Các phương pháp giải bài tập giải tích 12 nâng cao: Phần 1

Phần 1 tài liệu Giải bài tập giải tích 12 nâng cao do Nguyễn Đức Trí biên soạn cung cấp cho người đọc kiến thức cần nhớ và phương pháp giải các bài tập ứng dụng đạo hàm để khảo sát và vẽ đồ thị của hàm số, Mời các bạn cùng tham khảo.

NGUYEN oCfC CHI

I A I T I C H 12

XAXG CAO

THU VlifJ l\m BINH TH'JAN

NHA XUAT BAN DAI HOC Q U 6 c GIA HA

LCJl NOI DAU

G I A I B A I T A P D A I S O 12, duac bien soan v6i muc dich giiip hoc

sinh doi chieu va kiem tra lai cac ket qua khi thtfc hien giai cac bai tap trong sach giao khoa Muon the, cac em hay danh thcri gian nhat dinh de lam cac bai tap trong sach, sau do doi chieu va kiem tra lai ket qua thiTc hien

G I A I B A I T A P D A I S O 12, phu huynh c6 the suf dung de kiem tra con minh trong viec hoc tap va luyen tap cac kien thufc va ky nang

TJTNG D U N G D A O H A M D E K H A O S A T V A V E

D6 T H I C U A H A M S O

§1 TiNH DdN DIEU CUA HAM SO

ivdl D U I \ cAi^ I^ I O f

1, D i n h l i : Gia sil hkm so f c6 dao M m tren khoang I

• Neu f '(x) > 0, Vx e I t h i h ^ m so f dong bien tren khoang I

• N§'u f '(x) < 0, Vx e I t h i h ^ m so' f nghich bien t r e n khoang I

• Neu f '(x) = 0, Vx e I t h i h ^ m so f khong d6i t r e n khoang I

Chii y: Khoang I neu duoc thay bkng mot doan hoac mot niira khoang t h i

phai bd sung gia thiet " H a m so' lien tuc t r e n doan hoac niifa khoang do"

2 Vi#c x e t c h i e u b i e n t h i e n ciia h a m so' c6 dao h a m c6 the chuyen ve viec

xet dau dao h a m ciia h a m so' do

V§y hkm so d6ng bien t r g n moi khodng ( - 0 0 ; ~) vk ( 1 ; nghich bien

Vay h a m so' dong bien t r e n m6i khoang ( - 0 0 ; - 73 ) va (0; 73 ), nghich

bien tren m o i khoang ( - 73 ; 0) va (73 ; + 0 0 )

X + 1

H a m so xac d i n h t r e n M\{-11 (-2x - 2)(x + 1) - ( - x ' - 2x + 3) - 2 x ' - 2x - 2x - 2 + x ' + 2x - 3

._ - x ' - 2x - 5 (x +

Vay h a m so nghich bien t r e n m§i khodng ( - 0 0 ; - 1 ) va (-!;•+«)

, (Bdi 3 trang 8 SGK) Gidi

a) f(x) = x'^ - Gx"^ + 17x + 4

H a m so xac d i n h t r e n R

Ta CO f '(x) = 3x^ - 12x + 17

f '(x) > 0, Vx e R Vay h a m so' dong bien t r e n R b) f{x) = x^ + X - cosx - 4

H a m so xac d i n h t r e n R

• Ta c6 f '(x) = 3x^ + 1 + sinx > 0, Vx e R Vay h ^ m so dong bien t r e n R

(Bdi 4 trang 8 SGK) Gidi

ftx) = ax - x^

H a m so xAc dinh t r e n R

Ta c6: f '(x) = a - Sx^

* N6u a < 0 t h i f '(x) < 0, Vx e R H ^ m so nghich bi§'n t r g n

* Ng'u a = 0 t h i f '(x) < 0, Vx e R DAng thiJc chi xay ra k h i x = 0

Vay h a m so' nghich bien t r e n R

* Neu a > 0 t h i f '(x) = 0 cs x = ± j

-V3 Bang bien t h i e n

- /—; I — Vay a > 0 khong thoa dieu ki§n de toan

H ^ m so' dong bien t r e n

V£iy vdi a < 0, hkm so nghich bien t r e n R

5 (Bdi 5 trang 8 SGK) Gidi

fix) = - x^ + ax^ + 4x + 3

3

H ^ m s6' x^c d i n h t r e n R

f '(x) = x^ + 2ax + 4, c6 A' = a^ - 4

* Neu a^ - 4 < 0 hay - 2 < a < 2 t h i f '(x) > 0, Vx e R H ^ m so dong bien tren R

• Neu a = 2 t h i f '(x) = (x + 2f > 0,\fx* - 2 H a m so dong bien t r e n R

• Neu a = - 2 t h i f '(x) = (x - 2f > 0, Vx 2 H ^ m so dong bien t r e n M

• Neu a > 2 hoac a < - 2 t h i f '(x) = 0 c6 hai nghiem x i , X2 (xi < X2)

Bang bien t h i e n

f ' ( x )

f(x)

H a m so nghich bien t r e n (xj; X2) khong thoa de toan

Vay - 2 < X < 2 t h i h a m so da cho dong bien tren M

H a m so nghich bien t r e n m o i doan

Vay h a m so' nghich bien t r e n R

Chufng m i n h tuong tyf sinx > x, Vx < 0

x^

b) X e t hkm so g(x) = cosx + — - 1

H ^ m so l i e n tuc t r e n nufa khoang [0; +ao), g'(x) = - s i n x + x

Theo cdu a, g'(x) > 0, Vx > 0, do d6 h ^ m so g dong big'n t r e n [0; + « ) ,

Va t a CO g(x) > g(0), Vx > 0

x ' Nghia la cosx + 1 > 0 , V x > 0

2 TCr d6 suy r a v d i m p i x < 0, t a c6:

(-x)^

cos{-x) + 1 > 0 hay cosx + 1 > 0 , V x < 0

2 2 x"

Vay cosx > 1 - y , Vx 0

c) Xet h ^ m so h(x) = x - — - sinx

6

CO h'(x) = 1 - — - cosx Theo cau b) h'(x) < 0, Vx * 0

Do do h a m so h nghich bien t r e n K h(x) < h(0), Vx > 0 x^ x^

Xet h a m so' f(x) = sinx + tanx - 2x lien tuc tren niira khoang

Vay sinx + tanx - 2x > 0

hay sinx + tanx > 2x, Vx e

t + 5 a) N a m 1980, t a c6 t = 10

c) Toe dp t a n g dan so vac n a m 1990 1^

120 (20 + 5)'

Toe do tSng dan so' v^o n&m 2008 1^

120 (38 + 5)'

Vao n a m 1996 toc dp tang dan so ciia t h i tra'n l a 0,125

1 D i n h nghia: Gia suT h a m so' f xac dinh tren tap c M va XQ e y

* X Q dixac gpi la mot d i e m ciJ^c d a i ciia h a m so' f neu:

Ton t a i mot khoang (a; b) chiJa XQ sao cho (a, b) c 7

va f(x) < f ( x o ) vdi m p i x e (a, b) \

K h i do f(xo) dupe gpi l a gia t r i cufc d a i cua h a m so f

* Xo difpc gpi l a mot d i e m cxic t i e u cua h a m so' f neu:

Ton t a i mot khoang (a; b) chufa XQ sao cho (a; b) e D

v a f(x) > f(xo) v d i m p i x e (a; b) \

K h i do f(xo) dupe gpi l a g i a t r i cij^c t i e u ciia h a m so' f

* D i e m cifc dai va diem cUc tieu gpi chung la d i e m cu'c t r i

Gia t r i cUe dai va gia t r i cUc tieu gpi chung la cu'c t r i

* Neu X o l a diem C L C C t r i cua h a m so f t h i ( X Q ; f ( x o ) ) l a diem cu'c t r i ciia do

Gik sijf h a m so f lien tuc t r e n khoang (a; b) chiJa diem x o v a c6 dao h a m

t r e n cae khoang (a; X Q ) ( X Q ; b) K h i do

a) Neu f '(x) < 0 vdi m p i x e (a; X Q) v a f '(x) > 0 vdi m p i x e ( x o ; b) t h i h a m

so' f dat cvtc tieu t a i diem X Q

b) Neu f '(x) > 0 vdi mpi x e (a; X Q ) v a f '(x) < 0 vdi m p i x e ( X Q ; b) t h i h a m

so dat cue dai t a i diem X Q

4 Q u i t S c tim c\ic t r i

* Q u i t ^ c 1:

• T i m f '(x)

• T i m cac diem X i (i = 1, 2, ) t a i do dao h a m cua h a m so' hhng 0 hoac

h a m so' lien tuc nhung khong c6 dao h a m

• X e t da'u f '(x) neu f '(x) ddi da'u k h i x qua diem Xj t h i h a m so' dat cUc t r i

t a i X j

• f ' ( X )

X i

f ' ( x ) f(x)

Vay h a m so da cho dat cUc dai t a i diem x = - 3 , gia t r i eUe dai l a f(-3) :

H k m so dat cUc tieu t a i d i l m x = - 1 , gia t r i cUc tieu cua h a m so l a f l - 1 ) =

H a m so' dat ciTc tieu t a i diem x = 1, gia t r i ci/c tieu la f ( l ) = 2

Vay h a m so' dat cifc dai t a i x = - 1 , gia t r i cUc dai fi-1) = 1 va dat ciic tieu

t a i x = 0, gia t r i cifc tieu fiO) = 0

15 f) fix) = x^ - 3x + 3

H a m so dat cifc tieu t a i x = - 72 , gia t r i c\ic tieu la -2

H ^ m so dat c\ic dai t a i x = 72 , gia t r i cUc dai 1^ 2

b ) y = 78-x' - • •

H a m so xac d i n h va lien tuc t r e n [-2 72; 2 72 ] y' = , ~^ vdi moi x € (-2 72; 2 72 ); y' = 0 o x = 0

7 8 ^ Bang bien t h i e n

H a m so dat cifc dai t a i x = 0, gia t r i cxic dai: 2 72

c) Ap dung qui tAc 2

* Vi y " — + kn = 4 s i n = 4 s i n I 3J = - 4 s i n

= _ 4 ^ = - 2 V 3 < 0

Do d6 h a m so dat ciTc dai t a i c6c diem x = - — + k 7 t , k e Z

• • 6 Gia t r i c\Xc dai y ( - — + k 7 r ) = - - + k 7 i - sin

Ngoai r a y " = 2cosx + 4cos2x

• V i y"(k7i) = 2cosk7r + 4cos2k7i = 2cosk7i + 4 > 0, V k s Z

Do do h a m so da cho d a t cUc tieu t a i cac d i e m x = k n v a gid t r i cUc t i e u

y(k7t) = 3 - 2cosk7x - cos2k7r = 2 - 2cosk7r = 2 - 2 ( - l ) ' ' (k e Z)

Gidi he phiTOng t r i n h 3a + 2b = 0

a + b = l

a? + 2b = 0 -2a - 2b = - 2 3a + 2b = 0

• f "(0) = 6 > 0 Vay h k m so dat ciTc tieu t a i diem x = 0 va f(0) = 0

• f "(1) = - - 1 2 + 6 = - 6 < 0 Vay ham so dat cifc dai tai diem x = 1 va f(l) = 1

14 (Bcii 14 trang 17 SGK) Giai

l + a + b + c = 0 (3)

4a - 2b + c - 8 = 0 -4a + b + 12 = 0

a + b + c + l = 0

G i a i he phUcfng t r i n h :

a + b + c + l = 0

o ] - 4 a + b + 12 = 0 3a - 3b - 9 - 0

a + b + c + l = 0

o -12 + b + 12 = 0

a = 3 Kiem t r a l a i k e t qua

f(x) = X-' + 3x^ - 4

f ' ( x ) = 3x? + 6x, f '(x) = 0 o X =

a + b + c + l = 0 -4a + b + 12 = 0

H a m so dat ciTc t r i t a i dig'm x = - 2 gi^ t r i ciTc t r i 1^

f(-2) = - 8 + 12 - 4 = 0

Do f ( l ) = 1 + 3.1 - 4 = 0

Vay d i e m A ( l ; 0) thuoc do t h i ham so f

(x - m ) ' y' = O o x = m - l hoSc x = m + 1

Vay vdi moi gia t r i ciia m, h a m so da cho dat ciTc dai t a i diem x m - 1 v;

dat ciTc tieu t a i diem x = m + 1

§3 GIA TR! L6N NHAT, GIA TRj NHO NHAT

C U A HAM SO

I \ 6 l D U I \ C A l ^ r^iiof

1 D i n h n g h i a : Gia stf ham so f xac dinh t r e n tap 7 ('J cz R)

* Neu ton t a i diem XQ € V sao cho: f(x) < f(xo) vdi moi x,, e V

Thi so M = f(Xo) goi la gia t r i IdTn n h a t cua ham so f t r e n V

K i hieu M = maxf(x)

X 7

* Neu ton t a i diem x,, s 7 sao cho: f(x) > fixo) vdi moi x e 7

T h i so m - f(xo) goi 1^ gia t r i nho n h a t cua h^m so f t r e n 7

K i hieu m = m i n f ( x )

• r ' ^ ^ ^ i t a c tim gia t r i Idn n h a t , gia t r i nho n h a t

* T i m cac diem x,, x-,, x^ thuoc (a; b) t a i do h^m so f c6 dao ham bfing

0 hoac khong c6 dao ham

, T i n h f(xi), f ( x 2 ) , fix J , f(a), f(b)

* So s^nh cdc gia t r i t i m di^cfc _ So Idn nhat trong cac gia t r i do la gia t r i Idn nhat ciia f t r e n [a; b] _ So nho nhat trong cac gia t r i do la gia t r i nho nhat ciia f tren [a, b]

J B A I T A P

f(x) = sin''x + cos''x

Ham so xac dinh t r e n K

* f(x) = (sin^x + cos^x)^ - 2sin^xcos^x = 1 1 (2sinxcosx)

Bang bien thien

, (4x + 5)(x + 2) - ( 2 x ' + 5x + 4)

(x + 2)' ~ (x + 2)'

f (x) = 0 » 2x^ -i- 8x + 6 = 0 •» X = - 1 hoac x = -3 Bang bien thien:

t-5.1 1 + 4 "

f(2) 2 i =

-2 -2

Vay max f(x) = —

x u l 0 ; 2 l 2

Ham so khong dat gia t r i nho nhat tren nOfa khoang (0; 2J

18 (Bai 18 trang 22 SGK) Gidi

a) y = 2sin^x + 2sinx - 1 H^m so xdc dinh tren M Dat t = sinx, - 1 < t < 1

y = fit) = 2t^ + 2t - 1

Tim gid t r i Idn nhS't wk gid t r i nh6 nhS't cua ham so y = fit) tren

doan [ - 1 ; 11,.do cung la gid t r i nhd nhat cua ham so da cho trSu

f ( t ) = 4t + 2; f ( t ) = 0 o 4 t + 2 = 0< T > t =

Ta CO y = f(t) = - t ^ - - 1 + 5

2

1 1 l i f'(t) = - 2 t - - ; f'(t) = 0 c : > - 2 t - - = O c : t = - -

2 2 4 Bang bien t h i e n '*

Do do S(x) dat g i ^ t r i 16n nhat k h i M d v i t r i t r e n BC sao cho B M = ^ BC v^

gi^ t r i Idn n h a t ciia d i ^ n tich h i n h chuf nhat la —

8

Tren moi don v i dien tich ciia mat ho c6 n con ca t h i sau mot vu, so' ca t r e n moi don v i dien tich mat ho t r u n g binh can nang

f(n) = nP(n) = 480n - 20n''^ (gam)

Xet ham so f(x) = 480x - 20x^ x e (0; +x) 'Bien so n e N* dUOc thay bkng bien x e (0; + « ) )

f (x) = - 4 0 x + 480 f'(x) = 0 » - 4 0 x + 480 = 0 « X = 12 Bang bien t h i e n

_ f ' ( X ) + 0

f(12) = 480.12 - 20.12^ = 2880

Goi M(x; x^) la diem bat k y cua parabol (.f)

f(x) dat gia t r i nho n h a t t a i diem x = - 1 , gia t r i nho n h a t la f ( - l ) = 5

Do do khoang each A M dat gia t r i nho n h a t k h i M d v i t r i diem M o ( - l ; 1) va

khoang each nho n h a t AM,, = Vs

Van toe eua ca k h i bcfi ngUorc dong la (v - 6) knJgib

Thori gian de ca viiot khoang each 300 k m la 300

= 300 e 3v2 - 18v^ - v^

{ v - 6 ) ^

= 300 c. 2 X 1 ^ = 600 cv^ ^^-^^

( v - 6 ) ^ ( v - 6 ) ^ E'(v) = 0 <o V = 0 hoac v = 9 (v = 0 loai do v > 6)

g (Bai 26 trang 23 SGK) Gidi

S6' ngu'di n h i e m benh ke tii ngay xuat h i e n benh n h a n dau t i e n den ngay

y ii: fit) rz 90t - 3t^

y' = f "(t) = -6t + 90

y' = 0 « t = 15 Bang bien t h i e n

c) Toe dp t r u y e n b e ^ Idn hon 600

y > 600 c:^ sot - 3t^ > 600 e> t^ - 30t + 200 < 0 ci- (t - 15)^ < 5^

a) a) fix) = V 3 - 2 x

H a m so' xac d i n h t r e n [-3; IJ

- 1 f'(x) = = < 0 V d i moi x e

§4 DO TH! CUA HAM SO

- PHEP TjNH TIEN HE TOA Dp

I\6l D U I \ CAIV I«ICf

1 P h e p t i n h t i e n h# t o a dp v a c o n g thijfc c h u y e n hp tpa dp

- Trong m a t ph^ng toa do Oxy cho diem

Kxo; yo), goi I X Y la h$ toa do mdi goc I va

hai true I X , l Y theo t h i l t u c6 cung vectot dan

vi i , j vdi hai true Ox, Oy

Goi M la diem bat k i eiia m a t ph^ng

(x; y) l a toa do diem M doi vdi he tpa do Oxy

(X; Y) l a toa do diem M doi v6i he tpa dp I X Y

[x = X + X ( ,

l y = Y + y„

C^e he thijfc t r e n goi la cong thiJc chuyen he tpa dp t r o n g ph6p t i n h tien

theo vecto 0 1

2 Phifofng t r i n h c u a do t h i do'i vdfi hp tpa dp m d i

Gpi y - f(x) l a phuong t r i n h cua do t h i {'^) do'i v 6 i he tpa dp Oxy, k h i do

phJOng t r i n h ciia do t h i {'f) doi v6i he tpa dp I X Y la Y = f(X + x,,) - yo

Cong thuTc chuyen h§ toa dp trong phep t i n h tien vectcf 0 1 \h

Phi/otng t r i n h duorng Parabol doi vdi he tpa dp I X Y la:

Cong thufc chuyen he tpa dp t r o n g phep t i n h tien vectof 0 1 la

Phucng t r i n h Parabol do'i vdi he tpa dp I X Y \k:

Y - - = - ( X + 1 ) ' - ( X + l ) - 3

2 2 hay Y = i ( X ' + 2 X + 1 ) - X - 1 - 3 + -

Cong thufc chuyen h f tpa dp trong phep t i n h tien vectcf 0 1 l a

Phuorng t r i n h Parabol doi vdi he tpa dp I X Y 1^:

8 X^ + - X + 2 n

4 64 16

Y = X + A - 4 X ' - X - — - — = ^ Y = - 4 X 1

8 16 16 d) y = 2x2 - 5

D i n h I X , = 0

y, = - 5 T a C O 1(0; - 5 ) Cong thiJc chuyen he tpa dp trong phep t i n h tien vectof 0 1 la Phuong t r i n h Parabol doi vdi he tpa dp I X Y la:

Y - 5 - 2 X ' - 5 hay Y = 2 X '

X

y

X

Y - 5

30 (Bdi 30 trang 27 SGK) Giai

y = Y - 1 Phirang t r i n h diTcfng cong doi vdi h$ toa dp I X Y la

vay t r d n khoang (1; +oc) {'f) n k m phia t r e n (d)

^l (Bai 31 trang 27 SGK)' Gidi

fx = X - 2 Cong thilc chuyen he toa dp trong phep t i n h tien vector Ol la: _ y + 2

Phuong t r i n h diTdng cong Cf) doi vdi he toa dp I X Y la:

32. (Bai 32 trang 28 SGK) Gidi

a) Thifc h i f n ph6p t i n h t i e n vectcf O I vdi 1(1; 1) Cong thiJc chuyen h# toa do trong ph6p t i n h t i e n n^y 1^ X = X + 1

y = Y + 1 Phi/cfng t r i n h cua do t h i doi vdi hp toa dp I X Y 1^

+ 1 => Y = —

X + 1 - 1 X Dat Y = « x ) = -

Tap xac d i n h cua h a m so n^y 1^ 7 = R \: VX e 7, - X G 7

Thirc h i p n ph6p t i n h t i e n vectcf 0 1 vdi I ( - l ; 3)

Cong thiJc chuyen he toa dp trong phep t i n h t i e n n^y \k

Phiforng t r i n h ciia do t h i doi vdi he toa dp I X Y 1^:

Do d6 t a m doi xilng cua ham s6' da cho la I ( - l ; 3)

(Bai 33 trang 28 SGK) Giai

Cong thufc chuyen hp tpa dp trong phep t i n h t i e n vectcf 0 1 la

=> Y = aX + axo + b + - axo - b=:>Y=aX + —

X X Dat Y = f(x) = aX +

T|ip xdc dinh cua ham so n^y la 7 = K \1

Vx e 7, -X 6 ( /

«-X) = a(-X) + -X a x - = -f(x)

Vay hkm so Y = aX + — la ham so le nen nhan goc toa do I la tarn doi xiiCng

§5 DJdNG TIEM CAN CUA DO THj HAM SO

1 Dadng th^ng y = yo difoc goi la diforng tiem can ngang (goi t^t la ti$m can

ngang) cua do thi ham so y = f(x) neu: lira f(x) = yo hoac lim f(x) = yo

/ ^ 0 0

y = flx)

2 Du6ng thing x = XQ dUdc goi la ducmg tiem can diJng (gpi tit la ti$m can

drfng) ciia do thi ham so y = f(x) neu:

limf(x) = + 0 0 hoac limf(x) =+oo hoac

limf(x) =-co hoac limf(x) = -3o

3 Difcfng thing y = ax + b (a # 0) dUcfc gpi la dudng tiem can xien (goi tit la

ti^m can xien) cua do thi ham so' y = f(x) neu:

lim [f (x) - (ax + b)] = 0 hoac lim [f (x) - (ax + b)] = 0

3

y = fix)

X

Chii y: De xdc dinh cac he so a, b trong phifong trinh diTdng tiem can xien

y = ax + b, ta CO the sut dung

, f(x) Cong thiTc a = lim

f(x) , b = lim [f (x) - ax]

hoac a = lim , b = lim[f(x)-ax]

(Khi a = 0 thi ta c6 tiem can ngang la y = b)

34 (Bai 34 trang 35 SGK)

x - 2 a) y = ^ ^ 3x + 2 Gidi

Ham so' xac dinh tren R \

1 3

* Vi lim y = —

Nen difcrng thing y = — la tiem can ngang cua do thi ham so' (khi x ^ +co 3

va khi X - c o )

* Vi lim y = +co va lim y = - Q O

Nen ducfng thing x = — la difcfng tiem can dufng ciia do thi ham so

c) y = X + 2

Hkm so xdc dinh tren R \|

* V i lim y = + x va lim y = -oo

Nen dudng th^ng x = 3 1^ ti^m ogn diifng ciia do thi ham so' (khi x ^ 3

Ham so xac dinh tren K \j -—

i V i lim y = -00 vk lim y = +oo

Nen dxibng t h i n g x = - — la tifm can diJng ciia do thi hkm so (khi

7

4

1 7 Vay difdng thang y = - x - - la ti^m can xien ciia do thi ham so (khi

-Ham so xkc dinh tr§ri R \I

* Vi hm y = +oc vk lim y = -oo x-.<-n" x->i-i)*

Nen du6ng thing x = -1 la tiem can diltig cua do thi ham so (khi x ~> (-1)

Ham so xac dinh tren R 1101

x' + 2

b ) y =

x ' - 2 x Hkm so xac dinh tren R \; 21

• V i lim y = -oo vk lim y = +oo

c ) y =

Nen diiejng thing y = x + 2 la ti|m can xien cua do thi hkm so (khi x + »

va khi X - » -oo) x^ + X + 1 •

X ^ - l Hkm so xac dinh tren R \; 11

H a m so xac dinh tren (-00; - 1 ] [ 1 ; +00)

Tiem can xien c6 dang y = ax + b

V i l i m y = l i m ( x + Vx^ +1) = l i m

1 = 0

" ^ ^ x - V x ^ + l Vky dudng t h i n g y = 0 Ik t i ^ m can ngang cua d& t h i h k m so (khi x -> -00)

V a y dudng t h i n g y = - x - i Ik tigm c a n xien cua do t h i h k m so (khi x -> -oc)

3 7 (Bai 37, trang 36 SGK) Gidi

Diforng thSng y = - x la t i e m can xien ciia do t h i h ^ m so' (khi x -oo)

Phufong t r i n h m ddi v6i h f toa dp IXY

Vay h a m so' Y - la ham so' le nen do t h i nhan go'c 1(3; 4) lam tam

X doi xiifng

39 (Bai 39, trang 36 SGK) Gidi

C6ng thufc chuyen he toa dp theo phep t i n h t i e n 01

c) Phiforng t r i n h cua (f) doi vdi he toa dp IXY

Dat Y = fiX) = ^ , tap xac dinh 7 = K \: Vx € '/, - X e 7

• Elide 1: T i m tap xac d i n h cua hkm so

• B i f d c 2: X6t sir bien t h i e n cua hkm s6

* T i m gi(Ji h a n t a i v6 cifc vk gidi han v6 cifc (n^u c6) cua hkm so T i m

cac dudng t i g m can cua do t h i (neu c6)

* Lap bang bien t h i e n cua hkm so bao gom:

T i m dao hkm cua hkm so

- X6t dau dao hkm, x6t chieu bien t h i e n vk t i m cUc t r i cua hkm s6 (n§'u c6), dien k e t qua vko bang

• BvCdc 3: Ve do t h i cua h a m so

* Ve ckc dufcfng t i | m can cua do t h i (neu c6)

* Xac d i n h mpt so diem dac bi#t ciia do t h i (giao d i e m vdi cdc true toa

dp, neu phep toan khong philc tap)

* N h a n xet ve do t h i (chi ra true vk t a m doi xijfng cua do t h i (neu c6))

Neu hkm so y = f(x) c6 dao h a m cap hai t r e n mpt khoang chiia diem X Q ,

f "(xo) = 0 va f "(x) doi dau k h i x qua Xo t h i d i e m U(xo; f(xo) la mot d i e m

uo'n cua do t h i hkm so y = f(x)

* Ghi chu: Do t h i h a m so bac ba y = ax^ + bx^ + cx + d (a ;^ 0) h a m c6 mot

diem uon va diem do \k t a m doi xiifng ciia do t h i

3 H a m so' t r u n g phi^o^ng y = ax'* + bx^ + c (a 5* 0)

a) Khdo sat sU bii'n thien va ve do thi ham s6 y = + 3x^ - 4

1- H k m so xac d i n h t r e n R

2- Su bien t h i e n cua h k m so:

a) Gidi han t a i v6 ciTc

l i m y = -00 va l i m y = +00 b) Suf bi§'n t h i e n , ciTc t r i

Ta c6: y' = Sx^ + 6x = 3x(x + 2)

y' = 0 o 3x(x + 2) = 0 c : > x = 0 hokc x = - 2

+ Hkm so d6ng hi&'n t r e n m5i k h o a n g ( - o c ; -2) vk (0; +00), n g h i c h bie'i

4 1 (Bai 41, trang 44 SGK) Gidi

a) Khdo sat sU bien thien va ve y = -x^ + 3x^ - 1

42 (Bdi 42, trang 44 SGK) Gidi

a) Khdo sat sil bien thien va ve do thi y = ^ x ^ - x^ - 3 x - -

3 3 1- H a m so xac d i n h t r e n K

2- S u b i e n t h i e n ciia h a m so'

^) Gidri h a n t a i v 6 cifc:

• l i m y = -00 • l i m y = +00 b) S u b i e n t h i e n , cUc t r i :

• T a c6 y ' = x^ - 2 x - 3

y ' = 0 c:> X = - 1 h o a c x = 3

H a m so' d o n g b i e n t r e n cac k h o a n g ( - 0 0 ; - 1 ) v a (3; +00); n g h i c h b i e n t r e n

k h o a n g ( - 1 ; 3)

* H^m s? dat cifc dai tai x = - 1 , gi^ t r i cue dai y ( - l ) = 0

* Hkm so dat cUc ti§'u tai x = 3, gi& t r i cUc tieu y(3) =

c) Bang big'n thien

khi X qua x = 1 Vay diem U 1, - — 1^ V 1,

-d i l m u6'n cua -do thi

• Giao diem cua do thi vdi true tung

3)

0; -5

NMn xet: Do thi hkm so nhan U

l^m tam doi xiifng

b) Khdo sat sUbiin thi&n dS thi id vey = - 3x + 1

1 H^m so xac dinh tren K

2 Sif bi§'n thien cua h^m so'

a) Gi6i han tai v6 cifc

• limy = -00 , limy = +oo

b) Sir bien thien, cUc t r i

• y' = 3 x ^ - 3 , y' = 0 » x = ±l

• H^m so dong bien tren cac khoang (-oo; -1) vk (1; +oo), nghich bien tren

khoang (-1; 1)

• Ukm so dat ciTc dai tai x - - 1 , gia t r i cUc dai y ( - l ) = 3

• Ham so dat cifc tieu tai x = 1, gik t r i cufc tieu y(l) = - 1

c) Bang bien thien

y" = 0 tai X = 0 va doi dau tii am sang

dUcJng khi x qua x = 0 Vay diem U(0; 1)

la diem uon cua do thi

• Giao diem cua do thi vdi true tung (0; 1)

Nhdn xet: Do thi hkm so nhan diem U(0; 1)

Ikm tam doi xufng

c) Khdo sat sU biin thiin dS thi y = - g? + - 2x - -

3 3

1 H^m s6' xkc dinh tren K

2 Si^ bien thi§n cua ham so"

a) Gi(Ji han tai v6 cUc

• limy = +=» ; limy = -oo

b) S\i bien thien, cuTc t r i

, / = -x^ + 2x - 2 < 0, Vx 6 R

* Ham so nghich bien tren E

c) Bang bien thien

Nhdn xet: Do thi ham so nh$n U 1; - 24) 23 lam tfim doi xiifng

d) Khdo sat sU biin thiin vd ve do thi hdm so y = x^ - 3x^ + 3x + 1

1- H^m so xac dinh tren M

2- Sir bien thien cua h^m so'

a) Gi6i han tai v6 ciTc

• limy = -00 , limy = +oo

b) Sir bien thien, cire t r i

• y' = 3x^ - 6x + 3, y' > 0, Vx l

• H^m so dong bien tren M \

c) Bang bien thien

y' + 0 +

y

— 0 0 '