Chapter 2. Limits and Continuity 16( ) 16( 9) = = 001. Section 2.1 Rates of Change and Limits (pp ) Quick Review 2.1

Transcription

1 Capter Limits and Continuity Section. Rates of Cange and Limits (pp. 969) Quick Review..... f ( ) ( ) ( ) 0 ( ) f ( ) f ( ) sin π sin π 0 f ( ). < < < 6. < c c < < c 7. < < < < < c < d d < c < d d c < < d c 8 ( )( 6) 6, ( ), ( )( ) Section. Eercises.. Section. 6 y 6( ) 6( ), say 0. 0 t 6( 0. 0) 6( 9) 00. 6( ) 6() ft/sec Confirm Algebraically y 6( ) 6( ) t 6( 9 6 ) 96 6 ( 96 6 ) ft/sec y ft if 0, ten 96 t sec y 6( ) 6( ), say 0. 0 t 6( 0. 0) 6( ) 6( ) 6( 6) ft/sec Confirm Algebraically y 6( ) 6( ) t 6( 6 8 ) ( 8 6 ) ft/sec y ft if 0, ten 8 t sec. y 6() 6() 0 t 0 8 ft/sec. c c c ( ) c. y 6( ) 6( 0) t 0 6 ft/sec 6. c 9 c c c 9 Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

2 6 Section ( ) / ( ) ( ) ( ) ( ) ( 7) () () () 7 7 y y y 6 ( ) 6 0 y y y ( ) ( ) 0 0 y y ( ) 6 int int 0 / Note tat substitution cannot always be used to find its of te int function. Its use ere can be justified by te Sandwic Teorem, using g() () 0 on te interval (0, ). / / ( 6) ( 6) ( 8) 6.. (a) f() f() Te it appears to be. 6. (a) f() Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

3 Section f() Te it appears to be. 7. (a) f() f() Te it appears to be (a) f() f() Tere is no clear indication of a it. 9. (a) f() f() Te it appears to be approimately.. 0. (a) f() f() Te it appears to be 0.. You cannot use substitution because te epression is not defined at. Since te epression is not defined at points near, te it does not eist.. You cannot use substitution because te epression is not defined at 0. Since becomes arbitrarily large as approaces 0 from eiter side, tere is no (finite) it. (As we sall see in Section., we may write.) 0 Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

4 6 Section.. You cannot use substitution because te epression is not defined at 0. Since and, te left- and 0 0 rigt-and its are not equal and so te it does not eist.. You cannot use substitution because te ( ) 6 epression is not defined at 0. Since. ( ) for all 0, te it eists and is equal to ( 8 ) Algebraic confirmation: ( ) ( ) Algebraic confirmation: t t ( t)( t) t t t ( t )( t ) t t t Algebraic confirmation: 8 ( 8) ( 6 ) () 8 0 () t t t t 0 Algebraic confirmation: ( ) 0 0 ( )( ) 0 ( ) ( ) 0 ( ) ( 0) Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

5 Section ( ) 8 0 Algebraic confirmation: ( ) 8 6 ( 6 ) () 0 () 0 sin 0 Algebraic confirmation: sin sin 0 0 sin sin 0 Algebraic confirmation: sin sin ( ) 0 0 sin 0 0 Algebraic confirmation: sin sin sin sin sin 0 0 (sin 0)( ) sin 0 Algebraic confirmation: sin sin 0 0 sin 0 0 () 0 (). [0, 0] by [0, 0] 7 Algebraic confirmation: ( )( ) ( ) () () 7. Answers will vary. One possible grap is given by te window [.7,.7] by [, ] wit Xscl and Yscl. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

6 66 Section. 6. Answers will vary. One possible grap is given by te window [.7,.7] by [, ] wit Xscl and Yscl. (d) True (e) True 7. Since int 0 for in (0, ), int 0. 0 (f) True, since f( ) f( ). 8. Since int for in (, 0), int. 0 (g) True, since bot are equal to 0. () True 9. Since int 0 for in (0, ), int (i) True, since f( ) for all c in (, ). c 0. Since int for in (, ), int.. (a) f( ).. Since for 0,. > 0 Since for 0,. < 0. (a) True True (c) f( ) f( ) does not eist, because te left and rigt-and its are not equal. (d) f() 6. (a) gt ( ) t (c) False, since f( ) 0. 0 gt ( ) t (d) True, since bot are equal to 0. (e) True, since (d) is true. (c) gt ( ) does not eist, because te leftt and rigt-and its are not equal. (f) True (d) g( ) (g) False, since f( ) (a) f( ) () False, f( ), but undefined. (i) False, f( ) 0, but undefined. f( ) is f( ) is (c) f( ) f( ) (d) f(0) (j) False, since f( ) (a) ps ( ) s. (a) True False, (c) False, since f( ). since f( ). (c) ps ( ) s ps ( ) s (d) p() Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

7 9. (a) (c) 0. (a) (c) F( ) 0 F( ) 0 F( ) does not eist, because te left 0 and rigt- and its are not equal. (d) F(0) G ( ) ' G ( ) G ( ) (d) G() (d) 6. (a) Section. 67 g ( ) g ( ) f ( ) f ( ) 0 ( f( ) g( )) b f( ) g( ) b b 7 ( ) ( f( ) g( )) b f( ) g( ) b b ( 7) ( )... ( )( ) y, (c) ( )( ) y ( ) y, (d) (c) g( ) g( ) ( ) b b f( ) f( ) b 7 7 (d) b g ( ) g ( ) b 7. (a) [, 6] by [, ] ( )( ). y (a). (a) ( g ( ) ) g ( ) 6 8. (a) f( ) ; f( ) (c) No, because te two one-sided its are different. (c) f( ) f( ) 0 0 g ( ) g( ) 9 [, 6] by [, ] f( ) ; f( ) (c) Yes; te it is. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

8 68 Section. 9. (a) (c) c π (d) c π [, ] by [, 8] 6. (a). y f( ) ; f( ) does not eist. 60. (a) (c) No, because te left-and it does not eist.. (0, ) (, ) [.7,.7] by [.,.] (c) c f( ) 0; f( ) 0 (c) Yes; te it is (a) (d) c 0. y 6. (a). y. π ( π, 0) ( 0, π ) (c) c π (d) c π 6. (, ) (, ) (, ) (c) None (d) None 6. (a) y. π π π,, π π ( sin ) 0 0 Confirm using te Sandwic Teorem, wit g( ) and ( ). sin sin sin Because ( ) 0, te 0 0 Sandwic Teorem gives ( sin ) 0. 0 Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

9 ( sin ) 0 Confirm using te Sandwic Teorem, wit g() and (). sin sin. sin Because ( ) 0, te 0 0 Sandwic Teorem gives ( sin ) 0 0 sin 0 0 Confirm using te Sandwic Teorem, wit g() and (). sin sin. sin Because ( ) 0, te 0 0 Sandwic Teorem give sin 0. 0 cos 0 0 Confirm using te Sandwic Teorem, wit g() and (). cos cos. Section. 69 cos Because ( ) 0, te 0 0 Sandwic Teorem give cos (a) In tree seconds, te ball falls.9 ( ). m, so its average speed is 70. (a).. 7 m/sec. Te average speed over te interval from time t to time is y 9.( ) 9.() t ( ) 96.( ) Since ( 9.. 9) 9., te instantaneous speed is 9. m/sec. y gt 0 g( ) 0 g or. 6 0 Average speed m/sec (c) If te rock ad not been stopped, its average speed over te interval from time t to time t is y. ( ). ( ) t ( ) 8. ( ) 0. Since ( 0. ) 0, te instantaneous speed is 0 m/sec. 7. True; te definition of a it. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

10 70 Section. 7. True sin sin 0 0 sin 0 sin as C 7. B 7. E 76. C 77. (a) Because te rigt-and it at zero depends only on te values of te function for positive -values near zero. Area of OAP ( base)( eigt) ()(sin θ ) sinθ ( angle)( radius) Area of sector OAP θ () θ Area of OAT ( base)( eigt) ()(tan θ) tanθ (c) Tis is ow te areas of te tree regions compare. (d) Multiply by and divide by sin θ. (g) sin ( θ ) sin sin θ θ θ θ θ () If te function is symmetric about te y-ais, and te rigt-and it at zero is, ten te left-and it at zero must also be. (i) Te two one-sided its bot eist and are equal to. 78. (a) Te it can be found by substitution. f( ) f ( ) ( ) Te graps of y f( ), y 8., and y. are sown. Te intersections of y wit y and y are at.767 and.8, respectively, so we may coose any value of a in [.767, ) (approimately) and any value of b in (,.8]. One possible answer: a.7, b.8. (c) Te graps of y f( ), y 99., and y 0. are sown. Te intersections of y wit y and y are at.9867 and.0, respectively, so we may coose any value of a in [.9867, ), and any value of b in (,.0] (approimately). One possible answer: a.99, b.0. (e) Take reciprocals, remembering tat all of te values involved are positive. (f) Te its for cos θ and are bot equal to. Since sinθ is between tem, it must θ also ave a it of. 79. (a) f π π sin 6 6 Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

11 Section. 7 Te graps of y f( ), y 0., and y 07. are sown. Te intersections of y wit y and y are at 0.07 and 0.77, respectively, so we may coose any value π of a in 0. 07,, and any value of b 6 π in 077 6,., were te interval endpoints are approimate. One possible answer: a 0.0, b (c) Te graps of y f( ), y 09., and y 0. are sown. Te intersections of y wit y and y are at 0. and 0., respectively, so we may coose any value π of a in 0.,, and any value of b 6 π in 0 6,., were te interval endpoints are approimate. One possible answer: a 0., b Line segment OP as endpoints (0, 0) and ( a, a ), so its midpoint is 0 a 0 a a a,, and its slope is a 0 a. Te perpendicular bisector is te a 0 a a line troug, wit slope, so its a equation is a a y a, wic is equivalent to a y. Tus te a a y-intercept is b. As te point P approaces te origin along te parabola, te value of a approaces zero. Terefore, a 0 b. P O a 0 Section. Limits Involving Infinity (pp. 7077) Eploration Eploring Teorem. Neiter f ( ) or g( ) eist. In tis case, we can describe te beavior of f and g as by writing f ( ) and g ( ). We cannot apply te quotient rule because bot its must eist. However, from Eample, sin sin 0, so te it of te quotient eists.. Bot f and g oscillate between 0 and as, taking on eac value infinitely often. We cannot apply te sum rule because neiter it eists. However, (sin cos ) ( ), so te it of te sum eists.. Te it of f and g as do not eist, so we cannot apply te difference rule to f g. We can say tat f( ) g( ). We can write te difference as f( ) g( ) ln ( ) ln ( ) ln. We can use graps or tables to convince ourselves tat tis it is equal to ln.. Te fact tat te its of f and g as do not eist does not necessarily mean tat te its of f g, f g or f do not eist, just g tat Teorem cannot be applied. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

12 7 Section. Quick Review.. y y y Intercange and y. y f ( ).. [, ] by [ 8, 8] y e ln y Intercange and y. ln y f ( ) ln [ 6, 6] by [, ] y tan π π tan y, < y < Intercange and y. π π tan y, < < π π f ( ) tan, < <. 6. [ 6, 6] by [, ] q ( ) 7 r ( ) q ( ) r ( ) 7. (a) f( ) cos ( ) cos 8. (a) f( ) f cos ( ) e e. [ 6, 6] by [, ] y cot cot y, 0 < y <π Intercange and y. cot y, 0 < < π f ( ) cot, 0 < < π 9. (a) f / e ln ( ) ln ( ) f( ) ln ln f ln Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

13 Section (a) f( ) sin( ) ( sin ) sin. f sin sin Section. Eercises. (a) f( ) 0 f( ) (c) y 0. (a) f( ) f( ) (c) y (a) f( ) f( ). (c) No orizontal asymptotes.. (a) f( ) 0 f( ) 0 (c) y 0 (a) f( ) f( ) (c) y, y Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

14 7 Section (a) f( ) f( ) (c) y, y 9. 0 cos. So, for > 0 we ave cos 0. By te Sandwic Teorem, cos 0 ( 0) cos. So, for > 0 we ave cos 0. By te Sandwic Teorem, cos 0 ( 0) 0. sin. sin, so for < 0,. Terefore sin 0 by te Sandwic Teorem. (a) f( ) f( ) (c) y, y.. sin ( ), so for > 0, sin( ) sin( ) (a) f( ) f( ) (c) y Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

15 Section int csc int 0. y π ( ) sec ( ) Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

16 76 Section. An end beavior model for y is y y. y 0 An end beavior model for y is y y... Use te metod of Eample 9 in te tet. ( ) cos cos cos( 0) 0 0 ( ) cos cos cos( 0) 0 0. Note tat sin sin y. sin So, y 0. Similarly, y (a), Left-and it at is. Rigt-and it at is. Left-and it at is. Rigt-and it at is. (a) Left-and it at is. Rigt-and it at is. (a) Left-and it at is. Rigt-and it at is.. 6. sin sin Use y sin 0 ± 0 ± So, y 0 and y 0. sin sin y So, y 0 and y 0. (a), Left-and it at is. Rigt-and it at is. Left-and it at is. Rigt-and it at is. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

17 Section An end beavior model is. (d). (a) kπ, k any integer at eac vertical asymptote: Left-and it is. Rigt-and it is. 8. An end beavior model is 9. (a) 0. (a) None None.. (a).. (a) π nπ, n any integer If n is even: Left-and it is. Rigt-and it is. If n is odd: Left-and it is. Rigt-and it is. tan sin f( ) sin sin cos cos cos 0 at: a ( k ) π and b ( k ) π were k is any real integer. f( ), f( ), a a f( ), f( ). b b cot cos f( ) cos sin cos sin sin 0 at a kπ and b ( k ) π were k is any real integer. f ( ), f ( ), a a f ( ), f ( ). b b. An end beavior model is. (a). (a). (a). (a) y 0 y None None. (a) Te function y e is a rigt end beavior model because e 0. e e Te function y is a left end beavior model because e e An end beavior model is 0.. (c) Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

18 78 Section. 6. (a) Te function y is a rigt end beavior model because e e Te function y e is a left end beavior model because e e e ( e ) 0.. / Te grap of y f e f( ) f 0 0 f( ) f 0 is sown. 7. (a, b) Te function y is bot a rigt end beavior model and a left end beavior model because ln ln ± ± (a, b) Te function y is bot a rigt end beavior model and a left end beavior model because sin sin. ± ±. Te grap of y f ln f( ) f 0 0 f( ) f 0 0 is sown. 9. / Te grap of y f e f( ) f 0 f( ) f 0 0 is sown.. (a) sin Te grap of y f is sown. f( ) f 0 f( ) f 0 f( ) 0 f( ) ( ) (c) f( ) 0 0 (d) f( ) ( ) 0 0 Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

19 . (a) (c) (d) f( ) f( ) 0 0 f( ) f( ) 0 0 Section True. For eample, f( ) as y ± as orizontal asymptotes. 60. False; consider f( ). 6. A; ( ) approaces zero from te left, so approaces.. One possible answer: y y f() 6. One possible answer: 0 cos( ) 6. E; is undefined because 0 cos( ) as a vertical asymptote at C; let t. Ten t 0 as 0 ; and sin ( ) sin t 0 t 0 t / sin t 0 t t 6. D; 6. (a) Note tat fg f()g(). f as 0, f as 0, g 0, fg 7. Note tat f f f g g f g g ( ) ( ) ( ) f( ) g( ) ( ) ( ) g ( ) ( ) f( ) ( ) ( ) f As becomes large, g and f bot g approac. Terefore, using te above equation, f f g g must also approac. 8. Yes. Te it of ( f g) will be te same as te it of g. Tis is because adding numbers tat are very close to a given real number L will not ave a significant effect on te value of ( f g) since te values of g are becoming arbitrarily large.. Note tat fg f()g() 8. f as 0, f as 0, g 0, fg 8 (c) Note tat fg f()g() ( ). f as, f as, g 0, fg 0 (d) Note tat fg f ( ) g ( ). ( ) f, g 0, fg (e) Noting you need more information to decide. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

20 80 Section. 66. (a) Tis follows from < int, wic is true for all. Dividing by gives te result.. E; (b, c) Since, te ± ± Sandwic Teorem gives int int. 68. Tis is because as approaces infinity, sin continues to oscillate between and and doesn't approac any given real number. 69. ln ln, because ln ln ln 70. ln ( 0), since log ln ln ln 0. log (ln ) (ln 0) 7. ln. ln ln ( ) ln Since ln ( ) ln ln ln, ln ( ) ln ln ( ) ( ) ln ln ln ln But as, approaces, so ln approaces ln() 0. Also, as, ln approaces infinity. Tis means te second term above approaces 0 and te it is. Quick Quiz Sections. and.. D;. A; 6 ( )( ) ( ) f( ). (a) Domain: (, 0) ( 0, ) Range: (, ) f( ) cos( ) cos f( ), so f is odd. cos (c), cos cos 0 (d) For all > 0, cos. cos Terefore,. Since 0, it follows by te Sandwic Teorem tat cos 0. Section. Continuity (pp. 7886) Eploration Removing a Discontinuity. 9 ( )( ). Te domain of f is (, ) (, ) (, ) or all ±.. It appears tat te it of f as eists and is a little more tan.. f() sould be defined as ( )( )( ), 9 ( )( ), so ( )( ) f( ) for. ( )( ) 0 0 Tus,. 6 Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

21 Section. 8 0 ( ), so g is continuous at.. g g( ) Quick Review... (a) (c). (a) (c). ( ) ( ) 6 ( ) f( ) int ( ) f( ) f( ) f( ) does not eist, because te left- and rigt-and its are not equal. (d) f( ) int ( ) f ( ) ( ) ( ) f( ) ( ) f( ) does not eist, because te left- and rigt-and its are not equal. (d) f() ( f g)( ) f( g( )) f ( ) ( ) ( ) ( ), 0 6 ( g f )( ) g( f ( )) g,. Note tat sin ( g f)( ) g(f()) g( ). Terefore: g() sin, 0 ( f g)( ) f(g()) f(sin ) (sin ) sin, 0 6. Note tat ( g f)( ) g( f( )) f( ). Terefore, f( ) for > 0. Squaring bot sides gives f( ). Terefore, f( ), > 0. ( f g)( ) f( g( )) ( ), > ( )( ) 0 Solutions:, Solution: For, f() wen, wic gives. (Note tat tis value is, in fact,.) For >, f() wen 6 8, wic gives 6 0. Te discriminant of 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. or Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

22 8 Section. 0. A grap of f() is sown. Te range of f() is (, ) [, ). Te values of c for wic f () c as no solution are te values tat are ecluded from te range. Terefore, c can be any value in [, ). Section. Eercises. Te function y is continuous ( ) because it is a quotient of polynomials, wic are continuous. Its only point of discontinuity occurs were it is undefined. Tere is an infinite discontinuity at.. Te function y is continuous because it is a quotient of polynomials, wic are continuous. Its only points of discontinuity occur were it is undefined, tat is, were te denominator ( )( ) is zero. Tere are infinite discontinuities at and at.. Te function y is continuous because it is a quotient of polynomials, wic are continuous. Furtermore, te domain is all real numbers because te denominator,, is never zero. Since te function is continuous and as domain (, ), tere are no points of discontinuity.. Te function y is a composition ( f g)( ) of te continuous functions f() and g(), so it is continuous. Since te function is continuous and as domain (, ), tere are no points of discontinuity.. Te function y is a composition ( f g)( ) of te continuous functions f( ) and g(), so it is continuous. Its points of discontinuity are te points not in te domain, i.e., all <. 6. Te function y is a composition ( f g)( ) of te continuous functions f( ) and g(), so it is continuous. Since te function is continuous and as domain (, ), tere are no points of discontinuity. 7. Te function y is equivalent to, < 0 y, > 0. It as a jump discontinuity at Te function y cot is equivalent to cos y, a quotient of continuous functions, sin so it is continuous. Its only points of discontinuity occur were it is undefined. It as infinite discontinuities at kπ for all integers k. / 9. Te function y e is a composition ( f g)( ) of te continuous functions f() e and g ( ), so it is continuous. Its only point of discontinuity occurs at 0, were it is undefined. Since / e, tis may be considered an 0 infinite discontinuity. 0. Te function y ln ( ) is a composition ( f g)( ) of te continuous functions f () ln and g(), so it is continuous. Its points of discontinuity are te points not in te domain, i.e.,.. (a) Yes, f( ) 0. Yes, (c) Yes f( ) 0. (d) Yes, since is a left endpoint of te domain of f and f( ) f( ), f is continuous at.. (a) Yes, f(). Yes, f( ). Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

23 Section. 8 (c) No. (d) No. (a) No No, since is not in te domain.. Everywere in [, ) ecept for 0,,.. Since f() Since f(). f( ) 0, we sould assign f( ), we sould reassign 7. No, because te rigt-and and left-and its are not te same at zero.. (a) Not removable, it s an infinite discontinuity. (a) 8. Yes, Assign te value 0 to f (). Since is a rigt endpoint of te etended function and f( ) 0, te etended function is continuous at. 9. y Removable, assign te value 0 to f( ).. (a) All points not in te domain along wit 0, 0 is a removable discontinuity, assign f(0) 0. is not removable, te one-sided its are different.. (a) All points not in te domain along wit, 0. (a) Not removable, te one-sided its are different. (a) y Removable, assign te value to f (). is not removable, te one-sided its are different. is a removable discontinuity, assign f ().. For, 6. 9 ( )( ) f( ). Te etended function is y. For, f( ) ( )( ) ( )( ). Te etended function is y. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

24 8 Section sin Since, te etended function is 0 sin, 0 y., 0 sin sin Since ( ), te 0 0 sin, 0 etended function is y., 0 9. For (and > 0), ( )( ) f( ). Te etended function is y. 0. For (and ), 0 f( ) ( )( )( ) ( )( ) ( )( ). Te etended function is y.. Te domain of f is all real numbers. f is continuous at all tose points so f is a continuous function.. Te domain of g is all real numbers >. f is continuous at all tose points so g is a continuous function.. f is te composite of two continuous functions g were g ( ) and ( ).. f is te composite of two continuous functions g were g() sin and ( ).. f is te composite of tree continuous functions g k were g ( ) cos, ( ), and k ( ). 6. f is te composite of two continuous functions g were g() tan and ( ). 7. One possible answer: Assume y, constant functions, and te square root function 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 functions, and te cube root function are continuous. By te difference teorem, y is continuous. By te composite teorem, y is continuous. By te product teorem, y is continuous. By te sum teorem, y is continuous. Domain: (, ) 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 difference teorem, y is continuous. By te composite teorem, y is continuous. Domain: (, ) 0. One possible answer: Assume y and y are continuous. Use te product, difference, and quotient teorems. One also needs to verify tat te it of tis function as approaces is. Alternately, observe tat te function is equivalent to y (for all ), wic is continuous by te sum teorem. Domain: (, ) Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

25 Section. 8. One possible answer: y 6. y f() Solving, we obtain te solution... One possible answer: 7. Since f() f() a() 6a and f( ) 8, te function will be continuous if 6a 8. Tus a. 8. Since f( ) f( ) ( ) 7 and f( ) a, te function will be continuous if a 7. Tus a.. One possible answer: y 9. Since f( ) f( ) a( ) a and f( ) ( ), te function will be continuous if a. Tus a. y f() 0. Since f() f( ) and f( ) ( ) a a, te function will be continuous if a. Tus a.. One possible answer:. Consider f() e. f is continuous, f(0), and f () > 0.. By te e Intermediate Value Teorem, for some c in c (0, ), f(c) 0 and e c... (a) Luisa s salary is 0 $6,00 $6,00 ( 0. ) for te first year (0 t < ), $6,00(.0) for te second year ( t < ), $6,00 ( 0. ) for te tird year ( t < ), and so on. Tis corresponds to y 6,00 (. 0) int t. Solving, we obtain te solutions 0.7 and.. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

26 86 Section. [0,.8] by [000, 000] Te function is continuous at all points in te domain [0, ) ecept at t,,,.. (a) We require: , 0< 0., < 0., < f ( ) 0., < 0., < 660., < 6 7., 6<. Tis may be written more compactly as 0. int( ), 0 6 f ( ) 7., 6< [0, ] by [0, 9] Tis is continuous for all values of in te domain [0, ] ecept for 0,,,,,, 6.. False. Consider f( ) wic is continuous and as a point of discontinuity at (a) Te function is defined wen > 0, tat is, on (, ) (0, ). (It can be argued tat te domain sould also include certain values in te interval (, 0), namely, tose rational numbers tat ave odd denominators wen epressed in lowest terms.) [, ] by [, 0] (c) If we attempt to evaluate f( ) at tese values, we obtain f ( ) 0 0 (undefined) and f () 0 0 (undefined). Since f is undefined at tese values due to division by zero, bot values are points of discontinuity. (d) Te discontinuity at 0 is removable because te rigt-and it is 0. Te discontinuity at is not removable because it is an infinite discontinuity. (e) 0. False; if f as a jump discontinuity at a, ten f( ) f( ), so f is not a a continuous at a. 6. B; f( ) is not defined E; f ( ) 0 is te only defined option. 8. A; f(). 9. E; causes te denominator to be zero even after te rational epression is reduced. Te it is about.78, or e. 6. Tis is because f( a ) f( ). a 6. Suppose not. Ten f 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. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

27 Section Since te absolute value function is continuous, tis follows from te teorem about continuity of composite functions. 6. For any real number a, te it of tis function as approaces a cannot eist. Tis is because as approaces a, te values of te function will continually oscillate between 0 and. Section. Rates of Cange and Tangent Lines (pp. 879) Quick Review.. ( ) 8 y. a y b 9. Since y is equivalent to y, we use m. y [ ( )] 7 y 0. b 0 b 9 b Section. Eercises. m ( ) 7 7. (a) f f () f () ( ) m ( ) 6 f f () f ( ) 0 ( ) y [ ( )] y m 7 y ( ) 6 7 y y ( ) 9 y m y ( ) 8 y. (a). (a). (a) f f ( ) f ( 0) 0 f f ( ) f ( 0) 7 0 f f () 0 f ( ) e 0( ) f f () f () e e f f ( ) f ( ) ln 0 ln f f ( 0) f ( 00) 0 00 ln 0 ln 00 0 ln 00 ln 0. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

28 88 Section.. (a) 6. (a) π ( ) f π ( ) f f π π π π π ( ) f π ( ) f f 6 0 π π π π 6 f f( π ) f( 0) π 0 π π f f( π) f( π) 0 π ( π) π 7. We use Q ( 0, ), Q (, 7), Q ( 6., 7), Q ( 8, 0), and P (0, 60). 60 (a) Slope of PQ : Slope of PQ : Slope of PQ : Slope of PQ : Secant Slope PQ PQ 6 PQ 0 PQ 0 Te appropriate units are meters per second. Approimately 0 m/sec 8. We use Q (, 0), Q (, 7 8), Q (., 8 6), Q (., 9 7), and P (0, 80) (a) Slope of PQ : Slope of PQ: Slope of PQ: Slope of PQ: (a) Secant Slope PQ PQ PQ 6 PQ 6 Te appropriate units are meters per second. Approimately 6 m/sec y( ) y( ) ( ) ( ) ( ) Te tangent line as slope and passes troug (, y()) (, ) y [ ( )] y (c) Te normal line as slope and passes troug (, y( )) (, ). y [ ( )] 9 y (d) [ 8, 7] by [, 9] Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

29 Section. 89 y( ) y( ) 0. (a) [( ) ( )] [ ( )] ( ) Te tangent line as slope and passes troug (, y()) (, ). y ( ) y (c) Te normal line as slope and passes troug (, y()) (, ). y ( ) 7 y (d) [ 6, 6] by [ 6, ] y( ) y( ) ( ). (a) ( ) ( ) Te tangent line as slope and passes troug (, y()) (, ). y ( ) y (c) Te normal line as slope and passes troug (, y()) (, ). y ( ) y (d) [.7,.7] by [.,.] y( 0 ) y( 0). (a) ( ) ( ) ( ) Te tangent line as slope and passes troug (0, y(0)) (0, ). y ( 0) y (c) Te normal line as slope and passes troug (0, y(0)) (0, ). y ( 0) y (d) [ 6, 6] by [, ]. (a) Near, f (). f( ) f( ) ( ) Near, f (). f( ) f( ) ( ) Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.

30 90 Section.. Near, f () ( ). f( ) f( ) [ ( )] ( ). First, note tat f (0). f( 0 ) f( 0) ( ) 0 ( ) f( 0 ) f( 0) ( ) 0 No, te slope from te left is and te slope from te rigt is. Te two-sided it of te difference quotient does not eist. 6. First, note tat f (0) 0. f( 0 ) f( 0) 0 0 f( 0 ) f( 0) ( ) 0 0 ( ) Yes. Te slope is. 7. First, note tat f ( ) f( ) f( ) 6 0 ( ) ( ) ( ) ( ) ( ) f ( ) f ( ) 0 [ ( )] Yes. Te slope is. 8. No; te function is discontinuous at π because f( ) sin π sin ( / ) ( / ) π π but 9. (a) 0. (a) f π π cos. f ( a ) f ( a) [( a ) ] ( a ) a a a a ( a ) a Te slope of te tangent steadily increases as a increases. f ( a ) f ( a) a 6 a a ( a ) a( a ) aa ( ) a Te slope of te tangent is always negative. Te tangents are very steep near 0 and nearly orizontal as a moves away from te origin. Copyrigt 0 Pearson Education, Inc. Publising as Prentice Hall.