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(x) y = P3(x) y = P4(x) 3.2 The Factor Theorem and the Remainder Theorem 257 3.2 The Factor Theorem and the Remainder Theorem Suppose we wish to find the zeros of f (x) = x3 + 4x2 − 5x − 14. Setting f (x) = 0 results in the polynomial equation x3 + 4x2 − 5x − 14 = 0. Despite all of the factoring techniques we learned1 i... |
apply the Quadratic Formula to get x = −3 ± 2. The point of this section is to generalize the technique applied here. First up is a friendly reminder of what we can expect when we divide polynomials. √ 1and probably forgot 2pun intended 258 Polynomial Functions Theorem 3.4. Polynomial Division: Suppose d(x) and p(x) a... |
is a nonzero polynomial. The real number c is a zero of p if and only if (x − c) is a factor of p(x). The proof of The Factor Theorem is a consequence of what we already know. If (x − c) is a factor of p(x), this means p(x) = (x − c) q(x) for some polynomial q. Hence, p(c) = (c − c) q(c) = 0, so c is a zero of p. Conv... |
x + 7 x−2 x3+4x2− 5x −14 2x2 6x2 12x 7x 14 0 Now, let’s move things up a bit and, for reasons which will become clear in a moment, copy the x3 into the last row. x2 + 6x + 7 x−2 x3+4x2− 5x −14 2x2 12x 14 6x2 0 7x Note that by arranging things in this manner, each term in the last row is obtained by adding the two terms... |
write 2 in the place of the divisor and the coefficients of x3 + 4x2 − 5x − 14 in for the dividend. Then ‘bring down’ the first coefficient of the dividend. 2 1 4 −5 −14 2 1 4 −5 −14 ↓ 1 Next, take the 2 from the divisor and multiply by the 1 that was ‘brought down’ to get 2. Write this underneath the 4, then add to get 6.... |
lynomials, then write the dividend, quotient and remainder in the form given in Theorem 3.4. 1. 5x3 − 2x2 + 1 ÷ (x − 3) 2. x3 + 8 ÷ (x + 2) 3. 4 − 8x − 12x2 2x − 3 Solution. 1. When setting up the synthetic division tableau, we need to enter 0 for the coefficient of x in the dividend. Doing so gives 3 5 −2 ↓ 5 0 15 39 13... |
both sides by 2 and 2 distributing gives −12x2 − 8x + 4 = (2x − 3) (−6x − 13) − 35. At this stage, we have written −12x2 − 8x + 4 in the form (2x − 3)q(x) + r(x), but how can we be sure the quotient polynomial is −6x − 13 and the remainder is −35? The answer is the word ‘unique’ in Theorem 3.4. The theorem states that... |
, 2, and −3. So q(x) = 2x2 + 2x − 3. Theorem 3.4 tells us p(x) = (x − 1) 2x2 + 2x − 3. To find the remaining real zeros of p, we need to solve 2x2 + 2x − 3 = 0 for x. Since this doesn’t factor nicely, we use the quadratic formula to find that the remaining zeros are x = −1±. 2 √ 7 3.2 The Factor Theorem and the Remainder... |
one zero at a time until we are left with a quadratic, whose roots can always be found using the Quadratic Formula. Secondly, we 3 are found x = ± both factors of p. We can certainly put the Factor Theorem to the test and continue the synthetic division tableau from above to see what happens. 3 are zeros of p. The Fac... |
a factor of p(x) The point (c, 0) is an x-intercept of the graph of y = p(x) 3.2 The Factor Theorem and the Remainder Theorem 265 3.2.1 Exercises In Exercises 1 - 6, use polynomial long division to perform the indicated division. Write the polynomial in the form p(x) = d(x)q(x) + r(x). 1. 4x2 + 3x − 1 ÷ (x − 3) 2. 2x3... |
If p(c) = 0, factor p(x) = (x − c)q(x). 21. p(x) = 2x2 − x + 1, c = 4 22. p(x) = 4x2 − 33x − 180, c = 12 23. p(x) = 2x3 − x + 6, c = −3 24. p(x) = x3 + 2x2 + 3x + 4, c = −1 25. p(x) = 3x3 − 6x2 + 4x − 8, c = 2 26. p(x) = 8x3 + 12x2 + 6x + 1, c = − 1 2 27. p(x) = x4 − 2x2 + 4, c = 3 2 29. p(x) = x4 + x3 − 6x2 − 7x − 7,... |
c = 1 − √ 3 In Exercises 41 - 45, create a polynomial p which has the desired characteristics. You may leave the polynomial in factored form. 41. 42. 43. 44. 45. The zeros of p are c = ±2 and c = ±1 The leading term of p(x) is 117x4. The zeros of p are c = 1 and c = 3 c = 3 is a zero of multiplicity 2. The leading ter... |
− 19) 7. 3x2 − 2x + 1 = (x − 1) (3x + 1) + 2 8. x2 − 5 = (x − 5) (x + 5) + 20 9. 3 − 4x − 2x2 = (x + 1) (−2x − 2) + 5 10. 4x2 − 5x + 3 = (x + 3) (4x − 17) + 54 11. x3 + 8 = (x + 2) x2 − 2x + 4 + 0 12. 4x3 + 2x − 3 = (x − 3) 4x2 + 12x + 38 + 111 13. 18x2 − 15x − 25 = x − 5 3 (4x + 2) + 0 14. 4x2 − 1 = x − 1 2 15. 2x3 +... |
√ √ 7) x3 + (1 − √ 30. p(2 − 3) = 0, p(x) = (x − (2 − 3))(x − (2 + 3)) √ 28. p − 2 = 74 27 3 √ 7)x − 7)x2 + (1 − √ Polynomial Functions √ 7 31. x3 − 6x2 + 11x − 6 = (x − 1)(x − 2)(x − 3) 32. x3 − 24x2 + 192x − 512 = (x − 8)3 33. 3x3 + 4x2 − x − 2 = 3 x − 2 3 34. 2x3 − 3x2 − 11x + 6 = 2 x − 1 2 (x + 1)2 (x + 2)(x − 3) ... |
)(x − 1)(x − 117)2 where a can be any negative real number 46. p(x) = 5x2 − 6x − 4 3.3 Real Zeros of Polynomials 269 3.3 Real Zeros of Polynomials In Section 3.2, we found that we can use synthetic division to determine if a given real number is a zero of a polynomial function. This section presents results which will ... |
it into the largest (in absolute value) of the remaining coefficients, in this case | − 6| = 6. This yields M = 3 so it is guaranteed that all of the real zeros of f lie in the interval [−4, 4]. Whereas the previous result tells us where we can find the real zeros of a polynomial, the next theorem gives us a list of poss... |
the left hand side is an integer multiple of p, and the right hand side is an integer multiple of q. (Can you see why?) This means anpn is both a multiple of p and a multiple of q. Since p and q have no common factors, an must be a multiple of q. If we rearrange the equation as anpn + an−1pn−1q +... + a1pqn−1 + a0qn =... |
3.2. 3. Use synthetic division to find the real zeros of f, and state their multiplicities. Solution. 1. In Example 3.3.1, we determined all of the real zeros of f lie in the interval [−4, 4]. We set our window accordingly and get 3.3 Real Zeros of Polynomials 271 2. In Example 3.3.2, we learned that any rational zero o... |
we were wrong to think that −1 had multiplicity 3.) ± It is interesting to note that we could greatly improve on the graph of y = f (x) in the previous example given to us by the calculator. For instance, from our determination of the zeros of f and √ 6 their multiplicities, we know the graph crosses at x = − 2 ≈ −1.2... |
real zeros, one positive, and one negative. Our only hope at this point is to try and find the zeros of f by setting f (x) = x4 + x2 − 12 = 0 and solving. If we stare at this equation long enough, we may recognize it as a ‘quadratic in disguise’ or ‘quadratic in form’. In other words, we have three terms: x4, x2 and 12... |
techniques discussed in this section can help us find them exactly. In the latter case, we are forced to approximate, which in this subsection means we use the ‘Zero’ command on the graphing calculator. 3.3.2 For Those Wishing NOT to use a Graphing Calculator Suppose we wish to find the zeros of f (x) = 2x4 + 4x3 − x2 −... |
the number of positive and negative real zeros, not the actual value of the zeros. Lastly, Descartes’ Rule of Signs counts multiplicities. This means that, for example, if one of the zeros has multiplicity 2, Descsartes’ Rule of Signs would count this as two zeros. Lastly, note that the number of positive or negative ... |
bound for the real zeros of f. That is, there are no real zeros greater than c. If c < 0 is synthetically divided into f and the numbers in the final line of the division tableau alternate signs, then c is a lower bound for the real zeros of f. That is, there are no real zeros less than c. NOTE: If the number 0 occurs ... |
6. Let f (x) = 2x4 + 4x3 − x2 − 6x − 3. 1. Find all of the real zeros of f and their multiplicities. 2. Sketch the graph of y = f (x). Solution. 2, ± 1, ± 3 1. We know from Cauchy’s Bound that all of the real zeros lie in the interval [−4, 4] and that our possible rational zeros are ± 1 2 and ± 3. Descartes’ Rule of Si... |
possible zero, − 1 2. Synthetic division shows us it is not a zero, nor is it a lower bound (since the numbers in the final line of the division tableau do not alternate), so we proceed to −1. This division shows −1 is a zero. Descartes’ Rule of Signs told us that we may have up to three negative real zeros, counting m... |
lynomial, it may well be impossible5 to find their zeros exactly. The polynomial f (x) = x5 − x − 1 is one such beast.6 According to Descartes’ Rule of Signs, f has exactly one positive real zero, and it could have two negative real zeros, or none at all. The Rational Zeros 4This is apparently a bad thing. 5We don’t use... |
ection Method as reversing the sign diagram process: instead of finding the zeros and checking the sign of f using test values, we are using test values to determine where the signs switch to find the zeros. It is a slow and tedious, yet fool-proof, method for approximating a real zero. Our next example reminds us of the... |
The solution to p(x) < 0 is −∞, − 1 2 ∪ {1}. the solution to p(x) ≤ 0 is −∞, − 1 2 2 and x = 1. Hence, 3. To interpret this solution graphically, we set f (x) = 2x5 + 6x3 + 3 and g(x) = 3x4 + 8x2. We recall that the solution to f (x) ≤ g(x) is the set of x values for which the graph of f is below the graph of g (where... |
through the usual computations, we find x = 5 is the only rational zero. Using this, we factor f (x) = x3 − 7x2 + 9x + 5 = (x − 5) x2 − 2x − 1, and we find the remaining zeros by applying the Quadratic Formula to x2 − 2x − 1 = 0. We find three real zeros, x = 1 − 2 = −0.414..., x = 1 + 2 = 2.414..., and x = 5, of which o... |
. For instance, 1E − 11 corresponds to 0.00000000001, which is pretty close in the calculator’s eyes8to 0. 8but not a Mathematician’s 280 3.3.3 Exercises Polynomial Functions In Exercises 1 - 10, for the given polynomial: Use Cauchy’s Bound to find an interval containing all of the real zeros. Use the Rational Zeros The... |
(x) = 36x4 − 12x3 − 11x2 + 2x + 1 19. f (x) = 3x3 + 3x2 − 11x − 10 20. f (x) = 2x4 + x3 − 7x2 − 3x + 3 21. f (x) = 9x3 − 5x2 − x 22. f (x) = 6x4 − 5x3 − 9x2 23. f (x) = x4 + 2x2 − 15 24. f (x) = x4 − 9x2 + 14 25. f (x) = 3x4 − 14x2 − 5 26. f (x) = 2x4 − 7x2 + 6 27. f (x) = x6 − 3x3 − 10 28. f (x) = 2x6 − 9x3 + 10 29. ... |
2 + 5x 39. x3 − 7x2 = 7 − x 41. x3 + x2 = 11x + 10 3 43. 14x2 + 5 = 3x4 36. 9x2 + 5x3 = 6x4 38. x4 + 2x3 = 12x2 + 40x + 32 40. 2x3 = 19x2 − 49x + 20 42. x4 + 2x2 = 15 44. 2x5 + 3x4 = 18x + 27 In Exercises 45 - 54, solve the polynomial inequality and state your answer using interval notation. 45. −2x3 + 19x2 − 49x + 20 ... |
3 + 2097x2 − 1971x + 567. With the help of your classmates, find the x- and y- intercepts of the graph of f. Find the intervals on which the function is increasing, the intervals on which it is decreasing and the local extrema. Sketch the graph of f, using more than one picture if necessary to show all of the important ... |
− 49x + 20 All of the real zeros lie in the interval − 51 2, ±1, ±2, ± 5 Possible rational zeros are ± 1 There are 3 or 1 positive real zeros; there are no negative real zeros 2, 51 2, ±4, ±5, ±10, ±20 2 7. For f (x) = −17x3 + 5x2 + 34x − 10 All of the real zeros lie in the interval [−3, 3] Possible rational zeros are... |
mult. 1) 14. f (x) = x3 + 4x2 − 11x + 6 x = −6 (mult. 1), x = 1 (mult. 2) 15. f (x) = x3 − 7x2 + x − 7 x = 7 (mult. 1) 16. f (x) = −2x3 + 19x2 − 49x + 20 x = 1 2, x = 4, x = 5 (each has mult. 1) 17. f (x) = −17x3 + 5x2 + 34x − 10 √ x = 5 17, x = ± 2 (each has mult. 1) 18. f (x) = 36x4 − 12x3 − 11x2 + 2x + 1 2 (mult. 2)... |
(x) = 2x6 − 9x3 + 10 √ 3√ 20 2, x = 3 29. f (x) = x5 − 2x4 − 4x + 8 x = √ 2 (each has mult. 1) x = 2, x = ± 2 (each has mult. 1) 30. f (x) = 2x5 + 3x4 − 18x − 27 √ each has mult. 1) 31. f (x) = x5 − 60x3 − 80x2 + 960x + 2304 x = −4 (mult. 3), x = 6 (mult. 2) 32. f (x) = 25x5 − 105x4 + 174x3 − 142x2 + 57x − 9 x = 3 5 (... |
(−∞, − 3) ∪ ( √ √ 3, ∞) √ √ √ 54. (−∞, − 3 √ 3 ) ∪ ( 3 √ 2, ∞) 55. V (x) ≥ 80 on [1, 5 − 5] ∪ [5 + 5, ∞). Only the portion [1, 5 − 5] lies in the applied domain, however. In the context of the problem, this says for the volume of the box to be at least 80 cubic inches, the square removed from each corner needs to have... |
fies the two following properties 1. i2 = −1 2. If c is a real number with c ≥ 0 then √ √ −c = i c Property 1 in Definition 3.4 establishes that i does act as a square root2 of −1, and property 2 establishes what we mean by the ‘principal square root’ of a negative real number. In property 2, it is important to remember ... |
your answer in the form5 a + bi. 1. (1 − 2i) − (3 + 4i) 2. (1 − 2i)(3 + 4i) 3. 1 − 2i 3 − 4i √ √ −3 −12 4. Solution. (−3)(−12) 5. 6. (x − [1 + 2i])(x − [1 − 2i]) 1. As mentioned earlier, we treat expressions involving i as we would any other radical. We combine like terms to get (1 − 2i) − (3 + 4i) = 1 − 2i − 3 − 4i =... |
+ 2i] + [1 − 2i][1 + 2i] = x2 − x + 2ix − x − 2ix + 1 − 2i + 2i − 4i2 = x2 − 2x + 5 A couple of remarks about the last example are in order. First, the conjugate of a complex number a + bi is the number a − bi. The notation commonly used for conjugation is a ‘bar’: a + bi = a − bi. For example, 3 + 2i = 3 − 2i, 3 − 2i... |
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 ... |
�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 ‘exist... |
fficients. If the degree of f is n and n ≥ 1, then f has exactly n complex zeros, counting multiplicity. If z1, z2,..., zk are the distinct zeros of f, with multiplicities m1, m2,..., mk, respectively, then f (x) = a (x − z1)m1 (x − z2)m2 · · · (x − zk)mk. Note that the value a in Theorem 3.14 is the leading coefficient of... |
3.14 (and a student’s mettle!) would be to take the factored form of f (x) in the previous example and multiply it out9 to see that it really does reduce to the original formula f (x) = 12x5 − 20x4 + 19x3 − 6x2 − 2x + 1. When factoring a polynomial using Theorem 3.14, we say that it is factored completely over the com... |
z)n−1 +... + a2 (z)2 + a1z + a0 = anzn + an−1zn−1 +... + a2z2 + a1z + a0 = anzn + an−1zn−1 +... + a2z2 + a1 z + a0 = anzn + an−1zn−1 +... + a2z2 + a1z + a0 = anzn + an−1zn−1 +... + a2z2 + a1z + a0 = f (z) since (z)n = zn since the coefficients are real since z w = zw since This shows that z is a zero of f. So, if f is a ... |
i 0 0 1 64 ↓ 2 + 2i 8i −16 + 16i −64 1 2 + 2i 8i −16 + 16i 0 0 2. Since f is a fourth degree polynomial, we need to make two successful divisions to get a quadratic quotient. Since 2 + 2i is a zero, we know from Theorem 3.15 that 2 − 2i is also a zero. We continue our synthetic division tableau. 3.4 Complex Zeros and t... |
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. (I... |
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 a... |
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 f... |
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 ... |
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 (−... |
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 ... |
46 · i2 = 1 · (−1) = −1 25. i117 = i429 · i = 1 · i = i 26. i304 = i476 = 176 = 1 27. f (x) = x2 − 4x + 13 = (x − (2 + 3i))(x − (2 − 3i)) Zeros: x = 2 ± 3i 28. f (x) = x2 − 2x + 5 = (x − (1 + 2i))(x − (1 − 2i)) Zeros: x = 1 ± 2i 29. f (x) = 3x2 + 2x + 10 = 3 x − − 1 3 + Zeros: x = − 1 3 ± √ 29 3 i √ 29 3 i x − − 1 3 − ... |
+ 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... |
− (−3 − √ 5)) √ √ 5 x + i 5 √ 3)(x + √ 3) x − 2i √ 2 x + 2i √ 2 45. f (x) = x5 − x4 + 7x3 − 7x2 + 12x − 12 = (x − 1) x2 + 3 x2 + 4 √ √ = (x − 1)(x − i 3)(x + i 3)(x − 2i)(x + 2i) √ Zeros: x = 1, ± 3i, ±2i 46. f (x) = x6 − 64 = (x − 2)(x + 2) x2 + 2x + 4 x2 − 2x + 4 = (x − 2)(x + 2) x − −1 + i 3 x − −1 − i 3 x − 1 + i ... |
+ 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... |
, −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... |
2 − 1(3x − 2) = = 2x2 − 1 x2 − 1 2x2 − 1 3x − 2 · x2 − 1 3x − 2 = 2x2 − 1 x2 − 1 (x2 − 1) (3x − 2) A few remarks about Example 4.1.1 are in order. Note that the expressions for f (x), g(x) and h(x) work out to be the same. However, only two of these functions are actually equal. Recall that functions are ultimately set... |
.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 b... |
) 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... |
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,... |
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 rea... |
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 ... |
= 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. ... |
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 So... |
omial functions with leading coefficients a and b, respectively. If the degree of p(x) is the same as the degree of q(x), then y = a b is thea horizontal asymptote of the graph of y = r(x). If the degree of p(x) is less than the degree of q(x), then y = 0 is the horizontal asymptote of the graph of y = r(x). If the degre... |
lynomial is determined by its leading term. Applying this to the numerator and denominator of f (x), we get that as x → ±∞, f (x) = 2x−1 x = 2. This last approach is useful in Calculus, and, indeed, is made rigorous there. (Keep this in mind for the remainder of this paragraph.) Applying this reasoning to the general c... |
lying Theorem 4.2, y = 0 is the horizontal asymptote. Sure enough, we see from the graph that as x → −∞, f (x) → 0− and as x → ∞, f (x) → 0+. 2. The numerator of g(x), x2 − 4, has degree 2, but the degree of the denominator, x + 1, has degree 1. By Theorem 4.2, there is no horizontal asymptote. From the graph, we see t... |
set N (t) = 300 to get 500 − 450 1+3t = 300 and solve. Isolating the fraction gives 450 Clearing denominators gives 450 = 200(1 + 3t). Finally, we get t = 5 take 5 12 months, or about 13 days, for 300 students to have had the flu. 1+3t = 200. 12. This means it will 3. To determine the behavior of N as t → ∞, we can use... |
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 func... |
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 t... |
−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... |
��nd that as x → −∞, the graph of y = h(x) approaches the asymptote from below, and as x → ∞, the graph of y = h(x) approaches the asymptote from above. The graph of y = f (x) The graph of y = g(x) The graph of y = h(x) The reader may be a bit disappointed with the authors at this point owing to the fact that in Exampl... |
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 Po... |
2.1.5 in Section 2.1) to help in your interpretation.) 22. In Exercise 35 in Section 3.1, we fit a few polynomial models to the following electric circuit data. (The circuit was built with a variable resistor. For each of the following resistance values (measured in kilo-ohms, kΩ), the corresponding power to the load (... |
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, ... |
5 2 No holes in the graph Horizontal asymptote: y = − 7 2 As x → −∞, f (x) → − 7 2 − As x → ∞, f (x) → − 7 2 + 4. f (x) = x x2 + 1 Domain: (−∞, ∞) No vertical asymptotes No holes in the graph Horizontal asymptote: y = 0 As x → −∞, f (x) → 0− As x → ∞, f (x) → 0+ 6. f (x) = x3 + 1 x2 − 1 = x2 − x + 1 x − 1 Domain: (−∞,... |
∪ (−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) ∪ ... |
= x 318 Rational Functions 13. f (x) = − 3, ∞ 2x2 + 5x − 3 3x + 2 Domain: −∞, − 2 ∪ − 2 3 Vertical asymptote: x = − 2 3 As x → − 2 3 As x → − 2 3 No holes in the graph Slant asymptote: y = 2 As x → −∞, the graph is above y = 2 As x → ∞, the graph is below y = 2, f (x) → ∞, f (x) → −∞ 3 x + 11 + 9 3 x+ 11 3 x + 11 9 9 ... |
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) Horizonta... |
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... |
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 ... |
. 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 ... |
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) =... |
−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 estima... |
, 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... |
−. 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 ... |
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 t... |
−, 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 ... |
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 ... |
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... |
(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 ... |
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(... |
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+v... |
331 We end this section with an example that shows it’s not all pathological weirdness when it comes to rational functions and technology still has a role to play in studying their graphs at this level. Example 4.2.4. Sketch the graph of r(x) = x4 + 1 x2 + 1. Solution. 1. The denominator x2 + 1 is never zero so the do... |
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... |
2 + x − 12 x x2 + x − 12 8. f (x) = 4x x2 − 4 10. f (x) = 3x2 − 5x − 2 x2 − 9 12. f (x) = 14. f (x) = x2 − x 3 − x −x3 + 4x x2 − 9 16.18 f (x) = x2 − 2x + 1 x3 + x2 − 2x In Exercises 17 - 20, graph the rational function by applying transformations to the graph of y = 1 x. 17. f (x) = 19. h(x) = 1 x − 2 −2x + 1 x 18. g(... |
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 fu... |
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−, ... |
: 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... |
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