nguyen huu ngqc - classbook
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NGUYEN HUU NGQC
, r ' r ')
CAC D�NG TOAN VA PHllONG PHAP GIAI • I
GIAITI£D 12 (TVLU!N VA TRAC NGHitM)
(Tai bdn Ian tluc hai)
NHA XUAT BAN GIAO Dl)C Vl�T NAM
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Cong ty CPDV xullt ban Gilio dl}c �i Da Nang - Nha xullt ban Gilio dl}c Vi�t Nam giu quy�n cong b6 tac phAm
24-2010/CXB/814-2242/GD Ma s6: TZT36n0-ITS
Chuong I.
-
/
, ' �(ING DUNG DAO HAM DE
. .
KHAO sAT vA VE Do THI cuA HAM s6
§ 1. QUAN H� GIOA D�O HAM VA 51/ BIEN THIENcuAHAMs6
A. TOM TAT Li THUYET
I. TINH DON DI�U CUA HAM SO
1. Dien ki¢n can di ham so dan di¢n
Gia si'.r ham soy = f (x) c6 di;to ham tren khoang D.
a) Neu f (x) d6ng bien (tang) tren D th'i f' (x) � 0 vai m9i x E D.
b) Neu f(x) nghich bien (giam) tren D th1 f'(x) :SO vai m9i x ED.
2. Dien ki¢n du di ham so dan di¢n
Ham so y = f (x) c6 di;to ham tren khoang (a; b).
a) Neu f' (x) > 0 v6'i m9i x E (a; b) th'i f (x) d6ng bien (tang) tren
(a;b).
b) Neu f' (x) < 0 v6'i m9i x E (a; b) th1 f (x) ngh!ch bien (giam) tren
(a;b).
c) Neu f'(x)=O v6'i m9i xE(a;b)thl f(x) kh6ng d6i (ham hang)
tren (a; b ).
3
'
II. eve T� CUA HAM s6
1. l>ieu ki�n can d� ham so co c11c triGia si'r ham s6 y = f (x) d(;lt C\fC tri t(;li diem Xo. Khi d6, neu f c6 d(;lO ham t(;li Xo th} f' ( Xo) = 0.
2. l>ieu ki�n du d� ham s6 co c11c tri
�.fl/#,.Cho ham s6 y = f (x) lien tl}C tren (a; b) Chila X0 Va CO d(;lO ham trencac khoang (a; X
0) Va ( X
0; b ).
a) Neu f'(x)>O vm m9i xE(a;x.0) va f'(x)<O vm m9i
X E ( X0 ; b) thl f d(;lt C\fC d(;li t(;li diem X0•
b) Neu f'(x)<O vm m9i xE(a;x0) va f'(x)>Ovm m9i
X E ( x0 ; b) th} f d(;lt C\fC tieu t(;li diem X0•
�.fq.Z.
Cho ham s6 y = f(x) c6 d(;lo ham c!p mQt tren khoang (a;b) chua diem X
0, f' (x
0) = 0 Va f(x) c6 d(;lO ham Cdp 2 khac Q t(;li X
0•
a) Neu f 11 ( X0) < 0 thl ham s6 f d(;lt C\fC d(;li t(;li diem x
0•
b) Neu f" ( x0
) > 0 th} ham s6 f d(;lt C\fC tieu t(;li diem x0
•
111. GIA T� LON NHAT v A GIA T� NH6 NHAT CUA HAM s6
4
1. Dinh nghia
Cho ham s6 y = f(x) xac dinh tren D c R
a) Neu t6n t(;li x0
ED sao cho f(x):::; f(x0
) vm m9i x ED thl s6
M = f ( Xo) duqc g9i la gia trj lan nh�t cua ham s6 f tren D. Kf hi¢u :
M = maxf(x). xED
b) Neu t6n t(;li x0 ED sao cho f(x) � f(x
0) vm m9i x ED thl s6
m = f ( Xo) duqc g9i la gia ttj nho nhlt cua ham s6 f tren D. Kf hi¢u :
m = minf(x). xED
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2. Quy tac fim gia trj Ian nhat (GTLN) va gia tri oho nhat (GTNN)tren m<)t do�n
- Tim CaC diem X1, x2, .•• , Xm thu(>c ( a ; b) t�i d6 ham SO f CO dl;!O hambang ho�c khong c6 d�o ham. - Tinh f( XI ) , f ( X 2 ) , ... , f ( X m ) , f (a) va f ( b) .- So sanh cac gia ttj tlm duqc. S6 Ian. nhat trong cac gia trj d6 la gia trjIan. nhat cua f tren do�n [ a ; b], so nho nhat trong cac gia trj d6 la giatrj nho nhat cua f tren do�n [ a ; b].
B. cAc D�NG TOAN
D�ng 1. Khao sat tinh dan di�u cua ham so.
Lo(,li 1. DUNG DIEU KI�N DU DE KHAO SAT TINH DdN DI�U CUA HAMSO.
Vi di!, 1. Xac djnh chieu bifo thien cua ham so sau :
y = f (x) =
x2 + x-1.x+2
Giai
Ham so f xac djnh khi va chi khi x + 2 :;z!: 0 {::;> x :;z!: -2.
y' = f'(x) =
x2 +4x+3(x +2)2
X=-3 y' = 0 {::;,
x=-1 Bang bien thien
x - 00
f'(x)
-3 -2 - l
+ 0 0
+ 00
+
f(x) �� +oo +oo
-00 -00 ��
v �y ham so d6ng bien trong khoang ( -oo ; -3) va ( -1 ; + 00 ) ; nghjch bien trong khoang ( -3 ; -2) va ( -2 ; -1).
5
Vi df:l 2. Bi�n lu�n theo m tfnh tang, giam cua ham so sau:
6
y = f(x) = _!_x3 _ _!_m(m + l)x2 + m3x + m2 + 1.
3 2
Giai
D=IR y'=f'(x)=x2 -m(m+l)x+m3 , y' c6bi�tso �=m2 (m-d
I X = m-y =0¢::> X=m
+ Truo'llg h<
Lo<J,i 2. TIM DIEU KII;:N CUA THAM s6 DE HAM s6 LUON LUON TANG (GIAM) TREN JR (HOAC TREN TAP I c JR)
Huong dan: Dung dieu ki¢n can ve tfnh dan di¢u cua ham s6. 3
Vi du. Cho ham s6 y = f(x) = �-x2 +(m-l)x +m. Tim m de: . 3
a) Ham s6 lu6n luon tang tren Rb) Ham s6 luon luon giam tren ( 0 ; 2).
Giai
y' = f' (x) = x2- 2x + m -1.
a) Ham s6 luon luon tang tren JR khi va chi khix2 -2x + m -12: 0, \fx E JR
{:} �' ::S 0, (vl a= I > 0) {:}2-m::SO {:}m2:2
V�y ham s6 luon luon tang tren JR khi m 2: 2. b) Ham s6 luon luon giam tren (0; 2) khi va chi khi
x2 - 2x + m -1 ::S 0, \fx E ( 0 ; 2)
+ m = 2 : f' (x) ::;;;: (x -1)2 = 0 =} x = 1 E ( 0 ; 2)
+ m > 2 : f' (x) > 0, \fx E JR =} f' (x) > 0, \fx E ( 0 ; 2).+ m < 2 : f' (x) c6 hai nghi¢m Xi, x2 sao cho Xi ::SO< 2 ::S x2
{:} - {:} - {:}m<llaf'(O)<O
1m-l<O
af'(2)::SO m-1::SO -
V�y khi m ::SI ho�c m = 2 thl ham s6 lu6n luon giam tren (0; 2).
Lo<J,i 3. ONG Dl)NG TINH DdN DII;:U CUA HAM s6 DE CHUNG MINH BAT DANG THUC.
3
Vi du. Chung minh v6i moi x > 0 th'i x-� < sinx < x . . 6
Giai
+ Xet f(x) = sin x-x v6i Dr= (0; + oo ).
(I)
7
f'(x) = cosx-1::; 0, Vx =} f(x) < 0 tren Dr V�y ham s6 giam tren (0; + oo ). Theo dinh nghia ham s6 giam, ta c6 : Vx E Dr: x > Q=} f(x) < f(O)
=}f(x)<O =} sin x < x.
3 + Xet g (x) = sin x -x + � voi D g
= ( O ; + oo).6 x2 g'(x) = cosx-1 +-.2
D(lt h (x) = g' (x) =} h' (x) = g" (x) = -sin x + x > 0 (theo chung minh tren)Do d6 ham g(x) tang tren (0; + oo ). Vl v�y x > 0 =} g(x) > g(O)
x 3 =} sinx > x--6 V�y (1) dung voi m9i x > 0. I D�ng 2. Khao sat qrc tr! cua ham s6. Lof:li 1. TIM Cl/C TRJ CUA HAM so Dl/A V Ao DAU HI$U 1. Vt dlJ. Tim C1:fC ttj cua ham s6 y = 3x4 -4x3 + 1 ..
8
Giai
y' =f'(x)= 12x3 -12x2 = 12x(x-l) x=Oy' = 0 {::} X=l
Bang bie'n thien x - 00 0 f'(x) 0 0 +
+ 00
f(x) +�l�O�oo
Ham s6 c6 qrc ti�u t<;ti x = 1 va y CT = 0 .( T<;ti x = 0 f' (x) tri¢t tieu nhtmg kh6ng d6i diiu do d6 ham s6 kh6ng c6 Cl.!C ttj t<;ti x = 0).
Lo{li 2. TIM eve TRJ CUA HAM so DVA V Ao DAU HII;:U 2.
Vi du. Tim cue tri cua ham s6: y = f(x) = _!:_x +..!:..�12-3x2• . . . 2 2
Giai
Di�u ki¢n xac djnh: D = [-2; 2].
y'=f'(x)= ..!:.. [1- 3x l 2 �12-3x2
f'(x) = 0¢:;,�l2-3x2 =3x¢:;,lx2:0
=;,x=l.x=±l
- 18M�tkhac y" =f"(x)= � 3
<0,VxE[-2;2](12-3x2 )
°
=;, f" (1) < 0. V�y ham s6 d<;tt Cl.!C d<;ti t<;ti x = 1 va Yeo = 2.
Lo{li 3. TIM DIEU KII;:N CUA THAM so DE HAM so DAT eve TRr Hu'fmg d<in : Dung dieu ki¢n dn ve tfnh dan di¢u cua ham s6.
?
V'd Ch h' , f'() mx-+x+l-m Ti d.'h' 'd I u. . o am s6 y = x = . 1m m e am s6 at cue tn. x-1 · · ·
Giai
O f. ( ) x + I d l' h' h , b' A, A kh A , + m = : x = --: a.y a am n at 1en nen ong co cue tn. x-1 · ·
0 . f' ( ) = mx 2
-2mx + m -2 ( 1)+m:;c. x 2 ,m:;c (x-1)
Ham s6 c6 cl;lc tri khi phuang trlnh sau c6 hai nghi¢m phan bi¢t khac 1 g(x) = mx2 -2mx +m-2 =0
¢:;, ¢:;, =;, m > 0. !Di.' > 0 l2m > 0
g(l):;cO -2:;cOV�y khi m > 0 ham s6 c6 qrc trj.
9
Lo<J,i 4. ClfC TR� CO DIEU KI$N.
H ' d"' V'' h' h- ·· f( ) u(x) uong an : m am lIU t1 y = x = v (x)
ta thuang sir d1:1ng tfnh chat
sau de fim Cl;[C trj dO'n gian hO'n. G9i Xo la nghi�m cua f' (x) = 0 th! gia trj cua Cl;[C trj t�i Xo duqc tfnh
bai cong th(rc ,(· )- u(x0) _ u'(x0)t Xu - -- - ---'---'-
V ( x0) v'(x
0)
V'd Ch h' A' f( ) x2 -2x+m 'T') d:! h' A' , d · ' 1 '!-· o am so y = x = . 1 1m m e am so co cl;[c �· va x-m
Cl;[C tieu sao cho hai diem Cl;[C trj nam cung phfa d6i v6'i Ox.
Giai
'-f
'( )-x2 -2mx+m \-/y - x - . ? , vx :;r: m
(x-mt * Ham s6 d�t cl;[c tri khi phuO'ng tdnh g(x) = 0 c6 hai nghi�m khac m
{g(m) :;t O <=> {
m2 -2m2 + m :;t O <=> [m < 0 (1 )
�'>0 m2-m>O m>lb) Hoanh d(> cac Cl;[C trj la cac nghi�m X
i,X
2 khac m cua phtrO'ng tr1nh
x2 -2mx +m = 0.Tung d9 cua cac Cl;[C tri :
y1
=2x1 -2; y2
=2x2-2.
Su y ra : y I y 2 = ( 2 x 1 -2) ( 2 x 2 - 2) > 0 P -S + I > 0 <=> m -2m + l > 0 <=> m < I
Ket hqp v6'i (1 ) suy ra m < 0.
Lo<J,i 5 : BI$N LUAN ClfC TR�
Vi di!,. Cho ham so y = f ( x) = ( m +I) x 4 - 2 ( m -1) x 2• Tim m d� ham so d�t
3 Cl;[C trj. T�i g6c to� d9 la di�m Cl;[C d�i hay Cl;[C ti�u ?
lO
Giai
y' = f' ( x) = 4 ( m + 1) x 3 - 4 ( m -1) x
=4x[(m+ 1)x2 -(m-1)]
f'(x)=O<=> , [x = 0, \im (m+ l )x- =m-1
Ham so d�t 3 Cl.fC tri khi va chi khi : m+l :;t:O m-1:;t:Om-1 -->0m+l
<=>{::�I
m<-lvm>I
<=> [m < -I m>l
Khi d6 hoanh d¢ cac di�m qrc tf! la : x I = 0, Xz 3 = ±�m -ll . m+
y" = f" ( x) = 12 ( m + 1) x 2 - 4 ( m -1 )
=> f"(O) = - 4(m-l) + Neu m > I : f"(x) < 0 do d6 goc to� d¢ 0 la di�m qrc d�i.+ Neu m < 1 : f"(x) > 0 do d6 goc to� d¢ 0 la diem qrc tieu.
Lo<J,i 6. UNG Dl)NG eve TRJ DE CHUNG MINH BAT DANG THUC.
Vid�.Cht1ngminh: x-x3�
2f, \ixE(O;l).
Giai
Xet ham so y = f ( x) = x -x 3 v6"i D = ( 0; I). y'=f'(x)=l-3x2
•
J3 x=-
J3 y' = 0 <=>
x=--
Bang bien thien
11
x - 00
f'(x)
-J3 3
0 +
0
2.JJ V �Y 'v' X E ( 0 ; 1): f ( X) :S; Y max = -
9-
+
-J3 3
0
Dau "=" xay ra khi va chi khi x = � E ( 0; .1 ).
1 +�
D�ng 3. Tim gia trj km nhat va gia trj nho nhat cua ham so dl;Ca vao d,;lO ham.
Vi di!, 1. Tim gia trj Ian nhat va gia trj nho nhat cua ham so l I y = f(x) = cosx--cos2x +-
12
tren t�p D = [ 0 ; 21t] c IR.
Giai
2 2
Xet y' = f' ( x) = -sin x + sin 2x = 2 sin x ( cos x -i)
f'(x)=O�rsinx=� �r
x=k1t1t cos x = 2 x = ± 3 + k2 7t
Do xED nen y'=f'(x)=O�r::�:::�n3 3
Xet y" = f"(x) = -cos x + 2 cos2x = 4cos2 x -cosx -2 + x = 0 => f" ( x) = 1 > 0 => f . = l
min
+ x = 1t => f" ( x) = _ l < 0 => f = � 3 . 2 max 4
+ x = 1t => f" ( x) = 3 > 0 => f . = -1mm
51t "(
3 5 + X=-=>f x)=--<0=:>f =-3 2 max 4
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Mi;'lt khac: f (O) = 1; f(2n) = 1. V�y maxf( x)=max{�·� ·I·l}=� khi x= 1tvx=51t
D 4'4'' 4 3 3 min f ( x) = min { 1 ; - I ; 1 ; 1} = -1 khi x = n.
D
Vi d1:12. Cho phucmg trinh 2x2 + 2(m + I)x +m2 +4m + 3 =0 (I). G9i X i ,X2
la ha i nghi¢m cua (1) va di;'lt y =lx1 .X2 -2(x1 +x2 )I· nm gia tri km nhAt va gia tri nho nhAt cua y.
Giai
y t6n tl;li khi phucmg trinh ( 1) c6 nghi�m. tuc la !).' = -m 2 -6m-5 � 0 � -5 � m � -1 (2)
Theo gia thiet: y = lx1X 2 -2( X 1 + X 2 )I= IP-2SI = l1(m2 + 8m + 7 )I Xet hams6 f(m)=!(m2 +8m+7), D=[-5;-1]2 f'(m) =m+4=0 � m =-4 eD
x -oo -5 -4 -1 +oor(x) f(x) -4
�
-9Suy ra maxf = O,minf =-D D 2
Do y = lf(m)I nen
0 +
m;x y = max {lol ; 1-�I} = � khi m = -4 E (-5 ; -1] m�� y = min {lol ; 1-�I} = 0 khi m = -1 E [-5 ; -1].
+
13
C. BAI T!P
Bai 1. Khao sat tfnh d6ng bien, nghich bien va tim qrc d<;1i, qrc tieu cua cacham so sau:a) y=-x3 +3x2 -4 b) y=x 4 -8x2 +16.
Bai 2. Khao sat tf nh dcm di¢u va tlm qrc tri cua cac ham so :
) x 2 + 5 x -6 b) x 2 - 2x + 5a y= y =
x+l x-1Bai 3. Cho ham soy= f(x) = (x +2)�.
a) Khao sat qrc tri cua ham sob) Tim max f(x), min f(x).
[-2; 1] [-2; 1]
Bai 4. Cho ham so y = f ( x) = x + .J 12 - 3x 2 •
a) Khao sat qrc ttj cua ham sob) Tim max f ( x), min f ( x).
Bai 5. Cho ham soy= f (x) = .!.(m + l)x3 -(m + 2)x2 +(m +3) x, (m * -1). 3
Bi¢n lu�n tfnh tang, giam va tlm Cl,J'C tri cua ham so theo m. Bai 6. Cho ham so y=f(x)= 2x3 -3x2 +m, (m>O) . Tim maxf(x) va
[O;m]
min f(x). [O;m)
Bai 7. Cho ham so y=f(x)= x3 +(m-4h2 -4(m-lh+4m+l. Tim m de ham so a) luon luon tangb) luon luon giam trong (-2; 1] .
B'"8 Ch h' ,, f( ) mx-- x+m ( O) Tl d., 2 al . 0 am so y = X = · , m -:/= • 1m ill e y min = y max . mx-1
Bai 9. Cho ham so y = f ( x) = _!_ ( m + 1) x 3 - ( m + 2 )x 2 + ( m + 3 )x, ( m * -1).3
Tim m d� ham SO nh�n goc tO<;l d9 Jam di�m Cl!C ti�U.
Bai 10. Cho ham soy= f(x) = x2 -lx +m , (m * 0, m * I). Tim m de ham
x-m so d<;1t qrc d<;1i va qrc tieu sao cho y max .y min < 0 .
14
HUONG DAN GLiI v A oAP s6
Bai 1. a) y = -x3 +3x2 -4. T�p xac d!nh D = R
y'=-3x2 +6x, y'=O<=>[x=O x=2
y' > 0 <::::> 0 < x < 2 : Ham s6 tang trong (0 ; 2)
y' < O <::::> [x < O : Ham s6 giam trong { -oo; 0) va ( 2 ; + oo) x>2
Ham s6 d<;tt qtc tieu y min = -4 khi x = 0 va d<;tt qrc d�i y max = 0 khi x = 2.
b) y=x4 -8x2 +16=(x2 -4/T�p xac d!nh D = IR
y' = 4x (x2 -4) = 0 <::::> [x = O x=±2
Ham s6 tang trong cac khoang ( -2 ; 0) va ( 2 ; + oo) , giam trong cac khoang ( -oo ; -2) va ( 0 ; 2)
x
f'(x)
f(x)
-2 0 2 + 00
0 + 0 0 +
+oo 16 +oo�o�max�o�
min min
V �y ham s6 d<;tt qrc d<;ti y max = 16 t<;ti X = 0 ; d<;tt ClfC tieu y min = 0 t<;ti x=-2;x=2.
8,.2 ) x2 +5x-6 4 IO at .a y= =x+ ---x+l x+l
T�p xac d!nh D = R \ {-I} IO
y' = l+ 2 > 0(x+l)
Ham s6 tang trong hai khoang ( -oo ; -1) va ( -1 ; + oo) va khOng d<;tt C\!C tf!.
b) y = x2 - 2x+5 = x-1 +__i__
x-1 x-1
15
T�p xac dtnh D = JR\ {I} y' = l- 4 ') = (x+lHx-3)
= O<=>[x =-1 (x-lt (x-1)2 x = 3
Ta c6 ket qua theo bang sau x -00 -1 1
f'(x) + 0
3 +oo0 +
· f(x) +oo +c.o�4�
mm
Bai 3. y=f(x) =(x+2)� T�p xac dtnh D = ( -oo ; 1)
I f'( ) -3X y = x = � 2v1-x-y' = 0 <=> x = 0 E D x -00 -2 0 1 + 00
f'(x) + + 0f(x)
a) ymax = 2 khi x = 0 va khong c6 y min.
b) maxy = 2khix=Ova miny = O khi x =-2 hoac x=l.[-2;1] [-2;1]
Bai 4.y = f(x) = x+..Jl2-3x2
T�p xac dtnh D = [ -2 ; 2]
16
X, , f''( ) 1 3x ..J12-3x2 -3x et y = x = - =-r=====--..J12-3x 2 ..J12-3x2
y' = O<:::>..Jl2-3x2 = 3x<=>{x�O ')
2 <=>{x�O �x = l12-3x- = 9x x=±l
-36 4 Xet y" = f"(x) =
1�f"(I)=--<0 (12-3x2 )vl2-3x2 3
a) Ham so d<;1t qrc d<;1i y max = 4 khi x = 1 va khong d<;1t qrc tieu.b) max y = 4 khi x = 1 va min y = 2 khi x = ±2.
[-2;2] [-2;2]
Bai 5. y = ..!_(m + 1)x3 -(m +2)x2 +(m +3)x, (m * -1) 3
y' =(m+Ih2 -2(m+2h+m+3
[x
1 = I
y' =0<=> m+3 x2
=--,(m,t;-l) m+l
Vi x, * X2, Vm * -1 nen ham so d<;1t 2 Cl;l'C trj. y" = 2 ( m + l) x -2 ( m + 2)
" ( l) 2 0 H' "'d d . . l ' d · � · m + 3 y = - < : am so <;1t cl;l'c <;11 t<;11 x = va <;1t Cl!C tieu t<;11 x = --1
Xet hai tnrcmg hqp
+ Tru<'Jng hf/P 1 . m > -1 , khi d6 : I < m + 3 m+l
X - 00
f(x) +
1
0
m+3 m+ 1
0 +
f(x) ____-*ma� _ _____--*
-oo mm
+ Tru<'Jng hf/P 2 . m < -1 , khi d6 m + 3 < I m+l
x - 00
f(x)
m+3 m+l
0
1
+ 0
+oo
+ 00
+oo
f(x) + x
. ___,_::, max�
� min ___-- - oc
Bai 6.y = f(x) = 2x3 -3x2 +m, (m > O) D�t D =[0; m]
y' = f'(x) = 6x2 -6x = 0 <=> [x =0 x = 1
m+
17
,,
Ta ci1ng c6 : y max = m va y min
= m -1. + Tru&ng h<!JJ I . 0 <ms 1
� {:r:: :;m'-3m 2 +m
+ Tru&ng h<!JJ 2 . m > 1
{miny = m-1
� �xy=max{m;m(m-1H2m-1)}
3 - Neu 1 < m < - : max y = m.2 D
-Neu m2::�: maxy=m(m-lH2m-l).2 D
•
Bai 7. y = f( x) = x3 +(m-4)x 2 -4(m-t)x +4m + 1 y'= f(x)=3x2 +2(m-4h-4(m-l), VxE�
�I =(m+ 2)2
a) Ham s6 luon luon tang khi va chi khi y' 2:: 0, Vx E �<=>� I SO<=> (m + 2)2
s O �m=-2
b) Ham s6 luon luon giam trong (-2 ; l] khi va chi khiy' s 0, Vx E (-2 ; l]
+ Trumzg h<!JJ I. m = -2y' = 3(x - 2)2
� 0, Vx nen y' > 0, Vx E (-2 ; 1]. Truong hqp nay lo�i. + Trumzg h<!JJ 2. m -::1:--2y'sO, VxE(-2; 1]¢::> X1 S -2 < 1 S X2
<=> <=> <=> <=> m � 4 {3f'(-2)so
{3(3 2-8m)so
{m2::4
3f'(I)s0 3(-2- 2m) m2::-l 2
B,. 8 mx -x+m al . y=---mx-1
18
D=R\{�}.(m*O)
, in 2x2 -2mx-m2+1 g(x) y = 2 = 2 (mx-1) (mx-1) Ham s6 d�t qrc trt khi va chi khi g(x) = 0 c6 hai nghi¢m phfin bi¢t khac 1 , I' -, tac a m
{:��1::/ m*O (ilioA)
Vi h¢ s6 a:;:: m2 > 0 nen to� d¢ cac qrc trj la:
) D ·:! d . l I I 2 a 1em qrc �1 : x 1 == -- => y
1 == --
m m • • 1 3 b) Diem ct;rc tieu : x
2 == -+ 1 => y
2 == -
Ym in ==2Ymax
� � +2==2(� -2)
1 h '') => m ==-(t oa. 6
m m
Bai 9. y ==i(m + 1)x3 -(m +2)x2 +Cm +3)x
D=R y' =(m+l)x2 -2(m+2)x+m+3
y'=O�[:• :�+32 m+l
y" == 2(m+l)x- 2(m+2) =>y"(0==-2<0
m+3 V�y Xm in ==X2 ==--=>Ym in ==Y2 m+l Ham s6 nh� g6c to� d¢ lam diem ct;rc tieu khi
19
{ m+ 3
X2 =--= 0 m+l =>m=-3.
y2 =y(O)= O 2
2 2 2 B, . 10 _ x - x + m , _ x - mx + m ( )a1 . y - => y - 2 , x -::;:. m x-m (x-m)
Yeu du bai toan tucmg ducmg :
{y'(m)-:;:.O
{m-:/:.OAm-::/:.1
!:!,.'=m2 -m>0<=> m<Ovm>lYmax·Ymin <0 (2x 1 -2){2x2 -2)<0
( v'i x 1, x2 la hai nghi�m cua phucmg tr'inh x 2 - 2mx + m = 0)
V�y m > 1.
{m < 0 v m > I {m < 0 v m > l <=> <=> =>m>l
P-S+l<O 1-m<O
§ 2. DUONG TlfM �N.PHEP TINH Tlf N Hf TRIJC TO� De)
A. TOM TAT Li THUYET
I. DUONG TI:eM CAN CUA HAM so
G9i (C) la do thi cua ham s6 y = f (x) xac dinh tren K \ { Xo}.
1. Dtroog ti�m c�n dung ciia ham so
Duong thing (0 1 ): x = Xo duqc g9i la dtrang ti�m c� n dung cua do thi(C) neu ft nhat m¢t trong cac dieu ki¢n sau duqc thoa man :
20
lim f (x) = +oo ho�c lim f (x) = +oo x ..... x0 x ..... x�
lim f (x) = -oo ho�c lim f (x) = -oo x ..... � x ..... �