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(x) y = P3(x) y = P4(x) 3.2 The Factor Theorem and the Remainder Theorem 257 3.2 The Factor Theorem and the Remainder Theorem Suppose we wish to ﬁnd the zeros of f (x) = x3 + 4x2 − 5x − 14. Setting f (x) = 0 results in the polynomial equation x3 + 4x2 − 5x − 14 = 0. Despite all of the factoring techniques we learned1 in Intermediate Algebra, this equation foils2 us at every turn. If we graph f using the graphing calculator, we get The graph suggests that the function has three zeros, one of which is x = 2. It’s easy to show that f (2) = 0, but the other two zeros seem to be less friendly. Even though we could use the ‘Zero’ command to ﬁnd decimal approximations for these, we seek a method to ﬁnd the remaining zeros exactly. Based on our experience, if x = 2 is a zero, it seems that there should be a factor of (x − 2) lurking around in the factorization of f (x). In other words, we should expect that x3 + 4x2 − 5x − 14 = (x − 2) q(x), where q(x) is some other polynomial. How could we ﬁnd such a q(x), if it even exists? The answer comes from our old friend, polynomial division. Dividing x3 + 4x2 − 5x − 14 by x − 2 gives x2 + 6x + 7 x−2 x3 + 4x2 − 5x − 14 −x3 − 2x2 6x2 − 5x −6x2 − 12x) 7x − 14 − (7x − 14) 0 As you may recall, this means x3 + 4x2 − 5x − 14 = (x − 2) x2 + 6x + 7, so to ﬁnd the zeros of f, we now solve (x − 2) x2 + 6x + 7 = 0. We get x − 2 = 0 (which gives us our known zero, x = 2) as well as x2 + 6x + 7 = 0. The latter doesn’t factor nicely, so we
apply the Quadratic Formula to get x = −3 ± 2. The point of this section is to generalize the technique applied here. First up is a friendly reminder of what we can expect when we divide polynomials. √ 1and probably forgot 2pun intended 258 Polynomial Functions Theorem 3.4. Polynomial Division: Suppose d(x) and p(x) are nonzero polynomials where the degree of p is greater than or equal to the degree of d. There exist two unique polynomials, q(x) and r(x), such that p(x) = d(x) q(x) + r(x), where either r(x) = 0 or the degree of r is strictly less than the degree of d. As you may recall, all of the polynomials in Theorem 3.4 have special names. The polynomial p is called the dividend; d is the divisor; q is the quotient; r is the remainder. If r(x) = 0 then d is called a factor of p. The proof of Theorem 3.4 is usually relegated to a course in Abstract Algebra,3 but we can still use the result to establish two important facts which are the basis of the rest of the chapter. Theorem 3.5. The Remainder Theorem: Suppose p is a polynomial of degree at least 1 and c is a real number. When p(x) is divided by x − c the remainder is p(c). The proof of Theorem 3.5 is a direct consequence of Theorem 3.4. When a polynomial is divided by x − c, the remainder is either 0 or has degree less than the degree of x − c. Since x − c is degree 1, the degree of the remainder must be 0, which means the remainder is a constant. Hence, in either case, p(x) = (x − c) q(x) + r, where r, the remainder, is a real number, possibly 0. It follows that p(c) = (c − c) q(c) + r = 0 · q(c) + r = r, so we get r = p(c) as required. There is one last ‘low hanging fruit’4 to collect which we present below. Theorem 3.6. The Factor Theorem: Suppose p
is a nonzero polynomial. The real number c is a zero of p if and only if (x − c) is a factor of p(x). The proof of The Factor Theorem is a consequence of what we already know. If (x − c) is a factor of p(x), this means p(x) = (x − c) q(x) for some polynomial q. Hence, p(c) = (c − c) q(c) = 0, so c is a zero of p. Conversely, if c is a zero of p, then p(c) = 0. In this case, The Remainder Theorem tells us the remainder when p(x) is divided by (x − c), namely p(c), is 0, which means (x − c) is a factor of p. What we have established is the fundamental connection between zeros of polynomials and factors of polynomials. Of the things The Factor Theorem tells us, the most pragmatic is that we had better ﬁnd a more eﬃcient way to divide polynomials by quantities of the form x − c. Fortunately, people like Ruﬃni and Horner have already blazed this trail. Let’s take a closer look at the long division we performed at the beginning of the section and try to streamline it. First oﬀ, let’s change all of the subtractions into additions by distributing through the −1s. 3Yes, Virginia, there are Algebra courses more abstract than this one. 4Jeﬀ hates this expression and Carl included it just to annoy him. 3.2 The Factor Theorem and the Remainder Theorem 259 x2 + 6x + 7 x−2 x3 + 4x2 − 5x −14 −x3+ 2x2 6x2 − 5x −6x2+ 12x 7x −14 −7x+14 0 Next, observe that the terms −x3, −6x2 and −7x are the exact opposite of the terms above them. The algorithm we use ensures this is always the case, so we can omit them without losing any information. Also note that the terms we ‘bring down’ (namely the −5x and −14) aren’t really necessary to recopy, so we omit them, too. x2 + 6
x + 7 x−2 x3+4x2− 5x −14 2x2 6x2 12x 7x 14 0 Now, let’s move things up a bit and, for reasons which will become clear in a moment, copy the x3 into the last row. x2 + 6x + 7 x−2 x3+4x2− 5x −14 2x2 12x 14 6x2 0 7x Note that by arranging things in this manner, each term in the last row is obtained by adding the two terms above it. Notice also that the quotient polynomial can be obtained by dividing each of the ﬁrst three terms in the last row by x and adding the results. If you take the time to work back through the original division problem, you will ﬁnd that this is exactly the way we determined the quotient polynomial. This means that we no longer need to write the quotient polynomial down, nor the x in the divisor, to determine our answer. x3 −2 x3+4x2− 5x −14 2x2 12x 14 6x2 0 7x x3 260 Polynomial Functions We’ve streamlined things quite a bit so far, but we can still do more. Let’s take a moment to remind ourselves where the 2x2, 12x and 14 came from in the second row. Each of these terms was obtained by multiplying the terms in the quotient, x2, 6x and 7, respectively, by the −2 in x − 2, then by −1 when we changed the subtraction to addition. Multiplying by −2 then by −1 is the same as multiplying by 2, so we replace the −2 in the divisor by 2. Furthermore, the coeﬃcients of the quotient polynomial match the coeﬃcients of the ﬁrst three terms in the last row, so we now take the plunge and write only the coeﬃcients of the terms to get 2 1 4 −5 −14 14 12 0 7 2 1 6 We have constructed a synthetic division tableau for this polynomial division problem. Let’s rework our division problem using this tableau to see how it greatly streamlines the division process. To divide x3 + 4x2 − 5x − 14 by x − 2, we
write 2 in the place of the divisor and the coeﬃcients of x3 + 4x2 − 5x − 14 in for the dividend. Then ‘bring down’ the ﬁrst coeﬃcient of the dividend. 2 1 4 −5 −14 2 1 4 −5 −14 ↓ 1 Next, take the 2 from the divisor and multiply by the 1 that was ‘brought down’ to get 2. Write this underneath the 4, then add to get 6. 2 1 4 −5 −14 ↓ 2 1 2 1 4 −5 −14 ↓ 2 1 6 Now take the 2 from the divisor times the 6 to get 12, and add it to the −5 to get 7. 2 1 4 −5 −14 ↓ 2 12 1 6 2 1 4 −5 −14 ↓ 2 12 7 1 6 Finally, take the 2 in the divisor times the 7 to get 14, and add it to the −14 to get 0. 2 1 4 −5 −14 ↓ 2 12 14 7 1 6 2 1 4 −5 −14 ↓ 2 14 12 0 7 1 6 3.2 The Factor Theorem and the Remainder Theorem 261 The ﬁrst three numbers in the last row of our tableau are the coeﬃcients of the quotient polynomial. Remember, we started with a third degree polynomial and divided by a ﬁrst degree polynomial, so the quotient is a second degree polynomial. Hence the quotient is x2 + 6x + 7. The number in the box is the remainder. Synthetic division is our tool of choice for dividing polynomials by divisors of the form x − c. It is important to note that it works only for these kinds of divisors.5 Also take note that when a polynomial (of degree at least 1) is divided by x − c, the result will be a polynomial of exactly one less degree. Finally, it is worth the time to trace each step in synthetic division back to its corresponding step in long division. While the authors have done their best to indicate where the algorithm comes from, there is no substitute for working through it yourself. Example 3.2.1. Use synthetic division to perform the following polynomial divisions. Find the quotient and the remainder po
lynomials, then write the dividend, quotient and remainder in the form given in Theorem 3.4. 1. 5x3 − 2x2 + 1 ÷ (x − 3) 2. x3 + 8 ÷ (x + 2) 3. 4 − 8x − 12x2 2x − 3 Solution. 1. When setting up the synthetic division tableau, we need to enter 0 for the coeﬃcient of x in the dividend. Doing so gives 3 5 −2 ↓ 5 0 15 39 13 39 1 117 118 Since the dividend was a third degree polynomial, the quotient is a quadratic polynomial with coeﬃcients 5, 13 and 39. Our quotient is q(x) = 5x2 + 13x + 39 and the remainder is r(x) = 118. According to Theorem 3.4, we have 5x3 −2x2 +1 = (x−3) 5x2 + 13x + 39+118. 2. For this division, we rewrite x + 2 as x − (−2) and proceed as before −2 8 0 0 1 ↓ −2 4 −8 1 −2 4 0 We get the quotient q(x) = x2 − 2x + 4 and the remainder r(x) = 0. Relating the dividend, quotient and remainder gives x3 + 8 = (x + 2) x2 − 2x + 4. 3. To divide 4 − 8x − 12x2 by 2x − 3, two things must be done. First, we write the dividend in descending powers of x as −12x2 − 8x + 4. Second, since synthetic division works only. Our strategy is to ﬁrst divide for factors of the form x − c, we factor 2x − 3 as 2 x − 3 2. The tableau becomes −12x2 − 8x + 4 by 2, to get −6x2 − 4x + 2. Next, we divide by x − 3 2 5You’ll need to use good old-fashioned polynomial long division for divisors of degree larger than 1. 262 Polynomial Functions 3 2 −6 −4 2 ↓ −9 − 39 2 −6 −13 − 35 2 From this, we get −6x2 − 4x + 2 = x − 3 2. Multiplying
both sides by 2 and 2 distributing gives −12x2 − 8x + 4 = (2x − 3) (−6x − 13) − 35. At this stage, we have written −12x2 − 8x + 4 in the form (2x − 3)q(x) + r(x), but how can we be sure the quotient polynomial is −6x − 13 and the remainder is −35? The answer is the word ‘unique’ in Theorem 3.4. The theorem states that there is only one way to decompose −12x2 − 8x + 4 into a multiple of (2x − 3) plus a constant term. Since we have found such a way, we can be sure it is the only way. (−6x − 13) − 35 The next example pulls together all of the concepts discussed in this section. Example 3.2.2. Let p(x) = 2x3 − 5x + 3. 1. Find p(−2) using The Remainder Theorem. Check your answer by substitution. 2. Use the fact that x = 1 is a zero of p to factor p(x) and then ﬁnd all of the real zeros of p. Solution. 1. The Remainder Theorem states p(−2) is the remainder when p(x) is divided by x − (−2). We set up our synthetic division tableau below. We are careful to record the coeﬃcient of x2 as 0, and proceed as above. −2 0 −5 2 ↓ −4 2 −4 3 8 −6 3 −3 According to the Remainder Theorem, p(−2) = −3. We can check this by direct substitution into the formula for p(x): p(−2) = 2(−2)3 − 5(−2) + 3 = −16 + 10 + 3 = −3. 2. The Factor Theorem tells us that since x = 1 is a zero of p, x − 1 is a factor of p(x). To factor p(x), we divide 1 2 0 −5 ↓ 2 2 2 −3 3 2 −3 0 We get a remainder of 0 which veriﬁes that, indeed, p(1) = 0. Our quotient polynomial is a second degree polynomial with coeﬃcients 2
, 2, and −3. So q(x) = 2x2 + 2x − 3. Theorem 3.4 tells us p(x) = (x − 1) 2x2 + 2x − 3. To ﬁnd the remaining real zeros of p, we need to solve 2x2 + 2x − 3 = 0 for x. Since this doesn’t factor nicely, we use the quadratic formula to ﬁnd that the remaining zeros are x = −1±. 2 √ 7 3.2 The Factor Theorem and the Remainder Theorem 263 In Section 3.1, we discussed the notion of the multiplicity of a zero. Roughly speaking, a zero with multiplicity 2 can be divided twice into a polynomial; multiplicity 3, three times and so on. This is illustrated in the next example. Example 3.2.3. Let p(x) = 4x4 − 4x3 − 11x2 + 12x − 3. Given that x = 1 2, ﬁnd all of the real zeros of p. Solution. We set up for synthetic division. Since we are told the multiplicity of 1 continue our tableau and divide 1 2 into the quotient polynomial 2 is a zero of multiplicity 2 is two, we 1 2 1 2 12 −3 3 0 4 −4 −11 ↓ 4 −2 −12 ↓ 2 0 −12 4 2 −1 −6 6 0 −6 0 From the ﬁrst division, we get 4x4 − 4x3 − 11x2 + 12x − 3 = x − 1 2 second division tells us 4x3 − 2x2 − 12x + 6 = x − 1 2 have 4x4 − 4x3 − 11x2 + 12x − 3 = x − 1 2 4x2 − 12 = 0 and get x = ± 4x3 − 2x2 − 12x + 6. The 4x2 − 12. Combining these results, we 2 4x2 − 12. To ﬁnd the remaining zeros of p, we set √ 3. A couple of things about the last example are worth mentioning. First, the extension of the synthetic division tableau for repeated divisions will be a common site in the sections to come. Typically, we will start with a higher order polynomial and peel oﬀ
one zero at a time until we are left with a quadratic, whose roots can always be found using the Quadratic Formula. Secondly, we 3 are found x = ± both factors of p. We can certainly put the Factor Theorem to the test and continue the synthetic division tableau from above to see what happens. 3 are zeros of p. The Factor Theorem guarantees x − 3 and − 12 −3 3 0 −4 −11 −2 −12 2 −1 −6 6 0 −4 4 2 0 −12 12 3 0 3 3 0 This gives us 4x4 − 4x3 − 11x2 + 12x − 3 = x − 1 2 with the constant in front √ 4), or, when written p(x 264 Polynomial Functions We have shown that p is a product of its leading coeﬃcient times linear factors of the form (x − c) where c are zeros of p. It may surprise and delight the reader that, in theory, all polynomials can be reduced to this kind of factorization. We leave that discussion to Section 3.4, because the zeros may not be real numbers. Our ﬁnal theorem in the section gives us an upper bound on the number of real zeros. Theorem 3.7. Suppose f is a polynomial of degree n ≥ 1. Then f has at most n real zeros, counting multiplicities. Theorem 3.7 is a consequence of the Factor Theorem and polynomial multiplication. Every zero c of f gives us a factor of the form (x − c) for f (x). Since f has degree n, there can be at most n of these factors. The next section provides us some tools which not only help us determine where the real zeros are to be found, but which real numbers they may be. We close this section with a summary of several concepts previously presented. You should take the time to look back through the text to see where each concept was ﬁrst introduced and where each connection to the other concepts was made. Connections Between Zeros, Factors and Graphs of Polynomial Functions Suppose p is a polynomial function of degree n ≥ 1. The following statements are equivalent: The real number c is a zero of p p(c) = 0 x = c is a solution to the polynomial equation p(x) = 0 (x − c) is
a factor of p(x) The point (c, 0) is an x-intercept of the graph of y = p(x) 3.2 The Factor Theorem and the Remainder Theorem 265 3.2.1 Exercises In Exercises 1 - 6, use polynomial long division to perform the indicated division. Write the polynomial in the form p(x) = d(x)q(x) + r(x). 1. 4x2 + 3x − 1 ÷ (x − 3) 2. 2x3 − x + 1 ÷ x2 + x + 1 3. 5x4 − 3x3 + 2x2 − 1 ÷ x2 + 4 4. −x5 + 7x3 − x ÷ x3 − x2 + 1 5. 9x3 + 5 ÷ (2x − 3) 6. 4x2 − x − 23 ÷ x2 − 1 In Exercises 7 - 20 use synthetic division to perform the indicated division. Write the polynomial in the form p(x) = d(x)q(x) + r(x). 7. 3x2 − 2x + 1 ÷ (x − 1) 9. 3 − 4x − 2x2 ÷ (x + 1) 11. x3 + 8 ÷ (x + 2) 13. 18x2 − 15x − 25 ÷ x − 5 3 15. 2x3 + x2 + 2x + 1 ÷ x + 1 2 8. x2 − 5 ÷ (x − 5) 10. 4x2 − 5x + 3 ÷ (x + 3) 12. 4x3 + 2x − 3 ÷ (x − 3) 14. 4x2 − 1 ÷ x − 1 2 16. 3x3 − x + 4 ÷ x − 2 3 17. 2x3 − 3x + 1 ÷ x − 1 2 √ 19. x4 − 6x2 + 9 ÷ x − 3 18. 4x4 − 12x3 + 13x2 − 12x + 9 ÷ x − 3 2 20. x6 − 6x4 + 12x2 − 8 ÷ x + √ 2 In Exercises 21 - 30, determine p(c) using the Remainder Theorem for the given polynomial functions and value of c.
If p(c) = 0, factor p(x) = (x − c)q(x). 21. p(x) = 2x2 − x + 1, c = 4 22. p(x) = 4x2 − 33x − 180, c = 12 23. p(x) = 2x3 − x + 6, c = −3 24. p(x) = x3 + 2x2 + 3x + 4, c = −1 25. p(x) = 3x3 − 6x2 + 4x − 8, c = 2 26. p(x) = 8x3 + 12x2 + 6x + 1, c = − 1 2 27. p(x) = x4 − 2x2 + 4, c = 3 2 29. p(x) = x4 + x3 − 6x2 − 7x − 7, c = − √ 28. p(x) = 6x4 − x2 + 2, c = − 2 3 7 30. p(x) = x2 − 4x + 1, c = 2 − √ 3 266 Polynomial Functions In Exercises 31 - 40, you are given a polynomial and one of its zeros. Use the techniques in this section to ﬁnd the rest of the real zeros and factor the polynomial. 31. x3 − 6x2 + 11x − 6, c = 1 32. x3 − 24x2 + 192x − 512, c = 8 33. 3x3 + 4x2 − x − 2, c = 2 3 35. x3 + 2x2 − 3x − 6, c = −2 34. 2x3 − 3x2 − 11x + 6, c = 1 2 36. 2x3 − x2 − 10x + 5, c = 1 2 37. 4x4 − 28x3 + 61x2 − 42x + 9, c = 1 2 is a zero of multiplicity 2 38. x5 + 2x4 − 12x3 − 38x2 − 37x − 12, c = −1 is a zero of multiplicity 3 39. 125x5 − 275x4 − 2265x3 − 3213x2 − 1728x − 324, c = − 3 5 is a zero of multiplicity 3 40. x2 − 2x − 2,
c = 1 − √ 3 In Exercises 41 - 45, create a polynomial p which has the desired characteristics. You may leave the polynomial in factored form. 41. 42. 43. 44. 45. The zeros of p are c = ±2 and c = ±1 The leading term of p(x) is 117x4. The zeros of p are c = 1 and c = 3 c = 3 is a zero of multiplicity 2. The leading term of p(x) is −5x3 The solutions to p(x) = 0 are x = ±3 and x = 6 The leading term of p(x) is 7x4 The point (−3, 0) is a local minimum on the graph of y = p(x). The solutions to p(x) = 0 are x = ±3, x = −2, and x = 4. The leading term of p(x) is −x5. The point (−2, 0) is a local maximum on the graph of y = p(x). p is degree 4. as x → ∞, p(x) → −∞ p has exactly three x-intercepts: (−6, 0), (1, 0) and (117, 0) The graph of y = p(x) crosses through the x-axis at (1, 0). 46. Find a quadratic polynomial with integer coeﬃcients which has x = √ 29 5 3 5 ± as its real zeros. 3.2 The Factor Theorem and the Remainder Theorem 267 3.2.2 Answers 8 4 x + 81 + 283 8 2 x2 + 27 1. 4x2 + 3x − 1 = (x − 3)(4x + 15) + 44 2. 2x3 − x + 1 = x2 + x + 1 (2x − 2) + (−x + 3) 3. 5x4 − 3x3 + 2x2 − 1 = x2 + 4 5x2 − 3x − 18 + (12x + 71) 4. −x5 + 7x3 − x = x3 − x2 + 1 −x2 − x + 6 + 7x2 − 6 5. 9x3 + 5 = (2x − 3) 9 6. 4x2 − x − 23 = x2 − 1 (4) + (−x
− 19) 7. 3x2 − 2x + 1 = (x − 1) (3x + 1) + 2 8. x2 − 5 = (x − 5) (x + 5) + 20 9. 3 − 4x − 2x2 = (x + 1) (−2x − 2) + 5 10. 4x2 − 5x + 3 = (x + 3) (4x − 17) + 54 11. x3 + 8 = (x + 2) x2 − 2x + 4 + 0 12. 4x3 + 2x − 3 = (x − 3) 4x2 + 12x + 38 + 111 13. 18x2 − 15x − 25 = x − 5 3 (4x + 2) + 0 14. 4x2 − 1 = x − 1 2 15. 2x3 + x2 + 2x + 1 = x + 1 2 16. 3x3 − x + 4 = x − 2 3 17. 2x3 − 3x + 1 = x − 1 2 18. 4x4 − 12x3 + 13x2 − 12x + 9 = x − 3 2 19. x4 − 6x2 + 9 = x − 20. x6 − 6x4 + 12x2 − 8 = x + 2x2 + 2 + 0 + 38 9 − 1 4 3x2 + 2x + 1 3 2x2 + x − 5 2 3 x2 − 3x − 3 √ (18x + 15) + 0 3 x3 + √ 2 x4 − 4x3 + 4 2 x5 − √ √ √ 4x3 − 6x2 + 4x − 6 + 0 3 + 0 √ 2 x2 + 4x − 4 √ 2 + 0 21. p(4) = 29 23. p(−3) = −45 22. p(12) = 0, p(x) = (x − 12)(4x + 15) 24. p(−1) = 2 25. p(2) = 0, p(x) = (x − 2) 3x2 + 4 26. p − 1 2 = 0, p(x) = x + 1 2 8x2 + 8x + 2 268 27. p 3 2 29. p(− = 73 16 √ 7) = 0, p(x) = (x +
√ √ 7) x3 + (1 − √ 30. p(2 − 3) = 0, p(x) = (x − (2 − 3))(x − (2 + 3)) √ 28. p − 2 = 74 27 3 √ 7)x − 7)x2 + (1 − √ Polynomial Functions √ 7 31. x3 − 6x2 + 11x − 6 = (x − 1)(x − 2)(x − 3) 32. x3 − 24x2 + 192x − 512 = (x − 8)3 33. 3x3 + 4x2 − x − 2 = 3 x − 2 3 34. 2x3 − 3x2 − 11x + 6 = 2 x − 1 2 (x + 1)2 (x + 2)(x − 3) √ √ 35. x3 + 2x2 − 3x − 6 = (x + 2)(x + 36. 2x3 − x2 − 10x + 5 = 2 x − 1 2 (x + 3)(x − √ 5)(x − 3) √ 5) 37. 4x4 − 28x3 + 61x2 − 42x + 9 = 4 x − 1 2 2 (x − 3)2 38. x5 + 2x4 − 12x3 − 38x2 − 37x − 12 = (x + 1)3(x + 3)(x − 4) 39. 125x5 − 275x4 − 2265x3 − 3213x2 − 1728x − 324 = 125 x + 3 5 3 (x + 2)(x − 6) 40. x2 − 2x − 2 = (x − (1 − √ 3))(x − (1 + √ 3)) 41. p(x) = 117(x + 2)(x − 2)(x + 1)(x − 1) 42. p(x) = −5(x − 1)(x − 3)2 43. p(x) = 7(x + 3)2(x − 3)(x − 6) 44. p(x) = −(x + 2)2(x − 3)(x + 3)(x − 4) 45. p(x) = a(x + 6)2(x − 1)(x − 117) or p(x) = a(x + 6
)(x − 1)(x − 117)2 where a can be any negative real number 46. p(x) = 5x2 − 6x − 4 3.3 Real Zeros of Polynomials 269 3.3 Real Zeros of Polynomials In Section 3.2, we found that we can use synthetic division to determine if a given real number is a zero of a polynomial function. This section presents results which will help us determine good candidates to test using synthetic division. There are two approaches to the topic of ﬁnding the real zeros of a polynomial. The ﬁrst approach (which is gaining popularity) is to use a little bit of Mathematics followed by a good use of technology like graphing calculators. The second approach (for purists) makes good use of mathematical machinery (theorems) only. For completeness, we include the two approaches but in separate subsections.1 Both approaches beneﬁt from the following two theorems, the ﬁrst of which is due to the famous mathematician Augustin Cauchy. It gives us an interval on which all of the real zeros of a polynomial can be found. Theorem 3.8. Cauchy’s Bound: Suppose f (x) = anxn + an−1xn−1 +... + a1x + a0 is a \|an\|,..., \|an−1\| polynomial of degree n with n ≥ 1. Let M be the largest of the numbers: \|an\|. Then all the real zeros of f lie in in the interval [−(M + 1), M + 1]. \|an\|, \|a1\| \|a0\| The proof of this fact is not easily explained within the conﬁnes of this text. This paper contains the result and gives references to its proof. Like many of the results in this section, Cauchy’s Bound is best understood with an example. Example 3.3.1. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. Determine an interval which contains all of the real zeros of f. Solution. To ﬁnd the M stated in Cauchy’s Bound, we take the absolute value of the leading coeﬃcient, in this case \|2\| = 2 and divide
it into the largest (in absolute value) of the remaining coeﬃcients, in this case \| − 6\| = 6. This yields M = 3 so it is guaranteed that all of the real zeros of f lie in the interval [−4, 4]. Whereas the previous result tells us where we can ﬁnd the real zeros of a polynomial, the next theorem gives us a list of possible real zeros. Theorem 3.9. Rational Zeros Theorem: Suppose f (x) = anxn + an−1xn−1 +... + a1x + a0 is a polynomial of degree n with n ≥ 1, and a0, a1,... an are integers. If r is a rational zero of f, then r is of the form ± p q, where p is a factor of the constant term a0, and q is a factor of the leading coeﬃcient an. The Rational Zeros Theorem gives us a list of numbers to try in our synthetic division and that is a lot nicer than simply guessing. If none of the numbers in the list are zeros, then either the polynomial has no real zeros at all, or all of the real zeros are irrational numbers. To see why the Rational Zeros Theorem works, suppose c is a zero of f and c = p q in lowest terms. This means p and q have no common factors. Since f (c) = 0, we have p q +... + a1 + a0 = 0. p q p q + an−1 n−1 n an 1Carl is the purist and is responsible for all of the theorems in this section. Jeﬀ, on the other hand, has spent too much time in school politics and has been polluted with notions of ‘compromise.’ You can blame the slow decline of civilization on him and those like him who mingle Mathematics with technology. 270 Polynomial Functions Multiplying both sides of this equation by qn, we clear the denominators to get anpn + an−1pn−1q +... + a1pqn−1 + a0qn = 0 Rearranging this equation, we get anpn = −an−1pn−1q −... − a1pqn−1 − a0qn Now,
the left hand side is an integer multiple of p, and the right hand side is an integer multiple of q. (Can you see why?) This means anpn is both a multiple of p and a multiple of q. Since p and q have no common factors, an must be a multiple of q. If we rearrange the equation as anpn + an−1pn−1q +... + a1pqn−1 + a0qn = 0 a0qn = −anpn − an−1pn−1q −... − a1pqn−1 we can play the same game and conclude a0 is a multiple of p, and we have the result. Example 3.3.2. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. Use the Rational Zeros Theorem to list all of the possible rational zeros of f. Solution. To generate a complete list of rational zeros, we need to take each of the factors of constant term, a0 = −3, and divide them by each of the factors of the leading coeﬃcient a4 = 2. The factors of −3 are ± 1 and ± 3. Since the Rational Zeros Theorem tacks on a ± anyway, for the moment, we consider only the positive factors 1 and 3. The factors of 2 are 1 and 2, so the Rational Zeros Theorem gives the list ± 1 2, ± 3 2, ± 1, ± 3 or ± 1 2, ± 3 Our discussion now diverges between those who wish to use technology and those who do not. 3.3.1 For Those Wishing to use a Graphing Calculator At this stage, we know not only the interval in which all of the zeros of f (x) = 2x4 +4x3 −x2 −6x−3 are located, but we also know some potential candidates. We can now use our calculator to help us determine all of the real zeros of f, as illustrated in the next example. Example 3.3.3. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. 1. Graph y = f (x) on the calculator using the interval obtained in Example 3.3.1 as a guide. 2. Use the graph to shorten the list of possible rational zeros obtained in Example 3.
3.2. 3. Use synthetic division to ﬁnd the real zeros of f, and state their multiplicities. Solution. 1. In Example 3.3.1, we determined all of the real zeros of f lie in the interval [−4, 4]. We set our window accordingly and get 3.3 Real Zeros of Polynomials 271 2. In Example 3.3.2, we learned that any rational zero of f must be in the list ± 1 2, ± 3. From the graph, it looks as if we can rule out any of the positive rational zeros, since the graph seems to cross the x-axis at a value just a little greater than 1. On the negative side, −1 looks good, so we try that for our synthetic division. 2, ± 1, ± 3 −1 2 ↓ −2 −2 2 4 −1 −6 −3 3 0 3 2 −3 −3 We have a winner! Remembering that f was a fourth degree polynomial, we know that our quotient is a third degree polynomial. If we can do one more successful division, we will have knocked the quotient down to a quadratic, and, if all else fails, we can use the quadratic formula to ﬁnd the last two zeros. Since there seems to be no other rational zeros to try, we continue with −1. Also, the shape of the crossing at x = −1 leads us to wonder if the zero x = −1 has multiplicity 3. −1 −1 4 −1 −6 −3 3 0 2 ↓ −2 −2 2 ↓ −2 2 3 2 −3 −3 3 0 0 −3 0 2, which gives us x = ± Success! Our quotient polynomial is now 2x2 − 3. Setting this to zero gives 2x2 − 3 = 0, or √ 6 x2 = 3 2. Concerning multiplicities, based on our division, we have that −1 has a multiplicity of at least 2. The Factor Theorem tells us our remaining zeros, √ 6 ± 2, each have multiplicity at least 1. However, Theorem 3.7 tells us f can have at most 4 real zeros, counting multiplicity, and so we conclude that −1 is of multiplicity exactly 2 and √ 6 2 each has multiplicity 1. (Thus,
we were wrong to think that −1 had multiplicity 3.) ± It is interesting to note that we could greatly improve on the graph of y = f (x) in the previous example given to us by the calculator. For instance, from our determination of the zeros of f and √ 6 their multiplicities, we know the graph crosses at x = − 2 ≈ −1.22 then turns back upwards to touch the x−axis at x = −1. This tells us that, despite what the calculator showed us the ﬁrst time, there is a relative maximum occurring at x = −1 and not a ‘ﬂattened crossing’ as we originally 272 Polynomial Functions believed. After resizing the window, we see not only the relative maximum but also a relative minimum2 just to the left of x = −1 which shows us, once again, that Mathematics enhances the technology, instead of vice-versa. Our next example shows how even a mild-mannered polynomial can cause problems. Example 3.3.4. Let f (x) = x4 + x2 − 12. 1. Use Cauchy’s Bound to determine an interval in which all of the real zeros of f lie. 2. Use the Rational Zeros Theorem to determine a list of possible rational zeros of f. 3. Graph y = f (x) using your graphing calculator. 4. Find all of the real zeros of f and their multiplicities. Solution. 1. Applying Cauchy’s Bound, we ﬁnd M = 12, so all of the real zeros lie in the interval [−13, 13]. 2. Applying the Rational Zeros Theorem with constant term a0 = −12 and leading coeﬃcient a4 = 1, we get the list {± 1, ± 2, ± 3, ± 4, ± 6, ± 12}. 3. Graphing y = f (x) on the interval [−13, 13] produces the graph below on the left. Zooming in a bit gives the graph below on the right. Based on the graph, none of our rational zeros will work. (Do you see why not?) 2This is an example of what is called ‘hidden behavior.’ 3.3 Real Zeros of Polynomials 273 4. From the graph, we know f has two
real zeros, one positive, and one negative. Our only hope at this point is to try and ﬁnd the zeros of f by setting f (x) = x4 + x2 − 12 = 0 and solving. If we stare at this equation long enough, we may recognize it as a ‘quadratic in disguise’ or ‘quadratic in form’. In other words, we have three terms: x4, x2 and 12, and the exponent on the ﬁrst term, x4, is exactly twice that of the second term, x2. We may rewrite this as x22 + x2 − 12 = 0. To better see the forest for the trees, we momentarily replace x2 with the variable u. In terms of u, our equation becomes u2 + u − 12 = 0, which we can readily factor as (u + 4)(u − 3) = 0. In terms of x, this means x4 + x2 − 12 = x2 − 3 x2 + 4 = 0. 3, or x2 = −4, which admits no real solutions. Since We get x2 = 3, which gives us x = ± √ 3 ≈ 1.73, the two zeros match what we expected from the graph. In terms of multiplicity, 3 are factors of f (x). Since f (x) can the Factor Theorem guarantees x − 3 and x + be factored as f (x) = x2 − 3 x2 + 4, and x2 + 4 has no real zeros, the quantities x − 3 3 must both be factors of x2 − 3. According to Theorem 3.7, x2 − 3 can have at and x + √ most 2 zeros, counting multiplicity, hence each of ± 3 is a zero of f of multiplicity 1. √ √ √ √ √ The technique used to factor f (x) in Example 3.3.4 is called u-substitution. We shall see more of this technique in Section 5.3. In general, substitution can help us identify a ‘quadratic in disguise’ provided that there are exactly three terms and the exponent of the ﬁrst term is exactly twice that of the second. It is entirely possible that a polynomial has no real roots at all, or worse, it has real roots but none of the
techniques discussed in this section can help us ﬁnd them exactly. In the latter case, we are forced to approximate, which in this subsection means we use the ‘Zero’ command on the graphing calculator. 3.3.2 For Those Wishing NOT to use a Graphing Calculator Suppose we wish to ﬁnd the zeros of f (x) = 2x4 + 4x3 − x2 − 6x − 3 without using the calculator. In this subsection, we present some more advanced mathematical tools (theorems) to help us. Our ﬁrst result is due to Ren´e Descartes. Theorem 3.10. Descartes’ Rule of Signs: Suppose f (x) is the formula for a polynomial function written with descending powers of x. If P denotes the number of variations of sign in the formula for f (x), then the number of positive real zeros (counting multiplicity) is one of the numbers {P, P − 2, P − 4,... }. If N denotes the number of variations of sign in the formula for f (−x), then the number of negative real zeros (counting multiplicity) is one of the numbers {N, N − 2, N − 4,... }. A few remarks are in order. First, to use Descartes’ Rule of Signs, we need to understand what is meant by a ‘variation in sign’ of a polynomial function. Consider f (x) = 2x4 + 4x3 − x2 − 6x − 3. If we focus on only the signs of the coeﬃcients, we start with a (+), followed by another (+), then switch to (−), and stay (−) for the remaining two coeﬃcients. Since the signs of the coeﬃcients switched once as we read from left to right, we say that f (x) has one variation in sign. When 274 Polynomial Functions we speak of the variations in sign of a polynomial function f we assume the formula for f (x) is written with descending powers of x, as in Deﬁnition 3.1, and concern ourselves only with the nonzero coeﬃcients. Second, unlike the Rational Zeros Theorem, Descartes’ Rule of Signs gives us an estimate to
the number of positive and negative real zeros, not the actual value of the zeros. Lastly, Descartes’ Rule of Signs counts multiplicities. This means that, for example, if one of the zeros has multiplicity 2, Descsartes’ Rule of Signs would count this as two zeros. Lastly, note that the number of positive or negative real zeros always starts with the number of sign changes and decreases by an even number. For example, if f (x) has 7 sign changes, then, counting multplicities, f has either 7, 5, 3 or 1 positive real zero. This implies that the graph of y = f (x) crosses the positive x-axis at least once. If f (−x) results in 4 sign changes, then, counting multiplicities, f has 4, 2 or 0 negative real zeros; hence, the graph of y = f (x) may not cross the negative x-axis at all. The proof of Descartes’ Rule of Signs is a bit technical, and can be found here. Example 3.3.5. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. Use Descartes’ Rule of Signs to determine the possible number and location of the real zeros of f. Solution. As noted above, the variations of sign of f (x) is 1. This means, counting multiplicities, f has exactly 1 positive real zero. Since f (−x) = 2(−x)4 + 4(−x)3 − (−x)2 − 6(−x) − 3 = 2x4 − 4x3 − x2 + 6x − 3 has 3 variations in sign, f has either 3 negative real zeros or 1 negative real zero, counting multiplicities. Cauchy’s Bound gives us a general bound on the zeros of a polynomial function. Our next result helps us determine bounds on the real zeros of a polynomial as we synthetically divide which are often sharper3 bounds than Cauchy’s Bound. Theorem 3.11. Upper and Lower Bounds: Suppose f is a polynomial of degree n ≥ 1. If c > 0 is synthetically divided into f and all of the numbers in the ﬁnal line of the division tableau have the same signs, then c is an upper
bound for the real zeros of f. That is, there are no real zeros greater than c. If c < 0 is synthetically divided into f and the numbers in the ﬁnal line of the division tableau alternate signs, then c is a lower bound for the real zeros of f. That is, there are no real zeros less than c. NOTE: If the number 0 occurs in the ﬁnal line of the division tableau in either of the above cases, it can be treated as (+) or (−) as needed. The Upper and Lower Bounds Theorem works because of Theorem 3.4. For the upper bound part of the theorem, suppose c > 0 is divided into f and the resulting line in the division tableau contains, for example, all nonnegative numbers. This means f (x) = (x − c)q(x) + r, where the coeﬃcients of the quotient polynomial and the remainder are nonnegative. (Note that the leading coeﬃcient of q is the same as f so q(x) is not the zero polynomial.) If b > c, then f (b) = (b − c)q(b) + r, where (b − c) and q(b) are both positive and r ≥ 0. Hence f (b) > 0 which shows b cannot be a zero of f. Thus no real number b > c can be a zero of f, as required. A similar argument proves 3That is, better, or more accurate. 3.3 Real Zeros of Polynomials 275 f (b) < 0 if all of the numbers in the ﬁnal line of the synthetic division tableau are non-positive. To prove the lower bound part of the theorem, we note that a lower bound for the negative real zeros of f (x) is an upper bound for the positive real zeros of f (−x). Applying the upper bound portion to f (−x) gives the result. (Do you see where the alternating signs come in?) With the additional mathematical machinery of Descartes’ Rule of Signs and the Upper and Lower Bounds Theorem, we can ﬁnd the real zeros of f (x) = 2x4 + 4x3 − x2 − 6x − 3 without the use of a graphing calculator. Example 3.3.
6. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. 1. Find all of the real zeros of f and their multiplicities. 2. Sketch the graph of y = f (x). Solution. 2, ± 1, ± 3 1. We know from Cauchy’s Bound that all of the real zeros lie in the interval [−4, 4] and that our possible rational zeros are ± 1 2 and ± 3. Descartes’ Rule of Signs guarantees us at least one negative real zero and exactly one positive real zero, counting multiplicity. We try our positive rational zeros, starting with the smallest, 1 2. Since the remainder isn’t zero, we know 1 2 isn’t a zero. Sadly, the ﬁnal line in the division tableau has both positive and negative numbers, so 1 2 is not an upper bound. The only information we get from this division is courtesy of the Remainder Theorem which tells us f 1 is 2 on the graph of f. We continue to our next possible zero, 1. As before, the only information we can glean from this is that (1, −4) is on the graph of f. When we try our next possible zero, 3 2, we get that it is not a zero, and we also see that it is an upper bound on the zeros of f, since all of the numbers in the ﬁnal line of the division tableau are positive. This means there is no point trying our last possible rational zero, 3. Descartes’ Rule of Signs guaranteed us a positive real zero, and at this point we have shown this zero is irrational. Furthermore, the Intermediate Value Theorem, Theorem 3.1, tells us the zero lies between 1 and 3 2, since f (1) < 0 and f 3 2 8 so the point 1 = − 45 2, − 45 > 0. 8 1 2 2 4 −1 −6 3 5 ↓ 1 4 2 −3 − 21 8 2 5 3 2 − 21 4 − 45 8 1 2 4 −1 −6 −3 5 −1 ↓ 2 6 2 6 5 −1 −4 3 2 2 4 −1 −6 −3 99 ↓ 3 8 57 4 21 2 2 7 19 2 33 4 75 8 We now turn our attention to negative real zeros. We try the largest
possible zero, − 1 2. Synthetic division shows us it is not a zero, nor is it a lower bound (since the numbers in the ﬁnal line of the division tableau do not alternate), so we proceed to −1. This division shows −1 is a zero. Descartes’ Rule of Signs told us that we may have up to three negative real zeros, counting multiplicity, so we try −1 again, and it works once more. At this point, we have taken f, a fourth degree polynomial, and performed two successful divisions. Our quotient polynomial is quadratic, so we look at it to ﬁnd the remaining zeros. 276 − 1 2 4 −1 −6 5 4 2 ↓ −1 − 3 2 3 − 5 2 − 19 2 4 − 5 8 −3 19 8 −1 −1 Polynomial Functions 4 −1 −6 −3 3 0 2 ↓ −2 −2 2 ↓ −2 2 3 2 −3 −3 0 3 0 −3 0 √ 6 Setting the quotient polynomial equal to zero yields 2x2 − 3 = 0, so that x2 = 3 2, or x = ± 2. √ 6 Descartes’ Rule of Signs tells us that the positive real zero we found, 2, has multiplicity 1. Descartes also tells us the total multiplicity of negative real zeros is 3, which forces −1 to be a zero of multiplicity 2 and − √ 6 2 to have multiplicity 1. 2. We know the end behavior of y = f (x) resembles that of its leading term y = 2x4. This √ 6 means that the graph enters the scene in Quadrant II and exits in Quadrant I. Since ± 2 are zeros of odd multiplicity, we have that the graph crosses through the x-axis at the points. Since −1 is a zero of multiplicity 2, the graph of y = f (x) touches and √ − 6 2, 0 √ 6 2, 0 and rebounds oﬀ the x-axis at (−1, 0). Putting this together, we get y x You can see why the ‘no calculator’ approach is not very popular these days. It requires more computation and more theorems than the alternative.4 In general, no matter how many theorems you throw at a po
lynomial, it may well be impossible5 to ﬁnd their zeros exactly. The polynomial f (x) = x5 − x − 1 is one such beast.6 According to Descartes’ Rule of Signs, f has exactly one positive real zero, and it could have two negative real zeros, or none at all. The Rational Zeros 4This is apparently a bad thing. 5We don’t use this word lightly; it can be proven that the zeros of some polynomials cannot be expressed using the usual algebraic symbols. 6See this page. 3.3 Real Zeros of Polynomials 277 Test gives us ±1 as rational zeros to try but neither of these work since f (1) = f (−1) = −1. If we try the substitution technique we used in Example 3.3.4, we ﬁnd f (x) has three terms, but the exponent on the x5 isn’t exactly twice the exponent on x. How could we go about approximating the positive zero without resorting to the ‘Zero’ command of a graphing calculator? We use the Bisection Method. The ﬁrst step in the Bisection Method is to ﬁnd an interval on which f changes sign. We know f (1) = −1 and we ﬁnd f (2) = 29. By the Intermediate Value Theorem, we know that the zero of f lies in the interval [1, 2]. Next, we ‘bisect’ this interval and ﬁnd the midpoint is 1.5. We have that f (1.5) ≈ 5.09. This means that our zero is between 1 and 1.5, since f changes sign on this interval. Now, we ‘bisect’ the interval [1, 1.5] and ﬁnd f (1.25) ≈ 0.80, so now we have the zero between 1 and 1.25. Bisecting [1, 1.25], we ﬁnd f (1.125) ≈ −0.32, which means the zero of f is between 1.125 and 1.25. We continue in this fashion until we have ‘sandwiched’ the zero between two numbers which diﬀer by no more than a desired accuracy. You can think of the Bis
ection Method as reversing the sign diagram process: instead of ﬁnding the zeros and checking the sign of f using test values, we are using test values to determine where the signs switch to ﬁnd the zeros. It is a slow and tedious, yet fool-proof, method for approximating a real zero. Our next example reminds us of the role ﬁnding zeros plays in solving equations and inequalities. Example 3.3.7. 1. Find all of the real solutions to the equation 2x5 + 6x3 + 3 = 3x4 + 8x2. 2. Solve the inequality 2x5 + 6x3 + 3 ≤ 3x4 + 8x2. 3. Interpret your answer to part 2 graphically, and verify using a graphing calculator. Solution. 1. Finding the real solutions to 2x5 + 6x3 + 3 = 3x4 + 8x2 is the same as ﬁnding the real solutions to 2x5 − 3x4 + 6x3 − 8x2 + 3 = 0. In other words, we are looking for the real zeros of p(x) = 2x5 − 3x4 + 6x3 − 8x2 + 3. Using the techniques developed in this section, we get 1 1 − 1 2 2 −3 ↓ 2 −1 ↓ 2 2 1 ↓ −1 0 2 2 −1 6 −8 3 0 5 −3 −3 0 5 −3 −3 3 6 1 6 0 3 0 −3 0 6 The quotient polynomial is 2x2 + 6 which has no real zeros so we get x = − 1 2 and x = 1. 2. To solve this nonlinear inequality, we follow the same guidelines set forth in Section 2.4: we get 0 on one side of the inequality and construct a sign diagram. Our original inequality can be rewritten as 2x5−3x4+6x3−8x2+3 ≤ 0. We found the zeros of p(x) = 2x5−3x4+6x3−8x2+3 in part 1 to be x = − 1 2 and x = 1. We construct our sign diagram as before. 278 Polynomial Functions (−) 0 (+) 0 (+) − 1 2 −1 1 0 2, and we know p(x) = 0 at x = − 1
The solution to p(x) < 0 is −∞, − 1 2 ∪ {1}. the solution to p(x) ≤ 0 is −∞, − 1 2 2 and x = 1. Hence, 3. To interpret this solution graphically, we set f (x) = 2x5 + 6x3 + 3 and g(x) = 3x4 + 8x2. We recall that the solution to f (x) ≤ g(x) is the set of x values for which the graph of f is below the graph of g (where f (x) < g(x)) along with the x values where the two graphs intersect (f (x) = g(x)). Graphing f and g on the calculator produces the picture on the lower left. (The end behavior should tell you which is which.) We see that the graph of f is below the graph of g on −∞, − 1. However, it is diﬃcult to see what is happening near 2 x = 1. Zooming in (and making the graph of g thicker), we see that the graphs of f and g do intersect at x = 1, but the graph of g remains below the graph of f on either side of x = 1. Our last example revisits an application from page 247 in the Exercises of Section 3.1. Example 3.3.8. Suppose the proﬁt P, in thousands of dollars, from producing and selling x hundred LCD TVs is given by P (x) = −5x3 + 35x2 − 45x − 25, 0 ≤ x ≤ 10.07. How many TVs should be produced to make a proﬁt? Check your answer using a graphing utility. Solution. To ‘make a proﬁt’ means to solve P (x) = −5x3 + 35x2 − 45x − 25 > 0, which we do analytically using a sign diagram. To simplify things, we ﬁrst factor out the −5 common to all the coeﬃcients to get −5 x3 − 7x2 + 9x − 5 > 0, so we can just focus on ﬁnding the zeros of f (x) = x3 − 7x2 + 9x + 5. The possible rational zeros of f are ±1 and ±5, and going
through the usual computations, we ﬁnd x = 5 is the only rational zero. Using this, we factor f (x) = x3 − 7x2 + 9x + 5 = (x − 5) x2 − 2x − 1, and we ﬁnd the remaining zeros by applying the Quadratic Formula to x2 − 2x − 1 = 0. We ﬁnd three real zeros, x = 1 − 2 = −0.414..., x = 1 + 2 = 2.414..., and x = 5, of which only the last two fall in the applied domain of [0, 10.07]. We choose x = 0, x = 3 and x = 10.07 as our test values and plug them into the function P (x) = −5x3 + 35x2 − 45x − 25 (not f (x) = x3 − 7x2 + 9x − 5) to get the sign diagram below. √ √ 3.3 Real Zeros of Polynomials 279 (−) 1 + 0 (+) 0 √ 5 2 (−) 0 3 10.07 √ √ 2, 5). Since x measures the number of TVs in hundreds, We see immediately that P (x) > 0 on (1+ 2 corresponds to 241.4... TVs. Since we can’t produce a fractional part of a TV, we need x = 1 + to choose between producing 241 and 242 TVs. From the sign diagram, we see that P (2.41) < 0 but P (2.42) > 0 so, in this case we take the next larger integer value and set the minimum production to 242 TVs. At the other end of the interval, we have x = 5 which corresponds to 500 TVs. Here, we take the next smaller integer value, 499 TVs to ensure that we make a proﬁt. Hence, in order to make a proﬁt, at least 242, but no more than 499 TVs need to be produced. To check our answer using a calculator, we graph y = P (x) and make use of the ‘Zero’ command. We see that the calculator approximations bear out our analysis.7 7Note that the y-coordinates of the points here aren’t registered as 0. They are expressed in Scientiﬁc Notation
. For instance, 1E − 11 corresponds to 0.00000000001, which is pretty close in the calculator’s eyes8to 0. 8but not a Mathematician’s 280 3.3.3 Exercises Polynomial Functions In Exercises 1 - 10, for the given polynomial: Use Cauchy’s Bound to ﬁnd an interval containing all of the real zeros. Use the Rational Zeros Theorem to make a list of possible rational zeros. Use Descartes’ Rule of Signs to list the possible number of positive and negative real zeros, counting multiplicities. 1. f (x) = x3 − 2x2 − 5x + 6 2. f (x) = x4 + 2x3 − 12x2 − 40x − 32 3. f (x) = x4 − 9x2 − 4x + 12 4. f (x) = x3 + 4x2 − 11x + 6 5. f (x) = x3 − 7x2 + x − 7 6. f (x) = −2x3 + 19x2 − 49x + 20 7. f (x) = −17x3 + 5x2 + 34x − 10 8. f (x) = 36x4 − 12x3 − 11x2 + 2x + 1 9. f (x) = 3x3 + 3x2 − 11x − 10 10. f (x) = 2x4 + x3 − 7x2 − 3x + 3 In Exercises 11 - 30, ﬁnd the real zeros of the polynomial using the techniques speciﬁed by your instructor. State the multiplicity of each real zero. 11. f (x) = x3 − 2x2 − 5x + 6 12. f (x) = x4 + 2x3 − 12x2 − 40x − 32 13. f (x) = x4 − 9x2 − 4x + 12 14. f (x) = x3 + 4x2 − 11x + 6 15. f (x) = x3 − 7x2 + x − 7 16. f (x) = −2x3 + 19x2 − 49x + 20 17. f (x) = −17x3 + 5x2 + 34x − 10 18. f
(x) = 36x4 − 12x3 − 11x2 + 2x + 1 19. f (x) = 3x3 + 3x2 − 11x − 10 20. f (x) = 2x4 + x3 − 7x2 − 3x + 3 21. f (x) = 9x3 − 5x2 − x 22. f (x) = 6x4 − 5x3 − 9x2 23. f (x) = x4 + 2x2 − 15 24. f (x) = x4 − 9x2 + 14 25. f (x) = 3x4 − 14x2 − 5 26. f (x) = 2x4 − 7x2 + 6 27. f (x) = x6 − 3x3 − 10 28. f (x) = 2x6 − 9x3 + 10 29. f (x) = x5 − 2x4 − 4x + 8 30. f (x) = 2x5 + 3x4 − 18x − 27 3.3 Real Zeros of Polynomials 281 In Exercises 31 - 33, use your calculator,9 to help you ﬁnd the real zeros of the polynomial. State the multiplicity of each real zero. 31. f (x) = x5 − 60x3 − 80x2 + 960x + 2304 32. f (x) = 25x5 − 105x4 + 174x3 − 142x2 + 57x − 9 33. f (x) = 90x4 − 399x3 + 622x2 − 399x + 90 34. Find the real zeros of f (x) = x3 − 1 72 x + 1 72 by ﬁrst ﬁnding a polynomial q(x) with integer coeﬃcients such that q(x) = N · f (x) for some integer N. (Recall that the Rational Zeros Theorem required the polynomial in question to have integer coeﬃcients.) Show that f and q have the same real zeros. 12 x2 − 7 In Exercises 35 - 44, ﬁnd the real solutions of the polynomial equation. (See Example 3.3.7.) 35. 9x3 = 5x2 + x 37. x3 + 6 = 2x
2 + 5x 39. x3 − 7x2 = 7 − x 41. x3 + x2 = 11x + 10 3 43. 14x2 + 5 = 3x4 36. 9x2 + 5x3 = 6x4 38. x4 + 2x3 = 12x2 + 40x + 32 40. 2x3 = 19x2 − 49x + 20 42. x4 + 2x2 = 15 44. 2x5 + 3x4 = 18x + 27 In Exercises 45 - 54, solve the polynomial inequality and state your answer using interval notation. 45. −2x3 + 19x2 − 49x + 20 > 0 46. x4 − 9x2 ≤ 4x − 12 47. (x − 1)2 ≥ 4 49. x4 ≤ 16 + 4x − x3 48. 4x3 ≥ 3x + 1 50. 3x2 + 2x < x4 51. x3 + 2x2 2 53. 2x4 > 5x2 + 3 < x + 2 52. x3 + 20x 8 54. x6 + x3 ≥ 6 ≥ x2 + 2 55. In Example 3.1.3 in Section 3.1, a box with no top is constructed from a 10 inch × 12 inch piece of cardboard by cutting out congruent squares from each corner of the cardboard and then folding the resulting tabs. We determined the volume of that box (in cubic inches) is given by V (x) = 4x3 − 44x2 + 120x, where x denotes the length of the side of the square which is removed from each corner (in inches), 0 < x < 5. Solve the inequality V (x) ≥ 80 analytically and interpret your answer in the context of that example. 9You can do these without your calculator, but it may test your mettle! 282 Polynomial Functions 56. From Exercise 32 in Section 3.1, C(x) =.03x3 − 4.5x2 + 225x + 250, for x ≥ 0 models the cost, in dollars, to produce x PortaBoy game systems. If the production budget is $5000, ﬁnd the number of game systems which can be produced and still remain under budget. 57. Let f (x) = 5x7 − 33x6 + 3x5 − 71x4 − 597x
3 + 2097x2 − 1971x + 567. With the help of your classmates, ﬁnd the x- and y- intercepts of the graph of f. Find the intervals on which the function is increasing, the intervals on which it is decreasing and the local extrema. Sketch the graph of f, using more than one picture if necessary to show all of the important features of the graph. 58. With the help of your classmates, create a list of ﬁve polynomials with diﬀerent degrees whose real zeros cannot be found using any of the techniques in this section. 3.3 Real Zeros of Polynomials 283 3.3.4 Answers 1. For f (x) = x3 − 2x2 − 5x + 6 All of the real zeros lie in the interval [−7, 7] Possible rational zeros are ±1, ±2, ±3, ±6 There are 2 or 0 positive real zeros; there is 1 negative real zero 2. For f (x) = x4 + 2x3 − 12x2 − 40x − 32 All of the real zeros lie in the interval [−41, 41] Possible rational zeros are ±1, ±2, ±4, ±8, ±16, ±32 There is 1 positive real zero; there are 3 or 1 negative real zeros 3. For f (x) = x4 − 9x2 − 4x + 12 All of the real zeros lie in the interval [−13, 13] Possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±12 There are 2 or 0 positive real zeros; there are 2 or 0 negative real zeros 4. For f (x) = x3 + 4x2 − 11x + 6 All of the real zeros lie in the interval [−12, 12] Possible rational zeros are ±1, ±2, ±3, ±6 There are 2 or 0 positive real zeros; there is 1 negative real zero 5. For f (x) = x3 − 7x2 + x − 7 All of the real zeros lie in the interval [−8, 8] Possible rational zeros are ±1, ±7 There are 3 or 1 positive real zeros; there are no negative real zeros 6. For f (x) = −2x3 + 19x2
− 49x + 20 All of the real zeros lie in the interval − 51 2, ±1, ±2, ± 5 Possible rational zeros are ± 1 There are 3 or 1 positive real zeros; there are no negative real zeros 2, 51 2, ±4, ±5, ±10, ±20 2 7. For f (x) = −17x3 + 5x2 + 34x − 10 All of the real zeros lie in the interval [−3, 3] Possible rational zeros are ± 1 17, ± 2 There are 2 or 0 positive real zeros; there is 1 negative real zero 17, ±1, ±2, ±5, ±10 17, ± 10 17, ± 5 284 Polynomial Functions 8. For f (x) = 36x4 − 12x3 − 11x2 + 2x + 1 All of the real zeros lie in the interval − 4 Possible rational zeros are ± 1 18, ± 1 There are 2 or 0 positive real zeros; there are 2 or 0 negative real zeros 3, 4 12, ± 1 36, ± 1 3 9. For f (x) = 3x3 + 3x2 − 11x − 10 All of the real zeros lie in the interval − 14 3, 14 Possible rational zeros are ± 1 3, ± 10 3, ± 2 3, ±1, ±2, ±5, ±10 There is 1 positive real zero; there are 2 or 0 negative real zeros 3, ± 5 3 10. For f (x) = 2x4 + x3 − 7x2 − 3x + 3 All of the real zeros lie in the interval − 9 2, 9 Possible rational zeros are ± 1 2, ±1, ± 3 2, ±3 There are 2 or 0 positive real zeros; there are 2 or 0 negative real zeros 2 11. f (x) = x3 − 2x2 − 5x + 6 x = −2, x = 1, x = 3 (each has mult. 1) 12. f (x) = x4 + 2x3 − 12x2 − 40x − 32 x = −2 (mult. 3), x = 4 (mult. 1) 13. f (x) = x4 − 9x2 − 4x + 12 x = −2 (mult. 2), x = 1 (mult. 1), x = 3 (
mult. 1) 14. f (x) = x3 + 4x2 − 11x + 6 x = −6 (mult. 1), x = 1 (mult. 2) 15. f (x) = x3 − 7x2 + x − 7 x = 7 (mult. 1) 16. f (x) = −2x3 + 19x2 − 49x + 20 x = 1 2, x = 4, x = 5 (each has mult. 1) 17. f (x) = −17x3 + 5x2 + 34x − 10 √ x = 5 17, x = ± 2 (each has mult. 1) 18. f (x) = 36x4 − 12x3 − 11x2 + 2x + 1 2 (mult. 2), x = − 1 3 (mult. 2) x = 1 19. f (x) = 3x3 + 3x2 − 11x − 10 x = −2, x = 3± 69 (each has mult. 1) √ 6 3.3 Real Zeros of Polynomials 285 20. f (x) = 2x4 + x3 − 7x2 − 3x + 3 √ x = −1, x = 1 2, x = ± 21. f (x) = 9x3 − 5x2 − x √ x = 0, x = 5± 18 61 (each has mult. 1) 3 (each mult. 1) 22. f (x) = 6x4 − 5x3 − 9x2 x = 0 (mult. 2), x = 5± √ 12 241 (each has mult. 1) 23. f (x) = x4 + 2x2 − 15 x = ± 3 (each has mult. 1) 24. f (x) = x4 − 9x2 + 14 √ x = ± 2, x = ± 7 (each has mult. 1) 25. f (x) = 3x4 − 14x2 − 5 x = ± 5 (each has mult. 1) √ √ √ 26. f (x) = 2x4 − 7x2 + each has mult. 1) 27. f (x) = x6 − 3x3 − 10 √ x = 3 √ −2 = − 3 √ 2, x = 3 5 (each has mult. 1) 28. f
(x) = 2x6 − 9x3 + 10 √ 3√ 20 2, x = 3 29. f (x) = x5 − 2x4 − 4x + 8 x = √ 2 (each has mult. 1) x = 2, x = ± 2 (each has mult. 1) 30. f (x) = 2x5 + 3x4 − 18x − 27 √ each has mult. 1) 31. f (x) = x5 − 60x3 − 80x2 + 960x + 2304 x = −4 (mult. 3), x = 6 (mult. 2) 32. f (x) = 25x5 − 105x4 + 174x3 − 142x2 + 57x − 9 x = 3 5 (mult. 2), x = 1 (mult. 3) 33. f (x) = 90x4 − 399x3 + 622x2 − 399x + 90 each has mult. 1) 34. We choose q(x) = 72x3 − 6x2 − 7x + 1 = 72 · f (x). Clearly f (x) = 0 if and only if q(x) = 0 4 are the real zeros so they have the same real zeros. In this case, x = − 1 of both f and q. 6 and x = 1 3, x = 1 286 √ 35. x = 0, 5± 18 61 37. x = −2, 1, 3 39. x = 7 41. x = −2, 3± √ 6 69 √ 43. x = ± 5 45. (−∞, 1 2 ) ∪ (4, 5) 47. (−∞, −1] ∪ [3, ∞) Polynomial Functions 36. x = 0, 5± √ 241 12 38. x = −2, 4 40. x = 1 2, 4, 5 √ 42. x = ± 3 44 46. {−2} ∪ [1, 3] 48. − 1 2 ∪ [1, ∞) 49. [−2, 2] 50. (−∞, −1) ∪ (−1, 0) ∪ (2, ∞) 51. (−∞, −2) ∪ − √ √ 2 52. {2} ∪ [4, ∞) 2, 53.
(−∞, − 3) ∪ ( √ √ 3, ∞) √ √ √ 54. (−∞, − 3 √ 3 ) ∪ ( 3 √ 2, ∞) 55. V (x) ≥ 80 on [1, 5 − 5] ∪ [5 + 5, ∞). Only the portion [1, 5 − 5] lies in the applied domain, however. In the context of the problem, this says for the volume of the box to be at least 80 cubic inches, the square removed from each corner needs to have a side length of at least 1 inch, but no more than 5 − 5 ≈ 2.76 inches. √ 56. C(x) ≤ 5000 on (approximately) (−∞, 82.18]. The portion of this which lies in the applied domain is (0, 82.18]. Since x represents the number of game systems, we check C(82) = 4983.04 and C(83) = 5078.11, so to remain within the production budget, anywhere between 1 and 82 game systems can be produced. 3.4 Complex Zeros and the Fundamental Theorem of Algebra 287 3.4 Complex Zeros and the Fundamental Theorem of Algebra In Section 3.3, we were focused on ﬁnding the real zeros of a polynomial function. In this section, we expand our horizons and look for the non-real zeros as well. Consider the polynomial p(x) = x2 + 1. The zeros of p are the solutions to x2 + 1 = 0, or x2 = −1. This equation has no real solutions, but you may recall from Intermediate Algebra that we can formally extract the square roots of both −1 is usually re-labeled i, the so-called imaginary unit.1 sides to get x = ± The number i, while not a real number, plays along well with real numbers, and acts very much like any other radical expression. For instance, 3(2i) = 6i, 7i − 3i = 4i, (2 − 7i) + (3 + 4i) = 5 − 3i, and so forth. The key properties which distinguish i from the real numbers are listed below. −1. The quantity √ √ Deﬁnition 3.4. The imaginary unit i satis
ﬁes the two following properties 1. i2 = −1 2. If c is a real number with c ≥ 0 then √ √ −c = i c Property 1 in Deﬁnition 3.4 establishes that i does act as a square root2 of −1, and property 2 establishes what we mean by the ‘principal square root’ of a negative real number. In property 2, it is important to remember the restriction on c. For example, it is perfectly acceptable to say √ √ √ −4 = i 4 = i(2) = 2i. However, −(−4) = i −4, otherwise, we’d get √ 2 = 4 = −(−4) = i √ −4 = i(2i) = 2i2 = 2(−1) = −2, which is unacceptable.3 We are now in the position to deﬁne the complex numbers. Deﬁnition 3.5. A complex number is a number of the form a + bi, where a and b are real numbers and i is the imaginary unit. Complex numbers include things you’d normally expect, like 3 + 2i and 2 3. However, don’t forget that a or b could be zero, which means numbers like 3i and 6 are also complex numbers. In other words, don’t forget that the complex numbers include the real numbers, so 0 and π − 21 are both considered complex numbers.4 The arithmetic of complex numbers is as you would expect. The only things you need to remember are the two properties in Deﬁnition 3.4. The next example should help recall how these animals behave. 5 − i √ √ 1Some Technical Mathematics textbooks label it ‘j’. 2Note the use of the indeﬁnite article ‘a’. Whatever beast is chosen to be i, −i is the other square root of −1. 3We want to enlarge the number system so we can solve things like x2 = −1, but not at the cost of the established rules already set in place. For that reason, the general properties of radicals simply do not apply for even roots of negative quantities. 4See the remarks in Section 1.1.1. 288 Polynomial Functions Example 3.4.1. Perform the indicated operations. Write
your answer in the form5 a + bi. 1. (1 − 2i) − (3 + 4i) 2. (1 − 2i)(3 + 4i) 3. 1 − 2i 3 − 4i √ √ −3 −12 4. Solution. (−3)(−12) 5. 6. (x − [1 + 2i])(x − [1 − 2i]) 1. As mentioned earlier, we treat expressions involving i as we would any other radical. We combine like terms to get (1 − 2i) − (3 + 4i) = 1 − 2i − 3 − 4i = −2 − 6i. 2. Using the distributive property, we get (1 − 2i)(3 + 4i) = (1)(3) + (1)(4i) − (2i)(3) − (2i)(4i) = 3 + 4i − 6i − 8i2. Since i2 = −1, we get 3 + 4i − 6i − 8i2 = 3 − 2i − (−8) = 11 − 2i. 3. How in the world are we supposed to simplify 1−2i 3−4i? Well, we deal with the denominator 3 − 4i as we would any other denominator containing a radical, and multiply both numerator and denominator by 3 + 4i (the conjugate of 3 − 4i).6 Doing so produces 1 − 2i 3 − 4i · 3 + 4i 3 + 4i = (1 − 2i)(3 + 4i) (3 − 4i)(3 + 4i) = 11 − 2i 25 = 11 25 − 2 25 i 4. We use property 2 of Deﬁnition 3.4 ﬁrst, then apply the rules of radicals applicable to real radicals to get √ √ −3 √ √ −12 = i 3 i 12 = i2 √ √ 3 · 12 = − 36 = −6. 5. We adhere to the order of operations here and perform the multiplication before the radical to get (−3)(−12) = 36 = 6. √ 6. We can brute force multiply using the distributive property and see that (x − [1 + 2i])(x − [1 − 2i]) = x2 − x[1 − 2i] − x[1
+ 2i] + [1 − 2i][1 + 2i] = x2 − x + 2ix − x − 2ix + 1 − 2i + 2i − 4i2 = x2 − 2x + 5 A couple of remarks about the last example are in order. First, the conjugate of a complex number a + bi is the number a − bi. The notation commonly used for conjugation is a ‘bar’: a + bi = a − bi. For example, 3 + 2i = 3 − 2i, 3 − 2i = 3 + 2i, 6 = 6, 4i = −4i, and 3 + 5. The properties of the conjugate are summarized in the following theorem. 5 = 3 + √ √ 5OK, we’ll accept things like 3 − 2i even though it can be written as 3 + (−2)i. 6We will talk more about this in a moment. 3.4 Complex Zeros and the Fundamental Theorem of Algebra 289 Theorem 3.12. Properties of the Complex Conjugate: Let z and w be complex numbers = zw (z)n = zn, for any natural number n z is a real number if and only if z = z. Essentially, Theorem 3.12 says that complex conjugation works well with addition, multiplication and powers. The proof of these properties can best be achieved by writing out z = a + bi and w = c + di for real numbers a, b, c and d. Next, we compute the left and right hand sides of each equation and check to see that they are the same. The proof of the ﬁrst property is a very quick exercise.7 To prove the second property, we compare z + w and z + w. We have z + w = a + bi + c + di = a − bi + c − di. To ﬁnd z + w, we ﬁrst compute z + w = (a + bi) + (c + di) = (a + c) + (b + d)i so z + w = (a + c) + (b + d)i = (a + c) − (b + d)i = a − bi + c − di As such, we have established z +w = z + w. The proof for multiplication works similarly. The proof that the
conjugate works well with powers can be viewed as a repeated application of the product rule, and is best proved using a technique called Mathematical Induction.8 The last property is a characterization of real numbers. If z is real, then z = a + 0i, so z = a − 0i = a = z. On the other hand, if z = z, then a + bi = a − bi which means b = −b so b = 0. Hence, z = a + 0i = a and is real. We now return to the business of zeros. Suppose we wish to ﬁnd the zeros of f (x) = x2 − 2x + 5. To solve the equation x2 − 2x + 5 = 0, we note that the quadratic doesn’t factor nicely, so we resort to the Quadratic Formula, Equation 2.5 and obtain −(−2) ± −16 2 ± √ (−2)2 − 4(1)(5) 2(1) = = 2 ± 4i 2 = 1 ± 2i. 2 x = Two things are important to note. First, the zeros 1 + 2i and 1 − 2i are complex conjugates. If ever we obtain non-real zeros to a quadratic function with real coeﬃcients, the zeros will be a complex conjugate pair. (Do you see why?) Next, we note that in Example 3.4.1, part 6, we found (x − [1 + 2i])(x − [1 − 2i]) = x2 − 2x + 5. This demonstrates that the factor theorem holds even for non-real zeros, i.e, x = 1 + 2i is a zero of f, and, sure enough, (x − [1 + 2i]) is a factor of f (x). It turns out that polynomial division works the same way for all complex numbers, real and non-real alike, so the Factor and Remainder Theorems hold as well. But how do we know if a 7Trust us on this. 8See Section 9.3. 290 Polynomial Functions general polynomial has any complex zeros at all? We have many examples of polynomials with no real zeros. Can there be polynomials with no zeros whatsoever? The answer to that last question is �
�No.” and the theorem which provides that answer is The Fundamental Theorem of Algebra. Theorem 3.13. The Fundamental Theorem of Algebra: Suppose f is a polynomial function with complex number coeﬃcients of degree n ≥ 1, then f has at least one complex zero. The Fundamental Theorem of Algebra is an example of an ‘existence’ theorem in Mathematics. Like the Intermediate Value Theorem, Theorem 3.1, the Fundamental Theorem of Algebra guarantees the existence of at least one zero, but gives us no algorithm to use in ﬁnding it. In fact, as we mentioned in Section 3.3, there are polynomials whose real zeros, though they exist, cannot be expressed using the ‘usual’ combinations of arithmetic symbols, and must be approximated. The authors are fully aware that the full impact and profound nature of the Fundamental Theorem of Algebra is lost on most students studying College Algebra, and that’s ﬁne. It took mathematicians literally hundreds of years to prove the theorem in its full generality, and some of that history is recorded here. Note that the Fundamental Theorem of Algebra applies to not only polynomial functions with real coeﬃcients, but to those with complex number coeﬃcients as well. Suppose f is a polynomial of degree n ≥ 1. The Fundamental Theorem of Algebra guarantees us at least one complex zero, z1, and as such, the Factor Theorem guarantees that f (x) factors as f (x) = (x − z1) q1(x) for a polynomial function q1, of degree exactly n − 1. If n − 1 ≥ 1, then the Fundamental Theorem of Algebra guarantees a complex zero of q1 as well, say z2, so then the Factor Theorem gives us q1(x) = (x − z2) q2(x), and hence f (x) = (x − z1) (x − z2) q2(x). We can continue this process exactly n times, at which point our quotient polynomial qn has degree 0 so it’s a constant. This argument gives us the following factorization theorem. Theorem 3.14. Complex Factorization Theorem: Suppose f is a polynomial function with complex number coe
ﬃcients. If the degree of f is n and n ≥ 1, then f has exactly n complex zeros, counting multiplicity. If z1, z2,..., zk are the distinct zeros of f, with multiplicities m1, m2,..., mk, respectively, then f (x) = a (x − z1)m1 (x − z2)m2 · · · (x − zk)mk. Note that the value a in Theorem 3.14 is the leading coeﬃcient of f (x) (Can you see why?) and as such, we see that a polynomial is completely determined by its zeros, their multiplicities, and its leading coeﬃcient. We put this theorem to good use in the next example. Example 3.4.2. Let f (x) = 12x5 − 20x4 + 19x3 − 6x2 − 2x + 1. 1. Find all of the complex zeros of f and state their multiplicities. 2. Factor f (x) using Theorem 3.14 Solution. 1. Since f is a ﬁfth degree polynomial, we know that we need to perform at least three successful divisions to get the quotient down to a quadratic function. At that point, we can ﬁnd the remaining zeros using the Quadratic Formula, if necessary. Using the techniques developed in Section 3.3, we get 3.4 Complex Zeros and the Fundamental Theorem of Algebra 291 1 0 12 −20 ↓ 12 −14 ↓ 12 −8 ↓ −4 12 −12 19 −6 −2 6 −7 6 0 −2 12 6 −4 2 4 0 4 8 4 −4 0 12 Our quotient is 12x2 − 12x + 12, whose zeros we ﬁnd to be 1±i 2 know f has exactly 5 zeros, counting multiplicities, and as such we have the zero 1 and 1−i multiplicity 2, and the zeros − 1 2. From Theorem 3.14, we 2 with, each of multiplicity 1. 3, 1+i √ √ 3 3 3 2 √ 2. Applying Theorem 3.14, we are guaranteed that f factors as f (x) = 12 true test of Theorem
3.14 (and a student’s mettle!) would be to take the factored form of f (x) in the previous example and multiply it out9 to see that it really does reduce to the original formula f (x) = 12x5 − 20x4 + 19x3 − 6x2 − 2x + 1. When factoring a polynomial using Theorem 3.14, we say that it is factored completely over the complex numbers, meaning that it is impossible to factor the polynomial any further using complex numbers. If we wanted to completely factor f (x) over the real numbers then we would have stopped short of ﬁnding the nonreal zeros of f and factored f using our work from the synthetic division to write f (x) = x − 1 12x2 − 12x + 12, 2 or f (x) = 12 x − 1 x2 − x + 1. Since the zeros of x2 − x + 1 are nonreal, we call 2 x2 − x + 1 an irreducible quadratic meaning it is impossible to break it down any further using real numbers The last two results of the section show us that, at least in theory, if we have a polynomial function with real coeﬃcients, we can always factor it down enough so that any nonreal zeros come from irreducible quadratics. Theorem 3.15. Conjugate Pairs Theorem: If f is a polynomial function with real number coeﬃcients and z is a zero of f, then so is z. To prove the theorem, suppose f is a polynomial with real number coeﬃcients. Speciﬁcally, let f (x) = anxn + an−1xn−1 +... + a2x2 + a1x + a0. If z is a zero of f, then f (z) = 0, which means anzn + an−1zn−1 +... + a2z2 + a1z + a0 = 0. Next, we consider f (z) and apply Theorem 3.12 below. 9You really should do this once in your life to convince yourself that all of the theory actually does work! 292 Polynomial Functions f (z) = an (z)n + an−1 (
z)n−1 +... + a2 (z)2 + a1z + a0 = anzn + an−1zn−1 +... + a2z2 + a1z + a0 = anzn + an−1zn−1 +... + a2z2 + a1 z + a0 = anzn + an−1zn−1 +... + a2z2 + a1z + a0 = anzn + an−1zn−1 +... + a2z2 + a1z + a0 = f (z) since (z)n = zn since the coeﬃcients are real since z w = zw since This shows that z is a zero of f. So, if f is a polynomial function with real number coeﬃcients, Theorem 3.15 tells us that if a + bi is a nonreal zero of f, then so is a − bi. In other words, nonreal zeros of f come in conjugate pairs. The Factor Theorem kicks in to give us both (x − [a + bi]) and (x − [a − bi]) as factors of f (x) which means (x − [a + bi])(x − [a − bi]) = x2 + 2ax + a2 + b2 is an irreducible quadratic factor of f. As a result, we have our last theorem of the section. Theorem 3.16. Real Factorization Theorem: Suppose f is a polynomial function with real number coeﬃcients. Then f (x) can be factored into a product of linear factors corresponding to the real zeros of f and irreducible quadratic factors which give the nonreal zeros of f. We now present an example which pulls together all of the major ideas of this section. Example 3.4.3. Let f (x) = x4 + 64. 1. Use synthetic division to show that x = 2 + 2i is a zero of f. 2. Find the remaining complex zeros of f. 3. Completely factor f (x) over the complex numbers. 4. Completely factor f (x) over the real numbers. Solution. 1. Remembering to insert the 0’s in the synthetic division tableau we have 2 + 2
i 0 0 1 64 ↓ 2 + 2i 8i −16 + 16i −64 1 2 + 2i 8i −16 + 16i 0 0 2. Since f is a fourth degree polynomial, we need to make two successful divisions to get a quadratic quotient. Since 2 + 2i is a zero, we know from Theorem 3.15 that 2 − 2i is also a zero. We continue our synthetic division tableau. 3.4 Complex Zeros and the Fundamental Theorem of Algebra 293 2 + 2i 2 − 2i 0 0 8i 8i 1 ↓ 2 + 2i 1 2 + 2i ↓ 2 − 2i 8 − 8i 1 4 8 0 64 −16 + 16i −64 −16 + 16i 16 − 16i 0 0 Our quotient polynomial is x2 + 4x + 8. Using the quadratic formula, we obtain the remaining zeros −2 + 2i and −2 − 2i. 3. Using Theorem 3.14, we get f (x) = (x − [2 − 2i])(x − [2 + 2i])(x − [−2 + 2i])(x − [−2 − 2i]). 4. We multiply the linear factors of f (x) which correspond to complex conjugate pairs. We ﬁnd (x − [2 − 2i])(x − [2 + 2i]) = x2 − 4x + 8, and (x − [−2 + 2i])(x − [−2 − 2i]) = x2 + 4x + 8. Our ﬁnal answer is f (x) = x2 − 4x + 8 x2 + 4x + 8. Our last example turns the tables and asks us to manufacture a polynomial with certain properties of its graph and zeros. Example 3.4.4. Find a polynomial p of lowest degree that has integer coeﬃcients and satisﬁes all of the following criteria: the graph of y = p(x) touches (but doesn’t cross) the x-axis at 1 3, 0 x = 3i is a zero of p. as x → −∞, p(x) → −∞ as x → ∞, p(x) → −∞ 3, 0, we know
x = 1 Solution. To solve this problem, we will need a good understanding of the relationship between the x-intercepts of the graph of a function and the zeros of a function, the Factor Theorem, the role of multiplicity, complex conjugates, the Complex Factorization Theorem, and end behavior of polynomial functions. (In short, you’ll need most of the major concepts of this chapter.) Since the graph of p touches the x-axis at 1 3 is a zero of even multiplicity. Since we are after a polynomial of lowest degree, we need x = 1 3 to have multiplicity exactly 2. The Factor Theorem now tells us x − 1 is a factor of p(x). Since x = 3i is a zero and our ﬁnal answer is to 3 have integer (real) coeﬃcients, x = −3i is also a zero. The Factor Theorem kicks in again to give us (x−3i) and (x+3i) as factors of p(x). We are given no further information about zeros or intercepts so we conclude, by the Complex Factorization Theorem that p(x) = a x − 1 (x − 3i)(x + 3i) for 3 some real number a. Expanding this, we get p(x) = ax4 − 2a 9 x2 −6ax+a. In order to obtain integer coeﬃcients, we know a must be an integer multiple of 9. Our last concern is end behavior. Since the leading term of p(x) is ax4, we need a < 0 to get p(x) → −∞ as x → ±∞. Hence, if we choose x = −9, we get p(x) = −9x4 + 6x3 − 82x2 + 54x − 9. We can verify our handiwork using the techniques developed in this chapter. 3 x3 + 82a 2 2 294 Polynomial Functions This example concludes our study of polynomial functions.10 The last few sections have contained what is considered by many to be ‘heavy’ Mathematics. Like a heavy meal, heavy Mathematics takes time to digest. Don’t be overly concerned if it doesn’t seem to sink in all at once, and pace yourself in the Exercises or you’re liable to get mental
cramps. But before we get to the Exercises, we’d like to oﬀer a bit of an epilogue. Our main goal in presenting the material on the complex zeros of a polynomial was to give the chapter a sense of completeness. Given that it can be shown that some polynomials have real zeros which cannot be expressed using the usual algebraic operations, and still others have no real zeros at all, it was nice to discover that every polynomial of degree n ≥ 1 has n complex zeros. So like we said, it gives us a sense of closure. But the observant reader will note that we did not give any examples of applications which involve complex numbers. Students often wonder when complex numbers will be used in ‘real-world’ applications. After all, didn’t we call i the imaginary unit? How can imaginary things be used in reality? It turns out that complex numbers are very useful in many applied ﬁelds such as ﬂuid dynamics, electromagnetism and quantum mechanics, but most of the applications require Mathematics well beyond College Algebra to fully understand them. That does not mean you’ll never be be able to understand them; in fact, it is the authors’ sincere hope that all of you will reach a point in your studies when the glory, awe and splendor of complex numbers are revealed to you. For now, however, the really good stuﬀ is beyond the scope of this text. We invite you and your classmates to ﬁnd a few examples of complex number applications and see what you can make of them. A simple Internet search with the phrase ‘complex numbers in real life’ should get you started. Basic electronics classes are another place to look, but remember, they might use the letter j where we have used i. For the remainder of the text, with the exception of Section 11.7 and a few exploratory exercises scattered about, we will restrict our attention to real numbers. We do this primarily because the ﬁrst Calculus sequence you will take, ostensibly the one that this text is preparing you for, studies only functions of real variables. Also, lots of really cool scientiﬁc things don’t require any deep understanding of complex numbers to study them, but they do need more Mathematics like exponential, logarithmic and trigonometric functions. We believe
it makes more sense pedagogically for you to learn about those functions now then take a course in Complex Function Theory in your junior or senior year once you’ve completed the Calculus sequence. It is in that course that the true power of the complex numbers is released. But for now, in order to fully prepare you for life immediately after College Algebra, we will say that functions like f (x) = 1 x2+1 have a domain of all real numbers, even though we know x2 + 1 = 0 has two complex solutions, namely x = ±i. Because 1 x2 + 1 > 0 for all real numbers x, the fraction x2+1 is never undeﬁned in the real variable setting. 10With the exception of the Exercises on the next page, of course. 3.4 Complex Zeros and the Fundamental Theorem of Algebra 295 3.4.1 Exercises In Exercises 1 - 10, use the given complex numbers z and w to ﬁnd and simplify the following. Write your answers in the form a + bi. z + w 1 z z 1. z = 2 + 3i, w = 4i 3. z = i, w = −1 + 2i 5. z = 3 − 5i, w = 2 + 7i √ √ √ 7. z = 2. z = 1 2 i + zw z w zz 2 z2 w z (z)2 2. z = 1 + i, w = −i 4. z = 4i, w = 2 − 2i 6. z = −5 + i, w = 4 + 2i √ 3, w = −1 − i 8. z = 1 − i √ 10. z = − √ 2 2 + i √ In Exercises 11 - 18, simplify the quantity. √ √ −49 √ −9 11. 15. −16 12. 16. √ −9 √ √ −25 −4 13. 14. (−25)(−4) (−9)(−16) 17. −(−9) 18. − (−9) We know that i2 = −1 which means i3 = i2 · i = (−1) · i = −i and i4 = i2 · i2 = (−1)(−1) = 1. In Exercises 19 - 26, use this information to simplify
the given power of i. 19. i5 23. i15 20. i6 24. i26 21. i7 25. i117 22. i8 26. i304 In Exercises 27 - 48, ﬁnd all of the zeros of the polynomial then completely factor it over the real numbers and completely factor it over the complex numbers. 27. f (x) = x2 − 4x + 13 28. f (x) = x2 − 2x + 5 29. f (x) = 3x2 + 2x + 10 30. f (x) = x3 − 2x2 + 9x − 18 31. f (x) = x3 + 6x2 + 6x + 5 32. f (x) = 3x3 − 13x2 + 43x − 13 296 Polynomial Functions 33. f (x) = x3 + 3x2 + 4x + 12 34. f (x) = 4x3 − 6x2 − 8x + 15 35. f (x) = x3 + 7x2 + 9x − 2 36. f (x) = 9x3 + 2x + 1 37. f (x) = 4x4 − 4x3 + 13x2 − 12x + 3 38. f (x) = 2x4 − 7x3 + 14x2 − 15x + 6 39. f (x) = x4 + x3 + 7x2 + 9x − 18 40. f (x) = 6x4 + 17x3 − 55x2 + 16x + 12 41. f (x) = −3x4 − 8x3 − 12x2 − 12x − 5 42. f (x) = 8x4 + 50x3 + 43x2 + 2x − 4 43. f (x) = x4 + 9x2 + 20 44. f (x) = x4 + 5x2 − 24 45. f (x) = x5 − x4 + 7x3 − 7x2 + 12x − 12 46. f (x) = x6 − 64 47. f (x) = x4 − 2x3 + 27x2 − 2x + 26 (Hint: x = i is one of the zeros.) 48. f (x) = 2x4 + 5x3 + 13x2 + 7x +
5 (Hint: x = −1 + 2i is a zero.) In Exercises 49 - 53, create a polynomial f with real number coeﬃcients which has all of the desired characteristics. You may leave the polynomial in factored form. 49. 50. 51. 52. 53. The zeros of f are c = ±1 and c = ±i The leading term of f (x) is 42x4 c = 2i is a zero. the point (−1, 0) is a local minimum on the graph of y = f (x) the leading term of f (x) is 117x4 The solutions to f (x) = 0 are x = ±2 and x = ±7i The leading term of f (x) is −3x5 The point (2, 0) is a local maximum on the graph of y = f (x). f is degree 5. x = 6, x = i and x = 1 − 3i are zeros of f as x → −∞, f (x) → ∞ The leading term of f (x) is −2x3 c = 2i is a zero f (0) = −16 54. Let z and w be arbitrary complex numbers. Show that z w = zw and z = z. 3.4 Complex Zeros and the Fundamental Theorem of Algebra 297 3.4.2 Answers 1. For z = 2 + 3i and w = 4i z + w = 2 + 7i zw = −12 + 8i z2 = −5 + 12i 1 z = 2 13 − 3 13 = 12 13 + 8 13 i z = 2 − 3i zz = 13 (z)2 = −5 − 12i 2. For z = 1 + i and w = −. For z = i and w = −1 + 2i zw = 1 − i z w = −1 + i zz = 2 z + w = −1 + 3i zw = −2 − i 1 z = −i z = −i 4. For z = 4i and w = 2 − 2i z + w = 2 + 2i 4i 5. For z = 3 − 5i and w = 2 + 7i z w = 2 5 − 1 5 i zz = 1 zw = 8 + 8i z w = −1 + i zz = 16
z2 = 2i z)2 = −2i z2 = −1 w z = 2 + i (z)2 = −1 z2 = −16 z)2 = −16 z + w = 5 + 2i zw = 41 + 11i z2 = −16 − 30i 1 z = 3 34 + 5 34 i z w = − 29 53 − 31 53 i w z = − 29 34 + 31 34 i z = 3 + 5i zz = 34 (z)2 = −16 + 30i 298 Polynomial Functions 6. For z = −5 + i and w = 4 + 2i z + w = −1 + 3i zw = −22 − 6i z2 = 24 − 10i 1 z = − 5 26 − 1 26 i z w = − 9 10 + 7 10 i w z = − 9 13 − 7 13 i z = −5 − i zz = 26 (z)2 = 24 + 10i 7. For z = √ √ 2 − i 2 and zw = 4 z w = −i zz = 4 z2 = −4i w z = i (z)2 = 4i 8. For z = 1 − i √ 3 and w = −2i √ 3 zw = −4 z2 = −2 − 2i √ 3 9. For z = 1 2 + 3 2 i and + zz = 4 √ 3 2 i zw = −z)2 = −2 + 2i √ 3 z2 = − zz = 1 (z) 10. For and zw = 1 z w = −i zz = 1 z2 = −i w z = i (z)2 = i 11. 7i 12. 3i 13. −10 14. 10 3.4 Complex Zeros and the Fundamental Theorem of Algebra 299 15. −12 16. 12 17. 3 18. −3i 19. i5 = i4 · i = 1 · i = i 21. i7 = i4 · i3 = 1 · (−i) = −i 20. i6 = i4 · i2 = 1 · (−1) = −1 22. i8 = i4 · i4 = i42 = (1)2 = 1 23. i15 = i43 · i3 = 1 · (−i) = −i 24. i26 = i
46 · i2 = 1 · (−1) = −1 25. i117 = i429 · i = 1 · i = i 26. i304 = i476 = 176 = 1 27. f (x) = x2 − 4x + 13 = (x − (2 + 3i))(x − (2 − 3i)) Zeros: x = 2 ± 3i 28. f (x) = x2 − 2x + 5 = (x − (1 + 2i))(x − (1 − 2i)) Zeros: x = 1 ± 2i 29. f (x) = 3x2 + 2x + 10 = 3 x − − 1 3 + Zeros: x = − 1 3 ± √ 29 3 i √ 29 3 i x − − 1 3 − √ 29 3 i 30. f (x) = x3 − 2x2 + 9x − 18 = (x − 2) x2 + 9 = (x − 2)(x − 3i)(x + 3i) Zeros: x = 2, ±3i 31. f (x) = x3 +6x2 +6x+5 = (x+5)(x2 +x+1) = (x+5) Zeros: x = −5 32. f (x) = 3x3 − 13x2 + 43x − 13 = (3x − 1)(x2 − 4x + 13) = (3x − 1)(x − (2 + 3i))(x − (2 − 3i)) Zeros: x = 1 3, x = 2 ± 3i 33. f (x) = x3 + 3x2 + 4x + 12 = (x + 3) x2 + 4 = (x + 3)(x + 2i)(x − 2i) Zeros: x = −3, ±2i 34. f (x) = 4x3 − 6x2 − 8x + 15 = Zeros 4x2 − 12x + 10 + √ 29 2 x − − 5 2 − √ 29 2 35. f (x) = x3 + 7x2 + 9x − 2 = (x + 2) √ 29 Zeros: x = −2, x = − 5 2 9x2 − 3x + 3 36. f (x) = 9x3 + 2x + 1 = x
+ 1 3 √ 11 Zeros + √ 11 6 i √ 11 6 i 37. f (x) = 4x4 − 4x3 + 13x2 − 12x + 3 = x − 1 √ 2 Zeros: x = 1 2, x = ± 3i 2 4x2 + 12 = 4 x − 1 2 2 √ (x + i √ 3)(x − i 3) 300 Polynomial Functions 38. f (x) = 2x4 − 7x3 + 14x2 − 15x + 6 = (x − 1)2 2x2 − 3x + 6 √ 39 4 i √ 39 (x − 1)2 Zeros: x = 1, x = 3 3 4 + √ 39 4 i 4 ± 39. f (x) = x4 + x3 + 7x2 + 9x − 18 = (x + 2)(x − 1) x2 + 9 = (x + 2)(x − 1)(x + 3i)(x − 3i) Zeros: x = −2, 1, ±3i 40. f (x) = 6x4 + 17x3 − 55x2 + 16x + 12 = 6 x + 1 3 Zeros2 + 2 √ 2 x − −2 − 2 √ 2 41. f (x) = −3x4 − 8x3 − 12x2 − 12x − 5 = (x + 1)2 −3x2 − 2x − 5 − 1 3 + √ 14 3 i = −3(x + 1)2 x − Zeros: x = −1, x = − 1 √ 14 3 i √ 14 42. f (x) = 8x4 + 50x3 + 43x2 + 2x − x − (−3 + Zeros: x = − 1 5 43. f (x) = x4 + 9x2 + 20 = x2 + 4 x2 + 5 = (x − 2i)(x + 2i) x − i 4, x = −3 ± 2, 1 √ √ Zeros: x = ±2i, ±i 5 44. f (x) = x4 + 5x2 − 24 = x2 − 3 x2 + 8 = (x − √ 2 Zeros: x = ± 3, ±2i √ √ 5))(x
− (−3 − √ 5)) √ √ 5 x + i 5 √ 3)(x + √ 3) x − 2i √ 2 x + 2i √ 2 45. f (x) = x5 − x4 + 7x3 − 7x2 + 12x − 12 = (x − 1) x2 + 3 x2 + 4 √ √ = (x − 1)(x − i 3)(x + i 3)(x − 2i)(x + 2i) √ Zeros: x = 1, ± 3i, ±2i 46. f (x) = x6 − 64 = (x − 2)(x + 2) x2 + 2x + 4 x2 − 2x + 4 = (x − 2)(x + 2) x − −1 + i 3 x − −1 − i 3 x − 1 + i √ √ √ √ Zeros: x = ±2, x = −1 ± i 3 47. f (x) = x4−2x3+27x2−2x+26 = (x2−2x+26)(x2+1) = (x−(1+5i))(x−(1−5i))(x+i)(x−i) Zeros: x = 1 ± 5i, x = ±i 48. f (x) = 2x4 + 5x3 + 13x2 + 7x + 5 = x2 + 2x + 5 2x2 + x + 1 x − = 2(x − (−1 + 2i))(x − (−1 − 2i)) Zeros: x = −1 ± 2i 49. f (x) = 42(x − 1)(x + 1)(x − i)(x + i) 50. f (x) = 117(x + 1)2(x − 2i)(x + 2i) 51. f (x) = −3(x − 2)2(x + 2)(x − 7i)(x + 7i) 52. f (x) = a(x − 6)(x − i)(x + i)(x − (1 − 3i))(x − (1 + 3i)) where a is any real number, a < 0 53. f (x) = −2(x − 2i)(x + 2i)(x
+ 2) Chapter 4 Rational Functions 4.1 Introduction to Rational Functions If we add, subtract or multiply polynomial functions according to the function arithmetic rules If, on the other hand, we deﬁned in Section 1.5, we will produce another polynomial function. divide two polynomial functions, the result may not be a polynomial. In this chapter we study rational functions - functions which are ratios of polynomials. Deﬁnition 4.1. A rational function is a function which is the ratio of polynomial functions. Said diﬀerently, r is a rational function if it is of the form where p and q are polynomial functions.a r(x) = p(x) q(x), aAccording to this deﬁnition, all polynomial functions are also rational functions. (Take q(x) = 1). As we recall from Section 1.4, we have domain issues anytime the denominator of a fraction is zero. In the example below, we review this concept as well as some of the arithmetic of rational expressions. Example 4.1.1. Find the domain of the following rational functions. Write them in the form p(x) q(x) for polynomial functions p and q and simplify. 1. f (x) = 3. h(x) = 2x − 1 x + 1 2x2 − 1 x2 − 1 − 3x − 2 x2 − 1 Solution. 2. g(x) = 2 − 3 x + 1 4. r(x) = 2x2 − 1 x2 − 1 ÷ 3x − 2 x2 − 1 1. To ﬁnd the domain of f, we proceed as we did in Section 1.4: we ﬁnd the zeros of the denominator and exclude them from the domain. Setting x + 1 = 0 results in x = −1. Hence, 302 Rational Functions our domain is (−∞, −1) ∪ (−1, ∞). The expression f (x) is already in the form requested and when we check for common factors among the numerator and denominator we ﬁnd none, so we are done. 2. Proceeding as before, we determine the domain of g by solving x + 1 = 0. As before, we ﬁnd the domain of g is (−∞
, −1) ∪ (−1, ∞). To write g(x) in the form requested, we need to get a common denominator g(x) = 2 − 3 x + 1 (2x + 2) − 3 x + 1 = (2)(x + 1) (1)(x + 1 2x − 1 x + 1 = This formula is now completely simpliﬁed. 3. The denominators in the formula for h(x) are both x2 − 1 whose zeros are x = ±1. As a result, the domain of h is (−∞, −1) ∪ (−1, 1) ∪ (1, ∞). We now proceed to simplify h(x). Since we have the same denominator in both terms, we subtract the numerators. We then factor the resulting numerator and denominator, and cancel out the common factor. h(x) = = = = − 2x2 − 1 3x − 2 x2 − 1 x2 − 1 2x2 − 1 − 3x + 2 x2 − 1 (2x − 1)(x − 1) (x + 1)(x − 1) 2x − 1 x + 1 = = = 2x2 − 1 − (3x − 2) x2 − 1 2x2 − 3x + 1 x2 − 1 (2x − 1) (x − 1) (x + 1) (x − 1) 4. To ﬁnd the domain of r, it may help to temporarily rewrite r(x) as r(x) = 2x2 − 1 x2 − 1 3x − 2 x2 − 1 We need to set all of the denominators equal to zero which means we need to solve not only x2−1 = 0. We ﬁnd x = ±1 for the former and x = 2 x2 − 1 = 0, but also 3x−2 3 for the latter. Our domain is (−∞, −1) ∪ −1, 2 3, 1 ∪ (1, ∞). We simplify r(x) by rewriting the division as 3 multiplication by the reciprocal and then by canceling the common factor ∪ 2 4.1 Introduction to Rational Functions 303 r(x) = = 2x2 − 1 3x − 2 ÷ x2 − 1 x2 − 1 2x2 − 1 x2 − 1 x
2 − 1(3x − 2) = = 2x2 − 1 x2 − 1 2x2 − 1 3x − 2 · x2 − 1 3x − 2 = 2x2 − 1 x2 − 1 (x2 − 1) (3x − 2) A few remarks about Example 4.1.1 are in order. Note that the expressions for f (x), g(x) and h(x) work out to be the same. However, only two of these functions are actually equal. Recall that functions are ultimately sets of ordered pairs,1 so for two functions to be equal, they need, among other things, to have the same domain. Since f (x) = g(x) and f and g have the same domain, they are equal functions. Even though the formula h(x) is the same as f (x), the domain of h is diﬀerent than the domain of f, and thus they are diﬀerent functions. We now turn our attention to the graphs of rational functions. Consider the function f (x) = 2x−1 x+1 from Example 4.1.1. Using a graphing calculator, we obtain Two behaviors of the graph are worthy of further discussion. First, note that the graph appears to ‘break’ at x = −1. We know from our last example that x = −1 is not in the domain of f which means f (−1) is undeﬁned. When we make a table of values to study the behavior of f near x = −1 we see that we can get ‘near’ x = −1 from two directions. We can choose values a little less than −1, for example x = −1.1, x = −1.01, x = −1.001, and so on. These values are said to ‘approach −1 from the left.’ Similarly, the values x = −0.9, x = −0.99, x = −0.999, etc., are said to ‘approach −1 from the right.’ If we make two tables, we ﬁnd that the numerical results conﬁrm what we see graphically. x −1.1 −1.01 −1.001 −1.0001 f (x) (x, f (x)) (−1.1, 32) 32 (−1
.01, 302) 302 (−1.001, 3002) 3002 30002 (−1.001, 30002) f (x) x (x, f (x)) −28 −0.9 (−0.9, −28) −0.99 −298 (−0.99, −298) (−0.999, −2998) −0.999 −2998 −0.9999 −29998 (−0.9999, −29998) As the x values approach −1 from the left, the function values become larger and larger positive numbers.2 We express this symbolically by stating as x → −1−, f (x) → ∞. Similarly, using analogous notation, we conclude from the table that as x → −1+, f (x) → −∞. For this type of 1You should review Sections 1.2 and 1.3 if this statement caught you oﬀ guard. 2We would need Calculus to conﬁrm this analytically. 304 Rational Functions unbounded behavior, we say the graph of y = f (x) has a vertical asymptote of x = −1. Roughly speaking, this means that near x = −1, the graph looks very much like the vertical line x = −1. The other feature worthy of note about the graph of y = f (x) is that it seems to ‘level oﬀ’ on the left and right hand sides of the screen. This is a statement about the end behavior of the function. As we discussed in Section 3.1, the end behavior of a function is its behavior as x attains larger3 and larger negative values without bound, x → −∞, and as x becomes large without bound, x → ∞. Making tables of values, we ﬁnd x f (x) (x, f (x)) −10 ≈ 2.3333 ≈ (−10, 2.3333) −100 ≈ 2.0303 ≈ (−100, 2.0303) −1000 ≈ 2.0030 ≈ (−1000, 2.0030) −10000 ≈ 2.0003 ≈ (−10000, 2.0003) f (x) (x, f (x)) x 10 ≈ 1.7273 ≈ (10, 1.7273) 100 ≈ 1.9703 ≈ (100, 1.9703
) 1000 ≈ 1.9970 ≈ (1000, 1.9970) 10000 ≈ 1.9997 ≈ (10000, 1.9997) From the tables, we see that as x → −∞, f (x) → 2+ and as x → ∞, f (x) → 2−. Here the ‘+’ means ‘from above’ and the ‘−’ means ‘from below’. In this case, we say the graph of y = f (x) has a horizontal asymptote of y = 2. This means that the end behavior of f resembles the horizontal line y = 2, which explains the ‘leveling oﬀ’ behavior we see in the calculator’s graph. We formalize the concepts of vertical and horizontal asymptotes in the following deﬁnitions. Deﬁnition 4.2. The line x = c is called a vertical asymptote of the graph of a function y = f (x) if as x → c− or as x → c+, either f (x) → ∞ or f (x) → −∞. Deﬁnition 4.3. The line y = c is called a horizontal asymptote of the graph of a function y = f (x) if as x → −∞ or as x → ∞, f (x) → c. Note that in Deﬁnition 4.3, we write f (x) → c (not f (x) → c+ or f (x) → c−) because we are unconcerned from which direction the values f (x) approach the value c, just as long as they do so.4 In our discussion following Example 4.1.1, we determined that, despite the fact that the formula for h(x) reduced to the same formula as f (x), the functions f and h are diﬀerent, since x = 1 is in the domain of f, but x = 1 is not in the domain of h. If we graph h(x) = 2x2−1 x2−1 using a graphing calculator, we are surprised to ﬁnd that the graph looks identical to the graph of y = f (x). There is a vertical asymptote at x = −1, but near
x = 1, everything seem ﬁne. Tables of values provide numerical evidence which supports the graphical observation. x2−1 − 3x−2 3Here, the word ‘larger’ means larger in absolute value. 4As we shall see in the next section, the graphs of rational functions may, in fact, cross their horizontal asymptotes. If this happens, however, it does so only a ﬁnite number of times, and so for each choice of x → −∞ and x → ∞, f (x) will approach c from either below (in the case f (x) → c−) or above (in the case f (x) → c+.) We leave f (x) → c generic in our deﬁnition, however, to allow this concept to apply to less tame specimens in the Precalculus zoo, such as Exercise 50 in Section 10.5. 4.1 Introduction to Rational Functions 305 x h(x) (x, h(x)) ≈ (0.9, 0.4210) 0.9 ≈ 0.4210 0.99 ≈ 0.4925 ≈ (0.99, 0.4925) 0.999 ≈ 0.4992 ≈ (0.999, 0.4992) 0.9999 ≈ 0.4999 ≈ (0.9999, 0.4999) x h(x) (x, h(x)) ≈ (1.1, 0.5714) 1.1 ≈ 0.5714 1.01 ≈ 0.5075 ≈ (1.01, 0.5075) 1.001 ≈ 0.5007 ≈ (1.001, 0.5007) 1.0001 ≈ 0.5001 ≈ (1.0001, 0.5001) We see that as x → 1−, h(x) → 0.5− and as x → 1+, h(x) → 0.5+. In other words, the points on the graph of y = h(x) are approaching (1, 0.5), but since x = 1 is not in the domain of h, it would be inaccurate to ﬁll in a point at (1, 0.5). As we’ve done in past sections when something like this occurs,5 we put an open circle (also
called a hole in this case6) at (1, 0.5). Below is a detailed graph of y = h(x), with the vertical and horizontal asymptotes as dashed lines1 −2 −3 −4 −5 −6 −4 −3 −2 1 2 3 4 x Neither x = −1 nor x = 1 are in the domain of h, yet the behavior of the graph of y = h(x) is drastically diﬀerent near these x-values. The reason for this lies in the second to last step when we simpliﬁed the formula for h(x) in Example 4.1.1, where we had h(x) = (2x−1)(x−1) (x+1)(x−1). The reason x = −1 is not in the domain of h is because the factor (x + 1) appears in the denominator of h(x); similarly, x = 1 is not in the domain of h because of the factor (x − 1) in the denominator of h(x). The major diﬀerence between these two factors is that (x − 1) cancels with a factor in the numerator whereas (x + 1) does not. Loosely speaking, the trouble caused by (x − 1) in the denominator is canceled away while the factor (x + 1) remains to cause mischief. This is why the graph of y = h(x) has a vertical asymptote at x = −1 but only a hole at x = 1. These observations are generalized and summarized in the theorem below, whose proof is found in Calculus. 5For instance, graphing piecewise deﬁned functions in Section 1.6. 6In Calculus, we will see how these ‘holes’ can be ‘plugged’ when embarking on a more advanced study of continuity. 306 Rational Functions Theorem 4.1. Location of Vertical Asymptotes and Holes:a Suppose r is a rational function which can be written as r(x) = p(x) q(x) where p and q have no common zeros.b Let c be a real number which is not in the domain of r. If q(c) = 0, then the graph of y = r(x) has a hole at c, p(c) q(c). If q(
c) = 0, then the line x = c is a vertical asymptote of the graph of y = r(x). aOr, ‘How to tell your asymptote from a hole in the graph.’ bIn other words, r(x) is in lowest terms. In English, Theorem 4.1 says that if x = c is not in the domain of r but, when we simplify r(x), it no longer makes the denominator 0, then we have a hole at x = c. Otherwise, the line x = c is a vertical asymptote of the graph of y = r(x). Example 4.1.2. Find the vertical asymptotes of, and/or holes in, the graphs of the following rational functions. Verify your answers using a graphing calculator, and describe the behavior of the graph near them using proper notation. 1. f (x) = 2x x2 − 3 3. h(x) = x2 − x − 6 x2 + 9 Solution. 2. g(x) = x2 − x − 6 x2 − 9 4. r(x) = x2 − x − 6 x2 + 4x + 4 1. To use Theorem 4.1, we ﬁrst ﬁnd all of the real numbers which aren’t in the domain of f. To do so, we solve x2 − 3 = 0 and get x = ± 3. Since the expression f (x) is in lowest terms, there is no cancellation possible, and we conclude that the lines x = − 3 are vertical asymptotes to the graph of y = f (x). The calculator veriﬁes this claim, and from the √ − −, f (x) → −∞, as x → − graph, we see that as x → − 3, √ f (x) → −∞, and ﬁnally as x → 3, f (x) → ∞, as x →, f (x) → ∞. 3 and. Solving x2 − 9 = 0 gives x = ±3. In lowest terms g(x) = x2−x−6 x+3. Since x = −3 continues to make trouble in the denominator, we know the line x = −3 is a vertical asymptote of the graph of y
= g(x). Since x = 3 no longer produces a 0 in the denominator, we have a hole at x = 3. To ﬁnd the y-coordinate of the hole, we substitute x = 3 into x+2 x+3 and ﬁnd the hole is at 3, 5. When we graph y = g(x) using a calculator, we clearly see the 6 vertical asymptote at x = −3, but everything seems calm near x = 3. Hence, as x → −3−, g(x) → ∞, as x → −3+, g(x) → −∞, as x → 3−, g(x) → 5 6, and as x → 3+, g(x) → 5 6 x2−9 = (x−3)(x+2) (x−3)(x+3) = x+2 − +. 4.1 Introduction to Rational Functions 307 The graph of y = f (x) The graph of y = g(x) 3. The domain of h is all real numbers, since x2 + 9 = 0 has no real solutions. Accordingly, the graph of y = h(x) is devoid of both vertical asymptotes and holes. 4. Setting x2 + 4x + 4 = 0 gives us x = −2 as the only real number of concern. Simplifying, we see r(x) = x2−x−6 x+2. Since x = −2 continues to produce a 0 in the denominator of the reduced function, we know x = −2 is a vertical asymptote to the graph. The calculator bears this out, and, moreover, we see that as x → −2−, r(x) → ∞ and as x → −2+, r(x) → −∞. x2+4x+4 = (x−3)(x+2) (x+2)2 = x−3 The graph of y = h(x) The graph of y = r(x) Our next example gives us a physical interpretation of a vertical asymptote. This type of model arises from a family of equations cheerily named ‘doomsday’ equations.7 Example 4.1.3. A mathematical model for the population P, in thousands, of a certain species of bacteria, t days after it is introduced to an environment is given by
P (t) = 100 (5−t)2, 0 ≤ t < 5. 1. Find and interpret P (0). 2. When will the population reach 100,000? 3. Determine the behavior of P as t → 5−. Interpret this result graphically and within the context of the problem. 7These functions arise in Diﬀerential Equations. The unfortunate name will make sense shortly. 308 Solution. Rational Functions 1. Substituting t = 0 gives P (0) = 100 (5−0)2 = 4, which means 4000 bacteria are initially introduced into the environment. 2. To ﬁnd when the population reaches 100,000, we ﬁrst need to remember that P (t) is measured in thousands. In other words, 100,000 bacteria corresponds to P (t) = 100. Substituting for 100 P (t) gives the equation (5−t)2 = 100. Clearing denominators and dividing by 100 gives (5 − t)2 = 1, which, after extracting square roots, produces t = 4 or t = 6. Of these two solutions, only t = 4 in our domain, so this is the solution we keep. Hence, it takes 4 days for the population of bacteria to reach 100,000. 3. To determine the behavior of P as t → 5−, we can make a table t 4.9 4.99 4.999 4.9999 P (t) 10000 1000000 100000000 10000000000 In other words, as t → 5−, P (t) → ∞. Graphically, the line t = 5 is a vertical asymptote of the graph of y = P (t). Physically, this means that the population of bacteria is increasing without bound as we near 5 days, which cannot actually happen. For this reason, t = 5 is called the ‘doomsday’ for this population. There is no way any environment can support inﬁnitely many bacteria, so shortly before t = 5 the environment would collapse. Now that we have thoroughly investigated vertical asymptotes, we can turn our attention to horizontal asymptotes. The next theorem tells us when to expect horizontal asymptotes. Theorem 4.2. Location of Horizontal Asymptotes: Suppose r is a rational function and r(x) = p(x) q(x), where p and q are polyn
omial functions with leading coeﬃcients a and b, respectively. If the degree of p(x) is the same as the degree of q(x), then y = a b is thea horizontal asymptote of the graph of y = r(x). If the degree of p(x) is less than the degree of q(x), then y = 0 is the horizontal asymptote of the graph of y = r(x). If the degree of p(x) is greater than the degree of q(x), then the graph of y = r(x) has no horizontal asymptotes. aThe use of the deﬁnite article will be justiﬁed momentarily. Like Theorem 4.1, Theorem 4.2 is proved using Calculus. Nevertheless, we can understand the idea behind it using our example f (x) = 2x−1 x+1. If we interpret f (x) as a division problem, (2x−1)÷(x+1), 4.1 Introduction to Rational Functions 309 x+1 = 2 − 3 we ﬁnd that the quotient is 2 with a remainder of −3. Using what we know about polynomial division, speciﬁcally Theorem 3.4, we get 2x − 1 = 2(x + 1) − 3. Dividing both sides by (x + 1) gives 2x−1 x+1. (You may remember this as the formula for g(x) in Example 4.1.1.) As x becomes unbounded in either direction, the quantity 3 x+1 gets closer and closer to 0 so that the values of f (x) become closer and closer8 to 2. In symbols, as x → ±∞, f (x) → 2, and we have the result.9 Notice that the graph gets close to the same y value as x → −∞ or x → ∞. This means that the graph can have only one horizontal asymptote if it is going to have one at all. Thus we were justiﬁed in using ‘the’ in the previous theorem. x+1 ≈ 2x Alternatively, we can use what we know about end behavior of polynomials to help us understand this theorem. From Theorem 3.2, we know the end behavior of a po
lynomial is determined by its leading term. Applying this to the numerator and denominator of f (x), we get that as x → ±∞, f (x) = 2x−1 x = 2. This last approach is useful in Calculus, and, indeed, is made rigorous there. (Keep this in mind for the remainder of this paragraph.) Applying this reasoning to the general case, suppose r(x) = p(x) q(x) where a is the leading coeﬃcient of p(x) and b is the leading coeﬃcient of q(x). As x → ±∞, r(x) ≈ axn bxm, where n and m are the degrees of p(x) and q(x), respectively. If the degree of p(x) and the degree of q(x) are the same, then n = m so that r(x) ≈ a b, which means y = a b is the horizontal asymptote in this case. If the degree of p(x) is less than the degree of q(x), then n < m, so m − n is a positive number, and hence, r(x) ≈ a bxm−n → 0 as x → ±∞. If the degree of p(x) is greater than the degree of q(x), then n > m, and hence n − m is a positive number and r(x) ≈ axn−m, which becomes unbounded as x → ±∞. As we said before, if a rational function has a horizontal asymptote, then it will have only one. (This is not true for other types of functions we shall see in later chapters.) b Example 4.1.4. List the horizontal asymptotes, if any, of the graphs of the following functions. Verify your answers using a graphing calculator, and describe the behavior of the graph near them using proper notation. 1. f (x) = 5x x2 + 1 Solution. 2. g(x) = x2 − 4 x + 1 3. h(x) = 6x3 − 3x + 1 5 − 2x3 1. The numerator of f (x) is 5x, which has degree 1. The denominator of f (x) is x2 + 1, which has degree 2. App
lying Theorem 4.2, y = 0 is the horizontal asymptote. Sure enough, we see from the graph that as x → −∞, f (x) → 0− and as x → ∞, f (x) → 0+. 2. The numerator of g(x), x2 − 4, has degree 2, but the degree of the denominator, x + 1, has degree 1. By Theorem 4.2, there is no horizontal asymptote. From the graph, we see that the graph of y = g(x) doesn’t appear to level oﬀ to a constant value, so there is no horizontal asymptote.10 8As seen in the tables immediately preceding Deﬁnition 4.2. 9More speciﬁcally, as x → −∞, f (x) → 2+, and as x → ∞, f (x) → 2−. 10Sit tight! We’ll revisit this function and its end behavior shortly. 310 Rational Functions 3. The degrees of the numerator and denominator of h(x) are both three, so Theorem 4.2 tells −2 = −3 is the horizontal asymptote. We see from the calculator’s graph that as us y = 6 x → −∞, h(x) → −3+, and as x → ∞, h(x) → −3−. The graph of y = f (x) The graph of y = g(x) The graph of y = h(x) Our next example of the section gives us a real-world application of a horizontal asymptote.11 Example 4.1.5. The number of students N at local college who have had the ﬂu t months after the semester begins can be modeled by the formula N (t) = 500 − 450 1+3t for t ≥ 0. 1. Find and interpret N (0). 2. How long will it take until 300 students will have had the ﬂu? 3. Determine the behavior of N as t → ∞. Interpret this result graphically and within the context of the problem. Solution. 1. N (0) = 500 − 450 have had the ﬂu. 1+3(0) = 50. This means that at the beginning of the semester, 50 students 2. We
set N (t) = 300 to get 500 − 450 1+3t = 300 and solve. Isolating the fraction gives 450 Clearing denominators gives 450 = 200(1 + 3t). Finally, we get t = 5 take 5 12 months, or about 13 days, for 300 students to have had the ﬂu. 1+3t = 200. 12. This means it will 3. To determine the behavior of N as t → ∞, we can use a table. t N (t) 10 ≈ 485.48 100 ≈ 498.50 1000 ≈ 499.85 10000 ≈ 499.98 The table suggests that as t → ∞, N (t) → 500. (More speciﬁcally, 500−.) This means as time goes by, only a total of 500 students will have ever had the ﬂu. 11Though the population below is more accurately modeled with the functions in Chapter 6, we approximate it (using Calculus, of course!) using a rational function. 4.1 Introduction to Rational Functions 311 We close this section with a discussion of the third (and ﬁnal!) kind of asymptote which can be associated with the graphs of rational functions. Let us return to the function g(x) = x2−4 x+1 in Example 4.1.4. Performing long division,12 we get g(x) = x2−4 x+1. Since the term 3 x+1 → 0 as x → ±∞, it stands to reason that as x becomes unbounded, the function values g(x) = x − 1 − 3 x+1 ≈ x − 1. Geometrically, this means that the graph of y = g(x) should resemble the line y = x − 1 as x → ±∞. We see this play out both numerically and graphically below. x+(x) x −11 ≈ −10.6667 −10 −100 −101 ≈ −100.9697 −1000 ≈ −1000.9970 −1001 −10000 ≈ −10000.9997 −10001 g(x) x ≈ 8.7273 10 ≈ 98.9703 100 1000 ≈ 998.9970 10000 ≈ 9998.9997 x − 1 9 99 999 9999 y = g(x) and y = x −
1 as x → −∞ y = g(x) and y = x − 1 as x → ∞ The way we symbolize the relationship between the end behavior of y = g(x) with that of the line y = x − 1 is to write ‘as x → ±∞, g(x) → x − 1.’ In this case, we say the line y = x − 1 is a slant asymptote13 to the graph of y = g(x). Informally, the graph of a rational function has a slant asymptote if, as x → ∞ or as x → −∞, the graph resembles a non-horizontal, or ‘slanted’ line. Formally, we deﬁne a slant asymptote as follows. Deﬁnition 4.4. The line y = mx + b where m = 0 is called a slant asymptote of the graph of a function y = f (x) if as x → −∞ or as x → ∞, f (x) → mx + b. A few remarks are in order. First, note that the stipulation m = 0 in Deﬁnition 4.4 is what makes the ‘slant’ asymptote ‘slanted’ as opposed to the case when m = 0 in which case we’d have a horizontal asymptote. Secondly, while we have motivated what me mean intuitively by the notation ‘f (x) → mx+b,’ like so many ideas in this section, the formal deﬁnition requires Calculus. Another way to express this sentiment, however, is to rephrase ‘f (x) → mx + b’ as ‘f (x) − (mx + b) → 0.’ In other words, the graph of y = f (x) has the slant asymptote y = mx + b if and only if the graph of y = f (x) − (mx + b) has a horizontal asymptote y = 0. 12See the remarks following Theorem 4.2. 13Also called an ‘oblique’ asymptote in some, ostensibly higher class (and more expensive), texts. 312 Rational Functions Our next task is to determine the
conditions under which the graph of a rational function has a slant asymptote, and if it does, how to ﬁnd it. In the case of g(x) = x2−4 x+1, the degree of the numerator x2 − 4 is 2, which is exactly one more than the degree if its denominator x + 1 which is 1. This results in a linear quotient polynomial, and it is this quotient polynomial which is the slant asymptote. Generalizing this situation gives us the following theorem.14 Theorem 4.3. Determination of Slant Asymptotes: Suppose r is a rational function and r(x) = p(x) q(x), where the degree of p is exactly one more than the degree of q. Then the graph of the slant asymptote y = L(x) where L(x) is the quotient obtained by dividing y = r(x) has p(x) by q(x). In the same way that Theorem 4.2 gives us an easy way to see if the graph of a rational function r(x) = p(x) q(x) has a horizontal asymptote by comparing the degrees of the numerator and denominator, Theorem 4.3 gives us an easy way to check for slant asymptotes. Unlike Theorem 4.2, which gives us a quick way to ﬁnd the horizontal asymptotes (if any exist), Theorem 4.3 gives us no such ‘short-cut’. If a slant asymptote exists, we have no recourse but to use long division to ﬁnd it.15 Example 4.1.6. Find the slant asymptotes of the graphs of the following functions if they exist. Verify your answers using a graphing calculator and describe the behavior of the graph near them using proper notation. 1. f (x) = x2 − 4x + 2 1 − x 2. g(x) = x2 − 4 x − 2 3. h(x) = x3 + 1 x2 − 4 Solution. 1. The degree of the numerator is 2 and the degree of the denominator is 1, so Theorem 4.3 guarantees us a slant asymptote. To ﬁnd it, we divide 1 − x =
−x + 1 into x2 − 4x + 2 and get a quotient of −x + 3, so our slant asymptote is y = −x + 3. We conﬁrm this graphically, and we see that as x → −∞, the graph of y = f (x) approaches the asymptote from below, and as x → ∞, the graph of y = f (x) approaches the asymptote from above.16 2. As with the previous example, the degree of the numerator g(x) = x2−4 x−2 is 2 and the degree of the denominator is 1, so Theorem 4.3 applies. In this case, g(x) = x2 − 4 x − 2 = (x + 2)(x − 2) (x − 2) = (x + 2) (x − 2) 1 (x − 2) = x + 2, x = 2 14Once again, this theorem is brought to you courtesy of Theorem 3.4 and Calculus. 15That’s OK, though. In the next section, we’ll use long division to analyze end behavior and it’s worth the eﬀort! 16Note that we are purposefully avoiding notation like ‘as x → ∞, f (x) → (−x + 3)+. While it is possible to deﬁne these notions formally with Calculus, it is not standard to do so. Besides, with the introduction of the symbol ‘’ in the next section, the authors feel we are in enough trouble already. 4.1 Introduction to Rational Functions 313 so we have that the slant asymptote y = x + 2 is identical to the graph of y = g(x) except at x = 2 (where the latter has a ‘hole’ at (2, 4).) The calculator supports this claim.17 3. For h(x) = x3+1 x2−4, the degree of the numerator is 3 and the degree of the denominator is 2 so again, we are guaranteed the existence of a slant asymptote. The long division x3 + 1 ÷ x2 − 4 gives a quotient of just x, so our slant asymptote is the line y = x. The calculator conﬁrms this, and we �
��nd that as x → −∞, the graph of y = h(x) approaches the asymptote from below, and as x → ∞, the graph of y = h(x) approaches the asymptote from above. The graph of y = f (x) The graph of y = g(x) The graph of y = h(x) The reader may be a bit disappointed with the authors at this point owing to the fact that in Examples 4.1.2, 4.1.4, and 4.1.6, we used the calculator to determine function behavior near asymptotes. We rectify that in the next section where we, in excruciating detail, demonstrate the usefulness of ‘number sense’ to reveal this behavior analytically. 17While the word ‘asymptote’ has the connotation of ‘approaching but not equaling,’ Deﬁnitions 4.3 and 4.4 invite the same kind of pathologies we saw with Deﬁnitions 1.11 in Section 1.6. Rational Functions 314 4.1.1 Exercises In Exercises 1 - 18, for the given rational function f : Find the domain of f. Identify any vertical asymptotes of the graph of y = f (x). Identify any holes in the graph. Find the horizontal asymptote, if it exists. Find the slant asymptote, if it exists. Graph the function using a graphing utility and describe the behavior near the asymptotes. 1. f (x) = 4. f (x) = 7. f (x) = x 3x − 6 x x2 + 1 4x x2 + 4 10. f (x) = 13. f (x) = 3x2 − 5x − 2 x2 − 9 2x2 + 5x − 3 3x + 2 16. f (x) = x3 1 − x 2. f (x) = 5. f (x) = 8. f (x) = 3 + 7x 5 − 2x x + 7 (x + 3)2 4x x2 − 4 11. f (x) = 14. f (x) = 17. f (x) = x3 + 2x2 + x x2 − x − 2 −x3 + 4x x2 − 9
18 − 2x2 x2 − 9 3. f (x) = 6. f (x) = 9. f (x) = 12. f (x) = x x2 + x − 12 x3 + 1 x2 − 1 x2 − x − 12 x2 + x − 6 x3 − 3x + 1 x2 + 1 15. f (x) = −5x4 − 3x3 + x2 − 10 x3 − 3x2 + 3x − 1 18. f (x) = x3 − 4x2 − 4x − 5 x2 + x + 1 19. The cost C in dollars to remove p% of the invasive species of Ippizuti ﬁsh from Sasquatch Pond is given by C(p) = 1770p 100 − p, 0 ≤ p < 100 (a) Find and interpret C(25) and C(95). (b) What does the vertical asymptote at x = 100 mean within the context of the problem? (c) What percentage of the Ippizuti ﬁsh can you remove for $40000? 20. In Exercise 71 in Section 1.4, the population of Sasquatch in Portage County was modeled by the function P (t) = 150t t + 15, where t = 0 represents the year 1803. Find the horizontal asymptote of the graph of y = P (t) and explain what it means. 4.1 Introduction to Rational Functions 315 21. Recall from Example 1.5.3 that the cost C (in dollars) to make x dOpi media players is C(x) = 100x + 2000, x ≥ 0. (a) Find a formula for the average cost C(x). Recall: C(x) = C(x) x. (b) Find and interpret C(1) and C(100). (c) How many dOpis need to be produced so that the average cost per dOpi is $200? (d) Interpret the behavior of C(x) as x → 0+. (HINT: You may want to ﬁnd the ﬁxed cost C(0) to help in your interpretation.) (e) Interpret the behavior of C(x) as x → ∞. (HINT: You may want to ﬁnd the variable cost (deﬁned in Example
2.1.5 in Section 2.1) to help in your interpretation.) 22. In Exercise 35 in Section 3.1, we ﬁt a few polynomial models to the following electric circuit data. (The circuit was built with a variable resistor. For each of the following resistance values (measured in kilo-ohms, kΩ), the corresponding power to the load (measured in milliwatts, mW ) is given in the table below.)18 Resistance: (kΩ) Power: (mW ) 1.012 1.063 2.199 1.496 3.275 1.610 4.676 1.613 6.805 1.505 9.975 1.314 Using some fundamental laws of circuit analysis mixed with a healthy dose of algebra, we can derive the actual formula relating power to resistance. For this circuit, it is P (x) = 25x (x+3.9)2, where x is the resistance value, x ≥ 0. (a) Graph the data along with the function y = P (x) on your calculator. (b) Use your calculator to approximate the maximum power that can be delivered to the load. What is the corresponding resistance value? (c) Find and interpret the end behavior of P (x) as x → ∞. 23. In his now famous 1919 dissertation The Learning Curve Equation, Louis Leon Thurstone presents a rational function which models the number of words a person can type in four minutes as a function of the number of pages of practice one has completed. (This paper, which is now in the public domain and can be found here, is from a bygone era when students at business schools took typing classes on manual typewriters.) Using his original notation and original language, we have Y = L(X+P ) (X+P )+R where L is the predicted practice limit in terms of speed units, X is pages written, Y is writing speed in terms of words in four minutes, P is equivalent previous practice in terms of pages and R is the rate of learning. In Figure 5 of the paper, he graphs a scatter plot and the curve Y = 216(X+19). Discuss this equation with your classmates. How would you update the notation? Explain what the horizontal asymptote of the graph means. You should take some time to look at the original paper. Skip over the computations you don’t understand
yet and try to get a sense of the time and place in which the study was conducted. X+148 18The authors wish to thank Don Anthan and Ken White of Lakeland Community College for devising this problem and generating the accompanying data set. 316 4.1.2 Answers Rational Functions 1. f (x) = x 3x − 6 Domain: (−∞, 2) ∪ (2, ∞) Vertical asymptote: x = 2 As x → 2−, f (x) → −∞ As x → 2+, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 1 3 As x → −∞, f (x) → 1 3 + As x → ∞, f (x) → 1 3 − 3. f (x) = = x x2 + x − 12 x (x + 4)(x − 3) Domain: (−∞, −4) ∪ (−4, 3) ∪ (3, ∞) Vertical asymptotes: x = −4, x = 3 As x → −4−, f (x) → −∞ As x → −4+, f (x) → ∞ As x → 3−, f (x) → −∞ As x → 3+, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 5. f (x) = x + 7 (x + 3)2 Domain: (−∞, −3) ∪ (−3, ∞) Vertical asymptote: x = −3 As x → −3−, f (x) → ∞ As x → −3+, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 0 19As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 2. f (x) = 3 + 7x 5 − 2x − + 2 ) ∪ ( 5, f (x) → ∞, f (x) → −∞ Domain: (−∞, 5 2, ∞) Vertical asymptote: x = 5 2 As x → 5 2 As x →
5 2 No holes in the graph Horizontal asymptote: y = − 7 2 As x → −∞, f (x) → − 7 2 − As x → ∞, f (x) → − 7 2 + 4. f (x) = x x2 + 1 Domain: (−∞, ∞) No vertical asymptotes No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 6. f (x) = x3 + 1 x2 − 1 = x2 − x + 1 x − 1 Domain: (−∞, −1) ∪ (−1, 1) ∪ (1, ∞) Vertical asymptote: x = 1 As x → 1−, f (x) → −∞ As x → 1+, f (x) → ∞ Hole at (−1, − 3 2 ) Slant asymptote: y = x As x → −∞, the graph is below y = x As x → ∞, the graph is above y = x 19This is hard to see on the calculator, but trust me, the graph is below the x-axis to the left of x = −7. 4.1 Introduction to Rational Functions 317 7. f (x) = 4x x2 + 4 Domain: (−∞, ∞) No vertical asymptotes No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 9. f (x) = x2 − x − 12 x2 + Domain: (−∞, −3) ∪ (−3, 2) ∪ (2, ∞) Vertical asymptote: x = 2 As x → 2−, f (x) → ∞ As x → 2+, f (x) → −∞ Hole at −3, 7 5 Horizontal asymptote: y = 1 As x → −∞, f (x) → 1+ As x → ∞, f (x) → 1− 8. f (x) = 4x x2 − 4 = 4x (x + 2)(x − 2) Domain: (−∞, −2)
∪ (−2, 2) ∪ (2, ∞) Vertical asymptotes: x = −2, x = 2 As x → −2−, f (x) → −∞ As x → −2+, f (x) → ∞ As x → 2−, f (x) → −∞ As x → 2+, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 10. f (x) = = 3x2 − 5x − 2 x2 − 9 (3x + 1)(x − 2) (x + 3)(x − 3) Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) Vertical asymptotes: x = −3, x = 3 As x → −3−, f (x) → ∞ As x → −3+, f (x) → −∞ As x → 3−, f (x) → −∞ As x → 3+, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 3 As x → −∞, f (x) → 3+ As x → ∞, f (x) → 3− 11. f (x) = x3 + 2x2 + x x2 − x − 2 = x(x + 1) x − 2 Domain: (−∞, −1) ∪ (−1, 2) ∪ (2, ∞) Vertical asymptote: x = 2 As x → 2−, f (x) → −∞ As x → 2+, f (x) → ∞ Hole at (−1, 0) Slant asymptote: y = x + 3 As x → −∞, the graph is below y = x + 3 As x → ∞, the graph is above y = x + 3 12. f (x) = x3 − 3x + 1 x2 + 1 Domain: (−∞, ∞) No vertical asymptotes No holes in the graph Slant asymptote: y = x As x → −∞, the graph is above y = x As x → ∞, the graph is below y
= x 318 Rational Functions 13. f (x) = − 3, ∞ 2x2 + 5x − 3 3x + 2 Domain: −∞, − 2 ∪ − 2 3 Vertical asymptote: x = − 2 3 As x → − 2 3 As x → − 2 3 No holes in the graph Slant asymptote: y = 2 As x → −∞, the graph is above y = 2 As x → ∞, the graph is below y = 2, f (x) → ∞, f (x) → −∞ 3 x + 11 + 9 3 x+ 11 3 x + 11 9 9 14. f (x) = −x3 + 4x x2 − 9 = −x3 + 4x (x − 3)(x + 3) Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) Vertical asymptotes: x = −3, x = 3 As x → −3−, f (x) → ∞ As x → −3+, f (x) → −∞ As x → 3−, f (x) → ∞ As x → 3+, f (x) → −∞ No holes in the graph Slant asymptote: y = −x As x → −∞, the graph is above y = −x As x → ∞, the graph is below y = −x 15. f (x) = = −5x4 − 3x3 + x2 − 10 x3 − 3x2 + 3x − 1 −5x4 − 3x3 + x2 − 10 (x − 1)3 Domain: (−∞, 1) ∪ (1, ∞) Vertical asymptotes: x = 1 As x → 1−, f (x) → ∞ As x → 1+, f (x) → −∞ No holes in the graph Slant asymptote: y = −5x − 18 As x → −∞, the graph is above y = −5x − 18 As x → ∞, the graph is below y = −5x − 18 16. f (x) = x3 1 − x Domain: (−∞, 1) ∪ (1, ∞) Vertical asymptote: x = 1 As x → 1−, f (
x) → ∞ As x → 1+, f (x) → −∞ No holes in the graph No horizontal or slant asymptote As x → −∞, f (x) → −∞ As x → ∞, f (x) → −∞ 17. f (x) = = −2 18. f (x) = 18 − 2x2 x2 − 9 x3 − 4x2 − 4x − 5 x2 + x + 1 = x − 5 Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) No vertical asymptotes Holes in the graph at (−3, −2) and (3, −2) Horizontal asymptote y = −2 As x → ±∞, f (x) = −2 Domain: (−∞, ∞) No vertical asymptotes No holes in the graph Slant asymptote: y = x − 5 f (x) = x − 5 everywhere. 19. (a) C(25) = 590 means it costs $590 to remove 25% of the ﬁsh and and C(95) = 33630 means it would cost $33630 to remove 95% of the ﬁsh from the pond. (b) The vertical asymptote at x = 100 means that as we try to remove 100% of the ﬁsh from the pond, the cost increases without bound; i.e., it’s impossible to remove all of the ﬁsh. (c) For $40000 you could remove about 95.76% of the ﬁsh. 4.1 Introduction to Rational Functions 319 20. The horizontal asymptote of the graph of P (t) = 150t t+15 is y = 150 and it means that the model predicts the population of Sasquatch in Portage County will never exceed 150. 21. (a) C(x) = 100x+2000 (b) C(1) = 2100 and C(100) = 120. When just 1 dOpi is produced, the cost per dOpi is, x > 0. x $2100, but when 100 dOpis are produced, the cost per dOpi is $120. (c) C(x) = 200 when x = 20. So to get the cost per
dOpi to $200, 20 dOpis need to be produced. (d) As x → 0+, C(x) → ∞. This means that as fewer and fewer dOpis are produced, the cost per dOpi becomes unbounded. In this situation, there is a ﬁxed cost of $2000 (C(0) = 2000), we are trying to spread that $2000 over fewer and fewer dOpis. (e) As x → ∞, C(x) → 100+. This means that as more and more dOpis are produced, the cost per dOpi approaches $100, but is always a little more than $100. Since $100 is the variable cost per dOpi (C(x) = 100x + 2000), it means that no matter how many dOpis are produced, the average cost per dOpi will always be a bit higher than the variable cost to produce a dOpi. As before, we can attribute this to the $2000 ﬁxed cost, which factors into the average cost per dOpi no matter how many dOpis are produced. 22. (a) (b) The maximum power is approximately 1.603 mW which corresponds to 3.9 kΩ. (c) As x → ∞, P (x) → 0+ which means as the resistance increases without bound, the power diminishes to zero. 320 Rational Functions 4.2 Graphs of Rational Functions In this section, we take a closer look at graphing rational functions. In Section 4.1, we learned that the graphs of rational functions may have holes in them and could have vertical, horizontal and slant asymptotes. Theorems 4.1, 4.2 and 4.3 tell us exactly when and where these behaviors will occur, and if we combine these results with what we already know about graphing functions, we will quickly be able to generate reasonable graphs of rational functions. One of the standard tools we will use is the sign diagram which was ﬁrst introduced in Section 2.4, and then revisited in Section 3.1. In those sections, we operated under the belief that a function couldn’t change its sign without its graph crossing through the x-axis. The major theorem we used to justify this belief was the Intermediate Value Theorem, Theorem 3.1. It turns out the Intermediate Value Theorem applies
to all continuous functions,1 not just polynomials. Although rational functions are continuous on their domains,2 Theorem 4.1 tells us that vertical asymptotes and holes occur at the values excluded from their domains. In other words, rational functions aren’t continuous at these excluded values which leaves open the possibility that the function could change sign without crossing through the x-axis. Consider the graph of y = h(x) from Example 4.1.1, recorded below for convenience. We have added its x-intercept at 1 2, 0 for the discussion that follows. Suppose we wish to construct a sign diagram for h(x). Recall that the intervals where h(x) > 0, or (+), correspond to the x-values where the graph of y = h(x) is above the x-axis; the intervals on which h(x) < 0, or (−) correspond to where the graph is below the x-axis1 −2 −3 −4 −5 −6 −4 −3 −2 (+) (−) 0 (+) (+) − As we examine the graph of y = h(x), reading from left to right, we note that from (−∞, −1), the graph is above the x-axis, so h(x) is (+) there. At x = −1, we have a vertical asymptote, at, the graph is below the which point the graph ‘jumps’ across the x-axis. On the interval −1, 1 2 1Recall that, for our purposes, this means the graphs are devoid of any breaks, jumps or holes 2Another result from Calculus. 4.2 Graphs of Rational Functions 321 x-axis, so h(x) is (−) there. The graph crosses through the x-axis at 1 2, 0 and remains above the x-axis until x = 1, where we have a ‘hole’ in the graph. Since h(1) is undeﬁned, there is no sign here. So we have h(x) as (+) on the interval 1 2, 1. Continuing, we see that on (1, ∞), the graph of y = h(x) is above the x-axis, so we mark (+) there. To construct a sign diagram from this information, we not only need to denote the zero of h, but also the places not in the domain of h
. As is our custom, we write ‘0’ above 1 2 on the sign diagram to remind us that it is a zero of h. We need a diﬀerent notation for −1 and 1, and we have chosen to use ‘’ - a nonstandard symbol called the interrobang. We use this symbol to convey a sense of surprise, caution and wonderment - an appropriate attitude to take when approaching these points. The moral of the story is that when constructing sign diagrams for rational functions, we include the zeros as well as the values excluded from the domain. Steps for Constructing a Sign Diagram for a Rational Function Suppose r is a rational function. 1. Place any values excluded from the domain of r on the number line with an ‘’ above them. 2. Find the zeros of r and place them on the number line with the number 0 above them. 3. Choose a test value in each of the intervals determined in steps 1 and 2. 4. Determine the sign of r(x) for each test value in step 3, and write that sign above the corresponding interval. We now present our procedure for graphing rational functions and apply it to a few exhaustive examples. Please note that we decrease the amount of detail given in the explanations as we move through the examples. The reader should be able to ﬁll in any details in those steps which we have abbreviated. Steps for Graphing Rational Functions Suppose r is a rational function. 1. Find the domain of r. 2. Reduce r(x) to lowest terms, if applicable. 3. Find the x- and y-intercepts of the graph of y = r(x), if they exist. 4. Determine the location of any vertical asymptotes or holes in the graph, if they exist. Analyze the behavior of r on either side of the vertical asymptotes, if applicable. 5. Analyze the end behavior of r. Find the horizontal or slant asymptote, if one exists. 6. Use a sign diagram and plot additional points, as needed, to sketch the graph of y = r(x). 322 Rational Functions Example 4.2.1. Sketch a detailed graph of f (x) = 3x x2 − 4 Solution. We follow the six step procedure outlined above.. 1. As usual, we set the denominator equal to zero to get x2 − 4 = 0.
We ﬁnd x = ±2, so our domain is (−∞, −2) ∪ (−2, 2) ∪ (2, ∞). 2. To reduce f (x) to lowest terms, we factor the numerator and denominator which yields (x−2)(x+2). There are no common factors which means f (x) is already in lowest 3x f (x) = terms. 3. To ﬁnd the x-intercepts of the graph of y = f (x), we set y = f (x) = 0. Solving (x−2)(x+2) = 0 results in x = 0. Since x = 0 is in our domain, (0, 0) is the x-intercept. To ﬁnd the y-intercept, we set x = 0 and ﬁnd y = f (0) = 0, so that (0, 0) is our y-intercept as well.3 3x 4. The two numbers excluded from the domain of f are x = −2 and x = 2. Since f (x) didn’t reduce at all, both of these values of x still cause trouble in the denominator. Thus by Theorem 4.1, x = −2 and x = 2 are vertical asymptotes of the graph. We can actually go a step further at this point and determine exactly how the graph approaches the asymptote near each of these values. Though not absolutely necessary,4 it is good practice for those heading oﬀ to Calculus. For the discussion that follows, it is best to use the factored form of f (x) = 3x (x−2)(x+2). The behavior of y = f (x) as x → −2: Suppose x → −2−. If we were to build a table of values, we’d use x-values a little less than −2, say −2.1, −2.01 and −2.001. While there is no harm in actually building a table like we did in Section 4.1, we want to develop a ‘number sense’ here. Let’s think about each factor in the formula of f (x) as we imagine substituting a number like x = −2.000001 into f (x). The quantity 3x would be very close to
−6, the quantity (x − 2) would be very close to −4, and the factor (x + 2) would be very close to 0. More speciﬁcally, (x + 2) would be a little less than 0, in this case, −0.000001. We will call such a number a ‘very small (−)’, ‘very small’ meaning close to zero in absolute value. So, mentally, as x → −2−, we estimate f (x) = 3x (x − 2)(x + 2) ≈ −6 (−4) (very small (−)) = 3 2 (very small (−)) Now, the closer x gets to −2, the smaller (x + 2) will become, so even though we are multiplying our ‘very small (−)’ by 2, the denominator will continue to get smaller and smaller, and remain negative. The result is a fraction whose numerator is positive, but whose denominator is very small and negative. Mentally, f (x) ≈ 3 2 (very small (−)) ≈ 3 very small (−) ≈ very big (−) 3As we mentioned at least once earlier, since functions can have at most one y-intercept, once we ﬁnd that (0, 0) is on the graph, we know it is the y-intercept. 4The sign diagram in step 6 will also determine the behavior near the vertical asymptotes. 4.2 Graphs of Rational Functions 323 The term ‘very big (−)’ means a number with a large absolute value which is negative.5 What all of this means is that as x → −2−, f (x) → −∞. Now suppose we wanted to determine the behavior of f (x) as x → −2+. If we imagine substituting something a little larger than −2 in for x, say −1.999999, we mentally estimate f (x) ≈ −6 (−4) (very small (+)) = 3 2 (very small (+)) ≈ 3 very small (+) ≈ very big (+) We conclude that as x → −2+, f (x) → ∞. The behavior of y = f (x) as x → 2: Consider x → 2−. We imagine substituting x = 1.999999. Approximating f (x) as we did above
, we get f (x) ≈ 6 (very small (−)) (4) = 3 2 (very small (−)) ≈ 3 very small (−) ≈ very big (−) We conclude that as x → 2−, f (x) → −∞. Similarly, as x → 2+, we imagine substituting x = 2.000001 to get f (x) ≈ very small (+) ≈ very big (+). So as x → 2+, f (x) → ∞. 3 Graphically, we have that near x = −2 and x = 2 the graph of y = f (x) looks like6 y −3 −1 1 3 x 5. Next, we determine the end behavior of the graph of y = f (x). Since the degree of the numerator is 1, and the degree of the denominator is 2, Theorem 4.2 tells us that y = 0 is the horizontal asymptote. As with the vertical asymptotes, we can glean more detailed information using ‘number sense’. For the discussion below, we use the formula f (x) = 3x x2−4. The behavior of y = f (x) as x → −∞: If we were to make a table of values to discuss the behavior of f as x → −∞, we would substitute very ‘large’ negative numbers in for x, say for example, x = −1 billion. The numerator 3x would then be −3 billion, whereas 5The actual retail value of f (−2.000001) is approximately −1,500,000. 6We have deliberately left oﬀ the labels on the y-axis because we know only the behavior near x = ±2, not the actual function values. 324 Rational Functions the denominator x2 − 4 would be (−1 billion)2 − 4, which is pretty much the same as 1(billion)2. Hence, f (−1 billion) ≈ −3 billion 1(billion)2 ≈ − 3 billion ≈ very small (−) Notice that if we substituted in x = −1 trillion, essentially the same kind of cancellation would happen, and we would be left with an even ‘smaller’ negative number. This not only conﬁrms the fact that as x → −∞, f (x) → 0, it tells us that f (x) → 0
−. In other words, the graph of y = f (x) is a little bit below the x-axis as we move to the far left. The behavior of y = f (x) as x → ∞: On the ﬂip side, we can imagine substituting very large positive numbers in for x and looking at the behavior of f (x). For example, let x = 1 billion. Proceeding as before, we get f (1 billion) ≈ 3 billion 1(billion)2 ≈ 3 billion ≈ very small (+) The larger the number we put in, the smaller the positive number we would get out. In other words, as x → ∞, f (x) → 0+, so the graph of y = f (x) is a little bit above the x-axis as we look toward the far right. Graphically, we have7 y 1 −1 x 6. Lastly, we construct a sign diagram for f (x). The x-values excluded from the domain of f are x = ±2, and the only zero of f is x = 0. Displaying these appropriately on the number line gives us four test intervals, and we choose the test values8 x = −3, x = −1, x = 1 and x = 3. We ﬁnd f (−3) is (−), f (−1) is (+), f (1) is (−) and f (3) is (+). Combining this with our previous work, we get the graph of y = f (x) below. 7As with the vertical asymptotes in the previous step, we know only the behavior of the graph as x → ±∞. For that reason, we provide no x-axis labels. 8In this particular case, we can eschew test values, since our analysis of the behavior of f near the vertical asymptotes and our end behavior analysis have given us the signs on each of the test intervals. In general, however, this won’t always be the case, so for demonstration purposes, we continue with our usual construction. 4.2 Graphs of Rational Functions 325 (−) (+) 0 (−) (+) −2 0 2 −3 −1 1 3 y 3 2 1 −5 −4 −3 −1 1 3 4 5 x −1 −2 −3 A couple of notes are in order. First, the graph of y = f (x) certainly seems to
possess symmetry with respect to the origin. In fact, we can check f (−x) = −f (x) to see that f is an odd function. In some textbooks, checking for symmetry is part of the standard procedure for graphing rational functions; but since it happens comparatively rarely9 we’ll just point it out when we see it. Also note that while y = 0 is the horizontal asymptote, the graph of f actually crosses the x-axis at (0, 0). The myth that graphs of rational functions can’t cross their horizontal asymptotes is completely false,10 as we shall see again in our next example. Example 4.2.2. Sketch a detailed graph of g(x) = Solution. 2x2 − 3x − 5 x2 − x − 6. 1. Setting x2 − x − 6 = 0 gives x = −2 and x = 3. Our domain is (−∞, −2) ∪ (−2, 3) ∪ (3, ∞). 2. Factoring g(x) gives g(x) = (2x−5)(x+1) (x−3)(x+2). There is no cancellation, so g(x) is in lowest terms. 3. To ﬁnd the x-intercept we set y = g(x) = 0. Using the factored form of g(x) above, we ﬁnd 2 and x = −1. Since 2, 0 and (−1, 0). the zeros to be the solutions of (2x − 5)(x + 1) = 0. We obtain x = 5 both of these numbers are in the domain of g, we have two x-intercepts, 5 To ﬁnd the y-intercept, we set x = 0 and ﬁnd y = g(0) = 5 6, so our y-intercept is 0, 5. 6 4. Since g(x) was given to us in lowest terms, we have, once again by Theorem 4.1 vertical (x−3)(x+2), we proceed to our asymptotes x = −2 and x = 3. Keeping in mind g(x) = (2x−5)(x+1) analysis near each of these values. The behavior of y = g(x) as x → −2: As x → −2
−, we imagine substituting a number a little bit less than −2. We have g(x) ≈ (−9)(−1) (−5)(very small (−)) ≈ 9 very small (+) ≈ very big (+) 9And Jeﬀ doesn’t think much of it to begin with... 10That’s why we called it a MYTH! 326 Rational Functions so as x → −2−, g(x) → ∞. On the ﬂip side, as x → −2+, we get g(x) ≈ 9 very small (−) ≈ very big (−) so g(x) → −∞. The behavior of y = g(x) as x → 3: As x → 3−, we imagine plugging in a number just shy of 3. We have g(x) ≈ (1)(4) ( very small (−))(5) ≈ 4 very small (−) ≈ very big (−) Hence, as x → 3−, g(x) → −∞. As x → 3+, we get g(x) ≈ 4 very small (+) ≈ very big (+) so g(x) → ∞. Graphically, we have (again, without labels on the y-axis) y −3 −1 1 2 4 x 5. Since the degrees of the numerator and denominator of g(x) are the same, we know from Theorem 4.2 that we can ﬁnd the horizontal asymptote of the graph of g by taking the ratio of the leading terms coeﬃcients, y = 2 1 = 2. However, if we take the time to do a more detailed analysis, we will be able to reveal some ‘hidden’ behavior which would be lost otherwise.11 As in the discussion following Theorem 4.2, we use the result of the long division 2x2 − 3x − 5 ÷ x2 − x − 6 to rewrite g(x) = 2x2−3x−5 x2−x−6. We focus our attention on the term x−7 x2−x−6 as g(x) = 2 − x−7 x2−x−6. 11That is, if you use a calculator to graph. Once again, Calculus is the ultimate graphing power tool. 4.2 Graphs of Rational Functions
327 The behavior of y = g(x) as x → −∞: If imagine substituting x = −1 billion into x−7 x2−x−6, we estimate x−7 x2−x−6 ≈ −1 billion 1billion2 ≈ very small (−).12 Hence, g(x) = 2 − x − 7 x2 − x − 6 ≈ 2 − very small (−) = 2 + very small (+) In other words, as x → −∞, the graph of y = g(x) is a little bit above the line y = 2. The behavior of y = g(x) as x → ∞. To consider x−7 x2−x−6 as x → ∞, we imagine substituting x = 1 billion and, going through the usual mental routine, ﬁnd x − 7 x2 − x − 6 ≈ very small (+) Hence, g(x) ≈ 2 − very small (+), in other words, the graph of y = g(x) is just below the line y = 2 as x → ∞. On y = g(x), we have (again, without labels on the x-axis) y 1 −1 x and (3, ∞), and (−) on the intervals (−2, −1) and 5 6. Finally we construct our sign diagram. We place an ‘’ above x = −2 and x = 3, and a ‘0’ above x = 5 2 and x = −1. Choosing test values in the test intervals gives us f (x) is (+) on the intervals (−∞, −2), −1, 5 2, 3. As 2 we piece together all of the information, we note that the graph must cross the horizontal asymptote at some point after x = 3 in order for it to approach y = 2 from underneath. This is the subtlety that we would have missed had we skipped the long division and subsequent end behavior analysis. We can, in fact, ﬁnd exactly when the graph crosses y = 2. As a result of the long division, we have g(x) = 2 − x−7 x2−x−6 = 0. This gives x − 7 = 0, or x = 7. Note that x − 7 is the remainder when 2x2 − 3x − 5 is divided by x2 − x −
6, so it makes sense that for g(x) to equal the quotient 2, the remainder from the division must be 0. Sure enough, we ﬁnd g(7) = 2. Moreover, it stands to reason that g must attain a relative minimum at some point past x = 7. Calculus veriﬁes that at x = 13, we have such a minimum at exactly (13, 1.96). The reader is challenged to ﬁnd calculator windows which show the graph crossing its horizontal asymptote on one window, and the relative minimum in the other. x2−x−6. For g(x) = 2, we would need x−7 12In the denominator, we would have (1billion)2 − 1billion − 6. It’s easy to see why the 6 is insigniﬁcant, but to ignore the 1 billion seems criminal. However, compared to (1 billion)2, it’s on the insigniﬁcant side; it’s 1018 versus 109. We are once again using the fact that for polynomials, end behavior is determined by the leading term, so in the denominator, the x2 term wins out over the x term. 328 Rational Functions (+) (−) 0 (+) 0 (−) (+) −2 −9−8−7−6−5−4−3 −1 −2 −3 −4 Our next example gives us an opportunity to more thoroughly analyze a slant asymptote. Example 4.2.3. Sketch a detailed graph of h(x) = Solution. 2x3 + 5x2 + 4x + 1 x2 + 3x + 2. 1. For domain, you know the drill. Solving x2 + 3x + 2 = 0 gives x = −2 and x = −1. Our answer is (−∞, −2) ∪ (−2, −1) ∪ (−1, ∞). 2. To reduce h(x), we need to factor the numerator and denominator. To factor the numerator, we use the techniques13 set forth in Section 3.3 and we get h(x) = 2x3 + 5x2 + 4x + 1 x2 + 3x + 2 = (2x + 1)(x + 1)2 (x + 2)(x + 1) = 1
(2x + 1)(x + 1) (x + 2) (x + 1) 2 = (2x + 1)(x + 1) x + 2 We will use this reduced formula for h(x) as long as we’re not substituting x = −1. To make this exclusion speciﬁc, we write h(x) = (2x+1)(x+1), x = −1. x+2 3. To ﬁnd the x-intercepts, as usual, we set h(x) = 0 and solve. Solving (2x+1)(x+1) = 0 yields 2 and x = −1. The latter isn’t in the domain of h, so we exclude it. Our only x2, 0. To ﬁnd the y-intercept, we set x = 0. Since 0 = −1, we can use the x = − 1 intercept is − 1 reduced formula for h(x) and we get h(0) = 1 2 for a y-intercept of 0, 1. x+2 2 4. From Theorem 4.1, we know that since x = −2 still poses a threat in the denominator of the reduced function, we have a vertical asymptote there. As for x = −1, the factor (x + 1) was canceled from the denominator when we reduced h(x), so it no longer causes trouble there. This means that we get a hole when x = −1. To ﬁnd the y-coordinate of the hole, we substitute x = −1 into (2x+1)(x+1), per Theorem 4.1 and get 0. Hence, we have a hole on x+2 13Bet you never thought you’d never see that stuﬀ again before the Final Exam! 4.2 Graphs of Rational Functions 329 the x-axis at (−1, 0). It should make you uncomfortable plugging x = −1 into the reduced formula for h(x), especially since we’ve made such a big deal concerning the stipulation about not letting x = −1 for that formula. What we are really doing is taking a Calculus short-cut to the more detailed kind of analysis near x = −1 which we will show below. Speaking of which, for the discussion that follows, we will use
the formula h(x) = (2x+1)(x+1), x = −1. x+2 The behavior of y = h(x) as x → −2: As x → −2−, we imagine substituting a number a little bit less than −2. We have h(x) ≈ (very small (−)) ≈ very big (−) thus as x → −2−, h(x) → −∞. On the other side of −2, as x → −2+, we ﬁnd that h(x) ≈ very small (+) ≈ very big (+), so h(x) → ∞. (−3)(−1) (very small (−)) ≈ 3 3 The behavior of y = h(x) as x → −1. As x → −1−, we imagine plugging in a number a bit less than x = −1. We have h(x) ≈ (−1)(very small (−)) = very small (+) Hence, as x → −1−, h(x) → 0+. This means that as x → −1−, the graph is a bit above the point (−1, 0). As x → −1+, we get h(x) ≈ (−1)(very small (+)) = very small (−). This gives us that as x → −1+, h(x) → 0−, so the graph is a little bit lower than (−1, 0) here. 1 1 Graphically, we have y −3 x 5. For end behavior, we note that the degree of the numerator of h(x), 2x3 + 5x2 + 4x + 1, is 3 and the degree of the denominator, x2 + 3x + 2, is 2 so by Theorem 4.3, the graph of y = h(x) has a slant asymptote. For x → ±∞, we are far enough away from x = −1 to use the reduced formula, h(x) = (2x+1)(x+1), x = −1. To perform long division, we multiply out the numerator and get h(x) = 2x2+3x+1, x = −1, and rewrite h(x) = 2x − 1 + 3 x+2, x = −1. By Theorem 4.3, the
slant asymptote is y = 2x − 1, and to better see how the graph approaches the asymptote, we focus our attention on the term generated from the remainder, x+2 x+2 3 x+2. The behavior of y = h(x) as x → −∞: Substituting x = −1 billion into 3 −1 billion ≈ very small (−). Hence, h(x) = 2x−1+ 3 x+2, we get the x+2 ≈ 2x−1+very small (−). estimate This means the graph of y = h(x) is a little bit below the line y = 2x − 1 as x → −∞. 3 330 Rational Functions The behavior of y = h(x) as x → ∞: If x → ∞, then 3 x+2 ≈ very small (+). This means h(x) ≈ 2x − 1 + very small (+), or that the graph of y = h(x) is a little bit above the line y = 2x − 1 as x → ∞. Graphically we have y 4 3 2 1 −1 −2 −3 −4 x 6. To make our sign diagram, we place an ‘’ above x = −2 and x = −1 and a ‘0’ above x = − 1 2. 2, ∞ and h(x) is (−) on On our four test intervals, we ﬁnd h(x) is (+) on (−2, −1) and − 1 (−∞, −2) and −1, − 1 2. Putting all of our work together yields the graph below. (−) (+) (−) 0 (+) −4 −3 −2 −1 −1 −2 −3 −4 −5 −6 −7 −8 −9 −10 −11 −12 −13 −14 We could ask whether the graph of y = h(x) crosses its slant asymptote. From the formula h(x) = 2x − 1 + 3 3 x+2, x = −1, we see that if h(x) = 2x − 1, we would have x+2 = 0. Since this will never happen, we conclude the graph never crosses its slant asymptote.14 14But rest assured, some graphs do! 4.2 Graphs of Rational Functions
331 We end this section with an example that shows it’s not all pathological weirdness when it comes to rational functions and technology still has a role to play in studying their graphs at this level. Example 4.2.4. Sketch the graph of r(x) = x4 + 1 x2 + 1. Solution. 1. The denominator x2 + 1 is never zero so the domain is (−∞, ∞). 2. With no real zeros in the denominator, x2 + 1 is an irreducible quadratic. Our only hope of reducing r(x) is if x2 + 1 is a factor of x4 + 1. Performing long division gives us x4 + 1 x2 + 1 = x2 − 1 + 2 x2 + 1 The remainder is not zero so r(x) is already reduced. 3. To ﬁnd the x-intercept, we’d set r(x) = 0. Since there are no real solutions to x4+1 x2+1 = 0, we have no x-intercepts. Since r(0) = 1, we get (0, 1) as the y-intercept. 4. This step doesn’t apply to r, since its domain is all real numbers. 5. For end behavior, we note that since the degree of the numerator is exactly two more than the degree of the denominator, neither Theorems 4.2 nor 4.3 apply.15 We know from our attempt to reduce r(x) that we can rewrite r(x) = x2 − 1 + 2 x2+1, so we focus our attention x2+1 It should be clear that as x → ±∞, on the term corresponding to the remainder, x2+1 ≈ very small (+), which means r(x) ≈ x2 − 1 + very small (+). So the graph y = r(x) is a little bit above the graph of the parabola y = x2 − 1 as x → ±∞. Graphically. There isn’t much work to do for a sign diagram for r(x), since its domain is all real numbers and it has no zeros. Our sole test interval is (−∞, ∞), and since we know r(0) = 1, we conclude r(x) is (+) for all real numbers. At
this point, we don’t have much to go on for 15This won’t stop us from giving it the old community college try, however! 332 Rational Functions a graph.16 Below is a comparison of what we have determined analytically versus what the calculator shows us. We have no way to detect the relative extrema analytically17 apart from brute force plotting of points, which is done more eﬃciently by the calculator. y 6 5 4 3 2 1 −3 −1 1 2 x 3 As usual, the authors oﬀer no apologies for what may be construed as ‘pedantry’ in this section. We feel that the detail presented in this section is necessary to obtain a ﬁrm grasp of the concepts presented here and it also serves as an introduction to the methods employed in Calculus. As we have said many times in the past, your instructor will decide how much, if any, of the kinds of details presented here are ‘mission critical’ to your understanding of Precalculus. Without further delay, we present you with this section’s Exercises. 16So even Jeﬀ at this point may check for symmetry! We leave it to the reader to show r(−x) = r(x) so r is even, and, hence, its graph is symmetric about the y-axis. 17Without appealing to Calculus, of course. 4.2 Graphs of Rational Functions 333 4.2.1 Exercises In Exercises 1 - 16, use the six-step procedure to graph the rational function. Be sure to draw any asymptotes as dashed lines. 1. f (x) = 3. f (x) = 4 x + 2 1 x2 5. f (x) = 2x − 1 −2x2 − 5x + 3 7. f (x) = 4x x2 + 4 9. f (x) = 11. f (x) = 13. f (x) = 15. f (x) = x2 − x − 12 x2 + x − 6 x2 − x − 6 x + 1 x3 + 2x2 + x x2 − x − 2 x3 − 2x2 + 3x 2x2 + 2 2. f (x) = 5x 6 − 2x 4. f (x) = 6. f (x) = 1 x
2 + x − 12 x x2 + x − 12 8. f (x) = 4x x2 − 4 10. f (x) = 3x2 − 5x − 2 x2 − 9 12. f (x) = 14. f (x) = x2 − x 3 − x −x3 + 4x x2 − 9 16.18 f (x) = x2 − 2x + 1 x3 + x2 − 2x In Exercises 17 - 20, graph the rational function by applying transformations to the graph of y = 1 x. 17. f (x) = 19. h(x) = 1 x − 2 −2x + 1 x 18. g(x) = 1 − 3 x (Hint: Divide) 20. j(x) = 3x − 7 x − 2 (Hint: Divide) 21. Discuss with your classmates how you would graph f (x) = ax + b cx + d. What restrictions must be placed on a, b, c and d so that the graph is indeed a transformation of y = 1 x? 22. In Example 3.1.1 in Section 3.1 we showed that p(x) = 4x+x3 is not a polynomial even though its formula reduced to 4 + x2 for x = 0. However, it is a rational function similar to those studied in the section. With the help of your classmates, graph p(x). x 18Once you’ve done the six-step procedure, use your calculator to graph this function on the viewing window [0, 12] × [0, 0.25]. What do you see? 334 23. Let g(x) = x4 − 8x3 + 24x2 − 72x + 135 x3 − 9x2 + 15x − 7 Rational Functions. With the help of your classmates, ﬁnd the x- and y- intercepts of the graph of g. Find the intervals on which the function is increasing, the intervals on which it is decreasing and the local extrema. Find all of the asymptotes of the graph of g and any holes in the graph, if they exist. Be sure to show all of your work including any polynomial or synthetic division. Sketch the graph of g, using more than one picture if necessary to show all of the important features of the graph. Example 4.2.4 showed us that
the six-step procedure cannot tell us everything of importance about the graph of a rational function. Without Calculus, we need to use our graphing calculators to reveal the hidden mysteries of rational function behavior. Working with your classmates, use a graphing calculator to examine the graphs of the rational functions given in Exercises 24 - 27. Compare and contrast their features. Which features can the six-step process reveal and which features cannot be detected by it? 24. f (x) = 1 x2 + 1 25. f (x) = x x2 + 1 26. f (x) = x2 x2 + 1 27. f (x) = x3 x2 + 1 4.2 Graphs of Rational Functions 335 4.2.2 Answers 1. f (x) = 4 x + 2 Domain: (−∞, −2) ∪ (−2, ∞) No x-intercepts y-intercept: (0, 2) Vertical asymptote: x = −2 As x → −2−, f (x) → −∞ As x → −2+, f (x) → ∞ Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 2. f (x) = 5x 6 − 2x Domain: (−∞, 3) ∪ (3, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Vertical asymptote: x = 3 As x → 3−, f (x) → ∞ As x → 3+, f (x) → −∞ Horizontal asymptote: y = − 5 2 As x → −∞, f (x) → − 5 2 − As x → ∞, f (x) → − 5 2 + 3. f (x) = 1 x2 Domain: (−∞, 0) ∪ (0, ∞) No x-intercepts No y-intercepts Vertical asymptote: x = 0 As x → 0−, f (x) → ∞ As x → 0+, f (x) → ∞ Horizontal asymptote: y = 0 As x → −∞, f (x) → 0+ As x → ∞, f (x) →
0+ y 5 4 3 2 1 −7−6−5−4−3−2−1 −1 1 2 3 4 5 x −2 −3 −4 −5 y 3 2 1 −3−2−1 −2 −3 −4 −5 −6 −7 y 5 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x 336 Rational Functions 4. f (x) = = 1 x2 + x − 12 1 (x − 3)(x + 4) Domain: (−∞, −4) ∪ (−4, 3) ∪ (3, ∞) No x-intercepts y-intercept: (0, − 1 12 ) Vertical asymptotes: x = −4 and x = 3 As x → −4−, f (x) → ∞ As x → −4+, f (x) → −∞ As x → 3−, f (x) → −∞ As x → 3+, f (x) → ∞ Horizontal asymptote: y = 0 As x → −∞, f (x) → 0+ As x → ∞, f (x) → 0+ 2x − 1 (2x − 1)(x + 3) 2 ) ∪ ( 1 2, ∞) 5. f (x) = = − 2x − 1 −2x2 − 5x + 3 Domain: (−∞, −3) ∪ (−3, 1 No x-intercepts y-intercept: (0, − 1 3 ) −1, x = 1 f (x) = 2 x + 3 Hole in the graph at ( 1 2, − 2 7 ) Vertical asymptote: x = −3 As x → −3−, f (x) → ∞ As x → −3+, f (x) → −∞ Horizontal asymptote: y = 0 As x → −∞, f (x) → 0+ As x → ∞, f (x) → 0− 6. f (x) = = x x2 + x − 12 x (x − 3)(x + 4) Domain: (−∞, −4) ∪ (−4, 3) ∪ (3, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Vertical asymptotes
: x = −4 and x = 3 As x → −4−, f (x) → −∞ As x → −4+, f (x) → ∞ As x → 3−, f (x) → −∞ As x → 3+, f (x) → ∞ Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ y 1 −6 −5 −4 −3 −2 −1 1 2 3 4 x −1 y 1 −7 −6 −5 −4 −3 −2 −1 1 2 x −1 y 1 −6 −5 −4 −3 −2 −1 1 2 3 4 5 x −1 4.2 Graphs of Rational Functions 337 7. f (x) = 4x x2 + 4 Domain: (−∞, ∞) x-intercept: (0, 0) y-intercept: (0, 0) No vertical asymptotes No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 8. f (x) = 4x x2 − 4 = 4x (x + 2)(x − 2) Domain: (−∞, −2) ∪ (−2, 2) ∪ (2, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Vertical asymptotes: x = −2, x = 2 As x → −2−, f (x) → −∞ As x → −2+, f (x) → ∞ As x → 2−, f (x) → −∞ As x → 2+, f (x) → ∞ No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 9. f (x) = x2 − x − 12 x2 + 3 Domain: (−∞, −3) ∪ (−3, 2) ∪ (2, ∞) x-intercept: (4, 0) y-intercept: (0, 2) Vertical asymptote: x =

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