* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 3.3 more about zeros
Survey
Document related concepts
List of important publications in mathematics wikipedia , lookup
History of the function concept wikipedia , lookup
Non-standard calculus wikipedia , lookup
Factorization of polynomials over finite fields wikipedia , lookup
Fundamental theorem of calculus wikipedia , lookup
Horner's method wikipedia , lookup
System of polynomial equations wikipedia , lookup
Elementary mathematics wikipedia , lookup
Vincent's theorem wikipedia , lookup
Riemann hypothesis wikipedia , lookup
Transcript
pg170 [V] 170 G2 5-36058 / HCG / Cannon & Elich Chapter 3 jb 11-9-95 QC1 Polynomial and Rational Functions 3.3 MORE ABOUT ZEROS Why do we try to prove things anyway? I think because we want to understand them. We also want a sense of certainty. Mathematics is a very deep field. Its results are stacked very high, and they depend on each other a lot. You build a tower of blocks but if one block is a bit wobbly, you can’t build the tower very high before it will fall over. So I think mathematicians are concerned about rigor, which gives us certainty. But I also think proofs are so that we can understand. I guess I like explanations better than step-by-step rigorous demonstrations. William P. Thurston The problem that infected me with such virulence . . . concerned solving cubic equations and the answer had been known since Cardano published it in 1545. What I did not know was how to derive it. The sages who had designed the mathematics curricula . . . had stopped at solving quadratic equations. Questions by curious students about cubic and higher-order equations were deflected with answers such as “This is too advanced for you” or “You will learn this when you study higher mathematics,” thereby creating a forbidden-fruit aura about the subject. Mark Kac We now have considerable experience with graphs of polynomial functions and a feeling for the nature of their zeros. When we have completely factored a polynomial, we obviously know its zeros, but how do we find the zeros (or equivalently, the factors) of a nonfactored polynomial? Exact form answers are not always available, even theoretically (see the Historical Note, “There Is No Quintic Formula”). There are useful theorems about the nature of zeros, and we now have technological tools undreamed of by earlier mathematicians. This section adds to our arsenal of theorems about the nature of polynomial zeros, and then gives an informal introduction to Newton’s Method, an important tool for approximating zeros with great precision. Number of Zeros of Polynomial Functions How many zeros does a polynomial function of degree n have? Returning again to quadratic functions (degree 2, parabola as the graph), there can be two real zeros (when the parabola crosses the x-axis twice), a repeated real zero (when the parabola is tangent to the x-axis), or two nonreal conjugate zeros (when the parabola does not touch the x-axis). Such behavior is typical of polynomial functions in general. We explore some cubics in the following two examples. cEXAMPLE 1 Zeros of a family of cubics Consider the family ^ of cubic curves given by f ~x! 5 x 3 2 cx 1 2. Graph members of ^ for c 5 0, 1, 2, 3, 4. Decide what values of c give a cubic whose graph crosses the x-axis (a) just once, (b) exactly three times. (c) For what value of c does f have a repeated zero? Check by factoring. Solution (a) and (b) When c 5 0, the graph is a vertical shift of the cubic y 5 x 3, with only one real zero. When c 5 1 or 2, the graph has two turning points but crosses the x-axis only once, so there is only one real zero. At c 5 3, the graph appears to just touch the x-axis at the point (1, 0). See Figure 16. For values of c larger than 3, there are exactly three real zeros. (c) When c 5 3, it appears that there is a repeated zero at 1, and the other crossing looks like ~22, 0!. To check, we want to show that a factored form of x 3 2 3x 1 2 is ~x 2 1!2~x 1 2!. ~x 2 1!2~x 1 2! 5 ~x 2 2 2x 1 1!~x 1 2! 5 x 3 2 3x 1 2. b pg171 [R] G1 5-36058 / HCG / Cannon & Elich jb 3.3 11-13-95 QC2 More about Zeros 171 (0, 2) (0, 2) [– 2.5, 2] by [– 2.5, 4.5] (a) c = 0 [– 2.5, 2] by [– 2.5, 4.5] (b) c = 1 (0, 2) (0, 2) [– 2.5, 2] by [– 2.5, 4.5] (d) c = 3 [– 2.5, 2] by [– 2.5, 4.5] (c) c = 2 FIGURE 16 y 5 x 3 2 cx 1 2 cEXAMPLE 2 Zeros of a family of cubics, continued ^ of Example 1, f ~x! 5 x 3 2 cx 1 2. Consider the family (a) Find a value c for which 2 is a zero of f , and find the other zeros of f in exact form. (b) Find all zeros in exact form when c 5 21. Solution y (a) To find a value of c for which f ~2! 5 0, we substitute 2 for x and solve for c: 23 2 c · 2 1 2 5 0 2c 5 10, or c 5 5. 6 4 (0, 2) 2 (2, 0) x –4 –2 2 4 Thus, the cubic in ^ that has 2 as a zero is f ~x! 5 x 3 2 5x 1 2. We know then that ~x 2 2! is a factor, and we can use long division (or synthetic division, or simple factoring) to find the other factor: x 3 2 5x 1 2 5 ~x 2 2!~x 2 1 2x 2 1!. f(x) = x3 – 5x + 2 FIGURE 17 The remaining zeros of f are the roots of x 2 1 2x 2 1 5 0, which the quadratic formula gives as 21 6 Ï2 (Check.) See Figure 17. (b) For c 5 21, f ~x! 5 x 3 1 x 1 2 has one real zero and no turning points, and the x-intercept point appears to be (21, 0), which we can verify by substituting 21 for x: f ~21! 5 ~21!3 1 ~21! 1 2 5 0. Therefore, ~x 1 1! is a factor. pg172 [V] 172 G2 5-36058 / HCG / Cannon & Elich Chapter 3 jb 11-13-95 QC2 Polynomial and Rational Functions f ~x! 5 x 3 1 x 1 2 5 ~x 1 1!~x 2 2 x 1 2! As in part (a), the remaining zeros are the roots of x 2 2 x 1 2 5 0, which the quadratic formula gives as 1 62Ï7i . b Examples 1 and 2 remind us again of the utility of the factor and remainder theorems of Section 3.2, and they reinforce our ideas about the nature of zeros of cubic polynomials. A cubic always has three zeros, at least one real zero. There can be a repeated zero, and there is a possibility of two nonreal conjugate zeros. Fundamental Theorem of Algebra The general situation is summed up in the fundamental theorem of algebra and some of its consequences, called corollaries. These are stated in terms of the complex-number system and proofs necessarily involve complex numbers as well. The fundamental theorem was first proved by one of the greatest mathematicians of all times. See the Historical Note, “Carl Friedrich Gauss.” Fundamental theorem of algebra Suppose p is a polynomial function of degree n, n $ 1. There is at least one number c where p~c! 5 0; that is, p has at least one zero (which may be a nonreal complex number). Corollary 1: In the complex number system, p has exactly n zeros (counting multiplicities). Corollary 2: If the coefficients of p~x! are real numbers, then the graph of p can cross (or touch) the x-axis in at most n points. The fundamental theorem of algebra is what mathematicians call an existence theorem. For any given polynomial function of positive degree, the theorem states that zeros exist, but it provides no help for finding any particular zero. For linear and quadratic functions we can find the zeros exactly; for higher-degree polynomial functions, the situation becomes more difficult. For most of the problems that arise in applications we can only approximate zeros. Virtually all numerical techniques to do this are rooted in the locator theorem, our most fundamental tool. Mathematicians have found a number of theorems, however, that can help in the search for certain kinds of zeros. Gauss’ proof of the fundamental theorem applies to polynomial functions with both imaginary and real coefficients. We emphasize, however, that in this chapter we discuss only polynomial functions with real number coefficients. Nature of Zeros of Polynomial Functions The next theorem generalizes what we already know about quadratic polynomials. In certain situations zeros of quadratic functions come in pairs. For example, if f ~x! 5 x 2 2 4x 1 1, then the zeros of f are 2 1 Ï3 and 2 2 Ï3; if g~x! 5 x 2 2 2x 1 2, then the zeros of g are 1 1 i and 1 2 i. The numbers a 1 bi and a 2 bi are called complex conjugates. Certain kinds of zeros must occur in pairs in higher-degree polynomials, as well. pg173 [R] G5 5-36058 / HCG / Cannon & Elich clb 3.3 HISTORICAL NOTE 11-22-95 MP1 More about Zeros 173 CARL FRIEDRICH GAUSS (1777–1855) Called the “prince of 17 sides when he was almost 17 mathematicians,” Gauss is clearly himself. His construction was among the greatest mathematicians published before he turned 19. of all time. He contributed to all Gauss was the first to use i for areas of mathematics, as well as to Ï21 and thoroughly understood astronomy and physics, and we are the importance of complex still building directly on numbers in the solution of foundations he laid. equations. Although he left college Most of us could construct an before receiving his doctorate, he equilateral triangle or a square with submitted his dissertation and a compass and ruler. The Greeks attained the degree by the age of Mathematician, astronomer, also constructed a pentagon. Angle 21. His thesis established the and physicist Carl Friedrich Gauss. bisection allows us to double the fundamental theorem of algebra. number of sides, so it is theoretically This theorem so fascinated him that possible to construct regular polygons of n sides if he gave three different proofs during his lifetime. n is a power of 2 or n 5 2k · 3 or n 5 2k · 5, Returning to constructions, he proved in 1826 where k is any nonnegative integer. Gauss made that an n-gon is constructible for an odd prime n k the first significant progress in 2000 years when he only when n has the form 22 1 1 (for instance, 3, 5, and 17). discovered how to construct a regular polygon of Conjugate zeros theorem Let p be any polynomial function with real number coefficients. If the nonreal complex number a 1 bi is a zero of p, then a 2 bi is also a zero. Here b is not zero. If, in addition, p has integer coefficients and a 1 Ïb is a zero of p, then a 2 Ïb is also a zero. Here b is not a perfect square. Strategy: Since the coefficients of p are real numbers, the conjugate zeros theorem applies; if i is a zero, then 2i must also be a zero. The conjugate zeros theorem is illustrated in the next example. cEXAMPLE 3 Conjugate zeros Given that p~i! 5 0, find all zeros of the polynomial p~x! 5 2x 4 2 2x 3 1 x 2 2 2x 2 1. y Solution As the strategy suggests, both i and 2i are zeros, so both x 2 i and x 1 i are factors of p~x!. Thus p~x! has x 2 1 1 as a factor since 3 2 1 ~x 2 i!~x 1 i! 5 x 2 1 1. –1 x –3 –2 (0, – 1) 1 –1 2 3 Dividing p~x! by x 2 1 1, we find that the other factor is 2x 2 2 2x 2 1 and so p~x! 5 ~x 2 1 1!~2x 2 2 2x 2 1!. –2 –3 FIGURE 18 p~x! 5 2x 4 2 2x 3 1 x 2 2 2x 2 1 The remaining zeros of p are the roots of 2x 2 2 2x 2 1 5 0, which the quadratic formula gives as 1 62Ï3 . Therefore, p has two pairs of conjugate zeros: i and 2i, and 1 12 Ï3, 12 2 12 Ï3. The graph in Figure 18 shows the two real zeros and suggests, as we now know, that there are no others. b 1 2 pg174 [V] 174 G2 5-36058 / HCG / Cannon & Elich Chapter 3 clb 11-22-95 MP1 Polynomial and Rational Functions Approximating Real Zeros: Newton’s Method Without some initial information it may be effectively impossible to find zeros of a polynomial function. In Example 3, we are given the fact that i is a zero, which, with the conjugate zeros theorem, is enough for us to find all zeros in exact form. We know how to zoom in on zeros from a graph, but the process is slow, and it is difficult to get much precision. Many graphing calculators will immediately give approximations for all zeros of a polynomial function, to full calculator accuracy, at the press of a single key. How do they do it? A comprehensive answer is beyond the scope of this course, but most SOLVE routines essentially build on an iterative technique for approximation called Newton’s (or the Newton-Raphson) method. The idea behind the process has Tangent a simple graphical interpretation that we can describe with the aid of a few pictures. line (x0 , f(x 0)) For the pictures we need to speak of the “slope” or “direction” of a curve at a given point, another idea that is made precise in calculus is the derivative of a function. Tangent For our purposes, we will explain how to work with derivatives of polynomials of line degrees 3 and 4, but we could just as well use the numerical derivative in the form (c, 0) x already built-in in several graphing calculators. x2 x1 x0 y = f(x) First we look at a picture of a typical function near a zero we want to approximate. See Figure 19. This could be almost any function, under any degree of FIGURE 19 magnification. If we know that the number x 0 is near the desired zero (which we have indicated as c in Figure 19), then we want a mechanical procedure for getting a better approximation, a new number nearer to c than x0. The idea is that if we go to the point on the curve ~x0 , f ~x0!! and essentially “take aim” along the curve in the direction of what is called the tangent line to the curve, we should hit the x-axis at a point x 1 nearer c than x0 . Repeating the process from x 1 should take us nearer still, after which we could move to a point x 2 still nearer, and so forth. Fortunately, the general process (derived in calculus) is given in a simple formula that can be implemented easily on our calculators. We take the following information from calculus. Associated with every polynomial function f is a related function called the derivative of f , denoted by f 9. We give formulas for the derivatives of cubic and quartic functions, leaving explanations for later courses. (It may also be possible to use a built-in program of your calculator; see the Technology Tip following Example 4.) Function Derivative f ~x! 5 ax 1 bx 1 cx 1 d Example: f ~x! 5 x 3 2 4x 2 1 2x 2 1 f ~x! 5 Ax 4 1 Bx 3 1 Cx 2 1 Dx 1 E Example: f ~x! 5 2x 4 1 5x 3 2 x 1 17 f 9~x! 5 3ax 2 1 2bx 1 c f 9~x! 5 3x 2 2 8x 1 2 f 9~x! 5 4Ax 3 1 3Bx 2 1 2Cx 1 D f 9~x! 5 8x 3 1 15x 2 2 1 3 2 Having a way to write the derivative, we can describe the approximation process of Newton’s method. If x 0 is an approximation to the zero c, then the new approximation is given by x 1 5 x 0 2 f ~x 0!yf 9~x0!, where f 9~x 0! is the slope of the tangent line, the number we get when the derivative function (or the numerical derivative) of f is evaluated at x 0 . pg175 [R] G1 5-36058 / HCG / Cannon & Elich clb 3.3 11-22-95 MP1 More about Zeros 175 The process can be programmed very simply with almost any calculator, but we outline the steps as an algorithm that can be performed on your home screen. Then we illustrate with an example. Follow Example 4 with your calculator. Algorithm for Newton’s method 1. Make an initial guess, a reasonable approximation to the desired zero. Note: the better the first guess, the more efficient the method. 2. Store your guess in memory register x. (For example, if your initial guess is 1, 1 A x, ENTER.) 3. Evaluate the next approximation, using x 1 5 x 0 2 f ~x 0!yf 9~x0!, store it in memory register x and ENTER. This displays the new number and stores it for use in the next step. 4. Now press ENTER again and again, getting a new approximation with each repetition, until the display doesn’t change, indicating that we have the best approximation the calculator can deliver. cEXAMPLE 4 Approximating zeros The polynomial p~x!5x 3 2 5x 1 2 has a real zero between 0 and 1. Use Newton’s method, beginning with an initial guess of 0.5, to approximate the zero to ten decimal-place accuracy. Solution We would normally get our initial guess from a graph, but following directions, we will let x 0 5 0.5, and store: .5 A x ENT. For p~x! 5 x 3 2 5x 1 2, we have p9 5 3x 2 2 5, so we enter X 2 ~X3 2 5X 1 2!y~3X2 2 5! A X ENTER . The display (ten decimals) reads .4117647059, and as we repeatedly sequence of displayed numbers is as follows. ENTER, the .4142119097 .4142135624 .4142135624 This last number does not change when we continue to press ENTER. Thus, to ten decimal-place accuracy, the desired zero is .4142135624. This is the same polynomial function we considered in Example 2a, in which we found an exact form for the same zero, 21 1 Ï2. b TECHNOLOGY TIP r Numerical derivatives If your calculator has a built-in function for calculating a numerical derivative, as the TI-82 (MATH 8), TI-85 (2nd CALC F2), HP–38 (▪ CHARS, Choose , OK; you want X~F1~X!), or Casio fx 9700 (SHIFT dydx), you need not write out the derivative. Enter the function in your Y1 register (or on your SHIFT MEM list as f1). Then follow the first two steps of the algorithm. In place of having to write out both the function and its derivative in Step 3, simply enter X 2 Y1/nDER(Y1) A X and iterate. While this may not seem a significant advantage for polynomials, using X 2 Y1/nDER(Y1) A X will work with an appropriate initial guess for any function that has a reasonably smooth graph. pg176 [V] G2 176 5-36058 / HCG / Cannon & Elich Chapter 3 jb 11-13-95 QC2 Polynomial and Rational Functions Polynomials in Application In many applied problems, we are not interested in finding polynomial zeros in exact form; we may not care about the number of zeros. The nature of the problem may dictate that only certain zeros have physical meaning, as the next example illustrates. cEXAMPLE 5 Application leading to a polynomial Twin radio towers are to be erected on opposite sides of a 10-foot roadway, where 50-foot guy wires reach from the top of each tower to the base of the other. The wires must cross high enough to leave a 12-foot clearance above the road. How tall can the towers be, and how close to the side of the roadway can they be built? 50' 50' h h 24~x 1 5! 12 h . , or h 5 5 x x 2x 1 10 12' x x 10' Roadway Using the Pythagorean theorem with nACB, h 2 1 ~2x 1 10!2 5 50 2. (a) B 50 h D A x 12 10 E Solution We sketch a diagram to help us visualize the situation. See Figure 20. Because of the symmetry, all of the pertinent information is contained in the second part of the diagram, where x is the distance from the edge of the road to the tower of height h. From similar right triangles, nACB and nAED, we have x (b) FIGURE 20 C 242~x 1 5!2 1 4~x 1 5!2 5 2500, then expand x2 and simplify to get a polynomial equation, We substitute 24~x 1 5!yx for h, x 4 1 10x 3 2 456x 2 1 1440x 1 3600 5 0. Clearly, only positive values for x have any meaning, and from the picture, we estimate that x is probably between 5 and 15, so we look for zeros in that range. In a @5, 15# 3 @25, 50# window, we see just two vertical lines, but there appear to be two zeros. To get a picture that looks a little more like the polynomials we are familiar with, we need a much larger y-range. With a y-range of @28000, 7000#, we see the curve in Figure 21. We still see two zeros, one near 14 and one near 5.7. The corresponding heights are given by h < 32.6 and h < 45.2. How do we interpret two different answers? When we look more closely, it turns out that we can get at least a 12-foot clearance over the road if the distance 5.7 45.2' 14 32.6' [5, 15] by [– 8000, 7000] y = x 4 + 10x3 – 456x2 + 1440x + 3600 FIGURE 21 12' 10' 5.7 14' FIGURE 22 pg177 [R] G1 5-36058 / HCG / Cannon & Elich clb 3.3 11-27-95 MP2 More about Zeros 177 from the roadway is anything between 5.7 feet and 14 feet. The maximum tower height decreases as we move away from the road. See Figure 22. For maximum height, we can erect 45-foot towers 5.7 feet on each side of the roadway. b EXERCISES 3.3 Check Your Understanding Exercises 1–7 True or False. Draw graphs when helpful. 1. The function f ~x! 5 x 3 2 8x 2 1 5x 1 4 has three real zeros. 2. The graph of f ~x! 5 x 4 1 2x 2 1 1 crosses the x-axis. 3. The equation x 3 2 7x 1 5 5 0 has two negative and one positive root. 4. Since Ï3 is a zero of f ~x! 5 x 3 2 2x 2 Ï3, then 2Ï3 must also be a zero. 5. All zeros of f ~x! 5 x 3 2 8x 2 1 5x 1 8 lie between 21 and 8. 6. If f ~x! 5 2x 4 1 x 3 1 7x 2 2 x 2 6, then f ~x! , 15 for every x. 7. Based on what can be seen from the graph of f ~x! 5 x 3 2 40x 2 2 400x 1 1600 using @224, 50# 3 @25100, 17,000#, we can conclude that f has one negative and two positive zeros. Exercises 8–10 Fill in the blank so that the resulting statement is true. 8. For the family of functions f ~x! 5 x 3 1 cx 2 5, the value of c for which 21 is a zero is . 4 9. The number of real zeros of f ~x! 5 x 1 2x 3 2 . 2x 2 2 4x is 10. In applying Newton’s method, if f ~x! 5 x 3 2 2x 1 3, then f 9~x! 5 . Develop Mastery Exercises 1–4 Zeros from Graph (a) Graph y 5 p (x) and locate each zero between two consecutive integers. (b) Find an approximation (one decimal place) for the largest zero. 1. p~x! 5 x 3 2 3x 2 2 x 1 2 2. p~x! 5 x 3 2 3x 2 2 3 3. p~x! 5 x 3 2 2x 2 2 x 1 3 4. p~x! 5 x 3 1 2x 2 2 3x 2 2 Exercises 5–8 Locate Intersection Graph the two equations on the same screen and find the coordinates of the point of intersection (one decimal place). 6. y 5 x 2 x 3 5. y 5 x 3 2 3x y 5 23 y51 8. y 5 1 2 x 3 7. y 5 x 3 y5x11 y 5 x2 2 2 Exercises 9–12 Zeros to Function Find a polynomial function of lowest degree with integer coefficients, a leading coefficient of 1, and the given numbers as zeros. Give the result in standard (expanded) form. Use the conjugate zeros theorem, if needed. 9. 1 1 i, 1 2 Ï2 10. 1, Ï2 11. 2, 21 1 Ï3 12. Ï2, 1 2 i Exercises 13–16 Conjugate Zeros Theorem The given number is a zero of f. Find the remaining zeros. (Hint: Use the conjugate zeros theorem and long division.) 13. f ~x! 5 x 3 2 4x 2 1 3x 1 2; 1 2 Ï2 14. f ~x! 5 x 3 2 2x 2 2 9x 2 2; 2 1 Ï5 15. f ~x! 5 2x 3 2 9x 2 1 2x 1 1; 2 2 Ï5 16. f ~x! 5 2x 3 2 3x 2 2 4x 2 1; 1 1 Ï2 Exercises 17–24 Find All Zeros Assume that the domain of the variable is the set of complex numbers. Find all zeros in exact form. 17. f ~x! 5 x 3 2 4x 2 1 2x 2 8 18. f ~x! 5 4x 3 2 4x 2 2 19x 1 10 19. f ~x! 5 x 3 2 2.5x 2 2 7x 2 1.5 20. f ~x! 5 3x 3 2 1.5x 2 1 x 2 0.5 21. f ~x! 5 6x 4 2 13x 3 1 2x 2 2 4x 1 15 22. f ~x! 5 2x 4 1 3x 3 1 2x 2 2 1 23. f ~x! 5 4x 4 2 4x 3 2 7x 2 1 4x 1 3 24. f ~x! 5 4x 4 1 8x 3 1 9x 2 1 5x 1 1 Exercises 25–32 Exact Form Roots exact form. 25. 6x 3 2 2x 2 2 9x 1 3 5 0 26. 6x 3 2 x 2 2 13x 1 8 5 0 27. x 4 2 x 3 2 3x 2 1 x 1 2 5 0 28. x 4 2 x 3 2 8x 1 8 5 0 29. x 3 2 3x 1 2 5 0 30. 18x 3 1 27x 2 1 13x 1 2 5 0 ~14x 2 1 17x! 31. x 3 1 2 5 2 3 32. x 4 1 4x 3 2 5x 2 5 36x 1 36 Find all roots in pg178 [V] 178 G2 5-36058 / HCG / Cannon & Elich Chapter 3 clb 11-22-95 MP1 Polynomial and Rational Functions Exercises 33–34 Solution Set, Exact Form Find the solution set for (a) f ~x! $ 0 (b) f ~x 2 1! $ 0. Give answers in exact form. 33. f ~x! 5 x 3 1 x 2 2 11x 2 15 34. f ~x! 5 x 3 2 6x 2 1 4x 1 16 Exercises 43–46 Explore The equation defines a family of polynomial functions of degree 3. Experiment with several integer values of c and for each draw a graph. Describe the role that c appears to play. Include information about local extrema, number of real zeros, and any other graphical features you observe. Use complete sentences. 43. f ~x! 5 x 3 1 cx 2 2 4x 1 3 44. f ~x! 5 x 3 1 2x 2 1 cx 1 3 45. f ~x! 5 x 3 1 2x 2 2 4x 1 c 46. f ~x! 5 cx 3 1 2x 2 2 4x 1 3 Exercises 51–53 Explore Composition with Absolute Value (See Section 2.6) 51. Try several polynomial functions f of degree 3 and in each case draw a graph of f and then a graph of g~x! 5 f ~_ x _ !. For what functions f will the composition function g have (a) no zeros (b) two zeros (c) four zeros (d) six zeros 52. Draw a graph of f ~x! 5 _ x 3 1 x 2 2 2x _ . Explain why the graph of f cannot be the graph of a polynomial function. (Hint: Read “Smoothness and End Behavior” in Section 3.1.) 53. (a) Take any polynomial function f of degree 3 and draw the graph of g~x! 5 _ f ~x! _ . Use the graph to explain why g cannot be a polynomial function. (b) Is there a polynomial function f of degree 4 for which g~x! 5 _ f ~x! _ is also a polynomial function? Explain. 54. For each real number c, the graph of f ~x! 5 0.1~x 1 c!3 1 0.3~x 1 c!2 2 ~x 1 c! 2 8 is a horizontal translation of the graph of g~x! 5 0.1x 3 1 0.3x 2 2 x 2 8. For what integer values of c will f have a negative zero? 55. Explore The graph of f ~x! 5 x 3 2 6x 2 1 9x 2 6 has one real zero and local extrema at ~1, 22! and ~3, 26!. Draw a graph in a shifted decimal window to support this claim. Consider the family of functions g~x! 5 f ~x! 1 c. (a) For what integer values of c will g have three distinct real zeros? (b) For what integer values of c will g have a repeated zero? Justify algebraically. (c) For what integer values of c will g have a negative zero? 56. Explore Consider the family of functions f ~x! 5 0.3x 3 2 4x 1 c. Let P be the local maximum point and Q be the local minimum point. For what integer values of c will (a) P and Q be below the x-axis? (b) P be above and Q be below the x-axis? (c) P and Q be above the x-axis? Exercises 47–50 Explore For each real number k, f is a polynomial function of degree 3. Experiment with several integer values of k and determine the values of k for which f will have (a) three real zeros, (b) exactly one real zero, (c) one negative and no positive zeros. 48. f ~x! 5 x 3 1 kx 1 3 47. f ~x! 5 x 3 1 kx 2 2 3 50. f ~x! 5 x 3 1 kx 2 5 49. f ~x! 5 x 1 kx 1 8 Exercises 57–60 Newton’s method Use Newton’s method to find the largest zero of f (nine decimal places). See Example 4. 57. f ~x! 5 x 3 2 4x 1 2 58. f ~x! 5 x 3 1 4x 2 2 8 59. f ~x! 5 x 4 2 5x 2 1 3x 2 1 60. f ~x! 5 x 4 2 9x 2 1 8x Exercises 35–38 Evaluating Inverse (a) Graph the function to support the claim that f is either an increasing or decreasing function (tell which), so that f has an inverse, f 21. (b) Find f 21~3! (1 decimal place). (Hint: In x 5 f ~y! replace x by 3 and solve for y.) 35. f ~x! 5 2x 3 1 3x 2 4 36. f ~x! 5 x 3 2 3x 2 1 4x 1 5 37. f ~x! 5 2 2 x 1 x 2 2 2x 3 38. f ~x! 5 4 2 3x 2 2x 3 39. For f ~x! 5 x 3 2 3x 2 2 x 1 3 and g~x! 5 _ x _ , how many zeros does the function f 8 g have? Draw a graph. Does your answer contradict the corollaries to the fundamental theorem of algebra? Explain. 40. Repeat Exercise 39 for f ~x! 5 x 3 2 3x 2 2 4x 2 4. 41. Explore For f ~x! 5 x 3 1 cx 1 4, choose several integer values for c (positive and negative) and draw a graph of the corresponding function. What values of c give graphs with (a) turning points, (b) more than one x- intercept point? 42. Repeat Exercise 41 for f ~x! 5 x 3 1 cx 2 4. pg179 [R] G1 5-36058 / HCG / Cannon & Elich clb 3.3 61. A box with an open top is constructed from a rectangular piece of tin that measures 9 inches by 12 inches by cutting out from each corner a square of side x, and then folding up the sides as shown in the diagram. (a) If V denotes the volume of the box, find a formula for V as a function of x. What is the domain of the function? x x x xx 9 x 12 (b) What size corners should be cut out so that the box will have a capacity of 81 cubic inches? (Hint: There are two answers, both between 1 and 2.) 62. A storage tank consists of a right circular cylinder mounted on top of a hemisphere, as shown in the diagram. If the height of the cylindrical portion is 12 feet and the tank is to have a capacity of 1250p cubic feet, find the radius r of the cylinder to 3 significant digits. 11-22-95 MP1 More about Zeros 179 65. A storage tank has the shape of a cube. If one of the dimensions is increased by 2 and another by 3, while the third is decreased by 4, then the resulting rectangular tank will have a volume of 600 cubic units. What is the length of an edge of the original cube to 3 significant digits? 66. A rectangular storage container is 3 by 4 by 5 feet. (a) What is its capacity (volume)? (b) If we increase each of the dimensions by x feet in order to get a container with a capacity five times as large as the original, how large must x be, rounded off to 3 significant digits? 67. Two vertical poles, AB and CD, are connected by guy wires of lengths 30 ft and 40 ft, intersecting at point F, 12 feet from the ground. See the diagram. Find the heights u and v, of the poles and the distance d between the poles. D B 30 ft 40 ft F u v 12 ft A 12 r E C 68. In Example 5 we found that the two towers can be placed anywhere from 5.7 ft to 14 ft from the road. How far from the road should they be located if we want maximum clearance above the road? (Hint: Let u be the desired distance and v be the distance, _DE _ , from the right edge of the road to the guy wire AB.) Write v as a function of u. See diagram. y 63. An isosceles triangle has the dimensions shown in the diagram. If the area is equal to 10 square units, find the length of the altitude h (to the nearest tenth). B(0, h) Tower x+1 h 50 ft D C 12 v u 10 u E x+1 h A (2u + 10, 0) x Roadway 2x 64. What are the dimensions of a rectangle of area 7 square inches that has a diagonal 1 inch longer than the length of one of its sides? Give the result rounded off to 3 significant digits. 69. Solve the problem in Example 5 if the guy wires are 48 feet long. 70. In Exercise 69, how far from the road should the two towers be located so that there is maximum clearance above the road? See Exercise 68.