O C. PG.-3 #, 3b, 4, 5ce O C. PG.4 # Optios: Clculus O D PG.8 #, 3, 4, 5, 7 O E PG.3-33 #, 3, 4, 5 O F PG.36-37 #, 3 O G. PG.4 #c, 3c O G. PG.43 #, O H PG.49 #, 4, 5, 6, 7, 8, 9, 0 O I. PG.53-54 #5, 8 O I. PG.55-56 # O K O L WS WS O M PG.0- #3, 7, 8 O N PG.5-6 #, b, 7 O O PG.9 #, O P PG. #, 3
Check the powers of the umertor d deomitor. Review LIMITS THAT APPROACH INFINITY ) If the deomitor (bottom) is bigger power the it = 0. ) If the umertor (top) is bigger power the it = or -. 3) If powers re the sme the it = coefficiet of the highest power of umertor coefficiet of the highest power of deomitor EX#: 3 + 3 = EX# : 7 7 6 = EX#3 : 3 + 3 = EX#4 : 6 3 = EX#5 : 9 + 5 7 = EX#6 : 7 + 8 7 = HOLES IN THE GRAPH 0 0 Fctor d ccel or multiply by the cojugte d ccel, the plug i # EX#: + 3 40 8 = EX# : 5 5 8 64 = EX#3 : + = EX#4 : 7 + 7 4 4 = EX#5 : 3 3 3 = EX#6 : 4 =
RADICALS Optio B Limits ( You must first check tht the it eists o the side(s) you re checkig) If # mkes rdicl egtive, the it will ot eist t tht #. Whe we check t the brekig poit (the # tht mkes the rdicl zero) there re two possible swers: ) 0 if the it works from the side tht you re checkig. ) DNE if the it does ot work from the side tht you re checkig. EX #: 5 + 5 = EX # : 5 = EX #3 : 5 5 5 = EX #4 : 3 6 = EX #5 : 7 = EX #6 : = + 7 0 f ( ) = 3 5 < + < 6 PIECE - WISE FUNCTIONS The brekig poits re d. EX #: EX #4 : EX #7 : + = EX # : f f f + = EX #5 : = EX #8 : = EX #3 : f 0 f f ( ) = f ( ) = EX #6 : f = = EX #9 : = 4 f ASYMPTOTES (# 0 ) (Sice the poit DNE we hve to check poit tht is close o the side we re pprochig) There re three possible swers whe checkig er the brekig poit (the # tht mkes bottom = zero) ) If we get positive swer the it pproches ) If we get egtive swer the it pproches 3) DNE If we get positive swer o oe side d egtive swer o the other side, the the it DNE EX #: + 6 5 4 = EX # : = EX #3 : 8 + 8 7 7 7 7 = EX #4 : 0 3 = EX #5 : cos ( 0) 0 = EX #6 : t = π 3
TRIG. FUNCTIONS Trig. Idetities to kow : si = si cos cos = cos si FACTS : 0 0 si si b = 0 = b 0 cos cos b = 0 0 = 0 0 t = t = b b EX#: 0 si 7 = EX#4 : 0 6si t = EX# : 0 5 t 7 6 = EX#5 : 0 8si cos 3 = EX#3 : π si = EX#6 : 0 + si t 5 = Optio C Cotiuity of Fuctios Optio D Differetible Fuctios CONTINUITY Cotiuous fuctios hve o breks i them. Discotiuous fuctios hve breks i them (Asymptotes or Holes / Ope Circles). ** To check for cotiuity t, you must check left hd its s well s the vlue of the fuctio t tht poit f d right hd its f + f ( ). If ll three re equl the the fuctio is cotiuous t. If f = f ( ) + f ( ) the the fuctio is cotiuous t. If f is ot equl to either oe - sided it, the the fuctio is ot cotiuous (discotiuous) t. Cotiuous t Discotiuous t 4
Cotiuity / Differetibility Problem We must show fuctio is cotiuous before we discuss its differetibility. If fuctio is ot cotiuous, the it cot be differetible. = To show cotiuity, we must show tht f To show differetibility, we must show tht f + = f f = + f. = f ( ) d tht f eists. EX#: f ( ) = =, < 3 6 9, 3 At 3, f ( ) = ( 3) = 9 9 = 9 At 3 f f 3 3 + f () is cotiuous iff both hlves of the fuctio hve the sme swer t the brekig poit. 6 3 = f ( 3 ) = 9. Therefore f () is cotiuous., < 3 f ( ) = 6, 3 At 3 both hlves of the derivtive = 6. Therefore the fuctio is differetible. Sice both sides of f At 3, f () = ( 3) = 6 6 = 6 f () is differetible if d oly if the derivtive of both hlves of the fuctio hve the sme swer t the brekig poit. d f ( ) gree t 3, the f ( ) is cotiuous d differetible t = 3. EX# : b c d To be cotiuous grph c't hve y holes or symptotes. To be differetible grph c't hve y holes or symptotes or hrd poits. The grph is cotiuous t the poits d d. The grph is differetible t the poit oly. (d is hrd poit) 5
Optios E Idetermite Forms d L'Hôpitl's Rule Norml Forms Idetermite Forms 0 0 d, 0 0, 0,, 0 ***If it is i idetermite form, we covert it to orml form the use L'Hôpitl's Rule. L'Hôpitl's Rule: If = 0 0 or f g, the = f g f g ( ) EX#: + 3 0 si = EX# : = 0 EX#3 : e = 3 EX#4 : e = EX#5 : 3e 3 3 = 0 EX#6 : l = 0 + EX#7 : = EX#8 : 8 3 + 9 3 = 6
Optio F Rolle's Theorem d the Me Vlue Theorem (MVT) *Me-Vlue Theorem (Oly pplies if the fuctio is cotiuous d differetible) f(b) 4 Slope of tget lie = slope of lie betwee two poits f (c) = f (b) f () b f() b5 Accordig to the Me Vlue Theorem, there must be umber c betwee d b tht the slope of the tget lie t c is the sme s the slope betwee poits (, f ()) d (b, f (b)). The slope of sect lie from d b is the sme s slope of tget lie through c. 4 f(b) f() c b5 Use the Me Vlue Theorem to fid ll vlues of c i the ope itervl (, b) such tht f (c) = EX #: f ( ) = si [ 0, π ] EX # : f ( ) = + 7 [, ] f (b) f () b EX #3 : f ( ) = 3 [ 0, 5] 7
Summtio Formuls ) c = c i= ) i = + i= Optio G Riem Sums ( Fidig Are Usig Limits) 3) i = + 6 i= ( + ) 4) i 3 = + 4 i= Summtio Properties 5) i ± b i = i ± b i i= i= Are = f + b i b i=!## "## $!" # $# 6) k i = k i, where k is costt i= i= i= height width i = itervl, = # of subdivisios EX#: f ( ) = 3 [ 0,] subdivisios EX# : f ( ) = [, 3] subdivisios EX#3 : A) + +... = d B) d C) d D) d E) d 0 0 0 0 8
Optios H Averge Vlue d The ND Fud. Thm. of Clculus Averge Vlue (use this whe you re sked to fid the verge of ythig) If f is itegrble o the closed itervl [,b], the the verge vlue of f o the itervl is EX#: Averge vlue = b b f () d Fid the verge vlue of f () o the closed itervl. ) f ( ) = 0,3 [ ] b) f [ ] = 3, EX # : Fid verge ccelertio from [ 0, ] give v(t) = t 3 + t 5 d Fudmetl Theorem of Clculus (Whe tkig the derivtive of itegrl) Plug i the vrible o top times its derivtive mius plug i the vrible o bottom times its derivtive. EX #3 : ) d d d d dt f t = f 0 Evlute ech. d d t 3 dt = b) 0 f ( t) dt = f ( ) f ( 0) 0 = f 0 d d 0 t 3 dt = c) F dt = t 7 + = d) F F ( ) = F si = + t dt = = 9
f ( ) d = b b Optios I Improper Itegrls of the Form f d f = c d = f ( ) d + f ( ) F( ) b b = F( b) F b f d where c is udefied f b c d = b c c f d d + c + If f d = fiite # the the fuctio coverges. If f d = the the fuctio diverges. b c f d EX#: d = EX# : d = EX#3 : e d = 0 EX#4 : d = + 0 BONUS : d = 0 0
Sequece : Optio J Sequeces A sequece is ordered set of mthemticl objects. Sequeces of object re most commoly deoted usig brces. A sequece coverges if the it s pproches = 0. EX#: Simplify ech ) 0! 9! b) 7! 0! c) ( +)!! d) ( +)! ( + )! e) ( + 3)!! EX# : Fid the first five terms of the recursively defied sequece k+ = 5 k +0 ; = = 3 = 4 = 5 = EX#3 : ) = 3 + 5 7 Fid the it of ech sequece. b) = 7 8 4 c) = 6 + 8 9 + d) = 5
Optio K Ifiite Series ( th term/geometric Series) Ifiite series: = + + 3 + 4 +...+ +... Defiitio of Coverget d Diverget series : For the ifiite series, the th prtil sum is give by S = + + 3 + 4 +...+. If the sequece of prtil sums {S } coverges to S, the the series coverges. The it S is clled the sum of the series. S = + + 3 + 4 +...+ +... If {S } diverges, the the series diverges. Geometric series: r = + r + r + r 3 +...+ r +... 0 =0 ***Geometric series Test : A geometric series r coverges iff r <. A geometric series r diverges iff r. =m If geometric series coverges, it coverges to the Sum : S = r *** Test for Divergece ( th term test ) (Used to show immedite divergece) If = 0 the it my coverge. (bottom power is greter d we must proceed d use differet test, becuse this test cot prove covergece). If 0 the it diverges (top power is the sme or greter th the bottom power). =m EX #: Test for Covergece ) 3 b) =0 3 c) =0 =0 + EX # : A bll is dropped from height of 6 feet d begis boucig. The height of ech bouce is 3 4 the height of the previous bouce. Fid the totl verticl distce trvelled by the bll.
***Telescopig Series re coverget Fid the sum of the followig telescopig series. EX #3 : + ( Limit = 0 d the terms get smller s we pproch ) EX #4 : + EX #5 : = 4 ( +) Repetig decimls 0... 0.333333... 0.444... 0.833333... 3
Optio K Ifiite Series Itegrl Test/ p-series ***Itegrl Test : If f is positive, cotiuous d decresig for d = f, the d f d either both coverge or both diverge. EX #: Use Itegrl Test to test for covergece ) b) + + ***p-series: p ) Coverges if p > ) Diverges if 0 < p If p =, it is clled the hrmoic series. is the diverget hrmoic series EX # : Use p-series to test for covergece ) b) 3 c) 7 3 EX #3 : Review (Test ech series for covergece) ) 7 9 0 b) + 7 5 c) 6 6 + d) 7 4
***Direct Compriso Test Let 0 b for ll. Optio K Ifiite Series ( Compriso of Series) ) If b coverges, the lso coverges. ) If diverges, the b lso diverges. If coverges d b diverges. b C't tell Coverges Diverges b ***Limit Compriso Test Suppose the > 0, b > 0, d = L, where L is fiite d positive. The the two series d b b either both coverge or both diverge. Assume > 0 If b coverges, the b coverges. If b diverges, the diverges *** Limit test works well whe comprig "messy" lgebric series with p-series. I choosig pproprite p-series, you must choose oe with th term of the sme mgitude s the th term of the give series. Determie Covergece d Divergece for ech EX #: EX # : + 3 + EX #3 : EX #4 : + + 5 5
***Altertig Series Test Optio K Ifiite Series ( Altertig Series) Let > 0. The ltertig series d coverge if the followig two coditios re met. + ) = 0 (the bottom is bigger th the top) ) + for ll. (ech succeedig term is gettig smller th the precedig term) This test does ot prove divergece. If the codtios re ot met, usully bottom ot smller th the top, the the series diverges by th term test. EX #: ( ) + EX # : ( ) + + *Altertig Series Remider If coverget ltertig series stisfies the coditio +, the the bsolute vlue of the remider R ivolved i pproimtig the sum S by S is less th or equl to the first eglected term. Tht is, S S = R + EX #3 : Approimte the sum of the series by its first 6 terms (S 6 ) d fid the remider (error). ( ) +! 6
Absolute Covergece (Altertig Series) If the series coverges, the the series lso coverges. Defiitio ofabsolute d Coditiol Covergece ) is bsolutely coverget if coverges. ) is coditiolly coverget if coverges, but diverges. Determie if ech series is for bsolutely coverget, coditiolly coverget, or diverget EX #: EX # : EX #3 : + ( ) EX #4 : 3 + 8 9 EX #5 : ( ) EX #6 : + 7
***Rtio Test : Let Optio K Ifiite Series The Rtio d Root Test be series with ozero terms ) coverges bsolutely if + < ) diverges if + > or + = 3) The rtio test is icoclusive if + = EX : ( ) 5 ( +) 5 + 5 = ( +) 5 + 5 = ( +) 5 = 5 so series Coverges ***Root Test : Let be series with ozero terms ) coverges bsolutely if < ) diverges if > or = 3) The root test is icoclusive if = EX #: EX # :! =0 =0 + 3 EX#3 : e EX #4 : + 8
Optios L Tylor Polyomils d Approimtios Defiitio of th Tylor Polyomil d McLuri Polyomil ( c = 0 ) If f hs derivtives t c, the the polyomil: P ( ) = f c! is the th Tylor Polyomil for f t c. + f ( c) ( c) + f ( c) ( c) + f ( c) ( c) 3 +...+ f (c)! 3!! ( c) If c = 0, the + f ( 0) ( 0) + f ( 0) ( 0) + f ( 0) ( 0) 3 +...+ f (0) P ( ) = f 0!! is clled the th McLuri polyomil for f. Tylor series of f ( ) this is power series EX #: =0 f c Fid the Tylor Polyomils P 0, P, P, P 4 for f 3! ( c)! = l cetered t.! ( 0) EX # : Fid the fifth degree McLuri Series ( cetered t c = 0) for f ( ) = si 9
EX #3 : Fid the fourth degree McLuri Series for f ( ) = cos. EX #4 : Fid the fourth degree McLuri Series for f ( ) = e EX #5 : Fid the fourth degree Tylor Polyomil for f ( ) = si cetered t c = π 4. 0
Fuctio Power Series for Elemetry Fuctios Itervl of Covergece = ( ) + ( ) ( ) 3 + ( ) 4!+ ( ) ( ) +! 0 < < + = + 3 + 4 5 +!+ ( ) +! < < l = + ( )3 3 ( )4 +!+ 4 +! 0 < e = + +! + 3 3! + 4 4! + 5 +!+ 5!! +! < < si = 3 3! + 5 5! 7 7! + 9 9!!+ + ( +)! +! < < cos =! + 4 4! 6 6! + 8 8!!+ +! < <! rct = 3 3 + 5 5 7 7 + 9 9!+ + +! + rcsi = + 3 35 3 57 + + 3 4 5 4 6 7 +!+! +! + +! k 3 ( + ) k = + k + k k + k k + k k! 3! *The covergece t = ± depeds o the vlue of k. ( k ) k 3 4! 4 +! < < *
Optios L Tylor d McLuri Series We will fid other series by djustig kow series. We c djust by ddig/subtrctig, multiplyig/dividig, replcig(substitutig) or tkig derivtive/itegrl. Tylor Series to memorize : si = 3 3! + 5 5! 7 7! + 9 9!... cos =! + 4 4! 6 6! + 8 8!... e = + +! + 3 3! +! = + + + 3 + 4! EX #: Give cos =! + 4 4! 6 6! +... Fid ech: cos = si = si = EX # : Give + = + 3 +! Fid rct
EX #3 : Give e = + +! + 3 3! +! Fid e 0 d EX #4 : Fid si (si = cos ) Formuls for sig chges ( ) ( +) = + + + + + ( ) ( +) ( +) = + + + + ( ) ( +) ( +3) = + + + + ( ) ( +3) ( +4) = + + + + 3
Lgrge Error Boud Error = f ( ) P ( ) R ( ) where R ( ) f + ( z) ( c) + f + ( z) is the MAXIMUM of + ( +)! th derivtive of the fuctio EX : cos =! + 4 4!... c = 0 Approimte vlue of cos. cos P 4 ( ) f f 0. 0 5 5!! P 4 (.)! 995004667 Actul vlue of cos (.) = 995004653 = 0. 000000083 = cos ( ) = si ( ) = cos ( ) = si ( ) = cos ( ) = si ( The MAXIMUM of si is ) f f f 4 f 5 EX #: f =, f = 5, f = 7, f = ) Write d Degree Tylor Polyomil b) Use Tylor Polyomil to pproimte. c) Lgrge Error Boud t. 4
Power Series Whe fidig the itervl of covergece we use either Geometric Series Test, Root Test or Rtio Test. Fid the itervl of covergece, rdius of covergece d the ceter for ech emple below. EX #: 3 EX # : =0 =0 itervl of covergece : itervl of covergece : rdius of covergece : rdius of covergece : ceter : ceter : EX #3 : ( ) ( +) EX #4 : =0 =0 ( 5)! itervl of covergece : itervl of covergece : rdius of covergece : rdius of covergece : ceter : ceter : Whe fidig the itervl of covergece : If = 0 the the power series coverges from,. If = the the power series coverges t the ceter oly. 5
Optios M Differetil Equtios A differetil equtio is equtio ivolvig the derivtive(s) of ukow fuctio. FIRST ORDER DIFFERENTIAL EQUATIONS We will work with differetil equtios of the form f (, y) dy = 0 where y = y(). d + g, y These re kow s first order differetil equtios sice there is oly oe derivtive i the equtio, d it is first derivtive. SOLUTIONS OF DIFFERENTIAL EQUATIONS A fuctio y() is sid to be solutio of differetil equtio if it stisfies the differetil equtio for ll vlues of i the domi of y. Drw the slope field for ech EX #: Slope Fields dy d = y EX # : dy d = y - 0 - - - - 0 - - Here re the slope fields for the give differetil equtios. Sketch the solutio for the give poit. EX #3 : dy d = y ( 8 6 y) EX #4 : dy d = + y 6
Euler's Method New y = Old y + d dy d d = chge i (the step), dy = derivtive(slope) t the poit d EX #: y = y pssig thru ( 0, ), EX # : dy d = y pssig thru, 4 step of h = 0.. Fid f (0.3) = step of h = 0. 5. Fid f () = 7
Optio N Seprble Differetil Equtios (used whe you re give the derivtive DIFFERENTIAL EQUATIONS Seprtig Vribles EX #: ) d you eed to fid the origil equtio. We seprte the 's d y's d tke the itegrl). Fid the geerl solutio give dy d = + 4 b) y 3e y = 0 3y EX # : Fid the prticulr solutio of y dy l = 0 y() = 8 d dy EX #3 : AP Test 000 BC#6 d = y ) Fid the prticulr solutio y = f give f ( 0) = b) Drw slope field t poits idicted. - - 0 - - 8
Homogeous Differetil Equtios Differetil equtios of the form dy d = f y, where y = y(), re kow s homogeous differetil equtios. They c be solved usig the substitutio y = v where v is fuctio of. The substitutio will lwys reduce the differetil equtio to seprble differetil equtio. EX #: Solve the homogeous differetil equtios usig the substitutio y = v, where v is fuctio of. ) dy d = + y, y(3) = 3 y = v dy d = dv d + v dv d + v = + v dv + v = + v d dv d = + v dv + v = d l + v = l + C Fid C + v = C + = C 3 b) dy 3y y = d, y(0) = + y = C C = + y = y = y = 9
Suppose dy d + P ( )y = Q() where y = y(). ) Clculte the itegrtig fctor I() = e P Optio O The Itegrtig Fctor Method d ) Multiply the differetil equtio through by I().. You do ot eed costt of itegrtio. 3) Simplify the LHS d hece obti I()y = I()Q() d + c, where c is costt. 4) Itegrte to obti the geerl solutio. EX #: Solve the differetil equtio dy d + 3 y = EX : Solve the iitil vlue problem cos dy d Rewrite equtio to dy d y si cos cos dy si ycos d cos = cos si cos = si( ) cos cos dy d = ysi + si( ), y(0) =. I() = e t d ysi = si ( ) d (ycos ) = si ( ) d ycos = si( ) d ycos = cos() + c = e l(cos ) = cos At y(0) =, = + c, c = 3 ycos = cos() + 3 y = 3 cos() cos 30