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  • 8/7/2019 Bi ton Hm s

    1/13

    BI TON LIN QUAN N KHO ST HM SNH L VI T

    Phng trnh bc 2: ax2+bx+c=0 (a 0)

    X1+ x2=a

    b; x1x2=

    a

    c;

    axx

    = 21

    Phng trnh bc 3: ax3+bx2+cx+d=0

    X1+ x2+ x3=a

    b; x1x2x3=

    a

    d; x1x2+x2x3+x3x1=

    a

    c

    CC BI TON V NG BIN

    NGHCH BIN

    Cho hm s ( )xfy = c tp xc nh l min D.f(x) ng bin trn D ( ) Dxxf ,0' .f(x) nghch bin trn D ( ) Dxxf ,0' .(ch xt trng hp f(x) = 0 ti mt s hu hn im trn min D)

    Thng dng cc kin thc v xt du tam thc bc hai:

    f(x)=ax2+bx+c

    1. Nu 0 < th f(x) lun cng du vi a.

    2. Nu 0 = th f(x) c nghim 2b

    x a= v f(x) lun cng du vi a khi 2b

    x a .3. Nu 0 > th f(x) c hai nghim, trong khong 2 nghim f(x) tri du vi a, ngoi khong 2 nghim f(x) cng du vi a.

    So snh nghim ca tam thc vi s 0

    * 1 20

    0 0

    0

    x x P

    S

    >

    < < >

    < < > >

    * 1 20 0x x P < < >

    .

    hm s ( )y f x= c hai cc tr nm pha di trc honh 0. 0C CT

    C CT

    y yy y

    + .

    hm s ( )y f x= c cc tr tip xc vi trc honh . 0C CT y y = .

  • 8/7/2019 Bi ton Hm s

    2/13

    x1

    2

    0)(0

    S

    a f hoc >

    2

    0)(0

    S

    a f

    HM S BC 3

  • 8/7/2019 Bi ton Hm s

    3/13

    Hm s y = f(x) =ax3 + bx2 + cx+ d, vi (a 0)Min xc nh: D= R

    o hm: y=3ax2 +2bx+c, y=6ax +2b-Nu y c hai nghim phn bit th hm s c 2 cc tr-Nu y c nghim kp hm s n iu trn min xc nhMt s tnh cht

    Tnh cht 1: Hm s ng bin trn R

    >

    0

    0

    a

    Tnh cht 2: Hm s nghch bin trn R

    0Tnh cht 4: th nhn im un lm tm i xng. (Di tm i xng v gc ta , thy ym l hm s l )

    Theo cng thc di trc

    +=

    +=

    yYy

    xXx

    Thay x, y vo phng trnh hm s ta cY+yo=a(X+xo)3+ b(X+xo)2+c(X+xo)+d (1) Y=aX3+g(xo)X L hm s lTnh cht 5:(Bi ton 1) Tip tuyn ( h s gc ) max hay minTa phi da vo phng trnh y thao tc

    y=3ax2 +2bx+c

    B1: Rt 3a ra ngoi k=y=3a(xo+a

    b

    3)2 +

    a

    bac

    3

    2^3

    B2: a>0, th kmin =a

    bac

    3

    2^3 Khi xo=-

    a

    b

    3

    a

  • 8/7/2019 Bi ton Hm s

    4/13

    - Nu (1) c nghim x0 th (1) (x-xo)(ax2+b1x+c1)=0b, Xc nh iu kin ca tham s phng trnh c k no phn bit

    -on n0 x1 :(x-xo)(ax2+b1x+c1)=0

    =++=

    =

    011)(2 cxba xxg

    x ox

    + Mt nghim duy nht:

    =

    =

    0)(

    0

    x og

    g

    + Hai nghim phn bit:

    =

    0)(

    0

    x og

    g

    hoc

    =

    >

    0)(

    0

    x og

    g

    -Khc

    + Mt nghim Ct 0x ti 1 im

    >0* yCTyCD

    euHmsodondi

    >

    >

    0*

    0'

    0'

    y C y C D

    y

    y

  • 8/7/2019 Bi ton Hm s

    5/13

    +Hai nghim ( C)ct 0x ti hai im

    =

    =

    0)2(*)1(

    b i p h nn g h i . m2C 0'

    xyxy

    y

    +ba nghim phn bit (C ) ct 0x ti 3 im phn bit

    Hm s c cc i cc tiu v yCD*yCT

  • 8/7/2019 Bi ton Hm s

    6/13

    Tnh cht1: Hm s c cc tr vi mi gi tr ca tham s khi a 0Tnh cht2: Hm s c cc i v cc tiu

    y=0 c 3 nghim phn 02