CHAPTER (A) When x = 2, y = 6, so f( 2) = 6. (B) When y = 4, x can equal 6, 2, or 4.

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1 SECTION CHAPTER 3 Section No. A correspondence between two sets is a function only if eactly one element of te second set corresponds to eac element of te first set. 3. Te domain of a function is te set of all first components in te ordered pairs defining te function; te range is te set of all second components. 5. Te symbol f() does not denote multiplication of by f. 7. A function 9. Not a function (two range values correspond to domain values 3 and 5) 11. A function 13. A function; domain {, 3, 4, 5}; range {4, 6, 8, 10} 15. Not a function (two range values correspond to domain values 5 and 10) 17. A function; domain {Oio, Alabama, West Virginia, California}; range {Obama, McCain} 19. A function 1. Not a function (fails vertical line test since te y ais crosses te grap tree times.) 3. Not a function (fails vertical line test since te grap itself is vertical) 5. (A) A function. (B) Not a function, as long as tere is more tan one student in any Mat 15 class. 7. f() 3 5 (A) f(3) 3(3) 5 4 (B) f() 3 5 (C) f(3) + f() 4 + (3 5) 3 1 (D) f(3 + ) 3(3 + ) Common Error: f(3 + ) is not te same as f(3) + or f(3) + f(). 9. F(w) w + w (A) F(4) (4) + (4) (B) F( 4) ( 4) + ( 4) (C) F(4 + a) (4 + a) + (4 + a) (16 + 8a + a ) a 16 8a a a a 6a 8 (D) F( a) ( a) + ( a) (4 4a + a ) + 4 a 4 + 4a a + 4 a a + a 31. f(t) 3t (A) f( ) 3( ) 1 10 (B) f( t) 3( t) 3t (C) f(t) ( 3t ) + 3t (D) f( t) ( 3t ) + 3t 35. (A) Wen, y 6, so f( ) 6. (B) Wen y 4, can equal 6,, or F(u) u u 1 (A) F(10) (10) (10) 1 89 (B) F(u ) (u ) (u ) 1 u 4 u 1 (C) F(5u) (5u) (5u) 1 5u 5u 1 (D) 5F(u) 5(u u 1) 5u 5u Solving for te dependent variable y, we ave y 1 y + 1 Since + 1 is a real number for eac real number, te equation defines a function wit domain all real numbers.

2 10 CHAPTER 3 FUNCTIONS 39. Solving for te dependent variable y, we ave 3 + y 4 y y ± 4 Since eac positive number as two real square roots, te equation does not define a function. For 3 eample, wen 0, y ± 4 (0) ± 4 ±. 43. If + y 7, ten y + 7. Tis equation does not define a function. For eample, if 0, ten y 7, so y 7 or y Since f() is a polynomial, te domain is te set of all real numbers R, < <, (, ). 51. If te denominator of te fraction is zero, te function will be undefined, since division by zero is undefined. For any oter values of z, (z) represents a real number. Solve te equation 4 z 0; z 4. Tus, te domain is all real numbers ecept 4 or (, 4) (4, ); z < 4 or z > Te formula 7 3w is defined only if 7 + 3w 0, since te square root is only defined if te number inside is nonnegative. We solve te inequality: 7 + 3w 0 3w 7 w 7 3 Tus, te domain is all real numbers greater tan or equal to 7 3 or 7, Solving for te dependent variable y, we ave 3 y y 3 y 3 Since 3 is a real number for eac real number, te equation defines a function wit domain all real numbers. 45. Solving for te dependent variable y, we ave 3y + 1 3y 1 y 1 3 Since 1 is a real number for eac real 3 number, te equation defines a function wit domain all real numbers. 49. Since 3u + 4 is never negative, 3u 4 represents a real number for all replacements of u by real numbers. Te domain is R, < u <, (, ). 53. Te square root of a number is defined only if te number is nonnegative. Tus, t 4 is defined if t 4 0 and undefined if t 4 < 0. Note tat t 4 0 wen t 4, so te domain is all real numbers greater tan or equal to 4 or [4, ). u 57. Te fraction is defined for any value of u 4 u tat does not make te denominator zero, so we solve te equation u to find values tat make te function undefined. u u 4 Tis equation as no solution since te square of a number can't be negative. So tere are no values of u tat make te fraction undefined and te domain is all real numbers or (, ). 59. Tere are two issues to consider: we need to make certain tat so tat te number inside te square root is nonnegative, and we need to avoid -values tat make te denominator zero. First, wenever 4. So must be greater tan or equal to 4 to avoid a negative under te root. Also, 1 0 wen 1, so cannot be 1. Te domain is all real numbers greater tan or equal to 4 ecept 1, or [ 4, 1) (1, ); 4 < 1 or > 1.

3 SECTION Tere are two issues to consider: te number inside te square root as to be nonnegative, and te denominator of te fraction as to be nonzero. Since we just ave t inside te square root, t as to be greater tan or equal to zero. Net, we solve 3 t 0 to find any t-values tat make te denominator zero. 3 t 0 3 t 9 t Tus, t cannot be 9. Te domain is all real numbers greater tan or equal to zero ecept 9, or [0, 9) (9, ); 0 t < 9 or t > g() G() 8 4( + ) 67. Function f multiplies te square of te domain element by, ten adds 5 to te result. 71. F(s) 3s + 15 F( + ) 3( + ) + 15 F() 3() + 15 F( ) F() [3( ) 15] [3() 15] 73. g() g(3 + ) (3 + ) g(3) (3) g(3 ) g(3) [ (3 )] [ (3) ] 69. Function Z multiplies te domain element by 4, adds 5 to te result, ten divides tis result by te square root of te domain element. [6 3 15] [1] [ 9 6 ] [ 7] ( 6 ) (A) f() 4 7 f( + ) 4( + ) 7 f ( ) f( ) [4( ) 7] [4 7] Common Errors: f( + ) f() + f() or f( + ) f() + or Also note: f() 4 7 Parenteses must be supplied. (B) f() 4 7 f(a) 4a 7 f ( ) f( a) (47) (4a7) a a 4 7 4a 7 a 4 4a a 4( a) 4 a

4 104 CHAPTER 3 FUNCTIONS 77. (A) f() 4 (B) f() 4 f( + ) ( + ) 4 f(a) a 4 f ( ) f( ) [( ) 4] [ 4] f ( ) f( a) ( 4) (a 4) a a ( ) 4 4 4a 4 a a a 4 ( a )( a ) a (4 ) ( + a) + a (A) f() f( + ) 4( + ) + 3( + ) f ( ) f( ) [ 4( ) 3( ) ] [ 4 3] 4( ) 3( ) ( 843) (B) f() f(a) 4a + 3a f ( ) f( a) a a a a 4 34a 3a a 4 4a 33a a 4( a)( a) 3( a) a ( )[ 4( ) 3] a a a 4( + a) a + 3 ( 4 3 ) ( 4 3 )

5 SECTION (A) f() f( + ) f ( ) f( ) (B) f() ( ) ( ) ( ) ( ) ( ) ( ) 1 f(a) a f ( ) f( a) a a a a a a a ( ) ( a) ( a)( a) ( a) a ( a)( a) ( a )( a ) 1 a 83. (A) f() 4 (B) f() 4 f( + ) 4 4 f ( ) f( ) 4 4 4( ) ( ) 4 ( ) 4 ( ) f(a) 4 a 4 4 f ( ) f( a) a a a 4a 4 a( a) 4( a) a( a) 4 a

6 106 CHAPTER 3 FUNCTIONS 85. Given w widt and Area 64, we use A w to write A w 64 w. Ten P w + w w +. Since w must be positive, te domain of P(w) is w > 0. w w 87. Using te given letters, te Pytagorean teorem gives b b + 5 b b 5 (since is positive) Since b must be positive, te domain of (b) is b > Daily cost fied cost + variable cost C() $300 + ($1.75 per dozen dougnuts) (number of dozen dougnuts) C() Te cost is a flat $17 per mont, plus $.40 for eac our of airtime. 93. (A) S(0) 16(0) 0; S(1) 16(1) 16; S() 16() 16(4) 64; S(3) 16(3) 16(9) 144 (Note: Remember tat te order of operations requires tat we apply te eponent first ten multiply by 16.) S( ) S() 16( ) 16() 16(4 4 ) 16(4) (B) (64 16 ) (Note: Be careful wen evaluating S( + )! You need to replace te variable t in te function wit ( + ). S( + ) is not te same ting as S() +!) (C) 1: (1) 80 1: ( 1) : (0.1) ; 0.1: ( 0.1) : (0.01) : 0.01: ( 0.01) : (0.001) : ( 0.001) (D) Te smaller gets te closer te result is to 64. Te numerator of te fraction, S( + ) S(), is te difference between ow far an object as fallen after + seconds and ow far it's fallen after seconds. Tis difference is ow far te object falls in te small period of time from to + seconds. Wen you divide tat distance by te time (), you get te average velocity of te object between and + seconds. Part (C) sows tat tis average velocity approaces 64 feet per second as gets smaller.

7 SECTION From te above figures it sould be clear tat V lengt widt eigt (1 )(8 ). Since all distances must be positive, > 0, 8 > 0, 1 > 0. Tus, 0 <, 4 >, 6 >, or 0 < < 4 (te last condition, 6 >, will be automatically satisfied if < 4.) Domain: 0 < < From te tet diagram, since eac pen must ave area 50 square feet, we see Area (lengt)(widt) or 50 (lengt). Tus, te lengt of eac pen is 50 ft. I Te total amount of fencing 4(widt) + 5(lengt) + 4(widt gate widt) F() ( 3) F() Ten F(4) 8(4) F(6) 8(6) F() F(5) 8(5) F(7) 8(7) We note tat te pipeline consists of te lake section IA, and sore section AB. Te sore section as lengt 0. Te lake section as lengt, were 8 +, tus 64. A B 0 - cost of sore number of cost of lake number of Te cost of all te pipeline + section per milesore miles section per mile lake miles C() 10,000(0 ) + 15, From te diagram we see tat must be non-negative, but no more tan 0. Domain: 0 0

8 108 CHAPTER 3 FUNCTIONS Section 3-1. Te grap of a function f() is te set of all points wose first coordinate is an element of te domain of f and wose second coordinate is te associated element of te range. 3. Te grap of a function can ave one y intercept (wen y f(0) ) or none (if 0 is not in te domain). Te grap can ave any number of intercepts. 5. A function is increasing on an interval if for any coice 1, of values on tat interval, if > 1, ten f( ) > f( 1 ). 7. A function is defined piecewise if it is defined by different epressions for different parts of its domain. 9. (A) [ 4, 4) (B) [ 3, 3) (C) 0 (D) 0 (E) [ 4, 4) (F) None (G) None (H) None 11. (A) (, ) (B) [ 4, ) (C) 3, 1 (D) 3 (E) [ 1, ) (F) (, 1] (G) None (H) None 13. (A) (, ) (, ) (Te function is not defined at.) (B) (, 1) [1, ) (C) None (D) 1 (E) None (F) (, ] (, ) (G) [, ) (H) 15. f( 4) 3 since te point ( 4, 3) is on te grap; f(0) 0 since te point (0,0) is on te grap; f(4) is undefined since tere is no point on te grap at ( 3) 0 since te point ( 3, 0) is on te grap; (0) 3 since te point (0, 3) is on te grap; () 5 since te point (, 5) is on te grap. 19. p( ) 1 since te point (, 1) is on te grap; p() is undefined since tere is no point on te grap at ; p(5) 4 since te point (5, 4) is on te grap. 1. One possible answer: 3. One possible answer: 5. One possible answer: 7. f( ) 4 slope Te y intercept is f(0) 4, and te slope is. To find te intercept, we solve te equation f() 0 for. f() Te intercept is.

9 SECTION f() Te y intercept is f(0) 5 3, and te slope is 1. To find te intercept, we solve te equation f() 0 for. f() ( ) Te intercept is f() Te y intercept is f(0) 7.1, and te slope is.3. To find te intercept, we solve te equation f() 0 for. f() Te intercept is A linear function must ave te form f() m + b. We are given f(0) 10, ence 10 f(0) m(0) + b. Tus b 10. Since f( ), we also ave f( ) m( ) + b m Solving for m, we ave m m m 4 Hence f() m + b becomes f() A linear function must ave te form f() m + b. We are given f( ) 7, ence 7 f( ) m( ) + b. Tus b 7 + m. Since f(4), we also ave f(4) 4m + b. Substituting, we ave b 7 + m 4m + b 4m m 6m m m 3 3 b 7 + m Hence f() m + b becomes f() Te rational epression 3 1 is defined 4 everywere ecept at te zero of te denominator: Te domain of f is { }. A rational epression is 0 if and only if te numerator is zero: Te intercept of f is 4. Te y intercept of f is f(0) 3(0) 1 3. (0) 4

10 110 CHAPTER 3 FUNCTIONS 39. Te rational epression 3 is defined 4 5 everywere ecept at te zero of te denominator: Te domain of f is 5 4. A rational epression is 0 if and only if te numerator is zero: Te intercept of f is Te rational epression is defined ( ) everywere ecept at te zero of te denominator: ( ) 0 0 Te domain of f is { }. A rational epression is 0 if and only if te numerator is zero: Te intercept of f is 0. 4(0) Te y intercept of f is f(0) 0. (0 ) Te y intercept of f is f(0) 3(0) 4(0) Te rational epression is defined 9 everywere ecept at te zeros of te denominator: 9 0 ( 3)( + 3) 0 3 or 3 Te domain of f is { 3, 3}. A rational epression is 0 if and only if te numerator is zero: 16 0 ( 4)( + 4) 0 4 or 4 Te intercepts of f are ± Te y intercept of f is f(0) Te rational epression is defined 5 everywere ecept at te zeros of te denominator: 5 0 ( 5)( + 5) 0 5 or 5 Te domain of f is { 5, 5}. A rational epression is 0 if and only if te numerator is zero as no real solutions, ence f as no intercept. 0 7 Te y intercept of f is f(0) (A) For 1 < 0, f() + 1, so f( 1) For 0 1, f() + 1, so f(0) and f(1) (B).0 y (0, 1) (C) Te domain is te union of te intervals used in te definition of f: [ 1, 1]. From te grap, te range is [0, 1]. Te function is continuous on its domain. (-1, 0) 0. (1, 0) -1 1

11 SECTION (A) For 3 < 1, f(), so f( 3). f( 1) is not defined. For 1 <, f() 4, so f() 4. (B) 4 y (, 4) -4 4 (C) Te domain is te union of te intervals used in te definition of f: [ 3, 1) ( 1, ]. From te grap, te range is te set of numbers {, 4}. Te function is discontinuous at 1. (-3, -) (A) For < 1, f() +, so f( ) + 0. f( 1) is not defined. For > 1, f(), so f(0) 0. (B) 5 y -5 (-, 0) 5 (0, -) (C) Te domain is te union of te intervals used in te definition of f: (, 1) ( 1, ). From te grap, te range is R. Te function is discontinuous at (A) For <, f() 6, so f( 3) ( 3) 6 0. For < 3, f(), so f( ) f(0). For 3, f() 6 0, so f(3) and f(4) (B) 5 y (-3,, 0) -5 5 (3, -) (-, -) (0, -) (4, 4) (C) Te domain is te union of te intervals used in te definition of f. Te domain is terefore R. From te grap, te range is [, ). Te function is continuous on its domain (A) For <, f() 5 + 6, so f( 3) 5 ( 3) For 3, f() 1, so f( ) f(0) f(3) 1. For > 3, f() 3 7, so f(4) 3 (4) 7 5.

12 11 CHAPTER 3 FUNCTIONS (B) 5 y (4, (-, 1) (0, 1) (3, 1) -5 5 (-3, 3 ) 5 ) (C) Te domain is te union of te intervals used in te definition of f. Te domain is terefore R. From te grap, te range is R. Te function is continuous on its domain (A) For < 0, f() 3 + 4, so f( 1) 3 ( 1) f(0) is not defined. For 0 < <, f() 1 + 3, so f(1) 1 (1) f() is not defined. For >, f() 1, so f(3) 1 (3) 3. (B) (-1, 5 y 10 3 ) (1, 5 ) -5 5 (C) Te domain is te union of te intervals used in te definition of f. (, 0) (0, ) (, ). From te grap, te range is (, 4). Te function is discontinuous at 0 and. (3, 3 ) For 0, te grap is a line wit slope 1 and y intercept 1, tat is, y ( ) ( 3) ( 1) For > 0, te grap is a line wit slope 1 and y intercept 1, tat is, y 1. 0 Terefore, 1 if 0 f() 1 if For 1, te grap is te orizontal line y 3. For > 1, te grap is a line wit slope 1 ( 1) passing troug (1, 3). Te point-slope form yields y ( 3) 3( 1) or y 3. Terefore, 3 if 1 f() 3 if 1 An equally valid solution would be 3 if 1 f() 3 if 1 3

13 SECTION For <, te grap is te orizontal line y 3. Te function is not defined at or at 1. For < < 1, te grap is a line wit slope ( 4) wic would, if etended, pass troug (, ). 1 ( ) Te point-slope form yields y [ ( )], tat is, y. For > 1, te grap is te orizontal line y 1. Terefore, 3 if f() if 1 1 if If < 0, ten and f() ( ) 1. If 0, ten and f() 1 +. Terefore, 1 if 0 f() 1 if 0 Te function is defined for all real numbers; te domain is R. From te grap, te range is [1, ). Te function is continuous on its domain. y If < 0, tat is, <, f() ( ) +. If 0, tat is,, f(). Terefore, if f() if Te function is defined for all real numbers; te domain is R. From te grap, te range is [0, ). Te function is continuous on its domain (A) One possible answer: (B) Tis grap crosses te ais once. To meet te conditions specified a grap must cross te ais eactly once. If it crossed more times te function would ave to be decreasing somewere; if it did not cross at all te function would ave to be discontinuous somewere. -5 y 71. (A) One possible answer: (B) Tis grap crosses te ais twice. To meet te conditions specified a grap must cross te ais at least twice. If it crossed fewer times te function would ave to be discontinuous somewere. However, te grap could cross more times; in fact tere is no upper limit on te number of times it can cross te ais.

14 114 CHAPTER 3 FUNCTIONS 73. Graps of f and g 10 Grap of m() 0.5[ ] 0.5[ Grap of n() 0.5[ ] 0.5[ ] Graps of f and g Graps of f and g Grap of m() 0.5[ (0.3 4) ] 0.5[ ] Grap of m() 0.5[ (0.3 3) ] 0.5[ ] Grap of n() 0.5[ (0.3 4) ] 0.5[ ] Grap of n() 0.5[ (0.3 3) ] 0.5[ ] Te graps of m() sow tat te value of m() is always te larger of te two values for f() and g(). In oter words, m() ma[f(), g()]. 81. Since 100 miles are included, only te daily carge of $3 applies for mileage between 0 and 100. So if R() is te daily cost of rental were is miles driven, R() 3 if After 100 miles, te carge is an etra $0.16 for eac mile: te mileage carge will be 0.16 times te number of miles over 100 wic is 100. So te mileage carge is 0.16( 100) or Te $3 carge still applies so wen 100 te rental carge is , or R()

15 SECTION If 0 3,000, E() 00 Base + Commission on Sales Salary Over $3,000 If $3,000 < < 8,000, E() ( 3000) Tere is a point of discontinuity at 8,000. Salary + Bonus If 8,000. E() Summarizing, E() E(5,750) (5,750) $310 E(9,00) (9,00) $548 Common Error: Commission is not 0.04 (4% of sales) nor is it (base salary plus 4% of sales). 00 if 0 3, if 3,000 8, if 8,000 y 00 y y , , , , , f f f f f f f f f f (4) (0) 0 ( 4) (0) 0 (6) (1) 10 ( 6) ( 1) 10 (4) () 0 f rounds numbers to te tens place (5) (3) 30 (47) (5) 50 ( 43) ( 4) 40 ( 45) ( 4) 40 ( 46) ( 5) Since f() /10 rounds numbers to te nearest tent, (see tet eample 6) we try /100 f() to round to te nearest undredt. f(3.74) 37.9 / f(7.846) / f(-.8783) / A few eamples suffice to convince us tat tis is probably correct. (A proof would be out of place in tis book.) f() /100.

16 116 CHAPTER 3 FUNCTIONS 89. (A) C( ) (B) Te two functions appear to coincide, for eample $30 C(3.5) 4 f(3.5) However, 1 3 C(1) 15 f(1) C() $15 Te functions are not te same, terefore. In fact, f() C() at 1,, 3, 4, 5, y 91. On te interval [0, 10,000], te ta is On te interval (10,000, ), te ta is 0.03(10,000) ( 10,000) or , tat is, 1, Combining te intervals wit te above linear epressions, we ave 0.03 if 0 10,000 T() if 10, , A ta of 5.35% on any amount up to $19,890 is calculated by multiplying by te income, wic is represented by. So te ta due, t(), is if is between 0 and 19,890. For -values between 19,890 and 65,330, te percentage is computed only on te portion over 19,890. To find tat portion we subtract 19,890 from te income ( 19,890); ten multiply by to get te percentage portion of te ta. Te total ta is $1,064 plus te percentage portion, so t() 1, ( 19,890) if is between 19,890 and 65,330. Te ta for incomes over $65,340 is computed in a similar manner: $4,68 plus 7.85% of te portion over 65,330, wic is 65,330. We get $4, ( 65,330) if is over 65,330. Combined we get if 0 19,890 t() 1, ( 19,890) if 19,890 65,330 4, ( 65,330) if 65,330 or, after simplifying, if 0 19,890 t() if 19,890 65, if 65,330 t(10,000) (10,000) 535; te ta is $535 t(30,000) (30,000) $1, t(100,000) (100,000) $6, Section For eac point wit coordinates (, f()) on te grap of y f() tere is a corresponding point wit coordinates (, f() +k) on te grap of y f() + k. Since tis point is k units above te first point, eac point, and tus te entire grap, as been moved upward k units. 3. For eac point wit coordinates (, f()) on te grap of y f() tere is a corresponding point wit coordinates (, f()) on te grap of y f(). Since tis point is te reflection of te first point wit respect to te ais, te entire grap is a reflection of te grap of y f(). Similarly for y f( ).

17 SECTION Domain: Since 0, te domain is [0, ) Range: Since te range of f() is y 0, for (), y 0. Tus, te range of is (, 0]. 7. Domain: R Range: Since te range of f() is y 0, for g(), y 0. Tus, te range of g is (, 0]. 9. Domain: R; Range: R 11. Te grap of y () is te grap of y f() sifted up units. Te domain of is te domain of f, [, ]. Te range of is te range of f sifted up units, [0, 4]. 13. Te grap of y () is te grap of y g() sifted up units. Te domain of is te domain of g, [, ]. Te range of is te range of g sifted up units, [1, 3]. 15. Te grap of is te grap of f sifted rigt units. Te domain of is te domain of f sifted rigt units, [0, 4]. Te range of is te range of f, [, ]. Common Error: does not indicate sifting left. 17. Te grap of is te grap of g sifted left units. Te domain of is te domain of g sifted left units, [ 4, 0]. Te range of is te range of g, [ 1, 1]. 19. Te grap of is te grap of f reflected troug te ais. Te domain of is te domain of f, [, ]. Te range of is te range of f reflected troug te ais, [, ]. 1. Te grap of is te grap of g stretced vertically by a factor of. Te domain of is te domain of g, [, ]. Te range of f is te range of g stretced vertically by a factor of, [, ].

18 118 CHAPTER 3 FUNCTIONS 3. Te grap of is te grap of g srunk orizontally by a factor of 1. Te domain of is te domain of g srunk orizontally by a factor of 1, [ 1, 1]. 5. Te grap of is te grap of f reflected troug te y ais. Te domain of is te domain of f reflected troug te y ais, [, ]. Te range of is te range of f, [, ]. Te range of is te range of g, [ 1, 1]. 7. g( ) ( ) 3 + ( ) 3 ( 3 + ) g(). Odd 9. m( ) ( ) 4 + 3( ) m(). Even 31. F( ) ( ) F() ( 5 + 1) 5 1 Terefore F( ) F(). F( ) F(). F() is neiter even nor odd. 35. q( ) ( ) + ( ) 3 3. q() ( + 3) + 3 Terefore q( ) q(). q( ) q(). q() is neiter even nor odd. 33. G( ) ( ) G(). Even 37. g() Te graps of f() 3 (tin curve) and g() (tick curve) are sown g() 0.5(6 + ) Te graps of f() (tin curve) and g() (tick curve) are sown g() ( + 4) Te graps of f() (tin curve) and g() (tick curve) are sown

19 SECTION g() ( ). Te grap of f() g() and g() are sown. f() 45. Te grap of y is stretced vertically by a factor of 4 (or srunk orizontally by a factor of.) 47. Te grap of y is sifted left units. 49. m() 4 8 4( ) Te grap of y is sifted rigt units, stretced vertically by a factor of 4, and reflected troug te ais. 51. Te grap of y is reflected troug te ais and sifted up 3 units. 53. Te grap of y is sifted rigt 1 unit, stretced vertically by a factor of 3, and sifted up units. 55. Te grap of y is sifted up 3 units. 57. Te grap of y 3 is stretced vertically by a factor of and sifted up 1 unit.

20 10 CHAPTER 3 FUNCTIONS 59. Te grap of y is sifted left units. 61. Te grap of y is sifted rigt units, srunk vertically by a factor of 1, reflected troug te ais, and sifted up 4 units. 63. Te grap of y as been sifted units rigt. y ( ) 65. Te grap of y 3 as been sifted units down. y Te grap of y as been srunk vertically by a factor of 1 4. y 1 or y Te grap of y 3 as been reflected troug te y ais, (or te ais). y Te grap of y as been sifted left units and up units. y Te grap of y as been reflected troug te ais and sifted up 4 units. y Te grap of y as been reflected troug te ais and sifted rigt 1 unit and up 4 units. y 4 ( 1) 77. Te grap of y 3 as been srunk vertically by a factor of 1 and sifted rigt 3 units and up 1 unit. y 1 ( 3)3 + 1 or y 0.5( 3) (A) Te function f is a orizontal srink of y by a factor of 1/8, wile g is a vertical stretc of y 3 by a factor of. (B) Te graps are sown below in a standard window: tey are identical. 8 8 (C)

21 SECTION (A) Te graps are sown below. Te graps are different, so order is significant wen performing multiple transformations. ; (ii): (B) (i): y 5 y 5 Tese functions are different. In te second one, order of operations tells us to first multiply by 1, ten subtract 5. In te first, te parenteses indicate tat tis order sould be reversed. 83. f() is an odd function, since f( ) f(). g() is an even function, since g( ) g(). () is an even function, since ( ) ( ) (). m() 3 is an odd function, since m( ) ( ) 3 3 m(). n() is neiter even nor odd. n( ) and n( ). p() 3 is an odd function, since p( ) 3 3 p() 85. Te grap of y f( ) represents a orizontal sift from te grap of y f(). Te grap of y f() + k represents a vertical sift from te grap of y f(). Te grap of y f( ) + k represents bot a orizontal and a vertical sift but te order does not matter: Vertical first ten orizontal: y f() y f() + k y f( ) + k Horizontal first ten vertical: y f() y f( ) y f( ) + k Te same result is acieved; reversing te order does not cange te result. 87. Consider te grap of y. If a vertical sift is performed te equation becomes y + k. If a reflection is now performed te equation becomes y ( + k) or y k. If te reflection is performed first te equation becomes y. If te vertical sift is now performed te equation becomes y + k. Since y k and y + k differ (unless k 0), reversing te order canges te result. 89. Te grap of y f( ) represents a orizontal sift from te grap of y f(). Te grap of y f() represents a reflection of te grap of y f() in te ais. Te grap of y f( ) represents bot a orizontal sift and a reflection but te order does not matter: Sift first ten reflection: y f() y f( ) y f( ) Reflection first ten sift: y f() y f() y f( ) Te same result is acieved; reversing te order does not cange te result.

22 1 CHAPTER 3 FUNCTIONS 91. f() g() (A) E() 1 [f() + f( )] (B) O() 1 [f() f( )] E( ) 1 [f( ) + f{ ( )}] O( ) 1 [f( ) f{ ( )}] 1 [f( ) + f()] 1 [f( ) f()] 1 [f() + f( )] E(). 1 [f() f( )] O(). Tus, E() is even. (C) E() + O() 1 [f() + f( )] + 1 [f() f( )] Tus, O() is odd. 1 f() + 1 f( ) + 1 f() 1 f( ) f() Conclusion: Any function can be written as te sum of two oter functions, one even and te oter odd. 97. Te grap of te function C() 30,000 + f() is te same as te given grap of te function f() sifted up 30,000 units ($). 99. y ( 10) 3, y ( 10) 3, y ( 10) 3. Eac grap is a vertical sift of te grap of y 0.004( 10) V(t) C (C t) 0 t C t C 1 V 64(1 t) C V 16( t) not defined for t>1 0 4 not defined for t> 8 C 4 V 4(4 t) not defined for t > 4 C 8 V (8 t)

23 SECTION Eac grap is a portion of te grap of a orizontal translation followed by a vertical stretc (ecept for C 8) of te grap of y t. Te eigt of te grap represents te volume of water left in te tank, so we see tat for larger values of C, te water stays in te tank longer. We can conclude tat larger values of C correspond to a smaller opening. Section Te grap of a quadratic function f() a + b + c is a parabola wit verte at b/a, opening upward if a is positive, and opening downward if a is negative. 3. False. A quadratic function f() a + b + c as a maimum if and only if a is negative; if a is positive it as no maimum. 5. If a is positive te grap opens upward and as a minimum at te verte; if a is negative te grap opens downward and as a maimum at te verte. 7. f() ( + 3) 4 [ ( 3)] 4 Verte: ( 3, 4) ais: f() 5 3 Verte:, 5 ais: f() ( + 10) + 0 [ ( 10)] + 0 Verte: ( 10, 0) ais: Te grap of y is sifted rigt units and up 1 unit. 17. Te grap of y is sifted rigt units and down 3 units. 1. Te grap of y as been sifted to te rigt units and down 3 units. Tis is te grap of y ( ) 3, corresponding to te function m. 15. Te grap of y is sifted left 1 unit and reflected in te ais. 19. Te grap of y as been sifted to te rigt units. Tis is te grap of y ( ), corresponding to te function k. 3. Te grap of y as been reflected in te ais and sifted to te left 1 unit, corresponding to te function.

24 14 CHAPTER 3 FUNCTIONS 5. Begin by grouping te first two terms wit parenteses: f( ) 4 5 Find te number needed to complete te square 4? 5 4/ 4; add and subtract Factor parenteses, combine like terms 1 Te verte form is f( ) 1. Te verte is (, 1) and te ais is. 7. Begin by grouping te first two terms wit parenteses, ten factoring 1 out of tose two terms so tat te coefficient of is 1: ( ) 1 3 Find te number needed to complete te square 1? 3 / 1; add 1 inside te parenteses 1 1 3? We actually added 1(1), so add 1 as well Factor parenteses, combine like terms 1 1 Te verte form is ( ) 1 1. Te verte is ( 1, ) and te ais is Begin by grouping te first two terms wit parenteses, ten factoring out of tose two terms so tat te coefficient of is 1: m ( ) 6 Find te number needed to complete te square 6? 6/ 9; add 9 inside te parenteses 69? We actually added (9), so subtract 18 as well Factor parenteses, combine like terms 3 4

25 SECTION Te verte form is m ( ) 3 4. Te verte is (3, 4) and te ais is Begin by grouping te first two terms wit parenteses, ten factoring 1/ out of tose two terms so tat te coefficient of is 1: 1 7 f( ) 6 Find te number needed to complete te square 1 7 6? 6/ 9; add 9 inside te parenteses ? We actually added 1 (9) ; subtract 9 as well Factor te parenteses, combine like terms Te verte form is f( ) Te verte is ( 3, 8) and te ais is Begin by grouping te first two terms wit parenteses, ten factoring out of tose two terms so tat te coefficient of is 1: f( ) 1 90 Find te number needed to complete te square 1? 90 1 / 36 ; add 36 inside te parenteses ? We actually added (36); subtract 7 as well Factor te parenteses, combine like terms 6 18

26 16 CHAPTER 3 FUNCTIONS Te verte form is f( ) Te verte is (6, 18) and te ais is b 8 4; f( 4) ( 4) 8( 4) a (1) Te verte is ( 4, 8). Te coefficient of is positive, so te parabola opens up. Te grap is symmetric about its ais, 4. It decreases until reacing a minimum at ( 4, 8), ten increases. Te range is [ 8, ). 37. b 7 7 a ( 1) f Te verte is, 4. Te coefficient of is negative, so te parabola opens down. Te grap is symmetric about its ais,. It increases until reacing a maimum at,, ten decreases Te range is, 4.

27 SECTION b a (4) 8 4 f Te verte is 9, Te coefficient of is positive, so te parabola opens up. Te grap is symmetric about its ais, 9 4. It decreases until reacing a minimum at 9 19,, ten increases Te range is, b a ( 10) 0 f Te verte is 5, 149. Te coefficient of is negative, so te parabola opens down. Te grap is symmetric about its ais, 5. It increases until reacing a maimum at 5 149,, ten decreases. Te 149 range is,.

28 18 CHAPTER 3 FUNCTIONS 43. b 3 3 a (1) f Te verte is, 4. Te coefficient of is positive, so te parabola opens up. Te grap is symmetric about its ais, It decreases until reacing a minimum at,, ten increases. 4 9 Te range is, b a (0.5) f Te verte is, 9. Te coefficient of is positive, so te parabola opens up. Te grap is symmetric about its ais,. It decreases until reacing a minimum at, 9, ten increases. Te range is9,.

29 SECTION < 10 3 f() < 0 f() ( + 5)( ) < 0 Te zeros of f are 5 and. Plotting te grap of f(), we see tat f() < 0 for 5 < <, or ( 5, ) > 10 f() > 0 f() ( 3)( 7) > 0 Te zeros of f are 3 and 7. Plotting te grap of f(), we see tat f() > 0 for < 3 and > 7, or (, 3) (7, ) f() 8 0 f() ( 8) 0 Te zeros of f are 0 and 8. Plotting te grap of f(), we see tat f() 0 for 0 8 or [0, 8] f() f() ( + 5) 0 Te zeros of f are 5 and 0. Plotting te grap of f(), we see tat f() 0 for 5 0 or [ 5, 0 ] < + 1 < 0 ( 1) < 0 Since te square of no real number is negative, tese statements are never true for any real number. solution; is te solution set. No 57. < < 0 We attempt to find all real zeros of te polynomial b b 4ac a 1, b 3, c 3 a (1) ( 3) ( 3) 4(1)(3) 3 3 Te polynomial as no real zeros. Hence te statement is eiter true for all real or for no real. To determine wic, we coose a test number, say f() Find all real zeros of f() b b 4ac a 1, b 4, c 1 a ( 4) ( 4) 4(1)( 1) (1) ± , 4.36 Common Error: ± 0

30 130 CHAPTER 3 FUNCTIONS < 3 3 0? < 3(0) 3? 0 < 3 False. Te statement is never true for any real number. No solution. is te solution set. Plotting te grap of f() we see tat f() 0 for 5 and + 5, or (, 5 ] [ + 5, ) Te verte of te parabola is at (1, 4). Terefore te equation must ave form y a( 1) 4 Since te parabola passes troug (3, 4), tese coordinates must satisfy te equation 4 a(3 1) 4 8 4a a. Te equation is y ( 1) 4 y ( + 1) 4 y y Te verte of te parabola is at ( 1, 4). Terefore te equation must ave form y a( + 1) + 4 Since te parabola passes troug (1, ), tese coordinates must satisfy te equation. a(1 + 1) + 4 4a + 4 4a a 0.5 Te equation is y 0.5( + 1) + 4 y 0.5( + + 1) + 4 y y Notice tat te grap does not provide te eact coordinates of te verte, so we can't tell for certain wat tey are. We know tat f( 1) and f(3) are bot zero, so te ais of symmetry is alfway between 1 and 3. In oter words, te -coordinate of te verte is 1; te equation looks like f() a( 1) + k. Plug in 1: f( 1) a( 1 1) + k a( ) + k 4a + k 0 (since ( 1, 0) is on te grap) 4a + k 0 k 4a Substitute 4a in for k: f() a( 1) 4a Plug in 0: f(0) a(0 1) 4a a 4a 3a 3 (since (0, 3) is on te grap) 3a 3 a 1 f() ( 1) 4 or f() Notice tat te grap does not provide te eact coordinates of te verte, so we can't tell for certain wat tey are. We know tat f( 1) and f(5) are equal, so te ais of symmetry is alfway between 1 and 5. In oter words, te -coordinate of te verte is ; te equation looks like f() a( ) + k Plug in 1: f( 1) a( 1 ) + k a( 3) + k 9a + k 0 (since ( 1, 0) is on te grap) 9a + k 0 k 9a

31 SECTION Substitute 9a in for k: f() a( ) 9a Plug in 0: f(0) a(0 ) 9a a( ) 9a 4a 9a 5a.5 (since (0,.5) is on te grap) 5a.5 a 0.5 f() 0.5( ) or f() Te verte of te parabola is at (4, 8). Terefore te equation must ave form y a( 4) + 8 Since te intercept is 6, (6, 0) must satisfy te equation 0 a(6 4) a + 8 a Te equation is y ( 4) + 8 y ( ) + 8 y y Te verte of te parabola is at ( 5, 5). Terefore te equation must ave form y a( + 5) 5 Since te parabola passes troug (, 0), tese coordinates must satisfy te equation. 0 a( + 5) 5 0 9a a a a( ) + k a( + ) + k a a + a + k a (a) + (a + k) 71. Te verte of te parabola is at ( 4, 1). Terefore te equation must ave form y a( + 4) + 1 Since te y intercept is 4, (0, 4) must satisfy te equation. 4 a(0 + 4) a a a 0.5 Te equation is y 0.5( + 4) + 1 y 0.5( ) + 1 y Te equation is y 5( + 5) 5 y 5( ) 5 y y Te graps sown are f() , f() + 1, and f() Tese correspond to f() + k + 1 wit k 6, 0, and 4 respectively. Note tat all ave te same sape but a different verte. In fact, all tree are translations of te grap y. 79. Te graps of f() ( 1), g() ( 1) + 4, and () ( 1) 5 are sown. 10 It is clear tat f() as one intercept (at 1), g() as no intercepts and () as two intercepts. In general, for a > 0, te grap of f() a( ) + k can be epected to ave no intercepts for k > 0, one intercept at for k 0, and two intercepts for k <

32 13 CHAPTER 3 FUNCTIONS 81. Let one number. Ten te oter number is 30. Te product is a function of given by f() ( 30) 30. Tis is a quadratic function wit a > 0, terefore it as a minimum value at te verte of its grap (a parabola). Completing te square yields f() 30 1 ( 30) 15; ( 15) ( 15) 5 Tus te minimum product is 5, wen 15 and Tere is no "igest point" on tis parabola and no maimum product. 83. Find te verte, using a 1. and b 6.5: b ; P(6) 1.(6) 6.5(6) a ( 1.) Te company sould ire 6 employees to make a maimum profit of $3, (A) Find te first coordinate of te verte, using a 0.19 and b 1.: b a ( 0.19) Te maimum bo office revenue was tree years after 00, wic is 005. (B) Te function specifies yearly totals for revenue, so te domain sould be restricted to wole numbers. Te eact verte occurs at 3., so we needed to round down to (A) Since te four sides needing fencing are, y, + 50, and y, we ave + y + ( + 50) + y 50. Solving for y, we get y 100. Terefore te area A() ( + 50)y ( + 50)(100 ) (Since bot and 100 must be nonnegative, te domain of A() is ) (B) Tis is a quadratic function wit a < 0, so it as a maimum value at te verte: b 50 5 ; A(5) (5) 50(5) 5, , 50 5, 000 5, 65 a ( 1) Te maimum area is 5,65 square feet wen 5. (C) Wen 5, y Te dimensions of te corral are ten + 50 by y, or 75 ft by 75 ft. 89. According to Eample 7, te function describing te eigt of te sandbag is (t) 10,000 16t (since te initial eigt is 10,000 feet). We want to know wen it reaces ground level, so plug in zero for (t), ten solve for t. 0 10,000 16t 16t 10,000 t 65 t 5 It its te ground 5 seconds after it's dropped. 91. According to Eample 7, te function describing te eigt of te diver is (t) 0 16t, were 0 is te initial eigt in feet. (Tis initial eigt is te eigt of te cliff.) We know tat (.5) 0 since it takes.5 seconds to reac te water; we plug in.5 for t and 0 for (t), wic allows us to solve for (.5) Te cliff is 100 feet ig.

33 SECTION (A) Since d(t) is a quadratic function wit maimum value 484 wen t 5.5, an equation for d(t) must be of te form d(t) a(t 5.5) Since d(0) 0, 0 d(0) a(0 5.5) a a 16 Hence d(t) 16(t 5.5) (t 11t ) t + 176t t + 176t Since te grap of d must be symmetric wit respect to t 5.5, and d(0) 0, d(11) must also equal 0. Te distance above te ground will be nonnegative only for values of t between 0 and 11, ence te domain of te function is 0 t 11. (B) Solve 50 16t + 176t by graping Y1 50 and Y and applying a built-in routine From te graps t 1.68 sec and t 9.3 sec, to two decimal places. 95. (A) If coordinates are cosen wit origin at te center of te base, te parabola is te grap of a quadratic function () wit maimum value 14 wen 0. Te equation must be of form () a + 14 Since (10) 0 (wy?) 0 (10) a(10) a + 14 a 0.14 Hence () (B) Suppose te truck were to drive so as to maimize its clearance, tat is, in te center of te roadway. Ten alf its widt, or 4 ft, would etend to eac side. But if 4, () 0.14(4) ft. Te arc is only feet ig, but te truck is 1 feet ig. Te truck cannot pass troug te arc. (C) From part (B), if 4, () ft is te eigt of te tallest truck. (D) Find so tat () 1. Solve te equation (Te negative solution doesn t make sense) Te widt of te truck is at most (3.78) 7.56 feet

34 134 CHAPTER 3 FUNCTIONS 97. (A) Te entered data is sown ere along wit te results of te quadratic regression calculation. Te quadratic model for te skid mark lengt is L() (B) (C) Solve ( 1.) ( 1.) 4(0.06)( 14) (0.06) mp (te negative answer doesn t make sense) 99. (A) Beer consumption in 1960 is given as Solve (for convenience) 300 (300) 4( 6)(100) ( 6) 50 (discarding te negative answer) Tis represents te year 010. (B) Substitute 45 to obtain B(45) (45) (45) gallons 101. A profit will result if C() < R(). Solve < < 0 f() < 0 Find te zeros of f(). ( 8.4) ( 8.4) 4(0.04)(45) (0.04) 35 or 175 y Plotting te grap of f() we see tat f() < 0 for 35 < < 175. Te break-even points are terefore (35, R(35)) (35, 301) and (175, R(175)) (175, 55) Te revenue function is R() d() ( ) or R() R() as a maimum value at te verte of tis parabola, wic is given by b a ( 0.15) Ten p d(31) (31) $4.65 is te price wic maimizes te revenue.

35 SECTION (A) Te revenue function is R() d() ( ) Te domain is given by ( ) 0 or 0 50,000. Te cost function C() is given by C() Fied Cost + Variable Cost 4, Te domain is given by 0 or [0, ). Te company will break even wen R() C(). Solve , ,500 ( 3.15) ( 3.15) 4( )(4,500) ( ) 10,000 or 35,000. Te company will break even for sales of 10,000 or 35,000 gallons. (B) (C) (D) Te company makes a profit for tose sales levels for wic te grap of te revenue function is above te grap of te cost function, tat is, if te sales are between 10,000 and 35,000 gallons. Te company suffers a loss for tose sales levels for wic te grap of te revenue function is below te grap of te cost function, tat is, if te sales are between 0 and 10,000 gallons or between 35,000 and 50,000 gallons. Te profit function is given by P() R() C(). Tus P() ( ) (4, ) ,500 Te maimum value of tis function occurs at te verte of its parabola grap. Tis is given by te formulas (, k) were b a 3.15,500 ( ) and b k C 4a 4, , ( ) Tat is, te maimum profit is $10, wen,500 gallons are sold. Substitute,500 to find p d(,500) (,500) $1.9 per gallon. Section Te sum of two functions is found by adding te epressions for te two functions and finding te intersection of teir domains. 3. Answers will vary.

36 136 CHAPTER 3 FUNCTIONS 5. Te simplification may obscure values tat are not in te domain of one of te functions. 7. Construct a table of values of f() and g() from te grap, ten add to obtain (f + g)() f() g() (f + g)() Construct a table of values of f() and g() from te grap, ten multiply to obtain (fg)() f() g() (f g)() (f g)( 1) f[g( 1)]. From te grap of g, g( 1). From te grap of f, f[g( 1)] f(). 13. (g f)( ) g[f( )]. From te grap of f, f( ) 0. From te grap of g, g[f( )] g(0) From te grap of g, g(1) 0. From te grap of f, f[g(1)] f(0). 17. From te grap of f, f(). From te grap of g, g[f()] g( ) (f + g)( 3) f( 3) + g( 3) [ ( 3)] + 3 ( 3) (fg)( 1) f( 1)g( 1) [ ( 1)] 3 ( 1) Using values from te table, ( 7) 4 3. f g( ) f(g( )) f( 3 ( ) ) f( 5 ) 5 g, so f g(0) f g(0) f( ) 9, and 5. g f(1) g(f(1)) g( 1) g(1) 31 f g ( 7) f g( 7) f(4) 3. Similarly, f g (4) f g(4) f(6) (f + g)() f() + g() Domain: (, ) (f g)() f() g() 4 ( + 1) 3 1 Domain: (, ) (fg)() f()g() 4( + 1) Domain: (, ) f f ( ) 4 () g g ( ) 1 Domain: { 1}, or (, 1) ( 1, ) Common Error: f() g() Te parenteses are necessary.

37 SECTION (f + g)() f() + g() Domain: (, ) (f g)() f() g() ( + 1) 1 Domain: (, ) (fg)() f()g() ( + 1) 4 + Domain: (, ) f () ( ) f g g ( ) 1 Domain: (, ) (since g() is never 0.) 33. (f + g)() f() + g() Domain: (, ) (f g)() f() g() ( 1) Domain: (, ) (fg)() f()g() (3 + 5)( 1) Domain: (, ) f f () ( ) 3 5 Domain: { ±1}, or (, 1) ( 1, 1) (1, ) g g ( ) (f + g)() f() + g() + 3 (f g)() f() g() 3 (fg)() f()g() 3 ( )(3 ) 6 f f () ( ) g g ( ) 3 3 Te domains of f and g are: Domain of f { 0} (, ] Domain of g { + 3 0} [ 3, ) Te intersection of tese domains is [ 3, ]. Tis is te domain of te functions of f + g, f g, and fg. Since g( 3) 0, 3 must be ecluded from te domain of f g, so its domain is ( 3, ]. 37. (f + g)() f() + g() (f g)() f() g() + ( 4) (fg)() f()g() ( + )( 4) 8 f f () ( ) g g ( ) 4 Te domains of f and g are bot { 0} [0, ) Tis is te domain of f + g, f g, and fg. We note tat in te domain of f g, g() 0. Tus 4 0. To solve tis, we solve Common Error: 4 16 Hence, 16 must be ecluded from { 0} to find te domain of f g. Domain of f g { 0, 16} [0, 16) (16, ).

38 138 CHAPTER 3 FUNCTIONS 39. (f + g)() f() + g() (f g)() f() g() (fg)() f()g() f () ( ) f 6 6 g g ( ) Te domains of f and g are: Domain of f { + 6 0} { ( + 3)( ) 0} (, 3] [, ) Domain of g { } { (7 )(1 + ) 0} [ 1, 7] Te intersection of tese domains is [, 7]. Tis is te domain of te functions f + g, f g, and fg. Since g() (7 )(1 + ), g(7) 0 and g( 1) 0, ence 7 must be ecluded from te domain of f g, so its domain is [, 7). 41. (f + g)() f() + g() (f g)() f() g() (fg)() f()g() 1 1 f f ( ) 1 () 1 g g ( ) 1 Te domains of f and g are bot { 0} (, 0) (0, ). Common Error: Domain is not (, ). See below. Tis is terefore te domain of f + g, f g, and fg. To find te domain of f g, we must eclude from tis domain te set of values of for wic g() , 1 Hence, te domain of f g is { 0, 1, or 1} or (, 1) ( 1, 0) (0, 1) (1, ). 43. (f g)() f[g()] f( + 1) ( + 1) 3 Domain: (, ) (g f)() g[f()] g( 3 ) ( 3 ) Domain: (, ) 45. (f g)() f[g()] f( + 3) Domain: (, ) (g f)() g[f()] g( + 1 ) Domain: (, ) 47. (f g)() f[g()] f( 3 + 4) ( 3 + 4) 1/3 Domain: (, ) (g f)() g[f()] g( 1/3 ) ( 1/3 ) Domain: (, ) 49. (f g)() f[g()] f( 4) 4 Domain: { 4} or [4, ) (g f)() g[f()] g( ) 4 Domain: { 0} or [0, )

39 SECTION (f g)() f[g()] f 1 + Domain: { 0} or (, 0) (0, ) 1 (g f)() g[f()] g( + ) Domain: { } or (, ) (, ) 53. (f g)() f[g()] f( ) 4 Te domain of f is (, 4]. Te domain of g is all real numbers. Hence te domain of f g is te set of tose real numbers for wic g() is in (, 4], tat is, for wic 4, or. Domain of f g { } [, ] (g f)() g[f()] g( 4 ) ( 4 ) 4 Te domain of g f is te set of tose numbers in (, 4] for wic f() is in (, ), tat is, (, 4]. 55. (f g)() f[g()] f 5 5( ) Te domain of f is { 0}. Te domain of g is { }. Hence te domain of f g is te set of tose numbers in { } for wic g() is in { 0}. Tus we must eclude from { } tose numbers for wic 0, or 0. Hence te domain of f g is { 0, }, or (, 0) (0, ) (, ). 5 5 (g f)() g[f()] g Te domain of g f is te set of tose numbers in { 0} for wic f() is in { }. Tus we must 5 eclude from { 0} tose numbers for wic, or + 5, or 5. Hence te domain of g f is { 0, 5} or (, 0) (0, 5) (5, ) (f g)() f[g()] f 1 1 Te domain of f is { 0}. Te domain of g is { }. Hence te domain of f g is te set of tose numbers in { } for wic g() is in { 0}. Tus we must eclude from { } tose numbers 1 for wic 0; owever, tere are none. Hence te domain of f g is { } or (, ) (, ). (g f)() g[f()] g Te domain of g f is te set of tose numbers in { 0} for wic f() is in { }. Tus we must eclude from { 0} tose numbers for wic 1 ; owever, tere are none. Hence te domain of g f is { 0} or (, 0) (0, ).

40 140 CHAPTER 3 FUNCTIONS 59. (f g)() f[g()] f( 9 ) 5 ( 9 ) 5 (9 ) 16 Te domain of f is [ 5, 5]. Te domain of g is (, ). Hence te domain of f g is te set of tose real numbers for wic g() is in [ 5, 5], tat is, domain of f g is { 4 4} or [ 4, 4]. (g f)() g[f()] g( 5 ) 9 ( 5 ) 9 5, or 9 + 5, or 16, or 4 4. Hence te Te domain of g f is te set of tose numbers in [ 5, 5] for wic g() is real. Since g() is real for all, te domain of g f is [ 5, 5]. Common Error: Te domain of g f is not evident from te final form 34. It is not 34, 34 In Problems #61 troug 63, f and g are linear functions. f as slope and y intercept, so f() +. g as slope 1 and y intercept, so g(). 61. (f + g)() f() + g() ( + ) + ( ). Te grap of f + g is a straigt line wit slope 1 passing troug te origin. Tis corresponds to grap (d). 63. (g f)() g() f() ( ) ( + ) Te grap of g f is a straigt line wit slope 3 and y intercept 4. Tis corresponds to grap (a). 65. (f g)() f[g()] f( ) 1 ( ) (g f)() g[f()] g Graping f, g, f g, and g f, we obtain te grap at te rigt. Te graps of f and g are reflections of eac oter in te line y, wic is te grap of f g and g f (f g)() f[g()] f (g f)() g[f()] g Graping f, g, f g, and g f, we obtain te grap at te rigt. Te graps of f and g are reflections of eac oter in te line y, wic is te grap of f g and g f

41 SECTION f g() f[g()] f( 3 ) g f() g[f()] g Graping f, g, f g, and g f, we obtain te grap at te rigt. Te graps of f and g are reflections of eac oter in te line y, wic is te grap of f g and g f. 71. f g() f[g()] f( 3 + ) g f() g[f()] g( 3 ) ( 3 ) Graping f, g, f g, and g f, we obtain te grap at te rigt. Te graps of f g and g f are reflections of eac oter in te line y, wic is te grap of f g and g f. 73. If we let g() 7, ten () [g()] 4 Now if we let f() 4, we ave () [g()] 4 f[g()] (f g)() 79. If we let f() 1/, ten () 4f() + 3 Now if we let g() 4 + 3, we ave () 4f() + 3 g[f()] (g f)() 75. If we let g() 4 +, ten () g ( ) Now if we let f() 1/, we ave () g ( ) [g()] 1/ f[g()] (f g)(). 81. fg and gf are identical, since (fg)() f()g() g()f() (gf)() by te commutative law for multiplication of real numbers 77. If we let f() 7, ten () 3f() 5 Now if we let g() 3 5, we ave () 3f() 5 g[f()] (g f)() 83. Yes, te function g() satisfies tese conditions. (f g)() f(g()) f(), so f g f (g f)() g(f()) f(), so g f f 85. (f + g)() f() + g() (f g)() f() g() (fg)() f()g() 1 1 f f ( ) 1 () 1 g g ( ) 1 Te domains of f and g are bot { 0} (, 0) (0, ). Common Error: Domain is not (, ). See below. Tis is terefore te domain of f + g, f g, and fg. To find te domain of f g, we must eclude from tis domain te set of values of for wic g() 0.

42 14 CHAPTER 3 FUNCTIONS , 1 Hence, te domain of f g is { 0, 1, or 1} or (, 1) ( 1, 0) (0, 1) (1, ). 87. (f + g)() f() + g() (f g)() f() g() 1 1 (fg)() f()g() 1 1 (1) f f () ( ) 1 g g ( ) 1 owever, wen we eamine te domain of f g below Tis can be furter simplified Te domains of f and g are bot { 0} (, 0) (0, ) Tis is terefore te domain of f + g, f g, and fg. To find te domain of f g, we must eclude from tis domain te set of values of for wic g() Tis is true wen is negative. Te domain of f g is te positive numbers, (0, ). On tis domain,, so f g () Profit is te difference of te amount of money taken in (Revenue) and te amount of money spent (Cost), so P() R() C() 1 0 ( + 8,000) ,000 (Distribute!) ,000 we ave a profit function, but it's a function of te demand (), not te price (p). We were given 4,000 00p, so we can substitute 4,000 00p in for to get desired function:

43 SECTION P() P(4,000 00p) 18(4,000 00p) 7,000 3,600p 1 00 (4,000 00p) 8, (16,000,000 1,600,000p + 40,000p ) 8,000 7,000 3,600p 80, ,000p 00p 8,000 16, ,400p 00p Now grap te function and use te maimum feature to find te largest profit. Te maimum is (11, 9,00), so te largest profit occurs wen te price is $ We are given V(r) 0.1A(r) 0.1πr and r(t) 0.4t 1/3. Hence we use composition to epress V as a function of te time. (V r)(t) V[r(t)] 0.1π[r(t)] 0.1π[0.4t 1/3 ] 0.1π[0.16t /3 ] 0.016πt /3 We write V(t) 0.016πt /3 4 inces 93. c' R E' (A) We note: In te figure, triangles VCE and VC'E' are similar. Moreover r or r 4 c r E 4 inces R radius of cup 1 diameter of cup 1 (4) inces. Hence 1. We write r() 1. Section 3-6 V (B) Since V 1 3 πr and r 1, V 1 3 π π π3. We write V() 1 1 π3. (C) Since V() 1 1 π3 and (t) t, we use composition to epress V as a function of t. (V )(t) V[(t)] 1 1 π [(t)]3 1 π (4 0.5 t ) Te function will be one-to-one if and only if eac first component of te ordered pairs corresponds to eactly one second component.

44 144 CHAPTER 3 FUNCTIONS 3. If a function is not one-to-one, ten at least two input elements correspond to one output element. If te correspondence is reversed, ten te result cannot be a function. Eample: {(1, 3), (, 3)} wen reversed becomes {(3, 1), (3, )} wic is not a function. 5. Te result of composing a function wit its inverse is te identity function f(). Tis makes sense because te function and its inverse undo eac oter s operations. 7. Tis is a one-to-one function. All of te first coordinates are distinct, and eac first coordinate is paired wit a different second coordinate. If all of te ordered pairs are reversed, te situation is te same. 9. Tis is a function but it is not one-to-one. First coordinates 5 and are bot paired wit 4, and first coordinates 4 and 3 are bot paired wit 3. If te ordered pairs are reversed, te result is not a function since first coordinates 3 and 4 will eac be paired wit two different second coordinates. 11. Tis is not a function: 1 is a first coordinate tat is paired wit different second coordinates (as is 3). If te ordered pairs are reversed, te result is also not a function since 4 will be a first coordinate paired wit different second coordinates (as will ). 13. One-toone 15. Te range element 7 corresponds to more tan one domain element. Not one-to-one. 17. One-toone 19. Some range elements (0, for eample) correspond to more tan one domain element. Not one-to-one. 1. One-to-on 3. One-to-one 5. Assume F(a) F(b) 1 a b + 1 Ten 1 a 1 b 7. H() 4 Since H(1) 4(1) 1 3 and H(3) 4(3) (3) 3, bot (1, 3) and (3, 3) belong to H. H is not one-to-one f(g()) f g(f()) g(3 + 5) 1 3 (3 + 5) a b Terefore F is one-to-one. 9. Assume M(a) M(b) a 1 b 1 Ten a + 1 b + 1 a b M is one-to-one. 33. f(g()) f( 3 3 1) (( 3 3 1) + 1) 3 ( 3 3 ) 3 (3 ) 1 + f and g are not inverses since (f g)() is not. f and g are inverses f(g()) f g(f()) g ( 4) ( 4) f and g are inverses. 3( ) 4( ) 37. f(g()) f( 4 ) 4 + ( 4 ) g(f()) g(4 + ) as long as 0. f and g are inverses.

45 SECTION f(g()) f( 1 ) 1 ( 1 ) 1 (1 ) g(f()) g(1 ) 1 (1 ) as long as 0. f and g are not inverses since g f() is not. 41. From te grap: domain of f [ 4, 4] range of f [1, 5] Terefore: domain of f -1 [1, 5] range of f -1 [ 4, 4] 43. From te grap: domain of f [ 5, 3] range of f [ 3, 5] Terefore: domain of f -1 [ 3, 5] range of f -1 [ 5, 3] 45. Te grap of f() is a line; f is one-to-one. Te domain of f is terefore (, ) and te range of f is also (, ). Write y f() y 3 Solve y 3 for. 1 3 y Intercange and y: 1 3 y f -1 () 1 3 Te domain of f -1 is (, ) and te range of f -1 is (, ). 47. Te grap of f() is a line; f is one-to-one. Te domain and range of f are terefore (, ). Write y f() y 4 3 Solve y 4 3 for. y y 3 4 Intercange and y: 3 y 4 3 f -1 () 4 Te domain and range of f -1 are (, ).