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− 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 coeﬃcients. 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 polyn
omial 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 = an
zn + 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 + 2i 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 = i46 · 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 x
2 + 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 x2 − 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 polynomial 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 polynomial 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. Applying 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+3
t = 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 ﬁnd 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, k
Ω), 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 asym
ptotes 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 asympt
ote: 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 x
2 + 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 x2 + 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 = 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− y 1 −7−
6−5−4−3−2−1 y 5 4 3 2 1 −5 −4 −3 −2 −1 1 2 3 4 5 x −1 −2 −3 −4 −5 5 4 3 2 1 y −5 −4 −3 −2 −1 1 2 3 4 5 x −1 −2 −3 −4 −5 338 Rational Functions 10. f (x) = = 3x2 − 5x − 2 x2 − 9 (3x + 1)(x − 2) (x + 3)(x − 3) Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) x-intercepts: − 1 3, 0, (2, 0) y-intercept: 0, 2 9 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) = x2 − x − 6 x + 1 = (x − 3)(x + 2) x + 1 Domain: (−∞, −1) ∪ (−1, ∞) x-intercepts: (−2, 0), (3, 0) y-intercept: (0, −6) Vertical asymptote: x = −1 As x → −1−, f (x) → ∞ As x → −1+, f (x) → −∞ Slant asymptote: y = x − 2 As x → −∞, the graph is above y = x − 2 As x → ∞, the graph is below y = x − 2 12. f (x) = x2 − x 3 − x = x(x − 1) 3 − x Domain: (−∞, 3) ∪ (3, ∞) x-intercepts: (0, 0), (1, 0) y-intercept: (0, 0) Vertical asymptote: x = 3 As x → 3−
, f (x) → ∞ As x → 3+, f (x) → −∞ Slant asymptote: y = −x − 2 As x → −∞, the graph is above y = −x − 2 As x → ∞, the graph is below y = −9−8−7−7−6−5−4−3−2−1 −1 −3 −4 −5 −6 −7 −8 −4−3−2−1 1 2 3 4 x −2 −4 −6 y 8 6 4 2 −8−7−6−5−4−3−2−1 − 10 x −4 −6 −8 −10 −12 −14 −16 −18 4.2 Graphs of Rational Functions 339 13. f (x) = x3 + 2x2 + x x2 − x − 2 = x(x + 1) x − 2 x = −1 Domain: (−∞, −1) ∪ (−1, 2) ∪ (2, ∞) x-intercept: (0, 0) y-intercept: (0, 0) 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 14. f (x) = −x3 + 4x x2 − 9 Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) x-intercepts: (−2, 0), (0, 0), (2, 0) y-intercept: (0, 0) 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) → −∞ Slant asymptote: y = −x As x → −∞, the graph is above y = −x As x → ∞, the graph is below y = −x 15.
f (x) = x3 − 2x2 + 3x 2x2 + 2 Domain: (−∞, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Slant asymptote: y = 1 As x → −∞, the graph is below y = 1 As x → ∞, the graph is above 18 16 14 12 10 8 6 4 2 −9−8−7−6−5−4−3−2−1 −4 −6 −8 −10 y 7 6 5 4 3 2 1 −6−5−4−3−2−1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6 −7 y 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 340 16. f (x) = x2 − 2x + 1 x3 + x2 − 2x Domain: (−∞, −2) ∪ (−2, 0) ∪ (0, 1) ∪ (1, ∞) f (x(x + 2) No x-intercepts No y-intercepts Vertical asymptotes: x = −2 and x = 0 As x → −2−, f (x) → −∞ As x → −2+, f (x) → ∞ As x → 0−, f (x) → ∞ As x → 0+, f (x) → −∞ Hole in the graph at (1, 0) Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 17. f (x) = 1 x − 2 Shift the graph of y = to the right 2 units. 1 x 18. g(x) = 1 − 3 x Rational Functions y 5 4 3 2 1 −5 −4 −3 −2 −1 1 2 3 4 5 x −1 −2 −3 −4 −5 y 3 2 1 −1 1 2 3 4 5 x −1 −2 − Vertically stretch the graph of y = by a factor of 3. 3 x Reﬂect the graph of y = about the x-axis. Shift the graph of y = − up 1 unit. 3 x −6−5−4−3−2−1
−1 1 2 3 4 5 6 x −2 −3 −4 −5 4.2 Graphs of Rational Functions 341 19. h(x) = −2x + 1 x Shift the graph of y = down 2 units. = −2 + 1 x 1 x y 1 −3 −2 −1 1 2 3 x −1 −2 −3 −4 −3 −2 −1 1 2 3 4 5 x −1 20. j(x) = = 3 − 3x − 7 x − 2 Shift the graph of y = to the right 2 units. 1 x − 2 1 x Reﬂect the graph of y = about the x-axis. Shift the graph of y = − up 3 units. 1 x − 2 1 x − 2 342 Rational Functions 4.3 Rational Inequalities and Applications In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our ﬁrst example showcases the critical diﬀerence in procedure between solving a rational equation and a rational inequality. Example 4.3.1. 1. Solve x3 − 2x +. 2. Solve x3 − 2x +. 3. Use your calculator to graphically check your answers to 1 and 2. Solution. 1. To solve the equation, we clear denominators x3 − 2x + 1 x − 1 = x3 − 2x + 1 x − 1 · 2(x − 1(x − 1) 2x3 − 4x + 2 = x2 − 3x + 2 2x3 − x2 − x = 0 x(2x + 1)(x − 1, 1 expand factor Since we cleared denominators, we need to check for extraneous solutions. Sure enough, we see that x = 1 does not satisfy the original equation and must be discarded. Our solutions are x = − 1 2 and x = 0. 2. To solve the inequality, it may be tempting to begin as we did with the equation − namely by multiplying both sides by the quantity (x − 1). The problem is that, depending on x, (x − 1) may be positive (which doesn’t aﬀect the inequality) or (x − 1) could be negative (which would reverse the inequality). Instead of working by cases, we collect all of the terms on one side of the inequality with 0 on the other and make a sign diagram using the technique given on page 321
in Section 4.2. x3 − 2x + 1 x − 1 1 2 − x3 − 2x + x3 − 2x + 1 − x(x − 1) + 1(2(x − 1)) 2(x − 1) ≥ 0 ≥ 0 get a common denominator expand 2x3 − x2 − x 2x − 2 4.3 Rational Inequalities and Applications 343 Viewing the left hand side as a rational function r(x) we make a sign diagram. The only value excluded from the domain of r is x = 1 which is the solution to 2x − 2 = 0. The zeros of r are the solutions to 2x3 − x2 − x = 0, which we have already found to be x = 0, x = − 1 2 and x = 1, the latter was discounted as a zero because it is not in the domain. Choosing test values in each test interval, we construct the sign diagram below. (+) 0 (−) 0 (+) (+) − 1 2 0 1 We are interested in where r(x) ≥ 0. We ﬁnd r(x) > 0, or (+), on the intervals −∞, − 1 2 (0, 1) and (1, ∞). We add to these intervals the zeros of r, − 1 −∞, − 1 2, 2 and 0, to get our ﬁnal solution: ∪ [0, 1) ∪ (1, ∞). 3. Geometrically, if we set f (x) = x3−2x+1 and g(x) = 1 2 x − 1, the solutions to f (x) = g(x) are the x-coordinates of the points where the graphs of y = f (x) and y = g(x) intersect. The solution to f (x) ≥ g(x) represents not only where the graphs meet, but the intervals over which the graph of y = f (x) is above (>) the graph of g(x). We obtain the graphs below. x−1 The ‘Intersect’ command conﬁrms that the graphs cross when x = − 1 2 and x = 0. It is clear from the calculator that the graph of y = f (x) is above the graph of y = g(x) on −∞, − 1 2 as well
as on (0, ∞). According to the calculator, our solution is then −∞, − 1 ∪ [0, ∞) 2 which almost matches the answer we found analytically. We have to remember that f is not deﬁned at x = 1, and, even though it isn’t shown on the calculator, there is a hole1 in the graph of y = f (x) when x = 1 which is why x = 1 is not part of our ﬁnal answer. Next, we explore how rational equations can be used to solve some classic problems involving rates. Example 4.3.2. Carl decides to explore the Meander River, the location of several recent Sasquatch sightings. From camp, he canoes downstream ﬁve miles to check out a purported Sasquatch nest. Finding nothing, he immediately turns around, retraces his route (this time traveling upstream), 1There is no asymptote at x = 1 since the graph is well behaved near x = 1. According to Theorem 4.1, there must be a hole there. 344 Rational Functions and returns to camp 3 hours after he left. If Carl canoes at a rate of 6 miles per hour in still water, how fast was the Meander River ﬂowing on that day? Solution. We are given information about distances, rates (speeds) and times. The basic principle relating these quantities is: distance = rate · time The ﬁrst observation to make, however, is that the distance, rate and time given to us aren’t ‘compatible’: the distance given is the distance for only part of the trip, the rate given is the speed Carl can canoe in still water, not in a ﬂowing river, and the time given is the duration of the entire trip. Ultimately, we are after the speed of the river, so let’s call that R measured in miles per hour to be consistent with the other rate given to us. To get started, let’s divide the trip into its two parts: the initial trip downstream and the return trip upstream. For the downstream trip, all we know is that the distance traveled is 5 miles. distance downstream = rate traveling downstream · time traveling downstream 5 miles = rate traveling downstream · time traveling downstream Since the return trip upstream followed the same route as the trip downstream, we know that the distance traveled upstream is also
5 miles. distance upstream = rate traveling upstream · time traveling upstream 5 miles = rate traveling upstream · time traveling upstream We are told Carl can canoe at a rate of 6 miles per hour in still water. How does this ﬁgure into the rates traveling upstream and downstream? The speed the canoe travels in the river is a combination of the speed at which Carl can propel the canoe in still water, 6 miles per hour, and the speed of the river, which we’re calling R. When traveling downstream, the river is helping Carl along, so we add these two speeds: rate traveling downstream = rate Carl propels the canoe + speed of the river hour + R miles = 6 miles hour So our downstream speed is (6 + R) miles for the downstream part of the trip, we get: hour. Substituting this into our ‘distance-rate-time’ equation 5 miles = rate traveling downstream · time traveling downstream 5 miles = (6 + R) miles hour · time traveling downstream When traveling upstream, Carl works against the current. Since the canoe manages to travel upstream, the speed Carl can canoe in still water is greater than the river’s speed, so we subtract the river’s speed from Carl’s canoing speed to get: rate traveling upstream = rate Carl propels the canoe − river speed = 6 miles hour − R miles hour Proceeding as before, we get 4.3 Rational Inequalities and Applications 345 5 miles = rate traveling upstream · time traveling upstream 5 miles = (6 − R) miles hour · time traveling upstream The last piece of information given to us is that the total trip lasted 3 hours. If we let tdown denote the time of the downstream trip and tup the time of the upstream trip, we have: tdown+tup = 3 hours. Substituting tdown and tup into the ‘distance-rate-time’ equations, we get (suppressing the units) three equations in three unknowns:2    E1 (6 + R) tdown = 5 (6 − R) tup = 5 E2 tdown + tup = 3 E3 Since we are ultimately after R, we need to use these three equations to get at least one equation involving only R. To that end, we solve E1 for tdown by dividing both sides3 by the quantity (6+R) to get tdown = 5 6−R.
Substituting these into E3, we get:4 6+R. Similarly, we solve E2 for tup and get tup =. Clearing denominators, we get 5(6 − R) + 5(6 + R) = 3(6 + R)(6 − R) which reduces to R2 = 16. We ﬁnd R = ±4, and since R represents the speed of the river, we choose R = 4. On the day in question, the Meander River is ﬂowing at a rate of 4 miles per hour. One of the important lessons to learn from Example 4.3.2 is that speeds, and more generally, rates, are additive. As we see in our next example, the concept of rate and its associated principles can be applied to a wide variety of problems - not just ‘distance-rate-time’ scenarios. Example 4.3.3. Working alone, Taylor can weed the garden in 4 hours. If Carl helps, they can weed the garden in 3 hours. How long would it take for Carl to weed the garden on his own? Solution. The key relationship between work and time which we use in this problem is: amount of work done = rate of work · time spent working We are told that, working alone, Taylor can weed the garden in 4 hours. In Taylor’s case then: amount of work Taylor does = rate of Taylor working · time Taylor spent working 1 garden = (rate of Taylor working) · (4 hours) So we have that the rate Taylor works is 1 garden together, Taylor and Carl can weed the garden in just 3 hours. We have: 4 hours = 1 garden hour. We are also told that when working 4 2This is called a system of equations. No doubt, you’ve had experience with these things before, and we will study systems in greater detail in Chapter 8. 3While we usually discourage dividing both sides of an equation by a variable expression, we know (6 + R) = 0 since otherwise we couldn’t possibly multiply it by tdown and get 5. 4The reader is encouraged to verify that the units in this equation are the same on both sides. To get you started, the units on the ‘3’ is ‘hours.’ 346 Rational Functions amount of work done together = rate of working together · time spent working together 1 garden = (rate of working together) · (3 hours) garden From
this, we ﬁnd that the rate of Taylor and Carl working together is 1 garden hour. We are asked to ﬁnd out how long it would take for Carl to weed the garden on his own. Let us call this unknown t, measured in hours to be consistent with the other times given to us in the problem. Then: 3 hours = 1 3 amount of work Carl does = rate of Carl working · time Carl spent working 1 garden = (rate of Carl working) · (t hours) In order to ﬁnd t, we need to ﬁnd the rate of Carl working, so let’s call this quantity R, with units garden hour. Using the fact that rates are additive, we have: rate working together = rate of Taylor working + rate of Carl working hour + R garden hour = 1 garden garden hour 1 3 4 so that R = 1 12 garden hour. Substituting this into our ‘work-rate-time’ equation for Carl, we get: 1 garden = (rate of Carl working) · (t hours) 1 12 garden hour · (t hours) 1 garden = Solving 1 = 1 12 t, we get t = 12, so it takes Carl 12 hours to weed the garden on his own.5 As is common with ‘word problems’ like Examples 4.3.2 and 4.3.3, there is no short-cut to the answer. We encourage the reader to carefully think through and apply the basic principles of rate to each (potentially diﬀerent!) situation. It is time well spent. We also encourage the tracking of units, especially in the early stages of the problem. Not only does this promote uniformity in the units, it also serves as a quick means to check if an equation makes sense.6 Our next example deals with the average cost function, ﬁrst introduced on page 82, as applied to PortaBoy Game systems from Example 2.1.5 in Section 2.1. Example 4.3.4. Given a cost function C(x), which returns the total cost of producing x items, recall that the average cost function, C(x) = C(x) computes the cost per item when x items are x produced. Suppose the cost C, in dollars, to produce x PortaBoy game systems for a local retailer is C(x) = 80x + 150, x ≥ 0. 1. Find an
expression for the average cost function C(x). 2. Solve C(x) < 100 and interpret. 5Carl would much rather spend his time writing open-source Mathematics texts than gardening anyway. 6In other words, make sure you don’t try to add apples to oranges! 4.3 Rational Inequalities and Applications 347 3. Determine the behavior of C(x) as x → ∞ and interpret. Solution. 1. From C(x) = C(x) x, we obtain C(x) = 80x+150 x. The domain of C is x ≥ 0, but since x = 0 causes problems for C(x), we get our domain to be x > 0, or (0, ∞). 2. Solving C(x) < 100 means we solve 80x+150 x < 100. We proceed as in the previous example. 80x + 150 x < 100 80x + 150 x − 100 < 0 80x + 150 − 100x x 150 − 20x x < 0 common denominator < 0 If we take the left hand side to be a rational function r(x), we need to keep in mind that the applied domain of the problem is x > 0. This means we consider only the positive half of the number line for our sign diagram. On (0, ∞), r is deﬁned everywhere so we need only look for zeros of r. Setting r(x) = 0 gives 150 − 20x = 0, so that x = 15 2 = 7.5. The test intervals on our domain are (0, 7.5) and (7.5, ∞). We ﬁnd r(x) < 0 on (7.5, ∞). 0 (+) 0 (−) 7.5 In the context of the problem, x represents the number of PortaBoy games systems produced and C(x) is the average cost to produce each system. Solving C(x) < 100 means we are trying to ﬁnd how many systems we need to produce so that the average cost is less than $100 per system. Our solution, (7.5, ∞) tells us that we need to produce more than 7.5 systems to achieve this. Since it doesn’t make sense to produce half a system, our ﬁnal answer is [8, ∞). 3. When we apply Theorem 4.2
to C(x) we ﬁnd that y = 80 is a horizontal asymptote to the graph of y = C(x). To more precisely determine the behavior of C(x) as x → ∞, we ﬁrst use long division7 and rewrite C(x) = 80 + 150 x → 0+, which means C(x) ≈ 80 + very small (+). Thus the average cost per system is getting closer to $80 per system. x = 0, which is impossible, so we conclude that C(x) > 80 for all x > 0. This means that the average cost per system is always greater than $80 per system, but the average cost is approaching this amount as more and more systems are produced. Looking back at Example 2.1.5, we realize $80 is the variable cost per system − If we set C(x) = 80, we get 150 x. As x → ∞, 150 7In this case, long division amounts to term-by-term division. 348 Rational Functions the cost per system above and beyond the ﬁxed initial cost of $150. Another way to interpret our answer is that ‘inﬁnitely’ many systems would need to be produced to eﬀectively ‘zero out’ the ﬁxed cost. Our next example is another classic ‘box with no top’ problem. Example 4.3.5. A box with a square base and no top is to be constructed so that it has a volume of 1000 cubic centimeters. Let x denote the width of the box, in centimeters as seen below. height depth width, x 1. Express the height h in centimeters as a function of the width x and state the applied domain. 2. Solve h(x) ≥ x and interpret. 3. Find and interpret the behavior of h(x) as x → 0+ and as x → ∞. 4. Express the surface area S of the box as a function of x and state the applied domain. 5. Use a calculator to approximate (to two decimal places) the dimensions of the box which minimize the surface area. Solution. 1. We are told that the volume of the box is 1000 cubic centimeters and that x represents the width, in centimeters. From geometry, we know Volume = width × height × depth. Since the base of the box is a square, the width and
the depth are both x centimeters. Using h for the height, we have 1000 = x2h, so that h = 1000 x2 As for the applied domain, in order for there to be a box at all, x > 0, and since every such choice of x will return a positive number for the height h we have no other restrictions and conclude our domain is (0, ∞). x2. Using function notation,8 h(x) = 1000 2. To solve h(x) ≥ x, we proceed as before and collect all nonzero terms on one side of the inequality in order to use a sign diagram. 8That is, h(x) means ‘h of x’, not ‘h times x’ here. 4.3 Rational Inequalities and Applications 349 h(x) ≥ x 1000 x2 ≥ x 1000 x2 − x ≥ 0 1000 − x3 x2 ≥ 0 common denominator We consider the left hand side of the inequality as our rational function r(x). We see r is undeﬁned at x = 0, but, as in the previous example, the applied domain of the problem is x > 0, so we are considering only the behavior of r on (0, ∞). The sole zero of r comes when 1000 − x3 = 0, which is x = 10. Choosing test values in the intervals (0, 10) and (10, ∞) gives the following diagram. (+) 0 (−) 0 10 We see r(x) > 0 on (0, 10), and since r(x) = 0 at x = 10, our solution is (0, 10]. In the context of the problem, h represents the height of the box while x represents the width (and depth) of the box. Solving h(x) ≥ x is tantamount to ﬁnding the values of x which result in a box where the height is at least as big as the width (and, in this case, depth.) Our answer tells us the width of the box can be at most 10 centimeters for this to happen. 3. As x → 0+, h(x) = 1000 x2 → ∞. This means that the smaller the width x (and, in this case, depth), the larger the height h has to be in order to maintain a volume of 1000 cubic centimeters. As x → ∞, we ﬁnd h(x)
→ 0+, which means that in order to maintain a volume of 1000 cubic centimeters, the width and depth must get bigger as the height becomes smaller. 4. Since the box has no top, the surface area can be found by adding the area of each of the sides to the area of the base. The base is a square of dimensions x by x, and each side has dimensions x by h. We get the surface area, S = x2 + 4xh. To get S as a function of x, we. Hence, as a function of x, S(x) = x2 + 4000 substitute h = 1000 x. x2 The domain of S is the same as h, namely (0, ∞), for the same reasons as above. to obtain S = x2 + 4x 1000 x2 5. A ﬁrst attempt at the graph of y = S(x) on the calculator may lead to frustration. Chances are good that the ﬁrst window chosen to view the graph will suggest y = S(x) has the x-axis as a horizontal asymptote. From the formula S(x) = x2 + 4000 x, however, we get S(x) ≈ x2 as x → ∞, so S(x) → ∞. Readjusting the window, we ﬁnd S does possess a relative minimum at x ≈ 12.60. As far as we can tell,9 this is the only relative extremum, so it is the absolute minimum as well. This means that the width and depth of the box should each measure 9without Calculus, that is... 350 Rational Functions approximately 12.60 centimeters. To determine the height, we ﬁnd h(12.60) ≈ 6.30, so the height of the box should be approximately 6.30 centimeters. 4.3.1 Variation In many instances in the sciences, rational functions are encountered as a result of fundamental natural laws which are typically a result of assuming certain basic relationships between variables. These basic relationships are summarized in the deﬁnition below. Deﬁnition 4.5. Suppose x, y and z are variable quantities. We say y varies directly with (or is directly proportional to) x if there is a constant k such that y = kx. y varies inversely with (or is inversely proportional to) x if there is a constant k such
that y = k x. z varies jointly with (or is jointly proportional to) x and y if there is a constant k such that z = kxy. The constant k in the above deﬁnitions is called the constant of proportionality. Example 4.3.6. Translate the following into mathematical equations using Deﬁnition 4.5. 1. Hooke’s Law: The force F exerted on a spring is directly proportional the extension x of the spring. 2. Boyle’s Law: At a constant temperature, the pressure P of an ideal gas is inversely propor- tional to its volume V. 3. The volume V of a right circular cone varies jointly with the height h of the cone and the square of the radius r of the base. 4. Ohm’s Law: The current I through a conductor between two points is directly proportional to the voltage V between the two points and inversely proportional to the resistance R between the two points. 4.3 Rational Inequalities and Applications 351 5. Newton’s Law of Universal Gravitation: Suppose two objects, one of mass m and one of mass M, are positioned so that the distance between their centers of mass is r. The gravitational force F exerted on the two objects varies directly with the product of the two masses and inversely with the square of the distance between their centers of mass. Solution. 1. Applying the deﬁnition of direct variation, we get F = kx for some constant k. 2. Since P and V are inversely proportional, we write P = k V. 3. There is a bit of ambiguity here. It’s clear that the volume and the height of the cone are represented by the quantities V and h, respectively, but does r represent the radius of the base or the square of the radius of the base? It is the former. Usually, if an algebraic operation is speciﬁed (like squaring), it is meant to be expressed in the formula. We apply Deﬁnition 4.5 to get V = khr2. 4. Even though the problem doesn’t use the phrase ‘varies jointly’, it is implied by the fact that the current I is related to two diﬀerent quantities. Since I varies directly with V but inversely with R, we write I = kV R. 5. We write
the product of the masses mM and the square of the distance as r2. We have that F varies directly with mM and inversely with r2, so F = kmM r2. In many of the formulas in the previous example, more than two varying quantities are related. In practice, however, usually all but two quantities are held constant in an experiment and the data collected is used to relate just two of the variables. Comparing just two varying quantities allows us to view the relationship between them as functional, as the next example illustrates. Example 4.3.7. According to this website the actual data relating the volume V of a gas and its pressure P used by Boyle and his assistant in 1662 to verify the gas law that bears his name is given below. V P V P 48 46 44 42 40 38 36 34 32 30 28 26 24 29.13 30.56 31.94 33.5 35.31 37 39.31 41.63 44.19 47.06 50.31 54.31 58.81 23 61.31 22 21 20 19 18 17 16 15 14 13 12 64.06 67.06 70.69 74.13 77.88 82.75 87.88 93.06 100.44 107.81 117.56 1. Use your calculator to generate a scatter diagram for these data using V as the independent variable and P as the dependent variable. Does it appear from the graph that P is inversely proportional to V? Explain. 2. Assuming that P and V do vary inversely, use the data to approximate the constant of proportionality. 352 Rational Functions 3. Use your calculator to determine a ‘Power Regression’ for this data10 and use it verify your results in 1 and 2. Solution. 1. If P really does vary inversely with V, then P = k V for some constant k. From the data plot, the points do seem to lie along a curve like y = k x. 2. To determine the constant of proportionality, we note that from P = k V, we get k = P V. Multiplying each of the volume numbers times each of the pressure numbers,11 we produce a number which is always approximately 1400. We suspect that P = 1400 V. Graphing y = 1400 along with the data gives us good reason to believe our hypotheses that P and V are, in fact, inversely related. x The graph of the data The data with y = 1400 x 3. After performing a
‘Power Regression’, the calculator ﬁts the data to the curve y = axb where a ≈ 1400 and b ≈ −1 with a correlation coeﬃcient which is darned near perfect.12 In other words, y = 1400x−1 or y = 1400 x, as we guessed. 10We will talk more about this in the coming chapters. 11You can use tell the calculator to do this arithmetic on the lists and save yourself some time. 12We will revisit this example once we have developed logarithms in Chapter 6 to see how we can actually ‘linearize’ this data and do a linear regression to obtain the same result. 4.3 Rational Inequalities and Applications 353 4.3.2 Exercises In Exercises 1 - 6, solve the rational equation. Be sure to check for extraneous solutions. 1. 3. 5. x 5x + 4 = 3 = x2 − 3 x2 − x2 − 2x + 1 x3 + x2 − 2x = 1 2. 4. 6. 3x − 1 x2 + 1 = 1 2x + 17 x + 1 = x + 5 −x3 + 4x x2 − 9 = 4x In Exercises 7 - 20, solve the rational inequality. Express your answer using interval notation. 7. 1 x + 2 ≥ 0 4x x2 + 4 ≥ 0 ≥ 0 x3 + 2x2 + x x2 − x − 2 2x + 17 x + 1 > x + 5 10. 13. 16. 19. 8. 11. 14. 17. ≤ 0 x − 3 x + 2 x2 − x − 12 x2 + x − 6 x2 + 5x + 6 x2 − 1 −x3 + 4x x2 − 9 > 0 > 0 ≥ 4x 9. x x2 − 1 > 0 12. 15. 18. 3x2 − 5x − 2 x2 − 9 < 0 3x − 1 x2 + 1 1 x2 + 1 ≤ 1 < 0 x4 − 4x3 + x2 − 2x − 15 x3 − 4x2 ≥ x 20. 5x3 − 12x2 + 9x + 10 x2 − 1 ≥ 3x − 1 21. Carl and Mike start a 3 mile race at the same time. If Mike ran the race at 6 miles per hour and ﬁn
ishes the race 10 minutes before Carl, how fast does Carl run? 22. One day, Donnie observes that the wind is blowing at 6 miles per hour. A unladen swallow nesting near Donnie’s house ﬂies three quarters of a mile down the road (in the direction of the wind), turns around, and returns exactly 4 minutes later. What is the airspeed of the unladen swallow? (Here, ‘airspeed’ is the speed that the swallow can ﬂy in still air.) 23. In order to remove water from a ﬂooded basement, two pumps, each rated at 40 gallons per minute, are used. After half an hour, the one pump burns out, and the second pump ﬁnishes removing the water half an hour later. How many gallons of water were removed from the basement? 24. A faucet can ﬁll a sink in 5 minutes while a drain will empty the same sink in 8 minutes. If the faucet is turned on and the drain is left open, how long will it take to ﬁll the sink? 25. Working together, Daniel and Donnie can clean the llama pen in 45 minutes. On his own, Daniel can clean the pen in an hour. How long does it take Donnie to clean the llama pen on his own? 354 Rational Functions 26. In Exercise 32, the function C(x) =.03x3 − 4.5x2 + 225x + 250, for x ≥ 0 was used to model the cost (in dollars) to produce x PortaBoy game systems. Using this cost function, ﬁnd the number of PortaBoys which should be produced to minimize the average cost C. Round your answer to the nearest number of systems. 27. Suppose we are in the same situation as Example 4.3.5. If the volume of the box is to be 500 cubic centimeters, use your calculator to ﬁnd the dimensions of the box which minimize the surface area. What is the minimum surface area? Round your answers to two decimal places. 28. The box for the new Sasquatch-themed cereal, ‘Crypt-Os’, is to have a volume of 140 cubic inches. For aesthetic reasons, the height of the box needs to be 1.62 times the width of the base of the box.13 Find the dimensions of the box which will minimize the surface area
of the box. What is the minimum surface area? Round your answers to two decimal places. 29. Sally is Skippy’s neighbor from Exercise 19 in Section 2.3. Sally also wants to plant a vegetable garden along the side of her home. She doesn’t have any fencing, but wants to keep the size of the garden to 100 square feet. What are the dimensions of the garden which will minimize the amount of fencing she needs to buy? What is the minimum amount of fencing she needs to buy? Round your answers to the nearest foot. (Note: Since one side of the garden will border the house, Sally doesn’t need fencing along that side.) 30. Another Classic Problem: A can is made in the shape of a right circular cylinder and is to hold one pint. (For dry goods, one pint is equal to 33.6 cubic inches.)14 (a) Find an expression for the volume V of the can in terms of the height h and the base radius r. (b) Find an expression for the surface area S of the can in terms of the height h and the base radius r. (Hint: The top and bottom of the can are circles of radius r and the side of the can is really just a rectangle that has been bent into a cylinder.) (c) Using the fact that V = 33.6, write S as a function of r and state its applied domain. (d) Use your graphing calculator to ﬁnd the dimensions of the can which has minimal surface area. 31. A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius of the base and height) of the drum which would minimize the surface area. What is the minimum surface area? Round your answers to two decimal places. 32. In Exercise 71 in Section 1.4, the population of Sasquatch in Portage County was modeled t+15, where t = 0 represents the year 1803. When were there fewer by the function P (t) = 150t than 100 Sasquatch in Portage County? 131.62 is a crude approximation of the so-called ‘Golden Ratio’ φ = 1+ 2 14According to www.dictionary.com, there are diﬀerent values given for this conversion. We will stick with 33.6in3. 5 √ for this problem. 4.3 Rational Inequalities and Applications 355 In Exerc
ises 33 - 38, translate the following into mathematical equations. 33. At a constant pressure, the temperature T of an ideal gas is directly proportional to its volume V. (This is Charles’s Law) 34. The frequency of a wave f is inversely proportional to the wavelength of the wave λ. 35. The density d of a material is directly proportional to the mass of the object m and inversely proportional to its volume V. 36. The square of the orbital period of a planet P is directly proportional to the cube of the semi-major axis of its orbit a. (This is Kepler’s Third Law of Planetary Motion ) 37. The drag of an object traveling through a ﬂuid D varies jointly with the density of the ﬂuid ρ and the square of the velocity of the object ν. 38. Suppose two electric point charges, one with charge q and one with charge Q, are positioned r units apart. The electrostatic force F exerted on the charges varies directly with the product of the two charges and inversely with the square of the distance between the charges. (This is Coulomb’s Law) 39. According to this webpage, the frequency f of a vibrating string is given by f = T µ where T is the tension, µ is the linear mass15 of the string and L is the length of the vibrating part of the string. Express this relationship using the language of variation. 1 2L 40. According to the Centers for Disease Control and Prevention www.cdc.gov, a person’s Body Mass Index B is directly proportional to his weight W in pounds and inversely proportional to the square of his height h in inches. (a) Express this relationship as a mathematical equation. (b) If a person who was 5 feet, 10 inches tall weighed 235 pounds had a Body Mass Index of 33.7, what is the value of the constant of proportionality? (c) Rewrite the mathematical equation found in part 40a to include the value of the constant found in part 40b and then ﬁnd your Body Mass Index. 41. We know that the circumference of a circle varies directly with its radius with 2π as the constant of proportionality. (That is, we know C = 2πr.) With the help of your classmates, compile a list of other basic geometric relationships which can be seen as variations. 15Also known as the linear density. It is simply a measure of mass per
unit length. 356 4.3.3 Answers 1. x = − 6 7 2. x = 1, x = 2 4. x = −6, x = 2 5. No solution Rational Functions 3. x = −1 6. x = 0, x = ±2 √ 2 7. (−2, ∞) 9. (−1, 0) ∪ (1, ∞) 11. (−∞, −3) ∪ (−3, 2) ∪ (4, ∞) 8. (−2, 3] 10. [0, ∞) 12. −3, − 1 3 ∪ (2, 3) 13. (−1, 0] ∪ (2, ∞) 14. (−∞, −3) ∪ (−2, −1) ∪ (1, ∞) 15. (−∞, 1] ∪ [2, ∞) 17. (−∞, −3) ∪ −2 √ 2, 0 ∪ 2 19. [−3, 0) ∪ (0, 4) ∪ [5, ∞) √ 2, 3 16. (−∞, −6) ∪ (−1, 2) 18. No solution 20. −1, − 1 2 ∪ (1, ∞) 21. 4.5 miles per hour 22. 24 miles per hour 23. 3600 gallons 24. 40 3 ≈ 13.33 minutes 25. 3 hours 26. The absolute minimum of y = C(x) occurs at ≈ (75.73, 59.57). Since x represents the number of game systems, we check C(75) ≈ 59.58 and C(76) ≈ 59.57. Hence, to minimize the average cost, 76 systems should be produced at an average cost of $59.57 per system. 27. The width (and depth) should be 10.00 centimeters, the height should be 5.00 centimeters. The minimum surface area is 300.00 square centimeters. 28. The width of the base of the box should be ≈ 4.12 inches, the height of the box should be ≈ 6.67 inches, and the depth of the base of the box should be ≈ 5.09 inches; minimum surface area ≈ 164.91 square inches. 29. The dimensions are ≈ 7 feet by ≈ 14 feet; minimum amount of fencing required �
� 28 feet. 30. (a) V = πr2h (b) S = 2πr2 + 2πrh (c) S(r) = 2πr2 + 67.2 r, Domain r > 0 (d) r ≈ 1.749 in. and h ≈ 3.498 in. 31. The radius of the drum should be ≈ 1.05 feet and the height of the drum should be ≈ 2.12 feet. The minimum surface area of the drum is ≈ 20.93 cubic feet. 32. P (t) < 100 on (−15, 30), and the portion of this which lies in the applied domain is [0, 30). Since t = 0 corresponds to the year 1803, from 1803 through the end of 1832, there were fewer than 100 Sasquatch in Portage County. 4.3 Rational Inequalities and Applications 357 33. T = kV 36. P 2 = ka3 34. 16 f = k λ 37. 17 D = kρν2 39. Rewriting f = 1 2L T µ as f = √ 1 2 L T √ µ 35. d = km V 38. 18 F = kqQ r2 we see that the frequency f varies directly with the square root of the tension and varies inversely with the length and the square root of the linear mass. 40. (a) B = kW h2 (b) 19 k = 702.68 (c) B = 702.68W h2 16The character λ is the lower case Greek letter ‘lambda.’ 17The characters ρ and ν are the lower case Greek letters ‘rho’ and ‘nu,’ respectively. 18Note the similarity to this formula and Newton’s Law of Universal Gravitation as discussed in Example 5. 19The CDC uses 703. 358 Rational Functions Chapter 5 Further Topics in Functions 5.1 Function Composition Before we embark upon any further adventures with functions, we need to take some time to gather our thoughts and gain some perspective. Chapter 1 ﬁrst introduced us to functions in Section 1.3. At that time, functions were speciﬁc kinds of relations - sets of points in the plane which passed the Vertical Line Test, Theorem 1.1. In Section 1.4, we developed the idea that functions are processes - rules which match inputs
to outputs - and this gave rise to the concepts of domain and range. We spoke about how functions could be combined in Section 1.5 using the four basic arithmetic operations, took a more detailed look at their graphs in Section 1.6 and studied how their graphs behaved under certain classes of transformations in Section 1.7. In Chapter 2, we took a closer look at three families of functions: linear functions (Section 2.1), absolute value functions1 (Section 2.2), and quadratic functions (Section 2.3). Linear and quadratic functions were special cases of polynomial functions, which we studied in generality in Chapter 3. Chapter 3 culminated with the Real Factorization Theorem, Theorem 3.16, which says that all polynomial functions with real coeﬃcients can be thought of as products of linear and quadratic functions. Our next step was to enlarge our ﬁeld2 of study to rational functions in Chapter 4. Being quotients of polynomials, we can ultimately view this family of functions as being built up of linear and quadratic functions as well. So in some sense, Chapters 2, 3, and 4 can be thought of as an exhaustive study of linear and quadratic3 functions and their arithmetic combinations as described in Section 1.5. We now wish x and g(x) = x2/3, and the purpose of the to study other algebraic functions, such as f (x) = ﬁrst two sections of this chapter is to see how these kinds of functions arise from polynomial and rational functions. To that end, we ﬁrst study a new way to combine functions as deﬁned below. √ 1These were introduced, as you may recall, as piecewise-deﬁned linear functions. 2This is a really bad math pun. 3If we broaden our concept of functions to allow for complex valued coeﬃcients, the Complex Factorization Theorem, Theorem 3.14, tells us every function we have studied thus far is a combination of linear functions. 360 Further Topics in Functions Deﬁnition 5.1. Suppose f and g are two functions. The composite of g with f, denoted g ◦ f, is deﬁned by the formula (g ◦ f )(x) = g(f (x)), provided x is an element of the
domain of f and f (x) is an element of the domain of g. The quantity g ◦ f is also read ‘g composed with f ’ or, more simply ‘g of f.’ At its most basic level, Deﬁnition 5.1 tells us to obtain the formula for (g ◦ f ) (x), we replace every occurrence of x in the formula for g(x) with the formula we have for f (x). If we take a step back and look at this from a procedural, ‘inputs and outputs’ perspective, Deﬁntion 5.1 tells us the output from g ◦ f is found by taking the output from f, f (x), and then making that the input to g. The result, g(f (x)), is the output from g ◦ f. From this perspective, we see g ◦ f as a two step process taking an input x and ﬁrst applying the procedure f then applying the procedure g. Abstractly, we have f g x f (x) g ◦ f g(f (x)) In the expression g(f (x)), the function f is often called the ‘inside’ function while g is often called the ‘outside’ function. There are two ways to go about evaluating composite functions - ‘inside out’ and ‘outside in’ - depending on which function we replace with its formula ﬁrst. Both ways are demonstrated in the following example. Example 5.1.1. Let f (x) = x2 − 4x, g(x) = 2 − In numbers 1 - 3, ﬁnd the indicated function value. √ x + 3, and h(x) = 2x x + 1. 1. (g ◦ f )(1) 2. (f ◦ g)(1) 3. (g ◦ g)(6) In numbers 4 - 10, ﬁnd and simplify the indicated composite functions. State the domain of each. 4. (g ◦ f )(x) 5. (f ◦ g)(x) 6. (g ◦ h)(x) 7. (h ◦ g)(x) 8. (h ◦ h)(x) 9. (h ◦ (g ◦ f ))(x) 10. ((h ◦ g)
◦ f )(x) Solution. 1. Using Deﬁnition 5.1, (g ◦ f )(1) = g(f (1)). We ﬁnd f (1) = −3, so (g ◦ f )(1) = g(f (1)) = g(−3) = 2 5.1 Function Composition 361 2. As before, we use Deﬁnition 5.1 to write (f ◦ g)(1) = f (g(1)). We ﬁnd g(1) = 0, so (f ◦ g)(1) = f (g(1)) = f (0) = 0 3. Once more, Deﬁnition 5.1 tells us (g ◦ g)(6) = g(g(6)). That is, we evaluate g at 6, then plug that result back into g. Since g(6) = −1, (g ◦ g)(6) = g(g(6)) = g(−1) = 2 − √ 2 4. By deﬁnition, (g ◦ f )(x) = g(f (x)). We now illustrate two ways to approach this problem. inside out: We insert the expression f (x) into g ﬁrst to get (g ◦ f )(x) = g(f (x)) = g x2 − 4x = 2 − √ Hence, (g ◦ f )(x) = 2 − x2 − 4x + 3. (x2 − 4x) + 3 = 2 − x2 − 4x + 3 outside in: We use the formula for g ﬁrst to get (g ◦ f )(x) = g(f (x)) = 2 − f (x) + 3 = 2 − (x2 − 4x) + 3 = 2 − x2 − 4x + 3 We get the same answer as before, (g ◦ f )(x) = 2 − √ x2 − 4x + 3. To ﬁnd the domain of g ◦ f, we need to ﬁnd the elements in the domain of f whose outputs f (x) are in the domain of g. We accomplish this by following the rule set forth in Section 1.4, that is,
we ﬁnd the domain before we simplify. To that end, we examine (g ◦ f )(x) = 2 − (x2 − 4x) + 3. To keep the square root happy, we solve the inequality x2 − 4x + 3 ≥ 0 by creating a sign diagram. If we let r(x) = x2 − 4x + 3, we ﬁnd the zeros of r to be x = 1 and x = 3. We obtain (+) 0 (−) 0 (+) 1 3 Our solution to x2 − 4x + 3 ≥ 0, and hence the domain of g ◦ f, is (−∞, 1] ∪ [3, ∞). 5. To ﬁnd (f ◦ g)(x), we ﬁnd f (g(x)). inside out: We insert the expression g(x) into f ﬁrst to get √ (f ◦ g)(x) = f (g(x)) = + 32 − + 32 − 8 + 4 √ x + 3 362 Further Topics in Functions outside in: We use the formula for f (x) ﬁrst to get (f ◦ g)(x) = f (g(x)) = (g(x))2 − 4 (g(x)) √ √ x + 32 − same algebra as before Thus we get (f ◦ g)(x) = x − 1. To ﬁnd the domain of (f ◦ g), we look to the step before we did any simpliﬁcation and ﬁnd (f ◦ g)(x) = 2 − x + 3. To keep the square root happy, we set x + 3 ≥ 0 and ﬁnd our domain to be [−3, ∞). x + 32 − 4 2 − √ √ 6. To ﬁnd (g ◦ h)(x), we compute g(h(x)). inside out: We insert the expression h(x) into g ﬁrst to get (g ◦ h)(x) = g(h(x)) = g 2x − 2x x + 1 + 3 3(x + 1) x + 1 get common denominators + 2x x + 1 5x + 3 x + 1 outside in: We use the formula for g(x) �
�rst to get (g ◦ h)(x) = g(h(x)) = 2 − h(x) + 3 = 2 − = 2 − 2x x + 1 5x + 3 x + 1 + 3 get common denominators as before To ﬁnd the domain of (g ◦ h), we look to the step before we began to simplify: (g ◦ h)(x) = 2 − 2x x + 1 + 3 To avoid division by zero, we need x = −1. To keep the radical happy, we need to solve 2x x + 1 + 3 = 5x + 3 x + 1 ≥ 0 Deﬁning r(x) = 5x+3 x+1, we see r is undeﬁned at x = −1 and r(x) = 0 at x = − 3 5. We get 5.1 Function Composition 363 (+) (−) 0 (+) −1 − 3 5 Our domain is (−∞, −1) ∪ − 3 5, ∞. 7. We ﬁnd (h ◦ g)(x) by ﬁnding h(g(x)). inside out: We insert the expression g(x) into h ﬁrst to get x + 3 √ √ (h ◦ g)(x) = h(g(x)) = − √ √ = = outside in: We use the formula for h(x) ﬁrst to get (h ◦ g)(x) = h(g(x)) = √ 2 (g(x)) (g(x)) + − √ √ = = To ﬁnd the domain of h ◦ g, we look to the step before any simpliﬁcation: (h ◦ g)(x √ To keep the square root happy, we require x+3 ≥ 0 or x ≥ −3. Setting the denominator equal to zero gives 2 − x + 3 + 1 = 0 or x + 3 = 3. Squaring both sides gives us x + 3 = 9, or x = 6. Since x = 6 checks in the original equation, we know x = 6 is the only zero of the denominator. Hence, the domain of h ◦ g is [−3, 6) ∪ (6, ∞). √ √ 8. To �
�nd (h ◦ h)(x), we substitute the function h into itself, h(h(x)). inside out: We insert the expression h(x) into h to get (h ◦ h)(x) = h(h(x)) = h 2x x + 1 364 Further Topics in Functions 2x x + 1 + 1 2 2x x + 1 4x x + 1 2x x + 1 2x x + 1 4x 1 (x + 1) 2x 1 (x + 1) 4x 3x + 1 · (x + 1) (x + 1) + 1 4x x+1 · (x + 1) · (x + 1) + 1 · (x + 1) · (x + 1) · (x + 1) + x + 1 = = = = = outside in: This approach yields (h ◦ h)(x) = h(h(x)) = 2(h(x)) h(x) + 1 2x x + 1 2 2x x + 1 4x 3x + 1 = = + 1 same algebra as before To ﬁnd the domain of h ◦ h, we analyze (h ◦ h)(x) = 2x x + 1 2 2x x + 1 + 1 To keep the denominator x + 1 happy, we need x = −1. Setting the denominator 2x x + 1 + 1 = 0 gives x = − 1 3. Our domain is (−∞, −1) ∪ −1, − 1 3 ∪ − 1 3, ∞. 5.1 Function Composition 365 9. The expression (h ◦ (g ◦ f ))(x) indicates that we ﬁrst ﬁnd the composite, g ◦ f and compose x2 − 4x + 3. the function h with the result. We know from number 1 that (g ◦ f )(x) = 2 − We now proceed as usual. √ inside out: We insert the expression (g ◦ f )(x) into h ﬁrst to get (h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = h √ x2 − 4x + 3 2 − + 1 x2 − 4x + − √ √ x2 − 4x + 3 x2 − 4x + 3 x
2 − 4x + 3 = = outside in: We use the formula for h(x) ﬁrst to get (h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = 2 ((g ◦ f )(x)) ((g ◦ f )(x)) + 1 √ x2 − 4x + − √ √ x2 − 4x + 3 x2 − 4x + 3 x2 − 4x + 3 + 1 = = To ﬁnd the domain of (h ◦ (g ◦ f )), we look at the step before we began to simplify, (h ◦ (g ◦ f ))(x) = √ x2 − 4x + 3 2 2 − √ 2 − x2 − 4x + 3 + 1 √ For the square root, we need x2 − 4x + 3 ≥ 0, which we determined in number 1 to be (−∞, 1]∪[3, ∞). Next, we set the denominator to zero and solve: +1 = 0. x2 − 4x + 3 = 3, and, after squaring both sides, we have x2 − 4x + 3 = 9. To solve We get x2 − 4x − 6 = 0, we use the quadratic formula and get x = 2 ± 10. The reader is encouraged x2 − 4x + 3 +1 = 0. to check that both of these numbers satisfy the original equation, Hence we must exclude these numbers from the domain of h ◦ (g ◦ f ). Our ﬁnal domain for h ◦ (f ◦ g) is (−∞, 2 − 10, 1] ∪ 3, 2 + 10 ∪ 2 + x2 − 4x + 3 10, ∞. 10) ∪ (2 − 2 − 2 − √ √ √ √ √ √ √ 10. The expression ((h◦g)◦f )(x) indicates that we ﬁrst ﬁnd the composite h◦g and then compose that with f. From number 4, we have (h ◦ g)(x 366 Further Topics in Functions We now proceed as before. inside out: We insert the expression f (x) into h ◦ g ﬁrst to get ((h
◦ g) ◦ f )(x) = (h ◦ g)(f (x)) = (h ◦ g) x2 − 4x x2 − 4x) + 3 (x2 − 4x) + 3 x2 − 4x + 3 x2 − 4x + 3 = = outside in: We use the formula for (h ◦ g)(x) ﬁrst to get ((h ◦ g) ◦ f )(x) = (h ◦ g)(f (x)) = 4 − 2 3 − (f (x)) + 3 f (x)) + x2 − 4x) + 3 (x2 − 4x) + 3 x2 − 4x + 3 x2 − 4x + 3 We note that the formula for ((h ◦ g) ◦ f )(x) before simpliﬁcation is identical to that of (h ◦ (g ◦ f ))(x) before we simpliﬁed it. Hence, the two functions have the same domain, h ◦ (f ◦ g) is (−∞, 2 − 10, 1] ∪ 3, 2 + 10 ∪ 2 + 10, ∞. 10) ∪ (2 − √ √ √ √ It should be clear from Example 5.1.1 that, in general, when you compose two functions, such as f and g above, the order matters.4 We found that the functions f ◦ g and g ◦ f were diﬀerent as were g ◦ h and h ◦ g. Thinking of functions as processes, this isn’t all that surprising. If we think of one process as putting on our socks, and the other as putting on our shoes, the order in which we do these two tasks does matter.5 Also note the importance of ﬁnding the domain of the composite function before simplifying. For instance, the domain of f ◦ g is much diﬀerent than its simpliﬁed formula would indicate. Composing a function with itself, as in the case of ﬁnding (g ◦ g)(6) and (h ◦ h)(x), may seem odd. Looking at this from a procedural perspective, however, this merely indicates performing a task h and then doing it again - like setting the washing machine to do a
‘double rinse’. Composing a function with itself is called ‘iterating’ the function, and we could easily spend an entire course on just that. The last two problems in Example 5.1.1 serve to demonstrate the associative property of functions. That is, when composing three (or more) functions, as long as we keep the order the same, it doesn’t matter which two functions we compose ﬁrst. This property as well as another important property are listed in the theorem below. 4This shows us function composition isn’t commutative. An example of an operation we perform on two functions which is commutative is function addition, which we deﬁned in Section 1.5. In other words, the functions f + g and g + f are always equal. Which of the remaining operations on functions we have discussed are commutative? 5A more mathematical example in which the order of two processes matters can be found in Section 1.7. In fact, all of the transformations in that section can be viewed in terms of composing functions with linear functions. 5.1 Function Composition 367 Theorem 5.1. Properties of Function Composition: Suppose f, g, and h are functions. h ◦ (g ◦ f ) = (h ◦ g) ◦ f, provided the composite functions are deﬁned. If I is deﬁned as I(x) = x for all real numbers x, then I ◦ f = f ◦ I = f. By repeated applications of Deﬁnition 5.1, we ﬁnd (h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = h(g(f (x))). Similarly, ((h ◦ g) ◦ f )(x) = (h ◦ g)(f (x)) = h(g(f (x))). This establishes that the formulas for the two functions are the same. We leave it to the reader to think about why the domains of these two functions are identical, too. These two facts establish the equality h ◦ (g ◦ f ) = (h ◦ g) ◦ f. A consequence of the associativity of function composition is that there is no need for parentheses when we write h ◦ g ◦ f. The second property can also be ver
iﬁed using Deﬁnition 5.1. Recall that the function I(x) = x is called the identity function and was introduced in Exercise 73 in Section 2.1. If we compose the function I with a function f, then we have (I ◦ f )(x) = I(f (x)) = f (x), and a similar computation shows (f ◦ I)(x) = f (x). This establishes that we have an identity for function composition much in the same way the real number 1 is an identity for real number multiplication. That is, just as for any real number x, we have for any function. We shall see the concept of an identity take on great signiﬁcance in the next section. Out in the wild, function composition is often used to relate two quantities which may not be directly related, but have a variable in common, as illustrated in our next example. Example 5.1.2. The surface area S of a sphere is a function of its radius r and is given by the formula S(r) = 4πr2. Suppose the sphere is being inﬂated so that the radius of the sphere is increasing according to the formula r(t) = 3t2, where t is measured in seconds, t ≥ 0, and r is measured in inches. Find and interpret (S ◦ r)(t). Solution. If we look at the functions S(r) and r(t) individually, we see the former gives the surface area of a sphere of a given radius while the latter gives the radius at a given time. So, given a speciﬁc time, t, we could ﬁnd the radius at that time, r(t) and feed that into S(r) to ﬁnd the surface area at that time. From this we see that the surface area S is ultimately a function of time t and we ﬁnd (S ◦ r)(t) = S(r(t)) = 4π(r(t))2 = 4π 3t22 = 36πt4. This formula allows us to compute the surface area directly given the time without going through the ‘middle man’ r. A useful skill in Calculus is to be able to take a complicated function and break it down into a composition of easier functions which our last example illustrates. Example 5.1.3
. Write each of the following functions as a composition of two or more (nonidentity) functions. Check your answer by performing the function composition. 1. F (x) = \|3x − 1\| 2. G(x) = 2 x2 + 1 3. H(x Solution. There are many approaches to this kind of problem, and we showcase a diﬀerent methodology in each of the solutions below. 368 Further Topics in Functions 1. Our goal is to express the function F as F = g ◦ f for functions g and f. From Deﬁnition 5.1, we know F (x) = g(f (x)), and we can think of f (x) as being the ‘inside’ function and g as being the ‘outside’ function. Looking at F (x) = \|3x − 1\| from an ‘inside versus outside’ perspective, we can think of 3x − 1 being inside the absolute value symbols. Taking this cue, we deﬁne f (x) = 3x − 1. At this point, we have F (x) = \|f (x)\|. What is the outside function? The function which takes the absolute value of its input, g(x) = \|x\|. Sure enough, (g ◦ f )(x) = g(f (x)) = \|f (x)\| = \|3x − 1\| = F (x), so we are done. 2. We attack deconstructing G from an operational approach. Given an input x, the ﬁrst step is to square x, then add 1, then divide the result into 2. We will assign each of these steps a function so as to write G as a composite of three functions: f, g and h. Our ﬁrst function, f, is the function that squares its input, f (x) = x2. The next function is the function that adds 1 to its input, g(x) = x + 1. Our last function takes its input and divides it into 2, h(x) = 2 x. The claim is that G = h ◦ g ◦ f. We ﬁnd (h ◦ g ◦ f )(x) = h(g(f (x))) = h(g x2) = h x2 + 1 = 2 x2 + 1 =
G(x), so we are done. 3. If we look H(x) = √ x+1√ x−1 with an eye towards building a complicated function from simpler If we deﬁne x is a simple piece of the larger function. f (x)−1. If we want to decompose H = g ◦ f, then we can glean functions, we see the expression f (x) = the formula for g(x) by looking at what is being done to f (x). We take g(x) = x+1 x, we have H(x) = f (x)+1 √ √ x−1, so (g ◦ f )(x) = g(f (x)) = f (x) + 1 f (x(x), as required. 5.1 Function Composition 369 5.1.1 Exercises In Exercises 1 - 12, use the given pair of functions to ﬁnd the following values if they exist. (g ◦ f )(0) (g ◦ f )(−3) (f ◦ g)(−1) (f ◦ g) 1 2 (f ◦ f )(2) (f ◦ f )(−2) 1. f (x) = x2, g(x) = 2x + 1 2. f (x) = 4 − x, g(x) = 1 − x2 3. f (x) = 4 − 3x, g(x) = \|x\| 5. f (x) = 4x + 5, g(x) = √ x 4. f (x) = \|x − 1\|, g(x) = x2 − 5 √ 6. f (x) = 3 − x, g(x) = x2 + 1 7. f (x) = 6 − x − x2, g(x) = x √ x + 10 √ 8. f (x) = 3 x + 1, g(x) = 4x2 − x 9. f (x) = 3 1 − x, g(x) = 4x x2 + 1 11. f (x) = 2x 5 − x2, g(x) = √ 4x + 1 10. f (x) = x x + 5, g(x) = 2 7 −
x2 12. f (x) = √ 2x + 5, g(x) = 10x x2 + 1 In Exercises 13 - 24, use the given pair of functions to ﬁnd and simplify expressions for the following functions and state the domain of each using interval notation. (g ◦ f )(x) (f ◦ g)(x) (f ◦ f )(x) 13. f (x) = 2x + 3, g(x) = x2 − 9 14. f (x) = x2 − x + 1, g(x) = 3x − 5 15. f (x) = x2 − 4, g(x) = \|x\| 16. f (x) = 3x − 5, g(x) = 17. f (x) = \|x + 1\|, g(x) = √ x 19. f (x) = \|x\|, g(x) = √ 4 − x 21. f (x) = 3x − 1, g(x) = 1 x + 3 23. f (x) = x 2x + 1, g(x) = 2x + 1 x √ x √ x + 1 18. f (x) = 3 − x2, g(x) = 20. f (x) = x2 − x − 1, g(x) = √ x − 5 22. f (x) = 3x x − 1, g(x) = x x − 3 24. f (x) = 2x x2 − 4, g(x) = √ 1 − x 370 Further Topics in Functions In Exercises 25 - 30, use f (x) = −2x, g(x) = for the following functions and state the domain of each using interval notation. x and h(x) = \|x\| to ﬁnd and simplify expressions √ 25. (h ◦ g ◦ f )(x) 26. (h ◦ f ◦ g)(x) 27. (g ◦ f ◦ h)(x) 28. (g ◦ h ◦ f )(x) 29. (f ◦ h ◦ g)(x) 30. (f ◦ g ◦ h)(x) In Exercises 31 - 40, write the given function as a composition of two or
more non-identity functions. (There are several correct answers, so check your answer using function composition.) 31. p(x) = (2x + 3)3 33. h(x) = √ 2x − 1 35. r(x) = 37. q(x) = 39. v(x) = 2 5x + 1 \|x\| + 1 \|x\| − 1 2x + 1 3 − 4x 32. P (x) = x2 − x + 15 34. H(x) = \|7 − 3x\| 36. R(x) = 38. Q(x) = 40. w(x) = 7 x2 − 1 2x3 + 1 x3 − 1 x2 x4 + 1 41. Write the function F (x) = x3 + 6 x3 − 9 as a composition of three or more non-identity functions. 42. Let g(x) = −x, h(x) = x + 2, j(x) = 3x and k(x) = x − 4. In what order must these functions √ √ be composed with f (x) = x to create F (x) = 3 −x + 2 − 4? 43. What linear functions could be used to transform f (x) = x3 into F (x) = − 1 2 (2x − 7)3 + 1? What is the proper order of composition? In Exercises 44 - 55, let f be the function deﬁned by f = {(−3, 4), (−2, 2), (−1, 0), (0, 1), (1, 3), (2, 4), (3, −1)} and let g be the function deﬁned g = {(−3, −2), (−2, 0), (−1, −4), (0, 0), (1, −3), (2, 1), (3, 2)}. Find the value if it exists. 44. (f ◦ g)(3) 47. (f ◦ g)(−3) 45. f (g(−1)) 48. (g ◦ f )(3) 46. (f ◦ f )(0) 49. g(f (−3)) 5.1 Function Composition 371 50. (g ◦ g)(−2) 51. (g ◦ f
)(−2) 52. g(f (g(0))) 53. f (f (f (−1))) 54. f (f (f (f (f (1))))) 55. (g ◦ g ◦ · · · ◦ g) n times (0) In Exercises 56 - 61, use the graphs of y = f (x) and y = g(x) below to ﬁnd the function valuex) 56. (g ◦ f )(1) 59. (f ◦ g)(0) 57. (f ◦ g)(3) 60. (f ◦ f )(1) y = g(x) 58. (g ◦ f )(2) 61. (g ◦ g)(1) 62. The volume V of a cube is a function of its side length x. Let’s assume that x = t + 1 is also a function of time t, where x is measured in inches and t is measured in minutes. Find a formula for V as a function of t. 63. Suppose a local vendor charges $2 per hot dog and that the number of hot dogs sold per hour x is given by x(t) = −4t2 + 20t + 92, where t is the number of hours since 10 AM, 0 ≤ t ≤ 4. (a) Find an expression for the revenue per hour R as a function of x. (b) Find and simplify (R ◦ x) (t). What does this represent? (c) What is the revenue per hour at noon? 64. Discuss with your classmates how ‘real-world’ processes such as ﬁlling out federal income tax forms or computing your ﬁnal course grade could be viewed as a use of function composition. Find a process for which composition with itself (iteration) makes sense. 372 5.1.2 Answers 1. For f (x) = x2 and g(x) = 2x + 1, Further Topics in Functions (g ◦ f )(0) = 1 (f ◦ g)(−1) = 1 (f ◦ f )(2) = 16 (g ◦ f )(−3) = 19 (f ◦ g) 1 2 = 4 (f ◦ f )(−2) = 16 2. For f (x) = 4 − x and g(x) =
1 − x2, (g ◦ f )(0) = −15 (f ◦ g)(−1) = 4 (f ◦ f )(2) = 2 (g ◦ f )(−3) = −48 (f ◦ g) 1 2 = 13 4 (f ◦ f )(−2) = −2 3. For f (x) = 4 − 3x and g(x) = \|x\|, (g ◦ f )(0) = 4 (f ◦ g)(−1) = 1 (f ◦ f )(2) = 10 (g ◦ f )(−3) = 13 (f ◦ g) 1 2 = 5 2 (f ◦ f )(−2) = −26 4. For f (x) = \|x − 1\| and g(x) = x2 − 5, (g ◦ f )(0) = −4 (f ◦ g)(−1) = 5 (f ◦ f )(2) = 0 (g ◦ f )(−3) = 11 (f ◦ g) 1 2 = 23 4 (f ◦ f )(−2) = 2 5. For f (x) = 4x + 5 and g(x) = √ x, (g ◦ f )(0) = √ 5 (f ◦ g)(−1) is not real (f ◦ f )(2) = 57 (g ◦ f )(−3) is not real (f ◦ gf ◦ f )(−2) = −7 6. For f (x) = √ 3 − x and g(x) = x2 + 1, (g ◦ f )(0) = 4 (g ◦ f )(−3) = 7 (f ◦ g)(−1) = 1 (f ◦ g) 1 2 = √ 7 2 (f ◦ f )(2) = √ 2 (f ◦ f )(−2) = 3 − √ 5 5.1 Function Composition 373 7. For f (x) = 6 − x − x2 and g(x) = x √ x + 10, (g ◦ f )(0) = 24 (f ◦ g)(−1) = 0 (f ◦ f )(
2) = 6 (g ◦ f )(−3) = 0 (f ◦ g) 1 2 √ = 27−2 8 42 (f ◦ f )(−2) = −14 √ 8. For f (x) = 3 x + 1 and g(x) = 4x2 − x, (g ◦ f )(0) = 3 √ (g ◦ f )(−3f ◦ g)(−1) = 3 6 (f ◦ g) 1 2 = 3√ 12 2 √ (f ◦ f )(2) = 3 3 3 + 1 (f ◦ f )(−2) = 0 9. For f (x) = 3 1−x and g(x) = 4x x2+1, (g ◦ f )(0) = 6 5 (g ◦ f )(−3) = 48 25 (f ◦ g)(−1) = 1 (f ◦ g) 1 2 = −5 (f ◦ f )(2) = 3 4 (f ◦ f )(−2) is undeﬁned 10. For f (x) = x x+5 and g(x) = 2 7−x2, (g ◦ f )(0) = 2 7 (g ◦ f )(−3) = 8 19 (f ◦ g)(−1) = 1 16 = 8 143 (f ◦ g) 1 2 (f ◦ f )(2) = 2 37 (f ◦ f )(−2) = − 2 13 11. For f (x) = 2x 5−x2 and g(x) = √ 4x + 1, (g ◦ f )(0) = 1 (g ◦ f )(−3) = √ 7 (f ◦ g)(−1) is not real (f ◦ g) 1 2 = √ 3 (f ◦ f )(2) = − 8 11 (f ◦ f )(−2) = 8 11 12. For f (x) = √ 2x + 5 and g(x) = 10x x2+1, (g ◦ f )(0) = 5 √ 3 5 (f ◦ g)(−1) is not real (f ◦ f )(2) =
√ 11 (g ◦ f )(−3) is not real (f ◦ g) 1 2 = √ 13 (f ◦ f )(−2) = √ 7 13. For f (x) = 2x + 3 and g(x) = x2 − 9 (g ◦ f )(x) = 4x2 + 12x, domain: (−∞, ∞) (f ◦ g)(x) = 2x2 − 15, domain: (−∞, ∞) (f ◦ f )(x) = 4x + 9, domain: (−∞, ∞) 374 Further Topics in Functions 14. For f (x) = x2 − x + 1 and g(x) = 3x − 5 (g ◦ f )(x) = 3x2 − 3x − 2, domain: (−∞, ∞) (f ◦ g)(x) = 9x2 − 33x + 31, domain: (−∞, ∞) (f ◦ f )(x) = x4 − 2x3 + 2x2 − x + 1, domain: (−∞, ∞) 15. For f (x) = x2 − 4 and g(x) = \|x\| (g ◦ f )(x) = \|x2 − 4\|, domain: (−∞, ∞) (f ◦ g)(x) = \|x\|2 − 4 = x2 − 4, domain: (−∞, ∞) (f ◦ f )(x) = x4 − 8x2 + 12, domain: (−∞, ∞) 16. For f (x) = 3x − 5 and g(x) = √ x (g ◦ f )(x) = (f ◦ g)(x) = 3 √ 3x − 5, domain: 5 3, ∞ √ x − 5, domain: [0, ∞) (f ◦ f )(x) = 9x − 20, domain: (−∞, ∞) 17. For f (x) = \|x + 1\| and g(x) = √ x (g ◦ f )(x) = \|x + 1\|, domain: (−∞, ∞) √ √ (f ◦ g)(

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