Chde giai tich12-hki

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  • 1. T.s Nguyen Phu Khanh- a LatLHQCHU E 1: HAM SO AO HAMI.MIEN (TAP) XAC NH CUA HAM SO: D = {xR | y = f(x)R} Ham so Tap xac nhHam so Tap xac nh Ham soTap xac nh B(x ) > 0y = A(x ) A (x ) 0y = tgxx + k y = log A (x ) B(x ) 2 0 < A(x ) 1A (x ) a x y= B(x ) 0y = cot gxx k y= x x(a > 0)B(x )eA(x ) 0 arcsin xlog x y = 2 n A(x ) y= 1 x 1y=x > 0(n Z ) + arccos x ln xx D f (x ) g(x )y = 2 n +1 A(x ) y = [A(x )] B( x )A (x ) > 0 y= D = D f Dg(n Z ) + f (x ) g(x )II. MIEN (TAP) GIA TR CUA HAM SO: f(D) = {yR | y = f(x), xD}1.S ton tai nghiem cua phng trnh f(x)-y = 0, xDHam f(x)f(D): MGTHam f(x)f(D): MGT f (x ) a f (D ) = ( , a] a f (x ) bf (D ) = [a, b] f (x ) b f (D ) = [b,+ )a < f (x ) < b f (D ) = (a, b )2.anh gia bieu thc bang cac BT:* [A(x )] + a a a, x lam A(x ) xac nh.2* BT Cosi : a + b 2 ab . Bunhiacop sky : ac + bd (a 2)( + b 2 c2 + d 2 )III.HAM HP gof g o f la ham hp cua hai ham f : D fTf va g : D fZ * Tf D f = g o f : Dg o fZ * x D g o f : [g o f ](x ) = g[f (x )] va fog g o f{x | x D f f (x ) Dg } Tf D g ; * Dg o f = D f , {(Tf 0 ) (Tf Dg )}IV. HAM CHAN LE y=f(x) OI XNG QUA O: f ( x ) = f (x ) x D : f chan f ( x ) f (x ) : Ham khong chan khong le x D f (- x ) = f (x ) x D : f le V.GII HAN HAM SO:01.Phng phap 1: Kh dang vo nh0C s cua phng phap la lam xuat hien dang trong bieu thc ham cac tha so (x - x0), e roi gian c chnh cac tha so o cua t f (x )so va mau so tronglim vi cac chu y:x x 0 g(x ) Neu t va mau la cac a thc, s dung phep chia a thc t va mau cho (x - x0). Rieng ay ta dung thu thuat chia Hormer. Neu ch t hoac mau co cha can thc, ta nhan cho t va mau mot lng lien hp cua can thc o. A + B llh A B3A 3 B 3 A 3 AB + 3 B2 llh Neu t va mau eu co cha can thc, ta se nhan vao t va mau cung hai lng lien hp giao hoan tng ng. Khong loai tr cac kha nang s dung nhanh cac hang ang thc:1Trch t http://www.toanthpt.net-

2. T.s Nguyen Phu Khanh- a Lat LHQa2 b 2 = ( a b )( a + b ) a3 b 3 = ( a b ) ( a2 ab + b 2 )a4 b 4 = ( a2 + b 2 ) ( a b )( a + b )an b n = ( a b ) ( an 1 + an 2 b + an 3 b 2 + ... + ab n 2 + b n 1 ) e y rang viec bien oi s cap co the lam dang vo nh nay tr thanh dang vo nh khac. Chang han: lim f (x )g(x ) (dang 0 theo th t o) x 02. Phng phap 2: Kh dang vo nh PP1: at so mu ln nhat cua cac a thc thanh phan t va mau lam nhan t chung e kh vo nh. PP2: Dung cac nh ly gii han tng ng:1/ x Pn (x ) ~ an x n x + ax 2 + bx + c ~ x a ; (a > 0)2/ x ax + bx + c ~ x a ; (a > 0)2 b3 / ax 2 + bx + c ~ a x ++ (x ); vi a > 0 va lim (x ) = 0 2ax 3. Phng phap 3: Kh dang vo nh C s cua phng phap tm gii han nay la:1/S dung lng lien hp.b2/S dung bieu thc tiem can:ax 2 + bx + c ~ a x + + (x ) trong o: a > 0 va lim (x ) = 02ax 3/ S dung cac hang ang thc.4/ Khong dung ham so tng ng cho dang tong. 4.Phng phap 4: Gii han cua ham lng giac TH1: Khi x 0 (x tnh bang radian) sin u ( x )tgu ( x )lim= 1 hay sinu ( x ) ~ u ( x ) lim= 1 hay tgu ( x ) ~ u ( x ) u( x ) 0 u ( x)u( x ) 0 u ( x) 1 cos u ( x ) 112lim= hay 1-cos 2 u ( x ) ~ u ( x ) u( x ) 0 u ( x) 2 22 Khong loai tr nhan cac lng lien hp lng giac. ( 1 + sin u ) ( 1 sin u )llh( 1 + cos u ) ( 1 cos u ) llh TH2: Khix x 0 ham lng giac co dang vo nh (x tnh bang raian) x = x0 + t* at: t = x x0 x x 0 t 0* Khi:x x 0 t= x 0 x, t 0Ghi chu: khong s dung ham tng ng cho tong so.f (x ) g(x ) h(x ), x Vx 0 | {x 0 }5. Ham kep: lim g(x ) = L lim f (x ) = lim h (x ) = Lxx 0xx 0xx 0 lim f ( x ) = L lim f ( x ) = L x x0x x06. Ham cha gia tr tuyet oi: x x f ( x ) = 0 xx f ( x ) = 0limlim 0 0f (x 0 ) R, x 0 D7. Ham lien tuc: * hay lim y = 0 xlim0 f (x ) = f (x 0 ) xx 0 02 Trch t http://www.toanthpt.net - 3. T.s Nguyen Phu Khanh- a LatLHQ lim+ f (x ) = f (x 0 ) : lien tuc phailim+ f (x ) = lim f (x ) = f (x 0 ) 0xx* Lien tuc tai x0:xx 0 xx0 lim f (x ) = f (x 0 ) : lien tuc trai xx 08.Cong thc gii han: xlim log a x = + limsin x =1lim a = + x+ x + x 0 xx+ lim log a x = lim a = 0 tgx x x 0+ lim =1 x x 0 x lim e = + lim ln x = +x+x +( )lim U x = 0 x 0 lim e = 0 x+ a>1 lim ln x = + a>1x x 0 lim( ) =1sin U xex ln x + lim= + lim =0 x 0 U ( x)x+ xx + xx tgU ( x ) lim x.e = 0 lim x. ln x = 0 x 0+ lim =1x x 0 U ( x ) xlim a = 0 + lim log a x = x+ x +1 cos x 1 0 0 )v (x ) f : D f (D )4)Ham logarit:Cho: . Kha ao ham theo x va co hamu x y = f (x ) y = y(v ln u ) = u v ln u + v f 1 : f (D ) D ungc: . y x = f (y ) 1 1 1Ta co: yx = x y =x yy x 4. Bang tnh ao ham:Ham so f(x)ao ham f(x)Ham so f(x)ao ham f(x)n x ;u( )nn.x n 1 ( n 1 ; n.u .u) sinx cosxC0cosx -sinx1x1tgx = 1 + tg 2 xcos 2 x1 u x; ( u) ;2 x 2 u ex ex11 2 axaxlnaxx4 Trch t http://www.toanthpt.net - 5. T.s Nguyen Phu Khanh- a LatLHQ 1lnx x 1 1 cotgx sin 2 x( = 1 + cot g 2 x) logax x ln a5.ao ham cap cao:Khi can tnh ao ham cap (n): y(n) = f(n)(x), ngi ta s dung phng phap tnh quy nap bang ba bc c ban nh sau: Tnh y, y, y... e d oan cong thc cua: y(n) = f(n)(x)(1) Gia s (1) ung k 1 , tc la ta co: y(k) = f(k)(x)(2) Lay ao ham hai ve bieu thc (2) e chng minh:y(k+1) = f(k+1)(x); ung k 1Ket luan: Cong thc (1) la ao ham cap (n) can tm.6. ng dung cua ao ham: ao ham cua ham so y = f(x) tai mot iem f(x0) neu ton tai he so goc cua tieptuyen vi o th (C): y = f(x) tai iem o:tk = tg = f(x 0 ) (la y ngha hnh hoc cua ao ham) (C): y = f(x)x Neu mot ham f co ao ham tai x0 th ham f lien tuc tai iem x0. Nhng mot ham f lien tuc tai x0 th cha chac co ao ham tai iem x0. M(x0,y0) Mot ham f khong lien tuc tai x0 th khong co ao ham tai iem x0. Gia s ham f : y = f(x) co ao ham y=f(x) tren D, ta co: (h.1) f la ham hang tren D f(x ) = 0;x D (1) f ong bien tren D f(x ) 0;x D (2) f nghch bien tren D f(x ) 0;x D (3)e y trong (2) va (3), ao ham the hien mot ham so n ieu nghiem cach (ong bien hay nghch bien) trong D co the bang khongtai nhng gia tr ri rac cua bien so (xem h.2) nhng khong the triet tieu trong mot khoang tuy y cua (; ) D (xem h.3). yy A f(x0,1)=0 Af(x0,1)=0 C x0 (;) C DD f(x0,2)=0BBx x a 0 x0,1 x0,2 ba 0x0 b(h.2)(h.3) Neu ham f lien tuc tren [a;b] va f(a).f(b) < 0 th phng trnh f(x) = 0 co t nhat mot nghiem:x 0 (a; b ) . f lien tuc tren [ a;b] phng trnh f ( x ) = 0 Neu: f ( a ) f ( b ) < 0 co nghiem duy nhat x 0 [ a;b] f n ieu nghiem cach tren a;b[ ]y Bf(b) f(a)A (C) : y = f(x) a x (C) : y = f(x) 0 x0b a x0 x0 b B f(b)(h.6)5Trch t http://www.toanthpt.net- 6. T.s Nguyen Phu Khanh- a LatLHQ Gia s ham f : y = f(x) xac nh tren oan [a;b] Ham f at mot cc ai taix 0 (a; b ) , neu ton tai mot lan can V(x 0 ) (a; b ) sao cho: f (x ) < f (x 0 ); x x 0 . Ham f at mot cc tieu tai x 0 (a; b ) , neu ton tai mot lan can V(x 0 ) (a; b ) sao cho: f (x ) > f (x 0 ); x x 0 .*nh ly 1 Fermat: (ieu kien can e ham so f co cc tr)Neu ham f co ao ham tai V(x0) va at mot cc tr tai x0 o th ieu kien can la f(x0) = 0. yyf(x0)=0(C):y=f(x) (C):y=f(x)f(x0)>0 BB A Af(x0)0 f"(x0) 0 tren (a;b) o th (C) : y = f(x) lom trong (a;b) ve pha y dng.* nh ly 4: (ieu kien u th hai e mot ham co cc tr)Neu f(x0) = 0 trong V(x0) ong thi f(x0) # 0 th ham f co cc tr tai x0. Cu the: f(x0)=0f"(x0)>0f"(x0) 01. Nam cung pha vi truc hoanh y1 .y 2 < 02. Nam hai goc phan t:12Trch t http://www.toanthpt.net - 13. T.s Nguyen Phu Khanh- a Lat LHQ (I) va (III) (II) va (IV) y > 0 y > 0y > 0y > 0 x1 > 0; y1 > 0 hoacx 1 < 0 < x 2x1 < 0; y1 > 0 hoac x1 < 0 < x 2 x < 0; y < 0a > 0 va y = 0 VN x > 0; y < 0 a < 0 va y = 0 VN 2 2 y 22 yVII. NH THAM SO E HAM BAC 3 CAT TRUC HOANH TAI 1 HOAC 3 IEM: y = ax3 + bx 2 + cx + d PTH giao iem : ax3 + bx 2 + cx + d = 0 (*) (*) co nghiem ac biet x0(x x 0 )(ax 2 + bx + c) = 0 Co nghiem kepCo 1 nghiem Co 3 nghiem y = 0 ax 2 + bx + c = 0 ax 2 + bx + c = 0 co nghiem chungvo nghiem hoac nghiem kep y = 0 co 2 nghiem x x 0 ax 2 + bx + c = 0 nghiem kep 0 > 0 b ax 2 + bx + c = 0 nghiem x = x 0 = 2 a g(x 0 ) 0 (*) khong co nghiem ac biety = 3ax2 + 2 bx + c y max y min = 0 y 0 y > 0 y = 0 nghiem chungy > 0 y = 0y max y min < 0 y max y min < 0 y 0Ghi chu: PT bac 3: y=0 khong the co 3 nghiem phan biet y max y min > 0VIII.NH THAM SO E HAM BAC 3 CAT TRUC HOANH TAI 3 IEM CO HOANH O DNG (HAY AM): Hoanh oHoanh o dng Hoanh o am Ln hn Nho hn y > 0 y > 0 y > 0 y > 0af (0 ) < 0af (0 ) > 0 af ( ) < 0 af ( ) > 0x C > 0x C < 0 x > 0 x < 0 < x1 < x 2 x1 < x 2 < CT CTy max y min < 0 y max y min < 0y max y min < 0y max y min < 0 CHU E 4: GIA TR LN NHAT - NHO NHATI.GIA TR LN NHAT - NHO NHAT TREN OAN [a;b]:f lien tuc tren [a;b] co M[GTLN] va m[GTNN] cua f tren [a;b] m f (x ) M x [a; b]Tm gia tr cc tr cua f(x) tren [a;b] e tm maxf va minf.Chu y 1: maxf, minf f lien tuc tren [a; b] M = max {f (a), f (b ), fC , fCT } x[a; b ]1. m = min {f (a), f (b ), fC , fCT } x[a; b ]2. Dung MGT tm max, min:m y0 M .3. Dung BT Cosi, Bunhiacopsky. 13Trch t http://www.toanthpt.net- 14. T.s Nguyen Phu Khanh- a LatLHQChu y 2:1. Neu f(x) lien tuc trong khoang (a;b) co iem cc trx 0 (a; b ) . x x1 x2 + x x1x2+ y+00+y 0+0 max max yy min min2. f(x) tang hoac giam tren [a;b] x x0+ x x0 + y+y max y = f (b ) y + y min y = f (a) II.GIA TR LN NHAT - NHO NHAT CUA HAM BAC 2 TREN[; ] : ba>0 hoanh o nhx0 = 2a[ ]( ) Neu x 0 ; : min y = f x 0 ; max y = max f , f { ( ) ( )} Neu x 0 [ ; ] : so sanh f ( ) va f ( ) suy ra max y va min y.a0( I x0 ; f x0( ) ) : la iem uon cua ( C) : y = f ( x )f">0 x ( a; b ) : f ( x ) = 0 0(T)I(C)( ) i 2 0 f ( x ) khong oi dau khi x i qua x 0 ( I x0 ; f x0( ) ) : la iem uon cua ( C) : y = f ( x ) f"0 ( i4 ): f ( x ) khong oi dau khi x bang qua x 0hoac f x oi dau khi x i qua x (T) ( ) 0( I x0 , f x0 ( ) ) : la iem uon cua ( C) : y = f ( x )III.TIEM CAN: Tiem can ng x = x0 Tiem can ngang y = y0Tiem can xien y = ax+bya = lim x x lim y = lim y = y 0b = lim [y (ax + b )]x xx 0 x lim = x lim [y (ax