Chapter 1 Prerequisites for Calculus

Transcription

1 Section. Chapter Prerequisites for Calculus Section. Lines (pp. ) Quick Review.. + ( ) + () +. ( +). m. m ( ) ( ). (a) ( )? 6 (b) () ( )? 6. (a) 7? ( ) Yes (b) ( ) + 9 No Yes No Section. Eercises. Δ Δ. Δ ( ) Δ. Δ 8 ( ) Δ. Δ Δ 6. (a, c) B A (b) m ( ) 6. (a, c) 7. d ( ) + ( ) ( ) + ( ) 8. d ( ) + ( ) ( ) + ( ) (b) m ( ) ( ) 7. (a, c) B A (b) m

2 Section. 8. (a, c) (b) m (undefined) This line has no slope. 9. (a) (b). (a) (b). (a) (b). (a) (b). ( ) +. [ ( )] + ( + ) +. ( ) + 6. [ ( )] + ( + ) or m ( ) +. m ( ) +. m (undefined) ( ) Vertical line:. m ( ) [ ( )] + ( + ) + +. The line contains (, ) and (, ). m 6. The line contains (, ) and (, ). m (a) Slope: (b) -intercept: [, ] b [, ] (a) Slope: (b) -intercept:

3 Section (a) Slope: (b) -intercept:. + [, ] b [, ] (a) Slope: (b) -intercept:. (a) The desired line has slope and passes through (, ): ( ) + or. (b) The desired line has slope and passes through (, ): ( ) + or.. (a) The given equation is equivalent to +. The desired line has slope and passes through (, ): ( + ) + or. (b) The desired line has slope and passes through (, ): ( + ) + or +.. (a) The given line is vertical, so we seek a vertical line through (, ):. (b) We seek a horizontal line through (, ):.. (a) The given line is horizontal, so we seek a horizontal line through, :. (b) We seek a vertical line through, :. 9. m f( ) ( ) + 7 Check: f () () 6, as epected. 7 Since f( ), we have m 7 and b. 6. m ( ) f( ) ( ) + ( ) + Check: f () 6 (), as epected. Since f( ) +, we have m and b. 7. ( ) () 6 8. ( ) ( 8) ( + 8) i( ) This is the same as the equation obtained in Eample.. (a) When, we have, so c. c When, we have, so d. d (b) When, we have, so c. c When, we have, so d. d The -intercept is c and the -intercept is d.

4 Section.. (a) The given equations are equivalent to k + k and +, respectivel, so the slopes are k and. The lines are parallel when, so k. k (b) The lines are perpendicular when, k so k (a) m. 7. degrees/inch (b) m 9 6. degrees/inch m 7. degrees/inch (d) Best insulator: Fiberglass insulation Poorest insulator: Gpsum wallboard The best insulator will have the largest temperature change per inch, because that will allow larger temperature differences on opposite sides of thinner laers. p. Slope: k Δ Δ d atmospheres per meter At meters, the pressure is p.99() +.97 atmospheres.. (a) d(t) t (b) The slope is, which is the speed in miles per hour. (d) Suppose the car has been traveling mph for several hours when it is first observed at point P at time t. (e) The car starts at time t at a point miles past P.. (a),6.,77,8.669 (b) The slope is,6.. It represents the approimate rate of increase in earnings in dollars per ear. [99, ] b [, ] (d) When,,6.(),77,8.669,98. In, the construction workers average annual compensation will be about $, (a) (b) The slope is.68. It represents the approimate average weight gain in pounds per month. (d) When,.68() She weighs about 9 pounds. 7. False: m Δ and Δ, so it is undefined, or has no Δ slope. 8. False: perpendicular lines satisf the equation m m, or m m 9. A : m( ) + ( + ) + or ( + ). E.. D:. B: 7 ( ) 7 6 7

5 Section.. (a) 6,8,8 (b) The rate at which the median price is increasing in dollars per ear 6 7,6,6 (, ) (d) The median price is increasing at a rate of about $6 per ear in the Northeast, and about $7 per ear in the Midwest. It is increasing more rapidl in the Northeast.. (a) Suppose F is the same as C (, ) (, ) (, ) 6 Yes, F is the same as C. (b) It is related because all three lines pass through the point(, ) where the Fahrenheit and Celsius temperatures are the same.. The coordinates of the three missing vertices are (, ), (, ) and (, ), as shown below. (, ) (, ) (, ) 6 6 (, ) (, ) (, ) (, ) (, ) 6 6 Suppose that the vertices of the given quadrilateral are (a, b), (c, d ), (e, f ), and (g, h). Then the midpoints of the consecutive sides are W a+ c b+ d X c + e d + f,,,, Y e+ g f + h,, and Z g+ a h+ b,.when these four points are connected, the slopes of the sides of the resulting figure are: d+ f b+ d f b WX : c+ e a+ c e a f + h d+ f XY : h e+ g + d c e g c f + h h+ b f b ZY : e+ g g+ a e a h+ b b+ d WZ : g+ a + h d a c g c Opposite sides have the same slope and are parallel.

6 6 Section. 7. The radius through (, ) has slope. The tangent line is tangent to this radius, so its slope is. We seek the line of slope that passes / through (, ). ( ) (a) The equation for line L can be written as A B + C B, so its slope is A B. The perpendicular B line has slope and passes through (a, b), so A/ B A B its equation is A ( a) + b. (b) Substituting B A ( a) + b for in the equation for line L gives: A B B A a b + ( ) + C A + B ( a) + ABb AC ( A + B ) B a + AC ABb B a + AC ABb A + B Substituting the epression for in the equation for line L gives: B a + AC ABb A B C A + B + ABa AC A B ( + Bb ) CA ( + B) + A + B A + B AB a A C A Bb A B + + C B + C A + B ABb+ BC ABa B A + B A b + BC ABa A + B The coordinates of Q are B a + AC ABb A b + BC ABa,. A + B A + B Distance ( a) + ( b) B a + AC ABb a A + B A b + BC ABa + b A + B B a + AC ABb a( A + B ) A b + BC ABa b( A + B ) + A + B A + B AC ABb A a BC ABa Bb + A + B A + B A( C Bb Aa) B( C Aa Bb) + A + B A + B A ( C Aa Bb) ( A + B ) ( A + B )( C Aa Bb) ( A + B ) ( C Aa Bb) A + B C Aa Bb A + B Aa + Bb C A + B B ( C Aa Bb) + ( A + B ) Section. Functions and Graphs (pp. ) Eploration Composing Functions. g f, f g. Domain of : [, ] Range of : [, ] : : :

7 Section. 7. Domain of : [, ); Range of : (, ] :. ( ( )) ( ) ( ( )) ( ( )) ( ), Quick Review.. + Solution: [, ). ( ) > Solutions to ( ) :, Test : ( ) > ( ) > is true when <. Test : ( ) < ( ) > is false when < <. Test : ( ) > ( ) > is true when >. Solution set: (, ) (, ). 7 Solution set: [, 7]. or or 7 Solution set: (, ] [7, ). < 6 Solution to 6:, Test 6: ( 6) 6 > 6 < 6 is false when < Test : < 6 < 6 is true when < < Test 6 : 6 6 > 6 < 6 is false when >. Solution set: (, ) 6. 9 Solutions to 9 :, Test : 9 ( ) 9 6 7< 9 isfalse when <. Test : 9 9> 9 istruewhen < <. Test : < 9 is false when >. Solution set: [, ] 7. Translate the graph of f units left and units downward. 8. Translate the graph of f units right and units upward. 9. (a) f( ) 9 ( + )( ) or (b) f( ) 6 6 No real solution. (a) f( ) (b) f( ) No solution. (a) f( ) (b) Check: ; it checks. f( ) Check: 6+ 7 ; it checks.. (a) f( ) 8 7 (b) f( ) 7 8

8 8 Section. Section. Eercises d. Ad ( ) Ad ( ) ( ) in in in. hs () s m. m. S (e) 6 e 6 ( ft ) 6 ( ft ) ft. vr () r ( cm) ( 7 cm ) 6 cm. (a) (, ) or all real numbers (b) (, ] 6. (a) (, ) or all real numbers (b) [ 9, ) 6 7. (a) Since we require, the domain is [, ). (b) [, ) 8. (a) Since we require, the domain is (, ]. (b) (, ] 6 9. (a) Since we require, the domain is (, ) (, ). (b) Since can assume an value ecept, the range is (, ) (, ).. (a) Since we require, the domain is (, ]. (b) [, ). (a) Since we require, the domain is (, ) (, ). (b) Note that can assume an value ecept, so + can assume an value ecept. The range is (, ) (, ).

9 Section. 9. (a) Since we require, the domain is (, ) (, ) (b) Since > for all, the range is (, ). 7. (a) (, ) or all real numbers (b) Since / is equivalent to, the range is [, ). (a) (, ) or all real numbers (b) (, ) or all real numbers [.7,.7] b [.,.] 8. (a) This function is equivalent to, so its domain is [, ). (b) [, ) [.7,.7] b [.,.]. (a) Since we require, the domain is (, ]. (b) [, ) [.7,.7] b [, ] 9. (a) (, ) or all real numbers (b) (, ) or all real numbers [.7,.7] b [, 7]. (a) (, ) or all real numbers (b) The maimum function value is attained at the point (, ), so the range is (, ]. [ 9., 9.] b [.,.]. (a) Since ( ) >, the domain is (, ) (b) [., ) [.7,.7] b [.,.] 6. (a) Since we require 9 the domain is [, ] (b) [, ] [.7,.7] b [, ]. Even, since the function is an even power of.. Neither, since the function is a sum of even and odd powers of.. Neither, since the function is a sum of even and odd powers of ( + ).. Even, since the function is a sum of even powers of ( ).

10 Section.. Even, since the function involves onl even powers of. 6. Odd, since the function is a sum of odd powers of. 7. Odd, since the function is a quotient of an odd function ( ) and an even function ( ). 8. Neither, since, (for eample), ( ) / and (). 9. Neither, since, (for eample), ( ) is defined and () is undefined.. Even, since the function involves onl even powers of.. (a). (a). (a). (a) [.7,.7] b [, 6] (b) (, ) or all real numbers [, ) (b) (, ) or all real numbers [, ) [.7,.7] b [, 9] (b) (, ) or all real numbers (, ) or all real numbers. Because if the vertical line test holds, then for each -coordinate, there is at most one -coordinate giving a point on the curve. This -coordinate would correspond to the value assigned to the -coordinate. Since there is onl one -coordinate, the assignment would be unique. 6. If the curve is not, there must be a point (, ) on the curve where. That would mean that (, ) and (, ) are two different points on the curve and it is not the graph of a function, since it fails the vertical line test. 7. No 8. Yes 9. Yes. No. Line through (, ) and (, ): Line through (, ) and (, ): +, f( ) +, <, <, <. f( ), <,. Line through (, ) and (, ): + Line through (, ) and (, ): m, so ( ) + + +, < f( ) +, <. Line through (, ) and (, ): m, so ( ) Line through (, ) and (, ): m, so +, f( ) < +, < (b) (, ) or all real numbers [, )

11 Section.. Line through (, ) and (, ): Line through (, ) and (, ): Line through (, ) and (, ): m, so ( ) + +, < f( ), < +, < < 6. Line through (, ) and (, ): Line through (, ) and (, ): + Line through (, ) and (, ):, f( ) +, <, < T 7. Line through, and (T, ): m, so T ( T / ) T T T + T T, f( ) T T, < T T A, < T A, < T 8. f( ) T A, T < T A, T 9. (a). (a) The graph of f () is the graph of the absolute value function stretched verticall b a factor of and then shifted units to the left and units downward. (b) (, ) or all real numbers [, ). (a) f (g()) ( ) + + (b) g( f( )) ( + ) ( + + ) + + f (g()) + (d) g ( f ()) + + (e) g (g( )) [ ( ) ] (f) f ( f ()) ( + ) + +. (a) f (g()) ( ) + (b) g ( f ()) ( + ) f (g()) (d) g ( f ()) (e) g (g( )) ( ) (f) f ( f()) ( + ) (a) Since ( f g)( ) g( ), g( ) (b) Since ( f g)( ) +, we know that g ( ), so g g ( ) ( ). Since ( f g)( ) f, f( ). (d) Since ( f g)( ) f( ), f( ). The completed table is shown. Note that the absolute value sign in part (d) is optional. g() f () ( f g)( ) [ 9., 9.] b [ 6., 6.] Note that f( ) +, so its graph is the graph of the absolute value function reflected across the -ais and then shifted units right and units upward. (b) (, ) (, ] +,,,

12 Section.. (a) (b) [, ] b [, 8]. 89( 8). 7( 8) $ 9 million or $.9 billion (d) linear regression: (8) +. $9 million in 8.. (a) Because the circumference of the original circle was 8 and a piece of length was removed. 8 (b) r h 6 r (d) V r h 8 6 i ( 8 ) 6 6. (a) Note that mi,6 ft, so there are 8 + feet of river cable at $8 per foot and (,6 ) feet of land cable at $ per foot. The cost is C ( ) (, 6 ) (b) C( ) $,, C() $,7,8 C() $,86, C() $,, C() $,,7 C() $,78,79 C() $,,87 Values beond this are all larger. It would appear that the least epensive location is less than ft from point P. 7. False: + + ( ) + ( ) + ( ). 8. True: ( ) 9. B: Since 9 >, the domain is (, ) 6. A: 6. D: ( f g)( ) f( + )( ) ( + ) ( + ) () 9 6. C: Aw ( ) Lw L w Aw ( ) w 6. (a) Enter f( ) 7, g( ), (f g)() ( ()), and (g f)() ( ()) f g: g f: [, 7] b [, ] [, ] b [, ] Domain: [, ) Domain: [7, ) Range: [ 7, ) Range: [, ) (b) ( f g)( ) 7 ( g f)( ) 7 6. (a) Enter f( ), g( ), (f g)() ( ()), and (g f)() ( ()) f g: Domain: [, ) Range: (, ] g f: Domain: [, ] Range: [, ] (b) ( f g)( ) ( ), ( g f)( )

13 Section. 6. (a) Enter f( ), g( ) +, ( f g)() ( ()), and (g f )() ( ()). f g: g f: [, ] b [, ] Domain: [, ) Range: [, ) g f: [.7,.7] b [, ] Domain: (, ] [, ) Range: [, ) (b) ( f g)( ) ( + ) ( + ),, ( g f)( ) ( ) (a) Enter ( ) f( ),, + ( f g)() ( ()), and (g f )() ( ()). Use a decimal window such as the one shown. f g: Domain: (, ) (, ) Range: (, ) (, ) + (b) ( f g)( ) + + ( + ) ( ) ( ) ( ), + + 7, 7, + + ( g f)( ) + ( ) + ( + ) ( ) ( ), + 7, 7, 67. (a).. Domain: (, ) (, ) Range: (, ) (, ) (b)..

14 Section. 68. (a) (b) (b) 7. (a) [, ] b [, ] (b) Domain of : [, ) Domain of : (, ] Domain of : [, ] 69. (a). The functions,, and all have domain [, ], the same as the domain of + found in part (b). Domain of :[,) Domain of :(, ] (b). (d) The domain of a sum, difference, or product of two functions is the intersection of their domains. The domain of a quotient of two functions is the intersection of their domains with an zeros of the denominator removed.. 7. (a) Yes. Since ( f i g)( ) f ( ) i g( ) f () i g() ( f i g)(), the function ( f i g)() will also be even. 7. (a). (b) The product will be even, since ( f ig)( ) f( ) ig( ) ( f( )) i ( g( )) f( ) i g ( ) ( f i g)( ). Section. Eponential Functions (pp. 9) Eploration Eponential Functions.

15 Section.. >. <.. Section. Eercises. 6. < < for < ; > > > ; for. Quick Review.. Using a calculator, /.9.. Using a calculator, Using a calculator, for [, ] b [ 8, 6] Domain: (, ) Range: (, ). Domain: (, ) Range: (, ) ± 67 ±. 8 [, ] b [, 8] Domain: (, ) Range: (, ). 7. (. 7) $ (. 6) $.6 9. ( ) ( ) ab. c ac b 6 ab b i 8 c ac 6 a bc i 8 bc a a b c ab c 6 a 6 bc Domain: (, ) Range: (, ). 9 ( ) 6. 6 ( ) 7. 8 ( ) ( ) 9. -intercept:. -intercept:.. -intercept:.86 -intercept:.

16 6 Section.. -intercept:.6 -intercept:.. -intercept:. -intercept:.8. The graph of is increasing from left to right and has the negative -ais as an asmptote. (a). The graph of or, equivalentl,, is decreasing from left to right and has the positive -ais as an asmptote. (d). The graph of is the reflection about the -ais of the graph in Eercise. (e) 6. The graph of. or, equivalentl,, is the reflection about the -ais of the graph in Eercise. 7. The graph of is decreasing from left to right and has the line as an asmptote. (b) 8. The graph of. is increasing from left to right and has the line as an asmptote. (f) 9. (a), 9 8.,, , 9, , 8, 67 9.,,., 67 (b) One possibilit is,8(. ) n,8(. ) 967 thousand or,967,. (a) 7, 6 9., 7, 78. 7, 7, , 8 7, 88. 7, 9 7, 86. 7, 88 (b) One possibilit is 6,9 (. ) n 6,9 (. ) 78 thousand or 7,8,. Let t be the number of ears. Solving, (. 7) t,, graphicall, we find that t The population will reach million in about 9 ears.. (a) The population is given b P(t) 6 (. 7), t where t is the number of ears after 89. Population in 9: P(), Population in 9: P(),6 (b) Solving P(t), graphicall, we find that t The population reached, about 77 ears after 89, in (a) At () 66. t/ (b) Solving A(t) graphicall, we find that t 8.. There will be gram remaining after about 8. das.. Let t be the number of ears. Solving (. 6) t graphicall, we find that t.9. It will take about.9 ears. (If the interest is not credited to the account until the end of each ear, it will take ears.). Let A be the amount of the initial investment, and let t be the number of ears. We wish to solve A (. 6) t A, which is equivalent to. 6 t. Solving graphicall, we find that t.. It will take about. ears. (If the interest is credited at the end of each ear, it will take ears.) 6. Let A be the amount of the initial investment, and let t be the number of ears. We wish to solve t A. 6 + A, which is equivalent to t Solving graphicall, we find that t.9. It will take about.9 ears. (If the interest is credited at the end of each month, it will take ears months.) 7. Let A be the amount of the initial investment, and let t be the number of ears. We wish to solve. 6t A e. 6t A, which is equivalent to e. Solving graphicall, we find that t.9. It will take about.9 ears. 8. Let A be the amount of the initial investment, and let t be the number of ears. We wish to solve A (. 7) t A, which is equivalent to. 7 t. Solving graphicall, we find that t 9.6. It will take about 9.6 ears. (If the interest is credited at the end of each ear, it will take ears.)

17 Section Let A be the amount of the initial investment, and let t be the number of ears. We wish to solve 6t A. 7 + A 6, which is equivalent to 6t Solving graphicall, we find that t 9.8. It will take about 9.8 ears.. Let A be the amount of the initial investment, and let t be. 7t the number of ears. We wish to solve A e A,. 7t which is equivalent to e. Solving graphicall, we find that t 9.6. It will take about 9.6 ears.. After t hours, the population is P(t) t/. or, equivalentl, P(t) t. After hours, the population is P() 8.8 bacteria.. (a) Each ear, the number of cases is % % 8% of the previous ear's number of cases. After t ears, the number of cases will be C(t), ( 8. ) t. Solving C(t) graphicall, we find that t.9. It will take about.9 ears.... (b) Solving C(t) graphicall, we find that t.7. It will take about.7 ears. Δ Δ 8 Δ ratio Since Δ, the corresponding value of Δ is equal to the slope of the line. If the changes in are constant for a linear function, then the corresponding changes in are constant as well. 8. (a) When t, B e. There were bacteria present initiall. (b) When t 6, B e. 69( 6) 69.. After 6 hours, there are about 69 bacteria. Solving. e 69 t graphicall, we find that t.. The population will be after about hour. Since the population doubles (from to ) in about hour, the doubling time is about hour. 9. (a).8 (. 96) (b) Estimate:.8 (. 96), thousand or,,.,,,9,,. The estimate eceeds the actual b,., 9. or % (, )( ). (a).9 (. 78) (b) Estimate:.9 (. 78) 6,9 thousand or 6,9,. 6,9,,8, 7,. The estimate eceeds the actual b 7,., or.8% ( 6, 9)( )

18 8 Section.. False.. It is positive True. ( ). D. pe rt.t e.t ln ( e ) ln.t t ln..ears. A.. B. 6. C, f( ) e e + e + e ln( e ) ln (a) In this window, it appears the cross twice, although a third crossing off-screen appears likel. (b) change in Y change in Y 7 8 It happens b the time. Solving graphicall,.7667,,. (d) The solution set is approimatel (.7667, ) (, ). 8. Since f (). we have ka., and since f ( ). we have ka.. Dividing, we have ka ka.. a 9 a ± Since f() k a is an eponential function, we require a >, so a. Then ka. gives k., so k.. The values are a and k.. 9. Since f (). we have ka., and since f ( ) 6 we have ka 6. Dividing, we have ka ka. 6 a. a ±. Since f() k a is an eponential function, we require a >, so a.. Then ka. gives.k., so k. The values are a. and k. Quick Quiz (Sections..). C, m () D, g( ) ( ) f () () E.. (a) (, ) (b) (, ) e ln. 69 Section. Parametric Equations (pp. 6) Eploration Parametrizing Circles. Each is a circle with radius a. As a increases, the radius of the circle increases.

19 Section. 9. t :. t : t initial point: (, ) terminal point: (, ) t : t initial point: (, ) terminal point: (, ) t : t initial point: (, ) terminal point: (, ) t Let d be the length of the parametric interval. If d <, d ou get of a complete circle. If d, ou get the complete circle. If d >, ou get the complete circle but portions of the circle will be traced out more than once. For eample, if d the entire circle is traced twice. t initial point: (, ) terminal point: (, ). For t the complete circle is traced once clockwise beginning and ending at (, ). For t the complete circle is traced once clockwise beginning and ending at (, ). For t the half circle below is traced clockwise starting at (, ) and ending at (, ).

20 Section. Eploration Parametrizing Ellipses. a, b : a 7, b : a, b :. If a > b, then the major ais is on the -ais and the minor on the -ais. If a < b, then the major ais is on the -ais and the minor on the -ais.. t : a, b : t : a, b 6: t :. a, b : t : a, b : a 6, b : Let d be the length of the parametric interval. If d <, d ou get of a complete ellipse. If d, ou get the complete ellipse. If d >, ou get the complete ellipse but portions of the ellipse will be traced out more than once. For eample, if d the entire ellipse is traced twice.

21 Section.. t : t < : initial point: (, ) terminal point: (, ) t : For t, the entire graph described in part is drawn. The left branch is drawn from right to left across the screen starting at the point (, ). Then the right branch is drawn from right to left across the screen stopping at the point (, ). If ou leave out and, then the point (, ) is not drawn. t : initial point: (, ) terminal point: (, ) For < t, the right branch is drawn from right to left across the screen stopping at the point (, ). If ou leave out, then the point (, ) is not drawn. For t <, the left branch is drawn from right to left across the screen starting at the point (, ). If ou leave initial point: (, ) terminal point: (, ) Each curve is traced clockwise from the initial point to the terminal point. Eploration Graphing the Witch of Agnesi. We used the parameter interval [, ] because our graphing calculator ignored the fact that the curve is not defined when t or. The curve is traced from right to left across the screen. ranges from to.. t : < t : out, then the point (, ) is not drawn.. If ou replace cot t b cot t, the same graph is drawn ecept it is traced from left to right across the screen. If ou replace cot t b cot ( t), the same graph is drawn ecept it is traced from left to right across the screen. Quick Review.. m 8 ( ) When, we have 9, so the -intercepts are and. When, we have, so the 6 -intercepts are and.. When, we have, so the -intercepts are 6 and. When, we have, which has 9 no real solution, so there are no -intercepts.

22 Section. 6. When, we have +, so the -intercept is. When, we have, so the -intercepts are. (a) and. 7. (a) ()() +? Yes (b) ( ) ( ) + ( )? +? ( ) + ( )? +? No Yes [, ] b [, ] No initial or terminal point. (b) 9 t ( t) The parametrized curve traces all of the parabola defined b. 6. (a) 8. (a) 9() 8() + () ? Yes (b) 9() 8() + ( )? ? Yes 9( ) 8( ) + ( )? ? (a) + t t t (b) t t t + + t. (a) The equation is true for a. (b) The equation is equivalent to a a or a a. No Since a a is true for a and a a is true for a, at least one of the two equations is true for all real values of a. Therefore, the given equation a ± a is true for all real values of a. The equation is true for all real values of a. Section. Eercises. Graph. Window: [, ] b [, ], t. Graph (a). Window: [, ] b [, ], t. Graph (d). Window: [, ] b [, ], t. Graph (b). Window: [, ] b [, ], t Initial point: (, ) Terminal point: None (b) t ( t) 7. (a) The parametrized curve traces the left half of the parabola defined b (or all of the curve defined b ). [, ] b [, ] Initial point: (, ) Terminal point: None (b) t 8. (a) The parametrized curve traces all of the curve defined b (or the upper half of the parabola defined b ). No initial or terminal point. (b) sec t tan t The parametrized curve traces all of the parabola defined b.

23 Section. 9. (a) [, ] b [, ] Initial point: (, ) Terminal point: (, ) (b) t t + cos + sin. (a) The parametrized curve traces all of the ellipse defined b +. (b) + cos t + sin t. (a). (a) The parametrized curve traces the upper half of the circle defined b + (or all of the semicircle defined b ). Initial and terminal point: (, ) (b) + sin (t) + cos (t) The parametrized curve traces all of the circle defined b +. [.7,.7] b [.,.] Initial point: (, ) Terminal point: (, ) (b) t t + sin + cos The parametrized curve traces the right half of the ellipse defined b + (or all of the curve defined b ).. (a) Initial and terminal point: (, ) [, ] b [, ] Initial point: (, ) Terminal point: (, ) (b) + cos ( t) + sin ( t) The parametrized curve traces the upper half of the circle defined b + (or all of the semicircle (b) t t + sin + cos. (a) The parametrized curve traces all of the ellipse defined b +.. (a) defined b ). Initial and terminal point: (, ) [ 9, 9] b [ 6, 6] No initial or terminal point. (b) t 7 (t ) + + The parametrized curve traces all of the line defined b +.

24 Section. 6. (a). (a) (b) t ( t) + The parametrized curve traces the portion of the line defined b + to the left of (, ), that is, for. No initial or terminal point. (b) + t ( t) + The parametrized curve traces all of the line defined b (a) 8. (a) [, ] b [, ] Initial point: (, ) Terminal point: (, ) (b) t + The Cartesian equation is +. The portion traced b the parametrized curve is the segment from (, ) to (, ). Initial point: (, ) Terminal point: (, ) (b) t t + t + + ( ) ( ) 9. (a) The Cartesian equation is +. The portion traced b the curve is the segment from (, ) to (, ).. (a) Initial point: (, ) Terminal point: (, ) (b) t The parametrized curve traces the right portion of the curve defined b, that is, for. [, ] b [, ] No initial or terminal point, since the t-interval has no beginning or end. The curve is traced and retraced in both directions. (b) cos t cos t sin t sin t + The parametrized curve traces the portion of the parabola defined b + corresponding to.. (a) Initial point: None Terminal point: (, ) [ 6, 6] b [, 6] Initial point: (, ) Terminal point: None (b) t The parametrized curve traces the lower half of the parabola defined b (or all of the curve defined b +).

25 Section.. Using (, ) we create the parametric equations + at and + bt, representing a line which goes through (, ) at t. We determine a and b so that the line goes through (, ) when t. Since + a, a. Since + b, b. Therefore, one possible parametrization is + t, + t, t.. Using (, ) we create the parametric equations + at and + bt, representing a line which goes through (, ) at t. We determine a and b so that the line goes through (, ) at t. Since + a, a. Since + b, b. Therefore, one possible parametrization is + t, t, t.. The lower half of the parabola is given b + for. Substituting t for, we obtain one possible parametrization: t +, t, t. 6. The verte of the parabola is at (, ), so the left half of the parabola is given b + for. Substituting t for, we obtain one possible parametrization: t, t + t, t. 7. For simplicit, we assume that and are linear functions of t and that point (, ) starts at (, ) for t and passes through (, ) at t. Then f(t), where f() and f (). Since slope Δ Δ, t f (t) t + t. Also, g(t), where g() and g(). Since slope Δ Δ, t g(t) t + t. One possible parametrization is: t, t, t. 8. For simplicit, we assume that and are linear functions of t and that the point (, ) starts at (, ) for t and passes through (, ) at t. Then f(t), where f() and f (). Since slope Δ Δ ( ), t f(t) t + ( ) + t. Also, g(t), where g() and g(). Since slope Δ Δ t, g(t) t + t. One possible parametrization is: + t, t, t. 9. The graph is in Quadrant I when < <, which corresponds to < t <. To confirm, note that () and ().. The graph is in Quadrant II when <, which corresponds to < t. To confirm, note that () and ().. The graph is in Quadrant III when 6 <, which corresponds to t <. To confirm, note that ( ) and ( ).. The graph is in Quadrant IV when < <, which corresponds to < t <. To confirm, note that ( ) and ().. The graph of + + lies in Quadrant I for all >. Substituting t for, we obtain one possible parametrization: t, t + t +, t >.. The graph of + lies in Quadrant I for all. Substituting t for, we obtain one possible parametrization: t, t+, t >.. Possible answers: (a) a cos t, a sin t, t (b) a cos t, a sin t, t a cos t, a sin t, t (d) a cos t, a sin t, t 6. Possible answers: (a) a cos t, b sin t, t (b) a cos t, b sin t, t a cos t, b sin t, t (d) a cos t, b sin t, t 7. False. It is an ellipse. 8. True. Circle starting at (, ) and ending at (, ). 9. D.. C.. A.. E.

26 6 Section.. (a) The resulting graph appears to be the right half of a hperbola in the first and fourth quadrants. The parameter a determines the -intercept. The parameter b determines the shape of the hperbola. If b is smaller, the graph has less steep slopes and appears sharper. If b is larger, the slopes are steeper and the graph appears more blunt. The graphs for a and b,, and are shown.. (a) sec t and a for tan t in the identit sec t tan t gives b a. (b) (b) The graph is a circle of radius centered at (h, ). As h changes, the graph shifts horizontall. This appears to be the left half of the same hberbola. One must be careful because both sec t and tan t are discontinuous at these points. This might cause the grapher to include etraneous lines (the asmptotes of the hperbola) in its graph. The etraneous lines can be avoided b using the grapher's dot mode instead of connected mode. (d) Note that sec t tan t b a standard trigonometric identit. Substituting for sec t a and for tan t gives b a b. (e) This changes the orientation of the hperbola. In this case, b determines the -intercept of the hperbola, and a determines the shape. The parameter interval, gives the upper half of the hperbola. The parameter interval, gives the lower half. The same values of t cause discontinuities and ma The graph is a circle of radius centered at (, k). As k changes, the graph shifts verticall. Since the circle is to be centered at (, ), we use h and k. Since a radius of is desired, we need to change the coefficients of cos t and sin t to. cos t +, sin t, t. (a) (d) cos t, sin t +, t [, ] b [, ] No initial or terminal point. Note that it ma be necessar to use a t-interval such as [.7,.7] or use dot mode in order to avoid asmptotes showing on the calculator screen. (b) sec t tan t The parametrized curve traces the left branch of the hperbola defined b (or all of the curve defined b + ). add etraneous lines to the graph. Substituting b for

27 Section (a) No initial or terminal point. Note that it ma be necessar to use a t-interval such as [.7,.7] or use dot mode in order to avoid asmptotes showing on the calculator screen. (b) t t sec tan The parametrized curve traces the lower branch of the hperbola defined b (or all of the curve defined b + ). 7. Note that m OAQ t, since alternate interior angles formed b a transversal of parallel lines are congruent. OQ Therefore, tan t AQ, so cot t. tant Now, b equation (iii), we know that ( AQ) AB AO AQ AO ( AQ ) (cos t)( ) (cos t)( cot t) cos t. sin t 8. (a) If then the line is a vertical line and the first parametric equation gives, while the second will give all real values for since it cannot be the case that as well. Otherwise, solving the first equation for t gives t ( ) / ( ). Substituting that into the second equation gives + [( )/( )]( ) which is the point-slope form of the equation for the line through (, ) and (, ). Note that the first equation will cause to take on all real values, because ( ) is not zero. Therefore, all of the points on the line will be traced out. (b) Use the equations for and given in part (a), with t. Section. Functions and Logarithms (pp. 7 ) Eploration Testing for Inverses Graphicall. It appears that (f g)() (g f)(), suggesting that f and g ma be inverses of each other. (a) f and g: (b) f g: Then equation (ii) gives t AB t cos sin sin t cos t sin t sin t. The parametric equations are: g f: cot t, sin t, < t < Note: Equation (iii) ma not be immediatel obvious, but it ma be justified as follows. Sketch segment QB. Then OBQ is a right angle, so ΔABQ ΔAQO, which gives AB AQ. AQ AO

28 8 Section.. It appears that f g g f g, suggesting that f ma be the identit function. (a) f and g:. It appears that (f g)() (g f)(), suggesting that f and g ma be inverses of each other. (Notice that the domain of f g is (, ) and the domain of g f is (, ). (a) f and g: (b) f g: (b) f g: g f: g f:. It appears that ( f g)() (g f )(), suggesting that f and g ma be inverses of each other. (a) f and g: Eploration Supporting the Product Rule. The appear to be vertical translations of each other. (b) f g:. This graph suggests that each difference ( ) is a constant. g f:. ln (a) ln ln a + ln ln ln a Thus, the difference is the constant In a. Quick Review.. ( f g)() f (g()) f(). (g f )( 7) g( f ( 7)) g( )

29 Section. 9. ( f g)( ) f( g( )) f( + ) ( + ) /. ( g f)( ) g( f( )) g( ) ( ) + ( ) / +. Substituting t for, one possible answer is: t,, t. t 6. Substituting t for, one possible answer is: t, t, t <. 7. Section. Eercises. No, since (for eample) the horizontal line intersects the graph twice.. Yes, since each horizontal line intersects the graph onl once.. Yes, since each horizontal line intersects the graph at most once.. No, since (for eample) the horizontal line. intersects the graph twice.. Yes, since each horizontal line intersects the graph onl once. 6. No, since (for eample) the horizontal line intersects the graph at more than one point (, ) 8. Yes, the function is one-to-one since each horizontal line intersects the graph at most once, so it has an inverse function. 8, (. 67, ) 9. (a) No, the function is not one-to-one since (for eample) the horizontal line intersects the graph twice, so it does not have an inverse function. 9. (.8, ) (b) No points of intersection, since > for all values of.. (a). No, the function is not one-to-one since (for eample) the horizontal line intersects the graph more than once, so it does not have an inverse function. (.9, ) (b) No points of intersection, since e of. > for all values Yes, the function is one-to-one since each horizontal line intersects the graph onl once, so it has an inverse function.

30 Section... No, the function is not one-to-one since each horizontal line intersects the graph twice, so it does not have an inverse function. Verif. ( f f )( ) f ( ) ( f f)( ) f ( ) ( ). + ( + ) / Interchange and. Yes, the function is one-to-one since each horizontal line intersects the graph at most once, so it has an inverse function.. + Interchange and. f ( ). Verif. ( f f )( ) f + ( ) + ( f f)( ) f ( + ) ( + ). Interchange and. f ( ) / ( + ) / f ( ) ( + ) or + Verif. ( f f )( ) f( + ) ( f ( + ) ( + ) f )( ) f ( ) 6. +,, Interchange and. ( ) + f ( ) or ( ) / Verif. For (the domain of f ), ( f f )( ) f( ) ( ) + ( ) + For >, (the domain of f ), ( f f)( ) f ( + ) ( + )

31 Section. 7., Interchange and. f ( ) or / Verif. For (the domain of f ), ( f f )( ) f( ) ( ) For, (the domain of f ), ( f f)( ) f ( ) / 8., / / / ( ), / Interchange and. / f ( ) / Verif. For (the domain of f ), / / / ( f f )( ) f( ) ( ) for, (the domain of f ), / / / ( f f )( ) f ( ) ( ) 9. ( ), ( ), Interchange and. f ( ) or ( ) / Verif. For (the domain of f ), ( f f )( ) f( ) [( ) ] ( ) For (the domain of f ),. ( + + ), ( + ), + Interchange and. / f ( ) or Verif. For (the domain of f ), ( f f )( ) f( ) [( ) + ( ) + ] ( ) ( ) For (the domain of f ), ( f f)( ) f ( + + ) + + ( + ) + ( + )., >, > Interchange and. f ( ) or / Verif. For > (the domain of f ), ( f f )( ) f ( / ) For > (the domain of f ), ( f f)( ) f / ( f f)( ) f ( ( ) ) ( ) + ( )

32 Section.. Interchange and. f ( ) or / Verif. ( f f )( ) f ( / ) ( f f)( ) f / ( ) Interchange and. f ( ) Verif. ( f f )( ) f + + ( ) + ( ) ( ) + ( ) + ( f f)( ) f ( + ) ( + ) ( + ) ( + ) ( ) + + Interchange and. + + f ( ) Verif. ( f f + )( ) f ( + ) + ( ) ( + ) ( ) + ( f f)( ) f ( + ) + ( ) ( + ) ( )

33 Section.. Graph of f: t, e t Graph of f : e t, t Graph of : t, t. Graph of f: t, log t Graph of f : log t, t Graph of : t, t [ 6, 6] b [, ] 6. Graph of f: t, t Graph of f : t, t Graph of : t, t. Graph of f: t, sin t Graph of f : sin t, t Graph of : t, t [, ] b [, ] 7. Graph of f: t, t Graph of f t, t Graph of : t, t. Graph of f: t, tan t Graph of f : tan t, t Graph of : t, t [.,.] b [, ] 8. Graph of f: t, t Graph of f : t, t Graph of : t, t t. (. ) t ln(. ) ln t ln. ln ln t. 7 ln. Graphical support: 9. Graph of f: t, ln t Graph of f : ln t, t Graph of : t, t [.,.] b [, ]

34 Section.. t. e. t lne ln. t ln ln t ln. 97. Graphical support: 7. ln t+ ln t+ e e t+ e 8. ln( ) ln + ln ln( ) + ln + ln ln( ) e + ln + ln e e ( )( ) e e + e e + e e ( e + e ) e ( ) ( e ) e + e ± ( ) ()() () e ± ± ln 96. or 96. Graphical support: [, ] b [ 7, ] Domain: (, ) Range: (, ).. Domain: (, ) Range: (, ) ( + ) ( ) ( ) ( ) + ( ) ()() ± () ± ± log 6. or 6.. [, 6] b [, ] Domain: (, ) Range: (, ) Graphical support: Domain: (, ) Range: (, )

35 Section log ( ) log log log log log Interchange and. log f ( ) log Verif. ( f f )( ) f log log + log ( ) ( f f)( ) f + log + + log ( + ) log log ( ) log. ( ) log. log. log. log. log. Interchange and : log f Verif.. ( ) log. ( f f )( ) f log. +. log. log ( ) ( f f)( ) f +. log log. ( +. ) log. log (... )

36 6 Section.. (a) f( f( )) ( f( )) ( ), since (b) f( f( )) f / for all. + (. 98) ln( ) 7 (. 98) ln() ln( ). 98 e (. 9 )., or sometime during.. (a) f() (.7) ln() 6. (a) Amount 8 t/ t/ (b) 8 t/ 8 t/ t t 6 There will be gram remaining after 6 hours. 7. (. 7) t. 7 t ln(. 7 ) ln t ln. 7 ln ln t. 96 ln. 7 t It will take about.96 ears. (If the interest is paid at the end of each ear, it will take ears.) 8. 7, (. ),,. t 8 t 8 ln(. ) ln 8 t ln. ln ln( 8 / ) t. 8 ln. It will take about.8 ears. 9. (a) f(). + (.98) ln () t [, ] b [, ] (b) (.7) ln () (.7) ln (). trillion... trillion cubic feet. The estimate is an under estimate of. trillion cubic feet (. 7) ln( ). (. 7) ln( ) ln( ). 7 ln( ).. e. or sometime during.. (a) Suppose that f ( ) f ( ). Then m + b m + b so m m. Since m, this gives. (b) m+ b b m b m Interchange and. b m b f ( ) m The slopes are reciprocals. If the original functions both have slope m, each of the inverse functions will have slope. The graphs of the m inverses will be parallel lines with nonzero slope. (d) If the original functions have slopes m and m, respectivel, then the inverse functions will have (b). + (.98) ln () 6.79 trillion trillion The estimate eceeds the actual b.6 trillion slopes m and m, respectivel. Since each of m and m is the negative reciprocal of the other, the graphs of the inverses will be perpendicular lines with nonzero slopes.

37 Section.6 7. False. For eample, the horizontal line test finds three answers for.. False. Must satisf (f g)() (g f)(), not just (f g)().. C. ln () >, requiring + >, or >.. A. As ( + ), f(). 6. E. f( ) B. ( ) ( ) 8. (a) is a vertical shift (upward) of, although it s difficult to see that near the vertical asmptote at. One might use trace or table to verif this. (b) Each graph of is a horizontal line. The graphs of and a are the same. (d) e a, ln( e -- ) ln a, ln a, ln a ln ln a 9. If the graph of f () passes the horizontal line test, so will the graph of g() f () since it's the same graph reflected about the -ais. Alternate answer: If g( ) g( ) then f ( ) f ( ), f ( ) f ( ), and since f is one-to-one. 6. Suppose that g( ) g( ). Then f( ) f( ), f( ) f( ), and and since f is one-to-one. 6. (a) The epression a ( b c ) + d is defined for all values of, so the domain is (, ). Since b c attains all positive values, the range is (d, ) if a > and the range is (, d) if a <. (b) The epression a log b ( c) + d is defined when c >, so the domain is (c, ). Since a log b ( c) + d attains ever real value for some value of, the range is (, ). 6. (a) Suppose f ( ) f ( ). Then: a + b a b c + d + c + d ( a+ b)( c + d) ( a + b)( c + d) ac + ad+ bc + bd ac + ad + bc+ bd ad+ bc ad + bc ( ad bc) ( ad bc) Since ad bc, this means that. a + b (b) c + d c + d a + b ( c a) d + b d b + c a Interchange and : d b + c a d b f ( ) + c a a + b a As ±, f( ) c + d, so the horizontal c a asmptote is c ( c ). Since f() is undefined at d d, the vertical asmptote is. c c d b d (d) As ±, f ( ) +, so the c a c d horizontal asmptote is c ( c ). Since f ( ) a is undefined at, the vertical asmptote is c a. The horizontal asmptote of f becomes the c vertical asmptote of f and vice versa due to the reflection of the graph about the line. Section.6 Trigonometric Functions (pp. 6 ) Eploration Unwrapping Trigonometric Functions. (, ) is the circle of radius centered at the origin (unit circle). (, ) is one period of the graph of the sine function.

38 8 Section.6. The -values are the same in the interval t.. The -values are the same in the interval t.. The -values and the -values are the same in each interval.. tan t:. i 8 9. i 8. csc t: 6..6,.98 sec t: cot t: 7..98,.9 For each value of t, the value of tan t is equal to the ratio. For each value of t, the value of csc t is equal to the ratio. For each value of t, the value of sec t is equal to the ratio. For each value of t, the value of cot t is equal to the ratio. Quick Review.6. 8 i 6.. i or,. 97 or 8. f ( ) ( ) f () The graph is smmetric about the -ais because if a point (a, b) is on the graph, then so is the point ( a, b). 9. f( ) ( ) ( ) + ( ) f( ) The graph is smmetric about the origin because if a point (a, b) is on the graph, then so is the point ( a, b)..

39 Section.6 9 Section.6 Eercises. Arc length ( ) 8 7. Radius Angle radian or about 8.6 /. Angle radian or 6. Even. sec( ) sec( ) cos( ) cos( ) sin( ) sin( ) 6. Odd. tan( ) tan cos( ) cos( ) 7. Odd. csc( ) csc sin( ) sin( ) Since csc ( + ), the range ecludes numbers between and. The range is (, ] [, ). (d) [ -, ] b [ 8, 8]. (a) Period (b) Domain: (, ) Since sin ( + ), the range etends from + to +. The range is [, ]. (d) cos( ) cos( ) 8. Odd. cot( ) sin( ) sin( ) cot( ) 9. Using a triangle with sides:, 7 and 8; sin θ 8/7, tan θ 8/, csc θ 7/8, sec θ 7/, cot θ /8.. Using a triangle with sides:, and ; sin /, cos /, csc, sec, cot. (a) Period (b) Domain: Since csc ( + ) sin ( + ), we ( k ) require + k, or. This k requirement is equivalent to for integers k.. (a) Period k (b) Domain: We require + for odd integers k. ( k ) Therefore, for odd integers k. This 6 k requirement is equivalent to for odd 6 integers k. Since the tangent function attains all real values, the range is (, ). (d) [ -, ] b [ 8, 8]

40 Section.6. (a) Period (b) Domain: (, ) Range: Since sin +, the range is [, ]. (d). (a) The period of sec is, so the window should have length. One possible answer: [, ] b [, ] (b) The period of csc is, so the window should have length. One possible answer: [, ] b [, ] The period of cot is, so the window should have length. One possible answer: [, ] b [, ] 6. (a) The period of sin is, so the window should have length. One possible answer: [, ] b [, ] (b) The period of cos is, so the window should have length. One possible answer: [, ] b [, ] The period of tan is, so the window should have length. One possible answer: [, ] b [, ] 7. (a) Period (b) Amplitude. [, ] b [, ] 8. (a) Period. (a) Period / (b) Amplitude [, ] b [, ]. (a) Period 6 / (b) Amplitude [, ] b [, ]. (a) Period (b) Amplitude [, ] b [, ]. (a) Using a graphing calculator with the sinusoidal regression feature, the equation is. sin (68.6.9) +.8. [,.]b [.,.] (b) The frequenc is 68.6 radians per second, which is equivalent to 9. 9 ccles per second (Hz). The note is a G.. (a) b 6 8 (b) It s half of the difference, so a. 8 + k (d) The function should have its minimum at t (when the temperature is F) and its maimum at t 8 (when the temperature is 8 F). The value of h is + 8.Equation: sin ( ) 6 + (b) Amplitude, b [, ] 9. (a) Period (b) Amplitude [, ] b [, ]

41 Section.6. The portion of the curve cos between passes the horizontal line test, so it is one-to-one.. This equation is equivalent to sin, so the solutions in the interval < are 6 and. 6 [, ] b [, ] 6. The portion of the curve tan between / < < / passes the horizontal line test, so it is oneto-one. [In parametric mode, use T min / + ε and T ma / ε, where ε is a ver small positive number, sa..]. This equation is equivalent to cos, so the solution in the interval is cos. 9. Since the cosine function is even, the solutions in the interval < are.9 and.9.. The solutions in the interval < are 7 and 6 7. Since 6 [, ] b [, ] is in the range, of sin and sin., sin (. ) 8 radian or Since sin is the range 8 or, sin., of sin and radian 9. Using a calculator, tan ( ).7 radians or Using a calculator, cos ( 7. ) or radian. The angle tan (.).9 is the solution to this equation in the interval < <. Another solution in is tan (.) +.. The solutions are.9 and... The angle cos (.7).6 is the solution to this equation in the interval. Since the cosine function is even, the value cos (.7).6 is also a solution, so an value of the form ± cos ( 7. ) + k is a solution, where k is an integer. In < the solutions are cos (.7) and cos (.7) Since sin has period, the solutions are 7 all of the form + k or + k, where k 6 6 is an integer. 6. The equation is equivalent to tan, so the solution in the interval < < is tan ( ). Since the period of tan is, all solutions are of the form + k, where k is an integer. This is equivalent to + k, where k is an integer. 7. Note that Since sin 8 7 and, cos sin sin Therefore: sin 8, cos, tan, 7 7 cos 8 cot,sec, tan 8 cos 7 7 csc sin 8

42 Section.6 8. Note that +. / sin Since tan and / cos, we have sin and cos. In summar: sin, cos, tan cot,sec,csc tan cos sin 9. Note that r ( ) +. Then: sin, cos, tan, r r r r cot, sec, csc. Note that r ( ) +. Then: sin cos r r tan, cot, r r sec,csc. Let cos 7. Then θ and cos 7, so 7 sin cos sin cos Let sin 9. Then cos sin 9 9 sin tan sin tan cos 9/ / 88. (a) Amplitude 7 (b) Period 6 ( / 6) Horizontal shift 88 and sin 9, so. Therefore, (d) Vertical shift (e) f( ) 7 sin ( ) 6 +. (a) Highest : F Lowest : 7 F 6 + ( ) (b) Average F This average is the same as the vertical shift because the average of the highest and lowest values of sin is. cos( ) cos( ). (a) cot( ) cot( ) sin( ) sin( ) (b) Assume that f is even and g is odd. Then f ( ) f( ) f( ) so f is odd. The g( ) g ( ) g ( ) g situation is similar for g f. 6. (a) csc( ) csc( ) sin ( ) sin ( ) (b) Assume that f is odd. Then so f( ) f( ) f( ) f is odd. 7. Assume that f is even and g is odd. Then f( ) g( ) ( f( ))( g( )) f( ) g( ) so f(g) is odd. 8. If (a, b) is the point on the unit circle corresponding to the angle θ, then ( a, b) is the point on the unit circle corresponding to the angle (θ + ) since it is eactl half wa around the circle. This means that both tan (θ) and tan (θ + ) have the same value, b a. 9. (a) Using a graphing calculator with the sinusoidal regression feature, the equation is. sin ( ) (b) sin ( + ) +. False. It is because /B / implies the period B is.. False. The amplitude is /.. D.. B. The curve oscillates between and.. E.. A.

43 Section.6 6. (a) The graph is a sine/cosine tpe graph, but it is shifted and has an amplitude greater than. (b) Amplitude. (that is, ) Period Horizontal shift.78 that is, or.98 that is, 7 Vertical shift: sin + (sin ) cos (cos ) sin + + (sin ) (cos ) (sin + cos ) b (d) sin + tan a b sin( )cos tan a + b cos( )sin tan a a b sin( ) + cos( ) a + b a + b ( asin+ bcos ) a + b and multipling through b the square root gives the desired result. Note that the substitutions cos tan b a and a a + b sin tan b b depend on the requirement a a + b that a is positive. If a is negative, the formula does not work. 9. Since sin has period, sin ( + ) sin (). This function has period. A graph shows that no smaller number works for the period. Therefore, sin + cos sin (a) sin a + (b) See part (a). It works. (d) sin a + (sin a) cos (cos a) si + n (sin ) (cos ) + a a (sin cos ) a + a So, sin ( a) + cos( a) sin a (a) One possible answer: a + b sin + tan b a 6. Since tan has period, tan ( + ) tan. This function has period. A graph shows that no smaller number works for the period. 6. The period is. One possible graph: 6 [ -, 6 6 ] b [, ] (b) See part (a). It works.

44 Chapter Review 6. The period is 6. One possible graph: 8. Since is equivalent to +, the slope of the given line (and hence the slope of the desired line) is. ( ) + Quick Quiz (Sections..6). C.. D.. E.. (a) f( ) + + g ( ) + + (b) f g( ) f + + g f( ) g( ) ( ) + Chapter Review Eercises (pp. 6 7). ( ) + ( 6) 9. ( + ) m 6 8 ( ) ( + ) m ( ) Since + is equivalent to +, the slope of the given line (and hence the slope of the desired line) is. ( ). Since is equivalent to, the slope of the given line is and the slope of the perpendicular line is. ( + ) 9. Since + is equivalent to +, the slope of the given line is and the slope of the perpendicular line is. ( + ) The line passes through (, ) and (, ) m ( ). m ( ) f( ) ( + ) + f( ) + Check: f ( ) ( ) +, as epected.

45 Chapter Review. The line passes through (, ) and (, ). m ( ) 7 7 ( ) ( ) ( ) ( ) ( ) ( ) + Odd. ( ) sin ( ) + sin Neither even nor odd. ( ) + cos ( ) + cos Neither even nor odd 6. ( ) ( ) Smmetric about the origin. 6. Even 7. (a) The function is defined for all values of, so the domain is (, ). (b) Since attains all nonnegative values, the range is [, ) Smmetric about the -ais. Neither [, ] b [, ] 8. (a) Since the square root requires, the domain is (, ]. (b) Since attains all nonnegative values, the range is [, ). [ 9., 9.] b [, ] 9. (a) Since the square root requires 6, the domain is [, ]. Smmetric about the -ais. 9. ( ) ( ) + + () Even (b) For values of in the domain, 6 6, so 6. The range is [, ].. ( ) ( ) ( ) ( ) + + () Odd. ( ) cos( ) cos () Even. ( ) sec( )tan( ) sin ( ) sin cos ( ) cos sec tan ( ) [ 9., 9.] b [ 6., 6.] Odd

46 6 Chapter Review. (a) The function is defined for all values of, so the domain is (, ). (b) Since attains all positive values, the range is (, ).. (a) The function is defined for all values of, so the domain is (, ). (b) The function is equivalent to, which attains all nonnegative values. The range is [, ). [ 6, 6] b [, ]. (a) The function is defined for all values of, so the domain is (, ). (b) Since e attains all positive values, the range is (, ). [ 8, 8] b [, ]. (a) The logarithm requires >, so the domain is (, ). (b) The logarithm attains all real values, so the range is (, ). [, ] b [, ]. (a) The function is equivalent to tan, so we require k for odd integers k. The domain is given b k for odd integers k. (b) Since the tangent function attains all values, the range is (, ). [, ] b [, ] 6. (a) The function is defined for all values of, so the domain is (, ). (b) The cube root attains all real values, so the range is (, ). [ -, ] b [ 8, 8]. (a) The function is defined for all values of, so the domain is (, ). (b) The sine function attains values from to, so sin ( + ), and hence sin ( + ). The range is [, ]. [, ] b [, ] 7. (a) The function is defined for, so the domain is [, ]. (b) The function is equivalent to,, which attains values from to for in the domain. The range is [, ]. [ 6, 6] b [, ] [, ] b [, ]