56. Average Value of a Function (Calculator allowed) Relative humidity is a term used to describe the amount of water vapor in a mixture of air and water vapor. A storm is passing through an area and the relative humidity changes rapidly. The relative humidity in percentage is a differentiable function H of time t. The table below shows the relative humidity as recorded over a -hour period. t (hours) 4 6 8 H t (percent) 4 89 94 7 5 4 4 a. Use data from the table to find an approximation for H ( 6). Show the computations that lead to your answer. Indicate units of measure. () H ( 6) H( 4) H 6 H 6 6 4 H ( 8) H( 6) H 6 8 6 H ( 8) H( 4) 8 4 = % hr or = % hr or = % hr pt for difference quotient with units b. Approximate the average relative humidity in percent over the time period t using a trapezoidal approximation with subintervals of length Δt = hours. () 4 + 89 [ + ( 94) + ( 7) + ( 5) + ( 4) + 4 ] = 77 Average relative humidity 77 = 64.47% pt for trapezoidal method c. A meteorologist proposes the function R, given by R t humidity in the air at time t, where t is measured in hours and R t Using appropriate units, explain the meaning of your answer. () = 4 + te t, as a model for the relative is measured in percent. Find R ( 6). = te t t R t R 6 = 6e 6 6 + e t + e 6 ==.89 At t = 6, the relative humidity is decreasing by.89%. hr pt for R ( 6) pt for interpretation and units d. Use the function R defined in part c) to find the average relative humidity on the time interval t hours. () 4 + te t dt = 779.776 = 64.98% pt for integrand pt for limits and constant www.mastermathmentor.com - 56 - Illegal to post this document on the Internet
57. Average Value of a Function Tim is mowing a lawn with an uneven terrain forcing him to push the mower at a rate modeled by the piecewise function f. In the figure to the right, f ( t) = 9 t + t + for t and f is piecewise linear for < t 6. For 6 < t 5, f is modeled by f ( t) = 5 t 8. The function f is measured in yd min and t is measured in minutes. a. Find f ( 5). Indicate units of measure. () f ( 5) = 5 9 = yds min min. pt for f 5 and units b. For the time interval t 5, at what time t is f increasing at the greatest rate? Show the reasoning that supports your answer. (5) f increasing on [,) and ( 6,5] On (,), f t = t + t and f ( t) = 4 t + = t =. This is a maximum as f > on, On ( 6,5), f t f t = 5 t and f ( t) = 5 4t and f < on, so the maximum must occur at the endpoint t = 5. On (,5), f = 9 4 + = and f 5 = 5 5 = pt for f on (,) pt for f has a maximum at t = pt for f on ( 6,5) pt for f has a maximum at t = 5 Justification can also involve f ( x). f is increasing at its fastest rate at t =. c. Find the average number of square yards Tim cuts per minute on t 6. () 6 f ( t ) dt = 6 ( ) + = 5 yd min d. Tim adjusts his rate of cutting so that he cuts ( f ( t) + k) yd min wants to average 4 yd for t 6. () min 6 f t [ + k ] dt = 5 +k 5 +k = 5 k = 7. = 4 on t 6. Find the value of k if Tim pt for setup pt for value of k www.mastermathmentor.com - 57 - Illegal to post this document on the Internet
58. Average Value of a Function (Calculator Allowed) Because of its location in the center of North America, the Twin Cities region in Minnesota is subjected to some of the widest range of temperatures in the United States. The high temperatures over a year is modeled by the equation T H =.44sin πx 6.9 + 5.56 while the low temperatures over a year is modeled by the equation T L =.5sin πx 6.9 +.95. The graph of these functions is shown in the figure on the right. The variable x refers to the 5 th day of the month where represents December, represents January, and represents November and T is measured in degrees Fahrenheit. a. What is the average rate of change of the high temperature from January 5 to July 5? Specify units. () T H ( 7) T H 84. = =.48 degrees 7 6 month pt for units b. What is the average difference between the high and low temperatures over a year? Specify units. () ( T H T L ) dt = 5. = 9.6 degrees. and units c. During what month is the low temperature rising the fastest? Justify your answer. () T L =.5cos πx 6.9 π 6 T L =.5sin πx 6.9 π 6 T L rises the fastest at approximately t = 4 because T L switches from positive to negative. T L rises the fastest during March. pt for T L pt for numerical answer pt for justification d. The Newton family has programmed their air conditioner to run whenever the average high temperature is over 7 degrees. The cost of cooling the house accumulates at the rate of $4 per month for each degree the average high temperature is over 7 degrees. What is the total cost to the nearest dollar for cooling the house over a year? () T H =.44 sin πx 6.9 + 5.56 = 7 t = 5.65, 8.78 Cost = 4 8.78.44 sin πx 6.9 + 5.56 7 dt $9 pt for limits and 4 pt for integrand 5.65 www.mastermathmentor.com - 58 - Illegal to post this document on the Internet
59. Average Value of a Function (Calculator Allowed) A popular Comedy Club has two shows, the early show from 6 PM until 9 PM and the late show from 9 PM until midnight. The rate that people enter the club is given by R( t) = 4 t sin 9 8 t people per hour over the time interval t 6 where t = corresponds to 6 PM. When the early show is over at 9 PM, the club clears and people come in for the late show. The graph of R t is to the right. a. Find the difference between the number of people who come to the early show and those who come to the late show. Round to the nearest person. () 6 R ( t ) dt R( t) dt = 5.5 7. 54 pt for integrals b. The club hires an extra maitre d when more than 6 people per hour are entering the club. Find the average number of people entering the club during that period of time. () R( t) = 6 when t =.97 and t = 5.46 5.46.97 5.46 R( t) dt =.75 8.8.97 = 76. people hour. pt for values of t pt for integral c. The club brings in extra waiters when the number of people in the club is greater than 75. Write, but do not solve, an equation using an integral that solves for time t when extra waiters are first needed. () Since there were 7 people in the club during the early show, no extra waiter was needed. t R ( x ) > 75 d. The club cuts off all alcoholic drinks one hour before the late show ends. If the average early show customer spends $5 on drinks and the average late show customer spends $5 on drinks, what is the average amount of money spent on drinks per hour during the evening? () 5 5 R t 5 7. 5 dt 5 + 5 R t dt [ + 5 (.5 ) ] $7 hour pt for 5 R t 5 pt for 5 R t dt dt www.mastermathmentor.com - 59 - Illegal to post this document on the Internet
6. Straight Line Motion (Integrals) A particle moves along the y-axis with velocity at time t given by v( t) = e t 4. a. Find the acceleration of the particle at time t =. () = v ( t) = e t 4 pt for v = e = pt for a e a t a b. Is the particle speeding up or slowing down at t =? Give a reason for your answer. () v = e = e < a = e > Particle slowing down as v and a have opposite signs. with reason c. Find all values of t for which the particle changes direction. Justify your answer. () = when v( t) = e t 4 = and t 4 = so t =. < when t < and v( t) > when t > v t v t so the particle changes direction at t =. pt for v( t) = pt for t = with justification d. Find the total distance traveled by the particle over t 4. (4) 4 D = v ( t ) dt = e t 4 dt + e t 4 dt = t et 4 4 + et 4 t = e 4 + e4 4 = + e 4 + e4 4 + = e4 + e 4 4 OR x( t) = et 4 t x = e 4 x = = x( 4) = e4 4 Distance = x( 4) x x x = e4 [ ] 4 + e 4 = e4 + e 4 pt for limits pt for integrand pt for antidifferentiation pt for evaluation OR pt for antiderivative pt for x, x, and x pt distance between points pt for total distance www.mastermathmentor.com - 6 - Illegal to post this document on the Internet
6. Straight Line Motion (Integrals) Roller coaster A leaves the station at time t =. The velocity of the coaster is recorded for selected values of t over the interval t seconds, as shown in the table below. t (seconds) 4 6 8 v( t) (feet per second) 8 5 55 76 8 a. Find the average acceleration of roller coaster A over the time interval t. Indicate units of measure. () v v = 8 8 b. Using correct units, explain the meaning of = 6 ft sec v( t) dt in terms of the roller coaster s travel. Use a midpoint Riemann sum with subintervals of equal length to approximate v( t) dt. () Since the velocity is positive, v( t) dt represents the total distance in feet traveled by the roller coaster from t = to t =. v( t) dt represents the average velocity of the roller coaster over the first seconds in ft sec. Avg vel 4 + 5 + 76 = = 4 ft sec. pt for explanation pt for using v,v ( 6),v pt for value 4t + feet per second per second and its initial velocity is 4 feet per second. Which of the two coasters is traveling faster at t = seconds? Explain your answer. (4) c. Roller coaster B travels on a parallel track to coaster A. Its acceleration is given by a( t) = v B ( t) = 4 = 6 v B t 4t + dt = 4 + C C = 4 = 6 4t + + 4 = 6 49 + 4 = 76 v B 4( 4t +) dt = 6( 4t +) + C Coaster A is traveling faster at t = as 8 > 76. pt for 6( 4t +) pt for constant pt for using initial condition pt for v B and conclusion d. () pt for units in a and b. www.mastermathmentor.com - 6 - Illegal to post this document on the Internet
6. Straight Line Motion (Integrals) A searchlight is trained on the intersection of a building and a wall. At t =, the light travels along the wall in a straight line. For t, the velocity of the light is modeled by the piecewise function defined by the graph to the right. The graph is made of lines and a semi-circle. a. At what times in the interval < t <, if any, does the light change direction? Give a reason for your answer. () The light changes direction at t = 7 and t =6 because the velocity changes sign. with reason b. At what times in the interval t is the light farthest from the building? How far is the light from the building at this time? () t distance from wall The light is farthest from the building 7 + 9π at t = 7 and its greatest distance from 9π 6 the building is + 9π 9π pt for evaluation at t = 7, t = 6 pt for evaluation at t = c. Find the total distance the light travels during the time interval t. () D = v( t) dt = 9π + + 4 + 8 = 9π + 45 d. Write expressions for the light s acceleration a( t), velocity v( t), and distance x t that are valid to the time interval 6 < t < 9. (4) from the building a( t) = 4 v t 9 6 = = ( t 6) = 4 t x( t) = x( 6) + ( 4 u) du t = + 9π 6 = + 9π +4t t ( 84 6) = t +4t + 9π 6 t + ( 4u u ) 6 pt for a t pt for v t t du pt for 4 u 6 pt for x t www.mastermathmentor.com - 6 - Illegal to post this document on the Internet
6. Inverse Trig (Calculator Allowed) The cross-section of a hill is modeled by y = 6 π ( sin x ), measured in feet as shown in the figure to the right. a. Find dy. () dy = 6 x x 4 b. At the very top of the hill, there is a flagpole. The pole is supported by a wire that is tangent to y at x = and extends to the very top of the pole. Find the equation of the line that represents the wire. () x =,y = 6 π sin 4 = 79.87 Tangent line : m = 6 4 = 6 5 6 = 4 5 y 79.87 = 4 ( 5 x.5) y 79.87 = 6.968( x.5) or y = 6.968x +.7 pt for point on curve : (.5,79.87) pt for slope : 4 5 pt for tangent line c. Find the height in feet of the flagpole. () Height of hill: y = 6 π ( sin ) = π Height of hill and flagpole : 79.87 6.698.5 Height of flagpole :.7 π =5.8 ft. =.7 pt for height of hill d. Find the average height of the hill. () 6 π sin x = 4.777 pt for constant = 7.888 feet pt for limits www.mastermathmentor.com - 6 - Illegal to post this document on the Internet
64. Area and Volume Let R be the region in the first quadrant enclosed by the graphs of f x = cosπx and g( x) = 9x which intersect at point P, which has x-coordinate of, as shown in the figure to the right. a. Find the equation of the tangent line to the graph of f at point P. () = πsinπx f f x f = cos π = Tangent line : y = π x = πsin π = π = π pt for f b. Find the area of R. (4) Area = [ f ( x) g( x) ] = cosπx 9x = sinπx x π = π sin π = π 9 = π 9 pt for integrand pt for antiderivative of cosπx pt for antiderivative of 9x c. Write, but do not evaluate, an integral expression for the volume of the solid generated when R is rotated about the horizontal line y =. () π [( g ( x ) ) ( f ( x ) ) ] cosπx π 9x pt for limits and constant pt for integrand www.mastermathmentor.com - 64 - Illegal to post this document on the Internet
65. Area and Volume Let R be the region in the first quadrant enclosed by the graph of = 4 x f x the figure to the right., the x-axis and the lines x = and x =, as shown in a. Find the area of R. () A = 4 x A = 4x x = 4x + x A = 8 + ( 4 +) = 7 =.5 pt for integrand pt for antiderivative b. Write, but do not evaluate, a simplified integral expression that gives the volume of the solid generated when R is rotated about the line y = 4. () V = π 4 4 4 x V = π 6 x or π 6 x 4 pt for integrand pt for integrand simplified pt for limits and constant c. Region R is the base of a solid. The cross section of the solid taken perpendicular to the x-axis is a rectangle whose height is 4 times the length of its base in region R. Write, but do not evaluate, an integral expression that gives the volume of the solid. () Each rectangular section has area f ( x) Area = 4 4 x [ ] 4 f x [ ] Volume = 4 4 x or 4 4 x pts for integrand www.mastermathmentor.com - 65 - Illegal to post this document on the Internet
66. Area and Volume (Calculator Allowed) = e x. Let R be the region in the nd quadrant bounded Let f x by f ( x), the y-axis and the line y = x + 5. Let S be the region in the st quadrant bounded by f ( x), the y-axis and the lines y = x + and y = x + 5. a. Find the area of R. () e x = x + 5 at M =.74 and N =.57 [ ] A = e x ( x + 5) =.88.74 pt for limits M and zero pt for integrand b. Find the area of S. () e x = x + at P =.8.57 [ ] A = x + 5 ( x + ) + e x ( x + ).57.8.57 A = + e x ( x + ).8.57 A =.74 +.78 =.5 [ ] [ ] pt for limits N and P pt for integrand(s) c. Let k be the maximum value of f ( x). Write, but do not evaluate the integral expression that gives the volume of the solid generated when region R is rotated about the line y = k. () k = e e e x V = π e ( x + 5).74 pt for maximum k = e pt for constant and limits pt for integrand www.mastermathmentor.com - 66 - Illegal to post this document on the Internet
67. Area and Volume (Calculator Allowed) Let R be the region bounded by the curves y = ln x and y = x 4 to the right of the y-axis as shown in the figure to the right. a. Find the area of R. () ln x = x 4 x =.7 and.98..98 [ ] A = ln x x 4 = 4.666..7 pt for integrand pt for limits b. Find the volume generated when region R is rotated about the line y = 4. ().98 ( x 4 + 4) V = π ln x + 4 = 6.59π or 8.89..7 pts for integrand pt for limits, constant and answer c. Write, but do not evaluate the integral expression that gives the volume of the solid generated when region R is rotated about the y-axis. () y = ln x = ln x so ln x = y and x = ey y = x 4 so x = y + 4 V = π.749 ( y + 4 e y ) dy or by shells.98.98 [ ( )] V = π x ln x x 4.7 pts for integrand pt for limits, constant www.mastermathmentor.com - 67 - Illegal to post this document on the Internet
68. Area and Volume Let f be the function given by f ( x) = x 6x + 64x and let line l be the line tangent to the graph of f at x =, as shown in the figure to the right. Let R be the region bounded by the graph of f and the x-axis and let S be the region bounded by the graph of f, line l, and the x-axis. a. Find the equation of line l. () = 8 6( 4) + 64 = 7 = x x + 64 f = ( 4) + 64 = or y = x + 48 f f x line l : y 7 = x b. Find area of region S. (4) pt for f and f line l intersects the x axis at x = 4. A = ( x + 48) + 4 = 6x [ + 48x] 4 [ x + 48 ( x 6x + 64x) ] + 48x 6x + 6x x 4 4 = ( 96 9) + 96 4 + 8 4 = 96 + 9 = 8 OR A = ( 4) ( 48) + [ x + 48 ( x 6x + 64x) ] = 96 + 9 = 8 pt for limits of non triangular region pt for integrand of non triangular region pt for area of triangular region c. Region R is the base of a solid. The cross section of the solid taken perpendicular to the x-axis are semicircles. Write, but do not evaluate, an integral expression that gives the volume of the solid. () f ( x) = x( x 6x + 64) = x( x 8) So f intersects the x axis at x = 8. r = V = π x 6x + 64 x 8 x 6x + 64x or π 8 8 ( x 6x + 64 x) pt for limits and constant pt for integrand www.mastermathmentor.com - 68 - Illegal to post this document on the Internet
69. Area and Volume = x and g( x) =6 x. Let The functions f and g are given by f x R be the region bounded by the x-axis and the graphs of f and g, as shown in the figure to the right. a. Find the area of R. () 4 A = x + ( 6 x) OR A = x + 4 4 A = 5 x 5 A = 64 5 8 4 + ( 6x x 8 ) 4 +6 = 44 5 OR A = 5 x 5 +6 4 4 ( 8) pt for integral pt for antiderivative b. The region of R from x = to x = 4 is rotated about the line x = 4. Write, but do not evaluate, an integral expression the volume of the solid. () y = x x = y radius = 4 x = 4 y 8 V = π 4 y dy or π 6 8y + y 4 dy 8 pt for x = y pt for limits and constant c. The region R is the base of a solid. For each y, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose base lies in R and whose height is h( y). Write, but do not evaluate, an integral expression of the volume of the solid. () y = x x = y y = 6 x x = 8 y width = 8 y y 8 V = 8 y y h y dy pt for integrand pt for limits and constant www.mastermathmentor.com - 69 - Illegal to post this document on the Internet
7. Differential Equations Consider the differential equation dy = x + y. be the particular solution to the differential equation passing through the point (, ). Write an a. Let f x equation for the line tangent to the graph of f at (, ) and use it to approximate the root of f ( x). () Tangent line : m = + = y = ( x ) or y = x 5 x 5 = x = 5 =.6 so the root.6. pt for equation of tangent line pt for approximation of root b. Find the particular solution to the differential equation with initial condition f ( ) =. (5) dy = x + y y dy = ( x +) y = x + x + C or y = x + x + C = + C C = y = x + x + or y = ± x + x + Since f passes through (, ), y = x + x + pt for separation of variables pt for antiderivatives point for constant of integration pt for using initial condition pt for solution Max : 5 if no constant of integration Max : 5 if no separation of variables be a particular solution to the differential equation with the condition f ( ) =. Does c. Let y = f x f have a relative maximum, relative minimum, or neither at x =? Justify your answer. () dy + = (, ) = dy d y = y x + y d y + = (, ) So there is a relative maximum at x =. = pt for dy and d y at x = pt for relative maximum for incorrect answer based on y = x + x www.mastermathmentor.com - 7 - Illegal to post this document on the Internet
7. Differential Equations Consider the differential equation dy = y x. a. On the axes provided, sketch a slope field for the given differential equation at the 6 points indicated and for < x <, sketch the solution curve through (, ). () pt for zero and undefined slopes pt for non zero slopes pt for either the line with a hole at, or a V-shaped curve passing through,! b. Describe all points in the xy-plane for which dy =. () dy = y x = So y = x and x pt for description c. Find the particular solution y = f x f = 4. (5) to the given differential equation with the initial condition dy = y x x dy y = x ln y = ln x + C dy = y y = e ln x +C = Ce ln x = C x 4 = C C = 4 y = 4 x or y = 4x + 4, x pt for separation of variables pt for antiderivatives point for constant of integration pt for using initial condition pt for either solution Max: 5 if no constant of integration Max: 5 if no separation of variables www.mastermathmentor.com - 7 - Illegal to post this document on the Internet
7. Differential Equations Consider the differential equation dy = x y +. a) On the axes provided, sketch a slope field for the given differential equation at the 9 points indicated. () pts for sign of slope at each point and relative steepness of slope lines in rows and columns. b) Find d y in terms of x and y. Describe the region in the xy-plane in which all the solutions to the differential equation are concave down. () d y dy = = x y + d y = x + 4 y = x + 4y Solution curves will be concave down when x + 4 y < or below the curve y = x +. 4 pts for d y pt for description be a particular solution to the differential equation with the condition f ( ) =. Does c) Let y = f x f have a relative maximum, relative minimum, or neither at x =? Justify your answer. () dy = += (,) d y = x + 4y d y = 6 + 8 = > (, ) Thus, f has a relative minimum at x =. pt for justification d) Find the values of the constants m and b for which y = mx + b is a solution to the differential equation. () m = dy so m = x y + m = x mx + b + m = x mx b + m = x( m) + b pt for value of m So m = and m = b = m so b = m = = 4 pt for value of b www.mastermathmentor.com - 7 - Illegal to post this document on the Internet
7. Growth and Decay When a dam is built, the water from the river that is being dammed is blocked so that the dam can be built on a dry site. As the dam grows higher, residual water from the surrounding ground will collect behind the dam. At the Newton Dam, the water is feet high at the time the dam is finished and the river is unblocked. The height H of the water after the river is unblocked satisfies the differential equation dh = ( dt H ), H measured in feet and t measured in months. This model is in effect for the next years. a. Use the line tangent to the graph of H at t = to approximate the height of the water behind the dam after.5 months. () dh = dt t = H ( ) = Tangent line : y = 5 t + H.5 = 5 5 + =.5 ft. pt for dh at t = dt b. Find d H dt in terms of H. Use d H to determine whether your answer in part (a) is an underestimate or dt overestimate of the height of the water behind the dam at t =.5 () d H dt = dh = dt ( H ) = 44 H Since H >, d H > on [,.5 ] dt So the graph of H is concave up and.5 feet is an underestimate. pt for d H dt with reason c. Find the particular solution H( t) to the differential equation dh H dt with initial condition = H = and use it to find an expression for the height of the water after years. (5) H dh = dt ln H = t + C ln = + C C = ln H = e t + ln H = e t H = + e, t 4 H( 4) = + e t pt for separation of variables pt for antiderivatives point for constant of integration pt for using initial condition pt for solution and H 4 Max : 5 if no constant of integration Max : 5 if no separation of variables www.mastermathmentor.com - 7 - Illegal to post this document on the Internet
74. Growth and Decay A pool of water that consistently maintains the shape of a cube by movable walls is shown in the figure to the right. Let s be the height of the water in the pool, where s is a function of tine t, measured in minutes. The volume V of the water in the pool is changing at the rate of s where s is the height of the water, 6 measured in feet. a. Show that ds dt = 8 s. () V = s dv dt = ds s dt = s 6 ds dt = s 8s = 8 s pt for dv dt = s 6 pt for dv dt = ds s dt pt for showing result b. Given that s = 9 feet at time t =, solve the differential equation ds dt = s ds = 8 dt s = 8 t + C ( 9) = 8 + C C = 8 s = 8 8 t s = 7 t s = 7 t 8 s pt for separation of variables pt for antiderivatives point for constant of integration pt for using initial condition at t = pt for solving for s Max : 5 if no constant of integration Max : 5 if no separation of variables for s as a function of t. (5) c. At what time t is the pool empty? () 7 t = t = 4 minutes www.mastermathmentor.com - 74 - Illegal to post this document on the Internet
75. Growth and Decay (Calculator Allowed) The park service that administers a state park estimates that there are 495 deer in the park. They decide to remove deer according to the differential equation dp dt =.P. a. Show that the solution to the differential equation dp =.P is dt P = 495e.t, where t is measured in years and P is the population of deer. Use it to find the deer population in 5 years to the nearest deer. () dp dt =. dp =.dt ln P =.t + C P = Ce.t t =,P = 495 so P = 495e.t P( 5) = 495e.5 pt for dp dt =. pt for solving DEQ b. After this 5-year period, no human intervention is taken and the deer population grows again. From that time, the deer population increases directly proportional to 65 P, where the constant of proportionality is k. Find an equation for the deer population P( t) in terms of t and k for this 5-year period. (5) dp dt = k( 65 P) dp 65 P = kdt or dp P 65 = kdt ln P 65 = kt + C and P 65 = Ce kt P = 65 + Ce kt = so = 65 + C and C = 5 P P = 65 5e kt pt for dp dt = k ( 65 P ) pt for separation of variables pt for antiderivatives point for constant of integration pt for using initial condition at t = Max : 5 if no constant of integration Max : 5 if no separation of variables c. Using the growth model from part b), year later the deer population is 5. Find k. () P = 5 so 5 = 65 5e k 5e k = and k = ln or ln 5 or.54 5 pt for k d. Using the growth model from part b) and the value of k from part c), find lim P( t). () t lim 65 5e kt t = lim 65 5 = 65 t e kt www.mastermathmentor.com - 75 - Illegal to post this document on the Internet
76. L Hospital s Rule f ( x). Let f ( x) = asin ax + bcosbx b, g( x) = acosax + bsinbx a. Find lim x g( x). () lim x f lim g = + b b a + a = ( x) ( x) = a cosax b sinbx f x g x x a sin ax + b cosbx = a b pt for lim x = f x g x. Explain why lim =. () bx e x x a x a lim x e = bx ax a lim = x be bx lim x ( a ) x a a( a ) x a b e bx a a lim = x b e bx a( a ) ( a )... lim = x b a e bx = a a =... = lim x a... b a e bx x x a pt for lim x e = bx pt for showing that numerator eventually goes to zero pt for showing that denominator gets larger. The figure to the right shows the function f ( x) = + cosx. Find the limit, as x approaches, of the ratio of the area of the triangle to the total shaded area in the figure. (4) lim x Area of triangle Area under curve = x ( + cosx) ( + cosa)da + 4xsinx + cosx lim = 4 x 4 + 4cosx 8 = x x = x + x cosx 4x + sinx = pt for area of triangle pt for area under curve pt for ratio = www.mastermathmentor.com - 76 - Illegal to post this document on the Internet