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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. ... |
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 ... |
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, c... |
− 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.... |
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... |
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 i... |
.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 obt... |
’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... |
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 ... |
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... |
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 ... |
±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 deg... |
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 = ... |
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... |
(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 + 1... |
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... |
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... |
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 numerato... |
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... |
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 ... |
’ 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 vertic... |
. 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... |
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) ap... |
= 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. Veri... |
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... |
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 thousa... |
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 c... |
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, the... |
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 specificall... |
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. 11Tho... |
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 + ... |
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 exactl... |
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... |
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 ... |
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 ... |
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 formu... |
(−∞, 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)... |
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 bel... |
+, 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+, ... |
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−, ... |
(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 everyw... |
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 ... |
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 whe... |
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 Fun... |
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 wel... |
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... |
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, Theore... |
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 lo... |
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 ... |
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 bi... |
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 ... |
= 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’... |
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 formul... |
)) ≈ 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 (−)) = ... |
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) ... |
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 fi... |
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 presente... |
.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 + ... |
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 asymp... |
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 (... |
−∞, 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 asympto... |
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 = ... |
, 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... |
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 ... |
−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-... |
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 doma... |
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... |
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 ri... |
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 r... |
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... |
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 behavi... |
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, w... |
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.... |
→ 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 e... |
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 e... |
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 exp... |
‘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 arith... |
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 t... |
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. Wha... |
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 d... |
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] ∪ [... |
� 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 ... |
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 Sectio... |
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 ... |
◦ 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)(... |
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 −... |
�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... |
�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 ... |
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 ... |
◦ 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) ... |
‘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 orde... |
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 ... |
. 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 solu... |
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 i... |
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 1... |
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 +... |
)(−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 ... |
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 ◦... |
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... |
√ 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) = 3... |
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