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s such, use the decimal places to remind the reader that any result garnered from a calculator in this fashion is an approximation, and should be treated as such. 1.6 Graphs of Functions 107 1.6.2 Exercises In Exercises 1 - 12, sketch the graph of the given function. State the domain of the function, identify any inter...
mum at (4.55, −175.46) Increasing on [−2.84, 0.54], [4.55, ∞) Decreasing on (−∞, −2.84], [0.54, 4.55] 59. [−5, 5) 60. f (2) = 3 62. (−4, 0), (0, 0), (4, 0) 63. (0, 0) 65. [−4, 0] ∪ {4} 66. 3 68. [−2, 2) 69. [−4, −2], (2, 4] 71. f (−2) = −5, f (2) = 3 73. f (−2) = −5 75. No absolute maximum No absolute minimum Local max...
o graph g, we shift the graph of f down one unit by subtracting 1 from each of the y-coordinates of the points on the graph of f . Applying this to the three points we have speciﬁed on the graph, we move (0, 0) to (0, −1), (1, 1) to (1, 0), and (4, 2) to (4, 1). 124 Relations and Functions The rest of the points follow...
+ 3. We see from the graph that the range of m is (−∞, 3]. √ 6If we had done the reﬂection ﬁrst, then j1(x) = f (−x). Following this by a shift left would give us j(x) = j1(x + 3) = f (−(x + 3)) = f (−x − 3) = −x − 3 which isn’t what we want. However, if we did the reﬂection ﬁrst and followed it by a shift to the righ...
eft 3 2 units −−−−−−−−−−−−→ subtract 3 2 from each x-coordinate −2 y = m1(x Next, m2(x) = m1 graph of m1 by a factor of 2 will, according to Theorem 1.6, horizontally stretch the 3 −2 − 3 −2 y = m1(x) = x + 3 2 y 2 (−1, 1) (5, 2) −2 (−3, 0) −1 1 2 3 4 5 x −1 −2 horizontal scale by a factor of 2 −−−−−−−−−−−−−−−−−−→ mult...
y = f 7 − 2x 4 14. y = 5f (2x + 1) + 3 15. y = 2f (1 − x) − 1 17. y = f (3x) − 1 2 18. y = 4 − f (3x − 1) 7 The complete graph of y = f (x) is given below. In Exercises 19 - 27, use it and Theorem 1.7 to graph the given transformed function. y 4 3 2 1 (2, 2) (−2, 2) −4 −3 −2 −1 (0, 0) 2 3 4 x The graph for Ex. 19 - 27...
negative, we say the line is decreasing. A slope of 0 results in a horizontal line which we say is constant, and an undeﬁned slope results in a vertical line.2 Second, the larger the slope is in absolute value, the steeper the line. You may recall from Intermediate Algebra that slope can be described as the ratio ‘ ri...
o produce 0 PortaBoys. As mentioned on page 82, this is the ﬁxed, or start-up cost of this venture. 5. If we were to graph y = C(x), we would be graphing the portion of the line y = 80x + 150 for x ≥ 0. We recognize the slope, m = 80. Like any slope, we can interpret this as a rate of change. Here, C(x) is the cost in ...
. A plumber charges $50 for a service call plus $80 per hour. If she spends no longer than 8 hours a day at any one site, ﬁnd a linear function that represents her total daily charges C (in dollars) as a function of time t (in hours) spent at any one given location. 32. A salesperson is paid $200 per week plus 5% commi...
d(t) = 3t, t ≥ 0. 29. E(t) = 360t, t ≥ 0. 30. C(x) = 45x + 20, x ≥ 0. 31. C(t) = 80t + 50, 0 ≤ t ≤ 8. 32. W (x) = 200 + .05x, x ≥ 0 She must make $5500 in weekly sales. 33. C(p) = 0.035p + 1.5 The slope 0.035 means it costs 3.5¢ per page. C(0) = 1.5 means there is a ﬁxed, or start-up, cost of $1.50 to make each book. ...
2 So, for x < 2, \|x − 2\| = −x + 2 and our equation \|x − 2\| = x − 1 becomes −x + 2 = x − 1, which gives x = 3 2 . Since this solution satisﬁes x < 2, we keep it. Next, for x ≥ 2, \|x − 2\| = x − 2, so the equation \|x − 2\| = x − 1 becomes x − 2 = x − 1. Here, the equation reduces to −2 = −1, which signiﬁes we have no solut...
at (0, −1) and (0, 1) (Can you explain why?) so we get y −3 −2 −1 1 2 3 x f (x) = \|x\| x As we found earlier, the domain is (−∞, 0) ∪ (0, ∞). The range consists of just two y-values: {−1, 1}. The function f is constant on (−∞, 0) and (0, ∞). The local minimum value of f is the absolute minimum value of f , namely −1; t...
d. y 4 3 2 1 (2, 4) (1, 1) (−2, 4) (−1, 1) −2 −1 1 (0, 0) f (x) = x2 2 x Much like many of the absolute value functions in Section 2.2, knowing the graph of f (x) = x2 enables us to graph an entire family of quadratic functions using transformations. Example 2.3.1. Graph the following functions starting with the graph ...
dditional points. We see that the range of f is [−1, ∞) and we are done. 2. To get started, we rewrite g(x) = 6 − x − x2 = −x2 − x + 6 and note that the coeﬃcient of x2 is −1, not 1. This means our ﬁrst step is to factor out the (−1) from both the x2 and x terms. We then follow the completing the square recipe as above...
he units given to us in the statement of the problem are feet, we assume w and l are measured in feet. The area of the pasture, which we’ll call A, is related to w and l by the equation A = wl. Since w and l are both measured in feet, A has units of feet2, or square feet. We are given the total amount of fencing availa...
s0 where s is in meters, t is in seconds, v0 is the object’s initial velocity in meters per second and s0 is its initial position in meters. (a) What is the applied domain of this function? (b) Discuss with your classmates what each of v0 > 0, v0 = 0 and v0 < 0 would mean. (c) Come up with a scenario in which s0 < 0. (...
= g(x) is (x, g(x)). When we seek solutions to f (x) = g(x), we are looking for x values whose y values on the graphs of f and g are the same. In part 1, we found x = 3 is the solution to f (x) = g(x). Sure enough, f (3) = 5 and g(3) = 5 so that the point (3, 5) is on both graphs. In other words, the graphs of f and g ...
ties analytically using sign diagrams. Verify your answer graphically. 1. 2x2 ≤ 3 − x 3. x2 + 1 ≤ 2x Solution. 2. x2 − 2x > 1 4. 2x − x2 ≥ \|x − 1\| − 1 1. To solve 2x2 ≤ 3 − x, we ﬁrst get 0 on one side of the inequality which yields 2x2 + x − 3 ≤ 0. We ﬁnd the zeros of f (x) = 2x2 + x − 3 by solving 2x2 + x − 3 = 0 for...
2 ≤ \|x2 − 9\| < 9 29. x2 + x + 1 ≥ 0 31. x\|x + 5\| ≥ −6 12. x + 5 < \|x + 5\| 14. \|2x + 1\| ≤ 6 − x 16. \|3 − x\| ≥ x − 5 18. 16x2 + 8x + 1 > 0 20. 9x2 + 16 ≥ 24x 22. x2 + 1 < 0 24. x > x2 26. 5x + 4 ≤ 3x2 28. x2 ≤ \|4x − 3\| 30. x2 ≥ \|x\| 32. x\|x − 3\| < 2 33. The proﬁt, in dollars, made by selling x bottles of 100% All-Natural ...
1.287 indicates an increase in annual energy usage at the rate of 1.287 Quads per year. 4. To predict the energy needs in 2013, we substitute x = 2013 into the equation of the line of best ﬁt to get y = 1.287(2013) − 2473.890 ≈ 116.841. The predicted annual energy usage of the US in 2013 is approximately 116.841 Quads...
pproximately) 936 people per year. (b) According to the model, the population in 2010 will be 236,660. (c) According to the model, the population of Lake County will reach 250,000 sometime between 2024 and 2025. 2. (a) y = 796.8x − 1309762.5 with r = 0.8916 which indicates a reasonable ﬁt. The slope 796.8 indicates Lor...
, we know that Volume = width × height × depth. The key is to ﬁnd each of these quantities in terms of x. From the ﬁgure, we see that the height of the box is x itself. The cardboard piece is initially 10 inches wide. Removing squares with a side length of x inches from each corner leaves 10 − 2x inches for the width.5...
are in order. First, note that we purposefully did not label the y-axis in the sketch of the graph of y = f (x). This is because the sign diagram gives us the zeros and the relative position of the graph - it doesn’t give us any information as to how high or low the graph strays from the x-axis. Furthermore, as we hav...
and now use C(x) = .03x3 − 4.5x2 + 225x + 250, for x ≥ 0. As before, C(x) is the cost to make x PortaBoy Game Systems. Market research indicates that the demand function p(x) = −1.5x + 250 remains unchanged. Use a graphing utility to ﬁnd the production level x that maximizes the proﬁt made by producing and selling x Po...
≈ 35.255. Since x represents the number of TVs sold in hundreds, x = 3.897 corresponds to 389.7 TVs. Since we can’t sell 0.7 of a TV, we compare P (3.89) ≈ 35.254 and P (3.90) ≈ 35.255, so selling 390 TVs results in a (slightly) higher revenue. Since y represents the revenue in thousands of dollars, the maximum revenue...
erms to get 2 1 4 −5 −14 14 12 0 7 2 1 6 We have constructed a synthetic division tableau for this polynomial division problem. Let’s rework our division problem using this tableau to see how it greatly streamlines the division process. To divide x3 + 4x2 − 5x − 14 by x − 2, we write 2 in the place of the divisor and t...
the real zeros and factor the polynomial. 31. x3 − 6x2 + 11x − 6, c = 1 32. x3 − 24x2 + 192x − 512, c = 8 33. 3x3 + 4x2 − x − 2, c = 2 3 35. x3 + 2x2 − 3x − 6, c = −2 34. 2x3 − 3x2 − 11x + 6, c = 1 2 36. 2x3 − x2 − 10x + 5, c = 1 2 37. 4x4 − 28x3 + 61x2 − 42x + 9, c = 1 2 is a zero of multiplicity 2 38. x5 + 2x4 − 12x...
s, based on our division, we have that −1 has a multiplicity of at least 2. The Factor Theorem tells us our remaining zeros, √ 6 ± 2 , each have multiplicity at least 1. However, Theorem 3.7 tells us f can have at most 4 real zeros, counting multiplicity, and so we conclude that −1 is of multiplicity exactly 2 and √ 6 ...
f 1 is 2 on the graph of f . We continue to our next possible zero, 1. As before, the only information we can glean from this is that (1, −4) is on the graph of f . When we try our next possible zero, 3 2 , we get that it is not a zero, and we also see that it is an upper bound on the zeros of f , since all of the num...
+ 2x3 − 12x2 − 40x − 32 13. f (x) = x4 − 9x2 − 4x + 12 14. f (x) = x3 + 4x2 − 11x + 6 15. f (x) = x3 − 7x2 + x − 7 16. f (x) = −2x3 + 19x2 − 49x + 20 17. f (x) = −17x3 + 5x2 + 34x − 10 18. f (x) = 36x4 − 12x3 − 11x2 + 2x + 1 19. f (x) = 3x3 + 3x2 − 11x − 10 20. f (x) = 2x4 + x3 − 7x2 − 3x + 3 21. f (x) = 9x3 − 5x2 − x...
+ bi, where a and b are real numbers and i is the imaginary unit. Complex numbers include things you’d normally expect, like 3 + 2i and 2 3. However, don’t forget that a or b could be zero, which means numbers like 3i and 6 are also complex numbers. In other words, don’t forget that the complex numbers include the real...
ory, 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 ...
+ 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 ...
hould 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 ver...
l 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. ...
gree 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 ...
∞, 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 grap...
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) → ...
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-intercep...
) ∪ (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)...
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: ...
pproximate (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 ...
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 ﬁ...
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...
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 h...
)(−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 √...
numbers. Since the domains of f and g may not be all real numbers, we need the restrictions listed here. 2In other words, invertible functions have exactly one inverse. 380 Further Topics in Functions f . By Theorem 5.2, the domain of g is equal to the domain of h, since both are the range of f . This means the identit...
real numbers. 2 x+ 1 2 . To check this answer analytically, we ﬁrst check that f −1 ◦ f (x) = f −1 ◦ f (x) = f −1(f (x)) 1 f (x) + 2 1 − 2x 1 − 2x) + + x + 1 2 We now check that f ◦ f −1 (x) = x for all x in the range of f which is also all real numbers. (Recall that the domain of f −1) is the range of f .) f ◦ f −1 (...
and sold to achieve the maximum proﬁt? 5.2 Inverse Functions 395 26. Show that the Fahrenheit to Celsius conversion function found in Exercise 35 in Section 2.1 is invertible and that its inverse is the Celsius to Fahrenheit conversion function. 27. Analytically show that the function f (x) = x3 + 3x + 1 is one-to-one....
three test intervals. The sign diagram and accompanying graph are below. Note that the intervals on which f is (+) correspond to where the graph of f is above the x-axis, and where the graph of f is below the x-axis we have that f is (−). The calculator suggests something mysterious happens near x = 2. Zooming in shows...
x)1 (2 − x)2/3 3(2 − x) − x (2 − x)2/3 6 − 4x (2 − x)2/3 As before, when we set r(x) = 0 we obtain x = 3 2 . x (2 − x)2/3 x (2 − x)2/3 x (2 − x)2/3 = = = − − common denominator since 3√ √ u3 = ( 3 u)3 = u We now create our sign diagram and ﬁnd 3(2 − x)1/3 − x(2 − x)−2/3 ≤ 0 on 3 2 , 2 ∪ (2, ∞). To check this graphical...
usual steepness at x = −1 and x = 1 No cusps √ 2. f (x) = x2 − 1 Domain: (−∞, −1] ∪ [1, ∞) 0 (+) (+) 0 −1 1 No asymptotes Unusual steepness at x = −1 and x = 1 No cusps √ 3. f (x) = x 1 − x2 Domain: [−1, 1] (−) 0 0 (+) −1 0 0 1 No asymptotes Unusual steepness at x = −1 and x = 1 No cusps √ 4. f (x) = x x2 − 1 Domain: (...
rly to the graph of g. We summarize the basic properties of exponential functions in the following theorem.7 Theorem 6.1. Properties of Exponential Functions: Suppose f (x) = bx. The domain of f is (−∞, ∞) and the range of f is (0, ∞). (0, 1) is on the graph of f and y = 0 is a horizontal asymptote to the graph of f . ...
) = log2(x) is (0, ∞). Why? Because the range of f (x) = 2x is (0, ∞). In a way, this says everything, but at the same time, it doesn’t. For example, if we try to ﬁnd log2(−1), we are trying to ﬁnd the exponent we put on 2 to give us −1. In other words, we are looking for x that satisﬁes 2x = −1. There is no such real ...
g4(8) 31. log36 √ 4 36 34. log36 36216 37. log 3√ 105 40. log eln(100) 17. log6(216) 20. log8(4) 23. log 1 6 (216) 26. log(0.01) 29. log6(1) 32. 7log7(3) 35. ln e5 38. ln 1√ e 41. log2 3− log3(2) 3. 45/2 = 32 6. 10−3 = 0.001 9. log25(5) = 1 2 12. log(100) = 2 15. ln 1√ e = − 1 2 18. log2(32) 21. log36(216) 24. log36(36...
y and Intermediate Algebra. We ﬁrst extract two properties from Theorem 6.2 to remind us of the deﬁnition of a logarithm as the inverse of an exponential function. Theorem 6.3. (Inverse Properties of Exponential and Logarithmic Functions) Let b > 0, b = 1. ba = c if and only if logb(c) = a logb (bx) = x for all x and b...
le = log xy2 z Quotient Rule 3. We can certainly get started rewriting 4 log2(x) + 3 by applying the Power Rule to 4 log2(x) x4, but in order to use the Product Rule to handle the addition, we need to to obtain log2 rewrite 3 as a logarithm base 2. From Theorem 6.3, we know 3 = log2 23, so we get 4 log2(x) + 3 = log2 =...
its logarithmic form log2(129) = x. Either way, in order to get a reasonable decimal approximation to this number, we’d use the change of base formula, Theorem 6.7, to give us something more calculator friendly,1 say log2(129) = ln(129) ln(2) . Another way to arrive at this answer is as follows 2x = 129 ln (2x) = ln(1...
r x = ln(4) 2 = ln(2). (Can you explain the last equality using properties of logs?) As in the previous example, we need to be careful about choosing test values. Since ln(1) = 0, we choose ln 1 and ln(3). Evaluating,6 we get 2 , ln 3 2 r ln 1 2 = ln 1 2 = ln 1 2 = ln 1 2 4 ln 1 = 1 2 ) − 4 ln 1 e2 ln( 1 2 2 )2 − 4 ln ...
logarithms to rewrite the equation as log6 [(x + 4)(3 − x)] = 1. Rewriting this as an exponential equation, we get 61 = (x + 4)(3 − x). This reduces to x2 + x − 6 = 0, which gives x = −3 and x = 2. Graphing y = f (x) = ln(x+4) and y = g(x) = 1, we see they intersect twice, at x = −3 and x = 2. ln(6) + ln(3−x) ln(6) y ...
= 10 x x+1 . (a) Show that f −1 ◦ f (x) = x for all x in the domain of f and that f ◦ f −1 (x) = x for all x in the domain of f −1. (b) Find the range of f by ﬁnding the domain of f −1. x 1 − x (c) Let g(x) = and h(x) = log(x). Show that f = g ◦ h and (g ◦ h)−1 = h−1 ◦ g−1. (We know this is true in general by Exercise...
e in Calculus. 7Once you’ve had a semester of Calculus, you’ll be able to fully appreciate this very lame pun. 8Or deﬁne, depending on your point of view. 472 Exponential and Logarithmic Functions Using this deﬁnition of e and a little Calculus, we can take Equation 6.2 and produce a formula for continuously compounded...
ndicates the graph of y = T (t) is approaching its horizontal asymptote y = 350 from below. Physically, this means the roast will eventually warm up to 350◦F.12 The function T is sometimes called a limited growth model, since the function T remains bounded as t → ∞. If we apply the principles behind Newton’s Law of Coo...
c Functions This is all well and good, but the quadratic model appears to ﬁt the data better, and we’ve yet to mention any scientiﬁc principle which would lead us to believe the actual spread of the ﬂu follows any kind of power function at all. If we are to attack this data from a scientiﬁc perspective, it does seem to...
N0ekt. (a) Find the growth constant k. Round your answer to four decimal places. (b) Find a function which gives the number of yeast (in millions) per cc N (t) after t hours. (c) What is the doubling time for this strain of yeast? 6.5 Applications of Exponential and Logarithmic Functions 485 25. The Law of Uninhibited ...
section, there is no single diﬀerential equation which governs the enrollment growth. Thus there is no scientiﬁc reason to rely on a logistic function even though the data plot may lead us to that model. What are some factors which inﬂuence enrollment at a community college and how can you take those into account mathe...
that a point (x, y) is on the circle if and only if its distance to (h, k) is r. We express this relationship algebraically using the Distance Formula, Equation 1.1, as r = (x − h)2 + (y − k)2 By squaring both sides of this equation, we get an equivalent equation (since r > 0) which gives us the standard equation of a...
e form given in Equation 7.2. Here, x − h is x + 1 so h = −1, and y − k is y − 3 so k = 3. Hence, the vertex is (−1, 3). We also see that 4p = −8 so p = −2. Since p < 0, the focus will be below the vertex and the parabola will open downwards. y 5 4 3 2 1 −6 −5 −4 −3 −2 −1 1 2 3 4 x 7.3 Parabolas 507 The distance from t...
x (−3, 2) Focus (−6, 2) Directrix x = 0 Endpoints of latus rectum (−6, 8), (−6, −4) y 2 1 −5 −4 −3 −2 −1 x −1 −2 −7−6−5−4−3−2−1 −1 x −2 −3 −4 514 Hooked on Conics 4. (y + 4)2 = 4x Vertex (0, −4) Focus (1, −4) Directrix x = −1 Endpoints of latus rectum (1, −2), (1, −6) 5. (x − 1)2 = 4(y + 3) Vertex (1, −3) Focus (1, −2)...
= equation (x−h)2 9 + (y−1)2 √ a2 − b2, or 1 = b2 = 1, we get our ﬁnal answer to be (x−3)2 a2 + (y−k)2 8 = 1. √ 7.4 Ellipses 521 As with circles and parabolas, an equation may be given which is an ellipse, but isn’t in the standard form of Equation 7.4. In those cases, as with circles and parabolas before, we will need...
)2 9 + (y + 3)2 4 = 1 Center (1, −3) Major axis along y = −3 Minor axis along x = 1 Vertices (4, −3), (−2, −3) Endpoints of the Minor Axis (1, −1), (1, −5) Foci (1 + √ e = 5, −3), (1 − 5, −3) √ √ 5 3 7. (x + 2)2 16 + (y − 5)2 20 = 1 Center (−2, 5) Major axis along x = −2 Minor axis along y = 5 Vertices (−2, 5 + 2 Endpo...
tely (−3.39, 0)) and (2 + 29, 0) (approximately (7.39, 0)). To determine the equations of the asymptotes, recall that the asymptotes go through the center of the hyperbola, (2, 0), as well as the corners of guide rectangle, so they have slopes of ± b 2 . Using the point-slope equation 29 units to the left and right of ...
of the Conjugate Axis 6 17. Vertices (3, 2), (13, 2); Endpoints of the Conjugate Axis (8, 4), (8, 0) 18. Vertex (−10, 5), Asymptotes y = ± 1 2 (x − 6) + 5 In Exercises 19 - 28, ﬁnd the standard form of the equation using the guidelines on page 540 and then graph the conic section. 19. x2 − 2x − 4y − 11 = 0 20. x2 + y2 ...
in two variables with a1 = 3, a2 = − 1 2 and c = 0.1. We can also consider x = 5 to be a linear equation in two variables1 by identifying a1 = 1, a2 = 0, and c = 5. If a1 and a2 are both 0, then depending on c, we get either an equation which is always true, called an identity, or an equation which is never true, calle...
nition 8.2, we use subscripts to distinguish the diﬀerent variables. We have no idea how many variables may be involved, so we use numbers to distinguish them instead of letters. (There is an endless supply of distinct numbers.) As an example, the linear equation 3x1 −x2 = 4 represents the same relationship between the...
3) 4x − 9y + 2z = 5 Replace E1 with 1 2 E1 −−−−−−−−−−−−−→    E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Now it’s time to take care of the x’s in E2 and E3E1) 2 10x − z = 2 (E2) (E3) 4x − 9y + 2z = 5 Replace E2 with −10E1 + E2 −−−−−−−−−−−−−−−−−−→ Replace E3 with −4E1 + E3    2 y − 1 (E1) x + 3 (E2) (E3) 1 2 −15y...
blend which costs $6 per pound? 30. Skippy has a total of $10,000 to split between two investments. One account oﬀers 3% simple interest, and the other account oﬀers 8% simple interest. For tax reasons, he can only earn $500 in interest the entire year. How much money should Skippy invest in each account to earn $500 i...
an augmented matrix to transform the following system of linear equations into triangular form. Solve the system.    3x − y + z = 8 x + 2y − z = 4 2x + 3y − 4z = 10 Solution. We ﬁrst encode the system into an augmented matrix.    3x − y + z = 8 x + 2y − z = 4 2x + 3y − 4z = 10 Encode into the matrix −−−−−−−−−−−−−...
, −2) 30. At 9 PM, the temperature was 60◦F; at midnight, the temperature was 50◦F; and at 6 AM, the temperature was 70◦F . Use the technique in Example 8.2.3 to ﬁt a quadratic function to these data with the temperature, T , measured in degrees Fahrenheit, as the dependent variable, and the number of hours after 9 PM,...
n be represented by its position matrix, P : (x, y) ↔ P = x y Suppose we take the point (−2, 1) and multiply its position matrix by 3. We have 3P = 3 −2 1 = 3(−2) 3(1) = −6 3 which corresponds to the point (−6, 3). We can imagine taking (−2, 1) to (−6, 3) in this fashion as a dilation by a factor of 3 in both the horiz...
the product IkA to be deﬁned, k = 2; similarly, for AIk to be deﬁned, k = 3. We leave it to the reader to show I2A = A and AI3 = A. In other words, and 1 0 0 1 2 −10 0 −1 5 3 = 2 −10 0 −1 5 3 2 −10 0 −10 0 −1 5 3 While the proofs of the properties in Theorem 8.5 are computational in nature, the notation becomes quite i...
ix A = 1 2 3 4 write A as LU where L is a lower triangular matrix and U is an upper triangular matrix? 36. Are there any matrices which are simultaneously upper and lower triangular? 8.3 Matrix Arithmetic 8.3.2 Answers 1. For A = 2 −3 4 1 and B = 5 −2 8 4 595 3A = 6 −9 12 3 −B = −5 2 −4 −8 A2 = 1 −18 13 6 A − 2B = −8 1...
tant characteristics of invertible matrices and their inverses. Theorem 8.6. Suppose A is an n × n matrix. 1. If A is invertible then A−1 is unique. 2. A is invertible if and only if AX = B has a unique solution for every n × r matrix B. The proofs of the properties in Theorem 8.6 rely on a healthy mix of deﬁnition and...
d consistent system which is necessarily dependent. To solve this system, we encode it into an augmented matrix 0 4.375 1.25 −4.375     5.25 0 0 0 −5.625 −1.25 0 3.125 0 −3.125 0 0 0 0 0 10 0 0 0 −1.875 −5 0 1.875 5     and use the calculator to write in reduced row echelon form 608 Systems of Equations and Mat...
formula for det(A) which allows us to ﬁnd it by expanding along any row. Applying Deﬁnition 8.13 to the matrix A = 4 −3 1 2 we get 1We will talk more about the term ‘recursively’ in Section 9.1. 8.5 Determinants and Cramer’s Rule 615 det(A) = det 4 −3 1 2 = 4 det (A11) − (−3) det (A12) = 4 det([1]) + 3 det([2]) = 4(1)...
= 6 5 The reader is encouraged to solve this system for x and y similarly and check the answer. Our last application of determinants is to develop an alternative method for ﬁnding the inverse of a matrix.5 Let us consider the 3 × 3 matrix A which we so extensively studied in Section 8.5.1 A =   3 1 0 −1 1 2   2 5 ...
os of the characteristic polynomial just means that C itself had determinant zero which we already knew. Those two numbers are called the eigenvalues of C. The corresponding matrix solutions to CX = λX are called the eigenvectors of C and the ‘vector’ portion of the name will make more sense after you’ve studied vector...
and the right hand side. Then p(x) = 0 for all x in the open interval I. If p(x) were a nonzero polynomial of degree k, then, by Theorem 3.14, p could have at most k zeros in I, and k is a ﬁnite number. Since p(x) = 0 for all the x in I, p has inﬁnitely many zeros, and hence, p is the zero polynomial. This means there...
graphically, we sketch both equations and look for their points of intersection. The graph of x2 + y2 = 4 is a circle centered at (0, 0) with a radius of 2, whereas the graph of 4x2 + 9y2 = 36, when written in the standard form x2 4 = 1 is easily recognized as an ellipse centered at (0, 0) with a major axis along the ...
ﬁnd the most eﬃcient method to solve it. Sometimes you just have to try something. 8.7 Systems of Non-Linear Equations and Inequalities 643 We close this section discussing how non-linear inequalities can be used to describe regions in the plane which we ﬁrst introduced in Section 2.4. Before we embark on some examples...
y 2 √ 6. 1 + 7, −1 + y √ 7, 1 − √ 7, −1 − √ 7 4 3 2 1 −4−3−2−1 −1 1 2 3 4 x −2 −3 −4 √ 15 5 , ± √ 7. ± 2 10 5 √ 104−3−2−1 −1 1 2 3 4 x −2 −3 −4 8. (0, 1) 9. (0, ±1), (2, 0) 11. (3, 4), (−4, −3) 12. (±3, 4) 13. (−4, −56), (1, 9), (2, 16) 14. (−2, 2), (2, −2) 15. (3, 4) 16. Initially, there are 250000 49 ≈ 5102 bacteria....
an+1 = an + d to an+1 − an = d. Analogously, a geometric sequence is one in which we proceed from one term to the next by always multiplying by the same ﬁxed number r. If r = 0, we can rearrange the recursion equation to get an+1 = r, hence the name ‘common ratio.’ Some an sequences are arithmetic, some are geometric a...
erns Fail the Test. 658 Sequences and the Binomial Theorem 9.1.1 Exercises In Exercises 1 - 13, write out the ﬁrst four terms of the given sequence. 1. an = 2n − 1, n ≥ 0 3. {5k − 2}∞ k=1 5. xn n2 ∞ n=1 2. dj = (−1) j(j+1) 2 , j ≥ 1 4. 6. ∞ n=0 ∞ n2 + 1 n + 1 ln(n) n n=1 7. a1 = 3, an+1 = an − 1, n ≥ 1 8. d0 = 12, dm =...
r k ≥ 1, we let S denote the sum of the ﬁrst n terms. To derive a formula for S, we write it out in two diﬀerent ways S = S = (a + (n − 1)d) + (a + (n − 2)d) + . . . + (a + d) + a + . . . + (a + (n − 2)d) + (a + (n − 1)d) (a + d) + a If we add these two equations and combine the terms which are aligned vertically, we g...
est rate is 2.0%, term is 20 years 36. payments are $100, interest rate is 2.0%, term is 25 years 37. payments are $100, interest rate is 2.0%, term is 30 years 38. payments are $100, interest rate is 2.0%, term is 35 years 39. Suppose an ordinary annuity oﬀers an annual interest rate of 2%, compounded monthly, for 30 ...
suppose that the property we wish to prove is true for all k × k matrices. Let A be a (k + 1) × (k + 1) matrix. We have two cases, depending on whether or not the row R being replaced is the ﬁrst row of A. Case 1: The row R being replaced is the ﬁrst row of A. By deﬁnition, det(A) = n p=1 1pC a 1p 1p = (−1)(1+p) det A ...
! = (k + 2)(k + 1)(k)(k − 1)!. As a result, we have (k + 2)! (k − 1)! = (k + 2)(k + 1)(k)(k − 1)! (k − 1)! = (k + 2)(k + 1)(k) (k − 1)! (k − 1)! = k(k + 1)(k + 2) The stipulation k ≥ 1 is there to ensure that all of the factorials involved are deﬁned. 2. We proceed by induction and let P (n) be the inequality n! > 3n. ...
dditive relationship expressed in Theorem 9.3. For instance, and so forth. This relationship is indicated by the arrows in the + array above. With these two facts in hand, we can quickly generate Pascal’s Triangle. We start with the ﬁrst two rows, 1 and 1 1. From that point on, each successive row begins and ends with ...
we seek an angle γ so that β + γ = 90◦. We get γ = 90◦ − β = 90◦ − 37◦2817. While we could reach for the calculator to obtain an approximate answer, we choose instead to do a bit of sexagesimal7 arithmetic. We ﬁrst rewrite 90◦ = 90◦00 = 89◦600 = 89◦5960. In essence, we are ‘borrowing’ 1◦ = 60 from the degree place, an...
least one of which is positive and one of which is negative. 1. α = π 6 Solution. 2. β = − 4π 3 3. γ = 9π 4 4. φ = − 5π 2 1. The angle α = π 6 is positive, so we draw an angle with its initial side on the positive x-axis and rotate counter-clockwise (π/6) 12 of a revolution. Thus α is a Quadrant I angle. Coterminal an...

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