Section. Chapter Prerequisites for Calculus Section. Lines (pp. 9) Quick Review.. ( ) (). ( ). m 5. m ( ) 5 ( ) 5. (a) () 5 Section. Eercises.. (). 8 () 5. 6 5. (a, c) 5 B A 5 6 5 Yes (b) () () 5 5 No 6. (a) 7 () 5 7 5 Yes (b) () 5 9 No (b) m ( ) 6. (a, c) 5 7. ( ) ( ) ( ) ( ) A B 5 8. = ( ) ( ) ( ) ( ) 5 9 5 96 (b) m ( ( ) ) 7. (a, c) 5 B A 5 9. 7 7 7 (b) m 8. (a, c) 5. 5 5 5 5 A B 5
Section. 8. continue (b) m 5 (unefine) This line has no slope. 9. (a) (b). (a) (b). (a) (b). (a) (b). ( ). [ ()] ( ) 5. ( ) 6. [ ()] ( ) 7. m ( ) 6. The line contains (, ) an (5, ). m 5 5 5 7. (a) Slope: (b) -intercept: (c) [, ] b [, ] 8. (a) Slope: (b) -intercept: (c) 8. m ( ) 9. m (unefine) () Vertical line:. m ( ) [ ()] 9. [, ] b [, ] (a) Slope: (b) -intercept: (c) ( ).. or [, ] b [, ]. (a) Slope: (b) -intercept: (c).. 5. The line contains (, ) an (, 5). m 5 5 5 5 [, ] b [, ]
Section.. (a) The esire line has slope an passes through (, ): ( ) or. (b) The esire line has slope an passes through (, ): ( ) or.. (a) The given equation is equivalent to. The esire line has slope an passes through (, ): ( ) or. (b) The esire line has slope an 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, :. 5. m 9 7 f() 7 ( ) 7 Check: f(5) 7 (5) 6, as epecte. Since f() 7, we have m 7 an b. 6. m ( ) f() ( ) () Check: f(6) (6) 7, as epecte. Since f(), we have m an b. 7. () (6) 8. ( ) ( 8) ( 8) 8 6 9. (a).68 9. (b) The slope is.68. It represents the approimate average weight gain in pouns per month. (c) [5, 5] b [5, 5] () When,.68() 9. 9.. She weighs about 9 pouns.. (a),6.,77,58.669 (b) The slope is,6.. It represents the approimate rate of increase in earnings in ollars per ear. (c) [975, 995] b [,, 5,] () When,,6.(),77,58.669,98. In, the construction workers average annual compensation will be about $,98.. ( ) This is the same as the equation obtaine in Eample 5.. (a) When, we have, so c. c When, we have, so. (b) When, we have, so c. c When, we have, so. The -intercept is c an the -intercept is.. (a) The given equations are equivalent to k k an, respectivel, so the slopes are k an. The lines are parallel when, k so k. (b) The lines are perpenicular when k, so k.. (a) m 68 69..5.5.75 egrees/inch. (b) m 68. 58. 6. egrees/inch 6 5 (c) m 5 7. egrees/inch.7.7 () Best insulator: Fiberglass insulation Poorest insulator: Gpsum wallboar The best insulator will have the largest temperature change per inch, because that will allow larger temperature ifferences on opposite sies of thinner laers.
Section. 5. Slope: k p.9 9. 9.99 atmospheres per meter At 5 meters, the pressure is p.99(5) 5.97 atmospheres. 6. (a) (t) 5t (b) (, ) (, ) 6 (, ) (, ) 6 [, 6] b [5, ] (c) The slope is 5, which is the spee in miles per hour. () Suppose the car has been traveling 5 mph for several hours when it is first observe at point P at time t. (e) The car starts at time t at a point miles past P. 7. (a) 56,8,8 (b) The rate at which the meian price is increasing in ollars per ear (c) 7 5,6,6 () The meian price is increasing at a rate of about $56 per ear in the Northeast, an about $7 per ear in the Miwest. It is increasing more rapil in the Northeast. 8. (a) Suppose F is the same as C. 5. (, ) (, ) (a, b) Z 6 W (, ) (, ) (c, ) X 6 (b) 9 5 9 5 5 Yes, F is the same as C. (g, h) (e, f) Y Suppose that the vertices of the given quarilateral are (a, b), (c, ), (e, f ), an (g, h). Then the mipoints of the consecutive sies are W a Y e g, f h, an Z g c, b, X c e, f, a, h b. When these four points are connecte, the slopes of the sies of the resulting figure are: [9, 9] b [6, 6] It is relate because all three lines pass through the point (, ) where the Fahrenheit an Celsius temperatures are the same. 9. The coorinates of the three missing vertices are (5, ), (, ) an (, ), as shown below. (, ) 6 (, ) (, ) (5, ) 6 f b f WX: b c e a c e a f h f XY: h e g c e g c f h h b f ZY: b e g g a e a h b b WZ: h g a a c g c Opposite sies have the same slope an are parallel.
Section. 5 5. The raius through (, ) has slope. The tangent line is tangent to this raius, so its slope is /. We seek the line of slope that passes through (, ). ( ) 9 5 5. (a) The equation for line L can be written as A B C B, so its slope is A B. The perpenicular line has slope B A/B A an passes through (a, b), so its equation is B ( a) b. A (b) Substituting B ( a) b for in the equation for line L gives: A A B B A ( a) b C A B ( a) ABb AC (A B ) B a AC ABb B a C ABb A A B Substituting the epression for in the equation for line L gives: a C ABb A B A A B B C B C A(B a AC ABb) (A B ) A B A B AB B a A C A Bb A C B C A B A B Bb B C AB a A B A b BC ABa A B B a AC ABb A b BC ABa A B A B The coorinates of Q are,. (c) Distance ( a) ( b) B a AC A B A B b a A b BC A B A B a b B a AC ABb a(a B ) A b BC ABa b(a B ) A B A B A C A AB b B A a B C A AB a B B b A (C A B b B Aa) B (C A A a B Bb) A (C Aa Bb) B (C Aa Bb) (A B ) (A B ) (A B )(C Aa Bb) (A B ) (C A Aa B B b ) C Aa Bb A B A a Bb C A B
6 Section. Section. Functions an Graphs (pp. 9 9) Eploration Composing Functions. g f, f g. Domain of :[, ] Range of : [, ] : : : [.7,.7] b [,.] [.7,.7] b [,.] [.7,.7] b [,.]. Domain of : [, ); Range of :(, ] : 5. 6 Solutions 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 is false when. Test : 9 9 9 is true when. Test : 9 9 6 7 9 is false when. Solution set: [, ] 7. Translate the graph of f units left an units ownwar. 8. Translate the graph of f 5 units right an units upwar. 9. (a) f() 5 9 ( )( ) or (b) f() 6 5 6 No real solution. (a) f() 5 [, 6] b [, 6]. ( ()) () ( ()) ( ()) (), Quick Review.. 5 Solution: [, ). ( ) Solutions to ( ) :, Test : ( ) ( ) is true when. Test : ( ) ( ) is false when. Test : ( ) ( ) is true when. Solution set: (, ) (, ). 7 Solution set: [, 7]. 5 5 or 5 or 7 Solution set: (, ] [7, ) 5 5 (b) f() = No solution. (a) f() = 7 7 6 9 Check: 9 7 6 ; it checks. (b) f() 7 7 6 Check: 6 7 ; it checks.. (a) f() 8 7 (b) f() 7 8
Section. 7 Section. Eercises. Since A r, the formula is A, where A represents area an represents iameter. 8. (a) Since we require, the omain is (, ]. (b) (, ] (c). Let h represent height an let s represent sie length. h s s h s s h s h s [, ] b [, ] () None 9. (a) Since we require, the omain is (, ]. (b) [, ) (c) Since sie length an height must be positive, the formula is h s. s s s h s. S 6e, where S represents surface area an e represents ege length.. V r, where V represents volume an r represents raius. 5. (a) (, ) or all real numbers (b) (, ] (c) [5, 5] b [, ] () Smmetric about -ais (even) 6. (a) (, ) or all real numbers (b) [9, ) (c) [5, 5] b [, ] s () Smmetric about the -ais (even) 7. (a) Since we require, the omain is [, ). (b) [, ) (c) [.7,.7] b [6, 6] () None. (a) Since we require, the omain is (, ) (, ). (b) Since can assume an value ecept, the range is (, ) (, ). (c) [.7,.7] b [6, 6] () None. (a) (, ) or all real numbers (b) (, ) or all real numbers (c) [6, 6] b [, ] () None. (a) (, ) or all real numbers (b) The maimum function value is attaine at the point (, ), so the range is (, ]. (c) [6, 6] b [, ] () Smmetric about the -ais (even) () None [, ] b [, ]
8 Section.. (a) Since we require, the omain is (, ]. (b) [, ) (c) 8. (a) This function is equivalent to, so its omain is [, ). (b) [, ) (c) [, ] b [, ] () None. (a) Since we require, the omain is (, ) (, ). (b) Note that can assume an value ecept, so can assume an value ecept. The range is (, ) (, ). (c) [, ] b [, ] () None 5. (a) Since we require, the omain is [, ]. (b) Since will be between an, inclusive (for in the omain), its square root is between an, inclusive. The range is [, ]. (c) [, 5] b [, 8] () None 9. Even, since the function is an even power of.. Neither, since the function is a sum of even an o powers of.. Neither, since the function is a sum of even an o powers of ( ).. Even, since the function is a sum of even powers of ( ).. Even, since the function involves onl even powers of.. O, since the function is a sum of o powers of. 5. O, since the function is a quotient of an o function ( ) an an even function ( ). 6. Neither, since, (for eample), () / an (). 7. Neither, since, (for eample), () is efine an () is unefine. 8. Even, since the function involves onl even powers of. 9. (a) [.7,.7] b [.,.] () Smmetric about the -ais (even) 6. (a) This function is equivalent to, so its omain is all real numbers. (b) [, ) (c) [9., 9.] b [6., 6.] Note that f(), so its graph is the graph of the absolute value function reflecte across the -ais an then shifte units right an units upwar. (b) (, ) (c) (, ]. (a) The graph of f() is the graph of the absolute value function stretche verticall b a factor of an then shifte units to the left an units ownwar. [, ] b [, ] () Smmetric about the -ais (even) 7. (a) Since we require, the omain is (, ) (, ) (b) Since for all, the range is (, ). (c) [, 5] b [5, ] (b) (, ) or all real numbers (c) [, ). (a) [, ] b [, 5] () Smmetric about the -ais (even) [.7,.7] b [, 6] (b) (, ) or all real numbers (c) [, )
Section. 9. (a) [, ] b [, ] (b) (, ) or all real numbers (c) [, ). (a). Line through (, ) an (, ): m, so ( ) Line through (, ) an (, ): m, so, f(), 5. Line through (, ) an (, ): Line through (, ) an (, ): [.7, 5.7] b [, 9] (b) (, ) or all real numbers (c) (, ) or all real numbers. (a) [.5,.5] b [, ] (b) (, ) or all real numbers (c) [, ) 5. Because if the vertical line test hols, then for each -coorinate, there is at most one -coorinate giving a point on the curve. This -coorinate woul correspon to the value assigne to the -coorinate. Since there is onl one -coorinate, the assignment woul be unique. 6. If the curve is not, there must be a point (, ) on the curve where. That woul mean that (, ) an (, ) are two ifferent points on the curve an it is not the graph of a function, since it fails the vertical line test. 7. No 8. Yes 9. Yes. No. Line through (, ) an (, ): Line through (, ) an (, ):, f (),,,. f(),,. Line through (, ) an (, ): Line through (, ) an (5, ): m 5, so ( ) 5, f() 5 5, Line through (, ) an (, ): m, so ( ), f(),, 6. Line through (, ) an (, ): Line through (, ) an (, ): Line through (, ) an (, ):, f(),, 7. Line through T, an (T, ): m T (T/), T f () T T, T A, T A, T 8. f () T A, T T A, T T T,so T T T 9. (a) f (g()) ( ) 5 (b) g( f()) ( 5) ( 5) (c) f (g()) () g( f ()) (e) g(g()) [() ] (f) f ( f()) ( 5) 5
Section. 5. (a) f (g()) ( ) (b) g( f()) ( ) (c) f (g()) () g( f ()) (e) g(g()) ( ) (f) f ( f ()) ( ) 5. (a) Enter f() 7, g(), ( f g)() ( ()), an (g f )() ( ()) f g: g f: [, 7] b [, ] [, ] b [, ] Domain: [, ) Range: [7, ) (b) ( f g)() 7 Domain: [7, ) Range: [, ) (g f )() 7 5. (a) Enter f(), g(), ( f g)() ( ()), an (g f )() ( ()) f g: g f: [.7,.7] b [, ] Domain: (, ] [, ) Range: [, ) (b) ( f g)() ( ) ( ),, (g f )() ( ) 5. (a) Enter () f(),, ( f g)() ( ()), an (g f )() ( ()). Use a ecimal winow such as the one shown. f g: [9., 9.] b [6., 6.] Domain: (, ) (, ) Range: (, ) (, ) g f: [6, 6] b [, ] Domain: [, ) Range: (, ] g f: [9., 9.] b [6., 6.] Domain: (, ) (, ) Range: (, ) (, ) [.5,.5] b [,.] Domain: [, ] Range: [, ] (b) ( f g)() (), (g f )() 5. (a) Enter f(), g(), ( f g)() ( ()), an (g f)() ( ()). f g: [, ] b [, ] Domain: [, ) Range: [, ) (b) ( f g)() ( ) ( ), ( ) ( ) 7, 7, (g f )() ( ) ( ), ( ) ( ) 7, 7,
Section. 55. 58. [5, 5] b [, 5] We require (so that the square root is efine) an (to avoi ivision b zero), so the omain is (, ) (, ). For values of in the omain, an hence an can attain an positive value, so the range is (, ). (Note that grapher failure ma cause the range to appear as a finite interval on a grapher. [.5,.5] b [.55,.55] We require, so the omain is (, ) (, ) (, ). For values of in the omain, can attain an value in [, ) (, ), so can also attain an value in [, ) (, ). Therefore, can attain an value in (, ] (, ). The range is (, ] (, ). (Note that grapher failure can cause the intervals in the 56. range to appear as finite intervals on a grapher.) 59. (a) [5, 5] b [, 5].5 We require 9 (so that the fourth root is efine) an 9 (to avoi ivision b zero), so the omain is (, ). For values of in the omain, 9 can attain.5 an value in (, 9]. Therefore, 9 can attain an value in (, ], an can attain an value in 9,. The range is, or approimatel [.5, ). (Note that grapher failure ma cause the range to appear as (b).5 a finite interval on a grapher.) 57..5 6. (a) [.7,.7] b [.,.] We require 9, so the omain is (, ) (, ) (, ). For values of in the omain, 9 can attain an value in (, ) (, 9], so 9 can attain an value in (, ) (, 9]. Therefore, can attain an 9 value in (, ),. The range is 9 (, ), or approimatel (, ) [.96, ). 9 (Note that grapher failure can cause the intervals in the range to appear as finite intervals on a grapher.)
Section. 6. continue (b) () Since ( f g)() f (), f(). The complete table is shown. Note that the absolute value sign in part () is optional. g() f () ( f g)() 5 5, 6. (a),., 6. (a) Note that the ata in the table begins at. (We o not inclue the initial investment in the ata.) The power regression equation is 7.9.65.. (b) (b) 6. (a)... [, ] b [,, 8,] (c) When,,7. Accoring to the power regression equation, the investment will grow to approimatel $,7. () The linear regression equation is,577.97 77,75.5. When,,6. Accoring to the linear regression equation, the investment will grow to approimatel $,6. 65. (a) Because the circumference of the original circle was 8 an a piece of length was remove. (b) r 8 (c) h 6 r (b).. 6 6 6 6 6. 6. (a) Since ( f g)() g() 5 5, g(). (b) Since ( f g)(), we know that g( ), so g(). g( ) (c) Since ( f g)() f, f (). () V r h 8 6 (8 ) 6
Section. 66. (a) Note that mi,56 ft, so there are 8 feet of river cable at $8 per foot an (,56 ) feet of lan cable at $ per foot. The cost is C() 88 (,56 ) (b) C() $,, C(5) $,75,8 C() $,86,5 C(5) $,, C() $,,7 C(5) $,78,79 C() $,,87 Values beon this are all larger. It woul appear that the least epensive location is less than ft from point P. 67. (a) [, ] b [, ] (b) Domain of : [, ) Domain of :(, ] Domain of : [, ] (c) The functions,, an all have omain [, ], the same as the omain of foun in part (b). Domain of : [, ) Domain of : (, ] () The omain of a sum, ifference, or prouct of two functions is the intersection of their omains. The omain of a quotient of two functions is the intersection of their omains with an zeros of the enominator remove. 68. (a) Yes. Since ( f g)() f () g() f () g() ( f g)(), the function ( f g)() will also be even. (b) The prouct will be even, since ( f g)() f () g() (f ()) (g()) f () g() ( f g)(). Section. Eponential Functions (pp. 6) 5. [5, 5] b [, 5] 6. 5 for ; 5 for ; 5 for. Quick Review.. Using a calculator, 5 /.9.. Using a calculator,.79.. Using a calculator,.5.9.. 7 7.57 5. 5 5.888 6..567.567.8 7. 5(.75) 5 $6.58 8. (.6) $.6 6 9. ( ) 9 ( ) 6 9 b. a c 8 5 8 5 a b c a 6 b b 8 c a c 6 a b 8 b c c a a 6 b c 8 a b c 6 a b c 6 Section. Eercises. The graph of is increasing from left to right an has the negative -ais as an asmptote. (a). The graph of or, equivalentl,, is Eploration. Eponential Functions ecreasing from left to right an has the positive -ais as an asmptote. () [5, 5] b [, 5].... The graph of is the reflection about the -ais of the graph in Eercise. (e). The graph of.5 or, equivalentl,, is the reflection about the -ais of the graph in Eercise. (c) 5. The graph of is ecreasing from left to right an has the line as an asmptote. (b) 6. The graph of.5 is increasing from left to right an has the line as an asmptote. (f)
Section. 7. 6. [, ] b [8, 6] Domain: (, ) Range: (, ) -intercept:.585 -intercept: 8. [6, 6] b [, 6].86 7. [, ] b [, ] Domain: (, ) Range: (, ) -intercept: None -intercept: 9. [, ] b [, 8] Domain: (, ) Range: (, ) -intercept:.5 -intercept:. [6, 6] b [, 5].69 8. [6, 6] b [, 5].585 9. 5 [, ] b [8, ] Domain: (, ) Range: (, ) -intercept: None -intercept:. 9 ( ). 6 ( ). 8 ( ) 6. 7 ( ) 5. [6, 6] b [, 6].9.. 5 8 5 9 7 6
Section. 5. ratio 8.55.67 6.57 6.79. Let t be the number of ears. Solving 5,(.75) t,, graphicall, we fin that t 8.88. The population will reach million in about 9 ears.. (a) The population is given b P(t) 65(.75) t, where t is the number of ears after 89. Population in 95: P(5),5 Population in 9: P(5),65 (b) Solving P(t) 5, graphicall, we fin that t 76.65. The population reache 5, about 77 ears after 89, in 967. 5. (a) A(t) 6.6 t/ (b) Solving A(t) graphicall, we fin that t 8.5. There will be gram remaining after about 8.5 as. 6. Let t be the number of ears. Solving (.6) t 5 graphicall, we fin that t.9. It will take about.9 ears. (If the interest is not creite to the account until the en of each ear, it will take ears.) 7. Let A be the amount of the initial investment, an let t be the number of ears. We wish to solve A(.65) t A, which is equivalent to.65 t. Solving graphicall, we fin that t.. It will take about. ears. (If the interest is creite at the en of each ear, it will take ears.) 8. Let A be the amount of the initial investment, an let t be the number of ears. We wish to solve A..78.78.78 65 t A, which is equivalent to. 65 t. Solving graphicall, we fin that t.9. It will take about.9 ears. (If the interest is creite at the en of each month, it will take ears months.) 9. Let A be the amount of the initial investment, an let t be the number of ears. We wish to solve Ae.65t A, which is equivalent to e.65t. Solving graphicall, we fin that t.9. It will take about.9 ears.. Let A be the amount of the initial investment, an let t be the number of ears. We wish to solve A(.575) t A, which is equivalent to.575 t. Solving graphicall, we fin that t 9.65. It will take about 9.65 ears. (If the interest is creite at the en of each ear, it will take ears.). Let A be the amount of the initial investment, an let t be the number of ears. We wish to solve A. 575 65t 65 A, which is equivalent to. 575 65t 65. Solving graphicall, we fin that t 9.8. It will take about 9.8 ears.. Let A be the amount of the initial investment, an let t be the number of ears. We wish to solve Ae.575t A, which is equivalent to e.575t. Solving graphicall, we fin that t 9.6. It will take about 9.6 ears.. After t hours, the population is P(t) t/.5 or, equivalentl, P(t) t. After hours, the population is P() 8.85 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 fin that t.9. It will take about.9 ears. (b) Solving C(t) graphicall, we fin that t.75. It will take about.75 ears. 5. Since, the corresponing value of is equal to the slope of the line. If the changes in are constant for a linear function, then the corresponing changes in are constant as well. 6. (a) When t, B e. There were bacteria present initiall. (b) When t 6, B e.69(6) 69.5. After 6 hours, there are about 69 bacteria. (c) Solving e.69t graphicall, we fin that t.. The population will be after about hour. Since the population oubles (from to ) in about hour, the oubling time is about hour. 7. (a) Let represent 9, represent 9, an so on. The regression equation is P() 6.(.). [, ] b [, 9] (b) The regression equation gives an estimate of P() 6. million, which is not ver close to the actual population. (c) Since the equation is of the form P() P()., the annual rate of growth is about %. 8. (a) The regression equation is P().8(.9). [, ] b [5, ] (b) P(9) 6. million (c) Since the equation is of the form P() P().9, the annual rate of growth is approimatel.9%.
6 Section. 9. 5(.8) 9 769.7 million. (a) [5, 5] b [, ] In this winow, it appears the cross twice, although a thir crossing off-screen appears likel. (b) change in Y change in Y It happens b the time. (c) Solving graphicall,.7667,,. () The solution set is approimatel (.7667, ) (, ).. Since f ().5 we have ka.5, an since f ().5 we have ka.5. Diviing, we have ka ka. 5. 5 a 9 a Since f () k a is an eponential function, we require a, so a. Then ka.5 gives k.5, so k.5. The values are a an k.5.. Since f ().5 we have ka.5, an since f () 6 we have ka 6. Diviing, we have ka ka.5 6 a.5 5 7 8 a.5 Since f () k a is an eponential function, we require a, so a.5. Then ka.5 gives.5k.5, so k. The values are a.5 an k. Section. Parametric Equations (pp. 6 ) Eploration Parametrizing Circles. Each is a circle with raius a. As a increases, the raius of the circle increases. [.7,.7] b [.,.]. t : t : t : [.7,.7] b [.,.] [.7,.7] b [.,.] [.7,.7] b [.,.] t : t : [.7,.7] b [.,.] [.7,.7] b [.,.] Let be the length of the parametric interval. If, ou get of a complete circle. If, ou get the complete circle. If, ou get the complete circle but portions of the circle will be trace out more than once. For eample, if the entire circle is trace twice.
Section. 7. Eploration. a, b : Parametrizing Ellipses t initial point: (, ) terminal point: (, ) a, b : [, ] b [8, 8] t initial point: (, ) terminal point: (, ) a, b 5: [, ] b [8, 8] a, b 6: [, ] b [8, 8] t initial point: (, ) terminal point: (, ). a, b : [, ] b [8, 8] t 5 initial point: (, ) terminal point: (, ). For t the complete circle is trace once clockwise beginning an ening at (, ). For t the complete circle is trace once clockwise beginning an ening at (, ). For t the half circle below is trace clockwise starting at (, ) an ening at (, ). a 5, b : a 6, b : [9, 9] b [6, 6] [9, 9] b [6, 6] a 7, b : [9, 9] b [6, 6] [9, 9] b [6, 6]. If a b, then the major ais is on the -ais an the minor on the -ais. If a b, then the major ais is on the -ais an the minor on the -ais.
8 Section.. t : t : [6, 6] b [, ] Eploration Graphing the Witch of Agnesi. We use the parameter interval [, ] because our graphing calculator ignore the fact that the curve is not efine when t or. The curve is trace from right to left across the screen. ranges from to.. t : [6, 6] b [, ] [5, 5] b [, ] t : t : t : [6, 6] b [, ] t : [5, 5] b [, ] [6, 6] b [, ] Let be the length of the parametric interval. If, ou get of a complete ellipse. If, ou get the complete ellipse. If, ou get the complete ellipse but portions of the ellipse will be trace out more than once. For eample, if the entire ellipse is trace twice. 5. t : [5, 5] b [, ] For t, the entire graph escribe in part is rawn. The left branch is rawn from right to left across the screen starting at the point (, ). Then the right branch is rawn from right to left across the screen stopping at the point (, ). If ou leave out an, then the point (, ) is not rawn. [6, 6] b [, ] initial point: (5, ) terminal point: (5, ) t : For t, the right branch is rawn from right to left across the screen stopping at the point (, ). If ou leave out, then the point (, ) is not rawn. For t, the left branch is rawn from right to left t : [6, 6] b [, ] initial point: (5, ) terminal point: (5, ) [6, 6] b [, ] initial point: (, ) terminal point: (, ) Each curve is trace clockwise from the initial point to the terminal point. across the screen starting at the point (, ). If ou leave out, then the point (, ) is not rawn.. If ou replace cot t b cot t, the same graph is rawn ecept it is trace from left to right across the screen. If ou replace cot t b cot ( t), the same graph is rawn ecept it is trace from left to right across the screen. Quick Review.. m 8 5 5 5 ( ) 8 5 9
Section. 9... When, we have, so the -intercepts are 9 an. When, we have, so the -intercepts are 6 an. 5. When, we have, so the -intercepts are 6 an. When, we have, which has no real 9 solution, so there are no -intercepts. Section. Eercises. Graph (c). Winow: [, ] b [, ], t. Graph (a). Winow: [, ] b [, ], t. Graph (). Winow: [, ] b [, ], t. Graph (b). Winow: [5, 5] b [5, 5], t 5. (a) The resulting graph appears to be the right half of a hperbola in the first an fourth quarants. The parameter a etermines the -intercept. The parameter b etermines the shape of the hperbola. If b is smaller, the graph has less steep slopes an appears sharper. If b is larger, the slopes are steeper an the graph appears more blunt. The graphs for a an b,, an are shown. 6. When, we have, so the -intercept is. When, we have, so the -intercepts are an. 7. (a) () () + Yes (b) () () () No (c) () () Yes (b) (c) [, ] b [, ] [, ] b [, ] This appears to be the left half of the same hberbola. 8. (a) 9() 8() () 7 9 8 6 7 7 7 Yes (b) 9() 8() () 7 9 8 6 7 7 7 Yes (c) 9() 8() () 7 9 8 6 7 6 7 No 9. (a) t 5 (b) t 5 t 5 t t t t. (a) The equation is true for a. (b) The equation is equivalent to a a or a a. Since a a is true for a an 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. (c) The equation is true for all real values of a. [, ] b [, ] One must be careful because both sec t an tan t are iscontinuous at these points. This might cause the grapher to inclue etraneous lines (the asmptotes of the hperbola) in its graph. The etraneous lines can be avoie b using the grapher s ot moe instea of connecte moe. () Note that sec t tan t b a stanar trigonometric ientit. Substituting a for sec t an b for tan t gives a b. (e) This changes the orientation of the hperbola. In this case, b etermines the -intercept of the hperbola, an a etermines 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 iscontinuities an ma a etraneous lines to the graph. Substituting for sec t b an for tan t in the ientit sec t tan t gives a b a.
Section. 6. (a). (a) [6, 6] b [, ] The graph is a circle of raius centere at (h, ). As h changes, the graph shifts horizontall. (b) [.7,.7] b [.,.] Initial an terminal point: (, ) (b) cos t sin t The parametrize curve traces all of the ellipse efine b. [6, 6] b [, ] The graph is a circle of raius centere at (, k). At k changes, the graph shifts verticall. (c) Since the circle is to be centere at (, ), we use h an k. Since a raius of 5 is esire, we nee to change the coefficients of cos t an sin t to 5. 5 cos t, 5 sin t, t () 5 cos t, sin t, t 7. (a). (a) [.7,.7] b [.,.] Initial point: (, ) Terminal point: (, ) (b) sin t cos t [, ] b [, ] Initial point: (, ) Terminal point: (, ) (b) cos t sin t The parametrize curve traces the upper half of the circle efine b (or all of the semicircle efine b ). 8. (a). (a) The parametrize curve traces the right half of the ellipse efine b (or all of the curve efine b ). [9, 9] b [6, 6] Initial an terminal point: (, 5) (b) 5 sin t cos t [, ] b [, ] Initial an terminal point: (, ) (b) sin (t) cos (t) The parametrize curve traces all of the circle efine b. 9. (a). (a) The parametrize curve traces all of the ellipse efine b 5. [, ] b [, ] Initial point: (, ) Terminal point: (, ) (b) cos ( t) sin ( t) The parametrize curve traces the upper half of the circle efine b (or all of the semicircle efine b ). [, ] b [, ] No initial or terminal point. (b) 9t (t) The parametrize curve traces all of the parabola efine b.. (a) [, ] b [, ] Initial point: (, ) Terminal point: None
Section. (b) t (t) The parametrize curve traces the left half of the parabola efine b (or all of the curve efine b ). 5. (a) [, 5] b [, ] Initial point: (, ) Terminal point: None (b) t The parametrize curve traces all of the curve efine b (or the upper half of the parabola efine b ). 6. (a) [, 9] b [, ] No initial or terminal point. (b) sec t tan t The parametrize curve traces all of the parabola efine b. 7. (a) [, ] b [, ] No initial or terminal point. Note that it ma be necessar to use a t-interval such as [.57,.57] or use ot moe in orer to avoi asmptotes showing on the calculator screen. (b) sec t tan t The parametrize curve traces the left branch of the hperbola efine b (or all of the curve efine b ). 8. (a) 9. (a) [9, 9] b [6, 6] No initial or terminal point. (b) t 7 (t 5) The parametrize curve traces all of the line efine b.. (a) [6, 6] b [, ] No initial or terminal point. (b) t ( t) The parametrize curve traces all of the line efine b.. (a) [, ] b [, ] Initial point: (, ) Terminal point: (, ) (b) t The Cartesian equation is. The portion trace b the parametrize curve is the segment from (, ) to (, ).. (a) [, ] b [, ] Initial point: (, ) Terminal point: (, ) (b) t (t ) ( t) The Cartesian equation is. The portion trace b the curve is the segment from (, ) to (, ). [6, 6] b [5, ] No initial or terminal point. Note that it ma be necessar to use a t-interval such as [.57,.57] or use ot moe in orer to avoi asmptotes showing on the calculator screen. (b) sec t tan t The parametrize curve traces the lower branch of the hperbola efine b (or all of the curve efine b ).. (a) [6, 6] b [, 6] Initial point: (, ) Terminal point: None (b) t ( t) The parametrize curve traces the portion of the line efine b to the left of (, ), that is, for.
Section.. (a) [, 5] b [, ] Initial point: (, ) Terminal point: (, ) (b) t The parametrize curve traces the right portion of the curve efine b, that is, for. 5. (a) [, ] b [, ] No initial or terminal point, since the t-interval has no beginning or en. The curve is trace an retrace in both irections. (b) cos t cos t sin t sin t The parametrize curve traces the portion of the parabola efine b corresponing to. 6. (a) [, 5] b [, ] Initial point: None Terminal point: (, ) (b) t The parametrize curve traces the lower half of the parabola efine b (or all of the curve efine b ). 7. Using (, ) we create the parametric equations at an bt, representing a line which goes through (, ) at t. We etermine a an b so that the line goes through (, ) when t. Since a, a 5. Since b, b. Therefore, one possible parametrization is 5t, t, t. 8. Using (, ) we create the parametric equations at an bt, representing a line which goes through (, ) at t. We etermine a an b so that the line goes through (, ) at t. Since a, a. Since b, b 5. Therefore, one possible parametrization is t, 5t, t. 9. The lower half of the parabola is given b for. Substituting t for, we obtain one possible parametrization: t, t, t.. 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.. For simplicit, we assume that an are linear functions of t an that the point (, ) starts at (, ) for t an passes through (, ) at t. Then f(t), where f() an f(). Since slope, t f(t) t t. Also, g(t), where g() an g(). Since slope, t g(t) t t. One possible parametrization is: t, t, t.. For simplicit, we assume that an are linear functions of t an that the point (, ) starts at (, ) for t an passes through (, ) at t. Then f(t), where f() an f(). Since slope (), t f (t) t () t. Also, g(t), where g() an g(). Since slope, t g(t) t t. One possible parametrization is: t, t, t.. The graph is in Quarant I when, which correspons to t. To confirm, note that () an ().. The graph is in Quarant II when, which correspons to t 5. To confirm, note that () an (5). 5. The graph is in Quarant III when 6, which correspons to 5 t. To confirm, note that (5) an (). 6. The graph is in Quarant IV when, which correspons to t. To confirm, note that () an (). 7. The graph of lies in Quarant I for all. Substituting t for, we obtain one possible parametrization: t, t t, t. 8. The graph of lies in Quarant I for all. Substituting t for, we obtain one possible parametrization: t, t, t.
Section.5 9. Possible answers: (a) a cos t, a sin t, t (b) a cos t, a sin t, t (c) a cos t, a sin t, t () a cos t, a sin t, t. Possible answers: (a) a cos t, b sin t, t (b) a cos t, b sin t, t (c) a cos t, b sin t, t () a cos t, b sin t, t. Note that moaq t, since alternate interior angles forme b a transversal of parallel lines are congruent. Therefore, tan t O Q AQ,so cot t. tan t Now, b equation (iii), we know that AB (A Q) AO A Q AO (AQ) (cos t)() (cos t)( cot t) cos t. sin t Then equation (ii) gives AB sin t cos t sin t cos t sin t sin t. Section.5 Functions an Logarithms (pp. ) Eploration Testing for Inverses Graphicall. It appears that ( f g)() (g f )(), suggesting that f an g ma be inverses of each other. (a) f an g: (b) f g: (c) g f : [.7,.7] b [.,.] [.7,.7] b [.,.] [.7,.7] b [.,.]. It appears that f g g f g, suggesting that f ma be the ientit function. (a) f an g: The parametric equations are: cot t, sin t, t Note: Equation (iii) ma not be immeiatel obvious, but it (b) f g: [.7,.7] b [.,.] ma be justifie as follows. Sketch segment QB. Then OBQ is a right angle, so ABQ AQO, which gives AB A Q A Q AO.. (a) If then the line is a vertical line an the first parametric equation gives, while the secon 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 secon equation gives [( )/( )]( ) which is the point-slope form of the equation for the line through (, ) an (, ). 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 trace out. (b) Use the equations for an given in part (a), with t. (c) g f : [.7,.7] b [.,.] [.7,.7] b [.,.]. It appears that ( f g)() (g f )(), suggesting that f an g ma be inverses of each other. (a) f an g: (b) f g: [.7,.7] b [.,.] [.7,.7] b [.,.]
Section.5. continue (c) g f : [.7,.7] b [.,.]. It appears that ( f g)() (g f )(), suggesting that f an g ma be inverse of each other. (Notice that the omain of f g is (, ) an the omain of g f is (, ).) (a) f an g: 5. Substituting t for, one possible answer is: t,, t. t 6. Substituting t for, one possible answer is: t, t, t. 7. [, ] b [, ] (, 5) 8. (b) f g: [.7,.7] b [.,.] (c) g f : [.7,.7] b [.,.] 9. (a) [, ] b [, ] 8, (.67, ) [.7,.7] b [.,.] Eploration Supporting the Prouct Rule. The appear to be vertical translations of each other. [, ] b [, ] (.58, ) (b) No points of intersection, since for all values of.. (a) [, 8] b [, ]. This graph suggests that each ifference ( ) is a constant. [, 8] b [, ]. ln (a) ln ln a ln ln ln a Thus, the ifference is the constant ln a. Quick Review.5. ( f g)() f(g()) f (). (g f )(7) g( f(7)) g() 5. ( f g)() f(g()) f ( ) ( ) /. (g f )() g( f()) g( ) [, ] b [, ] (.9, ) (b) No points of intersection, since e for all values of. Section.5 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.5 intersects the graph twice. 5. 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. ( ) ( ) /
Section.5 5 7. [, ] b [, ] Yes, the function is one-to-one since each horizontal line intersects the graph at most once, so it has an inverse function. 8. [, ] b [, ] No, the function is not one-to-one since (for eample) the horizontal line intersects the graph twice, so it oes not have an inverse function. 9. [, ] b [, ] No, the function is not one-to-one since (for eample) the horizontal line 5 intersects the graph more than once, so it oes not have an inverse function.. [5, 5] b [, ] Yes, the function is one-to-one since each horizontal line intersects the graph onl once, so it has an inverse function.. [, ] b [, ] No, the function is not one-to-one since each horizontal line intersects the graph twice, so it oes not have an inverse function.. [9, 9] b [, ] Yes, the function is one-to-one since each horizontal line intersects the graph at most once, so it has an inverse function.. Interchange an. f (). Verif. ( f f )() f ( ) ( f f )() f ( ) ( ). 5 5 5 Interchange an. 5 f () 5 Verif. ( f f )() f 5 5 5 5 (5 ) ( f f )() f (5 ) 5 (5 ) 5. ( ) / Interchange an. ( ) / f () ( ) / or Verif. ( f f )() f ( ) ( ) ( ) ( f f)() f ( ) ( )
6 Section.5 6.,, Interchange an. f () or ( ) / Verif. For (the omain of f ), ( f f )() f( ) ( ) ( ) For, (the omain of f), ( f f)() f ( ) ( ) 7., Interchange an. f () or / Verif. For (the omain of f ), ( f f )() f() () For, (the omain of f ), ( f f)() f ( ) 8. /, / ( / ) /, / Interchange an. / f () / Verif. For (the omain of f ), ( f f )() f ( / ) ( / ) / for, (the omain of f ), ( f f )() f ( / ) ( / ) / 9. ( ), ( ), Interchange an. f () or () / Verif. For (the omain of f ) ( f f )() f ( ) [( ) ] () For (the omain of f), ( f f )() f (( ) ) ( ) ( ). ( ), ( ), Interchange an. f () or / Verif. For (the omain of f ), (f f )() f ( ) [( ) ( ) ] () () For (the omain of f), (f f )() f ( ) ( ) ( ).,, Interchange an. f () or / Verif. For (the omain of f ), ( f f )() f (/ ) For (the omain of f), (f f)() f /. Interchange an. f () or / Verif. ( f f )() f (/) ( f f)() f /
Section.5 7. ( ) Interchange an. f () Verif. ( f f )() f ( ) ( ) ( ) ( ) 5 5 ( f f)() f ( ) ( ) ( ) ( ) 5 5. ( ) Interchange an. f () Verif. ( f f )() f ( ) ( ) ( ) ( ) 5 5 ( f f)() f ( ) ( ) ( ) ( ) 5 5 5. Graph of f: t, e t Graph of f : e t, t Graph of : t, t [6, 6] b [, ] 6. Graph of f: t, t Graph of f : t, t Graph of : t, t [6, 6] b [, ]
8 Section.5 7. Graph of f: t, t Graph of f : t, t Graph of : t, t. [.5,.5] b [, ] 8. Graph of f: t, t Graph of f : t, t Graph of : t, t [.5,.5] b [, ] 9. Graph of f: t, ln t Graph of f : ln t, t Graph of : t, t. 5. 6. [, 5] b [7, ] Domain: (, ) Range: (, ) [5, ] b [5, 5] Domain: (, ) Range: (, ) [, 6] b [, ] Domain: (, ) Range: (, ) [.5,.5] b [, ]. Graph of f: t, log t Graph of f : log t, t Graph of : t, t [.5,.5] b [, ]. Graph of f: t, sin t Graph of f : sin t, t Graph of : t, t [, ] b [, ] Domain: (, ) Range: (, ) 7. (.5) t ln(.5) t ln t ln.5 ln ln t 5.75 ln.5 Graphical support: [, 8] b [, ] [, ] b [, ]. Graph of f: t, tan t Graph of f : tan t, t Graph of : t, t 8. e.5t ln e.5t ln.5t ln ln t ln.97.5 Graphical support: [6, 6] b [, ] [5, 5] b [, ]
Section.5 9 9. e e e e e (e e ) e () (e ) e e () ()( ) () e 5 ln 5.96 or.96 Graphical support: [, ] b [, 8]. 5 5 ( 5 ) () ( ) 5( ) 5 (5) ()() () 5 log 5.6 or.6 Graphical support:. log ( ) log log log log log Interchange an. log f () log Verif. ( f f )() f log log log [, ] b [, 8]. ln t e ln t e t e. ln( ) ln ln ln( ) ln ln e ln() ln ln e e ()() e ( ) ( f f )() f log log ( ) log log ( )
Section.5 5... 5. 5 log. (. ) log. 5 log. 5 log. 5 Interchange an : log. 5 f () log. 5 Verif. ( f f )() f log. 5 log. 5 log. 5 5. (a) f ( f()) (f ) ( ) ( ) (b) f( f()) f 6. (a) Amount 8 t/ (b) 8 t/ t/ 8 t/ t t 6, since for all / There will be gram remaining after 6 hours. 5. log. 5 5. log. 5 5 5 5 5 ( 5 ) 5 ( f f )() f 5. 5 log.. 5 5. 5 log. 5(. ) 5 log.. log. (. ) 7. 5(.75) t.75 t ln(.75 t ) ln t ln.75 ln ln t.96 ln.75 It will take about.96 ears. (If the interest is pai at the en of each ear, it will take 5 ears.) 8. 75,(.5) t,,.5 t 8 ln(.5 t ) ln 8 t ln.5 ln 8 ln( 8/ ) t.8 ln. 5 It will take about.8 ears. 9. (a) 59.85 66.896 ln (b) When 75, 9.9. About 9.9 million metric tons were prouce. (c) 59.85 66.896 ln 66.896 ln 99.85 ln 99. 85 66. 896 e 6 9 6 9.8.8 9 5 6.8 Accoring to the regression equation, Saui Arabian oil prouction will reach million metric tons when.8, in about.
Section.6 5. (a) 59.969 5.87 ln (b) When 85, 87.9. About 87.9 million metric tons were prouce. (c) 59.969 5.87 ln 5.87 ln 7.969 ln 7. 969 5. 87 e 7 5.. 9 8 6 9 7.8 Accoring to the regression equation, oil prouction will reach million metric tons when.8, in about 5. 5. (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 an. b m f () b m The slopes are reciprocals. (c) 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. () If the original functions have slopes m an, m respectivel, then the inverse functions will have slopes an m, respectivel. Since each of m an m is m the negative reciprocal of the other, the graphs of the inverses will be perpenicular lines with nonzero slopes. 5. (a) is a vertical shift (upwar) of, although it s ifficult 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. (c) The graphs of an a are the same. () e a, ln(e ) ln a, ln a, ln a ln ln a 5. If the graph of f () passes the horizontal line test, so will the graph of g() f() since it s the same graph reflecte about the -ais. Alternate answer: If g( ) g( ) then f ( ) f ( ), f ( ) f ( ), an since f is one-to-one. 5. Suppose that g( ) g( ). Then, f( ) f( ) f ( ) f ( ), an an since f is one-to-one. (b) The epression a log b ( c) is efine when c, so the omain is (c, ). Since a log b ( c) attains ever real value for some value of, the range is (, ). 56. (a) Suppose f ( ) f ( ). Then: a b a b c c (a b)(c ) (a b)(c ) ac a bc b ac a bc b a bc a bc (a bc) (a bc) Since a bc, this means that. (b) a b c c a b (c a) b b c a Interchange an : b c a f () b c a (c) As, f () a c b a c, so the horizontal asmptote is a c (c ). Since f() is unefine at c, the vertical asmptote is c. () As, f () c b a c, so the horizontal asmptote is c (c ). Since f () is unefine at a c, the vertical asmptote is a c. The horizontal asmptote of f becomes the vertical asmptote of f an vice versa ue to the reflection of the graph about the line. Section.6 Trigonometric Functions (pp. 5) Eploration Unwrapping Trigonometric Functions. (, ) is the circle of raius centere at the origin (unit circle). (, ) is one perio of the graph of the sine function. 55. (a) The epression a(b c ) is efine for all values of, so the omain is (, ). Since b c attains all positive values, the range is (, ) if a an the range is (, ) if a.
Section.6. The -values are the same in the interval t.. The -values are the same in the interval t.. The -values an the -values are the same in each interval. 5. tan t: csc t:. [, ] b [, ] Using trace, cos t an sin t are being compute for, 5,,, 6 egrees. Quick Review.6. 8 6..5 8 5. sec t: cot t:. 8 9. 5 8 5. 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. [, ] b [.5,.5].65,.98 6. [, ] b [.5,.5].98,.9 7. For each value of t, the value of cot t is equal to the ratio. Eploration Fining Sines an Cosines. The ecimal viewing winow [.7,.7] b [.,.] is square on the TI-8/8 an man other calculators. There are man other possibilities. [.7,.7] b [.,.]. Using the Ask table setting for the inepenent variable on the TI-8 we obtain, b [, ].785 or,.97 or 5 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).. Section.6 Eercises. Arc length 5 8 () 5 7. Raius.7 75 8 7
Section.6 7. Angle raian or about 8.65. Angle / raian or 5 6 5. (a) The perio of sec is, so the winow shoul have length. One possible answer: [, ] b [, ] (b) The perio of csc is, so the winow shoul have length. One possible answer: [, ] b [, ] (c) The perio of cot is, so the winow shoul have length. One possible answer: [, ] b [, ] 6. (a) The perio of sin is, so the winow shoul have length. One possible answer: [, ] b [, ] (b) The perio of cos is, so the winow shoul have length. One possible answer: [, ] b [, ] (c) The perio of tan is, so the winow shoul have length. One possible answer: [, ] b [, ] 7. Since 6 is in the range, of sin an sin 6.5, sin (.5) 6 raian or 6 8. 8. Since is the range, of sin an sin, sin raian or 8 5. 9. Using a calculator, tan (5).7 raians or 78.69.. Using a calculator, cos (.7).795 raian or 5.57.. (a) Perio (b) Amplitue.5 (c) [, ] b [, ]. (a) Perio (b) Amplitue (c), b [, ]. (a) Perio (b) Amplitue (c) [, ] b [, ]. (a) Perio / (b) Amplitue 5 (c) [, ] b [, ] 5. (a) Perio 6 / (b) Amplitue (c) [, ] b [5, 5] 6. (a) Perio (b) Amplitue (c) [, ] b [, ] 7. (a) Perio (b) Domain: Since csc ( ), we require sin ( ) k, or (k ). This requirement is equivalent to k for integers k. (c) Since csc ( ), the range eclues numbers between 5 an. The range is (, 5] [, ). (), b [8, 8] 8. (a) Perio (b) Domain: (, ) (c) Since sin ( ), the range etens from to 5. The range is [, 5]. (), b [8, 8] 9. (a) Perio (b) Domain: We require k for o integers k. Therefore, (k ) for o integers k. This 6 requirement is equivalent to k for o integers k. 6 (c) Since the tangent function attains all real values, the range is (, ). (), b [8, 8]. (a) Perio (b) Domain: (, ) (c) Range: Since sin, the range is [, ]. () [, ] b [, ]
Section.6. Note that 8 5 7. 8 Since sin an 7, cos sin 8 5. 7 7 8 Therefore: sin, cos 5 sin 8, tan, 7 7 c os 5 cot 5, sec 7, csc 7 tan 8 co s 5 sin 8. Note that 5. 5 Since tan 5/ sin an / c os,we 5 have sin an cos. In summar: 5 sin, cos 5, tan, cot, sec, tan 5 co s csc sin 5. Note that r () 5. Then: sin r 5, cos r 5, tan, cot, sec r 5, csc r 5. Note that r (). Then: sin r cos r, tan, cot, r r sec, csc 5. The angle tan (.5).9 is the solution to this 9. The solutions in the interval are 7 an 6. Since sin has perio, the solutions are 6 all of the form 7 k or k, where k 6 6 is an integer.. The equation is equivalent to tan, so the solution in the interval is tan (). Since the perio 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.. Let cos 7. Then an cos 7, so sin cos 7 sin cos 7 6.77. 7. Let sin 9. Then an sin 9, so cos sin 9 8 8. Therefore, tan sin 9 tan c sin 9/ 9.959. os 8 8 / 88. (a) Using a graphing calculator with the sinusoial regression feature, the equation is.5 sin (68.65.9).8. equation in the interval. Another solution in is tan (.5).. The solutions are.9 an.. 6. 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) 8.69 an cos (.7).. 7. This equation is equivalent to sin, so the solutions in the interval are 6 an 5. 6 8. 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 [,.] b [.5,.5] (b) The frequenc is 68.65 raians per secon, which is equivalent to 6 8.65 9.9 ccles per secon (Hz). The note is a G.. (a) b 6 (b) It s half of the ifference, so a 8 5. (c) k 8 55 () The function shoul have its minimum at t (when the temperature is F) an its maimum at t 8 (when the temperature is 8F). The value of h is 8 5. Equation: 5 sin 6 ( 5) 55 interval are.9 an.9.
Section.6 5 (e) [, ] b [, ] 5. (a) Amplitue 7 (b) Perio 65 ( / 65) (c) Horizontal shift () Vertical shift 5 6. (a) Highest: 5 7 6F Lowest: 5 7 F (b) Average 6 () 5F This average is the same as the vertical shift because the average of the highest an lowest values of sin is. 7. (a) cot () c os ( ) cos () cot () sin ( ) sin () (b) Assume that f is even an g is o. f( ) f( ) f( ) f Then so is o. The situation g ( ) g( ) g ( ) g is similar for g f. 8. (a) csc () csc () sin ( ) si n() (b) Assume that f is o. Then so f( ) f () f( ) f is o. 9. Assume that f is even an g is o. Then f ()g() ( f())(g()) f ()g() so ( fg) is o.. If (a, b) is the point on the unit circle corresponing to the angle, then (a, b) is the point on the unit circle corresponing to the angle ( ) since it is eactl half wa aroun the circle. This means that both tan () an tan ( ) have the same value, b a.. (a) Using a graphing calculator with the sinusoial regression feature, the equation is. sin (.9996.).9999. (b) sin ( ). (a) (b) Amplitue. (that is, ) (c) Perio Horizontal shift.785 that is, or 5.98 that is, 7 Vertical shift: sin (sin ) cos (cos ) sin (sin ) (cos ) (sin cos ) Therefore, sin cos sin.. (a) sin a (b) See part (a). (c) It works. () sin a (sin a) cos (cos a) sin (sin a) (cos a) (sin a cos a) So, sin (a) cos (a) sin a.. (a) One possible answer: a b sin tan b a (b) See part (a). (c) It works. () sin tan b a sin () cos tan b a cos () sin tan b a sin () a a cos () b a b b (a sin b cos ) a b an multipling through b the square root gives the esire result. Note that the substitutions cos tan b a a a an b sin tan b a b epen on the requirement a b that a is positive. If a is negative, the formula oes not work. 5. Since sin has perio, sin ( ) sin (). This function has perio. A graph shows that no smaller number works for the perio. [, ] b [, ] The graph is a sine/cosine tpe graph, but it is shifte an has an amplitue greater than. [, ] b [.5,.5]
6 Chapter Review 6. Since tan has perio, tan ( ) tan. This function has perio. A graph shows that no smaller number works for the perio. 9. Since is equivalent to, the slope of the given line (an hence the slope of the esire line) is. [, ] b [, 5] 7. The perio is 6. One possible graph: ( ). Since 5 is equivalent to 5, the slope of 5 the given line is 5 an the slope of the perpenicular line is 5., b [, ] 6 6 8. The perio is. One possible graph: 6 5 ( ) 5 9. Since is equivalent to, the slope of the given line is an the slope of the perpenicular line is. ( ), b [, ] 6 6 Chapter Review Eercises (pp. 5 5). ( ) (6) 9. ( ). 6. m 8 ( ) ( ) 6 5. 5 6. m 5 5 ( ) 5 5 5 7. 8. Since is equivalent to, the slope of the given line (an hence the slope of the esire line) is. ( ) 5 8. The line passes through (, 5) an (, ) m (5) 5 5 5. m () f () ( ) f () Check: f() (), as epecte.. The line passes through (, ) an (, ). m ( ) 7 7 ( ) 7 5. 7 6 7 [, ] b [, ] Smmetric about the origin. 6. [, ] b [, ] Smmetric about the -ais.
Chapter Review 7 7. [6, 6] b [, ] Neither 8. 9. (a) Since the square root requires 6, the omain is [, ]. (b) For values of in the omain, 6 6, so 6. The range is [, ]. (c) [.5,.5] b [.5,.5] Smmetric about the -ais. 9. () () () Even. () () 5 () () 5 () O. () cos() cos () Even. () sec () tan () sin ) c os (( sin ) cos sec tan () O ( ). () ( ) () () O. () sin () sin Neither even nor o 5. () cos () cos Neither even nor o 6. () () Even 7. (a) The function is efine for all values of, so the omain is (, ). (b) Since attains all nonnegative values, the range is [, ). (c) [9., 9.] b [6., 6.]. (a) The function is efine for all values of, so the omain is (, ). (b) Since attains all positive values, the range is (, ). (c) [6, 6] b [, ]. (a) The function is efine for all values of, so the omain is (, ). (b) Since e attains all positive values, the range is (, ). (c) [, ] b [5, 5]. (a) The function is equivalent to tan, so we require k for o integers k. The omain is given b k for o integers k. (b) Since the tangent function attains all values, the range is (, ). (c) [, ] b [, ] 8. (a) Since the square root requires, the omain is (, ]. (b) Since attains all nonnegative values, the range is [, ). (c), b [8, 8]. (a) The function is efine for all values of, so the omain is (, ). (b) The sine function attains values from to, so sin ( ), an hence sin ( ). The range is [, ]. (c) [9., 9.] b [, ] [, ] b [5, 5]