23

Click here to load reader

Giải tích 12 - Tính đơn điệu của hàm số

Embed Size (px)

Citation preview

  • 1. August 14 ,2009
    GII TCH 12
    Phn I : Tnhniucahms
    Sontheoschmigmcbnvnngcao
    Nhn space bar hay click chutxemdng hay trangktip
    Bintppps : vinhbinhpro

2. Phn I Tnhniucahms
http://my.opera.com/vinhbinhpro
3. TM TT GIO KHOA
I . nhngha:
Gi I lmtkhong ,mtonhocnakhong (a ; b) ; [ a ; b] ; ( - ; a]
; [b ; +)vflhmsxcnhtrn I.
* f(x)ngbintrnI
* f(x)nghchbintrnI
y
y
x
x
thhmsngbin
thhmsnghchbin
4. TM TT GIO KHOA
II .nhl
Cho hmsfcohmtrnkhong I
a) Hmsfngbintrnkhong I
b) Hmsfnghchbintrnkhong I
x
a
b
f(x)
f(x)
a
b
x
ngbin
f(x)
f (x)
nghchbin
Ch : ngthcf(x) = 0 chxyratimtshuhnimrirctrnkhong(a,b)
+
-
5. TM TT GIO KHOA
III . nhl (iukin)
Cho hmsfcohmtrnkhong I
a) Nu
thhmsfngbintrnkhong I
b) Nu
thhmsfnghchbintrnkhong I
c) Nu
thhmsf khngitrnkhong I
Ch :
1. Xttnhniucahms f trnmtonhocnakhongphib sung thmgithit Hmslintctrnon hay nakhong
6. TM TT GIO KHOA
Ch :
2. Vicchmsathc , hut , lnggic , m , logaritta
cthmrngnhlnhsau :
a) Nu :
Hmsfngbintrn I
b) Nu:
Hmsf nghchbintrn I
( f(x)= 0 chtimtshuhnimca I )
a
b
a
b
x
x
+0 +
-0 -
f(x)
f(x)
f(x)
f(x)
7. Phngphpgiibitonvtnhniucahms
Phngphp 1:
1.Bc 1 : TmminxcnhDcahms
2.Bc 2: Tnhf(x) vtmnghimcaphngtrnhf(x) = 0
3. Lpbngxtduf(x)
Tngktccktquvomtbnggilbngbinthin
8. Bitppdng
Bitp 1:Tmkhongniucahms :
Hngdn:
1. Tpxcnh :D = R
Xtduy
Bintppps: vinhbinhpro
9. Bitppdng
Bitp 1:Tmkhongniucahms :
Hngdn :
1. Tpxcnh :D = R
Xtduy
x

+
0
Xemlaxtduathc - Lp 10 vtrnhbygnli
_
_
- 6x
+
0
+
0
+
+
_
_
y
+
0
0
y
Ktlun : Khongngbin : (- ; 0 ) vnghchbin : (0 ; + )
Bintppps: vinhbinhpro
10. Bitppdng
Bitp 2 : Tmkhongniucahms:
Hngdn :
* Tmtpxcnh : Hmsxcnhkhi :
Xtduy -Duyphthucvo- x
*
-1
1
0
x
y
0

+
y
* Hmslintctrn [-1 , 1] nnhmsnghchbintrn [0,1] vngbintrn [-1,0]
11. Bitppdng
Bitp 3 : Tmkhongngbinvnghchbincahms :
Hngdn:
*Hmsxcnhkhi
Xtdu y .Do khngc qui tcxtdumtbiuthccchacnnnhcsinhcthdngcchgiibtphngtrnh
Hcsinhgii 2 bpt , tmnghimrisuyrakhong B v NB
Cchgiitrnthngtnnhiuthigianvihichnhxccao
12. B sung kinthc
1.im tihn : im
.
gilimtihncahmsnu
tif(x) bng 0 hay khngxcnh
2. Trongtpxcnh D cahmsf(x)
Giahaiimtiknhau
f(x) ginguynmtdu
P DNG
Hmschcmtimtihnx =1
x
-
2
0
Xtkhong(-,1)ly 1 gitrty - chnghnx=0vtnhy(0)
y(0)
1
_
y
0
y
y(0) > 0 suyray > 0 trn(-,1)
+
Tmduytrnkhong(1 , 2) - tngtnhtrn
13. TM TT : Giibitonvtnhniu
1. Bc 1 : Tmtp (min)xcnhcahms.
2. Bc 2: Tnhy(x) , Giiphngtrnhy(x) = 0
Nuy(x) lcchmsathc, phnthc thngthngthlpBNG XT DU y(x)
Nuy(x) lcchmskhngthngthng (vt , lnggic, m , logarit ,) th :
a) Tmimtihncahms.
b) Xcnhduy(x) trntngkhonghaiimtihnknhau I (hay khong (- , x) hay (x,+)) bngcchtnhy() (lmtgitrttachnthuckhongtrn ).Nu

  • y()> 0 => y(x) >0 , vimixthuc I

14. y()< 0 => y(x)