giải bài tập giải tích 12 nâng cao

515.076
^BlfCCHI
GI-103B
NANG
r
DVL.013442
m Nha
xuat
ban Dai
hoc
Quoc
gia Ha Noi
NGUYEN oCfC CHI
^UJOI
bai tdfL
I AI TICH 12
XAXG CAO
THU VlifJ l\m BINH TH'JAN
NHA XUAT BAN DAI HOC QU6c GIA HA
LCJl
NOI DAU
GIAI
BAI TAP DAI SO 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.
GIAI
BAI TAP DAI SO 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
CO
ban.
GlAl
BAI TAP
DAI
SO 12, cac dong nghiep c6 the suf dung de
tham khao.
Mong
diioc sii gop y
chan
thanh cua ban doc gan xa.
TAC
GIA
TJTNG
DUNG
DAO HAM DE

KHAO
SAT VA VE
D6 THI
CUA HAM SO
§1.
TiNH
DdN
DIEU
CUA HAM SO
ivdl
DUI\
cAi^ I^IOf
1,
Dinh
li: Gia sil hkm so f c6 dao Mm
tren
khoang I.
• Neu f '(x) > 0, Vx e I thi h^m so f dong bien
tren
khoang I
• N§'u f '(x) < 0, Vx e I thi h^m so' f nghich bien
tren
khoang I
• Neu f '(x) = 0, Vx e I thi h^m so f khong d6i
tren
khoang I
Chii
y: Khoang I neu
duoc
thay bkng mot doan

hoac
mot
niira
khoang thi
phai
bd sung gia
thiet
"Ham so'
lien
tuc
tren
doan
hoac
niifa
khoang do"
2.
Vi#c
xet
chieu bien thien
ciia ham so' c6 dao ham c6 the chuyen ve
viec
xet dau dao ham ciia ham so' do.
^ BAITAP
1.
(Bai 1 trang 7 SGK)
a) y = 2x^ + 3x^+1
Gidi
Ham
so' xdc
dinh tren

R
Ta c6; y' = Gx^ + 6x
y' = 0 o 6x^ + 6x = 0 o 6x(x + 1) = 0 o
Bang bien
thien:
X
= 0
x = -l
X
—00
-1
0
+0C
y'
+
0 0
+
y
—'—* — , —
Vay ham so' dong bien
tren
moi khoang (-oo; -1) va (0; +oo), nghich bien
tren
khoang (-1; 0)
b) y,= x^ - 2x^ + X + 1
Ham
so' xdc
dinh tren
K
Ta c6: y' = 3x^ - 4x + 1

y' = 0 <=> 3x^ - 4x + 1 = 0 o X = 1
hoac
x = -
Bang bien
thien
1
3
+00
V§y hkm so d6ng bien
trgn
moi khodng
(-00;
~) vk (1; nghich bien
3
tren
khoang
3
3
c) y = X + —
X
H^m
so xdc
dinh tren
M \1
Ta CO y' = 1 -
y'^Oc=>l 5- = 0c^x^ = 3ox = ±73
Bang bien
thien:
X
—00

-73
0
• J3
+00
y'
+
0
II -
0
+
y
-— ' ^
~—-————"
—' ——>
Vay ham so'
dong
bien
tren
m6i khoang
(-00;
- 73 ) va (0; 73 ), nghich
bien
tren
moi khoang (- 73 ; 0) va (73 ;
+00)
2
d) y = X
X
Ham
so \±c

dinh tren
R \)
Ta CO y' = 1 + ^ > 0, Vx e M \1
x''
Bang bien
thien
X
—00
0
+0C
y
+
II
+
y
Vay hkm so
dong
bien
tren
m6i khoang
(-00;
0) (0;
+00)
e) y = x^ - 2x^ - 5
Ham
so xac
dinh tren
M
Ta CO y' = 4x^ - 4x
^x = 0

y' = 0 <=> 4x'' - 4x = 0 o 4x(x'' - 1) = 0 o
Bdng bien
thien
x = ±1
Vay ham so
dong
bien
tren
m6i khoang (-1; 0) va (1;
+00),
nghich bien
tren
moi khoang
(-00;
-1) va (0; 1).
Oy = 74-x'
Ham
so' xac
dinh tren
[-2; 2]
Tac.
y-^^
y' = 0 » X = 0
Bang bien
thign
X
-2
0
+2
y'

+
0
y
2. (Bdi 2 trang 7 GSK)
• x-2
a) y = —7;
• X + 2
Ham
so xac
dinh tren
Gidi
1-2)
•Ta c6: y =
x + 2 - (x - 2)
(x + 2)^
Bdng bien
thien:
(x + 2)
J
> 0, Vx e !R\1-21
X
—00
-2
+00
y'
+
+
y
Vay ham so d6ng bien
tren

m6i khoang
(-00;
-2) va (-2;
+00)
-x'-2x + 3
b) y =
X +1
Ham
so xac
dinh tren
M\{-11
(-2x - 2)(x + 1) - (-x' - 2x + 3) -2x' - 2x - 2x - 2 + x' + 2x - 3
y =
(x + lf
(X +
._ -x' - 2x - 5
(x +
• Bang bien
thien
<
0, Vx e R\{-11
X
—00 —
+x
y'
- -
y
•
—
Vay ham so nghich bien

tren
m§i khodng
(-00;
-1) va
(-!;•+«)
, (Bdi 3 trang 8 SGK)
Gidi
a) f(x) = x'^ -
Gx"^
+ 17x + 4
Ham
so xac
dinh tren
R
Ta CO f '(x) = 3x^ - 12x + 17
f
'(x) > 0, Vx e R
Vay ham so'
dong
bien
tren
R
b) f{x) = x^ + X -
cosx
- 4
Ham
so xac
dinh tren
R
• Ta c6 f '(x) = 3x^ + 1 + sinx > 0, Vx e R

Vay h^m so
dong
bien
tren
R
. (Bdi 4 trang 8 SGK)
Gidi
ftx)
= ax - x^
Ham
so xAc
dinh tren
R
Ta c6: f '(x) = a - Sx^
* N6u a < 0 thi f '(x) < 0, Vx e R. H^m so nghich
bi§'n
trgn
* Ng'u a = 0 thi f '(x) < 0, Vx e R. DAng thiJc chi xay ra khi x = 0
Vay ham so' nghich bien
tren
R.
* Neu a > 0 thi f '(x) = 0 cs- x = ±
j-
V3
Bang bien
thien
x
—00
fa
~ V3

if
+00
f
'(x) 0
+
0
f(x)
"
———>
'—"
- /—; I— . Vay a > 0 khong thoa dieu
ki§n
de toan.
H^m
so'
dong
bien
tren
V£iy
vdi a < 0, hkm so nghich bien
tren
R.
5. (Bdi 5 trang 8 SGK)
Gidi
fix)
= - x^ + ax^ + 4x + 3
3
H^m
s6' x^c
dinh tren

R
f
'(x) = x^ + 2ax + 4, c6 A' = a^ - 4
* Neu a^ - 4 < 0 hay -2 < a < 2 thi f '(x) > 0, Vx e R. H^m so
dong
bien
tren
R
• Neu a = 2 thi f '(x) = (x + 2f >
0,\fx*
-2. Ham so
dong
bien
tren
R.
• Neu a = -2 thi f '(x) = (x - 2f > 0, Vx 2. H^m so
dong
bien
tren
M.
• Neu a > 2
hoac
a < -2 thi f '(x) = 0 c6 hai nghiem xi, X2 (xi < X2)
Bang bien
thien
X -_co Xi X2 +00
f'(x)
f(x)
Ham
so nghich bien

tren
(xj;
X2)
khong thoa de toan
Vay -2 < X < 2 thi ham so da cho
dong
bien
tren
M.
M LlJYEl^ TAP
6. (Bdi 6 trang 8 GSK)
Gidi
a) y = -x^ - 2x^ + 4x - 5
3
H^m
so' xdc
dinh lien
tuc
tri§n
R
y' = x^ - 4x + 4 = (x - 2f > 0, Vx e K
H^m
so'
dong
bien
tren
R
b) y = -lx^ +
6x2-9x-
-

3 3
H^m
so xdc
dinh
va
lien
tuc
tren
R
y' = -4x^ + 12x - 9 = -(2x - 3)^ < 0, Vx € R
H^m
so nghich bien
tren
K
c) y =
x'' - 8x + 9
x-5
Ham
so xac
dinh tren
R \}
(2x - 8)(x - 5) - (x' - 8x + 9) _ 2x' - ISx + 40 - x' + 8x - 9
y =
(x-5)^ (X - 5)^
x' -lOx + 31
(x-5)^
Bang bien
thien
>0,\fx^5
x

—00
5
+00
y'
+
+
y
Ham
so
dong
bien
tren
moi khoang (-00; 5) va (5; +00)
d)y=
N/2X-X'
Ham
so' xac
dinh tren
[0; 2]
rr. . , 1-X
Ta co: y = ,
V2x - x'
y' = 0 <» X = 1
Bang bien
thien
X
0
1
2
y'

+
0
y
Ham
so
dong
bien
tren
khoang (0; 1), nghich bien
tren
khoang (1; 2)
')y = Vx' -2x + 3
Ham
so xac
dinh tren
R (do x^ - 2x + 3 > 0)
, _ 2(x -1) _ x-1
~ 2N/X' - 2X + 3 Vx' - 2x + 3
y' = 0 O X = 1
Bang bien
thien
X
—00
1
+00
y'
0
+
V-—
y

• ->
Ham
so
dong
bien
tren
khoang (1; +00), nghich bien
tren
khoang (-00; 1)
f)y =
- 2x
x + 1
Ham
so' xac
dinh tren
R \)
-1
- 2x' - 4x - 2 -2x' - 4x - 3
BAng
bien
thien
(x + 1)'
(x + D'
<
0, Vx G M \I
-1
+00
y >•
Ham
so nghich bien

tren
m5i khoang (-00; -1) (-1; +00)
7.
iBdi
7
trang
8 SGK)
f(x)
= cos2x - 2x + 3
Ham
so' xac
dinh
tren
R
Gidi
f
'(x) =
-2sin2x
- 2 =
-2(sin2x
+ 1) < 0, Vx e K
f
'(x) = 0
<=> sin2x
= -1 o 2x = - ^ +
2k7t <:=>x
= -^
+k7r,
ke
Ham

so
nghich bien
tren
moi
doan
Vay
ham so'
nghich bien
tren
R. .
+
k7t;
- — + (k + 1)71
4 4
,
k e Z
8.
(Bai 8
trang
8 SGK)
Gidi
a)
X6t ham so' f(x) = x -
sinx
0;^
2
Hkm
so
lien
tuc

tren
Do
do ham so
dong
bi§'n
tren
Ta
CO fix) >
f(0).
Vx e
:
f '(x) = 1 - cosx > 0, Vx e , 0; -
V
2.
2]
(
Nghia
la X -
sinx
> 0, Vx e 0; -
V
2
j
hay
X >
sinx,
Vx e
0;-,
2)
Mat

khdc
x >
sinx,
Vx > — vi
sinx
< 1
2
Vay
x >
sinx,
Vx > 0
Chufng minh tuong
tyf
sinx
> x, Vx < 0
x^
b)
Xet hkm so g(x) = cosx + — - 1
H^m
so
lien
tuc
tren
nufa khoang
[0;
+ao),
g'(x) =
-sinx
+ x
Theo

cdu a, g'(x) > 0, Vx > 0, do d6 h^m so g
dong big'n
tren
[0; +«),
Va
ta CO g(x) > g(0), Vx > 0
x'
Nghia
la cosx + 1>0,
Vx>0
2
TCr
d6 suy ra
vdi mpi
x < 0, ta c6:
(-x)^
cos{-x) + 1 > 0 hay cosx + 1>0,
Vx<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 ham so h
nghich bien
tren
K h(x) < h(0), Vx > 0
x^
x^
^
X
sinx
< 0 => X <
sinx,
Vx > 0
6
6
x^
va
h(x) > h(0), Vx < 0 hay x >
sinx,
Vx < 0
6
J,
(Bai 9
trang
9 SGK)
Gidi
Xet
ham so' f(x) =
sinx
+

tanx
- 2x
lien
tuc
tren
niira
khoang
1
0;
Co f '(x) = cosx +
Do
cos^x +
cos^x
2 > cos X +
cos^x
-
2, Vx 6
>
2, Vx e
cos'x
Do
do ham so fix)
dong bien
tr§n
Ta
c6: f(x) > f(0;, Vx e
Vay sinx
+
tanx
- 2x > 0

hay sinx
+
tanx
> 2x, Vx e
10.
(Bai 10
trang
9 SGK)
Gidi
t
+ 5
a)
Nam 1980, ta c6 t = 10
^^^^^^
26.10.10
270
=
18
(nghin)
10
+ 5 15
S6 dan cua
thi
tran
nam 1980 la 18
nghin
ngif^i
Nam
1995, ta CO t = 25
26.25

+10 660
f(25)
=
=
22
25
+ 5 30
So dan cua
thi
tran
nam 1995 la 22
nghin
ngKbi
b)
Ta c6: f '(x) =
26(t
+ 5) -
(26t
+ 10)
120
(t
+
5)^
(t + 5f
Vay
ham so'
dong bien
tren
[0; +oo)
>

0, Vt e (0; +00)
c)
Toe dp
tang dan
so vac
nam 1990 1^
120
(20 +
5)'
Toe
do
tSng dan
so'
v^o n&m
2008
1^
120
(38 + 5)'
f'(20)
= -7—=^ ~ 0,192
f'(38)=
*
0,065
190
1
00
•
Ta CO —=
0,125
o (t +

5)^
= « t + 5 »
31
t = 26
(t
+ 5)'
0,125
Vao nam 1996
toc dp
tang dan
so
ciia thi
tra'n
la
0,125.
§2.
CaC TR!
CUA HAM
SO
1.
Dinh nghia:
Gia
suT ham
so' f
xac
dinh tren
tap
c
M
va

XQ
e y
*
XQ
dixac gpi
la
mot diem
ciJ^c
dai
ciia ham
so' f
neu:
Ton tai mot khoang
(a; b)
chiJa
XQ
sao cho (a, b) c 7
va f(x)
<
f(xo) vdi mpi
x e (a, b) \.
Khi
do f(xo)
dupe
gpi
la gia tri
cufc
dai cua
ham
so f

*
Xo
difpc gpi la mot diem cxic tieu
cua
ham
so' f
neu:
Ton tai mot khoang
(a; b)
chufa
XQ
sao cho (a; b) e D
va f(x) > f(xo) vdi mpi x e (a; b) \.
Khi
do
f(xo)
dupe
gpi
la gia tri
cij^c tieu ciia ham
so' f
* Diem
cifc
dai
va
diem cUc
tieu
gpi chung
la
diem

cu'c tri
Gia tri cUe dai
va gia
tri
cUc
tieu
gpi chung
la cu'c tri
* Neu
Xo la
diem
CLCC
tri
cua ham so f
thi (XQ; f(xo))
la
diem
cu'c
tri ciia
do
thi
ham
so f.
2.
Dieu ki^n
can de ham so dat
ci/c
tri
Gia
sijf

ham so f
dat
CLTC
tri
tai diem XQ. Khi d6, neu
f
c6
dao ham
tai
XQ
thi
f
'(xo)
= 0
3. Dieu ki^n
dii de ham so dat
ci^c
tri
Gik sijf
ham
so f
lien
tuc
tren
khoang
(a; b)
chiJa diem xo va
c6 dao ham
tren
cae

khoang
(a;
XQ)
(XQ;
b).
Khi
do
a) Neu
f
'(x)
< 0
vdi mpi
x e (a; XQ) va f
'(x)
> 0
vdi mpi
x e
(xo;
b)
thi
ham
so'
f
dat cvtc
tieu
tai diem
XQ.
b) Neu
f
'(x)

> 0
vdi mpi
x e (a; XQ) va f
'(x)
< 0
vdi mpi
x e
(XQ;
b)
thi
ham
so dat
cue
dai tai diem
XQ.
4. Qui
tSc
tim
c\ic tri
* Qui
t^c 1:
• Tim
f
'(x)
• Tim
cac
diem
Xi
(i = 1, 2, )
tai

do dao
ham
cua
ham
so'
hhng
0
hoac
ham
so'
lien
tuc
nhung khong
c6 dao
ham.
•
Xet
da'u
f
'(x) neu
f
'(x) ddi da'u khi
x
qua diem
Xj
thi ham
so'
dat
cUc
tri

tai
Xj.
•f'(X)
Xi
f'(x)
f(x)
X,
f(X)
I f(Xi)
cUc
tieu -
* Qui
tac 2
• Tim
f
'(x)
• Tim
cac
nghi^m
Xi
(i = 1, 2, ) cua
phupng
trinh
f
'(x)
= 0
• Tim
f
"(x) va
tinh

f
"(Xj)
+
Neu
f
"(xi)
< 0
thi ham s6' dat cUe dai tai diem Xj
+
Neu
f
"(Xi)
> 0
thi ham s6' dat cUc
tieu
tai diem Xj
JBAI
TAP
n.(Bai
11
trang
16 - 17 SGK)
a)f(x)= ix^
+
2x2
+ 3x- 1
Gidi
Ham
s6'
xac

dinh tren
K
f
(X)
=
x^
+ 4x + 3;
f
(X)
= 0 o
x^
+ 4x + 3 = 0 o
X
= -1
hoac
x =
Bdng bien
thien
X
1-00 -3 -1
-3
+0C
Vay
ham
so da cho
dat cUc dai tai diem
x = -3,
gia
tri
eUe dai la f(-3)

:
Hkm
so
dat cUc
tieu
tai
dilm
x =
-1,
gia
tri
cUc
tieu
cua
ham
so
la fl-1)
=
Cdch
2: Ham so xac
dinh tren
M
Ta
c6: • f
'(x)
=
x^
+ 4x + 3;
f
'(x)

=
0c:.x^
+ 4x + 3 = 0ox =
-l
hofic
x = -3;
f
"(x)
= 2x + 4
Vi
f
"(-1)
=
2(-l)
+ 4 = 2 > 0 nen
ham
so dat cUc
tieu
tai diem
x
7
f(-l)
=
3
f
"(-3)
= 2(-3) + 4 = -2 < 0 nen
ham
so dat cUe dai
tai diem

x
f(-3)
= -1
b) f(x)
= i
x^
-
x^
+ 2x - 10
3
Ham
so' xac
dinh tren
R
Ta
CO f
'(x)
=
x^
- 2x + 2 > 0, Vx e K
. Ham
so
dong
bien
tren
R, khong
c6 cUc
tri.
-1
7

3

=
-1;
=
-3,
c) f(x)
= X + —
X
Ham
so xac
dinh tren
f
'(x)
= 1 - =
R
\1
x^-1
2
•
f
'(X)
=
0«x'-l
= Oc:>X = ±l
Bang bien
thign
X
—oc
-1

(
)
1
+ 00
f'fx)
+
0
— —
0
+
f(x)
^ -2\
^2 ^
Vay ham so' dat
ciJtc
dai tai diem x =
-1,
gid tri
ciTc
dai la f(-l) = -2
Ham
so' dat
ciTc
tieu
tai diem x = 1, gia tri ci/c
tieu
la f(l) = 2
d) f(x) = IX
I
(x + 2)

Ham
so' xac
dinh
va
lien
tuc
tren
R
f-x(x + 2) v(Ji X < 0
|x(x + 2) vdi X > 0 "
• V6i X < 0, f '(x) = -2x - 2; f '(x) = 0 x = -1
• Vdi X > 0, f'(x) = 2x + 2 > 0
Bang bien
thien
Ta c6 f(x) =
X
—00
-1
0
f'(x)
+
0
_
fix)
Vay ham so' dat
cifc
dai tai x = -1, gia tri cUc dai fi-1) = 1 va dat ciic
tieu
tai
x = 0, gia tri

cifc
tieu
fiO) = 0
x^ x"
e) fix) = —+ 2
5 3
Ham
so' xac
dinh tren
M
f
'(x) = x^ - x^ = x^ (x^ - 1)
f
'(x) = 0 <^ x^ (x^ - 1) = 0 X = 0
hoac
x = ±1
Bang bien
thien
X
—00
-1
0
1
•
+C0
f
(x)
+
0
0

0
+
fix) 32
15
15
Vay ham so dat
ciTc
dai tai x = -1, gia tri cvtc dai fi-1)
28
32
15
va ham so
dat
cUc
tieu
tai x = 1, gia tri
ciTc tieu:
fil) =
15
f)
fix) =
x^ - 3x + 3
x-1
Ham
so' xac
dinh tren
M \|
(2x - 3)(x - 1) - (x' - 3x + 3) x' - 2x
f
'(x) =

(x -1)^
(x-1)^
f
'(x) =
0c:>x-2x
= 0<=>x = 0
hoac
x = 2
Bang bien
thien
X
—00
0 1 2
+ 00
f
(x)
+
0
0
+
fix)
; V$y ham so dat cUc dai tai diem x = 0, gia tri
ciTc
dai fiO) = -3 va dat cUc
tieu
tai diem x = 2, gid tri
cifc tieu
fi2) = 1
I.
(Bcii

12 trang 17 SGK)
Gidi
a) y = x V4-x^
Ham
so xac
dinh
va
lien
tuc
tren
[-2; 2]
y' = ^ ~ v6i moi X e (-2; 2); y' = 0 « x = ± V2
V4-x'
Bang bien
thien
x
-2 -72 72 2
y'
0
+
0
y
0
—*. -2-—-
^2—
~ ^0
Ham
so dat
cifc
tieu

tai x = - 72 , gia tri
c\ic tieu
la -2
H^m
so dat
c\ic
dai tai x = 72 , gia
tri
cUc dai 1^ 2
b)y = 78-x' - • •
Ham
so xac
dinh
va
lien
tuc
tren
[-2 72; 2 72 ]
y' = , ~^ vdi moi x € (-2 72; 2 72 ); y' = 0 o x = 0
78^
Bang bien
thien
x
-2sl2
0 272
y'
+
0
y
" 272

• •
Ham
so dat
cifc
dai tai x = 0, gia tri
cxic
dai: 2 72
c) Ap dung qui tAc 2
y = x - sin2x + 2
Ham
so xac
dinh tren
R
Ta c6: y' = 1 -
2cos2x
y' = 0 <=>
2cos2x
= 1 o
cos2x
= i<=>2x = ±|^ + k27x, k e Z
2 o
ox = ±—
+k7t,
keZ
6
Ngoai ra: y" = 4sin2x
* Vi y"
— + kn
=
4sin

=
4sin
I 3J
=
- 4sin
=
_4^
=-2V3
<0
Do d6 ham so dat
ciTc
dai tai c6c diem x = -—
+k7t,
keZ
• • 6
Gia tri
c\Xc
dai y(-
—
+ k7r) = - - + k7i - sin
• • 6 6
3
+
2
=
— + kn + — + 2
6 2
=
4sin^
=4.^

=273
>0
3 2
Do do h^m so dat
cUc
tieu tai cAc diem x = - + ku, k e Z
6
* Vi y"
— + k7t
=
4sin
- + k27I
k6
U
J
Gid
tri ciTc tieu y
—
+ kn
6
- + k27I
=
— + k7i - sin
6
7t ,* 73 -
=
-+k7t
+2
6 2
+

2
d) Ap dung qui tic 2
y = 3 -
2cosx
-
cos2x
Ham so xac dinh tren R
Ta c6: y' = 2sinx + 2sin2x = 2sinx +
2.2sinxcosx
= 2sinx(l +
2cosx)
y' = 0 <=>
sinx = 0
-1
+
2cosx
= 0
sinx
= 0
1
o
cosx =
2
X = kTI
X = ±— + 2k7i
3
Ngoai ra y" = 2cosx + 4cos2x
• Vi y"(k7i) = 2cosk7r + 4cos2k7i = 2cosk7i + 4 > 0, Vk s Z.
Do do ham so da cho dat cUc tieu tai cac diem x = kn va gid tri cUc tieu
y(k7t) = 3 - 2cosk7x - cos2k7r = 2 - 2cosk7r = 2 - 2(-l)'' (k e Z)

• Vi y"
±^ + k27:
3
=
2cos
±^ + k27r
3
+
4cos
3
0 271
=
2cos
—
. 4rr
+
4cos
— =
3 3
=
-2cos
—
-
4cos
—
= -
3 3
n
^
n

+
4cos
( ^)
7t + —
I
3;
I
3J
IS- =-3 < 0
3
Do do h^m so dat ciic dai tai cac diem x = ± — + 2k7T, k e Z va gia tri
3
cUc
dai la: y
^2Vk2;
= 3 - 2cos
f±2Vk27T^
- cos
^^Vk2.l
V
3 ,
I
3 )
V
3 j
=
3 -
2cos^
- cos^ = 3 +
2cos^

+ cos- = 3 +
3cos^
= ^
3 3 3 3 3 2
13. (BM 13 trang 17 SGK)
f(x) - ax^ + bx^ + cx + d
Giai
flam
so xac dinh tren K
f
'(x) = 3ax^ + 2bx + c
• Ta CO f(0) = 0 d = 0.
Ham so dat cifc tieu tai diem x = 0, nen:
f
(0) = 0
* Ta CO f(l) = 1 => a + b + c + d = 1 r.r> a + b = 1
Ham so dat cUc dai tai diem x = 1 nSn:
f
(1) = 0 =;
=> c = 0
3a + 2b + c = 0
3a + 2b = 0 .
Gidi
he phiTOng
trinh
3a + 2b = 0
a + b = l
a? + 2b = 0
-2a - 2b = -2
3a + 2b = 0

a = -2
b
= 3
|a = -2
Kiem
tra lai ket qua ^
Ta CO ham so f(x) = -2x-' + 3x^
t
f '(x) = -6x^ + 6x, f '(x) = 0 c=. X = 0
hoac
x = 1
f
"(x) = -12x + 6
• f "(0) = 6 > 0. Vay hkm so dat ciTc tieu tai diem x = 0 va f(0) = 0
• f "(1) = - -12 + 6 = -6 < 0. Vay ham so dat
cifc
dai tai diem x = 1 va f(l) = 1
14. (Bcii 14 trang 17 SGK) Giai
f(x) = x"* + ax^ + bx + c
Ham so xdc dinh tren K.
.f'(x) = 3x^ + 2ax + b
Ta CO f(-2) = 0 o -8 + 4a - 2b + c = 0 (1)
Hkm
so dat cufe tri tai diem x = -2, nen
f
'(-2) ^- 0 o 12 - 4a + b = 0 (2)
D6 thi ham so di qua diem A(l; 0) nen
l
+ a + b + c = 0 (3)
4a - 2b + c - 8 = 0

-4a + b + 12 = 0
a+b+c+l=0
Giai
he phUcfng
trinh:
a+b+c+l=0
o ]-4a + b + 12 = 0
3a - 3b - 9 - 0
a+b+c+l=0
o -12 + b + 12 = 0
a = 3
Kiem
tra lai ket qua
f(x) = X-' + 3x^ - 4
f
'(x) = 3x? + 6x, f '(x) = 0 o X =
a+b+c+l=0
-4a + b + 12 = 0
-9a+ 27 = 0
3 + c + 1 = 0
b
= 0
a = 3
c = -4
b
= 0
a = 3
r\\
/ n tt ^ ^ /
Ham so dat ciTc tri tai

dig'm
x = -2 gi^ tri ciTc tri 1^
f(-2) = -8 + 12 - 4 = 0
Do f(l) = 1 + 3.1 - 4 = 0
Vay
diem
A(l; 0)
thuoc
do thi ham so f.
15. (Bai 15 trang 17 SGK) Giai
x^ - m(m + l)x + m^ + 1
y =
x - m
Ham so' xac dinh tren R \)
1
Ta CO
y = X - m +
y' = 1-
(x - m)'
(chia
da
thufc)
-1
— , X m
X
- m
1
(x-m)'-l
(x-m)^
hay

(x - m + l)(x - m - 1)
y = ; 75 , X m
(x - m)'
y' = Oox = m- l
hoSc
x = m + 1
Bang
bien
thien
X
—oc
m - 1
m m + 1 +00
y'
+
0
-
0 +
y
^ -m^ + m - 2 \
\
^-m' + m +.2^
Vay vdi moi gia tri ciia m, ham so da cho dat ciTc dai tai
diem
x m - 1 v;
dat ciTc tieu tai
diem
x = m + 1.
§3. GIA TR! L6N
NHAT,

GIA TRj NHO
NHAT
CUA HAM SO
I\6l
DUI\
CAl^
r^iiof
1. Dinh nghia: Gia stf ham so f xac dinh tren tap 7 ('J cz R)
* Neu ton tai
diem
XQ
€ V sao cho: f(x) < f(xo) vdi moi x,, e V
Thi
so M =
f(Xo)
goi la gia tri
IdTn
nhat cua ham so f tren V.
Ki
hieu M =
maxf(x)
X. 7
* Neu ton tai
diem
x,, s 7 sao cho: f(x) >
fixo)
vdi moi x e 7
Thi
so m - f(xo) goi 1^ gia tri nho nhat cua h^m so f tren 7.
Ki

hieu m = minf(x)
•r'^^^itac
tim gia tri Idn nhat, gia tri nho nhat
* Tim cac
diem
x,, x-,, x^
thuoc
(a; b) tai do h^m so f c6 dao ham bfing
0
hoac
khong
c6 dao ham.
, Tinh f(xi), f(x2), fix
J,
f(a), f(b)
* So s^nh cdc gia tri tim
di^cfc
_ So Idn nhat trong cac gia tri do la gia tri Idn nhat ciia f tren [a; b]
_ So nho nhat trong cac gia tri do la gia tri nho nhat ciia f tren [a, b]
J
BAITAP
16. (Bdi 16 trang 22 SGK)
Giai
f(x) =
sin''x
+
cos''x
Ham so xac dinh tren K
* f(x) =
(sin^x

+
cos^x)^
-
2sin^xcos^x
= 1
1
(2sinxcosx)
= 1 -
isin^2x
< 1, Vx e
2
Ta
CO
f(0) = 1. Vay
maxf(x)
= 1
* fix) = 1 - -
sin^2x
> -, Vx e R
2 2
Ta CO f
= 1 - 1 = i. Vay minf (x) = ]-
2 2 ^^ ^ 2
17. (Bai 17 trang 22 SGK)
a) f(x) = x'^ + 2x - 5
Giai
Ham so' xac dinh tren [-2; 3].
f
(x) = 2x + 2;
f

(x) =
0o2x
+ 2 =
0<=>x
= -l
J^ng
bien
thien
[
X
—
X
-2
-1
3
+
x
• f'(x)
0
+
-5
-6 ^
^10
' f(-l) = (-1)' + 2(-l) - 5 = -6
' f(-2) = (-2)- + 2(-2) - 5 = -5
' f(3) = 3^ + 2.3 - 5 = 10
/'ay min f(x) = -6; max f(x) = 10
X6|
2: 3]
xt[~2;

31
3) f(x) = — + 2x^ + 3x - 4
3
lam so' xac dinh tren [-4; 0]
f'(x) = x^ + 4x + 3
f
(x) = 0ox- + 4x + 3 =
0ci>
x = -l
hoac
x = -3
Bang bien thien
X
—
00
-4
-3
-1
0 +or
f'(x)
+
0
0 •
+
fTx)
16 ^^^^
3
X
16
3

^
-4 '
^^^^
f(-4)
=
3
+
2(-4)' + 3.(-4)
3
«-3)
=
(-3)^
3
+
2.{-Zf
+ 3.(-3)
-
4 = -4
•
«0)
=
-4
fi:-i)
=
3
+
2{-\f + 3(-l) -
4 1^
3
1

Vay
minf(x)=-
—; max f(x) =-4
-'•01 3 xel-4;0)
C)
f(x)
:
X + —
X
n&m
so xac dinh tren (0;
+oc)
2 , Vx e (0; f(x) = 0
<=>
i'(x>
= l l =
''^-l
x-^
X'
X
= ±1
Bang bien thien
X
—00
-1
0 1
+x
f'(x)
+
0

-
0
+
f(x)
f(l)
= 1 + i = 2
Vay
min f(x) =2
Ham so khong dat gia tri Idn
nhat
tren (0; +«>)
d)
f(x) = -x' + 2x + 4
Ham so xac dinh tren [2; 4]
f
(X)
= -2x + 2;
f
(x)
= 0<=>-2x + 2 =
0ox=l
Bang bien thien
X
-X
1 2
4
f'(x)
+
0 -
f(x)

4
^^
f(2)
=
-2' + 2.2 + 4 = 4; m
=
-4' + 2.4 + 4 = -4
.
-4
Vay
minf(x)=-4;
max f(x) = 4
"£[2;
4] xe|2; 4)
e) fU) =
2x"
( 5x I 4
X
F 2
Ham so' xac
dinii
treu [0; 1|
,
(4x
+
5)(x + 2) - (2x'+ 5x
+
4)
fix)
= ——

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

+
+
+
f(xl
2
11
'^3
t-5.1 1
+
4 "
11
3
Vay
min f(xj = 2; max f(x) = —
x-io,
II
x-ao:
11
3
0 fix) = X - -
Ham so xac dinh tren (0; 2]
f'.(x)
= 1 + = > 0, Vx 6 (0; 2L
Bang bien thien
X
-00
0
2
+00
f'(x)

fix)
^^^^^^^^^^^^^^^^
f(2)
. 2 - i = -
2 2
Vay
max f(x) = —
xul0;2l
2
Ham so khong dat gia tri 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 tri Idn
nhS't
wk gid tri nh6
nhS't
cua ham so y = fit) tren
doan [-1; 11,.do cung la gid tri nhd
nhat
cua ham so da cho trSu
f(t)

= 4t + 2; f(t) = 0o4t + 2 =
0<T>t
=
Bang
bien thien
t
—
QC
-1
1
+
x
f'(t)
-
0 +
f(t)
-1
^ 3 ->3
2
f(-l)
= 2(-lf + 2(-l) - 1 = -1
f(_l).2(-i)^
+
2(-i)-l
= -|
2 2 2 2
f(l)
=
2(1)'^
+ 2.1 - 1 = 3

Vay min
f
(t)
= - - ; max
f
(t)
= 3
. t€(-l; II 2 II
3
Do d6 miny = —, maxy = 3
"•^^ 2 X€i
b) y =
cos^2x
-
sinxcosx
+ 4
Ham so xac dinh tren M
Ta
CO
y = 1 - sin^2x - - sin2x + 4 = -sin^2x - — sin2x + 5
2 2
Dat t = sin2x, -1 < t < 1
Ta CO y = f(t) =
-t^
- -1 + 5
2
1
1 li
f'(t)
= -2t - - ; f'(t) =

0c:>-2t-
- = Oc:.t =
2 2 4
Bang bien thien '*
t
-1
1
4
+1
f'(t)
+
0
f(t)
81
9
2
* 16 "
^ 7
2
f(-i)
= -(-i)2_ i(_i) + 5= y
I
"4
r_i]
. 4J
I 4,
Vay min/(t) . , max,f(t) =
• 7 81
Do do miny = -, maxy = —
xe» ^ 2 16

2
+
5
81
16
81
16
Gidt
9.
(Bdi 19 trang 22 SGK)
pat BM = x (0 < X < -)
2
Ta c6: AQMB = APNC
Nen BM = CN = X
Do do MN = a - 2x
QM
= BMtan60" = x x/3
Dien tich hinh chQ" nhat
MNPQ
la:
S(x) = MN.QM = (a - 2x).x N/3 = -2 x/3 x% a 73 x
Bai
toan qui ve tim gid tri Idn nhat cua S(x) tren
Ta c6: S'(x) =-4 ^/3 x + a
V3
S'(x) =
0«-4V3x
+ a^/3
=Oc>x=
^

4
Bang bien thien
X
0
a
4
a
2
k
S'(x)
+
0
i
S(x)
^a- ^
4j
+
a
Vs.
^a^
u
V3
=
-2V3 .
Vay max S(x) = —a^
xefo;^;
8
Do do S(x) dat gi^ tri 16n nhat khi M d vi tri tren BC sao cho BM = ^ BC v^
gi^
tri Idn nhat ciia di^n tich hinh chuf nhat la —

8
20. (Bai 20 trang 22 SGK) Gidi
Tren moi don vi dien tich ciia mat ho c6 n con ca thi sau mot vu, so' ca tren
moi don vi dien tich mat ho trung 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) = -40x + 480
f'(x) = 0 » -40x + 480 = 0 « X = 12
Bang bien thien
X
0 12
+»
_ f'(X)
+
0
b ^^^^
?ssn
^
f(12) =
480.12
-
20.12^
=
2880

Vay max f(x) =
2880
V$y
tren
(0; +oo) ham so f dat gid tri Idn
nhat
tai diem x = 12
Suy ra
tren
tap N' h^m so f dat giA tri
lorn nhat
tai diem n = 12
Vay.muS'n thu
hoach
duoc
nhieu
nhat
sau mpt vu thi
tren
moi don vi
diei,
tich tren
mat ho
phai
th^ 12 con cA.
^
LlIl-EI^
TAP
21. (Bai 21
trang

33 SGK)
Gidi
a) f(x)
x^ +1
H^m
so xac
dinh tren
K
, x^ + 1 - x.2x -x^ + 1
fx)
= =
fix)
= 0 o -x^ + 1 = 0 o X = ±1
Bang bien
thien
,
X
-x^
-1
1
f X
f'(x)
0
+
0
f(x)
* ~2
1
^ 2
Vay h^m so dat

ciJc
tieu
tai x = -1, gia tri
cifc
tilu
la f(-l) = - i, dat
cifc
2
dai
tai x = 1, gia tri
cifc
dai 1^ f(l) = -
2
3
b) f(x) =
x + 1
Ham
so x^c
dinh tren
K \1
,2^„ , ,^ ,.3 „„3
(x + 1)' (x + 1)' (x + 1)'
f'(x) =
Oc:>x
=
Ohoacx
=
Bang bien
thien
X

-00
3
2
I
0
,
f
(x)
0
+
+
0
+
fix)
27 ^
4
H^m
so dat cUc
tieu
tai diem x =
tilu
la
f
f I = ^ ,
-—, gia tri cUc
tieu
la f
I
2
, gid tri

ciic
c) i
c) f(x) = V5-x'
Ham
so xdc
dinh tren
[-
f'(x) =
-X
Vs^-x
f'(x) = 0 <=> X = 0
Bang bien
thien
; Vx €= (- N/5 ; N/5 )
X
-V5
0
+
v's
f'(x)
f(x)
+
*.f5
>•
H^m
so dat
cifc
dai tai diem x = 0, gid tri
ciic
dai la f(0) = Vs .

d) f(x) = X + x/x' -1
Ham
so xac
dinh tren
(-oo; -1] u [1; +oo)
X
f
(x) = 1 +
Bang bien
thien
7x^-1
vdi
X < -1
hoac
x > 1
X
-oo -11 +x
f'(x)
-
1"
II
f(x)
- —-—
f(x)
•—•
- —-—
Ham
so' nghich bien
tren
(-oo; -1),

dong
bien
tren
(1; +»)
Ham
so khong c6
cifc
tri
22. (Bai 22
trang
23 SGK)
Gidi
x^ + mx - 1
VKX)
= ;
X
-1
H&m
so xac
dinh tren
R \}
(2x + m)(x-l)-(x^+mx-l) _ x^ - 2x + 1 - m
O^TiT (x-1)^
x' - 2x + 1 - m = 0 (1)
x^l
(1)
CO
nghiem x ^ 1 khi va chi khi 1^ - 2.1 + 1 - m 0 o m ^ 0
H^m
so f

CO
cure
dai va cUc
tilu
khi va chi khi phifang
trinh
(1) c6 hai
nghi§m
phan biet kh^c 1.
f
(x) =
f
(x) = 0 o
\1
Nghia
1^
A'
= 1 - (1 - m) = m > 0
<=> m > 0
m
;^ 0
23. {Bdi 23
trang
23 SGK)
Gidi
X6t
hkm so G(x) =
0,75x^
-
0,025x\ > 0

Ta
CO
G'(x) =
0,075x'
+ l,5x
G'(x) = 0 »
-0,075x2
^ ^ ^ Q ^
x{-0,075x
+ 1,5) = 0
X
= C
hoSc
X = 20
Bang bien
thidn
X
— oo
0
20
G'(x)
-
+
0
G(x)
^^^^^^^ 100
Ta
CO
maxG(x) = 100, x e (0, +oo) -
Vay

lieu
lucfng thuoc can
tren
cho benh nhan de huyet ap giam nhieu nhat
la
20 mg. Khi do do giam huyet ap la 100.
24.
(Bai 24 trang 23 SGK)
Gidi
Goi M(x; x^) la diem bat ky cua parabol (.f)
Khoang
each
AM: AM = V(x + 3)^ + (x^ - 0)^
hay AM^ = (x + 3)^ + x^ = x^ + x^ + 6x + 9
Khoang
each
AM dat gia tri nho nhat khi va chi khi f(x) = AM^ dat gia tri
nho nhat
Ta c6: f(x) = x" + x" + 6x + 9
f'(x) = 4x'^ + 2x + 6 = 4x^' + Ay? - 4x=^ + 6x - 4x r '
=
4x2(x
+ 1) - 4x(x + 1) + 6(x + 1) = (x + l)(4r
f
(x)
= 0 o X + 1 = 0 (do 4x^ - 4x + 6 > 0) « X = -1
4x + 6)
X
—00
-1

+00
f'(x)
0
+
fix)
». 5
>
f(x) dat gia tri nho nhat tai diem x = -1, gia tri nho nhat la f(-l) = 5
Do do khoang
each
AM dat gia tri nho nhat khi M d vi tri diem Mo(-l; 1) va
khoang
each
nho nhat AM,, = Vs .
25,
(Bai 25 trang 23 SGK)
Gidi
Van
toe eua ca khi bcfi ngUorc dong la (v - 6)
knJgib
Thori
gian de ca
viiot
khoang
each
300 km la
300
V 6
gid
Nang luong

tieu
hao cua ca khi
wxidi
300 km la
E(v)= cvl
300
v-6
=
300 e.
v-6
(jun),
V > 6
Taeo
E'(v) =
300c.
3v^(v-6)-v^
(v-6)^
,3 1Q.,2
=
300 e.
3v2 - 18v^ - v^
{v-6)^
=
300 c. 2X1^= 600 cv^ ^^-^^
(v-6)^
(v-6)^
E'(v) = 0 <o V = 0
hoac
v = 9 (v = 0 loai do v > 6)
Bang bien

thien
"
V
6
9
+ C0
E'(v)
0
+
E(v)
— ».
72900
c
Vay de it
tieu
hao nSng lugng nhat thi v$n toe eiia ca (khi nufdc
diing
yen) la
9 km/gi6)
g. (Bai 26 trang 23 SGK)
Gidi
S6' ngu'di nhiem benh ke tii ngay xuat hien benh nhan dau
tien
den ngay
thu"
t la:
f(t)
= 45t^ - t^ t nguyen va t e [0; 25)
Xet ham so f(t) = 45t^ -
t'^

v6i t e [0,25]
CO
f
(t)
= 90t - 3t2 - 3t (30 - t)
a) Toe do truyen benh vao ngay
thuf
5 la:
f'(5) =
3.5(30
- 5) = 375 ngifdi/ngay
b) Xet ham so' xae
dinh
to'e do truyen benh
y ii: fit) rz 90t - 3t^
y' =
f
"(t)
= -6t + 90
y' = 0 « t = 15
Bang bien
thien
t
0
15
25
y'
+
0
y

—
>675
•
To'c dp truyen benh 16n nhat la vao ngay
thiJ
15
Toe dp do la
f'(15)
= 675 ngifdi/ngay
c) Toe dp truyen be^ Idn hon 600
y > 600 c:^ sot - 3t^ > 600 e>
t^
- 30t + 200 < 0 ci- (t - 15)^ < 5^
<=> -5 < t - 15 < 5 <=> 10 < t < 20
Vay tii ngay thuT 11 den ngay
thil
19, toe dp truyen benh Idn hon
600 ngUdi/ngay.
d) Ta CO
f
(t)
= 0 <:i> t = 0
hoac
t = 30
t
0
25
30
f'(t)
0

+
fit)
f'(t)
> 0, Vt e (0; 25)
Vay ham so f dong bien
tren
[0; 25]
27.
(Bdi 27 trang 29 SGK)
Gidi
a) a) fix) = V3-2x
Ham
so' xac
dinh
tren
[-3; IJ
-1
f'(x) = =
<
0. Vdi moi x e
V3-2x
Ham
so' f nghich bien
tren
doan [-3; 1;
3
-3
3
2
f(-3)

= J3 - 2( -3) = 3
=
x/3 -2.1 = 1
Vay
max f(x) = 3; min f(x) = 1
b) f(x) = X + V4 - x'
H^Tii
s6' xAc
dinh
li§n tuc
tr6n
[-2; 2]
f'(x)=
1- t
xe(-2;
2j
V4
-x'
f'(x)
= 0 o 1 -
7i~
=
0 C:> 74 - X^ = X <^
fo
4
Ta CO
f( ^f2 ) = V2 + vT^
^2sf2
f(-2)
= -2

+
Vr"4 = -2
ft2)
=
2 + vT^' = 2
Vay
max f(x) =2^,; min f(x) = -2
>'sl-2, 2]
xtl-2;2)
c) fix) =
sin''x
+ cos^x + 2
Ham
so xac
dinh tren
E
f(x)
=
sin^x
+ 1 -
sin^x
+ 2 =
sin''x
-
sin"x
+ 3
Dat
t =
sin''x,
0 < t < 1

Ta
CO g(t) = t' - t + 3
g(0)
= 3; g ^
Vay
max g(t) =3; min g(t) = —
Do
do max f(x) = 3; min f(x) =
11
d)
f(x) = X -
sin2x
H^m
so x^c
dinh lign
tuc
trgn
doan
f
(x)
= 1 - 2cos2x
71
f
(x)
= 0 « 1 - 2cos2x = 0 <=> cos2x = - = cos -
2 3
o 2x = ±~ +
k27r
<=> X = ±- +
kjt,

k e Z
o 6
71
Tt ^
y^i_
< X <
7i;
f
(x)
- 0 khi v4 chi khi x = - - hoSc x = g
571
hoac X =
Ta
c6 f
71
71
~
-
sin
—
6y
6 V
3;
6
71
3
71
6
71^
^571^

7t
. 71 7t
=
sm— = —
6
3 6
571
-
sm
71
73
2
571
.
71 STI
+
sm— =
3
73
2
.6
71
71 . 7t
=
sin(-7:)
= +
sm7c
=
2 2 2
i{n) = n -

sin27r
= n
57t 73
Vay
max f(x) = -~- + — va
min
f(x) = —
'
' 2
28.
(Bai 28
trang
24 SGK)
Gidi
Gpi
X (cm) la
chieu
d^i
hinh
chOt
nhat
0 < x < 20
Chieu
rong
hinh
chCi
nhat
1^ (20 - x) cm
Dien tich hinh chil nhat
la x(20 - x)cm^

Xet
ham so f(x) = x(20 - x) = -x'^ + 20x, x € (0; 20)
Taco
f'(x) =-2x + 20'
f'(x)
- 0 » X = 10
Bang
hi
en thien
X
0
10
20
f'(x)
+
0
f(x)
^
> 1 nn •
f(x)
II
III
-~ J
f(10)
= -10^ + 20.10 = 100
H^m
so dat gia
tri l<Jn nhat
tai x = 10, gia
tri

Idn
nhat
la 100 cm^
Vay hinh chiJ nhat
dat gia
tri lorn nhat
khi va chi khi
chieu
dai
hinh chijf nhat
=
chieu
rpng hinh
chOf
nhat
= 10 cm
Hay trong
tat ca c^c
hinh
chuf
nhat, hinh
vuong
c6
canh
tlai
10 cm c6
dien
tich lorn nhat.
§4.
DO TH! CUA HAM SO

-
PHEP
TjNH
TIEN HE TOA Dp
I\6l
DUI\
CAIV
I«ICf
1.
Phep
tinh
tien
h# toa dp va
cong
thijfc
chuyen
hp
tpa
dp
- Trong
mat
ph^ng
toa do Oxy cho
diem
Kxo;
yo), goi IXY la h$ toa do mdi goc I va
hai
true
IX,
lY

theo
thil
tu c6
cung vectot
dan
vi
i , j
vdi hai
true
Ox, Oy.
Goi
M la
diem
bat
ki eiia mat ph^ng
(x;
y) la toa do
diem
M doi
vdi
he
tpa
do Oxy
(X;
Y) la toa do
diem
M doi
v6i
he
tpa

dp
IXY
[x
= X + X(,
ly
= Y + y„
C^e he
thijfc
tren
goi la
cong
thiJc chuyen
he
tpa
dp
trong ph6p
tinh
tien
theo
vecto
01
2. Phifofng
trinh
cua
do
thi do'i
vdfi
hp
tpa
dp mdi

Gpi
y - f(x) la
phuong
trinh
cua do
thi
{'^) do'i v6i he tpa dp Oxy,
khi
do
phJOng
trinh
ciia
do
thi
{'f) doi v6i he tpa dp
IXY
la Y = f(X + x,,) - yo.
Ta
CO
y
y
Y
Y
M
y
y
y.o
A
0
Xo

X
MBAITAP
29.
(Bdi 29
trang
27 SGK)
a)
y =
2x^
- 3x + 1
3
Gidi
Dinh
I
y,
=2
4j
4
8
.
Ta CO I
r3
_1
Cong thuTc chuyen
h§ toa dp
trong phep
tinh
tien
vectcf
01 \h

Phi/otng
trinh
duorng Parabol
doi
vdi
he
tpa
dp
IXY la:
Y
- - =2
8
x
=
y
=
X +
—
4
Y-i
8
(
3^
2
(
3^
X
+ -
-
3

X +
-
+
1
l
4;
l
4;
hay
Y = 2
X'
+-X +
16
_
3X - -
+
1
+
1
4
8
Y
= 2X' +
3X+ 3X :=>Y
= 2X1
8
8
b)y
=
ix^-x-3

Dinh
I
X,
=1
1
, 7 . Ta CO I
y,
= 1^-1-3
=
^'2
2
2)
Cong thufc chuyen
he
tpa
dp
trong phep
tinh
tien
vectof
01 la
Phucng
trinh
Parabol do'i vdi
he
tpa
dp
IXY \k:
Y-
- = -(X+

1)'-(X+
l)-3
2
2
hay
Y=
i(X'+
2X+
1)-X-1
- 3 + -
=
X
+
1
2
2
1
2
Y
= -X^ + X+ - - X- - =^Y = Ax'
2
1
2
c)
y = X - 4x'
Dinh
I
X,
=
8

y.4-4
1^
16
.
Ta CO I
1_
1_
8'
16
Cong thufc chuyen
hf
tpa
dp
trong phep
tinh
tien
vectcf
01 la
Phuorng
trinh
Parabol
doi
vdi
he
tpa
dp
IXY 1^:
x2
X
=

y
=
x.l
8
Y.A
16
Y+
— = X+ A- 4
16
8
x.l
8
hay
Y = X + - - 4
8 2
n
X^
+ -X +
4
64
16
Y
= X+ A-4X'-X- — - — =^Y =
-4X1
8
16 16
d)
y =
2x2
- 5

Dinh
I
X,
= 0
y,
=-5
.
Ta CO
1(0;
-5)
Cong thiJc chuyen
he
tpa
dp
trong phep
tinh
tien
vectof
01 la
Phuong
trinh
Parabol
doi
vdi
he
tpa
dp
IXY la:
Y
- 5 - 2X' - 5

hay
Y = 2X'
X
y
X
Y-5
30. (Bdi 30 trang 27 SGK) Giai
a) f(x) = - 3x^ + 1
f'(x) = 3x^ - 6x. f"(x) = 6x - 6
f"(x) =
0cc6x-6
=
0c^x=l
X
= 1 ==. f(l) = 1^ - 3.1^ + 1 = -1
Vayld; !)
(x = X + l
b)
Cong
thiJfc
chuyen
he toa do trong
phep
tinh
tien
vecto
01 1^
y = Y-1
Phirang
trinh

diTcfng
cong
doi vdi h$ toa dp IXY la
Y - 1 = (X + If - 3(X
+1)2+1
hay Y = X^ + 3X2 + 3X + 1 - STX^ + 2X + 1) + 1 + 1
^ Y = X^ + 3X2 + 3X + 1 - 3X2 - 6X - 3 + 2 => Y = X^ - 3X
'
Taco
Y = f(X) =
X^ 3Xxdcdinhtr§nR
VX € R, -X e R
f(-X) = (-X)^ - 3(-X) = -X^ + 3X = -(X^ - 3X) = -f(X)
Vay h^m so Y = X^ - 3X la h^m
so'le,
nen goc toa do I la tam doi xufng
ciia do thi
c)
Phuong
trinh
tiep tuyen (d) vdi CiO tai 1(1; -1) 1^:
y + 1 ^ f'(l) (x-1)
hay y + 1 = -3(x - 1)
Vay (d): y = -3x + 2
Dat g(x) = -3x + 2, ta c6
f(x) - g{x) = x' - 3x2 _^ J _ (_3J^ + 2) = X^ - 3x2 + 3x - 1 = (X -
• Neu x <-1 thi \x) - g(x) < 0 <=> fix) < g(x)
Vay tren
khoang
(-oo;

1) {f) nkm phia difdi (d)
• Neu X > 1 thi f(x) - g(x) > 0 o f(x) > g(x)
vay
trdn
khoang
(1;
+oc)
{'f) nkm phia tren (d)
^l.
(Bai 31 trang 27 SGK)' Gidi
fx = X-2
Cong
thilc
chuyen
he toa dp trong
phep
tinh
tien
vector
Ol la: _ y + 2
Phuong
trinh
diTdng
cong
Cf) doi vdi he toa dp IXY la:
Y + 2 = 2 ^Y = -l
X-2+2
X
DatY = f(X) =
X

Tap x^c dinh ham so nay la 7 = R \1
VX G 7, -X e 7,- ft-X) = -1 = = -f(X)
X X
Do d6 h^ni so Y = - — 1^ h^m so 1§ nen do thi Cf)
nhSn
goc toa dp I 1^ tam
X
doi xijfng. f
32. (Bai 32 trang 28 SGK) Gidi
a)
Thifc
hifn
ph6p
tinh
tien
vectcf
OI vdi 1(1; 1)
Cong
thiJc
chuyen
h# toa do trong
ph6p
tinh
tien n^y 1^
X
= X + 1
y = Y + 1
Phi/cfng
trinh
cua do thi doi vdi hp toa dp IXY 1^

Y+ 1 =
2 1 V 2
+ 1 => Y = —
X+1-1 X
Dat Y = «x) = -
Tap xac dinh cua ham so n^y 1^ 7 = R \: VX e 7, -X G 7
2
Vay h^m so" Y = — la h^m so' 14 nen do thi ham so' nhan go'c tpa dp I \k
X
tam doi xijfng.
Do do tam doi xiing
ciaa
h^m s6' da cho 1^
diem
1(1; 1)
b)y =
3x - 2 3(x + 1) - 5
= 3-
x+1 x+1 x+1
Thirc hipn
ph6p
tinh
tien
vectcf
01 vdi I(-l; 3)
Cong
thiJc
chuyen
he toa dp trong
phep

tinh
tien n^y \k
Phiforng
trinh
ciia do thi doi vdi he toa dp IXY 1^:
Y + 3 = 3
fx = X-l
y = Y + 3
-^=>Y = -A
X-1+1 X
DatY=f(X)
=
-:J
A
Tap xac dinh cua ham s6' nay la 7 = R \1
Ta CO VX G 7, -X G 7
f(-X) = -| = I
=-f(X)
X X
Vay ham so' Y = - — la ham so' le nen do thi ham so nhan gdc toa dp I la
X
tam doi
xuTng
Do d6 tam 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

tinh
tien
vectcf
01 la
X
= X + X(,
y = Y + aXp + b
Phircmg
trinh
cua (C) doi vdi hp tpa dp IXY la
Y + axo + b = a(X + XQ) + b +
X + Xo - Xo
, a ?t 0 va c ;^ 0
=>
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
ax
=
-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
y. [
y = yo ^
y
=
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) =+00 hoac limf(x) =+oo hoac
limf(x)
=-co
hoac limf(x) =
-3o
o
y^
= flx)
yi
y
= fix)
II
y
II
0

=
fix)
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 -co)
*
Vi lim y = +co va lim y = -QO
Nen ducfng thing
x = — la
difcfng tiem
can
dufng ciia
do thi ham so
3
x
^
2^
va
khi X ->
K
3J
b)y
=
-2x-2
x +
3
Ham
so xac
dinh tren K

\1
*
Vi
lim
y = -2
Nen difcJng thing
y = -2 la
ti?m can ngang cua do
thi ham so
(khi
x ->
+oo
va
khi X -co)
*
Vi lim y = -oo va lim y = +oo
x-»(-3)"
x->(-3)*
Nen ducmg thing
x
=
-3 la
tiem can dilng cua do
thi
ham so (khi
x
-> (-3)"
va
khi
x ->

(-3)^)
c) y = X + 2
X - 3
Hkm so xdc dinh tren R \|
*
Vi 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
va
khi X 3*)
•
Vi Hm [y - (x + 2)] = 0
x-+±x
Nen
dirdng th^ng y = x + 2 Ik tiem can xien cua do thi ham so (khi x ->
+QO
va
khi X -ao)
x' - 3x + 4
d)y =
2x + l
Ham
so xac dinh tren K \j -—
i Vi lim y = -00 vk lim y = +oo
Nen
dxibng
thing x = -— la tifm can diJng ciia do thi hkm so (khi
2

f
1
vk khi x -»
"2,
)
*
Ti|m can xien c6 dang y = ax + b, ta c6:
,.
y ,. x^-3x + 4 ,. x^-3x + 4 1
a = lim
—
= hm _ -= hm
x txx
"-^-^ x(2x + l) 2x^ + X
'x^ - 3x + 4 X
2
b
= lim
X
=
lim
x->ix
2
2x + l
=
lim
2
-7x + 8
4x + 2
7

4
1 7
Vay
difdng thang y = - x - - la ti^m can xien ciia do thi ham so (khi
x -> -00 va khi x +oo)
x + 2
e)y=^
Hkm s6 xkc dinh tren R \; 1|
*
Vi lim y = +oo va lim y = -x
x-»(-l)-
x (-l)*
Nen
dudng thing x = -1 la tiem can dufng ctia do thi ham so (khi
X (-1)- va khi x -> (-If)
*
Vi lim y = -oo va lim y = +oc
x->
I
x-> I"
Nen
difdng thing x = 1 la tiem can dufng ciia do thi ham so' (khi x ^ 1
va
khi X 1*)
*
Vi lim y = 0
Nen
dircmg thing y = 0 Ik tiem
CSLQ
ngang ciia do thi hkm so (khi x +« va

khi
X -> -oo)
1^
«y= ^ - -
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)
va
khi X -> (-m
*
Vi lim y = 0
X->±C
N6n dutog thing y = 0 Ik ti?m ckn ngang cua do thi hkm so (khi x -> +oo
vk khi x -> -oo)
35.
(Bdi 35
trang
25 SGK) , Gidi
a, V = + X - 3
'X
Ham
so xac dinh tren R 1101
*
Vi lim y =
-<x>

vk lim y = -co
x »"'
x->0*
Nen
Avcdng thing y = 0 Ik ti^m c^n dufng ciia do thi hkm s6' (khi x -> 0"
vk
khi x 0^)
*Vi
lim[y-(x-3)l= lim^^ = 0
Nen
dufmg thing y = x - 3 Ik ti^m c$n xien cua d6 thi hkm so (khi x -> +oc
vk
khi X -oo)
x' + 2
b)y =
x'-2x
Hkm
so xac dinh tren R \; 21
•
Vi lim y = -oo vk
lim
y = +oo
x->(r x-»o*
N§n
diibng
thing x = 0 Ik ti^m ckn dufng ciia do thi hkm so (khi x -> 0"
vk
khi X -> 0*)
•
Vi lim y = -oo vk lim y = +oo

x~>2 x-»2*
Nen
dirdng thing x = 2 Ik ti|m ckn dufng cua do thi hkm so (khi x 2'
va
khi X ^ 2*)
•
Ti§m ckn xien c6 dang y = ax + b, ta c6
,.
y + 2
a = lim
—
= hm
x-»±x
X
x->±»(x
-2x)x
=
lim
x^+2
=
1
b
= lim[y - x] = lim
x^+2
t» x^ - 2x^
2x=*
+2
x' -2x
-
X

=
lim-
*±=t X - 2x
=
2
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
*
Vi lim y =
+00
vk
lira
y =
-00
Nen
diXbng
thing
x = -llk
ti$m
cgn
diifng
cua d6
thi

hkm
so (khi
x
->
(-IT
va
khi
X
(-1}*)
*
Vi
limy
=
-00
va limy
=
+00
x->r
X I*
NSn
dudng thing
x =: 1 la
tiem
can
dilTng ciia
do thi hkm so
(khi
x 1~
va khi
X -> 1"^)

*
Tiem
can
xien
c6
dang
y =: ax + b, ta c6
o
,.
y ,. x^ + x +1
a
=
lim
—
=
lim —5
= 1
x^.±« X
"-t* X -
X
b= lim[y-x]
= lim ^
1^
^
- X
1- 2x + l ^
=
lim = 0
x'
-1

d)y
=
Nen dudng thing
y = x Ik
tiem
can
xien
ciia
do thi
hkm
so'
(khi
x +00
vk
khi
x ->
-00)
x^
+ x + 1 ^-
-5x'
- 2x + 3
Hkm
so xkc
dinh tren
R \
*
Vi
lim
y = - -
x ±=c

5
Nen difdng thing
y = - ^ Ik
tiem
can
ngang
cua do
thi hkm
so
(khi
x ^
+co
o
vk
khi
X ->
-00)
*
Vi lim y =
+00
vk lim y =
-00
x->(-l)" x->(-l)'
N§n diremg
thSng
x = -1 Ik
tiem
can
dutng
cua do

thi hkm
so
(khi
x (-1)
vk
khi
X ^ (-ID
*
Vi
lim
y -
-00
vk
lim
y =
+00
3
Nen
diremg
thing
x = | Ik
tiem
can
dutng
cua do
thi hkm
so
(khi
x —
5

5
vk
khi
X
-)•
—)
5
36.
(Bai 36,
trang
36 SKG)
a)y
=
Vx^-1
Gidi
Hkm
so xac
dinh tren
[-00;
-1] u
[1;
+x>).
- Ti$m
can
xien
c6
dang
y = ax + b
=
limjl-4

=
l
*a=lim>:=lim:^
=
lim!rl
*
b =
lim[y
-
x]
=
lim[Vx2
- 1 -
x]
=
lim
, 0
"^*-Vx^-l
+
x
Vay
dUdng thing
y = x Ik
tiSm
can
xien
cua do thi hkm so
(khi
x ->
+00).

y
*
a =
lim—
=
lim
= lim
X
"^-^ X
-X,
1-
=
lim-Jl-^
= -1
*
b =
lini[y
+ x] =
lim(\/x^
-1 + x] =
lim-
-1
=
0
«7x^
-1 -X
V$y difdng thing
y = -x Ik
ti$m
can

xien
ciia
d6 thi
hkm
so
(khi
x -00).
b)
y = 2x +
Vx^
- 1
Ham
so xac
dinh tren
(-00;
-1]
[1;
+00).
Tiem
can
xien
c6
dang
y = ax + b
2x +Vx'
-1
a
=
lim—
=

lim
X-»+^X
X-»
+
X
=
lim
x-*+x
2
+
xJl-
= lim 2
+
=
3
b
=
lim[y
-
3x]
=
lim[2x
+
N/X'
- 1 - 3x]
=
lim[Vx^
- 1 -
x]
= li

im-
-1
=
0
'Vx'-l
+ x
Vay dudng thing
y = 3x Ik
ti^m
can
xien
cua do
thi hkm
so
(khi
x -»
+00).
y
*
a =
lim—
=
lim
X->-xv
x->-x
2
+
six'
-1
=

2 +
lim—^-
— = 2-1 = 1
*
b =
lim[y
- x] =
lim[x
+
Vx'
-1] =
lim
^= = 0
''^-"x-Vx'-l
Vay dudng thing
y = x la
ti|m
can
xien
cua do
thi hkm
so
(khi
x ->
-00).
c)
y = X +
vx^n
Hkm
so xkc

dinh
tr§n
R.
a
=
lim—
=
lim
1
+
4: x^+1
= lim
1
+
-
X.
1
+
=1+1=2
b
=
lim[y
-
2x]
=
lim[Vx'
+1
- x) =
lim-===i=
=

0
"-"'Vx^+l
+ x
Vsly dudng thing
y = 2x Ik
ti^m
can
xien
ciia
do
thi hkm
so
(khi
x
+<»)•
Vi
limy
=
lim(x
+
Vx^
+1) =
lim-
-1
=
0
"^^x-Vx^+l
Vky dudng thing
y = 0 Ik
ti^m

can
ngang
cua
d& thi hkm
so
(khi
x
->
-00).
d)
y =
Vx^
+ X + 1
Ham so'
xac
dinh tren
K.
*«
_ i™y Vx'
+
X + 1
a
=
lim—
=
lim-
= lim-
X
X
=

1)
*
b =
lim[y
- x] =
lim(Vx^
+ x + l - x) = lim . ^ + ^
''"^Vx^x
+ l + x
=
lim-
X
ill,
1
+ - + -^ +1
X
X
1
2
V^y
diremg
thing
y = x + ^
lli ti^m c$n
xi§n
cua
d6
thi hkm
so
(khi

x -> +x).
*
a =
lim^
= lim
,
„ 1 1
= lim-
x
X
= -1
*
b =
lim(y
+ x) =
lim(Vx^
+ x + l + x) =
lim-==^i
Vx^
+ X +
1
-X
=
lim-
x;
1
+
-X/1
+ 1- + 1 X
"

x^ x^
= lim-
1 1 1 1
X
X
J.
2
Vay dudng thing
y = -x - i Ik
tigm
can xien cua
do thi
hkm
so
(khi
x ->
-oc).
37.
(Bai 37,
trang
36 SGK)
Gidi
a)
y = X +
Vx^
-1 -
Hkm
so xkc
dinh tren (-oo;
-1] u

[1; +oo).
*
a =
lim-
= lim
°x
1
+
= lim
1
+
xJl-4-
x^
=1+1=2
*
b =
lim[y
- 2x] =
lim(Vx'
-1 - x) = lim , ^ = 0
x «o
"^""Vx^
- 1 + X
Dudng thing
y = 2x Ik
tifm
can
xien
cua do thi hkm so
(khi

x ->
+oo).
* limy
=
lim(x
+
Vx^
-1) = lim } =0
_x + Vx'-l
Dudng thing
y = 0 Ik
tif
m
can
ngang
ciia
do thi
hkm
so
(khi
x ->
-co).
b)
y = Vx'
-4X-I-3
Hkm
so xkc
dinh tren (-oo;
1] u [3; +-0).
4 3_

y
,•
Vx^-4x
+ 3 Y_x^_i
*
a =
lira—
= lim = lim = 1
X
"-"^"^
X
*
b =
lim(y
- x) =
lim(Vx^-
4x + 3 - x) = lim , ^^'^^
x^*x Vx^
- 4x
-I-
3
-I-
=
lim—F
-4X-H3
=
lim-
1
- - -H -J + 1
X

x"
X
xUl-^
+ 4+1
X
x^
=
lim
x-»+»|
1 4 3 ,
X
X
= -2
Dudng thing
y = x - 2 la
tiem
can
xien
cua d6 thi hkm so
(khi
x
- -> +00).
,.
y ,.
VX^-4X
+ 3 "^^1^
*
a =
lim-
= lim = lim

4 3
-xjl
+
x »X
x_^=lini|-Jl-i
+ A
X
X
= -1
-4x-3
b
=
lim(y
+ x) -
lim(Vx^
-4x-(-3
^-
x) = lim , =—
X—X
'Vx^-4x
+ 3
= lim-y-
lim
x-»-x|
-4 4- -
H
4 1
-Xjl ! ^
-X
X

X
'
A 3)
= 2
l-^
+ A-l
X
X
Bvtbng thing
y = -x + 2 Ik
tiem
can
xien
cua do thi
hkm
so
(khi
x ->
-00).
c)
y =
Vx^n
Hkm
so xkc
dinh tren
K.
=
lim
-^-lim,
1 + 4- = 1

X
" ^V X
*
a =
lim^
=
lim^^-
x->+<iox
x->+«i
X
I
4
b
=
lim(y
- x) =
lim(Vx^
+ 4 - x) = lim , = 0
x->4.:r
x-^^-x
"-"''VX
+4+X
Dirdng th^ng
y = x Ik
ti§m
c^n
xi§n
cua d6 thi
hkm
so

(khi
x +«).
*
a =
limi
-
lim^^^^^—-
=
lim-
^
= lim
x-*-xx
x-»-x
-1
b
=
lim(y
+ x) =
lim(Vx^
+4 + x) = lim ^ -
x->-«.
x->-x
""'""vx
+
4 —;
Diforng
thSng
y = -x la tiem can xien ciia do thi h^m so' (khi x -oo).
d)y =
X

+ X + 1
x^-1
H^m so xAc dinh tren K \; 11.
* Vi
lira
y =
-Ko
va lim y = -oo
x->(-l)" x->(-l)*
Nen
dUorng
thing x = -1 la tiem can
dufng
cua do thi ham so (khi x -> -1
va khi X -V).
* Vi limy = -oo va limy = +«
x^r
x->i*
Nen
diTdng
thing x = 1 la ti^m can diifng cua do thi ham so (khi x -> 1
va khi X -> 1*).
* limy = 1
X-»±X
Nen diforng thing y = 1 1^ ti^m ckn
ngang
cua do thi ham so (khi x +x
va khi X -oo).
38. (Bai 38, trang 36 SGK) Gidi
x' - 2x + 2

a) Ham so xac dinh tren E \.
* Vi limy =
-oo
va limy =
+oo
Nen
dudng
thang x = 3 la tiem can
dufng
cua do thi ham so (khi x -> 3
vk khi x ^ 3^).
5
Ta CO y = x + 1 +
x-3
lim[y - (x + 1)] = lim = 0
x^«x-3
Nen
dudng
thing y = x + 1 la tiem can xien cua do thi ham so (khi x +<x
va khi X -> -oo).
b) Toa do
giao
diem
I ciia hai tiem can la
nghiem
cua he
phu'ong
trinh:
y = X + 1 fy = 4
X

= 3 [x = 3
V$yl(3;4).
. fx = X + 3
Cong
thuTc
chuyen
h^ toa d6
theo
ph^p
tinh
tien 01
Phufong
trinh
m ddi v6i hf toa dp IXY.
*(X + 3)'
-2(X.+
3) + 2
Y + 4 =
X + 3-3
Y =
y = Y + 4
X' + 6X + 9 - 2X - 6 + 2
X
X' + 4X + 5 - 4X X' + 5
-4
X
X
Dat Y = fix) =
+5
, t^p x6c dinh = R \(0}.

VX G V, -X e 'J'
(-X)' +5 X' + 5
f(-X) =
-X
X
=
-f
(X)
X^ + 5
Vay ham so' Y - la ham so' le nen do thi nhan go'c 1(3; 4) lam tam
X
doi xiifng.
39. (Bai 39, trang 36 SGK). Gidi
+X-4
Hkm so xac dinh tren M \.
* Vi lim y = +C0 v^ lim y = -oo
x->(-2)- x->(-2)*
Nfen diffmg thing x = 2 1^
ti§m
can
dufng
ciia do thi ham so (khi x (-2)
v^ khi x ^
(-2)").
2
Ta CO y = X - 1 -
x + 2
Vi
lim[y - (x - 1)] = lim
x + 2

= 0
Nen
duforng
thing y = x - 1 Ik ti$m c^n xien ciia d6 thi hhm so (khi x -oo
vk khi X -> +oo).
b) Toa dp
giao
dieni ciia hai tiem can la
nghiem
he
phuong
trinh
x = -2 'x = -2
y = -3
y = X -1
C6ng
thufc
chuyen
he toa dp
theo
phep
tinh
tien 01
c) Phiforng
trinh
cua (f) doi vdi he toa dp IXY.
X
= X - 2
jy = Y-3
Y- 3 =

(X-2)'
+(X-2)-4
X-2 + 2
^ Y =
X'-4X + 4 + X- 2- 4
X
X' -3X-2 + 3X _
+ 3
X'-2
X ' * X
Dat Y = f(X) = ^—^, tap xac dinh = R \: VX € V, -X e V.
f(X)
=
k^ljzl
= =
_f
(X)
Vay ham
so'Y=
~^
-X
X
le nen nh$n goc tpa dp I(-2; -3) la tam doi xufng.
c) V = - 8x + 19
x-5
Ham so xac dinh tren R \.
r-s^
—
Vi
limy = -oo vk limy =

+00
Nen
du£fng
thing ^ 5 Ih
ti^ra
can
duTng
cua do thi h^m s6' (khi x 5
v& khi X -> 5*).
Tac6y-7X-3+^
x-5
Vi
limly - (x - 3)1 -
lim-i-
- 0
N§n difdng thdng y = x - 3 Ik ti$m can
xi&n
cua d6 thi h^n so (khi x -> +^
\k khi X -> -!»).
b) Toa do
giao
dilm I cua hai
ti§m
c$n Ik nghi^ra he phirong
trinh:
x=.5 fx = 5
ly
= x- 3'^ty = 2
Vgy 1(5; 2)
Cong

thu-
chuyen
he toa do
theo
ph6p
tinh
tien 01 F = + ^
ly
= Y + 2
c)
Phirong
trinh
m doi vdi h? toa dO IXY Ik:
Y
+ 2 =
(X +
5)'-8(X
+ 5) + 19
x + 5-5
^ Y = + IPX + 25 - 8X - 40 4 19 _ X' + 2X + 4 - 2X _ X^ + 4
X X ~ X
X^ + 4
Dat Y = fiX) = ^ , tap xac dinh 7 = K \: Vx € '/, -X e 7.
—A A
X^ + 4
Vay hkm so Y = ^ Ik hkm s6 1^ nen nhan g6c toa do 1(5; 2) lam tkm
doi xufng.
§6.
KHAO
SAT SI/

BIEN
THIEN
VA VE DO THj
MOT SO HAM DA
THlfC
1. Cac hvtdc khao sat bien thien va ve do thi cua ham so.
• Elide 1: Tim tap xac dinh cua hkm so.
•
Bifdc
2: X6t sir
bien
thien cua hkm s6.
* Tim gi(Ji han tai v6
cifc
vk gidi han v6
cifc
(n^u c6) cua hkm so. Tim
cac
dudng
tigm can cua do thi (neu c6).
* Lap
bang
bien
thien cua hkm so bao gom:
Tim dao hkm cua hkm so.
- X6t dau dao hkm, x6t
chieu
bien
thien vk tim cUc tri cua hkm s6 (n§'u
c6),

dien
ket qua vko
bang.
• BvCdc 3: Ve do thi cua ham so.
* Ve ckc
dufcfng
ti|m can cua do thi (neu c6).
* Xac dinh mpt so
diem
dac bi#t ciia do thi.
(giao
diem
vdi cdc true toa
dp, neu
phep
toan
khong
philc
tap).
* Nhan xet ve do thi (chi ra true vk tam doi xijfng cua do thi (neu c6)).
2. Ham so y = ax' + bx^ + cx + d (a 0).
*
Dang
do thi
M
1
/ ^
\
•y
V

\
Dang
1
Dang
2
* Dilm uon cua do thi
Neu hkm so y = f(x) c6 dao ham cap hai tren mpt
khoang
chiia
diem
XQ,
f
"(xo) = 0 va f "(x) doi dau khi x qua
Xo
thi
diem
U(xo; f(xo) la mot
diem
uo'n cua do thi hkm so y = f(x).
* Ghi chu: Do thi ham so bac ba y = ax^ + bx^ + cx + d (a ;^ 0) ham c6 mot
diem
uon va
diem
do \k tam doi xiifng ciia do thi.
3. Ham so' trung phi^o^ng y = ax'* + bx^ + c (a 5* 0).
*
Dang
do thi
\M/
\M2 ^

Dang
1
Dang
2
b)
1 BAITAP
40. (Bai 40, trang 43 SGK) Gidi
a) Khdo sat sU bii'n
thien
va ve do thi ham s6 y = + 3x^ - 4.
1-
Hkm so xac dinh tren R.
2- Su
bien
thien cua hkm so:
a) Gidi han tai v6
ciTc
limy = -00 va limy =
+00
b) Suf
bi§'n
thien,
ciTc
tri
Ta c6: y' = Sx^ + 6x = 3x(x + 2)
y' = 0 o 3x(x + 2) =
0c:>x
= 0
hokc
x = -2.

+
Hkm so
d6ng
hi&'n
tren
m5i khoang (-oc; -2) vk (0; +00), nghich bie'i
tren
khoang (-2; 0).
+
H^m so' dat c\ic dai tai diem x = -2, gid tri cUc dai 1^ y(-2) = 0, hkm SQ
dat
cifc
tieu
tai diem x = 0, gia tri
ciTc
tieu
la y(0) = -4.
c) Bang bien
thien
X
—00
-2
0
+00
y'
+
0 0
+
y
—cc^—

^ 0
-4-—
»+oo
3. Do thi
• Diem uon:
Ta c6 y" = 6x + 6, y" = 0 o x = -1.
y" = 0 tai X = -1 va y" doi dau tCf am sang
diTcfng
khi x qua x = -1. Vay diem U(-l; -2)
la
diem uon
ciia
do thi.
•
Giao
diem ciia do thi vdi true tung la (0; -4).
• y =
0c=>x^
+ 3x^-4 = 0o(x- l)(x^ + 4x + 4
o (x - l)(x + 2r = 0 o
-2 -1
ji
=
0
-2/
-4
••x
x = 1
x = -2
D6

thi c^t true hoanh tai (-2; 0) va (1; 0).
Nhdn
xet: Do thi nhan diem uon U(-l; -2) la tam doi xufng.
b)
PhiCang
trinh
tiep tuyen cua dd thi tai diem uon U(-l; -2) la
y + 2 = ft-lXx + 1) hay y + 2 = -3(x + ij => y = -3x - 5
fx
= X -1
c)
Cdng thiic chuyin he toa dd
theo
phep
tinh tien OU
ly
= Y-2
Phuong
trinh
do thi doi vdi he toa do UXY
Y
- 2 = (X - if + 3(X
-1)^-4
=
X^ - 3X^ + 3X - 1 + 3X^ - 6X + 3 - 4 = X^ - 3X - 2 => Y = X^ - 3X
Dat
Y = f(X) = X^ - 3X, tap xac
dinh
'J = R.
VX € -X e y.

f(-X) = (-X)' - 3(-X) = -X^* + 3X = -(X^ - 3X) = -f(X).
Vay ham so Y = X^ - 3X la ham le nen nhan goc toa do U(-l; -2) \km
tam
doi xufng.
Do do U(-l; -2) la tam doi xufng ciia do thi.
41. (Bai 41, trang 44 SGK)
Gidi
a)
Khdo sat sU bien thien va ve y = -x^ + 3x^ - 1.
1.
Ham so' xac
dinh
tren
M.
2. Sii bien
thien
ciia ham so',
a)
Gidi
han tai v6 cUc.
• lim = +CO vk lim = -00
b) SU bien
thien,
cUc tri
• Ta CO y' = -Sx^ + 6x = -3x(x - 2)
y' = 0 <=> -3x(x -2) =
0<=>x
= 0
hoac
x = 2.

+
Ham so
dong
bien
tren
khoang (0; 2), nghich bien
tren
m6i khoang
(-00; 0) va (2; +00).
+
Ham so' dat cUc
tieu
tai x = 0, gia tri cUc
tieu
y(0) = -1. Ham so' dat
ciTc
dai
tai x = 2, gia tri cUc dai y(2) = 3.
c) Bang bien
thien
x
—00
0
2
+00
y'
0
+
0
-

y
+00
__-*3-—
r
^ —00
3
y = m
1-
-••••ju
i
/I 1
J.1 2.
3
3. Do thi
• Diem uon:
Ta CO y" = -6x + 6, y" = 0 o x = 1.
y" = 0 tai x = 1 va y" doi dau tiT di/Ong
sang
am khi x qua x = 1. Vay diem U(l; 1)
la
diem uo'n cua do thi.
•
Giao
diem cua do thi va true tung la (0; -1).
Nhdn
xet: Do thi ham so nhan diem U(l; 1)
la
tam doi xufng.
b)
Bien

ludn
so nghiem cua phuang
trinh:
.
-x^
+ 3x^ - 1 = m (1)
Dat
y -x^ + 3x"'^ - 1 CO do thi (C) ve d cau a)
y = m CO do thi la dif6ng
thing
(d).
• Ve ducfng
thing
y = m, hoanh dp
giao
diem ciia (d) va (C) la nghiem cua (1).
* Ne'u m < -1
hoac
m > 3 thi phi^cfng
trinh
(1) c6 mot nghiem.
* Ne'u m = -1
hoac
m = 3 thi phiTcfng
trinh
(1) c6 hai nghiem.
* Ng'u -1 < m < 3 thi phiJcfng
trinh
(1) c6 ba nghiem.
42. (Bdi 42, trang 44 SGK)

Gidi
a)
Khdo sat sil bien thien va ve do thi y = ^x^ - x^ - 3x - - .
3 3
1-
Ham so xac
dinh
tren
K.
2- Su bien
thien
ciia ham so'.
^)
Gidri
han tai v6
cifc:
•
limy
= -00 •
limy
= +00
b) Su bien
thien,
cUc
tri:
• Ta c6 y' = x^ - 2x - 3
y' = 0 c:> X = -1
hoac
x = 3
Ham

so'
dong
bien
tren
cac khoang (-00; -1) va (3; +00); nghich bien
tren
khoang (-1; 3).