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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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