real zero between successive integers.. P() 9 6. P() 9 7. P() 8. P() 6 6 9 6 8 7 6 Find the smallest positive integer and largest negative integer that, by Theorem, are upper and lower bounds, respectively, for the real zeros of each of the polynomials given in Problems 9. 9. P(). P(). P() 9. P() 6 7. P(). P() B In Problems : (A) Find the smallest positive integer and largest negative integer that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). Also note the location of any zeros between successive integers. (B) Approimate to one decimal place the largest real zero of P() using the bisection method.. P() 6. P() 7. P() 8. P() 9. P() 7 9 7. P() 9 9. P(). P()
- Approimating Real Zeros of Polynomials In Problems : (A) Find the smallest positive integer and largest negative integer that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). (B) Approimate the real zeros of each polynomial to two decimal places.. P() 8. P(). P() 7 6. P() 8 7. P() 8. P() 9. P(). P() 6 9 C In Problems : (A) Find the smallest positive integer and largest negative integer that, by Theorem, are upper and lower bounds, respectively, for the real zeros of P(). Also note the location of any zeros between successive integers. (B) Approimate to two decimal places the largest real zero of P() using the bisection method.. P(). P() 7 8. P() 9. P() 9 In Problems : (A) Find the smallest positive integer multiple of and largest negative integer multiple of that, by Theorem, are upper and lower bounds, respectively, for the real zeros of each polynomial. (B) Approimate the real zeros of each polynomial to two decimal places.. P() 6. P() 7 7 7. P() 9, 8. P() 7, 9. P(),,. P() 7,. P(),7 7,87,. P() 9,8,67 8,. P().. 9,. P()..7 8.77,6, APPLICATIONS Epress the solutions to Problems as the roots of a polynomial equation of the form P() and approimate these solutions to one decimal place. Use a graphing utility, if available; otherwise, use the bisection method.. Geometry. Find all points on the graph of y that are unit away from the point (, ). [Hint: Use the distancebetween-two-points formula from Section -.] 6. Geometry. Find all points on the graph of y that are unit away from the point (, ). 7. Manufacturing. A bo is to be made out of a piece of cardboard that measures 8 by inches. Squares, inches on a side, will be cut from each corner, and then the ends and sides will be folded up (see the figure). Find the value of that would result in a bo with a volume of 6 cubic inches. 8. Manufacturing. A bo with a hinged lid is to be made out of a piece of cardboard that measures by inches. Si squares, inches on a side, will be cut from each corner and the middle, and then the ends and sides will be folded up to form the bo and its lid (see the figure). Find the value of that would result in a bo with a volume of cubic inches. in. 8 in. in. in. 9. Construction. A propane gas tank is in the shape of a right circular cylinder with a hemisphere at each end (see the figure). If the overall length of the tank is feet and the volume is cubic feet, find the common radius of the hemispheres and the cylinder. feet
Polynomial and Rational Functions. Shipping. A shipping bo is reinforced with steel bands in all three directions (see the figure). A total of. feet of steel tape is to be used, with 6 inches of waste because of a -inch overlap in each direction. If the bo has a square base and a volume of cubic feet, find its dimensions. y SECTION - Rational Functions Rational Functions Vertical and Horizontal Asymptotes Graphing Rational Functions Rational Functions Just as rational numbers are defined in terms of quotients of integers, rational functions are defined in terms of quotients of polynomials. The following equations define rational functions: f() 6 p() g() q() h() r() In general, a function f is a rational function if f() n() d() d() where n() and d() are polynomials. The domain of f is the set of all real numbers such that d(). If a and d(a), then f is not defined at a and there can be no point on the graph of f with abscissa a. Remember, division by is never allowed. It can be shown that: If f() n()/d() and d(a), then f is discontinuous at a and the graph of f has a hole or break at a. If a is in the domain of f() and n(a), then the graph of f crosses the ais at a. Thus: If f() n()/d(), n(a), and d(a), then a is an intercept for the graph of f. What happens if both n(a) and d(a)? In this case, we know that a is a factor of both n() and d(), and thus, f() is not in lowest terms (see Section -). Unless specifically stated to the contrary, we assume that all the rational functions we consider are reduced to lowest terms.