x is a square root function, not an exponential function. h(x) = 2 x h(5) = 2 5 h(5) = 32 h( 5) = 2 5 h(0) = 2 0

Transcription

1 Chaper 7 Eponenial Funcions Secion 7. Characerisics of Eponenial Funcions Secion 7. Page 4 Quesion a) The funcion y = is a polynomial funcion, no an eponenial funcion. b) The funcion y = 6 is an eponenial funcion. The base is greaer han 0 and he independen variable is he eponen. c) The funcion y = is a square roo funcion, no an eponenial funcion. d) The funcion y = 0.75 is an eponenial funcion. The base is greaer han 0 and he independen variable is he eponen. Secion 7. Page 4 Quesion a) For = 5, f() = 4 g() = 4 h() = f(5) = 4 5 g(5) = 4 h(5) = 5 f(5) = 04 g(5) = 04 h(5) = The funcion f() has he greaes value when = 5. b) For = 5, f() = 4 g() = 4 5 h() = f( 5) = 4 5 g( 5) = 4 h( 5) = 5 f( 5) = 04 g( 5) = 04 h( 5) = The funcion g() has he greaes value when = 5. 5 c) Any base raised o he eponen 0 is. f() = 4 g() = 4 f(0) = 4 0 g(0) = 4 0 h() = h(0) = 0 MHR Pre-Calculus Soluions Chaper 7 Page of 4

2 f(0) = g(0) = h(0) = Secion 7. Page 4 Quesion a) For y = 5, c > so he graph is increasing. The graph will pass hrough he poin (, 5). Graph B. b) For y =, c < so he graph is decreasing. The graph will pass hrough he poin 4 (, 0.5). Graph C. c) For y =, c < so he graph is decreasing. The graph will pass hrough he approimae poin (, 0.67). Graph A. Secion 7. Page 4 Quesion 4 a) There is a paern in he ordered pairs. y 0 9 As he value of increases by uni, he value of y increases by a facor of. Therefore, for his funcion, c =. Use he poin (, ) o check he funcion y = : Lef Side Righ Side y = = = The funcion equaion for he graph is y =. b) There is a paern in he ordered pairs. y As he value of increases by uni, he value of y decreases by a facor of 5. Therefore, for his funcion, c = 5. Use he poin (, 5) o check he funcion y = 5 : MHR Pre-Calculus Soluions Chaper 7 Page of 4

3 Lef Side Righ Side y 5 = 5 = 5 = 5 The funcion equaion for he graph is y = 5. Secion 7. Page 4 Quesion 5 a) The domain is { R}, and he range is {y y > 0, y The y-inercep is. The funcion is increasing. The equaion of he horizonal asympoe is y = 0. R}. b) The domain is { R}, and he range is {y y > 0, y The y-inercep is. The funcion is increasing. The equaion of he horizonal asympoe is y = 0. R}. c) The domain is { R}, and he range is {y y > 0, y The y-inercep is. The funcion is decreasing. The equaion of he horizonal asympoe is y = 0. R}. d) The domain is { R}, and he range is {y y > 0, y The y-inercep is. The funcion is decreasing. The equaion of he horizonal asympoe is y = 0. R}. MHR Pre-Calculus Soluions Chaper 7 Page of 4

4 Secion 7. Page 4 Quesion 6 a) The baceria in a Peri dish doubling heir number every hour represens growh. So, c >. b) The half-life of he radioacive isoope acinium-5 represens decay. So, c <. c) The amoun of ligh passing hrough waer decreases wih deph represens decay. So, c <. d) The populaion of an insec colony ripling every hour represens growh. So, c >. Secion 7. Page 4 Quesion 7 a) The funcion N = is eponenial since he base is greaer han zero and he variable is an eponen. b) i) For = 0, N = 0 = A he sar, person has he virus. ii) For =, N = = Afer day, people have he virus. iii) For = 4, N = 4 = 6 Afer 4 days, 6 people have he virus. iv) For = 0, N = 0 = 04 Afer 0 days, 04 people have he virus. Secion 7. Page 4 Quesion 8 a) If he populaion increases by 0% each year, he populaion becomes 0% of he previous year s populaion. So, he growh rae is 0%, or. wrien as a decimal. b) The domain is { 0, R}, and he range is {P P 00, P R}. MHR Pre-Calculus Soluions Chaper 7 Page 4 of 4

5 c) If he populaion decreases by 5% each year, he populaion becomes 95% of he previous year s populaion. So, he growh rae is 95%, or 0.95 wrien as a decimal. d) The domain is { 0, R}, and he range is {P 0 < P 00, P R}. Secion 7. Page 44 Quesion 9 a) The eponenial funcion ha relaes he amoun, L, as a percen epressed as a decimal, of ligh available o he deph, d, in 0-m incremens, is L = 0.9 d. b) c) The domain is {d d 0, d R}, and he range is {L 0 < L, L R}. d) 5 m is he same as.5 0-m incremens. For.5, L = 0.9 d = = The percen of ligh ha will reach Pera if she dives o a deph of 5 m is approimaely 76.8%. Secion 7. Page 44 Quesion 0 a) Le P represen he percen, as a decimal, of U-5 remaining. Le represen ime, in 700-million-year inervals. Then, he eponenial funcion ha represens he radioacive decay of kg of U-5 is P() =. MHR Pre-Calculus Soluions Chaper 7 Page 5 of 4

6 b) c) From he graph, i will ake 700-million-year inervals, or years, for kg of U-5 o decay o 0.5 kg. d) The sample in par c) will never decay o 0 kg, since P = 0 is he horizonal asympoe. Secion 7. Page 44 Quesion a) b) I will ake approimaely 64 years for he deposi o riple in value. c) The amoun of ime i akes for a deposi o riple does no depend on he value of he iniial deposi. Since each $ amoun invesed riples, i does no maer wha he iniial invesmen was. d) From he graph, he approimae doubling ime for his invesmen is 40 years. MHR Pre-Calculus Soluions Chaper 7 Page 6 of 4

7 From he rule of 7, he approimae doubling ime for his invesmen is or 4 years. Secion 7. Page 44 Quesion 7.75, Le P represen he world populaion, in billions. Le represen ime, in years, since 0. Then, he eponenial funcion ha represens world populaion over ime is P() = 7(.07). The populaion of he world will reach 9 billion in approimaely 0 years, or he year 0. Secion 7. Page 44 Quesion a) b) The poins (, y) on he graph y = 5 become he poins (y, ) on he graph of he inverse of he funcion. Thus, he domains and ranges are inerchanged. Also, he horizonal asympoe of he graph y = 5 becomes a verical asympoe of he graph of he inverse of he funcion. c) The equaion of he inverse of he funcion is = 5 y. MHR Pre-Calculus Soluions Chaper 7 Page 7 of 4

8 Secion 7. Page 45 Quesion 4 a) The funcion D = φ can be wrien as D =. The coarser he maerial, he greaer he diameer. Therefore, a negaive value of φ represens a greaer value of D. b) The diameer of fine sand is, or 0.5 mm. The diameer of coarse gravel is -5, or mm. Thus, fine sand is he diameer of coarse gravel. 56 Secion 7. Page 45 Quesion 5 a) Graph A() = (.78) 0.0 and A() = and deermine he poin of inersecion. The approimae doubling period is 4.7 years. φ b) Graph A = (.0) and A = and deermine he poin of inersecion. The approimae doubling period is 5 years. c) The resuls are similar, bu he coninuous compounding funcion gives a shorer doubling period by approimaely 0. years. Secion 7. Page 45 Quesion C a) Graph f() = and g() = on he same se of aes. Graph h() = separaely. MHR Pre-Calculus Soluions Chaper 7 Page 8 of 4

9 b) c) Eample: All hree funcions have he same domain, and each of heir graphs has a y-inercep. The funcions f() and g() have all key feaures in common. d) Eample: The funcion h() is he only funcion wih an asympoe, which resrics is range and resuls in no -inercep. Secion 7. Page 45 Quesion C a) b) c) No, he poins do no form a smooh curve. The locaions of he poins alernae beween above he -ais and below he -ais. d) Using echnology o evaluae f and 5 f For =, For = 5, resuls in an error: non-real answer. f( ) ( ) f ( ) f Boh values are undefined. f( ) ( ) 5 5 f ( ) 5 f ( ) 5 MHR Pre-Calculus Soluions Chaper 7 Page 9 of 4

10 e) Eample: Eponenial funcions are defined o only include posiive bases, because only posiive bases resul in smooh curves. Secion 7. Transformaions of Eponenial Funcions Secion 7. Page 54 Quesion Compare each funcion o he form y = a(c) b( h) + k. a) For y = (), a =. This is a verical srech by a facor of : choice C. b) For y =, h =. This is a horizonal ranslaion of unis o he righ: choice D. c) For y = + 4, k = 4. This is a verical ranslaion of 4 unis up: choice A. 5 d) For y =, b =. This is a horizonal srech by a facor of 5: choice B. 5 Secion 7. Page 54 Quesion Compare each funcion o he form y = a(c) b( h) + k. a) For y = 5, h =. This is a horizonal ranslaion of uni o he lef: choice D. b) For y = c) For y = d) For y =, a =. This is a reflecion in he -ais: choice A. 5 5, b =. This is a reflecion in he y-ais: choice B., k =. This is a verical ranslaion of unis down: choice C. 5 Secion 7. Page 54 Quesion Compare each funcion o he form y = a(c) b( h) + k. a) For f() = () 4, a =, verical srech by a facor of b =, no horizonal srech h = 0, no horizonal ranslaion k = 4, verical ranslaion of 4 unis down MHR Pre-Calculus Soluions Chaper 7 Page 0 of 4

11 b) For g() = 6 +, a =, no verical srech b =, no horizonal srech h =, horizonal ranslaion of unis o he righ k =, verical ranslaion of unis up c) For m() = 4() + 5, a = 4, verical srech by a facor of 4 and a reflecion in he -ais b =, no horizonal srech h = 5, horizonal ranslaion of 5 unis o he lef k = 0, no verical ranslaion ( ) d) For y =, a =, no verical srech b =, horizonal srech by a facor of h =, horizonal ranslaion of uni o he righ k = 0, no verical ranslaion e) For n() = ( 4) 5, a =, verical srech by a facor of and a reflecion in he -ais b =, horizonal srech by a facor of h = 4, horizonal ranslaion of 4 unis o he righ k =, verical ranslaion of unis up f) For y =, or y = a =, reflecion in he -ais ( ) b =, horizonal srech by a facor of h =, horizonal ranslaion of uni o he righ k = 0, no verical ranslaion 4 5 g) For y =.5(0.75), a =.5, verical srech by a facor of.5 b =, horizonal srech by a facor of h = 4, horizonal ranslaion of 4 unis o he righ MHR Pre-Calculus Soluions Chaper 7 Page of 4

12 k = 5, verical ranslaion of 5 unis down Secion 7. Page 55 Quesion 4 a) Since he graph has been refleced in he -ais, a < 0 and 0 < c <. The graph has also been ranslaed unis up, so k =. Choice C. b) The graph has also been ranslaed uni o he righ and unis down, so h = and k =. Choice A. c) Since he graph has been refleced in he -ais, a < 0 and c >. The graph has also been ranslaed unis up, so k =. Choice D. d) The graph has also been ranslaed unis o he righ and uni up, so h = and k =. Choice B. Secion 7. Page 55 Quesion 5 a) For y = (4) ( ), a =, verical srech by a facor of b =, reflecion in he y-ais h =, horizonal ranslaion of unis o he righ k =, verical ranslaion of unis up b) c) d) The domain is { R}, and he range is {y y >, y R}. The equaion of he horizonal asympoe is y =. The y-inercep is 4. MHR Pre-Calculus Soluions Chaper 7 Page of 4

13 Secion 7. Page 55 Quesion 6 a) i), ii) For y = () + 4, a =, verical srech by a facor of b =, no horizonal srech h = 0, no horizonal ranslaion k = 4, verical ranslaion of 4 unis up iii) iv) The domain is { R}, and he range is {y y > 4, y R}. The equaion of he horizonal asympoe is y = 4. The y-inercep is 6. b) i), ii) For m(r) = () r +, a =, reflecion in he -ais b =, no horizonal srech h =, horizonal ranslaion of unis o he righ k =, verical ranslaion of unis up iii) iv) The domain is {r r R}, and he range is {m m <, m R}. The equaion of he horizonal asympoe is m =. The m-inercep is 5 8, and he r-inercep is 4. c) i), ii) For y = (4), iii) a =, verical srech by a facor of b =, no horizonal srech h =, horizonal ranslaion of uni o he lef k =, verical ranslaion of uni up iv) The domain is { R}, and he range is {y y >, y R}. The equaion of he horizonal asympoe is y =. The y-inercep is 7. MHR Pre-Calculus Soluions Chaper 7 Page of 4

14 d) i), ii) For n(s) = 4, a =, verical srech by a facor of and a reflecion in he -ais b =, horizonal srech by a facor of 4 4 iii) h = 0, no horizonal ranslaion k =, verical ranslaion of unis down iv) The domain is {s s R}, and he range is {n n <, n R}. The equaion of he horizonal asympoe is n =. The n-inercep is 5 8. Secion 7. Page 55 Quesion 7 a) To obain he graph of y = f( ) +, he graph of f() mus be ranslaed unis o he righ and uni up: y =. b) To obain he graph of y = 0.5f( ), he graph of f() mus be verically sreched by a facor of 0.5, refleced in he -ais, and ranslaed unis o he righ: y = 0.5(5). c) To obain he graph of y = f() +, he graph of f() mus be refleced in he -ais, horizonally sreched by a facor of, and ranslaed uni up: y =. 4 d) To obain he graph of y = f ( ) 5, he graph of f() mus be verically sreched by a facor of, horizonally sreched by a facor of, refleced in he y-ais, and ranslaed uni o he righ and 5 unis down: y = ( ) (4) 5. MHR Pre-Calculus Soluions Chaper 7 Page 4 of 4

15 Secion 7. Page 56 Quesion 8 a) Map all poins (, y) on he graph of f() o ( +, y + ). b) Map all poins (, y) on he graph of f() o ( +, 0.5y). c) Map all poins (, y) on he graph of f() o, y. d) Map all poins (, y) on he graph of f() o ( +, y 5). MHR Pre-Calculus Soluions Chaper 7 Page 5 of 4

16 Secion 7. Page 56 Quesion 9 a) The number 0.79 represens he 79% of he drug remaining in he body of a dose aken. The number represens b) he decay rae of he dose aken. The dose decreases by 79% every h. c) The M-inercep represens he dose of 00 mg. d) For his siuaion, he domain is {h h 0, h R} and he range is {M 0 < M 00, M R}. Secion 7. Page 56 Quesion 0 a) Subsiue T i = 95 and T f = 0 ino 5 T() = ( T T )(0.9) T i f 5 T() = 75(0.9) 0. a = 75, verical srech by a facor of 75 b =, horizonal srech by a facor of 5 5 f : b) h = 0, no horizonal ranslaion k = 0, verical ranslaion of 0 unis up c) Subsiue = () 75(0.9) 0 T 00 5 T ( 00) 75(0.9) 0 T (00) The emperaure of he coffee afer 00 min is approimaely 9. C. d) The equaion of he horizonal asympoe is T = 0. This represens 0 C, he final emperaure of he coffee. MHR Pre-Calculus Soluions Chaper 7 Page 6 of 4

17 Secion 7. Page 56 Quesion a) For 5000 baceria, a = For an increase of 0%, c =.. For an increase ha happens every days, b =. Then, he ransformed eponenial funcion for his siuaion is P = 5000(.). b) a = 5000, verical srech by a facor of 5000 b =, horizonal srech by a facor of h = 0, no horizonal ranslaion k = 0, no verical ranslaion c) From he graph, he baceria populaion afer 9 days is approimaely 57. Secion 7. Page 56 Quesion a) The iniial percen of C-4 in an organism is 00%, so a = 00. For half-life, c =. Since he half-life of C-4 is abou 570 years, b =. Then, he ransformed 570 eponenial funcion ha represens he percen, P, of C-4 remaining afer years is 570 P = 00. b) From he graph, he approimae age of a dead organism ha has 0% of original C-4 is 05 years old. MHR Pre-Calculus Soluions Chaper 7 Page 7 of 4

18 Secion 7. Page 57 Quesion a) Le A represen he area covered by he baceria. Le represen ime, in hours. The doubling ime for he area is 0 h, so c = and b =. Since he iniial area was 0 00 cm 0, a = 00. Then, A = 00(). Subsiue = 4, A 00() () By Tuesday morning (4 h laer), he baceria covers an area of approimaely 57.8 cm. b) Deermine he surface area of Earh: 678 km = cm SA = 4πr = 4π( ) Graph A = 00() and A = I would ake hese baceria abou 555 h o cover he surface of Earh. Secion 7. Page 57 Quesion 4 a) Le P represen he fo populaion. Le represen ime, in years. The iniial fo populaion was 5 5 years ago, so a = 5 and h = 5. The populaion doubled in 5 years, so b = 5 and c =. Then, P = ( 5) 5() 5. Subsiue = 0, P 5() ( 5) 5 ( 05) 5 5() The fo populaion in 0 years will be abou 67. b) Eample: Disease or lack of food can change he rae of growh of he foes. Eponenial growh suggess ha he populaion will grow wihou bound, and herefore he fo populaion will grow beyond he possible food sources, which is no good if no conrolled. MHR Pre-Calculus Soluions Chaper 7 Page 8 of 4

19 Secion 7. Page 57 Quesion C Eample: The graph of an eponenial funcion of he form y = c has a horizonal asympoe a y = 0. Since y 0, he graph canno have an -inercep. Secion 7. Page 57 Quesion C a) Eample: For a funcion of he form y = a(c) b( h) + k, he parameers a and k can affec he -inercep. If a > 0 and k < 0 or a < 0 and k > 0, hen he graph of he eponenial funcion will have an -inercep. b) Eample: For a funcion of he form y = a(c) b( h) + k, he parameers a, h, and k can affec he y-inercep. The poin (0, y) on he graph of y = c ges mapped o (h, ay + k). Secion 7. Solving Eponenial Equaions Secion 7. Page 64 Quesion a) ( ) 8 ( ) b) 9 c) 8 d) Secion 7. Page 64 Quesion a) and 4 = b) 9 ( ) and 7 = c) and 4 8 d) 6 4 and 6 ( ) 4 MHR Pre-Calculus Soluions Chaper 7 Page 9 of 4

20 Secion 7. Page 64 Quesion a) 6 (4) 4 b) (4) c) 4(4 ) ( 4 d) 4(4) 4 4) Secion 7. Page 64 Quesion 4 a) ( ) 4 6 Equae he eponens. 4 = + 6 = 6 = b) 5 5 (5 ) Equae he eponens. = = c) 9 w w w w ( ) w w Equae he eponens. w + = w w = Check: Lef Side Righ Side = 4() = 4 + = = 4 6 = 4096 = 4096 Lef Side = Righ Side The soluion is =. Check: Lef Side Righ Side 5 5 = 5 = 5 ( ) = 5 = 5 6 = = Lef Side = Righ Side The soluion is =. Check: Lef Side Righ Side w + 9 w = + = 9 = 4 = 9 = 8 = 8 Lef Side = Righ Side The soluion is w =. MHR Pre-Calculus Soluions Chaper 7 Page 0 of 4

21 m m5 d) 6 6 m m5 (6 ) 6 6m m5 6 6 Equae he eponens. 6m = m + 5 4m = 7 m = 7 4 Check: Lef Side Righ Side m 6 m Lef Side = Righ Side The soluion is m = 7 4. Secion 7. Page 64 Quesion 5 a) 4 8 ( ) ( ) 6 9 Equae he eponens. 6 = 9 = 9 = Check: Graph y = 4 and y = 8 and find he poin of inersecion. The soluion is =. b) 7 9 ( ) ( ) 4 Equae he eponens. = 4 = 4 Check: Graph y = 7 and y = 9 and find he poin of inersecion. The soluion is = 4. y y4 c) 5 5 y y4 (5 ) (5 ) 6y y8 5 5 Equae he eponens. 6y = y + 8 4y = Check: Graph y = 5 and y = and find he poin of inersecion. MHR Pre-Calculus Soluions Chaper 7 Page of 4

22 y = 4 The soluion is y = 4. k k d) 6 4 k 5 k ( ) ( ) 8k 5k5 Equae he eponens. 8k = 5k + 5 k = 7 k = 9 Check: Graph y = 6 k and y = k + and find he poin of inersecion. The soluion is k = 9. Secion 7. Page 64 Quesion 6 a) Use sysemaic rial o solve = Mahemaical Reasoning.5 Try =. This gives a value greaer han. Try a lesser value Too low. The correc value is beween 0 and, bu much closer o 0. Try Close, bu oo high This is very close and a reasonable approimaion. The soluion is 0.. Check: Graph y = and y =.07 and find he poin of inersecion. The soluion is 0.. MHR Pre-Calculus Soluions Chaper 7 Page of 4

23 b) Use sysemaic rial o solve =... Mahemaical Reasoning.8 Try =. This gives a value greaer han. Try a lesser value..85 Too low. The correc value is beween and. Try Close, bu oo high This is very close and a reasonable approimaion. The soluion is.5. Check: Graph y = and y =. and find he poin of inersecion. The soluion is.5. c) Use sysemaic rial o solve 0.5 =... Mahemaical Reasoning Try = 4. This gives a value lesser han 0.5. Try a greaer value Too high. The correc value is beween 4 and. Try Too low Close, bu oo low This is very close and a reasonable approimaion. The soluion is.8. Check: Graph y = 0.5 and y =. and find he poin of inersecion. The soluion is.8. d) Use sysemaic rial o solve 5 = Mahemaical Reasoning Try = 0. This gives a value greaer han 5. Try a lesser value Too low. The correc value is beween 8 and 0. Try Close, bu oo high This is very close and a reasonable approimaion. The soluion is 8.9. MHR Pre-Calculus Soluions Chaper 7 Page of 4

24 Check: Graph y = 5 and y =.08 + and find he poin of inersecion. The soluion is 8.9. Secion 7. Page 64 Quesion 7 a) Graph y = 0(.04) 00 and find he -inercep. b) Graph y = -inercep. 0 and find he The soluion is The soluion is.66. c) Graph y = -inercep. 4 and find he d) Graph y = -inercep and find he The soluion is 5.8. The soluion is = 8. MHR Pre-Calculus Soluions Chaper 7 Page 4 of 4

25 e) Graph y = and find he -inercep. f) Graph y = 5 4 and find he -inercep. The soluion is.7. g) Graph y = 8 + and find he -inercep. The soluion is 4.4. h) Graph y = and find he -inercep. The soluion is.4. The soluion is.88. Secion 7. Page 64 Quesion 8 a) T 8 b) Graph R = 00(.7) poin of inersecion. and R = 00 and find he c) Subsiue T = 5. R 00(.7) T (.7) The relaive spoilage rae a 5 C is approimaely 64. The emperaure a which he relaive spoilage rae doubles o 00 is approimaely 5.6 C. MHR Pre-Calculus Soluions Chaper 7 Page 5 of 4

26 T 8 d) Graph R = 00(.7) of inersecion. and R = 500 and find he poin The maimum sorage emperaure is approimaely.0 C. Secion 7. Page 64 Quesion 9 Le N represen he number of baceria. Le represen ime, in hours. The iniial baceria coun is 000, so a = 000. The baceria double every 0.75 h, so b = 4 4 and c =. Then, N = 000(). Subsiue N = 000, N 000() () 6 () () Equae he eponens. 4 4 Afer h he baceria will be 000. Secion 7. Page 64 Quesion 0 Use he formula A = P( + i) n, where A = 7000, P = 6000, and i = A = P( + i) n 7000 = 6000( ) n 7 6 =.09n Graph y = and find he -inercep. Simionie would have o inves his money in a GIC for 4 years. MHR Pre-Calculus Soluions Chaper 7 Page 6 of 4

27 Secion 7. Page 65 Quesion a) Use he formula A = P( + i) n, where P = 000 and i = 0.0: A = 000(.0) n. b) Subsiue n = 6, he number of compounding periods in 4 years. A = 000(.0) n = 000(.0) 6 = The value of he invesmen afer 4 years is $7.79. c) Subsiue A = 000. A = 000(.0) n 000 = 000(.0) n =.0 n Graph y =.0 and find he -inercep. From he graph, i appears ha i will ake 5 compounding periods, or 8.75 years, for he invesmen o double in value. However, subsiuing n = 5 ino he original funcion A = 000(.0) n resuls in a value of $ So, i will ake 6 compounding periods, or 9 years, for he invesmen o acually double in value. Secion 7. Page 65 Quesion a) Le he iniial sample of Co-60 be, so a =. For half-life, c =. Since he half-life of Co-60 is abou 5. years, b =. Then, he eponenial funcion ha represens his 5. siuaion is m = 5. b) Subsiue = 6.5. m = = = = , where m is he amoun of Co-60 remaining afer years. MHR Pre-Calculus Soluions Chaper 7 Page 7 of 4

28 The fracion of a sample of Co-60 ha will remain afer 6.5 years is. c) Subsiue m = m Equae he eponens = 47.7 I will ake a sample of Co years o decay o Secion 7. Page 65 Quesion 5 of is original mass. a) Use he formula A = P( + i) n, where P = 500 and i = 0.0: A = 500(.0) n. b) Subsiue n = 0, he number of compounding periods in 5 years. A = 500(.0) n = 500(.0) 0 = The value of he invesmen afer 5 years is $ c) Subsiue A = 500. A = 500(.0) n 500 = 500(.0) n =.0 n Graph y =.0 and find he -inercep. I will ake 4 compounding periods, or 7 years, for he invesmen o riple in value. MHR Pre-Calculus Soluions Chaper 7 Page 8 of 4

29 Secion 7. Page 65 Quesion 4 Use he formula A = P( + i) n, where A = 0 000, i = 0.05, and n = 6. A = P( + i) n = P( ) P = 6.05 P = Glenn and Arlene will need o inves $ oday. Secion 7. Page 65 Quesion 5 a) i) 4 ( ) Equae he eponens. > + > b) i) Since he graph of y = is greaer han (above) he graph of y = 4 + when >, he soluion is >. ii) ( ) ( ) 4 6 Equae he eponens. 4 < 6 + > ii) Since he graph of y = 8 is less han (below) he graph of y = 7 + when >, he soluion is >. c) Eample: Solve he inequaliy >. Since he graph of y = is greaer han (above) he graph of y = when <, he soluion is <. MHR Pre-Calculus Soluions Chaper 7 Page 9 of 4

30 Secion 7. Page 65 Quesion (4 ) = 0 (4 ) + (4 ) = 0 (4 + )(4 ) = = 0 or 4 = 0 4 = 4 = = 0 Since he value 4 is always greaer han zero, here is no real value of for which 4 =. The real soluion is = 0. Secion 7. Page 65 Quesion = (4) = 4 4 ( 4 ) = 4 4 (0.75) = 4 4 = = 5 Equae he eponens. = 5 = 5 Subsiue = 5 ino ( ) ( ) The value of ( ) is approimaely 76.. Secion 7. Page 65 Quesion 8 Use he formula PMT = i = i PV ( i) n wih PMT = 8.90, PV = , and MHR Pre-Calculus Soluions Chaper 7 Page 0 of 4

31 i PMT PV n ( i) n ( 0.005) (. 005) n (.005) n n (.005) Graph y =.005 n and y = and find he poin of inersecion I will ake Tyseer 40 paymen periods, or 0 years, o pay off he morgage. Secion 7. Page 65 Quesion C a) You can epress 6 wih a base of 4 by wriing 6 as 4 and simplifying. 6 = (4 ) = 4 4 b) Eample: You can epress 6 wih a base of by wriing 6 as 4 and simplifying. 6 = ( 4 ) = 8 You can epress 6 wih a base of 4 by wriing 6 as and simplifying MHR Pre-Calculus Soluions Chaper 7 Page of 4

32 Secion 7. Page 65 Quesion C a) 6 8 ( ) ( ) b) Given equaion. Epress 6 and 8 as powers of. Apply he power of a power law. Equae he eponens. Isolae he -erm. Solve of. Chaper 7 Review Chaper 7 Review Page 66 Quesion a) For he populaion of a counry, in millions, ha grows a a rae of.5% per year, he graph would show he funcion y =.05: graph B. b) The graph of y = 0 conains he poin (, 0): graph D. c) For Tungsen-87, a radioacive isoope ha has a half-life of day, he graph would show he funcion y = : graph A. d) The graph of y = 0. conains he poin (, 5): graph C. Chaper 7 Review Page 66 Quesion a) b) The domain is { R}, and he range is {y y > 0, y R}. There is no -inercep. The y-inercep is. The funcion is decreasing for all values of. The equaion of he horizonal asympoe is y = 0. MHR Pre-Calculus Soluions Chaper 7 Page of 4

33 Chaper 7 Review Page 66 Quesion There is a paern in he ordered pairs. y As he value of increases by uni, he value of y decreases by a facor of 4. Therefore, for his funcion, c = 4. Use he poin (, 4) o check he funcion y = 4 : Lef Side Righ Side y 4 = 4 = 4 = 4 The funcion equaion for he graph is y = 4. Chaper 7 Review Page 66 Quesion 4 a) Since he ineres rae is.5% per year, each year he invesmen grows by a facor of 0.5%, or.05, as a decimal. b) Subsiue = 0. v =.05 =.05 0 =.76 The value of $ if i is invesed for 0 years will be $.8. c) Graph v =.05 and v = and find he poin of inersecion. I will ake approimaely.7 years for he value of he dollar invesed o reach $. MHR Pre-Calculus Soluions Chaper 7 Page of 4

34 Chaper 7 Review Page 66 Quesion 5 a) For y = (4) ( ) +, a =, verical srech by a facor of and refleced in he -ais b =, horizonal srech by a facor of h =, horizonal ranslaion of uni o he righ k =, verical ranslaion of unis up b) c) d) The domain is { R}, and he range is {y y <, y R}. The equaion of he horizonal asympoe is y =. The -inercep is. The y-inercep is 6. Chaper 7 Review Page 67 Quesion 6 a) Look for a paern in he poins. y = Transformed Funcion,, (0, ) (, ) (, ) (4, ) The ransformaion can be described by he mapping (, y) ( +, y). This represens a horizonal ranslaion of unis o he righ. MHR Pre-Calculus Soluions Chaper 7 Page 4 of 4

35 b) Look for a paern in he poins. y = Transformed Funcion,, (0, ) (0, ) (, ) (, ) The ransformaion can be described by he mapping (, y) (, y 4). This represens a verical ranslaion of 4 unis down. c) Look for a paern in he poins. y = Transformed Funcion 5,, (0, ) (, ) (, ) (0, ) The ransformaion can be described by he mapping (, y) (, y + ). This represens a reflecion in he -ais and a ranslaion of uni o he righ and unis up. Chaper 7 Review Page 67 Quesion 7 a) For f() = 5, sreched verically by a facor of 4, a = 4 sreched horizonally by a facor of and refleced in he y-ais, b = ranslaed uni up and 4 unis o he lef, k = and h = 4 The equaion of he ransformed funcion is y = 4(5) ( + 4) +. b) For g() =, sreched horizonally by a facor of 4, b = 4 sreched verically by a facor of and refleced in he -ais, a = ranslaed unis o he righ and uni down, h = and k = The equaion of he ransformed funcion is 4( ) y =. MHR Pre-Calculus Soluions Chaper 7 Page 5 of 4

36 Chaper 7 Review Page 67 Quesion 8 0 a) For T = 90, a = 90, verical srech by a facor of 90 b =, horizonal srech by a facor of 0 0 h = 0, no horizonal ranslaion k = 0, no verical ranslaion b) c) For his siuaion, he domain is { 0, R} and he range is {T 0 < T 90, T R}. d) Subsiue T =. 0 T = 90 0 = 90 0 Graph T = 90 and T = and find he poin of inersecion. The milk will keep fresh a C for approimaely. h. Chaper 7 Review Page 67 Quesion 9 a) 6 = 6 b) 6 6 c) Chaper 7 Review Page 67 Quesion 0 a) 5 = 7 5 = ( ) 5 = Equae he eponens. 5 = = = b) = 8 ( ) + = ( 5 ) = Equae he eponens. 6 = 5 5 = = MHR Pre-Calculus Soluions Chaper 7 Page 6 of 4

37 Chaper 7 Review Page 67 Quesion a) Graph y = and y = 5 and find he poin of inersecion. b) Graph y = and y = + and find he poin of inersecion. The soluion is 4.0. The soluion is 6.. Chaper 7 Review Page 67 Quesion a) Le he iniial sample of Ni-65 be, so a =. For half-life, c =. Since he half-life of Ni-65 is.5 h, b =. Then, he eponenial funcion ha represens his siuaion is.5 m =.5, where m is he amoun of Ni-65 remaining afer hours. b) Subsiue = 0. m = = = = 6 The fracion of a sample of Ni-65 ha will remain afer 0 h is 6. c) Subsiue m = 04 m Equae he eponens. 0.5 = 5 I will ake a sample of Ni-65 5 h o decay o of is original mass. 04 MHR Pre-Calculus Soluions Chaper 7 Page 7 of 4

38 Chaper 7 Pracice Tes Chaper 7 Pracice Tes Page 68 Quesion The funcions y =, y = Choice B., and y = 7 will all have he same y-value of when = 0: Chaper 7 Pracice Tes Page 68 Quesion ( 5) 4 To obain he graph of y =, ransform he graph of y = by a horizonal srech by a facor of 4 and a ranslaion of 5 unis o he righ and unis down: Choice C. Chaper 7 Pracice Tes Page 68 Quesion Le V represen he value of he car. Le represen ime, in years. The curren value is , so a = The value doubles every 0 years, so b = 0 and c =. Then, V = (). Subsiue = 0, V () () The value of he car 0 years ago was $5 000: Choice B. Chaper 7 Pracice Tes Page 68 Quesion (4 ) (( ) ) Choice A. 9 MHR Pre-Calculus Soluions Chaper 7 Page 8 of 4

39 Chaper 7 Pracice Tes Page 68 Quesion 5 Solve 0.75 = 0.8 by graphing. The glass should be approimaely. mm hick: Choice D. Chaper 7 Pracice Tes Page 68 Quesion 6 a) Look for a paern in he poins. y = 5 Transformed Funcion (0, ) (, ) (, 5) (, 7) (, 5) (, 7) The ransformaion can be described by he mapping (, y) (, y + ). This represens a horizonal ranslaion of unis o he lef and unis up: h = and k =. So, he equaion of he ransformed funcion is y = b) Look for a paern in he poins. y = Transformed Funcion (0, ) 9, (, ) (, 5) (, 4) (, 6) The ransformaion can be described by he mapping (, y) ( +, 0.5y 4). This represens a verical srech by a facor of 0.5, a reflecion in he -ais, and a ranslaion of uni o he lef and 4 unis down: a = 0.5, h =, and k = 4. So, he equaion of he ransformed funcion is y = 0.5() 4. Chaper 7 Pracice Tes Page 69 Quesion 7 a) b) MHR Pre-Calculus Soluions Chaper 7 Page 9 of 4

40 c) Chaper 7 Pracice Tes Page 69 Quesion 8 a) The base funcion for g() = () + 4 is f() =. For g() = () + 4, a =, verical srech by a facor of b =, no horizonal srech h =, horizonal ranslaion of unis o he lef k = 4, verical ranslaion of 4 unis down b) c) The domain is { R}, he range is {y y > 4, y R}, and he equaion of he horizonal asympoe is y = 4. Chaper 7 Pracice Tes Page 69 Quesion 9 a) ( 4) 9 ( 4) ( ) 4 Equae he eponens. = 4 = 4 4 b) ( ) ( ) 6 Equae he eponens. = + 6 = 8 4 c) ( ) ( ) Equae he eponens. 0 0 = = 6 = 8 MHR Pre-Calculus Soluions Chaper 7 Page 40 of 4

41 Chaper 7 Pracice Tes Page 69 Quesion 0 a) Graph y = and y =. and find he poin of inersecion. b) Graph y =.7 and y = 0. and find he poin of inersecion. The soluion is 9.7. The soluion is 0.. Chaper 7 Pracice Tes Page 69 Quesion a) If he growh rae remains consan a.77%, hen he populaion would have been muliplied by a facor of.077 afer year. b) Le he iniial percen of he populaion for Saskaoon in 00 be 00, so a = 00. For he growh rae from par a), c =.077. Then, he eponenial funcion ha represens his siuaion is P = 00(.077), where P is percen of he populaion years afer 00. c) For his siuaion, he domain is { 0, R} and he range is {P P 00, P R}. d) For Saskaoon s populaion o grow by 5%, subsiue P = 5. P = 00(.077) 5 = 00(.077).5 =.077 Graph y =.5 and y =.077 and find he poin of inersecion. I will ake approimaely 8. years for Saskaoon s populaion o grow by 5%. Chaper 7 Pracice Tes Page 69 Quesion a) MHR Pre-Calculus Soluions Chaper 7 Page 4 of 4

42 b) Subsiue P = 7. P H( P) 0 7 H (7) 0 7 H (7) 0 The hydrogen ion concenraion for a ph of 7.0 is [H+]. c) Deermine he hydrogen ion concenraion for a ph of 7.6. Subsiue P = 7.6. P H( P) H (7.6) 0 8 H (7.6).50 The equivalen range of hydrogen ion concenraion for a ph beween 7.0 and 7.6 is from [H+] o [H+]. Chaper 7 Pracice Tes Page 69 Quesion Use he formula A = P( + i) n, where A = 5000, P = 500 and i = 0.0. Solve 5000 = 500(.0) n = 500(.0) n =.0n 0 7 =.0n Graph y = and find he -inercep. I will ake approimaely 8 compounding periods, or 4.5 years, before Lucas has enough money o ake he vacaion he wans. MHR Pre-Calculus Soluions Chaper 7 Page 4 of 4

43 Chaper 7 Pracice Tes Page 69 Quesion 4 Deermine when a compuer purchased for $000 is worh 0% of is values, or $00. Subsiue V = 00. V ( ) Graph y = 0. and y = inersecion. and find he poin of I will ake abou 9.97 years, for he compuer o be worh 0% of is purchase price. MHR Pre-Calculus Soluions Chaper 7 Page 4 of 4