For Thought. 2.1 Exercises 80 CHAPTER 2 FUNCTIONS AND GRAPHS

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1 80 CHAPTER FUNCTIONS AND GRAPHS For Tougt. False, since {(, ), (, )} is not a function.. False, since f(5) is not defined.. True. False, since a student s eam grade is a function of te student s preparation. If two classmates ad te same IQ and onl one prepared ten te one wo prepared will most likel acieve a iger grade. 5. False, since ( + ) False, since te domain is all real numbers. 7. True 8. True 9. True ( ) ( ) 0. False, since 8, 8 and 8, 5 are two ordered pairs wit te same first coordinate and different second coordinates.. Eercises. relation. function. independent, dependent. domain, range 5. difference quotient 6. average rate of cange 7. Note, b πa is equivalent to a b π. Ten a is a function of b, and b is a function of a. 8. Note, b (5 + a) is equivalent to a b 0. So a is a function of b, and b is a function of a. 9. a is a function of b since a given denomination as a unique lengt. Since a dollar bill and a five-dollar bill ave te same lengt, ten b is not a function of a. 0. Since different U.S. coins ave different diameters, ten a is a function of b and b is a function of a.. Since an item as onl one price, b is a function of a. Since two items ma ave te same price, a is not a function of b.. a is not a function of b since tere ma be two students wit te same semester grades but different final eams scores. b is not a function of a since tere ma be identical final eam scores wit different semester grades.. a is not a function of b since it is possible tat two different students can obtain te same final eam score but te times spent on studing are different. b is not a function of a since it is possible tat two different students can spend te same time studing but obtain different final eam scores.. a is not a function of b since it is possible tat two adult males can ave te same soe size but ave different ages. b is not a function of a since it is possible for two adults wit te same age to ave different soe sizes. 5. Since in.5 cm, a is a function of b and b is a function of a. 6. Since tere is onl one cost for mailing a first class letter, ten a is a function of b. Since two letters wit different weigts eac under /-ounce cost cents to mail first class, b is not a function of a. 7. No 8. No 9. Yes 0. Yes. Yes. No. Yes. Yes 5. Not a function since 5 as two different second coordinates. 6. Yes 7. Not a function since as two different second coordinates. 8. Yes 9. Yes 0. Yes. Since te ordered pairs in te grap of 8 are (, 8), tere are no two ordered pairs wit te same first coordinate Coprigt 0 Pearson Education, Inc.

2 . FUNCTIONS 8 and different second coordinates. We ave a function.. Since te ordered pairs in te grap of + 7 are (, + 7), tere are no two ordered pairs wit te same first coordinate and different second coordinates. We ave a function.. Since ( + 9)/, te ordered pairs are (, ( + 9)/). Tus, tere are no two ordered pairs wit te same first coordinate and different second coordinates. We ave a function.. Since, te ordered pairs are (, ). Tus, tere are no two ordered pairs wit te same first coordinate and different second coordinates. We ave a function. 5. Since ±, te ordered pairs are (, ±). Tus, tere are two ordered pairs wit te same first coordinate and different second coordinates. We do not ave a function. 6. Since ± 9 +, te ordered pairs are (, ± 9 + ). Tus, tere are two ordered pairs wit te same first coordinate and different second coordinates. We do not ave a function. 7. Since, te ordered pairs are (, ). Tus, tere are no two ordered pairs wit te same first coordinate and different second coordinates. We ave a function. 8. Since, te ordered pairs are (, ). Tus, tere are no two ordered pairs wit te same first coordinate and different second coordinates. We ave a function. 9. Since, te ordered pairs are (, ). Tus, tere are no two ordered pairs wit te same first coordinate and different second coordinates. We ave a function. 0. Since +, te ordered pairs are (, + ). Tus, tere are no two ordered pairs wit te same first coordinate and different second coordinates. We ave a function.. Since (, ) and (, ) are two ordered pairs wit te same first coordinate and different second coordinates, te equation does not define a function.. Since (, ) and (, ) are two ordered pairs wit te same first coordinate and different second coordinates, te equation does not define a function.. Domain {,, 5}, range {,, 6}. Domain {,,, }, range {,, 8, 6} 5. Domain (, ), range {} 6. Domain {5}, range (, ) 7. Domain (, ); since 0, te range of + 5 is [5, ) 8. Domain (, ); since 0, te range of + 8 is [8, ) 9. Since, te domain of is [, ); range (, ) 50. Since, te domain of is [, ); Since is a real number wenever 0, te range is [0, ). 5. Since is a real number wenever, te domain of is [, ). Since 0 for, te range is [0, ). 5. Since 5 is a real number wenever 5, te domain of 5 is (, 5]. Since 5 0 for 5, te range is [0, ). 5. Since 0, te domain of is (, 0]; range is (, ); 5. Since 0, te domain of is (, 0]; range is (, ); g() () g() () Coprigt 0 Pearson Education, Inc.

3 8 CHAPTER FUNCTIONS AND GRAPHS 59. Since (, 8) is te ordered pair, one obtains f() 8. Te answer is. 60. Since (, 6) is te ordered pair, one obtains f() 6. Te answer is. 6. Solving + 5 6, we find Solving + 5, we find. 6. f() + g() f() g() 8 6 Te average rate of cange on [.9, ] is () (.9) ft/sec. Te average rate of cange on [.99, ] is () (.99) ft/sec. Te average rate of cange on [.999, ] is () (.999).999 ft/sec a a 66. w w 67. (a+) a (a 5) a 69. ( + + ) ( + ) ( 6 + 9) ( ) ft/sec Te average rate of cange is 0 0. million ectares per ear. 7. ( + ) + 7. ( ++ ) ( + + ) ( + ) ( + ) ( + + ) ( + ) ( + ) Te average rate of cange is 8, 000 0, $, 00 per ear. 78. Te average rate of cange as te number of cubic ards canges from to 0 and from 0 to 60 are $6 per d and $ per d, respectivel. 79. Te average rate of cange on [0, ] is () (0) 0 6 ft/sec. 0 0 Te average rate of cange on [, ] is () () ft/sec. 8. If 0. million ectares are lost eac ear and since ears, ten te forest will 0. be eliminated in te ear 8 ( ) f( + ) f() f( + ) f() f( + ) f() ( + ) ( + ) ( + ) Coprigt 0 Pearson Education, Inc.

4 . FUNCTIONS f( + ) f() ( + ) Let g() +. Ten we obtain g( + ) g() ( + ) + ( + ) ( + ) 9 ( + + ) 9 ( + + ) Difference quotient is + + ( + ) ( + ) ( + ) Let g(). Ten we get g( + ) g() ( + ) ( + ) Difference quotient is ( + + ) ( + ) ( ) ( ) Difference quotient is ( + ) + ( + ) Difference quotient is ( + ) ( + ) Difference quotient is Difference quotient is ( ) + + ( + + ( + + ) ) ( + + ) + Coprigt 0 Pearson Education, Inc.

5 8 CHAPTER FUNCTIONS AND GRAPHS 95. Difference quotient is 96. Difference quotient is 97. Difference quotient is + ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) + ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + + )( + ) ( + + )( + ) ( + ) ( + + ) ( + + )( + ) ( + + )( + ) ( + + )( + ) 98. Difference quotient is + ( + )( ) ( + )( ) ( ) ( + ) ( + )( ) ( + )( ) ( + )( ) 99. a) A s b) s A c) s d d) d s e) P s f) s P/ g) A P /6 ) d A 00. a) A πr b) A r π c) C πr d) d r e) d C π A g) d π 0. C n f) A πd 0. a) Wen d 00 ft, te atmosperic pressure is A(00).0(00) + atm b) Wen A.9 atm, te dept is found b solving.9 0.0d + ; te dept is d.9 0 ft. 0.0 (a) Te quantit C() (0.95)() $9.6 billion represents te amount spent on computers in te ear 00. (b) B solving 0.95n , we obtain n Tus, spending for computers will be $5 billion in 00. (a) Te quantit E() + C() [0.5() + ] $.6 billion represents te total amount spent on electronics and computers in te ear 00. (b) B solving (0.5n + ) + (0.95n + 5.8) 0.5n. n 9 we find tat te total spending will reac $0 billion in te ear 009 ( ). Coprigt 0 Pearson Education, Inc.

6 . FUNCTIONS 85 (c) Te amount spent on computers is growing faster since te slope of C(n) [wic is ] is greater tan te slope of E(n) [wic is 0.95]. 05. Let a be te radius of eac circle. Note, triangle ABC is an equilateral triangle wit side a and eigt a. A C B Tus, te eigt of te circle centered at C from te orizontal line is a + a. Hence, b using a similar reasoning, we obtain tat eigt of te igest circle from te line is a + a or equivalentl ( + )a. 06. In te triangle below, P S bisects te 90-angle at P and SQ bisects te 60-angle at Q. P d S a R 0 0 Q In te triangle SP R, we find P R SR d/. And, in te triangle SQR we get 6 P Q d. Since P Q P R + RQ, we obtain a d + 6d a ( 6 + )d d d a a. 07. Wen 8 and 0., we ave R(8.) R(8) 0., 950. Te revenue from te concert will increase b approimatel $,950 if te price of a ticket is raised from $8 to $9. If and 0., ten R(.) R() 0., 050. Te revenue from te concert will decrease b approimatel $,050 if te price of a ticket is raised from $ to $. 08. Wen r. and 0., we obtain. A(.5) A(.) Te amount of tin needed decreases b approimatel 6. in. if te radius increases from. in. to. in. If r and 0., ten A(.) A() Te amount of tin needed increases b about 8.6 in. if te radius increases from in. to in a d + 6 d 9 7 Coprigt 0 Pearson Education, Inc.

7 86 CHAPTER FUNCTIONS AND GRAPHS. If m is te number of males, ten m + m 6 m 6 m (6) males. ( + 6) + ( ) Te slope is 5 +. Te line is given 6 5. b + b for some b. Substitute te 6 coordinates of (, ) as follows: 6 Te line is given b 6 ( ) + b b ( 7)( + 6) 0 Te solution set is { 6, 7}.. Pop Quiz. Yes, since A πr were A is te area of a circle wit radius r.. No, since te ordered pairs (, ) and (, ) ave te same first coordinates.. No, since te ordered pairs (, 0) and (, 0) ave te same first coordinates.. [, ) 5. [, ) If a, ten a / $ per ear Te difference quotient is f( + ) f(). Linking Concepts ( + ) (a) Te first grap sows U.S. federal debt D versus ear debt Te inequalit is equivalent to < 9 < < < < < Te solution set is (, ) and te second grap sows population P (in millions) versus. ear population Tinking Outside te Bo XXII (0 + 5) ear Coprigt 0 Pearson Education, Inc.

8 . GRAPHS OF RELATIONS AND FUNCTIONS 87 (b) Te first table sows te average rates of cange for te U.S. federal debt 0 ear period ave. rate of cange Te second table sows te average rates of cange for te U.S. population 0 ear period ave. rate of cange (c) Te first table sows te difference between consecutive average rates of cange for te U.S. federal debt. 0-ear periods difference & & & & & Te second table sows te difference between consecutive average rates of cange for te U.S. population. 0-ear periods difference & & & & & (f) Te U.S. federal debt is growing out of control wen compared to te U.S. population. See part (g) for an eplanation. (g) Since most of te differences for te federal debt in part (e) are positive, te federal debts are increasing at an increasing rate. Wile te U.S. population is increasing at a decreasing rate since most of te differences for population in part (e) are negative. For Tougt. True, since te grap is a parabola opening down wit verte at te origin.. False, te grap is decreasing.. True. True, since f(.5) [.5]. 5. False, since te range is {±}. 6. True 7. True 8. True 9. False, since te range is te interval [0, ]. 0. True. Eercises. parabola. piecewise. Function includes te points (0, 0), (, ), domain and range are bot (, ) (d) For bot te U.S. federal debt and population, te average rates of cange are all positive. (e) In part (c), for te federal debt most of te differences are positive and for te population most of te differences are negative. Coprigt 0 Pearson Education, Inc.

9 88 CHAPTER FUNCTIONS AND GRAPHS. Function includes te points (0, 0), (, ), (, ), domain and range are bot (, ) 9. Function includes te points (0, 0), (±, ), domain is (, ), range is [0, ) Function 0 includes te points (, ), (0, 0), (, ), domain and range are bot (, ) 0. Function goes troug (0, ), (±, 0), domain is (, ), range is [, ) Function includes te points (, 0), (0, ), (, ), domain and range are bot (, ). Function includes te points (0, ), (±, 0), domain is (, ), range is (, ] Function 5 includes te points (0, 5), (±, 5), domain is (, ), range is {5} 6 -. Function includes te points (0, ), (±, ), domain is (, ), range is (, ] is not a function and includes te points (, 0), (, ), domain is {}, range is (, ) - Coprigt 0 Pearson Education, Inc.

10 . GRAPHS OF RELATIONS AND FUNCTIONS 89. Function + includes te points (0, ), (, ), (, ), domain is [0, ), range is [, ) 7. Function goes troug (0, 0), (, ), (, 9), domain and range is [0, ) 9. Function includes te points (0, ), (, 0), domain is [0, ), range is (, ] 8. Function goes troug (, 0), (, ), (, 9), domain [, ), and range [0, ), is not a function and includes te points (, 0), (, ±), domain is [, ), range is (, ) - 9. Function + goes troug (, 0), (, ), (8, ), domain (, ), and range (, ) is not function and includes te points (, 0), (0, ±), domain is (, ], range is (, ) 0. Function goes troug (, ), (, ), (8, 0), domain (, ), and range (, ) Coprigt 0 Pearson Education, Inc.

11 90 CHAPTER FUNCTIONS AND GRAPHS. Function, goes troug (0, 0), (, ), (, 8), domain (, ), and range (, ). Not a function, + goes troug (, 0), (0, ), (, 0), domain [, ], and range [, ] Function, goes troug (0, ), (, ), (, 0), domain (, ), and range (, ) 5. Function, goes troug (±, 0), (0, ), domain [, ], and range [0, ] Not a function, goes troug (, 0), (0, ), (, 0), domain [, ], and range [, ] 6. Function, 5 goes troug (±5, 0), (0, 5), domain [ 5, 5], and range [ 5, 0] Coprigt 0 Pearson Education, Inc.

12 . GRAPHS OF RELATIONS AND FUNCTIONS 9 7. Function includes te points (0, 0), (, ), (, 8), domain and range are bot (, ) 8. Function includes te points (0, 0), (±, ), domain is (, ), range is (, 0] Function includes te points (0, 0), (, ), (, 8), domain and range are bot (, ) 8. Function + includes te points (, 0), (0, ), (, ), domain is (, ), range is (, 0] Function includes te points (0, 0), (±, ), domain is (, ), range is [0, ) 5. Not a function, grap of includes te points (0, 0), (, ), (, ), domain is [0, ), range is (, ) - 0. Function includes te points (0, ), (, 0), (, ), domain is (, ), range is [0, ) is not a function and includes te points (, 0), (, ±), domain is [, ), range is (, ) - - Coprigt 0 Pearson Education, Inc.

13 9 CHAPTER FUNCTIONS AND GRAPHS 5. Domain is (, ), range is {±}, some points are (, ), (, ) 9. Domain is [, ), range is (, ], some points are (, ), (, 0), (, ) Domain is (, ), range is {, }, some points are (0, ), (, ) 0. Domain is (, ), range is (, ), some points are (, ), (, ), (, ) Domain is (, ), range is (, ] (, ), some points are (, ), (, ). Domain is (, ), range is [0, ), some points are (, ), (, ), (, ) Domain is (, ), range is [, ), some points are (, ), (, ). Domain is (, ), range is [, ), some points are (, ), (0, ), (, ) - - Coprigt 0 Pearson Education, Inc.

14 . GRAPHS OF RELATIONS AND FUNCTIONS 9. Domain is (, ), range is (, ), some points are (, ), (, ) 7. Domain [0, ), range is {,,, 5}, some points are (0, ), (, ), (.5, ) 5 -. Domain is [, ), range is [0, ), some points are (±, 0), (, ) 8. Domain is (0, 5], range is {,,, 0,, }, some points are (0, ), (, ), (.5, ) Domain is (, ), range is te set of integers, some points are (0, ), (, ), (.5, ) - 6. Domain is (, ), range is te set of even integers, some points are (0, 0), (, ), (.5, ) a. Domain and range are bot (, ), decreasing on (, ) b. Domain is (, ), range is (, ] increasing on (, 0), decreasing on (0, ) 50. a. Domain and range are bot (, ), increasing on (, ) b. Domain is (, ), range is [, ) increasing on (0, ), decreasing on (, 0) 5. a. Domain is [, 6], range is [, 7] increasing on (, ), decreasing on (, 6) b. Domain (, ], range (, ], increasing on (, ), constant on (, ) 5. a. Domain is [0, 6], range is [, ] increasing on (, 6), decreasing on (0, ) b. Domain (, ), range [, ), increasing on (, ), constant on (, ), decreasing on (, ) 5. a. Domain is (, ), range is [0, ) increasing on (0, ), decreasing on (, 0) b. Domain and range are bot (, ) increasing on (, /), decreasing on (, ) and ( /, ) Coprigt 0 Pearson Education, Inc.

15 9 CHAPTER FUNCTIONS AND GRAPHS 5. a. Domain is [, ], range is [0, ] increasing on (, 0), decreasing on (0, ) b. Domain is (, ), range is [, ) increasing on (, ), decreasing on (, ), constant on (, ) 55. a. Domain and range are bot (, ), increasing on (, ) b. Domain is [, 5], range is [, ] increasing on (, ), decreasing on (, ), constant on (, 5) 56. a. Domain is (, ), range is (, ] increasing on (, ), decreasing on (, ) b. Domain and range are bot (, ), decreasing on (, ) 60. Domain is (, ), range is [, ), increasing on (0, ), decreasing on (, 0), some points are (0, ), (, ) - 6. Domain is (, 0) (0, ), range is {±}, constant on (, 0) and (0, ), some points are (, ), (, ) 57. Domain and range are bot (, ) increasing on (, ), some points are (0, ), (, ) Domain and range are bot (, ), decreasing on (, ), some points are (0, 0), (, ) 6. Domain is (, 0) (0, ), range is {±}, constant on (, 0) and (0, ), some points are (, ), (, ) Domain is (, ), range is [0, ), increasing on (, ), decreasing on (, ), some points are (0, ), (, 0) 6. Domain is [, ], range is [0, ], increasing on (, 0), decreasing on (0, ), some points are (±, 0), (0, ) 5 Coprigt 0 Pearson Education, Inc.

16 . GRAPHS OF RELATIONS AND FUNCTIONS Domain is [, ], range is [, 0], increasing on (0, ), decreasing on (, 0), some points are (±, 0), (0, ) 68. Domain is (, ), range is (, ], increasing on (, ) and (0, ), decreasing on (, 0) and (, ), some points are (, ), (0, 0), (, ) Domain and range are bot (, ), increasing on (, ) and (, ), some points are (, 5), (0, ) Domain and range are bot (, ), decreasing on (, ), some points are (, ), (, ) Domain is (, ), range is (, ], increasing on (, ) and (, 0), decreasing on (0, ) and (, ), some points are (, 0), (0, ), (, ) f() 70. f() { for > for { for for > 7. Te line joining (, ) and (, ) is, and te line joining (, ) and (, ) is. Te piecewise function is { for f() for <. 7. Te line joining (, ) and (, ) is +, and te line joining (, ) and (, ) is. Te piecewise function is { + for f() for >. 7. Te line joining (0, ) and (, ) is, and te line joining (0, ) and (, ) is. Te piecewise function is { for 0 f() for < Te line joining (, ) and (, ) is, and te line joining (, ) and (, ) is +. Te piecewise function is { + for f() for >. 75. increasing on te interval [0.8, ), decreasing on (, 0.8] Coprigt 0 Pearson Education, Inc.

17 96 CHAPTER FUNCTIONS AND GRAPHS 76. increasing on te interval (, 0.7], decreasing on [0.7, ) 77. increasing on (, ] and [, ), decreasing on [, ] 78. increasing on [.5, 0) and [.5, ), decreasing on (,.5] and (0,.5] 79. increasing on [.7, 0) and [.7, ), decreasing on (,.7] and (0,.7] 88. Independent variable is time t in seconds, dependent variable is distance D from te pit D is increasing on te intervals (0, 0), (0, 60), (80, 00), (50, 70), (90, 0), and (0, 50); D is decreasing on te intervals (0, 0), (60, 80), (00, 0), (70, 90), (0, 0); D is constant on (0, 50) and (50, 00). D 80. increasing on (,.59], [.0,.0], and [.59, ), decreasing on [.59,.0] and [.0,.59] 8. increasing on [0, 50], and [70, ), decreasing on (, 0] and [50, 70] 8. increasing on (, 50], and [ 0, ), decreasing on [ 50, 0] 8. c, grap was increasing at first, ten suddenl dropped and became constant, ten increased sligtl 8. a t 89. Independent variable is time t in ears, dependent variable is savings s in dollars s is increasing on te interval [0, ]; s is constant on [,.5]; s is decreasing on [.5,.5]. 0 t 85. d, grap was decreasing at first, ten fluctuated between increases and decreases, ten te market increased s 86. b 87. Te independent variable is time t were t is te number of minutes after 7:5 and te dependent variable is distance D from te olodeck. D is increasing on te intervals [0, ] and [6, 5], decreasing on [, 6] and [0, 9], and constant on [5, 0]. D 90. Independent variable is time t in das, dependent variable is te amount, a, (in dollars) in te cecking account. a is decreasing on te interval [0, 0]; a is constant on [0, 5]; a is increasing on [5, 65] s t 6 t Coprigt 0 Pearson Education, Inc.

18 . GRAPHS OF RELATIONS AND FUNCTIONS In 988, tere were M(8) 565 million cars. In 00, it is projected tat tere will be M(0) 800 million cars. Te average rate of cange from 98 to 99 M() M() is 0 ear..5 million cars per 9. In developing countries and Eastern Europe, te average rate of cange of motor veicle M(0) M(0) ownersip is 6.5 million 0 veicles per ear. In developed countries, te average rate of cange of motor veicle ownersip from 990 to 00 is 0 million veicles per ear (see previous model). Ten veicle ownersip is epected to grow faster in developed countries. 9. Constant on [0, 0 ], increasing on [0, ) 9. Decreasing since a car as a lower mileage at ig speeds. 95. Te cost is over $5 for t in [5, ) C 5 t 96. If [w/00] < 86 ten [w/00] < 66/ Ten w/00 < 8 and te values of w lie in (0, 800). { 50 if 0 < < 97. f() 50 if 0 dollars f() cubic ards 0 { [ ] if 0 < 5 if < A mecanic s fee is $0 for eac alf-our of work wit an fraction of a alf-our carged as a alf-our. If is te fee in dollars and is te number of ours, ten 0[ ] For eample, coose an a, b satisfing a < b and grap te function defined b if a f() 5 a if a < < b 5b a if b 0. Since 0, te domain is [, ). Since is nonnegative, we find + is at least tree. Tus, te range is [, ). 0. Since an absolute value is nonnegative, te equation 55 9 as no solution. Te solution set is. 0. Note, we ave Te solution set is {55/}. 0. Rewrite witout te absolute values: 55 ±9 55 ± 9 6, 6 { 6 Te solution set is, 6 }. 05. Since we ave a perfect square ( 5) 0 te solution set is {5/}. 06. Appling te quadratic formula to w 5w 9 0, we find w 5 ± { } 5 ± 97 Te solution set is. Coprigt 0 Pearson Education, Inc.

19 98 CHAPTER FUNCTIONS AND GRAPHS Tinking Outside te Bo XXIII a) Consider te circle +( r) r were r > 0. Suppose te circle intersects te parabola onl at te origin. Substituting, we obtain. Linking Concepts (a) A grap of te benefit function is given below.,880 0,9 + ( r) r + ( r) 0 Tus, r 0 since te circle and te parabola as eactl one point of intersection. Te radius of te circle is /. b) Consider te circle + ( r) r were r > 0. If te circle intersects te parabola a onl at te origin, ten te equation below must ave eactl one solution, namel, 0. a + ( r) r ( ) + a r 0 Necessaril, we ave r 0 or a a. Tus, if r ten a /6. r. Pop Quiz. Domain [0, ), range (, ]. Domain [, ], range [0, ]. Range [, ). Increasing on [0, ) 5. Decreasing on (, ] (b) B(6) 80(6), 06 $09 (c) If B $, 880, ten 960a 5, 60, 880 a 960a 66, 0 a 69. At age 69 ears, te annual benefit is $,880. (d) Since te function is piecewise defined consisting of linear functions, te average rate of cange is te slope of te linear function. Ten te average rate of cange for ages 6-6 is 600. Te average rate of cange for ages 6-66 is 80. Te average rate of cange for ages is 960. (e) Yes, te answers to part (d) are te slopes of te tree lines in te piecewise function defining te benefit formula. (f) Note, B(6) $9000. Ten te total amount Bob epects to witdraw is (67.066(.0008) 6 6)$9000 $7, 800. (g) Note, B(70) $5, 80. Ten te total amount Bill epects to witdraw is (67.066(.0008) 70 70)$5, 80 $07, 700. Coprigt 0 Pearson Education, Inc.

20 . FAMILIES OF FUNCTIONS, TRANSFORMATIONS, AND SYMMETRY 99 For Tougt. False, it is a reflection in te -ais.. True. False, rater it is a left translation.. True 5. True 6. False, te down sift sould come after te reflection. 7. True 8. False, since teir domains are different. 9. True 0. True. Eercises. rigid. nonrigid. parabola. translation 5. reflection 6. identit 7. linear 8. constant 9. odd 0. even. f(), g(). f(), g() f(), g() , ( ) - 6., , f(), g() Coprigt 0 Pearson Education, Inc.

21 00 CHAPTER FUNCTIONS AND GRAPHS 8., ( )., -., 9. f(), g() f(), g() 5., -., - 6. f() +, g() -., - 7. g b 0. d. c. a. f. e ( 5) 8. ( + 7) - 9. ( 0) ( + 5) or 5. ( ( ) 6 ) or ( ) + 6 Coprigt 0 Pearson Education, Inc.

22 . FAMILIES OF FUNCTIONS, TRANSFORMATIONS, AND SYMMETRY ( + 6) 8 or 0 5. ( ) + ; rigt b, up b, domain (, ), range [, ) 9. 0, domain and range are bot (, ) ( + 5) ; left b 5, down b, domain (, ), range [, ) , domain and range are bot (, ) ; rigt b, up b domain (, ), range [, ) 5. 0, domain and range are bot (, ) ; left b, down b domain (, ), range [, ) , domain and range are bot (, ) Coprigt 0 Pearson Education, Inc.

23 0 CHAPTER FUNCTIONS AND GRAPHS , srink b /, reflect about -ais, up b 0, domain (, ), range (, 0] 57. +, rigt b, reflect about -ais, up b, domain [, ), range (, ] , stretc b, down b 00, domain (, ), range [ 00, ) , left b, reflect about -ais, down b, domain [, ), range (, ] , left b, reflect about -ais, srink b /, domain (, ), range (, 0] , left b, stretc b, reflect about -ais, up b, domain [, ), range (, ] , left b, srink b /, 56., rigt b, stretc b, domain (, ), range [0, ) reflect about -ais, up b, domain [, ), range (, ] 6-6. Smmetric about -ais, even function since f( ) f() Coprigt 0 Pearson Education, Inc.

24 . FAMILIES OF FUNCTIONS, TRANSFORMATIONS, AND SYMMETRY 0 6. Smmetric about -ais, even function since f( ) f() 6. No smmetr, neiter even nor odd since f( ) f() and f( ) f() 6. Smmetric about te origin, odd function since f( ) f() 65. Smmetric about, neiter even nor odd since f( ) f() and f( ) f() 66. Smmetric about, neiter even nor odd since f( ) f() and f( ) f() 67. Smmetr about, not an even or odd function since f( ) f() and f( ) f() 68. Smmetric about te -ais, even function since f( ) f() 69. Smmetric about te origin, odd function since f( ) f() 70. Smmetric about te origin, odd function since f( ) f() 7. No smmetr, not an even or odd function since f( ) f() and f( ) f() 7. No smmetr, not an even or odd function since f( ) f() and f( ) f() 7. No smmetr, not an even or odd function since f( ) f() and f( ) f() 7. Smmetric about te -ais, even function since f( ) f() 75. Smmetric about te -ais, even function since f( ) f() 76. Smmetric about te -ais, even function since f( ) f() 80. Smmetric about te -ais, even function since f( ) f() 8. e 8. a 8. g b 86. d 87. c 88. f ( ) (, ] [, ) 90., 9. (, ) (5, ) 9. [, ] 9. Using te grap of ( ) 9, we find tat te solution is (, ) Grap of is (, ] [, ) - - ( ) 9 sows solution 95. From te grap of 5, we find tat te solution is [0, 5] No smmetr, not an even or odd function since f( ) f() and f( ) f() 78. Smmetric about te -ais, even function since f( ) f() 79. Smmetric about te -ais, even function since f( ) f() Coprigt 0 Pearson Education, Inc. 5

25 0 CHAPTER FUNCTIONS AND GRAPHS 96. Grap of + sows solution is [, ) From te grap of + π 9, 97. Note, te points of intersection of and ( ) are ( ± (, ). Te solution set of ( ) > is, ) ( + ), we observe tat te solution set of + π 9 < 0 is (.6,.55). 0. From te grap of 5 + 6, Te points of intersection of and ( ) are (, ) and (, ). Te solution set of ( ) < is te interval (, ). - te solution set of > 0 is (0.0,.55) (.5, ). 0. a. Stretc te grap of f b a factor of. 99. From te grap of 5, we conclude tat te solution is ( 5, 5). 5 b. Reflect te grap of f about te -ais Grap of sows solution is [, ] Coprigt 0 Pearson Education, Inc.

26 . FAMILIES OF FUNCTIONS, TRANSFORMATIONS, AND SYMMETRY 05 c. Translate te grap of f to te left b - unit. g. Translate te grap of f to te rigt b -unit and up b -units. d. Translate te grap of f to te rigt b -units.. Translate te grap of f to te rigt b -units, stretc b a factor of, and up b -unit. 5 e. Stretc te grap of f b a factor of and reflect about te -ais a. Reflect te grap of f about te -ais. f. Translate te grap of f to te left b -units and down b -unit. b. Stretc te grap of f be a factor of. Coprigt 0 Pearson Education, Inc.

27 06 CHAPTER FUNCTIONS AND GRAPHS c. Stretc te grap of f b a factor of and reflect about te -ais. g. Translate te grap of f to te left b - units and reflect about te -ais d. Translate te grap of f to te left b -units.. Translate te grap of f to te rigt b - units, stretc b a factor of, and up b -unit. 5 5 e. Translate te grap of f to te rigt b -unit. 05. N() N() Yes, if te merit increase is followed b te cost of living raise ten te new salar becomes iger and is N ().05( + 000) If inflation rate is less tan 50%, ten <. Tis simplifies to <. After f. Translate te grap of f to te rigt b - units and up b -unit. squaring we ave < and so > 5%. 08. If production is at least 8 windows, ten Te need at least ( ) 8 56 rs..75 Coprigt 0 Pearson Education, Inc.

28 . FAMILIES OF FUNCTIONS, TRANSFORMATIONS, AND SYMMETRY (a) Bot functions are even functions and te graps are identical (b) One grap is a reflection of te oter about te -ais. Bot functions are odd functions. - - (c) Te second grap is obtained b sifting te first one to te left b unit.. If f() +, ten f(a + ) f(a) (a + ) + a a + a + a a +. Te grap is a lower semicircle of radius 6 wit center at te origin. Ten te domain is [ 6, 6] and te range is [ 6, 0].. Note / is equivalent to / or /. Ten te solution set is (, /] [/, ).. i 8 ( i ) 0 i i i a + b c a + b c (d) Te second grap is obtained b translating te first one to te rigt b units and units up i + i i 6 + 8i i i Tinking Outside te Bo XXIV - -8 Note, if (, + ) is suc an ordered pair ten te average is +/. Since te average is not a wole number, ten. Tus, te ordered pairs are (, 5), (9, 50), (99, 500), and (999, 5000). 0. Te grap of or equivalentl ( + ) can be obtained b sifting te grap of to te rigt b units.. Pop Quiz ( 9) Coprigt 0 Pearson Education, Inc.

29 08 CHAPTER FUNCTIONS AND GRAPHS. ( ) or. Domain [, ), range (, 5] 5. ( 6) + 6. Even function. Linking Concepts (a) From te grap of (7.67) / S it follows tat L must lie in (0, 0.7]. (d) Te variables wit negative coefficients ave maimum values. And, te variable D (wit a positive coefficient) as a minimum value. For Tougt. False, since f + g as an empt domain.. True. True. True 5. True, since A P /6. 6. True 7. False, since (f g) () 8. True 6 S we obtain S must lie in (0, 99.58]. (b) From te grap of D / False, since ( g) () True, since belongs to te domain if is a real number, i.e., if.. Eercises. sum. composition ( 9) 5 80 D 7. ( ) 8 8. ( ) / 0.. (a ) + (a a) a. (b ) (b b) b b we conclude D must lie in [9.97, ). (c) Given a portion of te grap of (7.6) / L (a )(a a) a a + a. (b )/(b b) 5. f + g {(, + ), (, 0 + 6)} {(, ), (, 6)}, domain {, } 6. f + {(, 0 + )} {(, )}, domain {} L 7. f g {(, ), (, 0 6)} {(, ), (, 6)}, domain {, } 8. f {(, 0 )} {(, )}, domain {} Coprigt 0 Pearson Education, Inc.

30 . OPERATIONS WITH FUNCTIONS f g {(, ), (, 0 6) {(, ), (, 0)}, domain {, } 0. f {(, 0 )} {(, 0)}, domain {}. g/f {(, /)} {(, )}, domain { }. f/g {(, /), (, 0/6) {(, /), (, 0)}, domain {, }. (f + g) () +, domain is [0, ). (f + ) () +, domain is [0, ) (, ) 5. (f ) (), domain is [0, ) (, ) 6. ( g) () +, domain is (, ) (, ) 7. (g ) (), domain is (, ) (, ) 8. (f ) (), domain is [0, ) (, ) ( ) g 9. (), domain is (0, ) f 0. ( f g ) (), domain is [0, ) (, ). {(, 0), (, 0), (, )}. {(, ), (0, 0)}. {(, )}. {(, 0)} 5. {(, ), (, )} 6. {(, 0)} 7. f() 5 8. g( ) 7 9. f() 5 0. ( ) 7. f(0.7) g(.95667) 9.7. (g f)() (g )(5) g() 5. ( f g)() ( f)(0) (9) 0 5. (f g )() (f g)() f() 5 6. ( g f)(0) ( g)( ) () ( ) a + 7. (f )(a) f ( ) a + (a + ) a 8. ( f)(w) (w ) (w ) + w w 9. (f g)(t) f ( t + ) (t + ) t (g f)(m) g (m ) (m ) + 9m 6m + 5. (f g)(), domain [0, ) 5. (g f)(), domain [, ) 5. (f )(), domain (, 0) (0, ) 5. ( f)(), domain (, ) (, ) 55. ( g)(), domain (0, ) 56. (g )(), domain (0, ) 57. (f f)() ( ), domain (, ) 58. (g g)(), domain [0, ) 59. ( g f)() ( ) domain (, ) 60. (f g )() f domain (0, ), ( ), 6. ( f g)() ( ) domain (0, ) (, ), Coprigt 0 Pearson Education, Inc.

31 0 CHAPTER FUNCTIONS AND GRAPHS ( ) 6. (g f)() g domain (, ) 6. F g 6. G g f 65. H g 66. M f g, 67. N g f 68. R f g 69. P g f g 70. C g 7. S g g 7. T 7. If g() and (), ten ( g)() g() f(). 7. If g() and (), ten ( g)() g() ( ). 75. If g() + 5 and (), ten ( g)() g() + 5 f(). 76. If g() and () + 5, ten ( g)() g() f(). 77. If g() and (), ten ( g)() g() f(). 8. ( + ) ( ) 87. Since m n and m, (n ). 88. Since u t + 9 and v u, v t Since w + 6, z w, and z 8, + 6 we obtain Since a b, c a + 5, and d c, we ave d b After multipling b + we ave ( ) + ( + ) + ( ) ( + ) Te domain of te original function is (, ) (, ) wile te domain of te simplified function is (, ). Te two functions are not te same. 78. If g() and (), ten ( g)() g() f(). 79. If g() and () + 5, ten - - ( g)() g() f(). 80. If g() + 5 and (), ten ( g)() g() + 5 f(). 8. ( + ) 6 8. ( ) ( ) After multipling b we ave ( + ) + ( ) ( + ) ( ) Te domain of te original function is (, ) (, ) wile te domain of te simplified function is (, ). Te two functions are not te same. Coprigt 0 Pearson Education, Inc.

32 . OPERATIONS WITH FUNCTIONS 97. Domain [0, ), range [, ) Domain [, ), range [ 7, ) Domain [ 6, ), range [0, ) Domain [, ), range [, ) Domain [, ), range [0, ) Domain [, ), range [, ) 99. P () 68 (0 + 00) Since 00/8 7., te profit is positive wen te number of trimmers satisfies P () (000 0 ) ( ) A d / 0. P A 0. (f f)() and (f f f)() 0.85 are te amounts of forest land at te start of 00 and 00, respectivel. 0. (f f)() () and (f f f)() 8 are te values of dollars invested in bonds after and 6 ears, respectivel. 05. Total cost is (T C)().05(.0) a) Te slope of te linear function is Using C() 00 00( 0), te cost before taes is C() b) T ().09 Coprigt 0 Pearson Education, Inc. c) Total cost of pallets wit ta included is (T C)().09( )

33 CHAPTER FUNCTIONS AND GRAPHS 07. Note, D d/0 d/0 L /00 00 d 0L 00 (6000) 0L 00 (6) L 00 () L. Epressing D as a function of L, we write D () L or D L. 08. Note, S (d/6) / d / /6 6(6500) d /. Epressing S as a function of d, we obtain S 6(6500) d / or S 0, 000d /. 09. Te area of a semicircle wit radius s/ is (/)π(s/) πs /8. Te area of te square is s. Te area of te window is W s + πs 8 (8 + π)s Te area of te square is A s and te area of te semicircle is S π ( s ). Ten s 8S π and A 8S π.. Addition and multiplication of functions are commutative, i.e., (f + g)() (g + f)() and (f g)() (g f)(). Addition, multiplication, and composition of functions are associative, i.e., and Q((f + g) + )() (f + (g + ))(), ((f g) )() (f (g ))(), ((f g) )() (f (g ))(). 5. Te difference quotient is ( + ) ( + ) 6. From te grap below, we conclude te domain is (, ), te range is (, ), and increasing on [, ).. Form a rigt triangle wit two sides of lengt s and a potenuse of lengt d. B te Ptagorean Teorem, we obtain d s + s. Solving for s, we ave s d.. Construct an equilateral triangle were te lengt of one side is d/ and te altitude is p/. B te Ptagorean Teorem, (p/) + (d/) (d/). Solving for p, we get p d.. If a coat is on sale at 5% off and tere is an additional 0% off, ten te coat will cost 0.90(.75) were is te regular price. Tus, te discount sale is.5% off and not 5% off. 7. Since 0, te domain is [, ). Since 5 as range (, 0], te range of f() 5 + is (, ]. 8. B te square root propert, we find ( ) 6 ± 6 ± 6. { } 6 ± 6 Te solution set is. Coprigt 0 Pearson Education, Inc.

34 .5 INVERSE FUNCTIONS 9. Since 5 >, te solution set is ( /5, ). 0. Since 7 9, te slope is. Tinking Outside te Bo XXV. (0.7)(0.7) (0.7) (0.7) 97.% or XXVI. Since 0 () + () and () + ( ), all integers at least 0 can be epressed in te form +. Note, tere are no wole numbers and tat satisf 9 +. Tus, 9 is te largest wole number N tat cannot be epressed in te form +.. Pop Quiz. A πr π(d/) or A πd. (f + g)() (f g)() 6. (f g)(5) f() 9 5. {(, 8 + 9)} {(, 7)} 6. Since (n m)() n() 5, we find {(, 5)}. 7. Since ( + j)() + +, te domain is [, ). ( ), 8. Since ( j)() + te domain is [, ). 9. Since (j )() +, te domain is (, ).. Linking Concepts a) Let n be te number of beads in one foot and suppose n is also te number of spaces. Since te sum of te diameters of te notces and te spaces is foot, we ave (n) d. Solving for n, we obtain n 6 d. b) If c is te area in square inces of a cross section of a bead, ten c ( ) d π πd 8. Tus, c πd 8 sq in. c) Note, a parallel bead is a alf-circular clinder. Since te lengt of a bead is inces and te area of a cross section of a bead is known (as in part (b)), te volume of a bead is given b v c πd πd () 8 cubic inces. d) Te volume v of glue on ft of floor is v v n. Tus, v v n πd cubic inces. 6 d 9πd e) Note, ft 78in and gal cubic inces. Let A be te number of square inces one gallon of glue will cover. Ten A v (9πd). Hence, A 5.6 πd square feet. f) For te square notc wose side is d inces in lengt, we find te corresponding values of n, c, v, v, and A as in parts (a)-(e). i) n 6 d. ii) c d square inces iii) v c d cubic inces iv) v v n d 6 7d cubic inces d v) As in part e), we ave For Tougt A v (7d). Tus, A. d square feet.. False, since te inverse function is {(, ), (5, 5)}. Coprigt 0 Pearson Education, Inc.

35 CHAPTER FUNCTIONS AND GRAPHS. False, since it is not one-to-one.. False, g () does not eist since g is not one-to-one.. True 5. False, a function tat fails te orizontal line test as no inverse. 6. False, since it fails te orizontal line test. ( ) 7. False, since f () + were False, f () does not eist since f is not one-to-one. 9. False, since is V -saped and te orizontal line test fails. 0. True.5 Eercises. one-to-one. invertible. inverse. smmetric 5. Yes, since all second coordinates are distinct. 6. Yes, since all second coordinates are distinct. 7. No, since tere are repeated second coordinates suc as (, ) and (, ). 8. No, since tere are repeated second coordinates as in (, ) and (5, ). 9. No, since tere are repeated second coordinates suc as (, 99) and (5, 99). 0. No, since tere are repeated second coordinates as in (, 9) and (, 9).. Not one-to-one. Not one-to-one. One-to-one. One-to-one 5. Not one-to-one 6. One-to-one 7. One-to-one; since te grap of sows is an increasing function, te Horizontal Line Test implies is one-to-one. 8. One-to-one; since te grap of 9 sows 9 is an increasing function, te Horizontal Line Test implies 9 is one-to-one. 9. One-to-one; for if q( ) q( ) ten 5 5 ( )( 5) ( )( 5) ( ) 0 0. Tus, if q( ) q( ) ten. Hence, q is one-to-one. 0. One-to-one; for if g( ) g( ) ten + + ( + )( ) ( + )( ) ( ) 0 0. Tus, if g( ) g( ) ten. Hence, g is one-to-one.. Not one-to-one for p( ) p(0).. Not one-to-one for r(0) r().. Not one-to-one for w() w( ).. Not one-to-one for v() v( ). 5. One-to-one; for if k( ) k( ) ten ( + 9) ( + 9) Coprigt 0 Pearson Education, Inc.

36 .5 INVERSE FUNCTIONS 5 Tus, if k( ) k( ) ten. Hence, k is one-to-one. 6. One-to-one; for if t( ) t( ) ten + + ( + ) ( + ) + +. Tus, if t( ) t( ) ten. Hence, t is one-to-one.. f {(5, ), (0, 0), (6, )}, f (5), (f f)(). f {(, ), (5, 0), ( 7, )}, f (5) 0, (f f)(). f {(5,.), (.99, )}, f (5)., (f f)() 5. Not invertible since it fails te Horizontal Line Test. 7. Invertible, {(, 9), (, )} Invertible, {(5, ), (6, 5)} 9. Not invertible Not invertible Invertible, {(, ), (, ), (, ), (7, 7)}. Invertible, {(, ), (, ), (6, ), (9, 7)}. Not invertible 6. Not invertible since it fails te Horizontal Line Test.. Not invertible 5. Not invertible, tere can be two different items wit te same price. 6. Not invertible since te number of ears (given as a wole number) cannot determine te number of das since our birt Invertible, since te plaing time is a function of te lengt of te VCR tape. 8. Invertible, since.6 km mile 7. Not invertible since it fails te Horizontal Line Test. 9. Invertible, assuming tat cost is simpl a multiple of te number of das. If cost includes etra carges, ten te function ma not be invertible. 0. Invertible, since te interest is uniquel determined b te number of das f {(, ), (5, )}, f (5), (f f)() Coprigt 0 Pearson Education, Inc.

37 6 CHAPTER FUNCTIONS AND GRAPHS 8. Not invertible since it fails te Horizontal Line Test a) f() is te composition of multipling b 5, ten subtracting. Reversing te operations, te inverse is f () 5 b) f() is te composition of multipling b, ten subtracting 88. Reversing te operations, te inverse is f () + 88 c) f () ( + 7)/ d) f () e) f () ( + 9) + 8 f) f () g) f() is te composition of taking te cube root of, ten subtracting 9. Reversing te operations, te inverse is f () ( + 9) ) f() is te composition of cubing, multipling te result b, ten subtracting 7. Reversing te operations, te inverse is f () + 7 i) f() is te composition of subtracting from, taking te cube root of te result, ten adding 5. Reversing te operations, te inverse is f () ( 5) + j) f() is te composition of subtracting 7 from, taking te cube root of te result, ten multipling b. Reversing te operations, te inverse is ( ) f () a) f() is te operation of dividing b. Reversing te operation, te inverse is f () b) f() is te operation of adding 99 to. Reversing te operation, te inverse is f () 99 c) f() is te composition of multipling b 5, ten adding. Reversing te operations, te inverse is f () ( )/5 d) f() is te composition of multipling b, ten adding 5. Reversing te operations, te inverse is f () 5 e) f() is te composition of dividing b, ten adding 6. Reversing te operations, te inverse is f () ( 6) 8 f) f() is te operation of taking te multiplicative inverse of. Since taking te multiplicative inverse twice returns to te original number, te inverse is taking te multiplicative inverse, i.e., f () / g) f() is te composition of subtracting 9 from, ten taking te cube root of te result. Reversing te operations, te inverse is f () + 9. ) f() is te composition of cubing, multipling te result b, ten adding. Reversing te operations, te inverse is f () ( ). i) f() is te composition of adding to, taking te cube root of te result, ten multipling b. Reversing te operations, te inverse is f () ( ) j) f() is te composition of adding to, taking te cube root of te result, ten subtracting 9. Reversing te operations, te inverse is f () ( + 9). Coprigt 0 Pearson Education, Inc.

38 .5 INVERSE FUNCTIONS 7 5. No, since te fail te Horizontal Line Test. 5. Yes, since te are smmetric about te line. 5. Yes, since te graps are smmetric about te line. 5. No since f() fails te Horizontal Line Test. 55. Grap of f 58. Grap of f 59. f () f () Grap of f f () Grap of f f () - Coprigt 0 Pearson Education, Inc.

39 8 CHAPTER FUNCTIONS AND GRAPHS 6. f () 69. Intercange and ten solve for. + for ( ) for f () ( ) + for 6. f () Intercange and ten solve for. for 0 + for 0 f () + for 0 7. Intercange and ten solve for. 65. f () ( + ) for 9 9 f () f () + for Intercange and ten solve for f () 68. Intercange and ten solve for f () 5 7. Intercange and ten solve for. + + f () + 7. Intercange and ten solve for ( ) 5 + f () Intercange and ten solve for ( ) 6 f () 6 Coprigt 0 Pearson Education, Inc.

40 .5 INVERSE FUNCTIONS Intercange and ten solve for. 76. Clearl f () f () 77. Intercange and ten solve for ( 5) 9 f () ( 5) Intercange and ten solve for ( 5) f () ( 5) 79. Intercange and ten solve for. ( ) 0 f () Intercange and ten solve for. 0 since domain of f is (, 0] f () 0 8. Note, (g f)() 0.5( + ) and (f g)() (0.5 ) +. Yes, g and f are inverse functions of eac oter. 8. Note, (g f)() 0.(0 5) + and (f g)() 0 5( 0. + ). Yes, g and f are inverse functions of eac oter. ( ) 8. Since (f g)() + and and (g f)() +, g and f are not inverse functions of eac oter. 8. Since (f g)() and and (g f)() ( ), g and f are not inverse functions of eac oter. 85. We find and (f g)() (f g)() (g f)() /( ) + + (g f)(). ( + ) / Ten g and f are inverse functions of eac oter. 86. We obtain and (f g)() (f g)() (g f)() /( ) ( ) (g f)() ( ) / Tus, g and f are inverse functions of eac oter. 87. We obtain (f g)() Coprigt 0 Pearson Education, Inc.

41 0 CHAPTER FUNCTIONS AND GRAPHS and 5 5 (f g)() ( ) (g f)() ( ) ( ) + (g f)(). Tus, g and f are inverse functions of eac oter. 88. We note and (f g)() ( + ) 7 (g f)() 7 +. Ten g and f are not inverse functions of eac oter. 89. and are inverse functions of eac oter and and are inverse functions of eac oter and C.08P epresses te total cost as a function of te purcase price; and P C/.08 is te purcase price as a function of te total cost. 9. V (), S() 9. Te grap of t as a function of r satisfies te Horizontal Line Test and is invertible. Solving for r we find, t r r t and te inverse function is r 7.89 t 0.9. If t 5.55 min., ten r rowers. 9. Solving for F, we obtain F 9C 5 F 9C and te inverse function is F 9C 5 + ; a formula tat can convert Celsius temperature to Fareneit temperature. 95. Solving for w, we obtain w V w V.96 and te inverse function is w V.96. If V 5 ft./sec., ten w 5 8, 80 lb..96 r V V were 97. a) Let V $8, 000. Te depreciation rate is r or r 0.9%. ( ) 8, 000 / , 000 Coprigt 0 Pearson Education, Inc.

42 .5 INVERSE FUNCTIONS b) Writing V as a function of r we find r ( r) 5 and V 50, 000( r) Let P 80, 558. Ten r ( V 50, 000 ) /5 V 50, 000 ( ), 0 / , 000 Te average annual growt rate is r 8.%. Solving for P, we obtain + r ( + r) 0 and P 0, 000( + r) Since g () + 5 we ave g f () ( P 0, 000 ) /0 P 0, 000 and f (), Likewise, since (f g)() 6 9, we get (f g) () Hence, (f g) g f. 00. Since ( f g g f ) () ( f (g g ) ) (f ()) f(f ()) and te range of one function is te domain of te oter function, we ave (f g) g f. 0. One can easil see tat te slope of te line joining (a, ( b) to (b, a) is, and tat teir midpoint is a + b, a + b ). Tis midpoint lies on te line wose slope is. Ten is te perpendicular bisector of te line segment joining te points (a, b) and (b, a) 0. It is difficult to find te inverse mentall since te two steps and + are done separatel and simultaneousl. 0. Dividing we get ( ) 0. f () or 5 f () (f g)() f(g()) f() + 5 (f g)() f()g()) Observe, te grap of f() 9 is a lower semicircle of radius centered at (0, 0). Ten domain is [, ], te range is [, 0], and increasing on [0, ] 08. No, two ordered pairs ave te same first coordinates Te solution set is {5.5}. 0. Te slope of te perpendicular line is /. Using m + b and te point (, ), we find () + b + b 5 b Te perpendicular line is + 5. Tinking Outside te Bo XXVII Since 60, , we ave eiter 0 and 5, or 5 and 0. In eiter case, 99. Coprigt 0 Pearson Education, Inc.

43 CHAPTER FUNCTIONS AND GRAPHS.5 Pop Quiz. No, since te second coordinate is repeated in (, ) and (, ).., since te order pair (5, ) belong to f.. Since f () /, f (8) 8/.. No, since f() f( ) or te second coordinate is repeated in (, ) and (, ). 5. f () + 6. Intercange and ten solve for ( + ) + ( + ) Te inverse is g () ( + ). ( j)() ( + 5) ( ) ( j)().5 Linking Concepts a) r 65 ( ( A P ) /n ) ( + 5) 5 b) For te first loan, ( (9 ) /0 r 65 ) or r 65.% annuall. For te second loan, ( (9 ) /0 r 65 ).9 00 or r 9% annuall. For te tird loan, ( (9 ) /0 r 65 ) or r 0.9% annuall.. c) If one borrowed $00 at an annual rate of r 65.% compounded dail, ten after one ear one will ave to pa back ( ) 65 $, e) It carges ig rates because of ig risks. For Tougt. False. False, since cost varies directl wit te number of pounds purcased.. True. True 5. True, since te area of a circle varies directl wit te square of its radius. 6. False, since k/ is undefined wen True 8. True 9. True 0. False, te surface area is not equal to (Surface Area ) k lengt widt eigt for some constant k..6 Eercises. varies directl. variation. varies inversel. varies jointll 5. G kn 6. T kp 7. V k/p 8. m k/m 9. C kr 0. V kr. Y k z. W krt/v. A varies directl as te square of r. C varies directl as D 5. varies inversel as 6. m varies inversel as m Coprigt 0 Pearson Education, Inc.

44 .6 CONSTRUCTING FUNCTIONS WITH VARIATION 7. Not a variation epression 8. Not a variation epression 9. a varies jointl as z and w 0. V varies jointl as L, W, and H. H varies directl as te square root of t and inversel as s. B varies directl as te square of and inversel as te square root of. D varies jointl as L and J and inversel as W. E varies jointl as m and te square of c. 5. Since k and 5 k 9, k 5/9. Ten 5/9. 6. Since kz and 0 k 00, k /0. So z/0. 7. Since T k/ and 0 k/5, k 50. Tus, T 50/. 8. Since H k/n and 9 k/( 6), k 5. So H 5/n. 9. Since m kt and 5 k 8, k. Tus, m t. 0. Since p k w and get k k, we. Ten p w.. Since k/ z and.9 k(.)/.5, we obtain k.7. Hence,.7/ z.. Since n k b and 8.95 k(.5) 5., we find k.6. So n.6 b. Since k and 9 k(), we obtain 9 ( ) 7/.. Since kz and 6 k, we obtain Since P k/w and / k, we find / k /6. Tus, P 6 /6. 6. Since H k/q and 0.0 k, we get 0.0 P Since A klw and 0 k()(5 ), we obtain A ( ) Since J kgv and k 8, we find J Since ku/v and 7 k 9/6, we find 8 /6 7/. 0. Since q k /j and 8 k 9/8, we get q 8 / Let L i and L f be te lengt in inces and feet, respectivel. Ten L i L f is a direct variation.. Let T s and T m be te time in seconds and minutes, respectivel. Ten T s 60T m is a direct variation.. Let P and n be te cost per person and te number of persons, respectivel. Ten P 0/n is an inverse variation.. Let n and w be te number of rods and te weigt of a rod, respectivel. 0, 000 Ten n is an inverse variation. w 5. Let S m and S k be te speeds of te car in mp and kp, respectivel. Ten S m S k /.6 0.6S k is a direct variation. 6. Let W p and W k be te weigt in pounds and kilograms, respectivel. Ten W p.w k is a direct variation. Coprigt 0 Pearson Education, Inc.