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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