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1 y. ( sincos ) (sin ) (cos ) + (cos ) (sin ) sin + cos cos y + ( )( ) ( + )( ) ( ) ( ) s [( t )( t + )] t t [ t ] t t s t + t t t ( t )( t) ( t + )( t) ( t ) t ( t ) y + + / / ( + + ) / / / Chapter Review y ( sin + cos ) cos + sin + ( sin ) + cos ( + )cos+ sin y ( sin cos ) cos + sin [ ( sin ) + cos ] ( )cos+ sin y tan sec 6 tan 6 sec tan y (tan cot ) sec ( csc ) sec + csc y sin + cos (sin + cos ) (cos sin ) (sin + cos ) sin cos (sin + cos ) y + sin cos (csc + sec ) csc cot + sec tan sec tan csc cot y 5 [( + )( )] 5 5 ( ) + ( 6 )( + ) r ( 5θ sec θ) θ θ θsecθ + 5θ secθ tanθ r tanθ θ θ θ + θ + sec θθ ( + θ+ ) tan θ( θ + ) ( θ + θ + ). 8. V r + 8r r + 6r r r A s + s s s+ s + s s Copyright Pearson Eucation, Inc. Publishing as Prentice Hall.

2 8 Chapter Review s + sin t t t + tan t cos t( + tan t) sec t( + sin t) ( + tan t) s + sin t t t + cost cos t( + cos t) ( sin t)( + sin t) ( + cos t) cost+ cos t+ sin t+ sin t ( + cos t) cost+ sin t+ ( + cos t) s t + t ( t t ) t t t t t + + y ( cos ) cos sin + cos sin + cos sin y sin u cosu + u u cscu sec u (sin u+ cos u) u () u y cot u cscu u u tan u sin u (cot ucsc u) u ( ) u y 5 [ ( )] ( ) y [ ( + )] ( + ) y t t t t t t+ t t + t t t t t t t y t t t t t + t + t y (sec tan cos ) tan sec y sin cot. () cos sin. Since y is efine for all an y cos sin, the function is ifferentiable for all.. Since y sin cos is efine for all real an y cos ( )( sin ) (cos )( ) sin, the function is ifferentiable for all real. cos. Since y is efine for all an y sin ( ) cos, which is efine ( ) for all, the function is ifferentiable for all. Copyright Pearson Eucation, Inc. Publishing as Prentice Hall.

3 + 5. Since y ( ) ( + 5) is efine for all an y ( )( ) ( + 5)( ), the ( ) ( ) function is ifferentiable for all y 5. Slope sec tan. y 6. Slope cos cos sin sin y ( sin ) cos Slope y Slope ( + sin ) ( + cos ) y (sec ) sec tan cos y (sec tan ) (sec tan ) tan + (sec ) sec sin + cos y. (csc ) csc cot y ( csccot ) ( csc cot ) cot + ( csc )( csc ) cos + sin y ( sin ) sin + cos sin + cos y (sin + cos ) cos + cos + ( sin ) cos sin Chapter Review 9 y ( cos ) ( cos + ( sin )) cos + sin y ( cos+ sin ) ( sin ) + ( sin+ cos ) sin + cos y, y 6, y, ( ) y, an the rest are all zero. y, y, 6 y, ( ) y, ( 5) y, an the rest are all zero. y ( 8 ) 6 At, y 8 ( ) an y 6( ). (a) Tangent: y ( ) or y + 6 (b) Normal: y ( ) or y + y ( + cotcsc ) csc + csc cot At, Copyright Pearson Eucation, Inc. Publishing as Prentice Hall.

4 5 Chapter Review y + cot csc + an y csc + csc cot + ()( ). (a) Tangent: y or y + + (b) Normal: y or y + y. (sin + cos ) cos sin At, y sin + cos an y cos sin. 8. (a) Tangent: Line is horizontal, so y. (b) Normal: Line is vertical, so y + At, y ( ) + an y 5 () (). 5. (a) Tangent: Line is horizontal, so y. (b) Normal: Line is vertical, so. 5. y 6( + ) or ( + ) ( + ). The points are (, ) an (, ). y 5. cos 6 cos, which is impossible. There are no points at which the tangent line has slope 6, so none. 5. (a) y (b) Yes, because both of the one-sie limits as are equal to f (). (c) No, because the left-han erivative at is + an the right-han erivative at is. 5. (a) For all m, since y sin an y m are both continuous on their omains, an they link up at the origin, where lim sin lim m, regarless of + the value of m. (b) For m only, since the left-han erivative at (which is cos ) must match the right-han erivative at (which is m). y 5 / 55. Note that is efine if an only if. The answers are y 6 6 ±. The points are (, ) an (, ). y ( ) or 9. The points are, an,. (a) For all (b) At (c) Nowhere y 5 / is efine if an only if. The answers are 56. Note that (a) For all (b) At Copyright Pearson Eucation, Inc. Publishing as Prentice Hall.

5 Chapter Review 5 (c) Nowhere 5. Note that lim f( ) lim ( ) an lim f( ) lim ( ). Since these + + values agree with f (), the function is continuous at. On the other han,, < f ( ), so the erivative is, < unefine at. (a) [, ) (, ] (b) At (c) Nowhere in its omain 6. The graph passes through (, 5) an has slope for < an slope.5 for >. y y f() 6. The graph passes through (, ) an has slope for <, slope for < <, an slope for < < 6. y 58. Note that the function is unefine at. (a) [, ) (, ] y f() (b) Nowhere (c) Nowhere in its omain 59. y y f () 9 / / 6. i. If f( ) + 9, then f ( ) 8 / an f ( ), which matches the given equation. 6. y 6. (a) iii (b) i (c) ii y f () 9 / ii. If f ( ), then 8 / f ( ), which contraicts the given equation. / iii. If f / ( ) + 6, then f ( ), which matches the given equation. / / iv. If f( ), then f ( ) an / f ( ), which contraicts the given equation. Answer is D: i an iii only coul be true. Note, however that i an iii coul not simultaneously be true. Copyright Pearson Eucation, Inc. Publishing as Prentice Hall.

6 5 Chapter Review 65. (a) [-, 5] by [-, 8] (c) f f + f ( ( )) ( ) ( ) f() + f () (b) t interval (c) 6. (a) [,.5] [.5, ] [,.5] [.5, ] [,.5] [.5, ] [,.5] [.5, ] [-, 5] by [-8, 8] avg. vel () Average velocity is a goo approimation to velocity. n n ( ) n ; ( n n ) n( n ) ; ( n n ) n( n)( n ) ;... an n n ( ) n( n)( n)( n) n n!. ( f( )) f ( ) (b) ( f ( )) f ( ) + f ( ) f() + f () + () (e) (f) 68. (a) (b) (c) f( ) f ( ) f( ) f () f() ( ) f( ) + f ( ) ( + ) f( ) ( + ) f () f() ( ) ( f ( ) f ( )) f ( ) f( ) + f( ) f ( ) f () f() ( )( 9) 6 [ f( ) g( )] f ( ) g ( ) ( ) 5 ( f( ) g( )) f ( ) g( ) + f( ) g ( ) ( )( ) + ( )( ) ( f( ) g( )) f ( ) g( ) + f( ) g ( ) ( )( ) + ( )( ) Copyright Pearson Eucation, Inc. Publishing as Prentice Hall.

7 Chapter Review 5 () (e) (f) f( ) g( ) f ( ) g( ) f( ) g ( ) ( g ( )) ( )( ) ( )( ) ( ) 9 f( ) g( ) f ( ) g( ) f( ) g ( ) ( g ( )) ( )( ) ( )( ) ( ) f( ) g( ) + f ( ) ( g( ) + ) f( ) g ( ) ( g ( ) + ) ( )( ) ( )( ) ( + ) Yes; the slope of f + g at is ( f + g)() f () + g (). The sum of two positive numbers must also be positive.. No; it epens on the values of f() an g(). For eample, let f() an g(). Both lines have positive slope everywhere, but ( f g)( ) has a negative slope at.. (a) s ( 6t 6t ) 6 t t t s ( 6 t) t t (b) The maimum height is reache when s, t which occurs at t sec. s (c) When t, 6, so the velocity is t 6 ft/sec. s () Since ( 6t. 6t ) 6 5. t, t t the maimum height is reache at 6 t. sec. The maimum height 5. 6 is s 9. 8 ft. 5.. (a) Solving 6 9 t, it takes sec. The 6 average velocity is 8 cm/sec... (a) s (b) Since vt () 98t, the velocity is t ( 98 ) 56 cm/sec. Since ν at () 98, the acceleration is t 98 cm / sec. V ( ) r ( ) (b) The marginal revenue is r ( ) ( 6+ 8) 6 ( )( ), 6 which is zero when or. Since the bus hols only 6 people, we require 6. The marginal revenue is when there are people, an the corresponing fare is p( ) $.. Copyright Pearson Eucation, Inc. Publishing as Prentice Hall.

8 5 Chapter Review (c) One possible answer: If the current riership is less than, then the propose plan may be goo. If the current riership is greater than or equal to, then the plan is not a goo iea. Look at the graph of y r( ). 5. (a) Since tan θ, we have θ (sec θ) 6. sec θ. At point t t A, we have θ an 6. sec 6. km/sec. t (b).6 ra revolution 6sec 8 revolutions sec ra min per minute or approimately 5. revolutions per minute. 6. (a) The graphs: Setting y ( + 6) ( ) 89, we ( + 6) ( + 6) get ±. Plugging these values back into the equation of the curve, we get ( ) ± ± Thus a The graph of the function inicates that the range is confine between the two points at which the graph has horizontal tangents. [6, 6] by [, ] Setting y ( + ) ( ) 8, we ( + ) ( + ) get ±. Plugging these values into the ( ± ) equation of the curve, we get ±. + Thus a. 9. [6., 6.] by [.,.] It appears that the erivative of y is y. cos( ) (b) Let y. The graphs of cos( ) y an y are shown below: (a) (b) k, where k is an o integer, [6., 6.] by [.,.] It again appears that the erivative of y is y.. The graph of the function inicates that the range is confine between the two points at which the graph has horizontal tangents. (c) Where it s not efine, at k, k an o integer () It has perio an continues to repeat the pattern seen in this winow. [, ] by [.5,.5] Copyright Pearson Eucation, Inc. Publishing as Prentice Hall.

9 Chapter Review 55 T r rl T l r r T r l T y () l l rl T r l l T rl T y ( ) rl T / ( ) rl T / rl T rl T y ( T) T rl ( T ) rl T rl T rl T Since y () r <, y () l <, an y ( ) <, increasing r, l, or woul ecrease the frequency. Since y ( T) >, increasing T woul increase the frequency. 8. y ( r) 8. (a) vt () s () t t (b) at () v () t 6t (c) Set vt () an solve for t: t ( t ) ( t )( t+ ) t or t The particle is at rest when t. () at () when t spee v( ) () (e) Towars the origin: s() () + 5 < v() () 5> The particle is left of the origin an it is moving to the right. 8. (a) y 5( ) (b) Yes; since f is ifferentiable at, it is continuous at. (c) Yes; since f is continuous on [, ], it takes on all values between f() an f() (Intermeiate Value Theorem). () 8. (a) g f( ) ( ) f ( ) f ( )( f( ) ) ( f ( ) ) f( ) ( f ( ) ) 9 6 (e) Since f(), the function g is not efine at. (b) [, ] by [, ] sin (cos ) ( sin )cos f ( ) (cos ) sin (cos ) (c) f ( ) where sin. On this interval, that is at, ±, ±. () The low turning points are f() f( ± ), while the high turning points are f ( ± ). The range is the interval,. Copyright Pearson Eucation, Inc. Publishing as Prentice Hall.