Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 41 pptx

11.2 Characterizing Polynomials 381 In this course we will mainly be working with the real number system as opposed to the complex number system, so unless otherwise stated, when we ask about the number of roots it is understood that we mean real roots. EXERCISE 11.4 Construct a polynomial equation with the specification given. (The answers to this exercise are not unique!) (a) a third degree polynomial equation with roots at x =−2, x = 3, and x = 0 (b) a second degree polynomial equation with a double root at x =−1 (c) a second degree polynomial equation with no roots (d) a fifth degree polynomial equation with roots only at x =−2, x = 3, and x = 0 Answers are provided at the end of the section. EXERCISE 11.5 Identify which of the following two polynomials is possible to construct, and construct it. (a) a fifth degree polynomial with no real zeros (b) a fourth degree polynomial with no real zeros Answer is provided at the end of the section. Finding the Zeros of a Polynomial Finding the zero of a linear function is simple. Finding the zeros of a quadratic is no problem when you use the quadratic formula. It is considerably harder to find the roots of a general cubic equation. A systematic procedure for finding a cubic’s roots does exist, but it is much more complicated than the quadratic formula and in practice is not often used. The story of the discovery of the cubic formula in Italy in the mid-1500s is one worth reading about in a math history book. It involves alleged lies about the discovery, a public competition to solve cubic equations being won by Nicolo Tartaglia and his “secret” method being passed on to Girolamo Cardano in confidence but then published anyway, and a final dispute from which one of the men is said to have been lucky to have escaped alive. 1 There is an algorithm for finding the zeros of fourth degree polynomials, but in the early 1800s the Norwegian mathematician Niels Abel (1802–1829) proved that a formula for finding the zeros of a fifth degree polynomial does not exist. For a general nth degree polynomial it can be difficult to find the zeros, and as is clear from the last paragraph, we cannot give a recipe for pinning them down exactly. We can, however, take advantage of the continuity of polynomials; if f(a)>0and f(b)<0,then there is a c somewhere between a and b such that f(c)=0. ◆ EXAMPLE 11.3 Find the zeros of f(x)=x 3 −2x 2 −8x. SOLUTION We can factor out an x from each term. x 3 − 2x 2 − 8x = x(x 2 − 2x − 8) Now we have a quadratic that we can factor. = x(x − 4)(x + 2) So, the zeros are x = 0, x = 4, and x =−2. ◆ 1 From Howard Eves, An Introduction to the History of Mathematics, and David Burton, A History of Mathematics. 382 CHAPTER 11 A Portrait of Polynomials and Rational Functions ◆ EXAMPLE 11.4 Find the zeros of f(x)=x 3 +3x 2 −3x −1. SOLUTION In this case, it is not clear how to factor f(x).Graph f on a graphing calculator. 10 –4 –3 –2 –1123 y x f(x) Figure 11.7 We see that there is a zero near x = 1. By zooming in, it looks like the zero might be exactly x = 1. To be sure, evaluate f at x = 1. f(1)=(1) 3 +3(1) 2 −3(1)−1=0, so x = 1 is a zero. There are two other zeros. By zooming in, we estimate that they are at x ≈−3.73 and x ≈−0.27. Let’s try to find the exact values. x = 1 is a zero of f(x);therefore, (x − 1) is a factor of f(x).Inother words, we can write x 3 + 3x 2 − 3x − 1 = (x − 1)p(x), where p(x) is a new polynomial, a polynomial of degree 2. To find p(x), we do long division of polynomials. Long division of polynomials is analogous to long division for numbers. Let’s review long division. Find 732 4 . Write denominator ) numerator. 4 goes into 7 one time. Write the 1 on the top and 1 • 4 under the 7. Subtract 4 from 7. Bring down the 3 from 732 to get 33. 4 goes into 33 eight times. Write the 8 on the top and 8 • 4 under the 33. Subtract 32 from 33. Bring down the 2 from 732 to get 12. 4 goes into 12 three times. Write the 3 on the top and 3 • 4 under the 12. Subtract 12 from 12. This gives zero. We've accounted for 732: 7 • 100 + 3 • 10 + 2, so we're done. 4 732 4 0 33 183 32 12 12 ͤ Find x 3 +3x 2 −3x−1 x−1 . Write denominator ) numerator. 11.2 Characterizing Polynomials 383 x times x 2 gives x 3 . Write the x 2 on the top and x 2 • (x –1) under the x 3 + 3x 2 . Subtract x 3 – x 2 from x 3 + 3x 2 . Bring down the –3x from above to get 4x 2 –3x. x times 4x gives 4x 2 . Write the 4x on the top and 4x • (x –1) under the 4x 2 –3x. Subtract 4x 2 –4x from 4x 2 –3x. Bring down the –1 from above to get x –1. x times 1 gives x. Write the 1 on the top and 1 • (x –1) under the x –1. Subtract x –1 from x –1. This gives zero. We've accounted for x 3 + 3x 2 –3x –1, so we're done. x–1 x 3 + 3x 2 –3x –1 x–1 x–1 x 2 + 4x +1 4x 2 – 3x 4x 2 – 4x x 3 – x 2 0 ͤ By doing long division as shown, we see that p(x) = x 2 + 4x + 1, so f(x)=x 3 + 3x 2 −3x −1=(x − 1)(x 2 + 4x + 1). We use the quadratic formula to find the roots of x 2 + 4x + 1 = 0 and find that x = −4 ± √ 16 − 4 2 = −4 ± √ 12 2 = −4 ± 2 √ 3 2 =−2± √ 3. Verify that the values of these two roots are approximately equal to our previous estimates from the graph. The zeros of f(x)=x 3 +3x 2 −3x −1are x = 1, x =−2+ √ 3, and x =−2− √ 3. Notice that if we had not known an exact zero (x = 1 in this case), then we could not have used long division and would have needed to approximate all three zeros graphically without learning their exact values. 2 ◆ EXERCISE 11.6 Find all the roots of f(x)=x 3 −1=0.(This is the cubic discussed in Section 7.3.) Characteristics of Polynomials and Differentiation of Polynomials Let’s return for a moment to the building blocks of polynomials, functions of the form y = x k for k = 0, 1, 2, 3, ,and look at them more closely. They are graphed below on the same set of axes. Two scales are used, one allowing us to compare the graphs for x near zero, the other facilitating comparison for large values of x. 2 Another alternative is to use what is called “Newton’s Method” for approximating roots. See the Appendix on Newton’s Method for details. 384 CHAPTER 11 A Portrait of Polynomials and Rational Functions 1 (1, 1) –1 1 –1 yy x x 2 x 4 x 2 x 3 x 3 x 3 x 4 x 5 x 5 x 5 x 6 x x x 10 100 –100 –10 Figure 11.8 Behavior for small x (a bug’s-eye view) The larger k is, the flatter the graph of x k is on the interval −1 <x<1. In other words, for |x| < 1, the higher the power to which x is raised, the smaller the magnitude of x k . 3 For instance,  1 2  5 is smaller than  1 2  2 . Behavior for large x (a bird’s-eye view) For |x| > 1, the magnitude of x k increases dramatically as k increases. For example, consider the values of y corresponding to x = 10 in each of the polynomials above. 10 5 is much greater than 10 2 . By summing up building block functions of the form x k and weighting them with coeffi- cients, we obtain a general nth degree polynomial P(x)=a 0 +a 1 x +a 2 x 2 +···+a n x n . Characteristics of such a polynomial are discussed next. A Bird’s-Eye View of Polynomials For x large enough in magnitude (zoom out a lot on your graphing calculator) the graph of an nth degree polynomial, P(x)=a 0 +a 1 x +a 2 x 2 +a 3 x 3 +···+a n x n ,isdominated by a n x n and can be characterized as follows: For n even, if a n is positive, then the graph reaches upward; . . . if a n is negative, then the graph reaches downward. . . . For n odd, if a n is positive, then the graph reaches down on the left and up on the right; . . . if a n is negative, the graph reaches up on the left and down on the right. . . . More precisely, let P(x) =a 0 +a 1 x + a 2 x 2 +a 3 x 3 +···+a n x n be a polynomial of degree n. 3 By “magnitude of x k ” we mean |x k |. We are referring to the size, without regard to the sign. 11.2 Characterizing Polynomials 385 Suppose n is even: If a n > 0, then lim x→∞ P(x)=+∞and lim x→−∞ P(x)=+∞. If a n < 0, then lim x→∞ P(x)=−∞and lim x→−∞ P(x)=−∞. Suppose n is odd: If a n > 0, then lim x→∞ P(x)=+∞and lim x→−∞ P(x)=−∞. If a n < 0, then lim x→∞ P(x)=−∞and lim x→−∞ P(x)=+∞. This bird’s-eye view tells us that a polynomial of odd degree will have neither an absolute maximum value nor an absolute minimum value. Its range is (−∞, ∞). On the other hand, a polynomial of even degree will have a global maximum but no global minimum if the leading coefficient is negative and vice versa if the leading coefficient is positive. A Bug’s-Eye View of Polynomials If we zoom in enough anywhere on the graph of any polynomial, the graph will eventually look like a straight line. Polynomial functions have the delightful property that they are locally linear everywhere. Another way to say this is that polynomials are differentiable everywhere. Their graphs are smooth; there are no sharp corners. Polynomials are continuous, locally linear, and easy to evaluate. This makes polynomial functions very user-friendly. In fact, polynomials are so delightful to work with that people use them to approximate local behavior of functions that are more difficult to handle. 4 Another perk is that polynomial functions are a pleasure to differentiate. We can compute the derivative of x n for n any positive integer, so we can differentiate any polynomial. d dx [x n ] = nx n−1 We can conclude that the derivative of an nth degree polynomial is a polynomial of degree n − 1. P(x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + ··· + a n x n P  (x) = a 1 + 2a 2 x 2 + 3a 3 x 3 + ··· + na n x n−1 Critical Points of a Polynomial The natural domain of polynomial functions is (−∞, ∞) and polynomials are differentiable everywhere; therefore, the critical points of a polynomial are simply the zeros of its derivative. The derivative of an nth degree polynomial is a polynomial equation of degree (n − 1). A polynomial of degree (n − 1) can have at most (n − 1) real roots. We conclude that a polynomial of degree n can have at most (n − 1) turning points. This makes sense graphically. If a polynomial of degree n has n zeros (the maximum possible), then between each of those zeros the polynomial must have a turning point so its graph can return to the x-axis. This means there would have to be (n − 1) turning points. Notice also that a polynomial of degree n may have fewer than n − 1 turning points. Consider the graphs of f(x)=x k in Figure 11.6; each of them had either zero or one turning point. 4 You might, for instance, try to approximate e x near x = 0 by a third degree polynomial. One of the treats awaiting you is the amazing discovery that you can actually express functions such as e x (and some of the trigonometric functions) as polynomials of infinite degree. 386 CHAPTER 11 A Portrait of Polynomials and Rational Functions By looking at the behavior of polynomials for x large in magnitude (taking a bird’s-eye view) we can argue that a polynomial of odd degree must have at least one zero and a polynomial of even degree must have at least one turning point. Complete this argument as an exercise. EXERCISE 11.7 Justify the assertion made above. Symmetry. Recall that a function is said to be even if f(−x) = f(x)for all x in its domain (the graph is symmetric about the y-axis). odd if f(−x) =−f(x)for all x in its domain (the graph is symmetric about the origin) y x f(x) = x 2 f even y x f(x) = x 3 f odd Figure 11.9 Looking back at the graphs of x k in Figure 11.6, we can see why the names “even” and “odd” were chosen for these types of symmetry. Functions of the form x k , where k is even, have even symmetry, and functions of the form x k , where k is odd, have odd symmetry. Any random polynomial may be even, odd, or neither. The following example is meant to illustrate criteria for a function to be even and to be odd. ◆ EXAMPLE 11.5 For each of the following, determine whether the function is even, odd, or neither even nor odd. (a) g(x) =−5x 4 +2x 2 +1 (b) h(x) = 2x 3 − x (c) f(x)=2x 3 −x +1 SOLUTION Our strategy is to look at g(−x) and see if it is equal to g(x), −g(x), or neither. (a) g(−x) =−5(−x) 4 + 2(−x) 2 + 8 =−5x 4 +2x 2 +8 =g(x ) So g(x) is even. 11.2 Characterizing Polynomials 387 (b) h(−x) =2(−x) 3 −(−x) =−2x 3 +x =−(2x 3 −x) =−h(x ) So h(x) is odd. (c) f(−x) =2(−x) 3 −(−x) +1 =−2x 3 +x+1 This is neither f(x)nor −f(x),sof is neither even nor odd. ◆ We can deduce the following: A polynomial with the variable raised to even powers only is even. A polynomial with no constant term and the variable raised to odd powers only, with no constant terms, is odd. Terminology: The polynomial f(x)=x 2 +x is a polynomial of even degree, but it is not an even function. Notice that f(x)=x(x +1), so the zeros of the polynomial are at 0 and −1; the graph of f is not symmetric with respect to the y-axis. Similarly, the polynomial g(x) = x 3 +1 is a polynomial of odd degree, but it is not an odd function. The graph of g has a y-intercept of 1, so it is not symmetric about the origin. Answers to Selected Exercises Answers to Exercise 11.4 (a) y = x(x +2)(x − 3) (b) y = (x + 1) 2 (c) y = x 2 + 1 (d) y = (x + 2)x 2 (x − 3) 2 Answer to Exercise 11.5 (b) y = x 4 + 3 PROBLEMS FOR SECTION 11.2 1. Determine whether or not the expression given is a polynomial. (a) 1 √ 2 x + √ 33x 2 + 19 11 (b) 2x 2 + 3x −1 + 5x 3 (c) 2x + x 1/2 + 5x 5 (d) 2 x + 2x 3 + 1 (e) 5 −1/2 x + 3 −1 x 2 + 1 π 2 −2 + 2 (f) (x 2 + 1) −1 388 CHAPTER 11 A Portrait of Polynomials and Rational Functions In Problems 2 through 8, below, construct a polynomial P(x) with the specified characteristics. Answers to these problems are not unique. 2. A second degree polynomial with zeros at x = 1 and x =−3. 3. A third degree polynomial with zeros at x =−1, 0, and 5. 4. A fourth degree polynomial with no zeros. 5. A fourth degree polynomial whose only zero is at x = √ 2. 6. A fifth degree polynomial with a zero of multiplicity two at x = 9 and zeros at x = 0, 3, and −e. 7. A third degree polynomial whose only zero is at x = π + 1, and whose y-intercept is 1. 8. A fourth degree polynomial whose only turning point is at (−3, 2). In Problems 9 through 14, construct a polynomial P(x) with the specified character- istics. Determine whether or not the answer to the problem is unique. Explain and/or illustrate your answer. 9. A fourth degree polynomial with zeros of multiplicity two at x = 2 and x =−3, and a y-intercept of −2. 10. A fifth degree polynomial with zeros of multiplicity two at x = 0 and x = π , and a zero at x =−2; lim x→∞ P(x)=∞. 11. A third degree polynomial whose only zero is at x =−1and such that lim x→∞ P(x)=∞. 12. The graph is a parabola with a vertex at (π,2)and a y-intercept of 0. 13. A fifth degree polynomial with a zero of multiplicity 3 at x = 0 and zeros at x = 1 and x =−2, and passing through the point (−1, 2). 14. A third degree polynomial with zeros at x = 1 and x = 2, a turning point at x =1, and a y-intercept of √ e. In Problems 15 through 21, find the (real) zeros of the polynomial given. 15. (a) f(x)=2x 3 +2x 2 −12x (b) g(x) =2x 3 +2x 2 +12x 16. (a) f(x)=−x 3 −x 2 −5x (b) g(x) =0.5x 4 − 0.5 11.2 Characterizing Polynomials 389 17. (a) P(x)= x 3 −x 2 −4x +4 (b) Q(x) = x 3 − x 2 + 4x − 4 Hint: Either guess a zero by observation or use a graphing calculator to guess a root; then use long division. 18. f(x)=−x 5 +16x 4 19. P(x)=x 4 −2x 3 −6x 2 +12x (Hint: At a certain point, guess a root.) 20. g(x) = 3x 3 + 3 21. h(x) = x 3 − 8 In Problems 22 through 24, P(x) is a polynomial with the characteristics specified. For each statement following the characteristic, determine whether the statement is definitely true; possibly true, but not necessarily true; or definitely false. Explain. 22. The graph of P(x) is symmetric about the origin. (a) The degree of P(x)is even. (b) The degree of P(x)is odd. (c) P(x) has no zeros. (d) P(x) has at least one zero. (e) The number of turning points of P(x)is of the form 2n where n is a nonnegative integer. 23. P(x) is a polynomial whose degree is even and nonzero. P(0)=0. (a) The graph of P(x) is symmetric about the origin. (b) The graph of P(x) is symmetric about the y-axis. (c) The graph of P(x) has at least one turning point. (d) The graph of P(x) has an even number of turning points. (e) The graph of P(x) has an odd number of turning points. 24. P(x) is a fifth degree polynomial. lim x→∞ P(x)=∞. (a) P(x) has five distinct real roots. (b) P(x) has no more than five roots. (c) P(x) has five turning points. (d) P(x) has four turning points. (e) P(x) has no more than four turning points. (f) P(x)has at least one real root. (g) lim x→−∞ P(x)=∞. 390 CHAPTER 11 A Portrait of Polynomials and Rational Functions 25. The graph of P(x) is given below P(x) x (a) The degree of P(x)is at least . (b) What is the sign of the leading coefficient of P(x)? 26. Below are graphs of polynomials, each with a zero at x = a. (a) For each graph determine whether the zero is a simple zero, a zero of even multiplicity, or a zero of odd multiplicity. (b) For each graph, determine whether the derivative at x = a is positive, negative, or zero. y x a (i) y x a (ii) y x a (iv) y x a (v) y x a (iii) . quadratic that we can factor. = x(x − 4)(x + 2) So, the zeros are x = 0, x = 4, and x =−2. ◆ 1 From Howard Eves, An Introduction to the History of Mathematics, and David Burton, A History of Mathematics. 382. difficult to handle. 4 Another perk is that polynomial functions are a pleasure to differentiate. We can compute the derivative of x n for n any positive integer, so we can differentiate any polynomial. d dx [x n ]. graphs of x k in Figure 11.6, we can see why the names “even” and “odd” were chosen for these types of symmetry. Functions of the form x k , where k is even, have even symmetry, and functions of