TIT : BI TP S NG BIN, NGHCH BIN CA HM S I. Mc ch yu cu 1. Kin thc: Hc sinh cn nm vng khi nim s ng bin nghch bin ca hm s, nh l v iu kin hm s n iu trn mt tp, qui tc xt tnh n iu ca hm s. 2. K nng: Rn luyn k nng xt s ng bin, nghch bin ca hm s, trong c tnh o hm, xt du o hm, lp bng bin thin. 3. T duy, tnh cch: Pht trin t duy logic, t duy bin chng, rn luyn tnh qui c cn thn. II. Phng tin: 1. Gio vin: Gio n, sch gio khoa, dng dy hc. 2. Hc sinh: dng hc tp. Hc k l thuyt v lm bi tp nh. III. Phng php: Kt hp cc phng php thuyt trnh, m thoi, gi m vn . IV. Tin trnh Thi gian Ni dung A. n nh lp (2 pht) B. Kim tra kin thc c (5 pht) HS nhc li qui tc xt tnh ng bin, nghch bin ca hm s v nh l. GV nhn mnh li cc bc xt tnh n iu ca hm s. C. Bi mi Bi 1. Xt s ng bin, nghch bin ca cc hm s sau: 1. y = 4 + 3x x2 ; 1 2. y = x3 + 3x2 7x 2; 3 3. y = x4 2x2 + 3; Hot ng

Gio vin cho bi tp 3 hs ln bng lm bi

1

Li gii: 1. y = 4 + 3x x2 TX: D = R y = 3 2x, xc nh vi mi x R 3 y =0x= 2 BBT

Gio vin cha bi lm ca hc sinh Nhn mnh cch xt du o hm

y

x

+

3 0 71

2 0 +

+

y

64 T bng BT suy ra hm s B trn (; 3/2), NB trn (3/2; +) 1 2. y = x3 + 3x2 7x 2 3 TX: D = R y = x2 + 6x 7, xc nh vi mi x R y = 0 x = 1; x = 7 y > 0 x (; 7) (1; +) y < 0 x (7; 1) BBT Gio vin cha bi, nhn mnh cch xt du o hm, xt du tam thc bc 2

x y y

0 0 +

+

3 Hm s B trn cc khong (; 7) v (1; +) Hm s NB trn (7; 1)

2

3. y = x4 2x2 + 3 TX: D = R y = 4x3 4x, xc nh vi mi x R y = 0 x = 0; x = 1 y > 0 x (1; 0) (1; +) y < 0 x (; 1) (0; 1) BBT

Gio vin bi v mnh xt du phng khong

cha nhn cch bng php

x y y +

1 0 2

0

1 0 +

+

2 Hm s B trn cc khong (1; 0) v (1; +) Hm s NB trn cc khong (; 1) v (0; 1) x2 2x ; 1 x 3. y = x2 x 20. Li gii 3x + 1 1. y = 1x TX: D = R \ {1} 4 y = > 0 vi mi x = 1 (1 x)2 BBT 2. y =

Bi tp 2. Tm cc khong n iu ca hm s: 1. y =

3x + 1 ; 1x

Gio vin cho bi 2 HS ln bng lm bi Gio vin cha bi, nhn mnh vic tch hai khong

x y y +

1 +

+

Hm s B trn cc khong (; 1) v (1; )

3

x2 2x 2. y = 1x TX: D = R \ {1} x2 + 2x 2 < 0 vi mi x = 1 y = (1 x)2 BBT

Gio vin cha bi, nhn mnh cch xt du tam thc bc 2

x y y

1

+

Hm s NB trn cc khong (; 1) v (1; +) 3. y = x2 x 20 TX: D = (; 4] [5; +) 2x 1 , xc nh trn (; 4) (5; +) y = 2 x2 x 20 1 y =0x= 2 y > 0 x > 5; y < 0 x < 4 BBT

Gio vin cha bi, n li cch gii BPT bc 2

x y y

4

1/2

5 +

+

Hm s B trn (5; +), NB trn (; 4) Bi tp 3. Chng minh cc bt ng thc: 1. tan x > x vi mi x (0; /2); x3 vi mi x (0; /2). 2. tan x > x + 3 Li gii 1. tan x > x vi mi x (0; /2) Xt hm s f (x) = tan x x trn ([0; /2) 1 1 = tan2 x 0 vi mi x [0; /2) f (x) = cos2 x Suy ra hm s f (x) B trn [0; /2). Do f (x) > f (0) = 0 vi mi x (0; /2) (PCM) 4

Gio vin cho bi v hng dn hc sinh lm bi Gio vin hng dn cch gii, nhn mnh vic m rng khong ang xt.

x3 vi mi x (0; /2) 2. tan x > x + 3 x3 Xt hm s f (x) = tan x x trn [0; /2) 3 f (x) = (tan x + x)(tan x x) 0 vi mi x [0; /2) (theo 1.) Suy ra f (x) B trn [0; /2) Do f (x) > f (0) = 0 trn [0; /2) (pcm) D. Cng c, hng dn bi tp v nh BTVN: 1d, 2bd, 3,4 (T10) Bi thm 1. Xt s B NB ca cc hm s sau: a) y = 2x3 + 3x2 + 1 b) y = 4 2x2 5 x c) y = 4 x2 x2 d) y = x+2 2. Tm a hm s sau B trn R: 1 y = x3 + ax2 + 4x + 3 3 3. Chng minh cc BT: a) sin x < x, x > 0 b) sin x > x, x < 0 c) sin x + tan x > 2x, x (0; /2)

HS nhc li qui tc xt s bin thin ca hm s Gio vin nhn mnh cc ch v xt du o hm, lp BBT, kt lun

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