( ) Chapter 1. Functions, Graphs, and Limits. 1.1 Functions. 0for all real numbers t the domain of ht () = t 2 + 1consists of all real numbers.

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1 Chapter Functions, Graphs, and Limits. Functions. ht () (t+ ) h( ) ( + ) h(0) (0+ ) h() (+ ) 7 4. f( ) f() f(0) 0 f( ) 0+ ( ) + 6. gu ( ) ( u+ ) g(0) (0+ ) g( ) ( + ) 0 ( ) g(8) (8+ ) g ( ) 4+ g( ) 4+ 6 g(0) g() if h ( ) + if > h() + 0 h() + 4 h(0) h( ) ( )( ) Since 0for ±, f() is defined only for ± and the domain does not consist of the real numbers. 4. The square root function only makes sense for non-negative numbers. Since t + 0for all real numbers t the domain of ht () t + consists of all real numbers. 6. f( ) The domain consists of all reals t f() t t t t t ( t )( t+ ) 0 if t and t. 0. hs () s 4 is defined only if s 4 0or equivalently ( s )( s + ) 0. This occurs when the factors ( s) and ( s + ) are zero or have the same sign. This happens when s or s and these values of s form the domain of h.. f( u) u f( ) ( ) f( u) (u+ 0) [ ] f( 5) ( 5) + 0 ( 0+ 0) 4 6. f( u) u f( + ) + 8. f( u) u, f ( ) 0. For f( ) +,

2 CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS f( + h) f( ) h (( + h) + ) (+ ) h + h+ h h h. For f ( ) 4,. The last equation has solution which is in the domains of f ( g ( )) and g( f( )). Thus f ( g ( )) gf ( ( )) only for. 8. f( ) + 5 f( + ) ( + ) f( + h) f( ) h (4( + h) ( + h) ) (4 ) h 4 + 4h hh 4+ h 4hhh 4h h 4. f ( ) +, g( ) f( g( )) ( ) + + and g( f( )) ( + ) So f ( g ( )) gf ( ( )) means but, by the quadratic formula, this last equation has no solutions. 40. f( ) (6) [ ] f( + ) ( + ) 6 ( + 6 6) 4 4. f( ) + f () f( ) 0 f( + 9) ( + 9) f( ) ( + ) 5 gu ( ) u h ( ) f ( ) 5 can be written as g[ h( ) ] with gu ( ) uand h ( ) f( g( )) / g( f( )) + + so f ( g ( )) gf ( ( )) means, Clearing denominators, multiplying, and collecting terms gives ( + )(+ ) (4)(4 ) ( + ) 6( ) f( ) + 4. g( u) u ( + 4) u h ( ) (a) Cq ( ) q 0q + 400q where q is the number of units. Thus C(0) (0) 0(0) + 400(0) $4,500. (b) The cost of manufacturing the 0 th unit is C(0) C (9) 4, (9) 0(9) 400 (9) 500 4, 500 4,9 $7

3 .. Functions t 54. (a) Ct () + 4t degrees Celsius, where t represents the number of hours after midnight. Thus t 4 at :00 p.m. and (4) C (4) + 4(4) (b) The difference in temperature between 6:00 p.m. ( t 8) and 9:00 p.m. ( t ) is C() C (8) () + + 4() 0 6 (8) + + 4(8) (a) f( n) +. The domain consists of n all real numbers n 0 (because of the denominator). (b) Since n represents the number of trials, n is a positive integer, like n,,, (c) For the third trial n, thus f () + 7 minutes. (d) f( n) 4, so + 4, or n n n (e) gets smaller and smaller as n n increases. Thus 0 as n and f( n) gets n closer and closer to. No, the rat will never be able to traverse the maze in less than minutes (a) f( ) 00 The domain consist of all 00, since denominators must not go to 0. (b) represents a percentage, so 0 00, so that f( ) 0, or better, 0 00, since books need not be distributed to more than the rural population. 600(50) (c) If 50, f (50) (d) With 00, 600(00) f (00) (e) With f( ) 50, , , 60% (a) N 9. / 00.4 elk per square kilometer. 0.7 (b) N 9. / m < when / 0.7 m > An animal of this species has an average mass of at least 48 kg. (c) Let A denote the area, in square kilometers, of the reserve. If the second species has average mass m, then since 00 is the 0.7 m A A number of animals per square kilometer. The first species has average mass m and so the estimated number per square kilometer is ( m) m A A Since there are A square kilometers on the reserve, there are approimately 60 of the heavier species. 6. (a) Ht ( ) 6t + 56 H() 6() feet. (b) During the third second, the ball travels

4 4 CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS 64. (a) H() H () 9 + 6() feet. (c) H (0) 56 feet. (d) Ht ( ) 0 when 6t , 6t 56, t 6, t 4. (Disregard t 4). Cq ( ) q + q+ 900 and qt ( ) 5 t, thus C q(t) C(5 t) [ ] (5 t) + 5t t + 5t (b) For t, Cq [ ()] 65() `+ 5() $6,600. (c) 65t + 5t+ 900, t + 5t 0,00 0 Divide by 5 to get smaller numbers, then 5t + t 404 (5t+ 0)( t4) 0 or t 4 hours. 0 Disregard t (a) R ( ) D ( ) ( ) P ( ) R ( ) C ( ) ( ) ( ) (b) Since P ( ) , the quadratic formula tells us P ( ) 0 when 5 and.. By evaluating P ( ) at a number of values, or by graphing the function, it is easy to 68. (a) see that P ( ) > 0, that is the commodity is profitable, for 5< <.. CEI Year/Sector -yr -yr 4-yr 4-yr Public Private Public Private Average annual increase For eample, in 00, the average increase in CEI for -yr public institutions over the given period was (b) Writing eercise, answers will vary. (c) Writing eercise, answers will vary. 70. VCI by year and education level: Year/Level of Education -yr Public -yr Private 4-yr Public 4-yr Private For each level of education, the VCI for 995 is not significantly different than that for 000. Within each year, the VCI for a - yr public school education is much larger than that for a 4-yr public school education,

5 indicating that even though the starting salaries are lower, there is a much bigger return on investment in the -yr case than in the 4-yr for a public institution education. The large difference between the VCI for a -yr public education and a -yr private education (similarly for 4 years of education) should not assumed to be significant. Since the starting salary figures represent a combination of both public and private school educations, you cannot assume there is no financial rewards from attending a private over a public school y + 4 is not defined when Graphing this polynomial and looking for intercepts yields. This could also have been found by inspection. The domain is all numbers such that. 74. f( ), g( ). f (4.8).8.90 g( f(4.8)) g (.90) The Graph of a Function. (a) polynomial (b) different (since is a non-integer power of.) (c) rational function (since ( )( + 7) + 4 is a polynomial.) 6. f( ) 0 when y 0, y 0 and the curve is symmetric with respect to the y ais. The general shape of this curve is like that of problem., but it is steeper. 4 (d) rational function (since + 9 (+ 9) ( ) and the numerator and denominator epand to polynomials.) 4. f( ) 0 when y 0, y 0 and the curve is symmetric with respect to the y ais. 8. f ( ). Note that the graph is a straight line. The slope is. The curve falls. 0 f( ) 0 4 5

6 6 CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS 4. f( ) +. Note the similarities between this graph and the one in eercise. The y-values here are the negatives of those in and the curve is translated (moved up) by unit. 0. f( ) ( )( + ) 0 when y, and y 0 or. when 0 5 f( ) f( ) + 8 ( + 4)( ) When or 4, y 0. When 0, y8. 6. if < f( ) if The -intercept is. The y-intercept is f (0). Some points on the curve are: 0 f( )

7 .. The Graph of a Function 7 The slopes are the same, namely, so the lines are parallel, but the y intercepts differ. These lines do not intersect.. y and y. Thus, ( ) ( ) 0 ( )( ) 0, and y 0. P(,0) is the point of intersection (really the point of contact) if f( ) + if > 0 when y 9. There are no intercepts. 4. y 8 and 5y Multiply the first equation by and the second by, to get y. Thus 8+ 6,, P(,). 0. y + 8 and y 6. (a) (0,)

8 8 CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS (b) (,0) and (,0) (c) Largest value of.5 at. (d) Smallest value of at. 8. (a) (0,0) (b) (,0), (0,0) and (.5,0) (c) Largest value of at and 4. (d) Smallest value of 4 at. (b) Ht ( ) will be 0 when t 0 (c) The maimum height appears to occur at t 5 when the object is at a height of 400 ft. 4. Sr () CR ( r ) S(0) CR and S( r) 0 when r R 0. Number of copies sold n ( ) 0( ) at dollars each and bought for $0 each. Revenue R ( ) 0 ( ) cost C ( ) 0 0( ), profit P ( ) R ( ) C ( ) 0( )( 0) Relevant values are 0. The maimum profit occurs at $6 per book.. (a) Ht ( ) 6t + 60t 6 t( t + 0) 6. (a) 7 if 0 w 60 if < w 8 if < w 06 if < w 4 9 if 4 < w 5 Pw ( ) 5 if 5 < w 6 75 if 6 < w 7 98 if 7 < w 8 if 8 < w 9 44 if 9 < w 0

9 .. The Graph of a Function 9 (b) 4. (a) C ( ) (b) C ( ) 90 + (c) 8. The verte of the parabola is the point (00, -5000) and this point should be included in the rectangle. One such viewing rectangle is [50,50] [-8000,5000]. 40. (a) Each y-value for y is the negative of the corresponding y-value of y. Hence the points on the graph of y are reflections across the -ais of the points of the graph of y. (b) If g ( ) f( ), the graph of g ( ) is the reflection across the -ais of the graph of f ( ). 0 y y The effect the additional term has is to pull the low point of the curve further to the right and down.

10 0 CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS If A > 0, the quantity B C B A + + A A 4A has no largest value and will be smallest when the squared term is as small as B possible which is 0 when A. Similarly, if A < 0, there is no smallest value and the largest value occurs at The etra terms create a valley further to the right and down from the low point of the original curve. 46. [Graph..46] no art file provided From the picture in the tet and the Pythagorean theorem, ( ) ( ) d + y y ( ) + ( y y) Note the absolute value signs are necessary if the point (, y) is to left of or below (, y ). Upon squaring the absolute value signs are not needed. 48. First note that B C B A + + A A 4A + B + B C B A + A A A 4A + B C + A A + B + C A A B A. 50. The function is defined for all values ecept for 0.68 and.68. These values can be determined with a graphing calculator or eactly by applying the quadrtaic formula to ± + 0to find 5.. Linear Functions m ( ) ( ) m 0 5

11 .. The Graph of a Function 6. 8 ( 5) m ( 7) The graph descends 5 units as you move units to the right of the origin. Thus m 5. The y intercept is (0,5) so b 5. 5 The equation of the line is y The graph descends units as you move.5 units to the right on the ais. Thus 6 m. The y intercept is (0,-) so.5 5 b. The equation of the line is 6 y y 4 or y m and b 5 5. y 5 Thus m 0 and b 5. There is no intercept. 4. y 6, m y-intercept b y + 5 ( + ) 5( y ) y+ 50 5y + y Thus m and b. The -intercept is 5 5.

12 CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS the point-slope formula yields y ( + ). 4. Writing + 5y in slope-intercept form yields y +. The slope of a perpendicular 5 5 line is m 5. Using the point, in the point-slope formula yields 5 y y + y y + y 0 5, y (a) Let denote the number of miles driven and C ( ) the corresponding cost (in dollars). C ( ) The slope is not defined and there is no y intercept. Thus. 6. y y 5 7 4, y y ( ) y The slope of the line through (, 5) and (, 4) is not defined. We deal with a vertical line. Its equation is.. Writing + y 5 in slope-intercept form yields 5 y +. Therefore, the given line has slope m. Using the point (,) in (b) The rental cost of a 50-mile trip is C (50) (c) , , 67., or rounded to 68 miles. 8. (a) Since 50 is the cost per week, then the value of the missed week is 50. Therefore the cost of the remaining weeks is 50 F ( ) + 50

13 .. The Graph of a Function (b) The fee after 5 weeks is (b) V (4),900(4) + 0,000, F (5) (5) + 50 $ N if,000 N 0, (a) FN ( ) 40N if 0,00 N 0, (a) Let denote the age in years of the machinery and V a linear function of. 5N if 0, 00 N 50, 000 At the time of purchase, 0 and (b) V (0) 0, 000. Ten years later, 0 and V (0),000. The slope of the line through (0,0000) and (0,000) is,0000,000 m, Thus V( ), , 000. V( ) 0 when 00 and V( ) is 9 00 valid for 0. 9

14 4 CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS C + 8 and solve for 5 C 5(75 8) 85 chirps. If 7 chirps are heard in 0 seconds then C (7) 74 chirps are heard per minute.then T (7) degrees F. Note the segments with positive slope indicate the hare or tortoise is moving at a constant rate. The horizontal segment corresponds to the hare taking a nap. 46. (a) Let denote the number of days since the reduced rate went into effect and N ( ) the corresponding number of vehicles qualifying for the reduced rate. Since the number of qualifying vehicles is increasing at a constant rate, N is a linear function of. Since N (0) 57 (when the program began) and N (0) 47 (0 days later), 4757 m. 00 N ( ) + 57 for 0 (b) In 4 days from now N (44) (a) We have two points (00, 97) and (500,0) m N ( 00) or N (b) N (00) 0.05(00) , , 9. (c) Writing eercise, answers will vary. 5. Let be the number of ounces of Food I, and y be the number of ounces of Food II. Then will be the number of gms of carbohydrate from the first food, and 5y the number of gms of carbohydrate from the second food. Similarly and y will be the number of gms of protein from the two foods. + 5y is the total number of gms of carbohydrate, which must equal 7, while + y is the total number of gms of protein, which must equal (a) From the data it is easy to see that every time the number of chirps increases by 5, there is a o F increase in temperature, Thus T is a linear function of C and the slope of this linear function is /5. Using this value, the point (0,8) and the point-slope formula T 8 ( C 0) or T C (b) Set T 75 in the formula from part (a)

15 .. The Graph of a Function (a) If a liter of beer is % alcohol, then it contains 0.0,000 0 ml of alcohol. If alcohol is metabolized at 0 ml per hour, then hours are required. (b) A T 0 (c) No one can have A ml of alcohol with fewer than T hours left in the party y and y are parallel because 54 9 and (a) L and D are not linearly related. (b) Planet D L L D Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto L D / or L D D (c) Writing eercise, answers will vary. 60. Let s say that two lines being parallel means that they have no points in common. First, we will prove that if they are parallel, then they have the same slope.. Let s say the lines are y m+ b and y m+ cand m m. With some work, we find that both of these lines share the same point c b, mc m b m m m m Therefore they are not parallel.

16 6 CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS This contradicts our assumption that they are parallel. Thus m m and they must have the same slope. Now we prove that if they have the same slope, they are parallel. Let s again suppose that they are not parallel, i.e., that they have a point in common, say ( ab, ) and let (, y), (, y ) be points on the two lines. Calculating the slopes of these lines, we find m y b and m y b. a a But these slopes are different because y y. Therefore it must be that the lines do not have any points in common, i.e. they are parallel..4 Functional Models. Let be the width, then is the length. The area is A ( ) square units. 4. Let and y be the two numbers with the smaller. Since y 8, y 8 and the sum is S ( ) Let denote the length and y the width of the rectangular playground. Let P be the number of meters of fencing required to enclose the playground, then P + y. Since the area is y, 600, y,600 and 7, 00 P ( ) +. The graph suggests that P() is minimized near 60 meters. If so,,600 y 60 meters and the playground 60 is a square. 8. The rectangular bo is closed. Let be the length (and width) and y the depth. Since the volume, y is,500 cu. in.,, 500 y and the surface area is 6,000 S ( ) + 4y Let r and h denote the radius and height of the can, respectively. The volume of the soda can is V πr h 6.89π and so 6.89 h r. The surface area consists of the top and bottom circles, each of area πr, and the curved side. The curved side can be flattened out to a rectangle having width the height of the can h, and length the circumference of the circular top which is πr. The total surface area of the can is then 6.89 S πr + πrh πr + πr r.78π πr + r

17 .. CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS 7. (a) Since S πrh+ πr, solving for h Sπr yields h. Substituting into πr V πr hyields π π S r r V r ( S πr ). πr (b) Using V πr h, we can epress h in V terms of V and r, h πr. Thus S πrh+ πr π V r + πr πr V + πr r 4. The surface area of the topless can is S πr + πrh 7π. Solving for h yields r + rh 7 7r h r The volume is π π 7 r r V r (7 r ). r 6. Let q be the amount of radium remaining and k a proportionality constant. Then R( q) kq. (Note R( q) kq where k > 0 is usually used because decay means that less and less radium is left as time goes on.) 8. Let q be the number of people who have caught the disease. Then nqis the number of people who have not yet caught the disease out of a total population of n people. If k is the proportionality constant then R( q) kq( nq ). 0. Let be the number of machines used. The setup cost is k while the operating k cost is where k, k constants of proportionality. The total cost is k C ( ) k +.. With N and A 00, Cowling s rule + suggests C mg while 4 Friend s rule suggests ()(00) C 64 mg Setting the formulas for Cowling s and Friend s rules equal to each other gives 6. (a) + N NA N N A or Solving for N gives N 5 or about year, month old. If N is smaller, Cowling s rule suggests the higher dosage. If N is larger, Friend s rule suggests the higher dosage. Winning Total price bid $,000 $,000 + (0.75)($,000) $,75 $5,000 $5,000 + (0.75)($5,000) $9,75 $00,000 $00, (0.75)($50, 000) + (0.0)($50,000) $,750 (b) Let denote the winning bid price. If is less than or equal to 50000, the total purchase price is dollars. If eceeds 50000, the first carries a premium of (0.75)(50000) 8750 dollars while the

18 8CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS remaining carries a 0% premium. The total price is then (0.0)( 50000) dollars. Summarizing, the total price, P(), is given by the function.75 if P ( ) if > (b) The slopes of the segments are 0., 0.5, 0.7 and 0.0. The slopes increase with the taable income indicating the shift into different ta brackets. That is, the more taable income the greater a portion of that income will be taed. 8. (a) Let denote taable income and f() the corresponding income ta. both in units of dollars. Then 0. if 6, < ( 6000) if 6, 000 7, 950 f( ), ( 7, 950) if 7, 950 < 67, 700 4, ( 67, 700) if 67, 700 < 4, 50 or equivalently 0. if 6, 000 < if 6, 000 7, 950 f( ) 0.7, 654 if 7, 950 < 67, , 684 if 67, 700 < 4, Let denote the length of a side of the square base and y the height of the bo.. The cost is given by C (cost per m of base top)(area of base and top) + (cost per m of sides)(area of sides). Thus C ( ) + (4 y) 4 + 4y. Since the volume is 50, y 50 or 50 y. It follows that 000 C ( ) Let denote the length of the side of one of the removed squares and V() the volume of the resulting bo. Then V( ) (area of base)(height) (8 )(8 ) 4 (9 )

19 .. CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS 9 From the graph, the value of producing a bo with greatest volume is estimated to be in. 4. Let denote the selling price in dollars of the book and P() the corresponding profit function. If is the price of the book then (5 ) is the number of $ decreases in the price of the book from $5. Since 0 more books (beyond 00) will be sold for each $ decrease, the total number of books sold at dollars is (5 ) The bookstore s revenue function is then R( ) (500 0 ) 5000 while the cost function is C ( ) (500 0 ) Since profit is revenue minus cost P ( ) R ( ) C ( ) The graph of P() suggests the profit is maimal when 4, that is, when the books are sold for $4 apiece. Note the verte of this parabola is located at supporting the graphical ( 0) estimate. 6. Let t be the time in hours since the truck was 00 miles due east of the car. After t hours the truck is 000t miles from the car s original location and the car has moved north 60t miles. These two distances form the legs of a right triangle the hypotenuse of which is the distance, D(t), between the car and truck. By the Pythagorean theorem Dt ( ) (60 t) + (000 t) 0 5t 0t Let be the number of additional trees planted. Then the number of trees will be 60 + and the average number of oranges per tree will be The yield is then y ( ) (400 4 )(60 + ). 4(00 )(60 + )

20 0CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS S ( ) D ( ) (4+ 9)( ) 0 so 9 or. The negative 4 value is not a valid level of production so e and p S( ) D( ) 9. e e e The number of additional trees for maimal yield appears to be 0 or 80 total trees. (b) 40. (a) Market equilibrium occurs when (b) S ( ) D ( ) so 80 or e (c) A market shortage occurs when the graph of S() is below the graph of D() or when <. A market surplus occurs when the graph of S() is above the graph of D() or when >. (c) A market shortage occurs when the graph of S() is below the graph of D() or when < 40. A market surplus occurs when the graph of S() is above the graph of D() or when > (a) Market equilibrium occurs when p 44. The supply is Sp ( ) and the demand 0 is Dp ( ) 60 p. Supply will equal demand when p 60 p or p + 0p Solving this quadratic gives p 0, 0. Disregarding the negative value, supply will equal demand when the blenders are priced at $0 apiece. At this price S(0) D (0) 40 blenders will be sold.

21 .. CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS 46. Let denote the time in hours the spy has been traveling. Then is the time the smugglers have been traveling (since 40 minutes is of an hour.) The distance the spy travels is the 7 kilometers while the corresponding distance traveled by the smugglers is 68. The smugglers will overtake the spy when Solving 7 for yields hours. This 96 6 corresponds to a distance of km which is beyond the 8.8 km to the border so the spy escapes pursuit. 48. (a) If is the number of tables produced then the manufacturer s cost function is C ( ) while the revenue function is R( ) 70. The break even point is where R( ) C( ) or Thus and 00. The manufacturer must sell 00 tables to break even. (b) Since profit is revenue minus cost, the profit function, P(), is P ( ) 70 ( ) For the profit to be $6,000, must satisfy Selling 50 tables yields a profit of $6,000. (c) Since P (50) 40(50) , there will be a loss of $,000 if only 50 tables are sold. (d) The overhead corresponds to the y intercept of the cost function. 50. Let be the number of checks that clear the bank. Then the first bank charges y dollars while the second one charges y dollars. Find the break even point by setting the two equal If less than 50 checks are written the second bank offers the better deal. If more than 50 checks will be written, the first bank is more economical. 5. Let be the number of additional days beyond 80 before the club takes all its glass to the recycling center. The rate at which the club collects glass is

22 CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS 4, pounds per day. Thus after 80 additional days the club will have collected a total of 4, pounds. The current price of cent per pound will decrease by /00 cent for each day. Thus the clubs revenue, in cents, in days will be (4, ). 00 (80 + )(00 ) From the graph, it appears the club should collect glass for 0 additional days to maimize revenue. 54. (a) The graph of S(q) is rising while that of D(q) is falling so a > 0 and c < 0. Both graphs intersect the p ais at positive values so b > 0 and d > 0. (b) aq+ b cq + d ( a c) q db db so qe and so a c d b adbc pe a + b ac ac (c) If a increases, the denominator in the formula for qe will grow larger and hence qe will decrease. Similarly, if d increases, the numerator in the formula for qe will grow larger and so q will increase. e.5 Limits. Yes the limit eists, because as 4. a y b, that is lim f ( ) b a lim f ( ) b but a + a The limit fails to eist. 6. Yes the limit eists because lim f ( ) lim f( ) b 8. a a+ lim ( + ) + a or lim f ( ) c and b c. lim lim + lim + ( ) lim ( 5 ) 5 lim / / lim ( + )( ) [ ] lim ( ) lim ( ) lim lim ( ) ( ) lim + + ( ) 8

23 .. CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS + lim( + ) 4. lim + lim( + ) lim + lim lim is not defined because the denominator 0 (while the numerator does not.) 9 ( + )( ) lim lim lim( + ) 6 Note that not matter how close is to. + 6 ( + )( ) lim lim lim( + ) 5 ( ) lim lim is not defined 0 0 because the denominator goes to 0 but the numerator does not ( + 5)( ) lim lim ( + )( ) lim + 5 lim + lim 9 ( + )( ) lim 9 + lim lim f( ) + lim f( ) lim ( ) lim f( ) lim ( ) f( ) ( + ) 6 lim f( ) lim lim f( ) lim f( ) 6+ lim f( ) lim 6 + lim f( ) lim f( ) 5 + lim f( ) lim lim f( ) lim 0 f( ) + lim f( ) lim + lim f( ) lim + 8. As one moves to the right along the ais, the graph approaches the horizontal line through. Similarly as one moves to the left through negative values, the graph 6

24 4CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS approaches the horizontal line through. Thus lim ( ) + f and lim f( ). 40. Let be the number of additional trees planted per acre. The number of oranges per tree will be and the number of trees per acre The yield per acre is # of oranges # of trees y ( ) tree acre 4(00 )(60 + ) The number of trees for optimal yield appears to be trees. 4. 5n lim lim 5+ n + n n + n The limit tells us that as more trials are conducted, the rat s traversal time will approach a minimum time of 5 minutes.. As approaches from the left, the curve approaches the point (, 4) so lim f( ) 4. From the right the curve approaches the point (,) so lim f( ). + Since the one-sided limits at are not equal, lim f ( ) does not eist. 4. As approaches from the left, the curve approaches the point (, ) so lim f( ). From the right the curve assumes larger and larger values as it nears so lim f( ) +. Since the one-sided + limits at are not equal, lim f ( ) does not eist The rational function is continuous at + so lim Answers will vary. The answer corresponding to each problem should include a sequence of numbers approaching the limiting value of from the right and left, along with the corresponding values of f ( ) /,000( ),050.00,05.4, /,000( ),05.7,05.7 Thus it appears, / lim,000( ), One-Sided Limits and Continuity The rational function is not continuous at since the denominator is 0 there. As approaches from the left, the numerator approaches + 4 8while the denominator approaches 0 through negative values. Thus + 4 lim. 0. lim + lim + lim ( )( + ) lim ( )( + ) lim ( )( + ) lim + + ( )

25 .. CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS lim lim lim 5 + if < f( ) + if lim f( ) lim lim f( ) lim ( + ) + + lim f ( ) f (0) 0 thus f ( ) is continuous at 0. Note that all polynomials are continuous. lim 4 8. lim f( ) lim 0 0 f () 6 thus f ( ) is continuous at. lim lim f( ) lim 6 0 which is not defined, thus f ( ) is not continuous at. lim. lim f( ) lim 4 f () 4 thus f ( ) is continuous at. 6. lim f( ) lim ( ) and as approaches 0 from the left, lim f( ) lim ( + ) 0 0 Hence the limit does not eist (since different limits are obtained form the left and right), and so f is not continuous at 0. if < f( ) + if then f ( ) and lim f() must be determined. As approaches from the right, lim f( ) lim ( ) + + and as approaches from the left, lim f( ) lim + ( )( + ) lim + lim ( ) Hence the limit eists and is equal to and so f is continuous at. f ( ), 8. f ( ) 5 is continuous for all values of. Polynomials are continuous everywhere. 0.. f( ) is not continuous at, 6 where the denominator is zero. f( ) is not continuous at + where the denominator is zero., 4. + if < 0 f( ) if 0 f (0) and lim f( ) must be determined. 0 As approaches 0 from the right, 4. f( ) is not continuous at ( + 5)( ) 5 or, where the denominator is zero.

26 6CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS f( ) ( ) ( )( + ) is not continuous at or, where the denominator is zero. if f( ) 9 if < is possibly not continuous at. As approaches from the right, lim + R R R lim f( ) lim and as approaches form the left, lim f( ) lim 4 Hence the limit does not eist and so f is continuous at. if f( ) + if > is possibly not continuous only at As approaches from the left, lim f( ) lim ( ) 5 and as approaches from the right, lim f( ) lim ( + ) Hence the limit as approaches eists and is equal to f( ). Thus there are no values of at which f() is not continuous. 0 if 0 < < R E ( ) if R if R< E( ) is not continuous at R since 44. Since the thickness is assumed to be a continuous function of, we would epect the thickness at the source to be 0.5( + ) 0.5 ( + ) lim lim ( + + 4) 0.5( + ) lim meters The graph is discontinuous at 6 and. What happened to cause these jumps is a writing eercise; answers will vary. 48. lim f( ) 4 4 f(4) 6A+ 8 6A+ 5 To be continuous at 4 we need 6A+ 5 or A. 50. f ( ) is a polynomial in the open interval, and thus f ( ) is continuous for all in the open interval. But at lim f( ) f () 4+ 8,

27 .. CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS 7 thus f () lim f( ). f( ) is not continuous on the closed interval. 5. Let f( ) ( + ) f ( ) is continuous at all and f (0), f () (+ ). By the root location property, there is at least one number 0< c such that f() c 0, and c is a solution. 54. Your weight in pounds increased (and/or decreased) continuously from your minimum weight at birth to your present weight, like from 7 pounds to 50 pounds. Your (present) height in inches is some number, like 65. But the intermediate value property, your weight must have been 65 pounds at least once in your lifetime. 56. Let s use some numbers for the purpose of the illustration. Suppose Nan is 60 inches tall at age 5 (say in 980) when Dan is 0 inches tall. Assume Nan is 70 inches tall at age (in 996) when Dan is 76 inches tall. Draw a continuous curve (it could be a straight line) from (980,60) to (996,70). This represents Nan s growth curve. Now draw a continuous curve from (980,0) to (996,75). This represents Dan s growth curve. The two curves cross at one point, say in 99 when they are both 66 inches tall. By the intermediate value property, 66 inches lies between 0 and 60 as well as between 0 and 75. Review Problems. The price months from now is 0 P ( ) 40+ dollars. Hence: + (a) The price 5 months from now is 0 P (5) 40 + $45 5+ (b) The change in price during the 5 th month is P(5) P(4) That is, the price decreases by $. (c) The price will be $4 when ( + ) 0 9 or 9 months from now. 0 (d) Since the term gets very small as + gets very large, the price 0 P ( ) will approach $40 in the long run. 4. (a) If f( ) + 4 then f( ) ( ) ( ) (b) If f( ) + then f( + ) + + ( + ) + +. (c) If f ( ) then f ( + ) f( ) ( + ) +.

28 8CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS 6. (a) Since the smog level Q is related to p by the equation Qp ( ) 0.5p and p is related to t by p() t 8+ 0.t it follows that the composite function [ ] ( t) Qpt ( ) t epresses the smog level as a function of the variable t. (b) The smog level years from now will be Qp [ ()].4+ 0.() units (b) Some points on the graph of y + 4 are shown below. Note that y + 4+ ( ) + 5 is a parabola with verte (, 5) and opening downward. y y (c) Set Qpt [ ()] equal to 5 and solve for t to get t.6 0.t t 6 or t 4 years from now. 8. (a) Some points on the graph of y + 8 ( + 4)( ) are shown below y y (a) Since m 5 and b 4, the equation of the line is y 5 4. (b) Since m and the point (,) is on the line, the equation of the line is y ( ) or y + 5. (c) The slope of the line is /0 m 0 9

29 .. CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS 9 and the y intercept is 0, so the equation of the line is y. 9 (d) Parallel lines have the same slope, so the desired line is of the form + y C where (5) + 4 C. Thus + y (e) y 5 7 or y The slope of the perpendicular line is m and its equation is of the form 5 y + C. 5 Since the line contains (,), C 5 and y (a) Note: In chapter, we will learn how to use calculus to get the optimal price. Without calculus, we could complete the square to get E( p) 50( p 8) +,00 from which it s clear that the greatest ependiture is $,00 and is generated when the price is p $8. (b) The function Dp ( ) 50p is linear with slope 50 and y intercept 800. It represents demand for 0 p 6. The total monthly ependiture is E( p) (price per unit)(demand) p( 50p+ 800) 50 pp ( 6) Since the ependiture is assumed to be non-negative, the relevant interval is 0 p 6. (c) The graph suggests that the ependiture will be greatest if p (a) Suppose C is the circulation of the newspaper and t is time measured in months. Further suppose t 0 represents three months ago. Then C is a linear funtion of t passing through the points (0, 00) and (, 4400). The slope is 400 and the C intercept is 00, hence C 400t + 00.

30 0CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS B 0 5, A the maimum profit occurs at a price of 5 dollars. 8. Let r denote the radius, h the height, and V the volume of the can. Then (b) Two months from now is represented by t 5 and the circulation at that time will be C 400(5) Let denote the selling price of a bookcase (in dollars). The manufacturer s cost and revenue functions are C ( ) 80(50 ) and R( ) (50 ) respectively. The profit function is then P ( ) R ( ) C ( ) (50 ) 80(50 ) + 0,000. V πr h To write h in terms of r, use the fact that the cost of constructing the can is to 80 cents. That is, 80 cost of bottom + cost of side where and cost of bottom (cost per in )(area) Hence πr cost of side (cost per in )(area) ( πrh) 4 πrh. 0 r 80 πr + 4 πrh, or h. πr 4 0. Assume the inventory to be maintained at the same level, continuously, over a 4-hour period. A discontinuity occurs when the inventory drops, say, at midnight the appropriate days. From the graph, or by using the formula for the verte of a parabola

31 .. CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS The graph is discontinuous at t 0, t 6 and t 4.. (a) The revenue is R( ) 75 and the cost is C ( ) 5 +,500 where is the number of kayaks sold. For the break even point, 75 5+,500 or 0. (b) The profit is P ( ) R ( ) C ( ) 50, 500 For P ( ),000, 5 50, 7 4. Let the power plant be at E, the opposite point at O, the point at which the cable 6. When 00 tables are sold the revenue is 00($70) $4,000 while the cost is 00($0) + A $6,000 + A where A is the overhead. Since revenue is equal to cost when 00 tables are sold, $4,000 $6,000 + A, and so the overhead, A, is $8, Let be the time in minutes for the hour hand to move form position to the position at which the hands coincide. The hour hand moves of a tick (distance between minute tick marks on the circumference) per minute while the minute hand moves one whole tick per minute. The minute has to cover 5 ticks before reaching position. Thus , 6.6, or 6 minutes and seconds. lim + lim( ) lim ( + ) reaches the opposite bank at P, and the factory at F. OP PF, 000 EP The cost of the cable in the river is C r and the cost of the cable on land is Cl 4(, 000 ) Thus the total cost is C 4(, 000 ) lim 0 lim lim lim lim + ( ) lim lim lim lim f ( ) 5 + is not continuous for < 0 since square roots of negative numbers do not eist in the real numbers.

32 CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS (a) + 5 g ( ) is not continuous at ( )( + ) and since the denominator in the definition of g ( ) is 0. + if < f( ) A if Then f ( ) is continuous everywhere ecept possibly at since + and A are polynomials. Since f() A, in order for f ( ) to be continuous at, A must be chosen so that lim f( ) A As approaches from the right, lim f( ) lim ( A ) A + + and as approaches from the left, lim f( ) lim ( + ) 5. lim f ( ) eists whenever A 5 or A 6. Furthermore, for A 6, lim f( ) 5, f () 6 5. Thus, f ( ) is continuous at only when A Since f( ) A 4, in order for f ( ) to be continuous at, A must be chosen so that lim f( ) A 4 As approaches from the right, lim f( ) lim ( A + ) A and as approaches from the left, lim f( ) lim + ( + )( ) lim + lim f ( ) eists whenever A 4 or A. Furthermore, for A, lim f( ), f( ) 4. Thus, f ( ) is continuous at when A. only (b) if < f( ) + A + if Then f ( ) is continuous everywhere ecept possibly at since + is a rational function and is a polynomial. A + The function is undefined at and which are the values of satisfying f( ) +, g( ) 5 + 4

33 .. CHAPTER. FUNCTIONS, GRAPHS, AND LIMITS 54. (a) g(.8) 5(.8) f( g(.8)) f(.9) (b) f ( ) +.0 g( f( )) g(.0) 5(.0) f( ) 0 f( ) 0 when 0 ( ) or ( 4)( + ) 0. The intercepts are then (4,0) and (,0). Since f (0), the y intercept is (0, ). The function is defined for all.