12. sothe isHorizontal one-to-one. SECTION AND 9. The graph of () () = 2−passes 3 is athe lineHorizontal with slopeLine 2. ItTest, passes Line INVERSE Test, so FUNCTIONS is one-to-one. SECTION 1.5 1.5 INVERSE FUNCTIONS AND LOGARITHMS LOGARITHMS Algebraic solution: If 1 6= 2 , then 21 6= 22 ¤ ¤ 51 49 ⇒ 21 − 3 6= 22 − 3 ⇒ (1 ) 6= (2 ), so is one-to-one. 7−4 will attain every height up to its maximum height twice: once on 13. (a) An arrow the−INVERSE way and again on way down. Thus, ¤ 47 ANDthe LOGARITHMS 51. = 6 ⇔ 7 − 4 = ln 6 ⇔ 7 − ln 6 = 4 ⇔ SECTION = 141.5 (7 ln 6)up,FUNCTIONS 60 (b) log − log log 5 ( = log − log8(5 ), so is not[by 2] 83−4 8 3equal not 28,16 may 1-1.LawPick even if 8160does The graph of () = − is1 )symmetric with 2respect -axis. any -values equidistant from 0 to find two √equal √to the 2 2 1 (b) ln(3 − 10) = 2 ⇔ 3 − 10 = ⇔ 3 = + 10 ⇔ = (22+ = () = 1 + + 3 ( 1)23 = +10) 3 ⇒ ( − 1)2 − 2 = 3 − log8− 5 1 = 2 + 3 ⇒ ( − =≥ log1) 8 20 ⇒ 10. equal 21. ⇒ function values. For example, (−1) = −15 and (1) = −15, so is not one-to-one. 14. is not 1-1. Eventually our feet stop growing and1 therefore, are two we have the same shoe size. 2 2 there −1 times 1at which 2 1 2 2 2 20 = 3 (2 − 1) − 3 . Interchange and : = ( − 1) − [by . So 2]() =√ ( − 1) − 3 . Note that the domain of −1 Law = log 8 53 52. (a) ln( − 1) = 3 ⇔ 2 − 1 = −1 ⇔ 32 = 1 + 3 3 ⇔ = ± 31 + 3 . 11. is() 1, so is=not 2 one-to-one. 15. (a) Because is.1-1,(0) (4)==17=and ⇔() (7) = 4. ≥=1.1 − sin 4==log 823 [by (7)] log Section 1.5 Inverse Functions and Logarithms 8 8 3 2 (b) − 3 + 21-1, = 0√ ⇔ (= − 1)( − 2) = 08. ⇔ = 1 or = 2 ⇔ = ln 1 or = ln 2, so = 0 or ln 2. −1 (b) Because is− 2the ⇔ (2) =Line 4 1 13 (8) 12. The graph of () = passes Horizontal Test, so is one-to-one. 1 1 ⇒ (2 + 3) = 4 − 1 22. = () = ln 2 2 −12 + by (9). Or: − ln 2 = ln⇒ 38. (a) − = 2 = = 2−13 = = 4 − 1 ⇒ 3 + 1 = 4 − 2 ⇒ + 3 −5 ln 2 ⇔ log 2 2 3 = − 5 ⇔ = 5 + log2 3. 2 53. (a) 2 = 3 16. we must determine such that () 3. By inspection, we see that 1, then (1) = Since is 1-1 ( is an 13. First, An arrow will attain everyheight up to height twice: once up, and the way down. Thus, 3 + on 1if the=way 3again + 3. 1 on 3 + 1 its=maximum 3 ) − 2) . Interchange : = . So −1 () . = 3 +ln(ln 1 =(4 ⇒ = ln 3 ln(ln3and ) 3 −1 −1 −1 ln 3 ln 3 (b) if 21−5 ==not 3 equal =− 3=by (9). Or: [by (9)] = 2 −⇔2 4⇔ −=2 =not a1-1. 1-1 then ((9). ()) , = so5+ ( (2)) = 2. increasing function), it[by has inverse, and does 2(9)] ,an2−5 ( equal (3) (2( ),=so Ifisln even 1 )4 may Or: ln 3 ⇔ −1.5) 2isln( = ln)34function, − 35 by = ⇔ ln ln 2 ln 2 2 2−1 1 39. ln 20 − 2 ln 2 = ln 20 − ln 2 [by Law 3] 23. = () = ⇒ ln = 2 − 1 ⇒ 1 + ln = 2 ⇒ = (1 + ln ). 17. First, we1-1. must determineour such that () = 4.and Bytherefore, inspection, we are see two that times if =at0,which then () = 4.theSince is 1-1size. ( is an 14. is not Eventually feet stop growing there we have same shoe (b) ln + ln( − 1) =ln(( − 1)) = 1 ⇔ ( − 1) = 1 ⇔ 2 −2 − = 0. The quadratic formula (with = 1, 20 −1 1 1 −1 (1 + ln ). So () = (1 + ln ). Interchange and : = increasing function), = ln 4it has an [byand Law 2] (4) = 0.2 −1√ 2 inverse, 15. (a) Because is 1-1, (4)gives = 7 ⇔ 4 , but we reject the negative root since the natural logarithm is not = −1, and = −) = 1 1 ±(7)1=+4. 2 −1 5itpasses 8. √ (2) ¤ FUNCTIONS 18. (b) (a) () is 1-1 because Line Because = isln 1-1, =1+21Horizontal = 24. =defined =for 2 4 ⇒ = − 2SECTION ⇒ 1.5 = INVERSE (1 − 2) ⇒ AND =LOGARITHMS⇒ ¤ 53 +⇔ = 1+ . Test. 0. So⇒(8) =the SECTION 1.5 INVERSE FUNCTIONS AND LOGARITHMS 53 2 1 + 2 1 − 2 2 3 40. ln Domain + 2−1 ln√ of −3 [−3 = ln3] = + Range ln − Law 3][−1 √ofln−1 .Range = 3] Domain ln [byof of −1 −1 3 (b) = = . 16. (a) First, must such that1√ () = 43. isBy we−1 see that range if =of1,csc then ).(1) = 3. Since is 1-1 ( is an 65. cscwe because csc 2 and ininspection, 0 2 ∪ √2 =determine (the ln(ln 1 4 = 2 −1 ln 2) 54. (a) ln = = 1]∪ ⇔ 3 ln (the ==range = Interchange = . Note that the range of and the ln ⇔of csc ln −1) 2==1 44.⇔ 65. (a)=ln(ln csc because = 3: 2⇔ and in 0 = ln csc −4=lnand Law . So 2 () −1 −1). −1 4 is [by 2 − 2= 2, it−1 1 − 2 is a 1-1 function, then 1 − 2 increasing has an=inverse, and is in(3) (2) 0. (c) arcsin Since1(0) arcsin). ( ()) = , so ( (2)) = 2. sin = 1 and (b) 1function), = 2 because (the range of −=2 1.2 If 2 2 because 2 arcsin). −1 1 (the range of sin = 1 and is in − (b) arcsin 1 = = ⇔ =).ln 2= ln[( 2)] ⇔ [by 2 (0 2 Law 2 ln domain of √ is 2] +ln ⇔ √= = ln + ⇔ 2 ln−13 −1must 4. By inspection, we = 0, then () = 4. Since is 1-1 ( is an (0) that ≈ −17. (d) sin Since (−17) 17. (a) First, we determine () see that if 66. (−1 because sin 2 and − is in − = −1 . √2 )≈=0,−4 such √ −= 4 4 2 2 −1 ln − is in − . 66. (a) sin (−1 2 ) = − because sin − 2 and = −1 √ −1 √ − = ln ⇔ ( − ) = ln ⇔ = 4 4 4 2 2 3 2 2 3 13 1 1 1 25. increasing = () = ln+( + 3) has + 3 and ⇒(4)ln[( = − 3. Interchange and : = − 3.[by So Laws −1 () = − 3. an inverse, −1 0. −⇒ ln( + 2) 41. ln( + √ 2) + 2) ]9]. ln This 3, 2] − 32592(=ln−it632) because is in [0 3 cos √32 19. (b) We solve function), = for cos :+953 6 = −= 32and= ⇒ = ++232. gives us2) a 2formula for the inverse function, that 6 (2 + 3 + 5 −1 32 = because cos 6 = 32 and 6 is in [0 ]. (b) cos − 6 √ √ √ 1 9 1⇒ − it −1 −1 5 theLine 5 ()temperature −Celsius −−45967 32 is, Because as a − function of ≥ −45967 ⇒ 55. ln because = the . of is = 0,the solution the is 18. (a) (a) () isFahrenheit 1-1 passes Horizontal Test. . 5 67. (a) cot = 5 cot = −domain and is in = (0 )− (the range of cot ).of⇒ 26. =the −3the + ln 1− ⇒ + original =+ −inequality 1≥ [multiply √3 temperature =1−because 6 ⇒ (1 + 5 6 ) = 1√ 6⇒ − 5 −1 −1 5 = ln( + 2) + ln [by Law 3] 1 + − 67. (a) cot 3 = because cot = − 3 and is in (0 ) (the range of cot ). 2 9 the function. of 6 6 + 3 + 2 . ⇒ 6 ≥ −27315, 0Domain ≥−1 −49167 domain inverse 3 5 sec [−1 (the range of sec−1 2= sec=3Range = 2 and is. in )..1 + (b) (b) of 3 because = [−3 3] ofthe 3−1 Range −1 ∪ = 0 2of √ 1+ 2 3] = Domain of −1 −1 3 ∪ (the range of sec 2by= 3] because sec−3 =2= 2 and is in 0 ). (b) sec each term ⇒ = ln ⇒ − − 1 ⇒ ( − 1) = − − 1 ⇒ = . ( + 2) 3 2 2 2 ln 3= ln 2 2 SECTION 2 [by AND ln −1≥ ln 3 ⇒ √ (b) ≥ 3 ⇒ 1] 1 − Law 1.5√ 2INVERSE 1 FUNCTIONS − 20 ¤ 53 ≥ ( + 1)( + 0 0−2) 0is in − .LOGARITHMS 2 √2 = 0.−1 (c) = Since = −1 68. arcsin(sin(54)) =(2) arcsin sin and − √20 =⇒− 4 because (0) = 2, ⇒ 1 − ⇒ = 20. (a) 1 − = = 1 − ⇒ = 1 − . 4 4 2 2 2 2 2 2 2− = −1 2 and − 2 68. (a) arcsin(sin(54)) is in − 2 2 . 13−1 − 2 2 = arcsin + 1+−1 √2 =−1− 4 because √ 4 1sin − ln 2 ⇒ 1 3 √ 0 −1 1 + ln 2 ⇒ 3 56. (a) 1 −1 lncsc 1the 3 −.21So ln 2is ⇒ 14. −12:5 ⇒ −1 () =0ln Interchange = and 0,=[see ∪3 (the range of csc 65. (b) (a) csc 2sin because = and in ). figure]. = (0) ≈ −17. (d) Let Since (−17) ≈ 4= ln 2 1− 2 −1 14−of the particle −14 13 5 the speed ThisLet gives us [see the figure]. (b) 1 formula 1 + 1of its mass , that is, = (). in terms 13 =sin (1 ln 2) + −1 2 2 5 3 3 (the range of arcsin). sin 2 = and in −⇒ 1 5 because cos sin= =91cos −issin 2 = cos 2 2 = 9 2+ 32. 19. (b) Wearcsin solve = ( This gives us a formula for the inverse function, that 13 √ −12 2 2 − 32 2 5 − 32) for2 : 5 = 9 5 − 3 +to2a publicly all positive, hence their since ln is3Learning. defined for 0, have +√ 1, + 23+ 2° sin = cos 2 = cos 2we − sin be3scanned, c= 20164 Cengage All Rights Reserved. May not+ copied, duplicated, accessible and website, in whole or in logarithms part. 13 .or posted 27. Note =cos that () + ( ≥ 0) ⇒ = 4 ⇒ 2, =or and √ 2 5 2 ⇒144 −1 12 ln 4−1 9 (b) 1 − 2 ln 3 ⇒ −2 2 ln −1 ⇒ 25 119 66. are (a) sin Fahrenheit (−1 2temperature ) == − 4 13because sin − 2 and − is in − = −1 . is, the as a function of the Celsius temperature . ≥ −45967 ⇒ + 32 ≥ −45967 ⇒ − = − = defined. 4 4 2 2 5 13 169 169 169 12 2 5 2 144 25 = = 119 √ 169 − 169 √ = 132 − 13 169 2 9 −1 0−27315, of−the −⇔ 3 =the 3in We ≥ −49167 ⇒− ≥ domain inverse function. ln 10 lnof 57 () = ln( − 3) is (ln 3 ∞). −1 3 32 = because cos 32⇔ [0≥ ]. 57. (b) (a) musthave : 3. Thus, the domain 5 cos 6 () So ( 0).log From and 6 3= =and6 is ln [by (10)]4 ≈ .1430677 (b) [by (10)] ≈ 3680144 42. Interchange (a) log5 10 = 3 57 = 4 ln 5 ln 3 2 √ √ 5 √ 2 2 2 5 2 2 −1 5 and of cot−1 −1 −1 −1 2 = ln( + 3). − 0 − 67. (a) cot 3 = because cot = − 3 is in (0 ) (the range ). 0 0 0 sin 2 cos+ 2 3), = ln( = ≤ = − 3 ⇒ = + 3 ⇒ = ln( so () 69. Let sin−1 − . 3) Then⇒ ≤ ⇒ cos ≥ 0, so cos(sin ) = = 1 = 1 − 21. − 0 . √ 6− 2 6 6 −1 ln ln 1 − ⇒ = 20. (b) = ⇒ 1 − ⇒ = 1 − ⇒ = 2reflections about the line = −1 2 the graph, we see that and are . 2 2 2 2 2 2 and 43. functions, log215 ⇒= cos ≥ = 2 69. To Letgraph =−1sin Then − we ≤use ≤ 0, so cos(sin ) .= cos =−1 1− sin = 1 − . 2 log 50 1these − 2. 2 ln 15 3 ln 50 ∪ (the range of sec because 2 and is in 0 ). is R. (b) sec + 33 0 ⇒ sec3 =−3, which is true for any real , so the domain of −1 Now 2 = 3 2 2 −1 + (). This formula gives us the speed of the particle in terms of its mass , that is, = −1all approach −∞ as √ √ These graphs → 0 , and they all pass through the 70. Let = sin−1 . Then sin = , so from the triangle (which ln 300 300 = −1 68. (a) arcsin(sin(54)) = arcsin −1 2 = − because sin − 2 and − is in − 2 2 . 70. Let =(9), sin . Then sinand =ln( , so from the triangle (which 4 4 4 58. (a) By = 300 ) = 300. point (1 0). Also, they are we allAll− increasing, and all approach ∞ as → ∞.or posted to a publicly accessible website, in whole or in part. illustrates Learning. see that − 0), cthe ° Cengage Rights Reserved. scanned, copied, duplicated, 28. (b) =Let () =2016 1case + ⇒[see − 1 May ⇒not be− = ln( −or 1) ⇒ −1 5 figure]. = sin illustrates the case 13 ln 0), wethe see = that 300 300 300 increase (b) calculator gives bases = 300 and extremely an error message ) since is larger than most calculators can evaluate. TheAfunctions with larger slowly, for andln( the ones with −1 √ ) = tan = . tan(sin and = − ln( 1). Interchange : = − ln( − 1). 2 2 5 −1 −−1 2 cos − sin 1 − cos 2 sin = cos 2 = √ ) = tan = . tan(sin √ smaller bases do so13somewhat more The functions with large bases 2=quickly. 3 59. We see that the graph of =1− () 2 + + 1 is increasing, so is 1-1. −1 2 +144 5graph, 2 the From we see that and are So −1 () = − ln( −=1). 12 25 119 −as 13→ = . − 169 = 169 approach the 0+169 -axis more closely 13 2 + + 1 and use your CAS to solve the equation for . Enter = about 3 + reflections the line = . 44. We that −1 graph of ln is the reflection of the graph of about the 71. Using Let see = tanthe . Then tan = , sosolutions from the involving triangle (which Derive, we get two (irrelevant) imaginary expressions, −1 √ 71. Let = tan−1 . Then tan −1 = , so from the triangle (which 69. line Let = .that Then − ≤of log ≤that ≥ 0, so )10 =cos = 1 − sin2 = 1 − 2 . ⇒ is thecos reflection of cos(sin the graph of = sin , and graph illustrates the see 10 to 2we 2 as well as the onecase which can0),be simplified the following: illustrates the case 0), we see that of 10 increases −1same line. √ √ more =√ about The graph than(1 that−1), 3 29. sin(tan Reflectthe the graph of√ line2. = . The pointsquickly (−1 −2), ) =−sin 3 √ the −1 4 about 27 −1 = () = − 20 − 3 + 272 − 20 + 3 2 −1 1 + , 2so 6 √ =sin = . +from 70. sin(tan Let = sin) =.sin Then the triangle (which 2 + ∞ note 1 → as (−2 → −1), ∞ more slowly of √ √ log 10reflected (22),. Also and (3 3)that on are to (−1 1), (2than 2), ln and.(3 3) 4 −we 2 + where = 3 27 40 16. illustrates the3 case 0), see that on −1 . [continued] . tan(sin−1 ) = tan = √ 1 72. Let = arccos . Then cos − = ,2 so from the triangle (which 72. Let = arccos . Then cos All= , Reserved. so from thenottriangle (whichor duplicated, or posted to a publicly accessible website, in whole or in part. cc 2016 ° Cengage Learning. Rights 2016 Cengage Learning. Allsee Rights Reserved. May May not be be scanned, scanned, copied, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. illustrates° the case 0), we that illustrates the case 0), we see that sin(2 ) = sin 2 =Reserved. 2 sin Maycos scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. c 2016 arccos ° Cengage Learning. All Rights not be −1arccos ) = sin sin(2 sin the costriangle √ (which 71. Let = tan . Then tan √ =2,=so22from = 2(√1 − )() = 2 √1 − 2 1− 2 )() = 2 1 − 2 2( see illustrates the case 0),=we that 73. −1 73. sin(tan ) = sin = √1 + 2 . The graph of sin−1 is the reflection of the graph of The graph of sin−1 is the reflection of the graph of sin about the line = . sin about the line = . 72. Let = arccos . Then cos = , so from the triangle (which illustrates the case 0), we see that sin(2 arccos ) = sin 2 = 2 sin cos √ √ = 2( 1 − 2 )() = 2 1 − 2 1 c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ° −1 to a publicly accessible website, in whole or in part.
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