Gio n t chnTon 12 nm hc 2010-2011Tit PPCT : 1S NG BIN,NGHCH BIN CA HM S Ngy son :20/8/2010 Ngy dy :22/8/2010A). MC TIU : 1)Kin thc:: T a ra nh l v tnh ng bin v nghch bin trn mt khang I. Gip hc sinh thng hiu iu kin (ch yu l iu kin ) hm s ng bin hoc nghch bin trn mt khang , mt an hoc mt na khang .2) K nng:Gip hsinh vn dng thnh tho nh l v iu kib ca tnh xt chiu bin thin ca hm s. Lm c cc bi tp SGk v cc bi tp trong SBT v cc bi tp khc . 3)T duy:T gic, tch cc trong hc tp.Sng to trong t duy. T duy cc vn tan hc, thc t mt cch logc v h thng.B). PHNG PHP GING DY :S dng cc phng php dy hc c bn sau mt cch linh hat nhm gip hc sinh tm ti , pht hin chim lnh tri thc : Gi m , vn p . Pht hin v gii quyt vn . T chc an xen hat ng hc tp cc nhn hoc nhm..C) Chun b 1. Chun b ca gio vin : Chun b cc phiu tr li trc nghim , phiu hc tp . Chun b bng ph trnh by cc nh l v gii hn. Chia 4 nhm, mi nhm c nhm trng.2. Chun b ca hc sinh :Cn n li mt s kin thc o hm hc . dng hc tp : thc k , compa, my tnh cm tay Kin thc hc v hm s D). TIN TRNH DY HC :A. Bi c :Xt chiu bin thin ca hm s: 2( ) 8 + + f x x xB. Bi mi :CC DNG BI TP DNG 1. BI TP XT CHIU BIN THIN CA MT HM S CHO TRC V LP BNG BIN THIN CA HM S A). Phng php.S dng iu kin ca tnh n iu ca hm s.B). Bi tp.Bi 1. 1). Xt chiu bin thin ca cc hm s sau:a). y = 2x3+3x2+1 ; b). y = x - 2x ; c). y = 24 x ; d). y =22 31x xx ++;2). Ty theo m xt chiu bin thin ca hm s: y = 4x3 + (m+3)x2 +mxBi 2. Kho st chiu bin thin ca cc hm s sau:a). y = 231xx++b). 28 y x x + +c). y = 21 x x x + + Chn bi:Xt chiu bin thin ca hm s y = 24 x Gii :Hot ng ca gio vin v hc sinh Ni dung ghi bngCu hi 1 gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011Tm tp xc nh ca hm s Cu hi 2 Tnh o hm ca hm s Cu hi 3 Cho o hm bng 0 v tm nghim o hmCu hi 4 Xt chiu bin thin ca hm s Cu hi 5 Kt lun tnh n iu ca hm s Hm s cho xc nh trn tp hp D = [-2;2]Ta c: 2'4 xyx' 0 0 y x Chiu bin thin ca hm s cho trong bng sauX -20 2 +y+ 0- 2 y 00Hm s ng bin trn mi khong [ ] [ 2;0] ch bin trn0;2 v ngh , DNG 2. BI TP TM GI TR CA THAM S MT HM S CHO TRC NG BIN, NGHCH BIN TRN MT KHONG CHO TRCA). Phng php.S dng iu kin ca tnh n iu ca hm s.S dng nh l du tam thc bc hai.B). Bi tp. 1). Tm cc gi tr ca tham s a hm s :f(x) = 3 214 33x ax x + + + ng bin trn R.2). Xc nh m hm s sau lun nghch bin trn R : y = (m -3)x (2m+1)cosxChn bi: Tm cc gi tr ca tham s a hm s :f(x) = 3 214 33x ax x + + + ng bin trn R.Hot ng ca gio vin v hc sinhNi dung ghi bngCu hi 1 Tm tp xc nh ca hm s Cu hi 2 Tnh o hm ca hm s Cu hi 3 Hm s ng bin trn R khi no ?Cu hi 4 Kt lun ?Hm s cho xc nh trn tp hp D = RTa c: 2' 2 4 y x ax + +0' 0, 2 2' 0ay x R a> 'Hm s ng bin trn R l :2 2 a . V. CNG C V DN D :1). Cng c: Nu quy trnh xt tnh n iu ca hm s .2). Dn d : Chun b cc bi tp phn luyn tp V. RT KINH NGHIM T BI DY : Tit PPCT : 2S NG BIN,NGHCH BIN CA HM S gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011
Ngy son :28/8/2010 Ngy dy :29/8/2010D). TIN TRNH DY HC : I) Bi c :Xt chiu bin thin ca hm s: f). 211xyx x+ +g). 22 3 y x x + +h). y = x3 6x2 +17x +4II)Bi mi :DNG 3 : BI TP S DNG CHIU BIN THIN CA HM S CHNG MINH BT NG THC Phng php.S dng kin thc sau : Du hiu mt hm s n iu trn on . f (x) ng bin trn on[ ] ; a bth f(a)[ ] ( ) ( ) , ; f x f b x a b f(x) nghch bin trn on[ ] ; a bth f(a)[ ] ( ( ) , ; f x f b x a b S dng bng bin thin.B). Bi tp.Bi 5. Chng minh cc bt ng tht sau:a). sinx < x, vi mi x > 0 ; sinx > x ,vi mi x< 0.b). cosx > 1 -22x vi mi x 0 ;c). sinx>x -36x, vi mi x > 0 ; sinx x+ 33x ,0;2x _ ,Bi 6.Cho x, y l hai s dng thay i tha mn ng thc x+y = 54(1) Hy chng minh bt ng thc:4 15 (2)4 x y+ Chn bi: Chng minh rng: sinx + tanx > 2x ,0;2x _ ,Hot ng ca gio vin v hc sinh Ni dung ghi bngCu hi 1 Xt tnh lin tc ca hm s trn khang no? Cu hi 2 Tnh o hm ca hm s Cu hi 3 Hm s ng bin trn R khi no ?Cu hi 4 t f(x)= sinx + tanx -2x Ta c f(x) lin tc trn 0;2_
,Ta c: 22 21 1' cos 2 cos 2 0, i x0;cos cos 2y x x mx x _ + > + > ,Do hm s ng bin trn 0;2_
, gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011 Kt lun ?v ta c f(x) > f(0), x0;2 _ , Hay sinx + tanx > 2x 0;2x _ , DNG 4*.BI TP TM IU KIN PHNG TRNH,BT PHNG TRNH C NGHIM THA IU KIN CHO TRC,HOC GII PHNG TRNH, BT PHNG TRNH.A). Phng php.S dng iu kin ca tnh n iuS dng nh l v gi tr trung gian ca hm s lin tc.S dng cc mnh sau f(x) l hm s lin tc trn.Khi :a). f(x) vi mi x maxf(x).b). f(x) vi mi x minf(x)c). f(x) c nghim minf(x).d). f(x) c nghim maxf(x).B). Bi tp.Bi 7.Tm m phng trnh:22 x mx + + =2x+1(1) c hai nghim thc phn bit.Bi 8. Tm m phng trnh: mx- 3 x m+1 (*) c nghim.Bi 9 . nh t sao cho phng trnh 2sin 1sin 2xtx++c ng 2 nghim thuc on[ ] 0;Bi 10 : Gii h phng rnh : 221212x yyy xx +' +Bi 11 : Tm m phng trnh: x3 mx -1 = 0 c nghim duy nht V. CNG C V DN D :1). Cng c: Hm s lin tc trn [a;b] v c o hm dng hc m trn khang (a;b) th ng bin hoc nghch bin trn [a;b].2). Dn d : Chun b cc bi tp phn luyn tp 3). Bi tp lm thm : Tm m phng trnh c hai nghim thc phn bit : 22 2 1 x mx x + + + p s :92mV. RT KINH NGHIM T BI DY : gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011Ngy son: 3/9/2010Ngy dy: 4/9//2010. Tit3Cc tr hm s.I. Mc tiu.- Kin thc: cng c cc quy tc tm cc tr ca hm s, bng bin thin ca hm s.- k nng: rn k nng xt s bin thin; hc sinh vn dng thnh tho cc quy tc tm cc tr vo gii quyt tt bi ton tm cc tr hm s v cc bi ton c tham s.- T duy -thi : ch ng, sng to, t duy logc.II. Thit b.- GV: gio n, h thng bi tp b tr.- HS:kin thc c v s bin thin, cc quy tc tm cc tr.III. Tin trnh.1. n nh t chc.2. Kim tra bi c.GV: nu cc quy tc tm cc tr hm s?HS: tr li ti ch.3. Bi mi.Hot ng GV Hot ng HS Ghi bngGV: nu vn Gi 7: nu quy tc p dng trong 7?Tm nghim ca phng trnh trong HS: gii quyt cc bi tp, ch k nng din t. 7: HS ch ra c quy tc 2; cc nghim trong [0; ] v so snh tm ra cc tr.Bi 1.Tm im cc tr ca cc hm s sau:1. y = 2x3 3x2 + 42. y =x(x 3) 3. 1y xx +4. 2x 2x 3yx 1 +5. y = sin2x6. 2xy10 x7.[ ]2y sin x 3 cos x trong 0; 8. xy sinx2 +Hng dn7. Ta c y = 2sinxcosx +3 sinxtrong [0; ], y= 0 sinx = 0 hoc cosx = -32 x= 0; x = ; x= 56gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011[0; ]?hi: hm s c cc tr ti x = 1 khi no?cn lu HS khi tm ra gi tr ca m phi kim tra li.GV kim tra k nng ca cc HS.hm s khng c cc tr khi no?HS cn ch ra c: x = 1 l mt nghim ca phng trnh y = 0.HS gii bi ton c lp khng theo nhm.khi phng trnh y = 0 v nghim.mt khc y = 2cos2x + 3 cosx nn ta c y(0) > 0 nn x = 0 l im cc tiu.tng t y( ) >0 nn x = l im cc tiu.y(56) 0 v ix -1. Do hm s lun ng bin . Hm s khng c cc tr. q + Nu q > 0 th: 2212 1'( ) 0( 1)1x qx x qf xxx q
+ + + +Lp bng bin thin xem hm t cc ti ti gi tr x no.Dng 3. Tm iu kin hm s c cc trBi ton: Tm m hm s c cc tr v cc tr tho mn mt tnh cht no .Phng phpB1: Tm m hm s c cc tr.B2: Vn dng cc kin thc khc Ch : gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011 Hm s 3 2ax( 0) y bx cx da + + + c cc tr khi v ch khi phng trnh y = 0 c hai nghim phn bit. Cc tr ca hm phn thc ( )( )p xyQ x. Gi s x0 l im cc tr ca y, th gi tr ca y(x0) c th c tnh bng hai cch: hoc 0 00 00 0( ) '( )( )hoc y(x )( ) '( )P x P xy xQ x Qx V d . Xc nh m cc hm s sau c cc i v cc tiu23 21 x 2 4. y = ( 6) 1 . y =3 2mx ma x mx m x bx+ + + + +Hng dn.a. TX: R 2' 2 6 y x mx m + + + . hm s c cc tr th phng trnh: 22 6 0 c 2 nghim phn bit x mx m + + + 23' 6 02mm mm> > < b. TX:{ } \ 2
2 22 22(2 )( 2) ( 2 4) 4 4 4'( 2) ( 2)m s c cc i, cc tiu khi' 0 hai nghim phn bit khc -2 4 4 4 0' 0 4 4 4 004 8 4 4 0 0x mx x mx m x x myx xH y c x x mmmm m+ + + + + + + + + + + > > III). Dy hc bi mi : I . KIN THC CN NH.1.nh ngha. Gi s hm s f(x) xc nh trn tp hp D( ) D R .Nu tn ti im0x D sao cho f(x)0( ) i mi x D f x v .Th s M = f(0) x c gi l gi tr ln nht ca hm s f trn D, k hiu l M =max ( )oxDf xNu tn ti im 0x D sao cho0( ) ( ) i mi x D f x f x v .Th s m = 0( ) f x c gi l gi tr nh nht ca hm s , k hiu l m = min ( )oxDf x.2.Nu vic tm gi tr ln nht, nh nht ca hm s m khng ni r tm trn tp no th ta nn hiu vic tm gi tr ln nht, nh nht trn tp xc nh ca hm s.II. BI TP C BN DNG 1. BI TP TM GI TR LN NHT,NH NHT CA HM S LIN TC TRN MT ONA). Phng php. Gi s hm s y = f(x) lin tc trn on[ ] ; . a b
Tnh [ ]'1 2 i0 m cc nghim x , .. c ana;b c trong cc x ny t x xthu ho 'm ti y khng xc nh.Tnh f(a), f(b), 1 2( ), ( ),.......... ( )nf x f x f xm = [ ]{ }1;min ( ) min ( ), ( ), ( ),......... ( )na bf x f a f b f x f x
M =max{ } ) ( ),..., 1 ( ), ( ), ( xn f x f b f a fB). Bi tp.Bi 1. Tm gi tr ln nht, nh nht ca hm s sau:a). f(x) =[ ]22 5 n-2 ; 3x x tr + b). f(x) =[ ]322 3 4 n on- 4 ; 03xx x tr + + c). f(x) =[ ]4 22 5 n on-2 ; 3 x x tr +d). f(x) =[ ]5 4 35 5 2 n on-1 ; 2 ; x x x tr + +e). f(x) = 22 3 i 1 gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011g). 212 2xyx x +h). y =21xx+ (x>0)Bi 2.1). Tm gi tr ln nht, nh nht ca hm s : a). 2( ) 4 f x x x + b). 2( ) 2 5 f x x x + 2). Tm gi tr ln nht, nh nht ca hm s: [ ]6 2 3( ) 4(1 ) n an-1;1 f x x x tr + Ch : Chng ta c th t t = 2x,[ ] ( )0;1 t v a vic tm gi tr ln nht, nh nht ca hm s ban u trn on[ ] 1;1 v vic tm gi tr ln nht, nh nht ca hm s : g(t) [ ]3 34(1 ) n0;1. t t tr + DNG 2. BI TP TM GI TR LN NHT, NH NHT CA MT HM S XC NH TRN MT TP HP NH LP BNG BIN THIN.A). Phng php.Lp bng bin thin ca hm s trn tp .Da vo bbt tm gi tr ln nht, nh nht.B). Bi tp.Bi 3. 1). Tm gi tr ln nht, nh nht ca hm s sau :a). y =x + 1n khong (0;+ ) trx b) y =x ( ]1n na khong0,2 trxBi 4 . 1). Tm gi tr ln nht, nh nht ca hm s : y = x 1 - 2 x2). Tm gi tr ln nht, nh nht ca hm s : y = x + 26 6 x x +Bi 5 .Cho t gic li ABCD vi AB = a, BC = b, CD = c, DA = d trong a, b, c, d l cc hng s . Chng minh t gic ABCD c din tch ln nht khi n ni tip c trong ng trn.Bi 6 . Bi 24 : Cho tm ba hnh ch nht c cnh l a v b ( vi b < a). Tnh cnh hnh vung m ta ct b t bn gc ca tm ba to nn mt hnh ch nht khng c np c th tch ln nht.Bi 7 .Ngi ta dng tm kim loi g mt thng hnh tr trn xoay c hai y vi th tch cho trc. Hy xc nh kch thc ca hnh tr vt liu tn t nht.Bi 8 . Cho hm s y = x4 6mx2 + m2 vi x [ ] 2;1 . Tm v bin lun theo m gi tr ln nht ca y.Bi 9 . Tm v bin lun theo a gi tr ln nht v gi tr nh nht ca hm s ( )2 21 x a x ayx +vi V. CNG C V DN D :1). Cng c: Nhc li ni dung nh ngha v nhn xt ca nh ngha Nu quy trnh tm GTLN,GTNN ca hm s .Khng th b qua tnh lin tc ti im x0 2). Dn d :bi tp Tm m phng trnh sau c nghim thc : 34 21 . 1 2 1 x mx x + + V. Rt kinh nghim .Tit PPCT : 8Ngy son :8/10/2010Ngy dy :9/10/2010
GI TR LN NHT- GI TR NH NHTA). MC TIU : HS nm c :gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-20111)Kin thc: n li tnh ng bin v nghch bin lp 10 hc .T a ra nh l v tnh ng bin v nghch bin trn mt khang I.Gip hc sinh thng hiu iu kin (ch yu l iu kin ) hm s ng bin hoc nghch bin trn mt khang , mt an hoc mt na khang .p dng lm cc v d SGK.2) K nng:Gip hc sinh vn dng thnh tho nh l v iu kib ca tnh n iu xt chiu bin thin ca hm s.Lm c cc bi tp SGk v cc bi tp trong SBT v cc bi tp khc . 3)T duy:T gic, tch cc trong hc tp.Sng to trong t duy.T duy cc vn tan hc, thc t mt cch logc v h thng.B). PHNG PHP GING DY :Gi m , vn p . Pht hin v gii quyt vn .T chc an xen hat ng hc tp cc nhn hoc nhm.C). TIN TRNH DY HC :1) n nh l p2)Bi c (xen bi mi)3)Bi miDNG 3.BI TP TM GI TR LN NHT CA HM S NH VO T BIN PH A).Phng php.Gi s ta cn tm gi tr ln nht,nh nht ca hm s y=f(x) trn tp.Khi ta c th lm theo cc bc sau:t t =(x), xD ta hm s y=f(x) v hm s y=g(t);a bi ton : tm min ,max ca f(x) trn D v vic tm min,max ca y= g(t) trnB).Bi tp.Bi 5. Tm gi tr ln nht, nh nht ca hm s :a). y = 22sin 2sin 1 x x + ; b). y = 2cos 2 sin cos 4 x x x +c). y = 4 2sin cos 2 x x + +Bi 6. 1). Tm gi tr ln nht, nh nht ca hm s:a). y = 1 sin 1 cos x x + + +b).y =21 4 2 3 4 1 x x x x + + + + + +3). Tm gi tr ln nht, nh nht ca hm s :a).y = cos3x 6cos2x + 9cosx + 5. b). y = sin3x cos2x + sinx + 2 c). 2sin 1sin sin 1xyx x++ +d). y = cosx (1+sinx)( ) 0 2 x e).( )4 2310x xy xx x+ + >+f)*. y = sin2008x+ cos2008xg).3 39sin cos sin cos4y x x x x + + h)*.222 11 2x xyx x+ +i). y = 342sin sin n an [0 ;]3x x tr DNG 4. BI TP TM GI TR LN NHT,B NHT CA HM S BNG PHNG PHP S DNG BT NG THCA).Phng php.S dng nh ngha v gi tr ln nht, nh nht ca hm s S dung cc bt ng thcgv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011B).Bi tp.Bi 7. Tm gi tr ln nht ,nh nht ca hm s1). y = 2x + 532 x+ trn (2;+)2). y = 1 1 1004 x x x + + + Bi 8. Tm gi tr ln nht, nh nht ca hm s : y = 5sin 3cos (*) x x + DNG 5. BI TP TM GI TR LN NHT ,B NHT CA HM S BNG PHNG PHP S DNG MIN GI TRA).Phng php.Gi s hm s y=f(x) xc nhtrn D.Gi G l tp gi tr ca hm s trn D .Khi : G= } {/ ng trnh :y=f(x) c nghim x D y Rph nh vy nu coi y l tham s, tm iu kin cn v ca y phng trnh y =f(x) c nghim trn D,t ta tm c tp G.B).Bi tp.Bi 9 . Tm gi tr ln nht, b nht ca hm s:a). y =2211x xx x+ + +b). y =2sin 3cos 12sin cos 3x xx x+ +c). cos 2sin cos 3xyx x++ +Bi 10. Tm m gi tr ln nht ca hm s : y = 22( 1) ( 1) 3 1ng nh hn 20081m x m x mkhx x+ + + + +V. CNG C V DN D :1). Cng c: Nhc li ni dung nh ngha v nhn xt ca nh ngha Nu quy trnh tm GTLN,GTNN ca hm s .Khng th b qua tnh lin tc ti im x0 2). Dn d :Chun b cc bi tp phn luyn tp 3). Bi tp lm thm : Tm GTLN,GTNN ca hm s :a). 2( ) 4 f x x x + b). 2( ) 2 5 f x x x + c). 2sin 1sin sin 1xyx x++ +d). y = cosx (1+sinx)( ) 0 2 x e).( )4 2310x xy xx x+ + >+f)*. y = sin2008x+ cos2008x g).3 39sin cos sin cos4y x x x x + + h)*.222 11 2x xyx x+ +V. Rt kinh nghim .Ngy son : 15/10/2010 Ngy dy :16/10/2010 Tit PPCT : 9KHO ST S BIN THIN V V TH HM A THC (Bi tp hm bc 4)I). MC TIU : 1)Kin thc:HS nm c : gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011Khi nim kho st hm s l g ? Bit c cc bc kho st hm s Kho st c cc hm c bn .bc 4 .Lm cc bi tp lin quan ti hm vc ba, bc bn .p dng lm cc v d SGK.2) K nng: Rn cho hc sinh cc k nng:Kho st c hm s bc 4 trng phng Thc hin cc bc kho st hm s V nhanh v ng th Lm c cc bi tp SGk v cc bi tp trong SBT v cc bi tp khc . 3)T duy:T gic, tch cc trong hc tpSng to trong t duy.T duy cc vn tan hc, thc t mt cch logc v h thng.II). PHNG PHP GING DY :Gi m , vn p . Pht hin v gii quyt vn .T chc an xen hat ng hc tp cc nhn hoc nhm.III). PHNG TIN DY HC :1. Chun b ca gio vin :Chun b cc cu hi gi m .Chun b phn mu vmt s dng khc .2. Chun b ca hc sinh :Cn n li mt s kin thc o hm hc . dng hc tp : thc k, compa, my tnh cm tay Kin thc hc v hm s IV).TIN TRNH DY HC :1) Bi c :Cu hi 1 : Hy nu cc bc kho st hm s?Cu hi 2 : Cc bc kho st hm s:Hot ng2 . CC BI TON LIN QUAN N HM BC4.B i 1(Bt43 sgk)Yu cu HS kho st nhanh th hm s bc bn trng phng?.NuPPdavothbinluns nghim ca PT?? Nu cch vit PTTT ti 1 im?xyy'-+0-01 -1-0 0-1-2-1-4 -3 -2 -1 1 2 3-6-5-4-3-2-11xyb) * Nu m < 2 th phng trnh c 2 nghim.* Nu m = 2 th phng trnh c 3 nghim.* Nu 2 < m < 1 th phng trnh c 4 nghim.* Nu m = 1 th phng trnh c 2 nghim.* Nu m > 1 th phng trnh v nghim.B i 2(Bi 47/SGK) .Cho hm s y = x4 (m+1)x2 + m (Cm)1). Kho st s bin thin v v th hm s khi m = 22). Chng minh rng thi hm s lun i qua hai im c nh vi mi gi tr ca m .gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011Hngdn v nh: bi tp Cho hm s(Cm).1)Kho st s bin thin v v th (C) khi m=3.2)Gi A l giao im ca (C) v trc tung. Vit phng trnh tip tuyn ca (C) ti ANgy son : 22/10/2010Ngy dy :23/10/2010 gv : V Qunh Ph Trng THPT Yn Thnh 2Hot ng ca gio vin Ni dung ghi bng Cu 1 : hc sinh t lm (bi c)Cu hi 1 Tm tp xc nh ca hm s Cu hi 2 Lp phng trnh honh giao im ca hai th. Cu hi 3 Hm s cho ct trc hanh ti ba im phn bit ?Cu 2Kho st : y = x4 3x2 + 2 Tp xc nh D = R S bin thin a). Gii hn ca hm s ti v cc
lim limx xy v y + + +b).Hm s khng c tim cn c). Bng bin thin : y = 4x3 -6x = 0 0 262x yx y
t
x - 62062+y -0+0- 0+ y +C + CT CT th : f(x)=x^4-3x^2+2-8 -6 -4 -2 2 4 6 8-8-6-4-22468xyTr li cu hi 2 th ca hm s i qua im( )0 0; xy khi v ch khi y0 = x04 (m+1)x02 + m( )2 4 20 0 0 01 0 x mx x y + th i qua im( )0 0; xy vi mi gi tr ca m khi v chi khiphng trnh (2)nghim ung mi gi tri ca m, tc l :
20 0 04 20 00 0 01 0 1 1 0 00x x xvy yx x y ' ' ' Vy vi mi gi tr ca tham s m thi lun i qua hai im c nh (-1 ;0) v (1 ;0).Gio n t chnTon 12 nm hc 2010-2011Tit PPCT : 10KHO ST S BIN THIN V V TH HM A THC (Bi tp hm bc 4)I). MC TIU : 1)Kin thc:HS nm c : Khi nim kho st hm s l g ? Bit c cc bc kho st hm s Kho st c cc hm c bn .bc 4 .Lm cc bi tp lin quan ti hm vc ba, bc bn .p dng lm cc v d SGK.2) K nng: Rn cho hc sinh cc k nng:Kho st c hm s bc 4 trng phng Thc hin cc bc kho st hm s V nhanh v ng th Lm c cc bi tp SGk v cc bi tp trong SBT v cc bi tp khc . 3)T duy:T gic, tch cc trong hc tpSng to trong t duy.T duy cc vn tan hc, thc t mt cch logc v h thng.II). PHNG PHP GING DY :Gi m , vn p . Pht hin v gii quyt vn .T chc an xen hat ng hc tp cc nhn hoc nhm.III). PHNG TIN DY HC :1. Chun b ca gio vin :Chun b cc cu hi gi m .Chun b phn mu vmt s dng khc .2. Chun b ca hc sinh :Cn n li mt s kin thc o hm hc . dng hc tp : thc k, compa, my tnh cm tay Kin thc hc v hm s IV).TIN TRNH DY HC :1) Bi c :Cu hi 1 : Hy nu cc bc kho st hm s?Cu hi 2 : Cc bc kho st hm s:Hot ng2 . CC BI TON LIN QUAN N HM BC4.Cho hm s y = -x4 - 4mx2 +2m . 1). Kho st s bin thin v v th hm s vi m = 12.2). Tm cc gi tr ca m sao cho hm s c ba im cc tr .Hot ng ca gio vin Ni dung ghi bngCu hi 1 Tm tp xc nh ca hm s Cu hi 2 S bin thin ca hm s Tr li cu hi 1Tp xc nh : D = R Tr li cu hi 2 a). Gii hn ca hm s ti v cc lim limx xy v y + b). Bng bin thin Ta c :' 34 4 y x x = 0( )24 4 0 0 x x x gv : V Qunh Ph Trng THPT Yn Thnh 2f ( x ) = - x ^ 4 - 2 x ^ 2 + 3- 8 - 6 - 4 - 2 2 4 6 8- 8- 6- 4- 22468xyGio n t chnTon 12 nm hc 2010-2011Cu hi 3 Cc im c bit ca th . x - 0 + y+0 -3 y - -
Hm s ng bin trn khang( ) ;0 ( ) nghch bin trn khong0;+ v Hm s t cc i ti im x = 0 ; Gi tr cc i ca hm s l y(0) = 3 Tr li cu hi 3Giao im ca th vi trc tung l im( ) 0; 3.Giao im ca th vi trc honh :y= 0 1x tVy th ct trc honh ti hai im (-1 ; 0)v( 1 ; 0) V th : Nhn xt : th cn nhn trc tung lm trc i xng.2)m0V. CNG C V DN D :1). Cng c :Cc bc kho st hm bc 3 . Cc dng th thng gp hm bc 3 (6 dng) . im Un l tm i xng ca th .2). Dn d :Lm cc bi tp SGK , cc bi tp trong SBT . ti liu pho to 3). Bi tp lm thm : V. Rt kinh nghim gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011.Tit PPCT : 11Ngy son : 5/11/2010Ngy dy :6/11/2010 KHO ST S BIN THIN V V TH HM A THC (Bi tp hm bc 3)I). MC TIU : HS nm c :1)Kin thc:Khi nim kho st hm s l g ? Bit c cc bc kho st hm s . 2) K nng: Rn cho hc sinh cc k nng:Kho st c hm s Bc 3 Lm c cc bi tp SGk v cc bi tp trong SBT v cc bi tp khc . 3)T duy:T gic, tch cc trong hc tp.Sng to trong t duy.T duy cc vn tan hc, thc t mt cch logc v h thng.II). PHNG PHP GING DY :S dng cc phng php dy hc c bn sau mt cch linh hat nhm gip hc sinh tm ti , pht hin chim lnh tri thc :Gi m , vn p . T chc an xen hat ng hc tp cc nhn hoc nhm.III). PHNG TIN DY HC :1. Chun b ca gio vin :Chun b cc cu hi gi m .Chun b phn mu vmt s dng khc .Chun b cc phiu tr li trc nghim, phiu hc tpChia 4 nhm, mi nhm c nhm trng.2. Chun b ca hc sinh : dng hc tp : thc k, compa, my tnh cm tay IV).TIN TRNH DY HC : Bi c Cu hi 1 : Hy nu cc bc kho st hm s? Bi mi :1: Lp bng bin thin ca hm s v v th ca hm s .3 23 1 y x x +p n :Hm s c tp xc nh l R . 1. S bin thin ca hm s Gii hn ca hm s ti v cc + + + lim limx xy v yTa c : y = 3x2 6x = 3x ( x 2 ).y = 0 0 cx =2x ho Bng bin thinx-0 2+y + 0- 0 +1+y- -3Hm s ng bin tr n (-;o)v (3;+) hm s nghich bin trn (0;3)Hm s t cc i ti x=0 ycd=y(0)=1Hm s t cc tiu ti x=3 ,yct=y(2)=-3 thgv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011
f(x)=x^3-3x^2+1-8 -6 -4 -2 2 4 6 8-8-6-4-22468xy Cu2Cho hm s3 2y x 3x 1 + c th (C)a. Kho st s bin thinv v th (C).b. Dng th (C) , xc nh k phng trnh sau c ng 3 nghim phn bit 3 2x 3x k 0 + .HD: b. (1)pt3 2x 3x 1 k 1 + y l pt honh im chung ca (C) v ng thng(d): y k 1 Cn c vo th , ta c :Phngtrnhc ba nghim phn bit 1 k 1 3 0 k 4 < < < pt:yy0= 0f (x ) .(xx0)Ket qua: y= 0 ;y= 9(x3)Bi ton 2: Tm iu kin hm s nghch bin R x Ph ng php gii: Tm iu kin R x 0 y, V d : Cho hm s y=31(a2-1) x3+(a-1)x2-2x+1Tm a hm s nghch bin trn R.gii: hm s nghch bin trn R th:y R x 0 y=(a2-1)x2 +2(a-1)x-2 R x 0 '0 0 a' ' + + >0 5) 3(12m 1) 9(2m 0 32 ' ' + +0 15 36m 1) 4m 9(4mm2 36m2-6 0 66m66 Kt lun:Vy 66m66 l nhng gi tr cn tm. V. CNG C V DN D : Cng c :Cc dng thng gp hm bc 3 (3 dng) . 2). Dn d :xem l i b i t p gi i 3). Bi tp lm thm : 1/Tm gi tr ca tham s a hm s3 21( ) ax 4 33f x x x + + +ng bin trn R.2/ Tm m hm s y = x3 3mx2 + ( m - 1)x + 2 t cc tiu ti x = 23/ Tm m hm s 3 23 2. V i gi trno ca m thhm s c C , CT? y x mx +V. Rt kinh nghim Ngy son : 26/11/2010Ngy dy :27/11/2010 KHO ST S BIN THIN v V T CA HM STit PPCT : 13 I. Mc tiu:1. Kin thc:Nm chc s kho st v v th hm s. gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011 2. K nng: Bit vn dng s kho st v v th hm s tin hnh kho st v v th cc hm s n gin v c bn nht trong chng trnh ton THPT. l cc hm a thc, phn thc hu t quen thuc. Bit cch phn loi cc dng th ca cc hm s cc hm phn thc dng ' '+ ax bya x b+. 3. T duy, thi :Xy dng t duy logc, bit quy l v quen. Cn thn, chnh xc trong tnh ton, lp lun. II. Chun b phng tin dy hc:1. Thc tin: HS nm c cch kho st tnh n iu, tm im cc tr v lp bng bin thin ca mt hm s.2. Phng tin: SGK, sch bi tp, bt, thc k v h thng v d, bi tp.III. Gi v phng php dy hc: Kt hp linh hot cc phng php: Vn p - gi m, pht hin v gii quyt vn .IV. Tin trnh t chc bi hc:1. n nh t chc lp.2. Kim tra bi c:H: Hy nu s kho st hm s bc bn dng trng phng v mt s im lu khi v th hm s bc bn dng trng phng?3. Bi mi:Hot ng 1.II. Kho st mt s hm s a thc v phn thc.3. Hm s dng ( 0, 0)ax by c ad bccx d+ +V d 1: Kho st s bin thin v v th hm s21xyx ++22 1xyx+Hot ng ca GV Hot ng ca HS Ni dungH1: Hy cho bit hm s trn c dng nh th no?GV gi mt HS ng dy davoskhost hmskhost hm cho.TL1: Hm s trn c dng ax bycx d++HS da vo s kho st hm s kho st hm choBng bin thin: 3. th: Ta c: 20 21xxx + + th ct Ox ti im (2;0)(0) 2 y th ct trc tung ti im (0;2) th hm s: Lu : th nhn giao im ca hai tim cn l tm i xng.Hot ng caGVHot ng caHSNi dunggv : V Qunh Ph Trng THPT Yn Thnh 2x-1 +yy -- -1-1+Gio n t chnTon 12 nm hc 2010-2011H1: Hychobit hms trn c dng nh th no?GV gi mt HS ng dy da vo s kho st hm s kho st hm cho.TL1: Hm s trn c dng ax bycx d++HS da vo s kho st hm s kho st hm choKho st v v th hm s:1. Tp xc nh:R\-0,5. S bin thin: Chiu bin thin:
'2 22 1 2( 2) 5(2 1) (2 1)x xyx x+ + + Bng bin thin: 3. th: th hm s: 54321-1-2-3-4-5-8 -6 -4 -2 2 4 6 8h y ( )=-12g x ( )=12f x ( ) =x-22 x+1Lu : th nhn giao im ca hai tim cn l tm i xng.4. Cng c.- GV nhn mnh li mt ln na s kho st v v thhm s.- GV nhn mnh mt s im dng th ca hm phn thc v lu khi kho st, v thi hm s bc bn trng phng: th hm s c mt tim cn ng, mt tim cn ngang v nhn giao im ca hai ng tim cn lm trc i xng.5 Dn d.Hng dn HS lm bi tp 3, 6 trang 43, 44, SGK Gii tch 12.V. Rt kinh nghim................................................................................................Ngy son : 2/11/2010Ngy dy :3/11/2010 KHO ST S BIN THIN v V T CA HM STit PPCT : 14 I. Mc tiu: -Gip Hs n li v nm chc s kho st v v thihm s i vi bc 1/ bc1. -Gip Hs Ren luyn cac ky nng tinh toan, tinh cn thn chinh xac trong qua trinh giai toan, rn luyn k nng v th gv : V Qunh Ph Trng THPT Yn Thnh 2xyy
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-+122 4 6 8 -2 -4 -6 -82468-2-4-6-8xyGio n t chnTon 12 nm hc 2010-2011II . Chun b:- Gv: Phiu hc tp v mt s bi tp lm thm.- Hs: n li cc bc kho st v v thihm s.III. Tin trnh:1. n nh lp: KT s s:2. Bi c: a) Trnh by cc bc kho st v v thihm s? b) Cach tim cac ng tim cn cua thihm s..3.Bi mi:Hoat ong cua HS Hoat ong cua GVNoi dung bai giang Lp nhan xet ket qua.MX:D= R\{ } 1 y= ( )211 x+< 0x D TC:x=1; TCN:y = 1Lap b thien+K luaniem ac bieto th:Giao iem 2 tiem canla tamxng cua o th. Goi hoc sinh len bang giai 2 cau a, bGVnhan manh lai tnh chathamso nhat bien ve tnh y', tm tiem can, dang o th.S dung pttt co dang g?Cantmcacyeuto nao e lap pttt.Bai1:a/ Ksat hso: y = 21xx++ (C). b/ Lappttieptuyencua (C) tai giemcua(C) vi Oya/ MX:D= R\{ } 1 b/ Giemcua(C) vaOyla M(0;2)Pt ttuyen tai Mco dang:yy0 = f (x0). (xx0) y2 = 1(x0) y= x+2a/ MX:D= R\{ } 1 y= ( )241 x+> 0, x DTC:x=1 ;TCN: y = 2Lap bang bien thien+Ket luan. iem ac bietGoi 1 hoc sinh len bangkhaosat ham so, 2 hoc sinh tm iem nguyen.Theogthietta co:1 11 21 4xxx+ t
+ t
+ t ,kluan cac iem thoa btoan.Luycachtmiem nguyenLpnhanxet kiem tra ket quaGoi hoc sinh len bang khao sat va 1 hoc sinh tm iem nguyen.Bai 2:a/ Ksat hso: y= 2 21xx+ (C). b/ Tm cac iem M tren o th (C) co toa o nguyen. o th:b/ M(x;y) (C) y = 241 x+Theogthiet taco:1 11 21 4xxx+ t
+ t
+ t gv : V Qunh Ph Trng THPT Yn Thnh 2xyGio n t chnTon 12 nm hc 2010-2011243+ xyb) M(x, y)(C)432yx +X, Y z ta phai co 4 2 2 12 22 4x xxx+ + t
t
+ t MGoi hoc sinh nhan xetKiem tra ket quaNeuphngphaptm iemM(C) co toa o nguyenGoi hoc sinh tmpttt vi (C) ve t 1 iemNeuphngphaptm tiep tuyen vi (C)Neu phng phap bo dau gia tr tuyet oi,kluan cac iem thoa btoan.Bai 3: Cho hso: y = 3 22xx ++ (C). a/ Khao sat ham so. b/ Tm cac iem tren th (C) co to la so nguyen. c/ Da vao th (C), => th cua so: y = 3 22xx + a) D = R \ {2} , 2402y' x D(x ) > +c) y = 23223f(x) neuxf(x) neux vax'< Phanoth(C) ngvi 23xgi laiLay /x phan (C) vi23x 0pt0 3 . 18 6 4 . 4ln . 2 ln ln x x x0 183232. 4ln ln 2 ,_
,_
x x t t =0 ,32ln> ,_
txKQ : S = 2 eb.6 2 . 4 22 2cos sin +x x0 6 2 . 4 22 2cos cos 1 + x x0 6 2 . 422 22coscos + xxt t =0 , 22cos> txKQ : Phng trnh c mt h nghim x = Z k k + ,2-Nhn xt - TL : Da vo tnh cht1 cos 02 x
2 2 12cos x 2 1 tIV. Cng c, dn d - Y/c HS nm c cch gii phng trnh, bt phng trnh, h phng trnh, h bt phng trnh m v lgarit,- Y/c HS v lm th cc bi tp v hm s m v lgarit trong SBT.V. Rt kinh nghimNgy son: 18/1/2011Ngy ging: 20/1/2011Tit22 LUYN TP V TCH PHNI. Mc tiu:gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011Cng c cho HS v:- nh ngha v ngha ca tch phn, cc phng php tnh tch phn.- Rn luyn cho HS k nng tnh ton, kh nng phn tch, t duy,..II. Chun b :- GV: Gio n, bi tp.- HS: SGK, n li cc kin thc v tch phn.III. Tin trnh.1. n nh lp2. Kim tra bi c:Cu hi:Nu cch tnh tch phn bng nh ngha? Sau tnh tch phn I =131( 1) x dx- Gi mt HS ln bng - Gi mt HS khc nhn xt- GV nhn xt li 3. Ni dung bi mi:H1: Cha bi tp. Hot ng ca GV Hot ng ca HSBi 1. Tnh cc tch phn saua)161x dx b) 40sin 2x dxc)20Cos2x dxd) 402 x dx - Nu bi tp- Gi mt HS ln bng - Gi mt HS khc nhn xt- GV nhn xt li - Nu HS khng bit gii th HD HS giiBi 2. Tnh cc tch phn saua)150(3 2) x dx b) 1 230( )2xdxx
c)210xxe dxd)3121xx e dxBi 1 HS ln bng giia) 1616 16 1 32 21 112 2.63 423 3xdx x dx x _ , b) 420 01sin2 cos 2 12xdx x _ ,c) 2 220 01 cos 2cos2xxdx dx + 20sin22 4 4x x _ + ,d) 402 x dx = ( ) ( )2 40 22 2 x dx x dx + 2 42 20 22 2 42 2x xx x _ _ + , ,Bi 2- Mt HS ln bng giia) t 13 2 33u x du xdx dx du 0 2, 1 1 x u x u .Khi ta c:gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011e)12 lnexdxx+ f)21 lneedxx x +- Nu bi tp- Gi mt HS ln bng - Gi mt HS khc nhn xt- GV nhn xt li - Neeus HS khng bit gii th HD HS giiBi 3. Tnh cc tch phn saua) 20cos x x dxc) 21(2 1)ln x x dx +- Nu bi tp- Gi mt HS ln bng - Gi mt HS khc nhn xt- GV nhn xt li - Nu HS khng bit gii th HD HS gii+ Nhc li cng thc tch phn tng phn+p dngcngthctnhcctch phn1 15 40 21(3 2)3x dx u du 1523315 15u b) t 3 22 3 u x du x dx 0 2, 1 1 x u x u Khi ta c:1 230( )2xdxx 123duu12ln ln 23 3u - HS khc nhn xtBi 3- Mt HS ln bng giia) t cos sinu x du dxdv xdx v x ' ' T ta c ( )2 2200 0cos sin sin x x dx x x xdx = 20cos 12 2x + c) t ( )21ln2 1u x du dxxdv x dxv x x ' ' + +T ta c ( ) ( )( )222211 1(2 1)ln ln 1 x x dx x x x x dx+ + + 22156ln2 6ln 2 .2 2xx _ + , IV. Cng c, dn d - Y/c HS nm c cc phng php tnh tch phn.- Nhn mnh cc dng bi tp v phng php gii.- n tp cc vn v ng dng ca tch phn trong vic tnh din tch v th tch.- Gi sau tip tc luyn tp v tch phnV. Rt kinh nghimNgy son: 25/2/2011Ngy ging: 27/1/2011Tit 23 LUYN TP V TCH PHNI. Mc tiu:gv : V Qunh Ph Trng THPT Yn Thnh 2Gio n t chnTon 12 nm hc 2010-2011Cng c cho HS v: - nh ngha v ngha ca tch phn, cc ng dng ca tch phn trong tnh din tch hnh phng v th tch ca vt th trn xoay.- Rn luyn cho HS k nng tnh ton, kh nng phn tch, t duy,..II. Chun b :- GV: Gio n, bi tp.- HS: SGK, n li cc kin thc v tch phn v ng dng ca tch phn.III. Tin trnh.1. n nh lp2. Ni dung bi mi:Hot ng 1: n tp kin thcHot ng ca GV Hot ng ca HS- Cho hm s f(x)lin tctrn on[a; b]. Vit cng thc tnh din tch hnh thang cong gii hn bi cc ng ( ), ,y f x x a x b v trc honh?* Phng php gii tonBc1.Lpbngxt duhmsf(x) trn on [a; b].Bc 2.Da vo bng xt du tnh tch phn ( )baf xdx.- Nucngthctnhthtchkhi trn xoay do hnh phng gii hn bi cc ng[ ] ( ) 0, ; = " y f x x a b , y 0 =,x a =vx b (a b) =
