text stringlengths 235 3.08k |
|---|
18.3: Analyzing the values of the sine and cosine functions. The conclusion is that after one complete counterclockwise rotation, the values of sin(θ) and cos(θ) range over the interval [−1, 1]. As the ball moves through the four quadrants, we have indicated the “order” in which these function values are assumed by la... |
the domain values. 0. Similar reasoning shows that as the ball moves in the fourth quadrant, the slopes of the hypotenuses of the triangles are always non-positive, varying from 0 to ANY negative value. In other words, the range of values for tan(θ) on the domain −π/2 < θ 0 will be − z < < z ∞ 0. ≤ ≤ ≤ On your calcula... |
+y2 = 1. But, since the x coordinate is cos(θ) and the y coordinate is sin(θ), we have (cos(θ))2 + (sin(θ))2 = 1. It is common notational practice to write (cos(θ))2 = cos2(θ) and (sin(θ))2 = sin2(θ). This leads to the most important of all trigonometric identities: Important Fact 18.2.1 (Trigonometric identity). For ... |
, we have cos(θ) = cos(θ+2πn), sin(θ) = sin(θ+ 1, 2πn), and tan(θ) = tan(θ + nπ). 2, ± ± ± 18.2. IDENTITIES Next, we draw an angle θ and its negative in the same unit circle picture in standard central position. We have indicated the points Pθ and P−θ used to define the circular functions. It is clear from the picture i... |
�+ π) = − sin(θ), and cos(θ + π) = − cos(θ). Important Fact 18.2.6. For any angle θ, we have sin(π − θ) = sin(θ) and cos(π − θ) = − cos(θ). 5π 6 For example, we have sin = sin 2. This calculation leads to a computational observation: Combining Table 17.1 with the previous two identities we can compute the EXACT value o... |
given a circular function z = f(θ), the graph consists of all pairs (θ, f(θ)), where θ varies over a domain of allowed values. We will record and discuss these graphs below; a graphing device will painlessly produce these for us! There is a point of possible confusion that needs attention. We purposely did not use the... |
. 1 and −1 cos(θ) sin(θ) ≤ ≤ ≤ ≤ By Fact 18.2.3, we know that the values of the sine and cosine repeat themselves every 2π radians. Consequently, if we know the graphs of the θ sine and cosine on the domain 0 2π, then the picture will repeat for ≤ 0, etc. θ the interval 2π 4π, −2π θ ≤ ≤ ≤ ≤ ≤ 244 CHAPTER 18. TRIGONOMET... |
ine. Instead, we will utilize an identity. Given an angle θ, place it in central standard position in the unit circle, as one of the four cases of Figure 17.16. For example, we have pictured Case I in this figure. Since the sum of the angles in a triangle is 180◦ = π radians, we know that θ, π/2, and π 2 − θ are the thr... |
close to vertical line as θ gets close to − π 2 z-axis θ = π 2 Figure 18.14: The behavior of tan(θ) as θ approaches asymptotes. As we have already seen, unlike the sine and cosine circular functions, the tangent function is NOT defined for all values of θ. Since tan(θ) = sin(θ) cos(θ), here are some properties we can i... |
= 1000. 18.4. TRIGONOMETRIC FUNCTIONS Conclude that as θ “approaches π 2 from below”, the values of tan(θ) are becoming larger and larger. This says that the function values become “unbounded”. Likewise, imagine the case when θ is slightly bigger than − π 2, say θ =. Then the cal, and culation of tan(θ) involves divid... |
which lie at the heart of studying all kinds of periodic behavior. We have no desire to “reinvent the wheel”, so let’s use our previous work on the circular functions to define the trigonometric functions. 18.4.1 A Transition Given a real number t, is there a sensible way to define cos(t) and sin(t)? The answer is yes a... |
refer to these as the basic trigonometric functions. If we are working with radian measure and t is a real number, then there is no difference between evaluating a trigonometric function at the real number t and evaluating the corresponding circular function at the angle of measure t radians. Example 18.4.2. Assume th... |
to mean y = (sin t2) + (2t + 1). ), As a rule, whenever you see an expression involving sin( or tan( ” is in units of RADIANS, unless otherwise noted. When computing values on your calculator, MAKE SURE YOU ARE USING RADIAN MODE! ), we assume “ ), cos( · · · · · · · · · · · ·!!! CAUTION!!! 250 CHAPTER 18. TRIGONOMETRI... |
reaching implications in all of our lives. 5 10 20 25 30 15 time Figure 19.1: The depth of a salmon as a function of time. 19.1 A special class of functions Beginning with the trigonometric function y = sin(x), what is the most general function we can build using the graphical techniques of shifting and stretching? hor... |
the sign of C: If C is negative, the effect of shifting C units right is the same as shifting |C| units left. If the domain of sin(x) is 0 2π, then the domain of sin(x − C) is 0 x − C 2π, again by Facts 13.3.1. Rewriting this, the ≤ domain of sin(x − C) is C 2π + C and the graph will go through precisely one period on... |
some constant c > 0. We know that y = sin(x) is a 2π-periodic function and observe that horizontally dilation still results in a periodic function, but the period will typically NOT be 2π. For future purposes, it is useful to rewrite the equation for the horizontally stretched curve in a way more directly highlighting... |
five step procedure” one can follow, assuming we are given A, B, C, and D. It is a good idea to follow Example 19.1.3 as you read this procedure; that way it will seem a lot less abstract. 1. Draw the horizontal line given by the equation y = D; this line will + D into symmetrical upper 2π B (x − C) “split” the graph of... |
+ B, D) there will be a minima (valley); i.e. at the point (C + 3 4B,D − A). It is now possible to roughly sketch the graph on C + B by connecting the points described. Once the domain C this portion of the graph is known, the fact that the function is periodic C + 2B, tells us to simply repeat the picture in the inte... |
then “connect the dots” to get a rough sketch on the domain 11 35. ≤ ≤ t d(t) graph will oscillate inside this strip 30 25 20 15 10 maxima (17, 25) (11, 19) (35, 19) (23, 19) (29, 13) minima y = 25 y = 19 y = 13 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 Figure 19.9: Visualizing the maximum and minimum over ... |
, there are two ways in which we might encounter a trigonometric type function that is not in this standard form: • • The constants A or B might be negative. For example, y = −2 sin(2x− + 4 are examples that fail to be in 7) − 3 and y = 3 sin standard form. 2 x + 1 − 1 We might use the cosine function in place of the s... |
IDAL BEHAVIOR 259 This function is now in the standard form of Definition 19.1.1, so it is a sinusoidal function with phase shift C = 2 − 2π, mean D = 4, amplitude A = 3 and period B = 4π. (iii) Start with y = 2 cos(3x+1) −2, then here are the steps to put the equation in standard form. A key simplifying step is to use ... |
reference point”. This “reference point” is a location where the graph crosses the mean line y = D on its way up from a minimum to a maximum. The funny thing is that the phase shift C is NOT unique; there are an infinite number of correct choices. One choice that will work 260 CHAPTER 19. SINUSOIDAL FUNCTIONS is C = (x-... |
is June 21, which is day 172 of the year, so C = (day with max daylight) − B 4 = 172 − 366 4 = 80.5. In summary, this shows that d(t) = 3.7 sin 2π 366 (t − 80.5) + 12. A rough sketch, following the procedure outlined above, gives this graph 366; we have included the mean line y = 12 for on the domain 0 reference. ≤ ≤ ... |
(−7) + 3 = 1.686 feet. Finally, graphically, the last question amounts to determining how many times the graph of d(t) crosses the line y = 4 on the domain [0,10]. This can be done using Figure 19.13. A simultaneous picture of the two graphs is given, from which we can see the salmon is exactly 4 feet below the surface... |
that, it bounces up and down between its minimum height of 10 cm and a maximum height of 26 cm, and its height h(t) is a sinusoidal function of time t. It first reaches a maximum height 0.6 seconds after starting. (a) Follow the procedure outlined in this section to sketch a rough graph of h(t). Draw at least two compl... |
oidally. (a) Find a formula for the function y = h(t) that computes the height of the tide above low tide at time t. (In other words, y = 0 corresponds to low tide.) (b) What is the tide height at 11:00 a.m.? Problem 19.5. Your seat on a Ferris Wheel is at the indicated position at time t = 0. Start 53 feet Let t be th... |
b) Is the voltage output of the circuit ever equal to zero? Explain. (c) The function V(t) = 2p(t), where p(t) = 3 sin(5πt − 3π) + 1. Put the sinusoidal function p(t) in standard form and sketch the graph for 0 1. Label the coordinates of the extrema on the graph. ≤ ≤ t (b) Use the graph sketches to help you find the th... |
ways: with your function from part (c), and using some common sense (where is point A after one second?). (e) Is the function you found in (c) a sinu- soidal function? Explain. 266 CHAPTER 19. SINUSOIDAL FUNCTIONS Chapter 20 Inverse Circular Functions An aircraft is flying at an altitude 10 miles above the elevation of... |
(θ) cross z = 1 2? The first thing to notice is that these two graphs will cross an infinite number of times, so there are infinitely many solutions to Example 20.1.1! However, notice there is a predictable spacing of the crossing points, which is just a manifestation of the periodicity of the sine function. In fact, if w... |
resulting function is one-toone? All of these questions must be answered before we can come to grips with any understanding of the inverse functions. Using the graphs of the circular functions, it is an easy matter to arrive at the following qualitative conclusions. Important Fact 20.1.2. None of the circular function... |
tan θ? 20.2 Inverse Circular Functions Except for specially chosen angles, we have not addressed the serious problem of FINDING values of the inverse rules attached to the circular function equations. (Our previous examples were “rigged”, so that we could use Table 17.1.) To proceed computationally, we need to obtain ... |
f(θ). In addition, the function should be one-to-one on this restricted domain. The function f(θ) should be “continuous” on this restricted domain; i.e. the graph on this domain could be traced with a pencil, without lifting it off the paper. In the case of z = sin(θ), the principal domain − π π 2 satisfies our criteri... |
z domain − π ≤ ≤ 2 ≤ 1, then sin−1(z) is the unique angle θ in the principal θ π 2 with the property that sin(θ) = z. ≤ (ii) If −1 z domain 0 ≤ 1, then cos−1(z) is the unique angle θ in the principal π with the property that cos(θ) = z. ≤ ≤ θ ≤ (iii) For any real number z, tan−1(z) is the unique angle θ in the princip... |
� 2 π 2 π θ-axis tangent function restricted to principal domain y-axis principal domain on the unit circle θ x-axis Figure 20.7: Principal domain for tan(θ). (a) If − π θ 2 ≤ θ π 2, then sin−1(sin(θ)) = θ. ≤ π, then cos−1(cos(θ)) = θ. (b) If 0 ≤ ≤ (c) If − π 2 < θ < π 2, then tan−1(tan(θ)) = θ. We have been very expli... |
.9386 or cos2(θ) = 0.06136 cos(θ) = 0.2477. cos(θ) = 0.9688 or Finally, we use the inverse cosine function to arrive at our two acute angle solutions: cos(θ) = 0.9688 cos(θ) = 0.2477 θ = cos−1(0.9688) = 14.35◦ θ = cos−1(0.2477) = 75.66◦ ⇒ ⇒ h α 32 d Example 20.3.2. A 32 ft ladder leans against a building (as shown belo... |
. A picture of the situation is shown in Figure 20.9(a). After a look at the picture, three right triangles pop out and beg to be exploited. We highlight these in the Figure 20.9(b), by imposing a coordinate system and labeling the various sides of our triangles. We will work in radian units and label θ to be the requi... |
example of this, illustrating the reasoning required. Example 20.3.4. A rigid 14 ft pole is used to vault. The vaulter leaves and returns to the ground when the tip is 6 feet high, as indicated. What are the angles of the pole with the ground on takeoff and landing? 276 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 14 ft α 6... |
nothing to do with triangles and so we 20.4. HOW TO SOLVE TRIGONOMETRIC EQUATIONS 277 need to free ourselves from the necessity of relying on such a geometric picture. There are two general strategies for finding solutions to the equations c = sin(θ), c = cos(θ), and c = tan(θ): • • The first strategy is summarized in P... |
π 366 (t − 80.5) t ≤ ≤ Solution. To begin, we want to roughly sketch the graph of y = d(t) on the domain 0 366. If you apply the graphing procedure discussed in Chapter 19, you obtain the sinusoidal graph below on the larger domain −366 732. (The reason we use a larger domain is so that the “pattern” that will arise in... |
of the year. 172. ≤ ≤ t 1You can get the actual data from the naval observatory at this world wide web address: <http://tycho.usno.navy.mil/time.html> 20.4. HOW TO SOLVE TRIGONOMETRIC EQUATIONS 279 2. Symmetry Solution. To find another solution to the equation 14 = d(t), we will use symmetry properties of the graph of ... |
about 14 hours of daylight. ≤ ≤ t 280 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS 20.5 Summary • • • The inverse sine function, sin−1 x, is defined as the inverse of the sine function, sin x, restricted to the domain −π/2 π/2. x ≤ ≤ Inverse cosine and inverse tangent are defined similarly. To find the solutions to an equation... |
�degree mode”. (c) Find four values of x that satisfy the equation 5 sin(2x2 + x − 1) = 2. (d) Find four solutions to the equation 5 tan(2x2 + x − 1) = 2 Problem 20.2. For each part of the problem below: • • • Sketch the graphs of f(x) and g(x) on the same set of axes. Set f(x) = g(x) and find the principal and symmetry... |
410oF and 425oF. Problem 20.5. The temperature in Gavin’s oven is a sinusoidal function of time. Gavin sets his oven so that it has a maximum temperature of 300◦F and a minimum temperature of 240◦. Once the temperature hits 300◦, it takes 20 minutes before it is 300◦ again. Gavin’s cake needs to be in the oven for 30 ... |
pose coordinates as pictured. r= 100 yards 0.025 rad/sec Michael starts here Tiff starts here 0.03 rad/sec (a) Where is each runner located (in xy-coordinates) after 8 seconds? (b) How far has each runner traveled after 8 seconds? (c) Find the angle swept out by Michael after t seconds. (d) Find the angle swept out by ... |
the domain − π t 2 ≤ ≤ π 2, what is the domain D and range R of y = 27 sin( 2π 366 (t − 80.5)) + 45. How many solutions does the equation 40 = 27 sin( 2π 366 (t − 80.5)) + 45 have on the domain D and what are they? (d) If y = cos(x) on the domain 0 π, what is the domain D and range R of y = 4 cos(2x + 1)−3? How many s... |
onometric functions: (a) Find four solutions of tan(2x2 + x −1) = 5. (b) Solve for x: tan−1(2x2 + x − 1) = 0.5 284 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS Appendix 285 286 CHAPTER 20. INVERSE CIRCULAR FUNCTIONS Appendix A Useful Formulas Abbreviations inch = in feet = ft yard = yd mile = mi millimeter = mm centimeter = ... |
= c = 2.99792 density of water = 1 g/cm3 × × 108 m/s2 mass of earth = 5.9736 1024 kg × earth’s equatorial radius = 3,960 mi = 6.38 acceleration of gravity at earth’s surface = 32 ft/sec2 = 9.8 m/s2 × 106 m Algebra axay = ax+y (ax)y = axy a0 = 1 x = ax bx = ax−y a b ax ay n√a = a1/n (a + b)(c + d) = ac + ad + bc + bd (... |
the air is 9.740 million kg, more than the mass of the Tower. Answer 1.5 (a) 5.5 min/mi = 5:30 pace. (b) 14 2 Adrienne. 3 ft/sec. (c) × ≤ × ≤ ≥ (d) 1500 $56000 and (taxes) = 0.15 (c) (John’s Salary) (John’s Salary). ≤ 1800. (e) (cost red Porsche) > 3 Answer 1.6 (a) (John’s Salary) = $56000 and (taxes) = $0. (b) (John’... |
�F, rate of change = 2◦F/hr. (b) 58◦F. (c) Initial time = 4.5pm, Initial temperature = 54◦F, Final time = 6.25pm, Final temperature = 26◦F, rate of change = -16◦F/hr. Answer 1.17 (a) t = 9. (b) a = 1/8. (c) x = −3a/4. (d) t > 3. (e) x+2 x(x+1). Answer 2.1 (a) d = √2, ∆x = 1 = ∆y = 1. ∆x = −1, ∆y = −2. d = √10t2 + 2t + ... |
vessel does not catch the ferry before Edmonds. p Answer 1.11 About 5 × 106 times around the equator. Answer 2.5 (a) d(t) = (65.3)t (b) 227 minutes, 168.4 miles (c) t = 80.86 seconds. Answer 1.12 (a) Radii are about r = 0.2416, 2.317, 4.184 and 9.167 cm at the indicated times. (b) No. Answer 1.13 (a) 140 million gallo... |
2xy + y2, etc. (c) Answer and steps correct. Answer 2.12 (a) x = 1+β2 α2 ±q (b) x = αβ α+β (c) x = β αβ+1 Answer 2.13 (a) 5t2 + 6t + 5 (b) 2t2 + 4t (c) √5t2 + 4t + 4 2 t2 −1 (d) Answer 3.1 (a) (x+3)2+(y−4)2 = 9 (b) (x−3)2+ 1 16 (c) Draw a vertical and horizontal line through (1, 1). On the vertical line, a circle of r... |
pose a coordinate system so that the tractor is at the origin at t = 0 seconds. With this coordinate system, the south edge of the sidewalk is modeled by ys = 100; the north edge is modeled by, yn = 110. (b) t = 32 minutes. (c) t = 52 minutes. (d) 20 minutes. Answer 3.6 (a)] The equation for eastward travel from Kingst... |
line y = 2x + 1 2 y = − 11 4 x + 17 4 − 11 4 y = −2x + 1 −,000 y = 0 1 2 0 0 1 17 4 1 1 (0, 1) (− 1 2, 0) (3, − 4) (−1,7) (0, 1) ( 1 2, 0) (0, 1) (−2, 0) 1,000 (0, 1,000) (0, 1,000) 0 (0, 0) (1, 0) x = 3 Undef None (3, 3) (3, −2) y = x − 14 1 −14 (5, − 9) (0, −14) Answer 4.5 (a) y = 6,850(x − 1970) + 38,000 (b) y = 8,... |
4◦F. (F − 32) and F = 9 5 C + 32. In Oslo, the Answer 4.10 At closest 26.22471828 miles from Paris. the point, she will be Answer 4.11 Impose coordinates with Angela’s initial location as the origin. Angela is closest to Mary at (18.8356,41.4383); this takes approximately 3.8 seconds. Answer 4.12 (a) Impose coordinates... |
6400 2 (b) x = 4, 8. (c) x = 36. (d) Answer 6.1 (a) 0, 2, 3. (b) x = Intersect at (4,4) and (− 4 3, 4 3 ± ). Area is 16 3. 4, x = 0, no solution. (c) Answer 6.2 (a) | − 0.5x − 1| = −0.5x − 1 0.5x + 1 −2 if x ≤ if x > −2 y 3 2.5 2 1.5 1 0.5 -4 -2 2 x 4 ± Answer 6.3 (a) x = 4 and x = − 22 equation has no solutions. 3 (b)... |
-intercepts= 1 6 (b) ( 1 (3 − √93),5) and ( 1 6 6 No, Yes. (e) ( √33). y-intercept = −2. (3 + √93),5). (c) None. (d) Yes, No, 1 + √2 + 3(1 + √2)). 1 + √2, − 2 − 3 (3 ± p p (4 − x)2 + 1. (c) The Answer 5.7 (b) f(x) = 30000x + 50000 (0., 206155), (0.5, 197003), table of values (x,f(x)) will be: (1, 188114), (1.5, 179629)... |
2 (c) y = −0.5x2 + 3.5x − 4. (d) No solution. 3 x2 + 5 3 x. (b) y = 1.125x2 −1.5x−1.625. 10t 2502 + (12(t − 25))2 4002 + (250 − 9(t − 175 3 ))2 d(t) = p q Answer 6.9 25, t if 0 if 25 if 175 ≤ ≤ 3 ≤ ≤ t ≤ t ≤ 175 3, 205 3. y = d(t) = 18t 902 + 182(t − 5)2 902 + (90 − 18(t − 10))2 90 − 18(t − 15) p p 5 10 ... |
: 8.038 minutes. 73 ft. wide: 116.205 minutes. Answer 6.12 (a) 0 x and 2 ≤ 4. Decreasing: −2 ≤ 6. (b) Increasing: −6 x 2 and 4 y x ≤ ≤ ≤ ≤ (c) ≤ −2 x 6. ≤ ≤ ≤ y = 4 + 4 − 6. (e) 2 (d) 2 y ≤ ≤ 2x + 12 4 − (x + 2)2 4 − (x − 2)2 −2x + 12 p p y ≤ ≤ 4. if −6 if −4 if 0 if 4 ≤ ≤ ≤ ≤ x x − Answer 6.13 (a) 0. (b) a =... |
of the parabola is (1, − 4). |x2 − 2x − 3| = x2 − 2x − 3 −x2 + 2x + 3 x2 − 2x − 3 ≤ −1 if x if −1 < x < 3 3 if x ≥ Answer 7.7 (a) 100 ft. (d) When x = 54.81 ft. or x = 570.19 ft.; i.e. at (54.81,39.04) and (570.19, − 64.04). (c) (625, − 125). (b) 156.25 ft. Answer 7.8 (a) No, since f(1) = 1 (b) The points (1,1 +... |
) = (400 − 8.944t,50 + 4.472t). Study the distance SQUARED from M(t) to T (t). Michael and Tina will be closest when t = 26.641 sec; they are 54.3 ft at that instant. 6 Answer 7.18 −2α2 x = ± √4α4 − 4α 2α α = −x2 ± √x4 − 8x 4x Answer 7.19 (d) There are two possible values for α: If α = 8 + 2√17, then the unique solutio... |
= 1. (c) f(g(x)) = x, √x √x (d) f(g(x)) = 6(x − 4)2 + 5, f(f(x)) = 81x + 20, g(f(x)) = x. f(f(x)) = 6(6x2 +5)2 +5, g(f(x)) = 6x2 +1. (e) f(g(x)) = 8x+21, f(f(x)) = 4(4x3 − 3)3 − 3, g(f(x)) = 2x. (f) f(g(x)) = 2x3 + 1, f(f(x)) = 4x+3, g(f(x)) = (2x+1)3. (g) f(g(x)) = 3, f(f(x)) = 3, g(f(x)) = 43. (h) f(g(x)) = −4, f(f(... |
1(y) = 2+4y 3y. = 4 3 }; range=f= {y|y = 0}. Answer 9.2 Answer 9.3 (c) f−1(y) = 3+√17+8y − 17 8 }. 4 on the domain {y|y ≥ Answer 9.4 Only (B) is one-to-one on the entire domain. Answer 9.5 Answer 9.6 (a) h = f(x) = −2x2 + 124x. 31 − 1 2 √3844 − 2h. (c) x = g(h) = -3 -2 -1 1 2 x 3 Answer 9.7 (a) (1) -0.5 -1 Answer 8.2 (... |
); (b) 3.555; 19.8; 0.0729. x = (1/3) log10(y). Answer 12.2 (a) I(d) = I ◦ (0.94727)d. (b) 85 meters. Answer 12.3 (a) y = 13e1.09861t. (b) y = 2e−2.0794t. Answer 12.4 (a) 9.9 years. (b) 34.66%. ) = 3( y+2 (4) f(f−1(y)) = f( y+2 3 3 f−1(f(x)) = f−1(3x − 2) = 3x−2+; = x. 3 t 30 hours; Range: 0 Answer 9.8 (a) w = f(t) = 2... |
x = −0.7658 + 2kπ or x = 3.9074 + 2kπ. (f)-0.61. (d) (c) 0.3552. (g) sin(x) = −0.6931; Answer 10.3 (a) 1.64 formulas are identical. × 106 cells. (b) True. (c) The two Answer 12.9 (d) 8.617 weeks. Answer 10.4 (a) 261.31 Hz. (b) 440 Hz. (c) 27.5 Hz. (d) 16.35 Hz. Answer 12.10 31.699250014 days Answer 10.5 (a) 29 = 512. ... |
25 20 15 10 5 -2 -1 1 2 3 x 4 Answer 13.5 (a) y 4 3 2 1 -4 -3 -2 -1 1 2 3 x 4 -1 -0.5 0 0 0.5 1 1.5 2 (b) No. (c) -0.5 -1 -1.5 10 5 0 -12 -10 -8 -6 -4 -2 0 2 4 -5 -10 0 if x ≤ if −2 if 0 if x ≤ ≥ −2x + 2 0 y = f(−x) = y 4 3 2 1 -4 -3 -2 -1 1 2 3 x 4 y = −f(x) -1 0 −2x − 2 x − 2 0 y 1 if x ≤ if −1 i... |
2}; zero at x = 0; horizontal asymptote y = 2; vertical asymptote x = 1; graph below: = 1}; range={y|y -4 -2 y 15 10 5 -5 -10 -15 2 4 x Answer 14.2 (a) y = 0.2x−10.4 ft. (c) 10 ft. x−10. (b) x = 9.143 ft.; x = 9.916 Answer 14.3 (a) m(t) = 35t + 200. 1987. (d) r(t) = 35t+200 30t−50. (e) 7 6. (b) k(t) = 30t − 50. (c) An... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.