Calculus - Chapter 2 Solutions

Transcription

1 Calculus - Chapter Solutions. a. See graph at right. b. The velocity is decreasing over the entire interval. It is changing fastest at the beginning and slowest at the end. c. A = ( )(5) = 450 feet d. A = [( )(5) + (85 + 8)(3) + (8+ 70)(5) +!!!!!!!!!!(70 + 6)(0) + (6 + 48)(0) + ( )(5) +!!!!!!!!!!( )(0) + (35 + 5)(5) + (5 +9)(0) +!!!!!!!!!!(9 +)(0) + ( + 4)(0) + (4 + 0)(9)] = !feet e. No, they had to walk = feet.. See graph at right. a. Yes, we are not given information about the intermediate values. f() b. A = (5 + 0)(3) + (0 + )(3) + ( + 5)(3) +!!!!!!! (5 + 3)(3) + (3 + 6)(3) = 63.5!un c. The 3 represents the equal heights, the ½ is used to find the area of a trapezoid. 3 d. ( ) = A e. The first and last y-values occur only once while all other y-values are added twice. f. ( g() + g(4) + g(6) + g(8) + g(0) ) = ( + () + (5) + (4) + 3) = 6!un 3. a b c. (4! 0 " 3) + (4!" 3) +!+ (4! 4 " 3) + (4! 5 " 3) = "3 + +! d. (3! " ) + (3! " ) + 3(3! 3 " ) + 4(3! 4 " ) = a. He moved the inde p by half steps instead of by. b. He squared after adding each value instead of adding each square. c = = 50 Chapter Solutions

2 4 5. Sample answers: a. " 3! 4 i b.! i c.! f (i) i= 6 i= a. See graph at right. b. The answer should be between 36 and 48. Using trapezoids: (4 + 8)() + (8 + )(5) + ( + 6)(6)! $ " # +(6 + 0)(4) + (0 + 8)(7) + (8 + 6)(4) % & ( ) = 389 = y 7. a. See graph at right. Horizontal asymptote: y = 0 Vertical asymptote: = 4 b. y = + 4 = f! () Yes, f passes the horizontal line test, and f! has at most one output per unit. c. See graph at right. Horizontal asymptote: y = 4 Vertical asymptote: = 0 It is a reflection of f() across the line y =, and the asymptotes are reversed from those of f. 8. This is a sphere with two connected cones cut out of the sphere s interior. V = 4 3!r3 " # 3!r3 = 4 3! 33 " # 3! 33 = 36! " 8! = 8!!un ! + 3.5! + 4.5! + 5.5! " 5.384, which is very reasonable. 0. There is a triangle below the -ais: A ( f,!! ) = ( ) ("5)(0) = "5!un. a. (!4) + (!3) + (!) + (!) = 60 b. (!4) 3 + (!3) 3 + (!) 3 + (!) = 0 c.!3 +! +! = d. sin(!5) + sin(!4) +!+ sin(4) + sin(5) = 0, since sin(!i) =! sin(i) so all terms cancel out ecept for sin 0 = 0. Calculus Solutions

3 . a. even b. odd c. nether d. odd 3. a. When the velocity changes from positive to negative: t = 8 seconds b. When the v(t) is the greatest: t = seconds c. This is the area A(v, 0! t! ) = " 3 " 0 = 7 ft (by adding every secs) d. (i) ft s (ii) 0 ft s (iii)!.5 ft s 4. a () + 3 = 7 b. 7 c.! = 3 d. It is not continuous, or jumps, at =. 5. One way to do this is by realizing: # 0 & " ( j! ) = " j + " (!) = % " j (! 9 = 579 j= j= j= $ j= ' (A calculator could be used at any of these steps.) 6. One method is to use the sequential function g(i) = g(i! ) + " cos (#( i) ) with g(3) = 0. The sum is g() = 4. It is not hard to get this answer by canceling out terms such that only the last term is left. 7. a. A ( f,!! 5) " f () + f (.5) +!+ f (4.5) = b. For the rectangle, base = and height = f (), so A = bh = f (). A ( f,!! 5) " f () + f (.5) +!+ f (4.5) c. i. 0 to 5 gives 6 total rectangles. ii: iii. d. 4 3, the same. 8 The width of every rectangle. The guarantees that the first height is evaluated at. Since sigma notation evaluates integers only, ( + i) argument allows for evaluation at every unit from along the domain. 8. a.! 4 f (4 + 4i) b.! 3 f (4 + 3 i) c. 8! 9 f ( i) 3 d. The epanded sum could not be factored and simplified if the widths were not equal. Chapter Solutions 3 8

4 9. a. A = 4() + 5(3) + 0() + () + () = 67!un b. A = 4() + 5() + 0() + () + () = 84!un 4! c. f (i) 0. They are the same, only the indices are changed.. a. 300 calories, the area under the rate function for the first hour. b. This is the full area: = 75 calories total calories c. = 75! cal hr total time ". f (i! 4) 3.! cos = a cos(b " 0)!! + 3 = a cos(0)!! = a! a = 3! cos(4" )! 4 = a cos(b # " )!!! 4 = 3 cos("b)!!3 = 3 cos("b) cos("b) =! In order for there to be one peak, b = a. b. c.. 5. a. As! + 3 or! " 3,!y! 3 : there is a hole at 3, 3 As! ",!y! ", and as! "#,!y! "#. 4 Calculus Solutions

5 6. This is a trapezoid with vertices (0, 5), (0, 7), (, 5), and (, 7). A ( f, 5! y! 7) = " + = 3!un 7. If f were not strictly increasing or decreasing, then it would repeat some y-value for two different -values and it would not pass the horizontal line test, so f! would not be a function. So if f! is a function and f is continuous, then f must be strictly increasing or decreasing. 8. A! " f #" #" #" " " " ( ( ) + f ( 3 ) + f ( 6 ) + f ( 0) + f ( 6 ) + f ( 3 ) + f ( )) Since cos() is an even function, compute the area on " # 0,! $ % and double it. 9. a. A ( f,!! 5) " 4 f () + 4 f (.5) + 4 f (.5) +!+ 4 f (4.75) " b. A f,!! 5 # 4 f + i " c should be more accurate because it comes from twice as many rectangles (of equal spacing). 30. a n =,!! = 7 3 b. n = 0,!! = 8 0 = 4 5 = [! # f (a +! #i] :!A f, 3 " " 0 :!A f, " # # 6 3.! 3 f 3 + i or 3! 3 f i 3 i= n n$ % = % 3 f 3 + i 3 n" = [! $ f (a +! $i] 3. # [! " f (a +! " i] is the best answer. i= 33. Xavier is incorrect. 34. a. f (.5 + i) 3! % = % 5 f " i b. It is a pile of thin cylinders, with increasing diameters, stacked sideways. c. V! "( 0.5) + "(.5) + "(.5) + "( 3.5) = 8"! 5.33!un 3 Chapter Solutions 5

6 35. k() = g ( f ()) = log = log / = log 36. The slope gradually increases and approaches! as approaches. Afterwards it gradually decreases but stars positive. Towards!" and! the slope approaches. As! "#, y! "# ; as! ", y! " ; as! ", y! # ; as! +, y! "#. The end behavior function is y = The argument cos! p 0 repeats after 4 values of p: ( ) = 0 cos! + cos! + cos 3! = 0 0 " $ cos! p + cos! = 0 # 0 = 0 p= 38. See sample graph at right. The inverse of an odd function is odd, but it may not be a function. 39. This is a hemisphere with a radius of 3. V = 4 ( 3! " ) 33 = 8!!un 3 " For left rectangles: A( f, 3!! 7) " # f 3 + i 4!un 7 8 " 6.56 For right rectangles: A( f, 3!! 7) " # f 3 + i 3!un For midpoint rectangles: A( f, 3!! 7) " f i 5 i= 7 # " 6.59!un 4. a.! (3 + i) b. " i cos i! c. " i= 5 i= 4 5 y f (i! ) 4. a. The data lies on the line Price = (year-978), from which we can predict a price of $,500 for a 988 Mustang, which seems like a very reasonable prediction according to the data. b. The data lie on the line Price = (year-990), which gives an estimate of $3,00 for a 988 Mustang. c. No. d. This can help them make a better prediction of a fair price if they are willing to compromise away from he lines found in parts (a) and (b). 6 Calculus Solutions

7 43. a. As the year approaches, 988 from the right, we predict the asking price to be $3,00. b. The left and right limits would have to be equal. 44. Yes, it looks like lim! f () " No, because a limit is a prediction based upon surrounding values of, but not the actual value of f(). 46. a. As approaches 5, f() approaches 6. b. As approaches 3 from the right, g(h) approaches!". 47. See graph at right. a. As! +, y! ". As! ", y! ". b. lim!" 3 # # = " 48. No; the function could have left and right limits which are equal but where there is a hole. 49. a. As approaches from the right, + + 3approaches 3. b. As we let time (or the amount of time that passes) approach infinity, a soda s temperature approaches room temperature. 50. a. For f, D = { : < 3} and for g, D = { :! 3}. b. See graph at right. c. In order to be continuous the functions must have the same value at = 3. f (3) is undefined and g(3) =!. { }. d. For f, R = (!", "), and for g, R = y : y! " 5. a. See graph at right. Graph is scaled by s. As! "#, y! "# ; as! ", y! 7 ; as! +, y! 7 ; as! ", y! ". Solution continues on net page. Chapter Solutions 7

8 5. Solution continued from previous page. b. See graph at right. Graph is scaled by 4 s. Vertical asymptote: = ; slant asymptote: y = + 7 As! "#, y! "# ; as! ", y! "# ; as! +, y! " ; as! ", y! ". c. The factor cancels out, causing a hole. d. Part (a): The function simplifies to b() = + 6 Part (b): Epanding the numerator and using long division (we do not need the + 7 remainder):! + 5! 6 Therefore, b() = + 7.!!!!!!!!7! 6 5. lim!"# f () = 0,! lim f () = "0.5,! lim f () = 0.5,! lim f () = 0!" "!" +!# = ( +.5)! 4.5, so the verte is (.5, 4.5), and the line is y =! 7 ( +.5)! a. f! () = 3 b. f ( f! ()) = c. f! ( f ()) = " 55. a. A( f, 0!! " ) # $ 6 sin " + i " #.03!un 6 5 b. Each rectangle becomes a thin cylinder. 5 " c. V! $ 6 #" # sin " + i " = % " 6! 4.935!un3 56. a. Point c: speed! 4 ft/sec b. Point a: velocity!.3 ft/sec c. Speed is the rate of travel in any direction and is never negative. Velocity is the rate of travel in the (fied. positive direction, and it may be negative. 8 Calculus Solutions

9 57. a.! b. 0 c. d. e. lim!" f () equals a constant and that constant is the y-value of the horizontal asymptote. f. See graph at right. lim!" f () = # lim!"# f () = " $ y!! " 58. a. f() approaches! on each side: lim f () = #,! lim f () = #!0 "!0 b. Yes. + c. lim!0 f () = " or the limit does not eist. 59. a. The right-hand and left-hand limits are different, so the limit at does not eist. b. Again, the limits differ from the right and left. 60. See graph at right. The right and left limits differ, lim f () = "!0 " and lim f () =, so lim f () does not eist.!0 +!0 6. a lim!a " lim f ()!a + lim f ()!a f (a) DNE DNE.5 4!!! DNE 6. a. No, continuity ensures that lim!a f () eists. b. No, it does not make sense. c. Not necessarily, because lim!a f () and f(a) can both eist but may not be equal. d. See the Math Note in Lesson..3 for a list of conditions. 63. a. 3 b.!" or DNE c. d. e. undefined f.! or DNE g. 4 h. DNE i. No, there is a vertical asymptote. Chapter Solutions 9

10 64. a. Yes, based on the continuity of temperature. b. 65. a. See graph at right. The temperature decreases toward a constant value. b. As time approaches infinity, the temperature of the coffee approaches room temperature. temp. time 66. a. D = { : is real},!r = { y : y! } b. D = { :! "3},!R = { y : y! "} 67. See sample graph at right. The graph has horizontal asymptotes: y = and y =. 68. A line. # f (! + i) 69. A f,! " " 6 7 $ = $ (! + i) cos(! + i) #!4.66!un 7 # 70. A f,! " " 6 This is a different result. 5 f (! + i 5 $ ) = cos (! + i ) $! + i #!3.49!un 7. a. We can rewrite the function as approach 0 and f ()!. b. 5 c. Does not eist. d. (!5)!3 (!5)+5 =!8! " = " As gets bigger and bigger 3 and e. (m!3)!3 (m!3)+5 = m m+8 = m m+4 f. +h! 3 +h+5 g. These are the horizontal asymptote. 0 Calculus Solutions

11 7. a. It is the slope. b. When t is between and 4 seconds and between 8 and 0 seconds. c. 40 ft/s (between t = 5 and 6) d. The car s velocity is constant. e. This is the area: A! 40() () + 0() (.5) + 60(.5) () + 40() = 490 ft 73. Sample graphs: a. b. c. 74. For =, the third condition. For = 0, all conditions. For =, the st and 3 rd conditions. For =, the nd and 3 rd conditions. These are the four basic types of discontinuities: all conditions fail, or there is no limit and the st and 3 rd fail, or f(a) does not eist and the nd and 3 rd fail, or just the 3 rd fails. 75. As g goes continuously from ( 4, 0) to (, 3), it must cross the -ais somewhere and have a root. It cannot jump the -ais if it is continuous lim = 0 therefore lim!0 "!0 +!0. 0 = 0 by definition. 3. lim!0 = a. True b. Yes, the rate of the car as it slows is continuous; therefore, all intermediate values between 65 and 0 must occur. 78. If f (a) < y < f (b) then f cannot hop over the line y = y ; it must cross it. So any y-value between f(a) and f(b) must be reached at least once. 79. The function is not continuous between and. Chapter Solutions

12 a. 0 seconds: 40 ft 0 s = 4 ft/s, 5 seconds: b. 50, 48, 46, 44, 4, 40, 38, 36, 34, 3 (in feet) Velocity is decreasing in a linear manner. 30 ft 5 s = 46 ft/s c. Since velocity is linear, the height will be quadratic a. " 3 f! + 3 i b.! f 6 + i Since f is continuous, f () =!4, and f (3) =, it must have a root between and 3 by the Intermediate Value Theorem. 84. a.! f ( + i) b.! 3 f + i c. 3! 00 f + i a. See graph at right. The y-ais is scaled by 5 s. 0 rectangles. b. { :!3 " " } c. A ( f,!3 " " ) # 9.375!un 86. Total books sold = 300! ! = 3900 Average = 3900 books = days 7 books/day 8. a. lim!0 y = 9 b. lim!3 f () = " c. Only the limit in part (a) eits. Infinity is not a value for a limit that eists. Calculus Solutions

13 88. a See graph at right. b. At =,, and 3. y 89. a. When the epression is evaluated at =, the result is undefined. b. There is a hole. c. lim! + = + = 4 = d. It simplifies the fraction to a form that can be evaluated. 90. a. Factor and cancel; hole b. Substitute in a value like.9; vertical asymptote c. Multiply by the conjugate of the denominator; hole d. (a) and (c) have limits that eist because the graphs have holes. (b) has a vertical asymptote and therefore the limit does not eist. 9. a. lim " (")(+) = lim = lim ( + ) = ( + ) = 4! "! "! b. lim!" #7 +4 # = lim!" # # = = c. lim 3 "7!0 3 3 = (0) 3 "7 +4 "5 3(0) 3 = 7 +4(0)"5 5 " d. lim! " = lim " = lim! ( +)( ")! e. lim 5!4 " "4 + = + = : This cannot be simplified. The graph has a vertical asymptote. Substituting in values of 3.9, 3.99, 3.999, etc. can show that the graph approaches negative infinity and therefore the limit does not eist. f. lim "!"# + = lim!"# " +! lim! lim = "#!"#!"# Again, there is a vertical asymptote and therefore the limit does not eist. 9. a. H.A. at y = 3 ; lim!" V.A. at = 5 ; lim!5 b. H.A. at y = ±! ; lim!" c. H.A. none; lim!±" f () = 3 ; lim!"# f () = 3 f () = " ; lim f () = "# + " f () = # ;!5 lim!"# f () = " $ ; No V.A. or holes f () = " ; V.A. none; hole at = ; lim! f () = "4 Chapter Solutions 3

14 93. Graph must be odd, go through the origin and have horizontal asymptotes at 5 and a. undefined b. DNE; vertical asymptote c. d. e. f. DNE; Right and left limits are not equal. g. 4 h. DNE; lim! i. j. 4 k. No. lim!4 f () " f (4) f () " f () 95. Let y = f (). This is a semicircle of radius. a. Sample answer: A! 3 = 3 f (" 4 3 ) + f " 4 ( 3 ) + f ( 3 ) + f ( 3 ) ( )! 4.50!un 5 b. Sample answer: A! () ( f () + f (0) + f (0) + f () ) = ( )! 7.464!un c. The average from parts (a) and (b) is un. (The eact answer is!.) 96. The limit as approaches 3 of f() appears to eist because the limit appears to agree from the left and the right. Both the left and right limits approach 7. This does not appear to be the case for g(). The left limit approaches 7, while the right limit approaches I must be true, II must be false, III could be true. 98. Try eact numbers to find the answer. I. log 4 = greater than II. log 4 =! less than III. log 4 = between 0 and IV. log = equal to = 3 3(0+) + 3 = = 7 = = 3 3(y+) 3 = 3 3y+3 = 3y + 3! = 3y! 3 = y The intercepts are (4, 0) and ( 0,! 3 ). 4 Calculus Solutions

15 00. f () g() =!9!+8 = (+3)(!3) (!3)(!3) = +3 (! 3) g() f () =!+8 = (!3)(!3) = (!3)!9 (+3)(!3) (+3) H.A. at y = 0, V.A. at = 3 H.A. at y =, V.A. at =!3, hole at (3, 0) 0. This is a cone with a radius of 6 and a height of 6. V = 3! " 6 " 6 = 7!!un 3 0. a. Distance measurements over time. b. v(0) = 0, street has a constant slope, and that the street has a straight path. 03. a. A quadratic should be best. c. Velocity increases due to gravity s acceleration of the ball. d. If the speed was constant, then the distance graph would be linear. e. It is slowest at the beginning and fastest at the end. At fast speeds the curve is steeper, and a slow speeds it is flatter. f. This is an estimation of the slope at t = 4 seconds, e.g. as f (5)! f (3) 5! a. See graph at right. b. min = 35 mph, ma = 50 mph c. This is area under the curve: A = 35+50! =.5 mph miles per hour 05. a. Even, because cos(3) = cos(!3). hours b. Even, because! =! (!). 06. a. = y +! = y! = y f! () =! c. g! ( f ()) = (( + )! ) + 3 = (! ) + 3 = 4! b. = y! 3 +! = y! 3 (! ) = y! 3 (! ) + 3 = y g! () = (! ) + 3!for! " Chapter Solutions 5

16 07. The slope is always positive and smoothly decreases at first to almost zero then increases, then decreases to almost zero, then increases.,!!! < < 08. a. " First we see! < 0!"! $ <!"!! < <. So, h() =!! # %$!!!!!,!!!!!otherwise b. D = {: is real}, R = {y: y 0} c. e.g., 8 left-side rectangles: A(h,! " " 3) # $ h!+ i = 6!un 3 d. A(h,! " " 3) # $ 6 h!+ i = 7 6 3!un 09. a. a = 6, b = 0, n = 8 b. a = 4, b = 5, n = a. The polynomial in the denominator has a larger degree than the polynomial in the numerator, so the limit is 0 as! ". b. The big terms are in the numerator and denominator. We can rewrite this as! " " " = ". The limit as! " is. c. The polynomial in the numerator has a larger degree than the denominator, so the limit is!. { } { }. a. h() = ( + )! 3 + =! +,!!D = : " b. k() = ( + ) + = 4 + 3,!!D = : is real. a. Eric Samantha Kirt Lisa Joshua collision time 0 sec sec sec 9 sec 34 sec?? distance 0 ft 4 ft 569 ft 990 ft 357 ft 760 ft See graph at right. b. Quadratic, approimately d(t) =.7t t! c. 760 =.7t t! =.7t t! t "! !4(.7)(! ) " 38.7 sec (.7) d. The slope must be estimated at 38.7s, as about 90.8 ft/s. She is not eligible ft sec! 60 sec min! 60 min hr! mi 580 ft! 6.9 mph e. The car is traveling fastest at the time of impact. The steeper the graph, the faster she travels. 6 Calculus Solutions y

17 3. a. Marlayna zoomed in closely at = 0. b. Most functions appear to be locally linear. Sample functions: y =! sin, y = tan, y = (! 3)( + 3) c. Answers vary. Sample functions: y =, or y = 4. See sample graph at right. 5. a. v(0)! 0.7 m/s,!v()! ".5 m/s,!v()! m/s b. Yes, at about t = 0.5,.5, and 3.75 seconds, where the graph levels off and changes direction. c. For 0.5 < t <.5 and 3.75 < t < 4.75, approimately. 6. This is the area under the rate curve. Using trapezoids: A ( rate, 0! t! 30) " 4 # # # # # # 4+6 " f 0 + i! " f 0 + i! " ! 0 f 0 + i " As the number of rectangles increases, the area approaches the actual area a. lim "9!3 "3 = lim (+3)("3) = lim ( + 3) = 6!3 "3!3 b. lim!" 6 # 4 # = lim!" 6#/ 4#/ c. lim #!" = lim # = lim!"!" d. lim 3 ##!" ## = lim!" # # # = lim!" 6 4 = = 0 (eponential decay) # = lim!" # = #" e. lim!" sin DNE; The function continuously oscillates between and. f. lim sin = 0 because the numerator stays between and, but the denominator!" becomes larger and larger. 9. a. 0 =!6t + 500,!!6t = 500,!!t = 500 b. h(5.0)!h(5) 5.0!5 " 98.4! =!60 ft/sec 6 " 5.59 sec Chapter Solutions 7

18 0. Possible graphs: a. c. b. Not possible; the limit is a condition for continuity.. It is a cylinder with a hole. h "! ( r inside ) h =!(3)( ) "!(3)( ) =! " 3! = 9!!un 3 V =! r outside. b. Not quite linear, only approimately linear. The function is still a curve. 3. The answers below refer to the slopes of the lines when the calculator is zoomed in, not the actual equations of the lines. a. y = :!! =! " y =!4,! = 0 " y = 0,!! = " y = 4 b. y = sin :! =! " y #!,! = 0 " y =,! = " y #! c. y =! cos :! =! " y #!,! = 0 " y = 0,! = " y = d. y = 3 :! =! " y =,! = 0 " y = 0,! = " y = e. y = :! =! " y =!,! = 0 " not linear, = " y = f. y = + :! =! " y =,! = 0 " y =,! =, y = 9 g. Answer will vary, but it is a sharp point on the curve y = sin y = b. The approimation is good near = 0; the linearization is a good estimate near the point of linearization. c. Under on!", 0 and over on ( 0,!). d. ; Zoom in on the graph at = 0. e. 0; Zoom in on the graph at = 0. 8 Calculus Solutions

19 5. a. Even: sin = (! sin ) = sin(!) b. Odd: (!) + (!) 3!(!) = +! 3 + =! + 3! 6. a. See graph at right. D = [0,!),!!R = [0,!) b. Yes. = 0+40 c. A f, 0!! 8 " " " 3 = 590 Which represents the approimate amount of customers in this 8-hour period. f() 7. a. The plant grows at a relatively steady rate until around 4 months when there is a sharp increase in growth. Around 4.5 months the rate starts to level off again. b. When 4! t! 5 (approimately). By looking at where the slope is steepest. c.! 3 ft/month, by estimating the slope of the tangent line at = 3. 4 d. Take the total growth over total time:! 8!ft!.45 ft/month 5.5!months 8. It is a stack of sideways cylinders, each with a height of : V =! " +! " 3 +! " +! " = 8!!un 3 9. The slopes are negative reciprocals of each other. 30. f (!3) must be lim "9!"3 +3 = lim (+3)("3) = lim ( " 3) = "6!"3 +3!"3 # Therefore f () =!9 %!!for! "!3 +3 $. &%!!!6!!!! for! =!3 3. a. lim!" 9 #3+ 4 # = lim!" b. lim +!" = = 0 c. lim #3#4!" (#3) = lim!" 9 #3+ (4 #) = lim!" / $ #3#4 / #6+9 = lim!" 9# 3 + Chapter Solutions 9 4# # 3 #4 # 6 +9 = # 4 9 = lim!" 3 4 = 0 "4 d. lim! + " = lim (+)(") = lim + ) = 4 but lim " ( + ) = "4! + "! +(! " Therefore the limit does not eist.

20 3. a. 0 < <! b.! < <! c.! < <! d. 0 < <! e. 0 < <! 33. a.! = b"a n b. It decreases (in inverse proportion). " b#a 34. A f, a!! b n# $ f a + b#a n n i n! 35. b. 0 0 left = " f n n i, 0 right =! f ( 0 n n i) n i= c. Left would be over, right would be under. This is determined by where the graph is increasing or decreasing. d. E.g., for n = 0, A! 405 ft; As n increases, the right and left endpoint approimations approach each other. e. Use as many rectangles as possible. Left or right endpoint does not matter. Since as n approaches infinity, we get eact area either way. 36. See graph at right. a. The functions intersect at (, 4). The area should be approimated with rectangles or trapezoids for 0 and added to the area of the triangle for 4. b. Area = 6 3!un y + f ( 5 ) + f 3 ( 5 ) + f 4 ( ( 5 ) + f () ) 37. f (0) + f a. V avg = v(0.5)!v(0) 0.5!0 = 3(0.5)!3(0) 0.5 = =.5mph b. The distance is to school is 0.75 miles, so 0.75 must equal the triangular area under the velocity curve:!0.75 = (t)(!t)!.5 =!t 0.75 = t " t 0 Calculus Solutions

21 39. f (b)! f (a) b!a 3 b!a " 40. a. b!a f (a b!a " 9 b!a " 0 i) 4 i) b. b!a f (a i) c. b!a 0 f (a + 4. See graph at right. 4. a. lim ( " 4 " 5) = 4 " 4 " 5 = "5!4 + b. lim "4" ("6)(+) = lim = lim ( " 6) = "8!" +!" +!" c. lim "4" ("6)(+) = lim = lim ( " 6) = 0!6 +!6 +!6 d. 0; The denominator becomes larger and larger, resulting in an infinitely small fraction. e. lim +6" 7 +6"7 = lim = lim ( + 6) = "#!"#!"#!"# f. lim # 7 #7#0 # 0!" = lim =!" 43. At = 3 and at = 4, all conditions fail. 44. a lim f ()!a f(a) Continuous at f(a)? yes yes 3 none none no 4 none no 45. a. b. c. d. 7 ( (3 )!3+3)! ( )!+3 = 8! ( 8 + ) = 9 ( (. )!.+3)!( ( )!+3) ( = )! ( (.0 )!.0+3)!( ( )!+3) = = ! ( 8 + ) = 7.0 Chapter Solutions