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SOLUTIONS for Exam # 1

SOLUTIONSforExam# 2 1012f(x) 120 21g(x)20 12 1(f g)(x)10212(g f)(x) pointsgClassifythefunctionsas even,odd,or (x) (x) (x)2X2x x3X22x2+ pointsgStatethedomainof thefunctionf(x) =xpx (2;1)orx >2We have thatx 2 0forthesquareroot andalsothedenominatorshouldnotbe0, 26= pointsgGiventhefunctionsf(x) =x3x 1andg(x) =1x2 1, ndthecompositionf gandstateitsdomain.(f g)(x) =1x2 13(1x2 1) 1=13 (x2 1)=14 x2=1(2 x)(2+x).D=fall realx6= 1; pointsgIn each part, ndtheinversef 1(x)if theinverseexists.(a)f(x) = cos(2x) ,0 x 2We have that0 2x forwhich intervalcos 1is de ,y= cos(2x)is equivalent tocos 1y= 2xandthereforex=12cos ,f 1(x) =12cos 1x.(b)f(x) = sin(2x) ,0 x 2 Thefunctionsin(2x)is notone-to-onefor0 2x (we have thatsin(2x) = sin( 2x)) andthereforethereis pointsgCompletetheidentity tan 1x =1p1+x2tan 1 x1x 1+ pointsgSolve forxwithoutusinga ln 1x + ln(2x4) = ln 8We have that2 ln 1x + ln(2x4) = ln 1x 2(2x4) = ln 2x2 .So,ln 2x2 = ln 8andtherefore2x2= 8which givesx2= 4andx= ,thedomainforln 1x isx <0andthereforex= 2shouldbe ,thesolutionisx= !

6. f 10 points g Complete the identity using the triangle method. cos ¡ tan¡1 x = p 1 1+x2 tan−1 x 1 x Ö1+x2 7. f 10 points g Solve for x without using a calculating utility.

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Transcription of SOLUTIONS for Exam # 1

1 SOLUTIONSforExam# 2 1012f(x) 120 21g(x)20 12 1(f g)(x)10212(g f)(x) pointsgClassifythefunctionsas even,odd,or (x) (x) (x)2X2x x3X22x2+ pointsgStatethedomainof thefunctionf(x) =xpx (2;1)orx >2We have thatx 2 0forthesquareroot andalsothedenominatorshouldnotbe0, 26= pointsgGiventhefunctionsf(x) =x3x 1andg(x) =1x2 1, ndthecompositionf gandstateitsdomain.(f g)(x) =1x2 13(1x2 1) 1=13 (x2 1)=14 x2=1(2 x)(2+x).D=fall realx6= 1; pointsgIn each part, ndtheinversef 1(x)if theinverseexists.(a)f(x) = cos(2x) ,0 x 2We have that0 2x forwhich intervalcos 1is de ,y= cos(2x)is equivalent tocos 1y= 2xandthereforex=12cos ,f 1(x) =12cos 1x.(b)f(x) = sin(2x) ,0 x 2 Thefunctionsin(2x)is notone-to-onefor0 2x (we have thatsin(2x) = sin( 2x)) andthereforethereis pointsgCompletetheidentity tan 1x =1p1+x2tan 1 x1x 1+ pointsgSolve forxwithoutusinga ln 1x + ln(2x4) = ln 8We have that2 ln 1x + ln(2x4) = ln 1x 2(2x4) = ln 2x2 .So,ln 2x2 = ln 8andtherefore2x2= 8which givesx2= 4andx= ,thedomainforln 1x isx <0andthereforex= 2shouldbe ,thesolutionisx= !

2 0px2+ 9 3x=limx!0px2+ 9 3xpx2+ 9 + 3px2+ 9 + 3=limx!0(x2+ 9) 32xpx2+ 9 + 3=limx!0x2xpx2+ 9 + 3=limx!0xpx2+ 9 + 3=0p0 + 9 + 3= !+1 1 5x x=limx!+1 1 5x x5( 5)= limx!+1 1 5x x5! 5=e 5=1e5 !+1(3 2x)3x(7 +x+ 9x2)=limx!+1(3 2x)31x3x(7 +x+ 9x2)1x3=limx!+1 3 2xx 37+x+9x2x2=limx!+1 3x 2 37x2+1x+ 9=(0 2)30 + 0 + 9= 89 !0tan(3x)sin x2 =limx!0 tan(3x)3x x2sin x2 ! 3xx2 = limx!0sin(3x)3xcos(3x) limx!0x2sin x2 ! limx!0312 = (1)(1)(6)= 6 pointsgA positive number"= 0:01anda limitL= 4forthefunctionf(x) =9x2 43x 2ata=23aregiven:limx!239x2 43x 2= 4:Finda number such thatjf(x) Lj< "if0<jx aj< .We havejf(x) Lj= 9x2 43x 2 4 = (3x 2)(3x+2)3x 2 4 =j3x+ 2 4j=j3x 2j= 3 x 23 <3 for0<jx aj= x 23 < .So,if we set such that3 "= 0:01, forexample = 0:0033,we shallhave thatjf(x) 4j<0:01if0< x 23 < .Bonus !0 1 cos(2x)xsin(2x)+xsin (1=x) = limx!01 cos(2x)xsin(2x) + limx!0xsin (1=x) .For the rstlimitwe havelimx!01 cos(2x)xsin(2x)=limx!01 cos(2x)xsin(2x)1 + cos(2x)1 + cos(2x)=limx!

3 01 cos2(2x)xsin(2x)(1 + cos(2x))=limx!0sin2(2x)xsin(2x)(1 + cos(2x))= limx!0sin(2x)x(1 + cos(2x))= limx!0sin(2x)2x21 + cos(2x)= (1)21 + 1= thesecondlimitwe shallusetheinequalities 1 sin (1=x) 1andtherefore jxj xsin (1=x) !0asx!0by theSqueezingTheoremwe havethatlimx!0(xsin (1=x)) = ,limx!0 1 cos(2x)xsin(2x)+xsin (1=x) = 1 + 0 = 1.


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