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# PRECALCULUS ALGEBRA MAC 1140

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This 7 page Class Notes was uploaded by Nathen Fadel on Thursday September 17, 2015. The Class Notes belongs to MAC 1140 at Florida State University taught by Staff in Fall. Since its upload, it has received 34 views. For similar materials see /class/205607/mac-1140-florida-state-university in Calculus and Pre Calculus at Florida State University.

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Date Created: 09/17/15

Section 38 Polynomial and Rational Inequalities Must understand 33 Solving Polynomial lnequ es by Graphing Recall To graph the polynomlal l Ends behavlur 2 Zeros and melr mulllpllelly lr me pulynumlal ls nutmlly raeleree We need to factor ll rst a Determlrle leeal behavlur near zeros CROSS EVEN multlpllclty or TOUCH ODD multlpllclty A Cenneelme graph To solve me correspondlng polynomlal unequallw we need an extra step 5 Observe me graph anerlne eulme lnlervals en Whlch me lnequallly ls sallsrlee Exemsel 38 1aPTSolve 32 2 3z 7 23 7 at lt 0 0 none of these 3 l O 00 5 U 012 U 300 0 g U 300 0 gnol U 23 0 mvU2l3 Exemsez O 381BPTSolve 2139 7 I S 3 O 00 1 U 3700 O 1 0 None of these 0 wv glUllym fx and xpxqx x f0 PKqx rm px and qx px and qx px and qx px and rm conclusion The ranuna functer fx 2 andme pu ynumwa fxpxqx qx assume we same 5 ng Where bum are We Lde ned Notice Aunuugn may assume we same 5 ng Wherever de ned mew graph are NOT me same The rauun functer mews up andfurms asympmtes nearme Items at em Then the steps for solving rational Inequalmes are Transfurmt e ratmna mequahtytu eqmva ent puxynumax mequahty 2 vae we puxynumax mequahty by grapmng a mnenun s unde ned atmese pumts Examplez 382aPTSolve 0 None of these 0 590 0 ever U 500 0 45 0 Peer U 500 A matrix is in echelon form when 1 Each row containing a nonzero number has the number 1 appearing in the row s first nonzero column Such an entry will be referred to as a leading one 2 The column numbers of the columns containing the first nonzero entries in each of the rows strictly increases from the first row to the last row Each leading one is to the right of any leading one above it 3 Any row which contains all zeros is below the rows which contain a nonzero entry The three conditions above will ensure that the entries below the leading ones in each row which contains a non zero entry are all zeros A matrix is in reduced echelon form when in addition to the three conditions for a matrix to be in echelon form the entries above the leading ones in each row which contains a nonzero entry are all zero s Note that if a matrix is in Reduced Row Echelon Form then it must also be in Echelon form To Determine if a Matrix is in Reduced Row Echelon Form Circle the first nonzero entry in each row of the matrix Then verify that 1 any row with m nonzero entry is at the bottom of the matrix 2 the circled entries are all 1 s and will be referred to as leading one s 3 each leading one is to the right of any leading one above it and 4 all entries above and below the leading one s are zeros that is all other entries in the same column as a circled 1 are zeroes f conditions 14 above are satisfied then the matrix is in Reduced Row Echelon Form If conditions 14 above are satisfied with the possible modification to condition 4 that entries below the leading one s but not necessarily above the leading one s are zeroes then the matrix is in Echelon Form Examples of Matrices M in Echelon Form 3910121012 0011 111113013 0105 005002102 391010 1020 0000 010010 1 0101 00010 12 In Echelon Form but not Reduced Row Echelon Form 1 0 1 2 1 0 2 1 1 1 3 1 2 3 0 1 1 1 1 2 0 0 1 0 1 2 0 0 0 1 0 0 1 ln Reduced Row Echelon Form 1 0 3 1 0 1 1 2 0 0 1 1 0 0 0 0 0 1 1 O 1 O 1 O 2 O 1 0 2 1 O 1 1 O O 1 O 0 0 1 0 2 O O O 1 O O O 1 0 0 0 0 Any matrix can be put in an equivalent Echelon Form using elementary row operations Such a matrix is not unique For instance the two elementary row equivalent matices below are both in echelon form 1 2 3 1 0 1 0 1 2 ref 2132 1 l0 1 2 However the equivalent matrix in Reduced Row Echelon Form is unique Sectinn 111 Sequences Sequence Mutations Dennmon A sequence 5 a hst er n mtE y many eruereu numbers whese mmEES are pusmve mtegers startth hum 1 1 ntntenn formula We use an te uehete the sequence 111112413 The f mu a m the ur y brackets QWES the funnu a fur rth term Exercise1 1111aPTThe rst four terms of the sequence WMT are Exercisez check If formulas give the ngnt rst 4 terms 1111cPT The n term nfthe sequence 1 53 O 1quot2 4quot O 11I ln O 1quot quot 1 2 Recursive formula A sequence can be uehheu recurswew re except the gWEn rst term each term at the sequence rs uehheu by a furmu a rhvewrhg the prevreus terms Exercised 1111bPTlThe rst tour xerms ut39 the x39urs39n39e sequenc even by a 211 quot are D 21312

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