Chapter 2 Limits and Continuity. Section 2.1 Rates of Change and Limits (pp ) Section Quick Review 2.1

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1 Section. 6. (a) N(t) t (b) days: 6 guppies week: 7 guppies (c) Nt () t t t ln ln t ln ln ln t Tere will be guppies ater ln days, or ater nearly 9 days. (d) Because it suggests te number o guppies will continue to double indeinitely and become arbitrarily large, wic is impossible due to te inite size o te tank and te oygen supply in te water. 66. (a) y.77. (b) y. 77( ). 99 Te estimate is 99 less tan te actual number. (c) y m b m.77 Te slope represents te approimate annual increase in te number o doctorates earned by Hispanic Americans per year. 67. (a) y (767.6) (. 98) (b) (767.6) (. 98) 9,8 tousand or 9,8, 9,8, 9,9,, Te prediction is less tan te actual by,. 7, 8 (c). 98 or % ( 9, 8)( ) 68. (a) m (b) y (c) y (d) 69. (a) (, ) > (b) (, ) all real numbers (c) ( ) ln( ) ln( ) ln( ) e e. 78 (d) y ln( ) ln( y) ln( y) e y y e (e) ( )( ) ( ( )) ( e ) ln( e ) ln( e ) ( ) ( )( ) ( ( ) ( ln( )) ( ln( )) ln( ) e e ( ) 7. (a) (, ) all real numbers (b) [, ] cos () oscillates between and (c) (d) Even. cos ( θ ) cos ( θ) (e).6 Capter Limits and Continuity Section. Rates o Cange and Limits (pp. 9 69) Quick Review.. ( ) ( ) ( ) ( ). ( ). ( ) sin sin. ( )

2 Section.. < < < 6. < c c < < c 7. < < < < < 8. c < d d < c< d d c< < d c 9. 8 ( )( 6) 6,. ( ), ( )( ) Section. Eercises. Δ y Δ 6 () 6 () 8 t t sec. Δ y Δ 6 ( ) 6 ( ) 6 t t sec. Δ y Δ 6 ( ) 6 ( ), say. t 6(. ) 6( 9) 6(. 9 6) 6() t.. sec Conirm Algebraically Δy Δ 6( ) 6( ) t 6( 9 6 ) 96 6 i, ten Δ y Δ t 96 t sec. Δ y Δ 6( ) 6( ) t say. 6(. ) 6( ). 6( 6. 8) 6( 6) t sec ( 96 6) t sec Conirm Algebraically Δy Δ 6( ) 6( ) t 6( 6 8 ) ( 8 6) t sec i, ten Δ y Δ t 8 t sec. ( ) c c c c 6. c 9 c c c 9 7. ( ) / Grapical support: ( ) 8. ( ) ( ) ( ) Grapical support: ( 7) () () () 7 7 Grapical support:

3 Section. y. y. y 6 ( ) 6 y Grapical support: y y y ( ) ( ) y ( ) 6 Grapical support:. int int / Note tat substitution cannot always be used to ind its o te int unction. Its use ere can be justiied by te Sandwic Teorem, using g() () on te interval (, ). Grapical support:. You cannot use substitution because te epression is not deined at. Since te epression is not deined at points near, te it does not eist. 6. You cannot use substitution because te epression not deined at. Since becomes arbitrarily large as approaces rom eiter side, tere is no (inite) it. (As we sall see in Section., we may write.) 7. You cannot use substitution because te epression is not deined at. Since and, te let- and rigt-and its are not equal and so te it does not eist. 8. You cannot use substitution because te epression ( ) 6 is not deined at. Since 9. ( ) or all, te it eists and is equal to ( 8 ) 8 8. is / /. ( 6) ( 6) ( 8) 6 Grapical support: Algebraic conirmation: ( ) ( ).. Grapical support: t t t t Algebraic conirmation: t t ( t )( t ) t t r t t t

4 Section Algebraic conirmation: 8 6 ( 8) ( 6) 8 6 ( ) 8 ( ) Algebraic conirmation: ( ) 8 6 ( 6 ) 6( ) ( ). sin Algebraic conirmation: sin sin (). Algebraic conirmation: ( ) ( ) ( ) ( ) ( ). sin Algebraic conirmation: sin sin sin ( ) ( ) 6. ( ) 8 sin Algebraic conirmation: sin sin ( ) sin

5 Section sin Algebraic conirmation: sin sin sin sin sin ( ) (sin ) ( ) sin sin Algebraic conirmation: sin sin sin sin sin sin () () 9. Answers will vary. One possible grap is given by te window [.7,.7] by [, ] wit Xscl and Yscl.. Answers will vary. One possible grap is given by te window [.7,.7] by [, ] wit Xscl and Yscl.. Since int or in (, ), int.. Since int or in (, ), int.. Since int or in (, ), int... Since int or in (, ), int. 6. Since or <,. 7. (a) True (b) True (c) False, since ( ). (d) True, since bot are equal to. (e) True, since (d) is true. () True (g) False, since ( ). () False, ( ), but ( ) is undeined. (i) False, ( ), but ( ) is undeined. (j) False, since ( ). 8. (a) True (b) False, since ( ). (c) False, since ( ). (d) True (e) True () True, since ( ) ( ). (g) True, since bot are equal to. () True (i) True, since ( ) or all c in (, ). 9. (a) ( ) c (b) ( ) (c) ( ) does not eist, because te let- and rigtand its are not equal. (d) (). (a) gt ( ) t (b) gt ( ) t t (c) gt ( ) does not eist, because te let- and rigtand its are not equal. (d) g( ). Since or >,.

6 6 Section.. (a) ( ) (b) ( ) (c) ( ) (d) (). (a) ps ( ) s (b) ps ( ) s (c) ps ( ) s (d) p( ). (a) F( ) (b) F( ) (c) F( ) does not eist, because te let- and rigtand its are not equal. (d) F (). (a) G ( ). y 6. y 7. y ' (b) G ( ) (c) G ( ) (d) G() (c) (b) (d) ( )( ), ( )( ) ( ), (c) g ( ) g( ) g (d) ( ) ( ) ( ) 9 g ( ) ( ( ) ) ( ). (a) ( ( ) g( )) ( ) g( ) ( ) ( ) b b b 7 ( ) (b) ( ( ) g ( )) ( ) g ( ) ( )( ) b b b ( 7) ( ) (c) g ( ) g ( ) ( ) b b (d) ( ) ( ) b 7 7 b g ( ) g ( ) b. (a) [, 6] by [, ] (b) ( ) ; ( ) (c) No, because te two one-sided its are dierent.. (a) [, 6] by [, ] (b) ( ) ; ( ) (c) Yes. Te it is.. (a) 8. y (a) ( )( ) 9. (a) ( g ( ) ) ( g ( )) ( ) 6 (b) ( ) ( ) ( )( ) [, ] by [, 8] (b) ( ) ; ( ) does not eist. (c) No, because te let-and its does not eist.

7 Section. 7. (a) 8. (a) [.7,.7] by [.,.] (b) ( ) ; ( ) (c) Yes. Te it is.. (a) [.7,.7] by [.,.] (b) (, ) (, ) (, ) (c) None 9. (d) None [ π, π] by [, ] (b) (, ) (, ) ( sin ) 6. (a) (c) c (d) c [ π, π] by [, ] (b),, 6. Conirm using te Sandwic Teorem, wit g ( ) and ( ). sin sin sin Because ( ), te Sandwic Teorem gives ( sin ). 7. (a) (c) c (d) c ( sin ) [, ] by [, ] (b) (, ) (, ) (c) c (d) c Conirm using te Sandwic Teorem, wit g() and (). sin sin. sin Because ( ), te Sandwic Teorem gives ( sin )

8 8 Section sin Conirm using te Sandwic Teorem, wit g() and (). sin sin. sin Because ( ), te Sandwic Teorem give sin. cos Conirm using te Sandwic Teorem, wit g() and (). cos cos. cos Because ( ), te Sandwic Teorem give cos. 6. (a) In tree seconds, te ball alls.9 (). m, so its average speed is.. 7 m/sec. (b) Te average speed over te interval rom time t to time is y 9.( ) 9.() 96.( ) t ( ) Since ( 9.. 9) 9., te instantaneous speed is 9. m/sec. 6. (a) y gt g( ) g or. 6 (b) Average speed m/sec (c) I te rock ad not been stopped, its average speed over te interval rom time t to time t is y. ( ). ( ) 8. ( ) t ( ). Since (. ), te instantaneous speed is m/sec. 6. True. Te deinition o a it. 66. True sin sin l im sin sin as. 67. C. 68. B. 69. E. 7. C. 7. (a).... () (b).... () Te it appears to be.

9 Section (a).... () (b).... () Tere is no clear indication o a it. 7. (a).... () (b).... () Te it appears to be approimately.. 7. (a).... () (b).... () Te it appears to be. 7. (a) Because te rigt-and it at zero depends only on te values o te unction or positive -values near zero. (b) Area o OAP base eigt ( )( ) ( ) (sin ) sin ( angle)( radius) Area o sector OAP () Area o OAT ( base)( eigt) ( tan ) (tan ) (c) Tis is ow te areas o te tree regions compare. (d) Multiply by and divide by sin θ. (e) Take reciprocals, remembering tat all o te values involved are positive. () Te its or cos θ and are bot equal to. Since sin is between tem, it must also ave a it o. sin ( ) sin sin (g) () I te unction is symmetric about te y-ais, and te rigt-and it at zero is, ten te let-and it at zero must also be. (i) Te two one-sided its bot eist and are equal to. 76. (a) Te it can be ound by substitution. ( ) ( ) ( ) (b) Te graps o y (), y.8, and y. are sown. Te intersections o y wit y and y are at.767 and.8, respectively, so we may coose any value o a in [.767, ) (approimately) and any value o b in (,.8]. One possible answer: a.7, b.8. (c) Te graps o y (), y.99, and y. are sown. Te intersections o y wit y and y are at.9867 and., respectively, so we may coose any value o a in [.9867, ), and any value o b in (,.] (approimately). One possible answer: a.99, b (a) 6 sin 6 (b) Te graps o y (), y., and y.7 are sown. Te intersections o y wit y and y are at.7 and.77, respectively, so we may coose any value o a in. 7,, 6 and any value o b in 6,. 77, were te interval endpoints are approimate. One possible answer: a., b.77.

10 6 Section. 77. Continued (c) Te graps o y (), y.9, and y. are sown. Te intersections o y wit y and y are at. and., respectively, so we may coose any value o a in.,, 6 and any value o b in 6,., were te interval endpoints are approimate. One possible answer: a., b Line segment OP as endpoints (, ) and (a, a ), so its a a midpoint is a a,, and its slope is a a. Te perpendicular bisector is te line troug a a a, wit slope, so its equation is a y a a a, wic is equivalent to y a a. Tus te y-intercept is b a. As te point P approaces te origin along te parabola, te value o a approaces zero. Tereore, a b. P a Section. Limits Involving Ininity (pp. 7 77) Eploration Eploring Teorem. Neiter ( ) or g ( ) eist. In tis case, we can describe te beavior o and g as by writing ( ) and g( ). We cannot apply te quotient rule because bot its must eist. However, rom Eample, sin sin,. Bot and g oscillate between and as, taking on eac value ininitely oten. We cannot apply te sum rule because neiter it eists. However, (sin cos ) ( ), so te it o te sum eists.. Te it o and g as do not eist, so we cannot apply te dierence rule to g. We can say tat ( ) g( ). We can write te dierence as ( ) g ( ) ln( ) ln( ) ln. We can use graps or tables to convince ourselves tat tis it is equal to ln.. Te act tat te its o and g as do not eist does not necessarily mean tat te its o g, g or do not eist, just tat Teorem cannot be applied. g Quick Review.. y y y Intercange and y. y ( ) [, ] by [ 8, 8]. y e ln y Intercange and y. ln y ( ) ln [ 6, 6] by [, ] so te it o te quotient eists.

11 Section. 6. y tan tan y, < y< Intercange and y. tan y, < < ( ) tan, < < [ 6, 6] by [, ]. y cot cot y, < < Intercange and y. cot y, < y < ( ) cot, < < 7. (a) ( ) cos ( ) cos (b) cos ( 8. (a) ( ) e ) e (b) e / ln( ) ln( ) 9. (a) ( ) (b) ln / ln / ln. (a) ( ) sin ( ) ( sin ) sin (b) sin / sin Section. Eercises. [ 6, 6] by [, ]. ) 8 7 q ( ) 7 r ( ) 6. ) q ( ) r ( ) (a) ( ) (b) ( ) (c) y

12 6 Section... (a) ( ) (b) ( ) (c) y. (a) ( ) (b) ( ) (c) y, y 6. (a) ( ) (b) ( ) (c) y. (a) ( ) (b) ( ) (c) y, y 7. (a) ( ) (b) ( ) (c) No orizontal asymptotes. (a) ( ) (b) ( ) (c) y, y

13 Section (a) ( ) (b) ( ) (c) y 9. cos. So, or > we ave cos. By te Sandwic Teorem, cos ( ).. cos. So, or > we ave cos. By te Sandwic Teorem, cos ( ).. sin.. sin ( )... 6.

14 6 Section int ( / ) sec. y ( ) An end beavior model or y is y y. int 9. csc. y An end beavior model or y is. y y. Use te metod o Eample in te tet. cos cos cos( ) cos cos cos( )

15 Section. 6. Note tat y sin sin. So, sin y. Similarly, y. sin. Use y sin i sin ± ± So, y and y. sin sin 6. y So, y and y. 7.. (a), (b) Let-and it at is. Rigt-and it at is. Let-and it at is. Rigt-and it at is.. (a), (b) Let-and it at is. Rigt-and it at is. Let-and it at is. Rigt-and it at is. 8. (a) k, k any integer (b) at eac vertical asymptote: Let-and it is. Rigt-and it is.. (a) (b) Let-and it at is. Rigt-and it at is. 9. (a) n, n any integer (b) I n is even: Let-and it is. Rigt-and it is. I n is odd: Let-and it is. Rigt-and it is. (a) (b) Let-and it at is. Rigt-and it at is.

16 66 Section. tan sin. ( ) sin sin cos cos cos at: a ( k ) and b ( k ) were k is any real integer. ( ), ( ), ( ), a a b ( ). b cot cos. ( ) cos sin cos sin sin at a k and b ( k ) were k is any real integer. ( ), ( ), ( ), a a b ( ). b. An end beavior model is. (a) 6. An end beavior model is.. (c) 7. An end beavior model is. (d) 8. An end beavior model is 9. (a) (b) None. (a) (b) None. (a) (b) y. (a) (b) y. (a) (b) None. (b). (a) Te unction y e is a rigt end beavior model e because. e e (b) Te unction y is a let end beavior model e e because. 6. (a) Te unction y is a rigt end beavior model e e because. (b) Te unction y e is a let end beavior model e because e e ( e ). 7. (a, b) Te unction y is bot a rigt end beavior model and a let end beavior model because ln ln ± ±. 8. (a, b) Te unction y is bot a rigt end beavior model and a let end beavior model because sin sin ± ±. 9. Te grap o y e / is sown. ( ) ( ). (a) (b) None

17 Section. 67. Te grap o y e / ( ) ( ). is sown.. (a) ( ) (b) ( ) (c) ( ) (d) ( ). One possible answer: y y () 6. One possible answer: Te grap o y ln is sown. ( ) ( ). 7. Note tat ( )/ ( ) ( ) g( ) ( )/ g ( ) g ( )/ g ( ) g ( ) ( ) ( )/ g ( ). Te grap o y sin ( ) ( ). (a) ( ) (b) ( ) ( ) (c) ( ) (d) ( ) ( ) is sown. As becomes large, and bot approac. g g / Tereore, using te above equation, must also g/ g approac. 8. Yes. Te it o ( g) will be te same as te it o g. Tis is because adding numbers tat are very close to a given real number L will not ave a signiicant eect on te value o ( g) since te values o g are becoming arbitrarily large. 9. True. For eample, ( ) orizontal asymptotes. 6. False. Consider () /. 6. A. approaces zero, so as y ± as approaces.

18 68 Section. cos( ) cos( ) 6. E. undeined 6. C. 6. D. 6. (a) Note tat g ()g(). as, as, g, g (b) Note tat g ()g() 8. as, as, g, g 8 (c) Note tat g ()g() ( ). as, as, g, g (d) Note tat g ( ) g( ) ( )., g, g (e) Noting you need more inormation to decide. 66. (a) Tis ollows rom < int, wic is true or all. Dividing by gives te result. (b, c) Since, te Sandwic Teorem ± ± gives int int. e 67. For > < e, <, so < <. Since bot and approac zero as, te Sandwic Teorem states tat e must also approac zero. 68. Tis is because as approaces ininity, sin continues to oscillate between and and doesn't approac any given real number. 69. ln, because ln ln. ln ln ln ln ln ln ln ( ), Since 7. log log (ln ) /(ln ) ln. 7. ln( ) ln Since ln( ) ln ln ln, ln ln ln ln( ) ln ln ln But as, / approaces, so ln( /) approaces ln(). Also, as, ln approaces ininity. Tis means te second term above approaces and te it is. Quick Quiz Sections. and. 6. D. were. (.) A. < is not done since were.... E. cos cos. (a) ( ) (b) For >, cos. Tus cos and / cos. Because, cos (c) cos, cos (d) For all >, cos. cos Tereore,. Since, it ollows by te Sandwic Teorem tat cos.

19 Section. 69 Section. Continuity (pp ) Eploration Removing a Discontinuity. 9 ( )( ). Te domain o is (, ) (, ) (, ) or all ±.. It appears tat te it o as eists and is a little more tan.. () sould be deined as ( )( )( ), 9 ( )( ) ( )( ),so ( ) or. Tus, ( )( ). 6. g ( ) g( ), so g is continuous at. Quick Review. ( ) ( ) 6. ( ). (a) ( ) int ( ) (b) ( ) ( ) (c) ( ) does not eist, because te let- and rigtand its are not equal. (d) ( ) int ( ). ( g)( ) ( g( )) ( ), ( ) 6 ( g )( ) ( g( )) g,. Note tat sin ( g )( ) g( ()) g( ). Tereore: g() sin, ( g)( ) (g()) (sin ) (sin ) or sin, 6. Note tat ( g g )( ) ( ( )) ( ). Tereore, ( ) or >. Squaring bot sides gives ( ). Tereore, ( ), >. ( g)( ) ( g( )) ( ), > 7. 9 ( )( ) 8. Solutions:,. (a) ( ) ( ) ( ) (b) ( ) ( ) (c) ( ) does not eist, because te let- and rigtand its are not equal. (d) () Solution:. 9. For, () wen, wic gives. (Note tat tis value is, in act,.) For >, () wen 6 8, wic gives 6. Te discriminant o tis equation is b ac ( 6 ) ()(). Since te discriminant is negative, te quadratic equation as no solution. Te only solution to te original equation is.

20 7 Section.. 8. Te unction y cot is equivalent to y cos, a quotient sin o continuous unctions, so it is continuous. Its only points o discontinuity occur were it is undeined. It as ininite discontinuities at k or all integers k. 9. Te unction y e / is a composition ( g)( ) o te A grap o () is sown. Te range o () is (, ) [, ). Te values o c or wic () c as no solution are te values tat are ecluded rom te range. Tereore, c can be any value in [, ). Section. Eercises. Te unction y is continuous because it is a ( ) quotient o polynomials, wic are continuous. Its only point o discontinuity occurs were it is undeined. Tere is an ininite discontinuity at.. Te unction y is continuous because it is a quotient o polynomials, wic are continuous. Its only points o discontinuity occur were it is undeined, tat is, were te denominator ( )( ) is zero. Tere are ininite discontinuities at and at.. Te unction y is continuous because it is a quotient o polynomials, wic are continuous. Furtermore, te domain is all real numbers because te denominator,, is never zero. Since te unction is continuous and as domain (, ), tere are no points o discontinuity.. Te unction y is a composition ( g)( ) o te continuous unctions () and g(), so it is continuous. Since te unction is continuous and as domain (, ), tere are no points o discontinuity.. Te unction y is a composition ( g)( ) o te continuous unctions ( ) and g(), so it is continuous. Its points o discontinuity are te points not in te domain, i.e., all <. 6. Te unction y is a composition ( g)( ) o te continuous unctions ( ) and g(), so it is continuous. Since te unction is continuous and as domain (, ), tere are no points o discontinuity. 7. Te unction y is equivalent to y, <, >. It as a jump discontinuity at. continuous unctions () e and g ( ), so it is continuous. Its only point o discontinuity occurs at, were it is undeined. Since / e, tis may be considered an ininite discontinuity.. Te unction y ln ( ) is a composition ( g)( ) o te continuous unctions () ln and g(), so it is continuous. Its points o discontinuity are te points not in te domain, i.e., <.. (a) Yes, ( ). (b) Yes,. (c) Yes (d) Yes, since is a let endpoint o te domain o and ( ) ( ), is continuous at.. (a) Yes, (). (b) Yes, ( ). (c) No (d) No. (a) No (b) No, since is not in te domain.. Everywere in [, ) ecept or,,.. Since ( ), we sould assign (). 6. Since ( ), we sould reassign (). 7. No, because te rigt-and and let-and its are not te same at zero. 8. Yes, Assign te value to (). Since is a rigt endpoint o te etended unction and ( ), te etended unction is continuous at. 9. (a) (b) Not removable, te one-sided its are dierent.

21 Section. 7. (a) (b) Removable, assign te value to ().. (a) (b) Not removable, it s an ininite discontinuity.. (a) (b) Removable, assign te value to ( ).. (a) All points not in te domain along wit, (b) is a removable discontinuity, assign (). is not removable, te one-sided its are dierent.. (a) All points not in te domain along wit, (b) is not removable, te one-sided its are dierent. is a removable discontinuity, assign (). 9 ( )( ). For, ( ). Te etended unction is y. 6. For, ( ) ( )( ) ( )( ). Te etended unction is y. 7. Since sin, te etended unction is sin, y,. 8. Since sin sin (), te etended unction is sin, y,. 9. For (and > ), ( ) ( )( ). Te etended unction is y... For (and ), ( ) ( )( )( ) ( )( ) ( )( ). Te etended unction is y.. Te domain o is all real numbers. is continuous at all tose points so is a continuous unction.. Te domain o g is all real numbers >. is continuous at all tose points so g is a continuous unction.. is te composite o two continuous unctions g were g ( ) and ( ).. is te composite o two continuous unctions g were g() sin and ().. is te composite o tree continuous unctions g k were g ( ) cos, ( ), and k ( ). 6. is te composite o two continuous unctions g were g() tan and ( ).

22 7 Section. 7. One possible answer:. One possible answer: Assume y, constant unctions, and te square root unction are continuous. By te sum teorem, y is continuous. By te composite teorem, y is continuous. By te quotient teorem, y is continuous. Domain: (, ) 8. One possible answer: Assume y, constant unctions, and te cube root unction are continuous. By te dierence teorem, y is continuous. By te composite teorem, y is continuous.. One possible answer: y By te product teorem, y is continuous. By te sum teorem, y is continuous. Domain: (, ) y () 9. Possible answer: Assume y and y are continuous. By te product teorem, y is continuous. By te constant multiple teorem, y is continuous. By te dierence teorem, y is continuous. By te composite teorem, y is continuous. Domain: (, ). One possible answer: Assume y and y are continuous. Use te product, dierence, and quotient teorems. One also needs to veriy tat te it o tis unction as approaces is. Alternately, observe tat te unction is equivalent to y (or all ), wic is continuous by te sum teorem. Domain: (, ). One possible answer: y y (). One possible answer:. Solving, we obtain te solutions.7 and.. 6. Solving, we obtain te solution..

23 Section We require tat a ( ) : a() 6a 8 a 8. Solve at ( ) ( ) 7 ( ) a at. 7 a 6 a a 9. Solve at ( ) ( ) ( ) a at a( ) a. Solve at ( ) ( ) a a a. Consider () e. is continuous, (), and () >.. By te Intermediate Value Teorem, or e some c in (, ), (c) and e c c.. (a) Sara s salary is $6, $6, (. ) or te irst year ( t < ), $6,(.) or te second year ( t < ), $6, (. ) or te tird year ( t < ), and so on. Tis corresponds to int t y 6, (. ). (b) [,.8] by [, ] Te unction is continuous at all points in te domain [, ) ecept at t,,,.. (a) We require:., <., <., < ( ),. <., < 66., < 6 7., 6<. Tis may be written more compactly as. int( ), ( ) 6 7., 6< (b) [, ] by [, 9] Tis is continuous or all valyes o in te domain [, ] ecept or,,,,,, 6.. False. Consider () / wic is continuous and as a point o discontinuity at.. True. I as a jump discontinuity at a, ten ( ) ( ) so is not continuous at a. a a 6. B. ( ) is not deined. 7. E. ( ) is te only deined option. 8. A. (). 9. E. causes a zero to be in te denominator. 6. (a) Te unction is deined wen >, tat is, on (, ) (, ). (It can be argued tat te domain sould also include certain values in te interval (, ), namely, tose rational numbers tat ave odd denominators wen epressed in lowest terms.) (b) [, ] by [, ]

24 7 Section. 6. Continued (c) I we attempt to evaluate ( ) at tese values, we obtain ( ) (undeined) and ( ) (undeined). Since is undeined at tese values due to division by zero, bot values are points o discontinuity. (d) Te discontinuity at is removable because te rigt-and it is. Te discontinuity at is not removable because it is an ininite discontinuity. (e) Te it is about.78, or e. 6. Tis is because ( a ) ( ). a 6. Suppose not. Ten would be negative somewere in te interval and positive somewere else in te interval. So, by te Intermediate Value Teorem, it would ave to be zero somewere in te interval, wic contradicts te ypotesis. 6. Since te absolute value unction is continuous, tis ollows rom te teorem about continuity o composite unctions. 6. For any real number a, te it o tis unction as approaces a cannot eist. Tis is because as approaces a, te values o te unction will continually oscillate between and. Section. Rates o Cange and Tangent Lines (pp. 87 9) Quick Review.. Δ ( ) 8 Δ y. Δ a Δ y b. m ( ) 7 7. m ( ) 6 ( ). y [ ( )] y 6 6. m y ( ) 6 7 y 7. y ( ) 9 y 8. m / y ( ) 8 y 9. Since y is equivalent to y, we use m. y [ ( )] 7 y. b b 9 b Section. Eercises. (a) Δ Δ () () (b) Δ Δ () ( ) ( ). (a) Δ Δ ( ) ( ) (b) Δ Δ ( ) ( ) (a) Δ Δ ( ) ( ) e. ( ) (b) Δ Δ () () e e 8. 68

25 Section. 7. (a) Δ Δ ( ) ( ) ln ln. 6 (b) Δ Δ ( ) ( ) ln ln 99 ln.. ln. (a) Δ Δ ( / ) ( / ). 7 ( / ) ( / ) / (b) Δ Δ ( / ) ( / 6). 6 ( / ) ( / 6) / 6. (a) Δ Δ ( ) ( ). 67 (b) Δ Δ ( ) ( ) ( ) 7. We use Q (, ), Q (, 7), Q (6., 7), Q (8, ), and P (, 6). 6 (a) Slope o PQ: 6 7 Slope o PQ: Slope o PQ : 6. 6 Slope o PQ : 8 Secant Slope PQ PQ 6 PQ PQ Te appropriate units are meters per second. (b) Approimately m/sec 8. We use Q (, ), Q (7, 8), Q (8., 6), Q (9., 7), and P (, 8). 8 (a) Slope o PQ: 8 8 Slope o PQ: Slope o PQ : 8. 6 Slope o PQ : Secant Slope PQ PQ PQ 6 PQ 6 Te appropriate units are meters per second. (b) Approimately 6 m/sec y 9. (a) ( ) y ( ) ( ) ( ) ( ) (b) Te tangent line as slope and passes troug (, y( )) (, ) y[ ( )] y (c) Te normal line as slope (, y( )) (, ). (d) y [ ( )] 9 y [ 8, 7] by [, 9] y. (a) ( ) y ( ) [( ) ( )] [ ()] ( ) and passes troug (b) Te tangent line as slope and passes troug (, y()) (, ). y( ) y

26 76 Section.. Continued (c) Te normal line as slope (, y()) (, ). (d) y ( ) 7 y [ 6, 6] by [ 6, ] y. (a) ( ) y ( ) ( ) ( ) ( ) and passes troug (b) Te tangent line as slope and passes troug (, y()) (, ). y( ) y (c) Te normal line as slope and passes troug (, y()) (, ). (d) y ( ) y y. (a) ( ) y ( ) ( ) ( ) ( ) (b) Te tangent line as slope and passes troug (, y()) (, ). y( ) y (c) Te normal line as slope (d) (, y()) (, ). y ( ) y [ 6, 6] by [, ]. (a) Near, (). and passes troug ( ) ( ) ( ) (b) Near, (). ( ) ( ) ( ). Near, () ( ). ( ) ( ) [ ( )] ( ) [.7,.7] by [.,.]

27 Section. 77. First, note tat (). ( ) ( ) ( ) ( ) ( ) ( ) ( ) No, te slope rom te let is and te slope rom te rigt is. Te two-sided it o te dierence quotient does not eist. 6. First, note tat (). ( ) ( ) ( ) ( ) ( ) ( ) Yes. Te slope is. 7. First, note tat ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ( )] Yes. Te slope is. 8. No. Te unction is discontinuous at because ( ) sin sin ( / ) ( / ) cos. but 9. (a) ( a ) ( a ) [( a ) ] ( a ) a a a a ( a ) a (b) Te slope o te tangent steadily increases as a increases.. (a) ( a ) ( a ) a a a ( a ) a( a ) aa ( ) a (b) Te slope o te tangent is always negative. Te tangents are very steep near and nearly orizontal as a moves away rom te origin.. (a) ( a ) ( a ) a a ( a) ( a ) a ( )( a ) ( a )( a ) ( a ) (b) Te slope o te tangent is always negative. Te tangents are very steep near and nearly orizontal as a moves away rom te origin.. (a) ( ) ( ) [ ( ) ] ( a a 9 a 9a ) 9a a 9 a a ( a) a (b) Te slope o te tangent steadily decreases as a increases.

28 78 Section.. Let (t).9 t. ( ) ( ) [. ( ) ] [ 9.()] ( ) 9. 6 Te object is alling at a speed o 9.6 m/sec.. Let (t) t. ( ) ( ) ( ) 6 ( 6 ) 6 Te rocket s speed is 6 t/sec.. Let () r r, te area o a circle o radius r. ( ) ( ) ( ) ( ) ( 6 ) 6 Te area is canging at a rate o 6 in / in.,tat is, 6 square inces o area per inc o radius. 6. Let () r r. ( ) ( ) ( ) ( ) ( ) ( 6 ) 6 Te volume is canging at a rate o 6 in / in., tat is, 6 cubic inces o volume per inc o radius. s 7. ( ) s ( ) 86. ( ) 86. ( ) (.7. 86) 7. Te speed o te rock is.7 m/sec. s 8. ( ) s ( ). ( ). ( ) (. 76. ). 76 Te speed o te rock is.76m/sec. 9. First, in te slope o te tangent at a. ( a ) ( a ) [( a ) ( a ) ] ( a a) a a a a a a ( a ) a Te tangent at a is orizontal wen a, or a. Te tangent line is orizontal at (, ( )) (, ).. First, ind te slope o te tangent at a. ( a ) ( a ) [ ( a ) ( a ) ] ( aa ) aa a a a a ( a ) a Te tangent at a is orizontal wen a, or a. Te tangent line is orizontal at (, ( )) (, 7).

29 Section. 79. (a) From Eercise, te slope o te curve at a, is ( a ) ( a ). Te tangent as slope wen, a. Note tat y( ) wic gives ( a ), so a or and y( ), so we need to ind te equations o lines o slope passing troug (, ) and (, ), respectively. At : y( ) y At : y( ) y (b) Te normal as slope wen te tangent as slope, so we again need to ind lines troug (, ) and (, ), tis time using slope. At : y ( ) y At :y ( ) y Tere is only one suc line. It is normal to te curve at two points and its equation is y.. Consider a line tat passes troug (, ) and a point (a, 9 a ) on te curve. Using te result o Eercise, tis line will be tangent to te curve at a i its slope is a. ( 9a ) a a 9a a( a) a a a a a ( a ) ( a ) a or a At a ( or ), te slope is ( ). y ( ) y At a ( or ), te slope is () 6. y6( ) y 6 8. (a) ( ) ( 99 99) (b) (. 9. ) ( ) (c) ( ) ( ) billion years billion years billion years 7. billion $. years $. billion year $ 6. billion year (d) y [, ] by [, ] (e) 99 to 99:(. 77( ). ( ) 6. ) (. 77( ). ( ) 6. ) $. billion to : (.77().()6.) (.77().()6.) $. billion year billion year to : (.77().() 6.) (.77().() 6.) $. billi on year (). 77(.). (.) 6. (. 77(.). ( ) 6. ) billion $.. year (g) One possible reason is tat te war in Iraq and increased spending to prevent terrorist attacks in te U.S. caused an unusual increase in deense spending.

30 8 Section.. (a). (a) ( ) ( ) e (b) e [, ] by [, ] (b) PQ 7. PQ PQ 6.. True. Te normal line is perpendicular to te tangent line at te point. 6. False. Tere's no tangent at because is undeined at. 7. D E. () () 9. C. ( a ) ( a ) y () y m( ) y y( ) y. A. From 9, m m m y ( ) y (c) Tey re about te same. (d) Yes, it as a tangent wose slope is about e.. (a) ( ) ( ) (b) (c) Tey re about te same. (d) Yes, it as a tangent wose slope is about ln. /. Let ( ) ( ) ( ) ( ). Te grap o y is sown. Te let- and rigt-and its are and, respectively. Since tey are not te same, te curve does not ave a vertical tangent at. No.

31 Section. 8. Let () / ( ) ( ) ( ). Te grap o y is sown. Yes, te curve as a vertical tangent at because ( ) ( ).. Let () / ( ) ( ) ( ). Te grap o y is sown. 8. Tis unction does not ave a tangent line at te origin. As te unction oscillates between y and y ininitely oten near te origin, tere are an ininite number o dierence quotients (secant line slopes) wit a value o and wit a value o. Tus te it o te dierence quotient doesn t eist. Te dierence quotient is ( ) ( ) sin wic oscillates between and ininitely oten near zero. 9. Let () sin. Te dierence quotient is ( ) () sin( ) sin(). A grap and table or te dierence quotient are sown. Yes, te curve as a vertical tangent at because ( ) ( ). 6. Let () / ( ) ( ) ( ). Te grap o y is sown. Te let- and rigt-and its are and, respectively. Since tey are not te same, te curve does not ave a vertical tangent at. No. 7. Tis unction as a tangent wit slope zero at te origin. It is sandwiced between two unctions, y and y, bot o wic ave slope zero at te origin. Looking at te dierence quotient, ( ) ( ), so te Sandwic Teorem tells us te it is. Since te it as is about., te slope o y sin at is about.. Quick Quiz Sections. and.. D. () (). E. ( ) ( ) were ( ) ( ) 7. B. ( a ) ( a ) 9 ( ) ( 9 ) y 9 9( ) y y ( ) y( ) y

32 8 Capter Review. (a) () () () 6 9 (b) ( ) ( ) ( ) 6 ( 9 6 ) (c) ( ) ( ) ( ) (d) were Capter Review Eercises (pp. 9 97). ( ) ( ) ( ) ( ). ( ) ( ). No it, because te epression is undeined or values o near.. No it, because te epression 9 is undeined or values o near. ( ). ( ) ( ) ( ) ( ) 6. ± 7 ± 7. An end beavior model or is. 8 Tereore sin sin () 9. Multiply te numerator and denominator by sin. csc sin sin csc sin ( ). e sin e sin 7. Let, were is in,. Ten int ( ) int 7 int ( 6 ) 6, because 6 is in (6, 7). Tereore, int ( ) / 7 / 7. Let, were is in,. Ten int ( ) int 7 int ( 6 ), because 6 is in (, 6). Tereore, int ( ) 7 / 7 /. Since ( e ) e, and e e or all, te Sandwic Teorem gives e cos. e cos. Since te epression is an end beavior model or bot sin sin and cos, cos.. Limit eists. 6. Limit eists. 7. Limit eists. 8. Limit does not eist. 9. Limit eists.. Limit eists.. Yes. No. No. Yes. (a) g ( ) (b) g(). (c) No, since g ( ) g( ). (d) g is discontinuous at (and at points not in te domain). (e) Yes, te discontinuity at can be removed by assigning te value to g().

33 Capter Review 8 6. (a) k( ). (b) k( ) (c) k() (d) No, since k( ) k( ) (e) k is discontinuous at (and at points not in te domain). () No, te discontinuity at is not removable because te one-sided its are dierent. 7. (a) Vertical asymptote: (b) Let-and it Rigt-and it: 8. (a) Vertical asymptotes:, (b) At : Let-and it ( ) Rigt-and it ( ) At : Let-and it ( ) Rigt-and it ( ) 9. (a) At : Let-and it ( ) ( ) Rigt-and it ( ) ( ) At : Let-and it ( ) ( ) Rigt-and it ( ) ( ) At : Let-and it ( ) ( ) Rigt-and it ( ) ( ) (b) At : Yes, te it is. At : Yes, te it is. At : No, te it doesn t eist because te two one-sided its are dierent. (c) At : Continuous because ( ) te it. At : Discontinuous because () te it. At : Discontinuous because te it does not eist.. (a) Let-and it ( ) () () Rigt-and it ( ) ( ) () () (b) No, because te two one-sided its are dierent. (c) Every place ecept or (d) At. Since () is a quotient o polynomials, it is continuous and its points o discontinuity are te points were it is undeined, namely and.. Tere are no points o discontinuity, since g() is continuous and deined or all real numbers.. (a) End beavior model:, or (b) Horizontal asymptote: y (te -ais). (a) End beavior model:, or (b) Horizontal asymptote: y. (a) End beavior model:, or (b) Since te end beavior model is quadratic, tere are no orizontal asymptotes. 6. (a) End beavior model:, or (b) Since te end beavior model represents a nonorizontal line, tere are no orizontal asymptotes. e 7. (a) Since, e e a rigt end beavior model is e. e e (b) Since, a let end beavior model is.

34 8 Capter Review 8. (a, b) Note tat ± ln ± ln and sin < < or all. ln ln ln Tereore, te Sandwic Teorem gives sin. Hence ± ln ln sin sin ± ln ± ln, so ln is bot a rigt end beavior model and a let end beavior model. 9. ( ) ( )( ) ( ) 8. Assign te value k 8.. ( ) sin sin () Assign te value k.. One possible answer: y y (). One possible answer: y. ( ) ( ) Va Va ( a ) H a H a a a H H ( a ) H( a) ah Sa. ( ) Sa ( ) 6 ( a ) 6a 6a a 6 6a ( a 6 ) a ya 6. ( ) ya ( ) [( ) a ( a ) ] ( a a) a a a a a a ( a ) a 7. (a) ( ) ( ) [( ) ( )] ( ) ( ) (b) Te tangent at P as slope and passes troug (, ). y( ) y y () (c) Te normal at P as slope and passes troug (, ). y ( ) y. ( / ) ( ) / /

35 Capter Review 8 8. At a, te slope o te curve is ( a ) ( a ) [( a ) ( a )] ( a a) a a a a a a ( a ) a Te tangent is orizontal wen a, at 9 a or. Since, te point were tis 9 occurs is,. 9. (a) p( ).( ) 7e 8 Peraps tis is te number o bears placed in te reserve wen it was establised. (b) pt ( ) t t. t 7e (c) Peraps tis is te maimum number o bears wic te reserve can support due to itations o ood, space, or oter resources. Or, peraps te number is capped at and ecess bears are moved to oter locations... int( ), <. (a) ( ), (Note tat we cannot use te ormula ().. int, because it gives incorrect results wen is an integer.) (b) 7, 9, 98 6 PQ. 7, 9 6, 66 PQ 7, 9 6, 69 7 PQ 7 (c) We use te average rate o cange in te population rom to wic is 7, (d) y , rate o cange is 9 tousand because rate o cange o a linear unction is its slope.. Let A ( ) and B g( ). Ten A B and c c A B. Adding, we ave A, so A, wence B, wic gives B. Tereore, ( ) c and g ( ). c. (a) 9 All not equal to or. (b), (c) 9 y (d) Wen 9, ( ). and are discontinuous. (e) Yes. It is continuous at every point in its domain. [, ] by [, ] is discontinuous at integer values o :,,,..., 9.. (a) [, ] by [, 8] (b) PQ PQ 7, 9, , 9,

36 86 Section.. (a) ( ) a ( ) a a (b) ( ) ( ) 8 (c) For, is continuous. For, we ave ( ) ( ) ( ) as long as a ±.. (a) g ( ) (b) ( ) g ( ) Capter Derivatives Section. Derivative o a Function (pp. 99 8) Eploration Reading te Graps. Te grap in Figure.b represents te rate o cange o te dept o te water in te ditc wit respect to time. Since y is measured in inces and is measured in days, te derivative dy would be measured in inces per day. Tose d are te units tat sould be used along te y-ais in Figure.b.. Te water in te ditc is inc deep at te start o te irst day and rising rapidly. It continues to rise, at a gradually decreasing rate, until te end o te second day, wen it acieves a maimum dept o inces. During days,,, and 6, te water level goes down, until it reaces a dept o inc at te end o day 6. During te sevent day it rises again, almost to a dept o inces.. Te weater appears to ave been wettest at te beginning o day (wen te water level was rising astest) and driest at te end o day (wen te water level was declining te astest).. Te igest point on te grap o te derivative sows were te water is rising te astest, wile te lowest point (most negative) on te grap o te derivative sows were te water is declining te astest.. Te y-coordinate o point C gives te maimum dept o te water level in te ditc over te 7-day period, wile te -coordinate o C gives te time during te 7-day period tat te maimum dept occurred. Te derivative o te unction canges sign rom positive to negative at C, indicating tat tis is wen te water level stops rising and begins alling. 6. Water continues to run down sides o ills and troug underground streams long ater te rain as stopped alling. Depending on ow muc ig ground is located near te ditc, water rom te irst day s rain could still be lowing into te ditc several days later. Engineers responsible or lood control o major rivers must take tis into consideration wen tey predict wen loodwaters will crest, and at wat levels. Quick Review. ( ) ( )... Since y y or y y <,. y y 8. ( )( ) ( ) ( ) 8. Te verte o te parabola is at (, ). Te slope o te line troug (, ) and anoter point (, ) on te parabola is ( ). Since, te slope o te line tangent to te parabola at its verte is. 6. Use te grap o in te window [6, 6] by [, ] to ind tat (, ) is te coordinate o te ig point and (, ) is te coordinate o te low point. Tereore, is increasing on (, ] and [, ). 7. ( ) ( ) ( ) ( ) ( ) 8. ( ) ( ) 9. No, te two one-sided its are dierent (see Eercise 7).. No, is discontinuous at because ( ) does not eist.