CHUYÊN ĐỀ TOÁN - s7ab974ecf8ca26a6.jimcontent.coms7ab974ecf8ca26a6.jimcontent.com/download/version/1431223040/mo… · Các ví dụ đều năm trong sách giáo khoa cơ bản

Trn Mu T-TMT- CLB Gia S Bi Thi Khoa Hc: https://www.facebook.com/giasubaithikhoahoc?fref=ts.

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VIC HC NH I THUYN NGC NC-KHNG TIN T S LI.

CHUYN TON:

PHONG CCH LM CHUN CHO 1 BI TON KHO ST HM S, HNG DN TRNH BY P V C TH Y CC BC KHO ST. Chuyn i gii p thc mc khi cc em trnh by ton

kho st hm s, y cc trng hp v mi loi hm s.

Cc v d u nm trong sch gio khoa c bn. Cc em hy c v cho kin ng gp v ch gio qua

nick fb: Gia s Bi Thi: MC LC:

- Hm s bc 3.2->7. - Hm trng phng7->9. - Hm s dng a thc bc nht trn bc nht9->11 - Hm s dng a thc bc 2 trn bc nht.11->HT.

Trn Mu T-TMT


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CC V D C TH CHO PHONG CCH TRNH BY. Hm bc 3. Bi 1: KHO ST V V TH HM S Y =3 + 3 4 Li gii: 1. TX: D = R 2. S bin thin.

- Chiu bin thin: + ao hm y' = 3 + 6

y' = 0 0

2

x

x

Ta c du ca y:

: + Do : Hm s ng bin trn khong (,2)(0,+) Hm s nghch bin trn (-2,0).

- Cc tr ca hm s: + Hm s t cc i ti x = -2, yC = 0. + Hm s t cc tiu ti x = 0 , yCT = -4

- Gii hn hm s:

+ 3 2 3

3

3 4lim lim( 3 4) lim (1 )x x x

y x x xx x

+ 3 2 3

3

3 4lim lim ( 3 4) lim (1 )x x x

y x x xx x

- th hm s khng c tim cn. - Bng bin thin.


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3. th ca hm s: - Giao im ca th vi cc trc ta : + Giao diem cua ham so voi truc Ox y = 0 x = -2 ; x = 1 + Giao diem cua ham so voi truc Oy x = 0 y = -4

- Nhan xet: + Do thi ham so nhan diem uon E( 0 ; -4 ) lam tam doi xung

- Ve do thi ham so:

Li bnh: Phng php trnh bay trn l cch trnh by chun v y nht cho bi ton, khi c cc em c th thy hi di, nhng khi lm thnh tho v nh cc bc th cc em thng lm bi ny trong vng 5 pht l xong. Ngoi ra khi v th cc em c th tm thm cc im ngoi nhp v th cho d khng cn trnh by vo bi lm, trn l cch trnh by rt chun ri.


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Bi 2: KHO ST HM S V V TH HM S Y =

+ 2

Li gii:

1. TX: D = R. 2. S bin thin : - Chiu bin thin:

+ o hm: y = + 2 2

y=0< => + 2 2 = 0 1 3

1 3

Ta c du ca y:

+ Do : Hm s ng bin trn khong (-,1 + 3)(1 + 3,+)

Hm s nghch bin trn khong (1 3,1 + 3) - Cc tr ca hm s:

+Hm s t cc i ti x = 1 3 v yC = y(1 3) = 2 + 23

+Hm s t cc tiu ti x = 1 + 3 v yCT = y(1 + 3) = 2 23 - Gii hn hm s:

+ 3 2 3

2 3

1 2 1 1 2 2lim lim( 2 ) lim ( )

3 3 3 3x x xy x x x x

x x x

+3 2 3

2 3

1 2 1 1 2 2lim lim( 2 ) lim ( )

3 3 3 3x x xy x x x x

x x x .

- Hm s khng c tim cn. - Bng bin thin.

3. th ca ca hm s:


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- th giao vi cc trc:

+ Giao vi ox:

+ 2

= 0 c 3 nghim , do th ct ox

ti 3 im phn bit.

+ Giao vi oy: x= 0 => y =

, do hm s ct trc oy ti im ( 0,

)

- Hm s nhn (-1,2) lm im un. - Ta c th hm s .

Li bnh: Bi ton ny a ra nhm cung cp cho cc em v vic 1 s bi ton cho s liu khng c p th cc em khng nht thit phi tm ra kt qu, v d nh phn tm giao im ca th vi ox, cc em ch cn ni n ct trc ox ti 3 im phn bit l c, v phn cc tr hm s cng vy

cc em khng cn tm ra kt qu l yC = y(1 3) = 2 + 23 yCT =

y(1 + 3) = 2 23,ch cn vit

yC = y(1 3)yCT = y(1 + 3) cng c cc em .


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Bi 3: KHO ST V V TH CA HM S y= 3 + 3 + 1 Li gii:

1. TX: D=R 2. S bin thin: - Chiu bin thin:

+ o hm: y = 3 6 + 3 y = 0 3 6 + 3 = 0 x = 1. Suy ra y = 3( 1) 0, . Do hm s lun ng bin trn R.

- Hm s khng c trc tr. - Gii hn:

+ 23 2 3

3

3 3 1lim lim ( ) l3 3 1 im (1 )x x x

y xx

x xx x

x

+ 23 2 3

3

3 3 1lim lim( ) l3 3 1 im (1 )x x x

y xx

x xx x

x

- Hm s khng c tim cn. - Bng bin thin.

3. th ca hm s. - th giao vi cc trc ta .

+ Giao vi ox: 3 + 3 + 1 = 01, 01m

+ Giao vi oy: x = 0=> y = 1, do th ct oy ti im (0,1) - th h/s nhn im ( 1, 2) lm tm i xng. - Ta c th ca hm s.


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Li bnh: Trn l 3 bi ton vi 3 trng hp hon ton khc nhau, cc em thy c s trnh by y v theo tun t nh th, l 3 bi ton in hnh cho vic kho st hm s bc 3. Sau y chng ta chuyn sang hm s mi.


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HM TRNG PHNG. Bi 4. KHO ST V V TH HM S y = + 2 2 Li gii:

1. TX: D=R 2. S bin thin: - Chiu bin thin:

+ o hm: y = 4 + 4

y = 0 4 + 4 = 0

0

1

1

x

x

x

Ta c du ca y:

+ Do : Hm s ng bin trn khong (-,1)(0,1) Hm s nghch bin trn khong (-1,0) v (1,+)

- Cc tr ca hm s. + Hm s t cc i ti 2 im c honh x = -1, yC=-1.V x = 1,yC=-1 +Hm s t cc tiu ti im c honh x=0, yCT=-2.

- Gii hn hm s:

+4 4

2 4

22 2)2 2

lim lim ( lim ( 1 )x x x

y xxx x

x

+4 4

2 4

22 2)2 2

lim lim ( lim ( 1 )x x x

y xxx x

x

- Hm s khng c tim cn. - Bng bin thin:


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3. th ca hm s. - Giao diem cua do thi voi cac truc toa do

+ Giao diem cua ham so voi truc Ox y = 0 Vo nghiem, ham so khong cat truc Ox

+ Giao diem cua ham so voi truc Oy x = 0 y = -2, do o ham so ct oy to im (0,-2) - Nhan xet + Ham so da cho la ham so chan nen do thi nhan truc tung lam truc doi xung -Ve do thi ham so.

Li bnh: Cc bi ton hm trng phng u c cch trnh by nh th, cc em trnh by v lm nhiu th s thnh tho v lm rt nhanh, cc trng hp s liu khng p th cc em cng lm theo hng dn ca anh trang th 5 ca chuyn ny. Tip theo ta sang dng hm s khc.


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HM S DNG A THC BC NHT TRN BC NHT.

Bi 5: KHO ST V V TH CA HM S y =

.

Li gii:

1. TX: D = R1

\2

.

2. S bin thin: - Chiu bin thin:

+ o hm:

y =

() , ta thy y >0, .

+ Do hm s lun ng bin trn khong (-,

)(

, +).

- Hm s khng c cc tr. - Gii hn hm s:

1 1( ) ( )

2 2

1 1( ) ( )

2 2

2 1lim lim

2 1 2

2 1lim lim

2 1 2

2lim lim

2 1

2lim lim

2 1

x x

x x

x x

x x

xy

x

xy

x

xy

x

xy

x

- Tim cn ca hm s:

Do: 1

lim lim2x x

y y

, nn hm s c tim cn ngang l t y=

.

Do:

1( )

2

1( )

2

lim

lim

x

x

y

y

nn hm s c tin cn ng l t x = -


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- Bng bin thin:

3. th hm s: - th giao vi cc trc ta .

+ Trc ox: ti im (2,0) + Trc oy ti im (2,0)

- th nhn giao im (

,

) lm tm i xng.

- Ta c th ca hm s.

Li bnh: Trn l cch trnh by ca dng hm s trn, cc em hon ton lm tun t nh th cho cc bi ton khc.


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HM S DNG A THC BC 2 TRN BC NHT

Bi 6: KHO ST V V TH HM S y =

Li gii:

Ta c: y =

= + 1 +

1. TX: D = R \ 1 .

2. S bin thin: - Chiu bin thin:

+ o hm:

y = 1

().

y = 0 1

()= 0

0

2

x

x

Ta c du ca y:

+ Do : Hm s ng bin trn khong (-,2)(0,+). Hm s nghch bin trn khong (-2,0).

- Cc tr ca hm s: + Ham so dat cuc dai tai diem x = -2; gia tri cuc dai cua ham so la y = -2 + Ham so dat cuc tieu tai diem x = 0; gia tri cuc tieu cua ham so la y = 2

- Gii hn ca hm s:

( 1)

2

( 1)

(

2

1) ( 1

2

)

2

2 2

1

2

lim lim

lim lim

lim lim

lim li

2

1

2 2

1

2m

2

1

x x

x x

x x

x x

y

y

y

x x

x

x x

x

x x

x

xy

x

x


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- Tim cn ca th hm s. th hs khng c tim cn ngang.

Do: ( 1 )

( 1 )

l im

l im

x

x

y

y

nn th hm s c tim cn ng x = -1.

Gi tim cn xin l y = a. x+b.

Trong a =

2 2 21

(lim li

)m

1x xxy

x

x

x x

b =

2 2 2lim( . ) lim( l

2)

1im 1

1x x xx x x

y ax

xx

x

Do th hs c tim cn xin y = x +1 - Bng bin thin:

3. th ca hm s. - Giao diem cua do thi voi cac truc toa do

+ Giao diem cua ham so voi truc Ox y = 0 Vo nghiem, ham so khong cat truc Ox + Giao diem cua ham so voi truc Oy x = 0 y = 2

- Nhan xet + Do thi ham so nhan giao diem C (-1;0) cua 2 tiem can lam tam doi xung

- Ve do thi ham so:


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Li bnh: Nh vy chng ta tri qua cc v d tip cn c phng php v cch trnh by loi ton kho st hm s. Mong cc em ng gp kin qua nick fb: Gia S Khoa Hc. https://www.facebook.com/profile.php?id=100008341510957. -HT-

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CHUYÊN ĐỀ TOÁN - s7ab974ecf8ca26a6.jimcontent.coms7ab974ecf8ca26a6.jimcontent.com/download/version/1431223040/mo… · Các ví dụ đều năm trong sách giáo khoa cơ bản